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Page 1: Ordinary Differential Equations: A Systems Approach

Ordinary Differential Equations:A Systems Approach

Bruce P. Conrad

November 24, 2010

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c©2010 Bruce P. Conrad

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Chapter 1

First-Order Equations

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1.1 Introduction

A differential equation is a relation involving an unknown function andsome of its derivatives. For example,

dydt

= y + et

is a differential equation that asks for a function, y = f (t), whose derivativeis equal to the function plus et. By differentiating, you can verify that afunction such as y = tet meets this specification.

Differential equations are a source of fascinating mathematical prob-lems, and they have numerous applications.

A mathematical model is a mathematical construction, such as a differ-ential equation, that simulates a natural or engineering phenomenon. Mostapplications of differential equations take the form of mathematical mod-els. For example, consider the problem of determining the velocity v of afalling object.

~?v

Newton’s second law of motion tells us that the net force on the object isequal to the product of its mass, m, and its acceleration, dv

dt . This law is adifferential equation,

mdvdt

= F,

Ignoring air resistance, for an object falling close to the earth’s surface theforce is F = mg, directed downward, where g is approximately 9.80 metersper second per second. Thus the differential equation

mdvdt

= mg

is a mathematical model corresponding to a falling object.To solve the differential equation, cancel the mass and note that v is an

antiderivative of the constant g; thus v = gt + C, where C is an arbitraryconstant.

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1.1. INTRODUCTION 5

We have just solved a differential equation: The solution is not a singlefunction, but a family of functions depending on an arbitrary constant C.To determine the equation of motion of this particular falling object, weneed to refine our model.

For example, an initial condition, specifying the velocity when t = 0,will determine the equation of motion. If we assume that the object wasfalling from rest, so that v = 0 when t = 0, then we know that the solutionthat we seek has the property v(0) = 0. We have knowledge of the starting,or initial value of v. It is easy to infer from this initial condition that theconstant C is equal to 0, and v = gt.

What if we don’t ignore air resistance? It is known that the force of air re-sistance is directed oppositely to the motion, with magnitude proportionalto the square of the velocity (assuming the velocity is much less than thespeed of sound). In other words, air resistance = k v2, where the constantof proportionality k is the drag coefficient). Combining air resistance andgravitation, we obtain the differential equation model

dvdt

= g− km

v2. (1.1)

It may be tempting to integrate as we did before:

v =∫ (

g− km

v2)

dt.

Since the “unknown function” v appears in the integrand here, there is noway to calculate this integral without first knowing the answer! Our objec-tive is to manage this quandary.

Terminology. The order of a differential equation is the order of the high-est derivative of the unknown function occurring in the equation. The dif-ferential equations describing the velocity of a falling object that we justconsidered above were first order. In the related second order equation,y′′ = g, the unknown function represented by the variable y is the distancethe object has fallen. The velocity would be v = y′. Including air resistance,we get y′′ = g− k(y′)2/m, another second order equation.

A differential equation involving only derivatives with respect to a sin-gle independent variable is called an ordinary differential equation, orODE. The falling body models that we just considered are ODEs, in whichthe independent variable is t. A differential equation that involves partialderivatives with respect to two or more independent variables is a partial

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differential equation, or PDE. As an example, here is the second order PDEthat models the vibration of a guitar string:

∂2y∂t2 − c2 ∂2y

∂x2 = 0.

The unknown function y represents the displacement of a point on thestring x centimeters from the bridge at time t, and c is a constant relatedto the tension and density of the string.

In any differential equation, a dependent variable is a variable that rep-resents an unknown function. A solution of a differential equation is afunction that can be substituted for the dependent variable to produce anidentity.

Example 1.1.1 The ODEdydt

= ky (1.2)

can be as a model for population growth, compound interest on savings accounts,etc. (see section 1.2). Show that y = Cekt, where C is an arbitrary constant, is asolution of this ODE.

SOLUTION. If y = Cekt then dydt = Ckekt and ky = Ckekt as well. Hence

equation (1.2) becomes an identity.

Differential equations typically have infinite families of solutions, butwe often need just one solution from the family. We refer to a single solutionof a differential equation as a particular solution to emphasize that it is oneof a family.

The general solution of a differential equation is the family of all itssolutions. The general solution of an ODE on an interval (a, b) is a familyof all solutions that are defined at every point of the interval (a, b). Findingthe general solution of an ODE requires two steps: calculation and verifica-tion. The calculation step is exemplified by our solution of the falling bodyequation v′ = g. Unless a mistake was made in the integration, the familyof solutions we found will satisfy the ODE.

The verification step is to show that all solutions of the ODE belong tothis family. The following theorem from calculus is useful for this purpose:

Theorem 1.1 (Equal derivatives theorem) Let f1(t) and f2(t) be defined anddifferentiable on an interval (a, b) (infinite endpoints are permitted), and assume

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1.1. INTRODUCTION 7

that f ′1(t) = f ′2(t) for all t ∈ (a, b). Then there is a constant C such that f1(t) =f2(t) + C for all t ∈ (a, b).

Example 1.1.2 Verify that the family of solutions v = gt + C for y′ = g is thegeneral solution.

SOLUTION. Let y0(t) = gt. If y1(t) is another solution of the differentialequation, then y′1(t) = y′0(t) = g. It follows from the equal derivativestheorem that y1(t) ≡ y0(t) + C. In other words, every solution of the ODEbelongs to the family y = gt + C.

Because ODEs typically have families of solutions, they are frequentlycoupled with additional information (called constraints) to single out a so-lution of interest. The constraint that we have already encountered, andshall frequently encounter in the future is an initial condition, specifyingthe value of the solution at an initial time. An ODE coupled with an initialcondition is called an initial value problem, or IVP. The motion of a bodyfalling from rest with air resistance would be modeled by the IVP,

dvdt

= g− kv2; v(0) = 0.

Solution by computer There are many techniques available to find so-lutions of ODEs. However, many ODEs have solutions that can’t be ex-pressed in terms of the familiar elementary functions you worked with incalculus courses. Since the invention of the first computer, solutions of in-tractable ODEs and PDEs have been calculated by computers.

These calculations are done by numerical means; that is, the computerdoes not work with a formula for a solution, but calculates a table of valuesthat give a close approximation of a solution of the differential equationfrom the differential equation itself. As a simple example, consider theODE of the form y′ = f (t). The solution,

y =∫

f (t) dt + C,

is helpful if a formula for the antiderivative of f (t) is available. Withoutsuch a formula, we might turn to a numerical method of evaluating theintegral, such as the rectangle rule, the trapezoidal rule, or Simpson’s rule.Numerical methods for solving ODEs of the form y′ = f (t, y), where theright side involves y as well as t, are generalizations of these “rules.”

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A numerical method cannot compute a family of solutions; it can onlyapproximate one solution at a time. The user is expected to provide con-straints to single out the solution to be approximated. A program that isdesigned to approximate solutions of ODEs with initial conditions as theconstraints is called an IVP solver.

While there are programs that are primarily IVP solvers, computer al-gebra systems (CAS), such as Maple, Mathematica, and Matlab include sub-routines that are IVP solvers, as well as the capability to find a formula forthe general solution of practically any differential equation for which thereis an established method of solution. These routines follow rules for ma-nipulating formulas instead of performing numerical calculations. Thereare advanced calculators that incorporate IVP solvers and CAS as well.

Qualitative study of an ODE Figure 1.1 displays the graphs of severalsolutions of the ODE v′ = g− kv2 representing the motion of a falling objectwith air resistance. These graphs appear to have a common asymptote,v = 100. We can explain this feature as follows: Let v∞ =

√g/k, so that

g = kv2∞. The ODE can be rewritten as

v′ = k(v2∞ − v2)

For v < v∞ we see that v′ > 0. This means that v is increasing. Whenv > v∞, v′ < 0, and v is decreasing. In both cases, v tends toward v∞, as infigure 1.1.

An explanation of the behavior of solutions of a given ODE obtained byanalyzing the equation itself, without referring to a formula for the generalsolution, is called a qualitative study. A qualitative study may result in abetter understanding of the physical phenomenon represented by the ODEthan a formula for the general solution could. For example, we see in fig-ure 1.1 that regardless of the initial velocity, the velocity of a falling objecttends to a limiting velocity (known as the terminal velocity ).

Our qualitative study of the falling object correlates well with the physics.For v < v∞, the gravitational force is greater than the air resistance, sothe net force is directed downward and causes the object to accelerate. Ifv > v∞ the air resistance is dominant, and the object decelerates. Finally,if v = v∞ gravity and air resistance balance each other, and the speed isconstant.

Exercises

For each of the differential equations in problems 1 – 13,

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1.1. INTRODUCTION 9

(i) determine the order,

(ii) determine whether the differential equation is an ODE or a PDE.

1. y′′ + y = 0.See answer

2. y′ = (x + y)4

3. (y′)2 + xy = ex.See answer

4. ∂y∂t + 6y ∂y

∂t = ∂3y∂x3 .

5. y′ = t2 + y2.See answer

6. y(100) − y = 0

7. y′ = sin y.See answer

8. y2 + (y′)2 = 1.

9. y3 + (y′)3 = 1See answer

10. t2y′′ + ty′ + (t2 − 1)y = 0.

11. ∂2u∂x∂y = 0.

See answer

12. ∂u∂t = ∂2u

∂x2 .

13. d3udt3 + 3t du

dt − u cosh t = 1t2+1 .

See answer

14. Show that y = C sec(t), is a family of solutions of

y′ − tan(t)y = 0.

and find solutions that satisfy the following initial conditions.

(a) y(0) = 0.

(b) y(π/4) = 1.

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10 CHAPTER 1. FIRST-ORDER EQUATIONS

15. Show that y = C sin(t) + D cos(t) is a family of solutions of

y′′ + y = 0,

and find all solutions that satisfy the following constraints

(a) y(0) = 0.

(b) y(0) = 0, y(π) = 0.

(c) y(0) = 0, y(π/6) = 1.

(d) y(0) = 1, y′(0) = −1.

See answer

16. Show that y = 2(√

t− 1) + Ce−√

t is a solution of

y′ +y

2√

t= 1,

and use it to solve the IVP that couples this ODE with the initial con-dition y(1) = 0.

17. Show that y = Cet2is a solution of the ODE y′ = 2ty, and solve

the IVP y′ = 2ty; y(0) = 3.See answer

18. Let y(t) ={

0 if t ≤ 0t2 if t > 0.

Show that y(t) is a solution of the IVP

y′ = 2√

y; y(1) = 1. Also show that

y = t2 for all real t

is not a solution! Hint: remember that√

t2 = |t|.

19. Let y(t) be a differentiable function defined on the interval (−1, 1)with the property that for all t,

t2 + y(t)2 = 1

(there are two such functions). Show that y(t) is a solution of the ODE

yy′ + t = 0.

See answer

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1.2. LINEAR GROWTH AND DECAY 11

The following problems require showing a given family of functions tobe the general solution of a given differential equation. This entails veri-fying that every member of the family is a solution, and that every solutionbelongs to the family. In each case, the equal derivatives theorem can be used.

20. Verify that y(t, C) = (t + C)e−t is the general solution of the ODE

ddt(ety) = 1

by showing that for every value of the constant C, y(t, C) satisfies theODE, and that every solution has the form y(t, C) for some C.

21. Verify that y = t ln t + Ct is the general solution of

ty′ − yt2 =

1t

on the interval (0, ∞). Hint: (ty′ − y)/t2 = ddt

( yt

).

See answer

1.2 Linear Models for Growth and Decay

An ODE used to simulate a scientific or engineering phenomenon is a typeof mathematical model of the phenomenon. Developing mathematical mod-els involves theorizing, computation, and testing. Theorizing starts byidentifying a variable, or variables that describe the phenomenon (for afalling body, the variables might be the velocity, the distance fallen, orboth). The ODE then describes how the variable or variables change withtime. The ODE may include parameters such as masses, friction constants,etc. These are constants in a given system, but can be varied to make themodel applicable to other systems.

Nature is complicated, so a model may not reflect every aspect of thephenomenon under study. For example, a model used to compute the tra-jectory of a projectile might ignore air resistance, so that the only force tobe considered is gravitational.

The computation and testing stage of the modeling process involvesfinding a solution of the ODE and comparing it with observed data. If thecomputed solution and the data don’t agree, the model must be modifiedor replaced.

The derivative of a quantity y with respect to time gives its absoluterate of change. It often happens that the relative rate of change, rather than

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the absolute rate of change, fits naturally into a mathematical model . Therelative rate of change of y is defined to be the absolute rate of change,expressed as a fraction of y. The distinction between absolute and relativerates of change can be seen by comparing simple and compound interest.Simple interest is calculated as a percentage of the original deposit, anddoes not change as long as no further deposits are made. The absolute rateof change of value of this account is the dollar amount of interest earnedper unit time. For example, if the account pays 4% interest and $4000 isinitially deposited, then interest payments of 0.04($4000) = $160 will bepaid each year for the life of the account. The value y of the account isdetermined by the ODE

dydt

= 160,

with y(0) = 4000. This indicates a constant absolute rate of change. After tyears, the value will be

4000 +∫ t

0160 dt = $(4000 + 160t).

With compound interest, the interest previously accumulated in the ac-count is included in calculating the interest to be paid. The relative rateof change is the absolute growth rate expressed as a percentage of the ac-count’s current balance. Banks quote this rate as the Annual PercentageRate, and often use the abbreviation “APR.” The formula for the relativerate of change of the balance y is

1y

dydt

.

Assuming that the bank uses “continuous compounding,” as most do, therelative rate of change of the account balance is equal to the APR, which isconstant. If the APR is k, then the ODE

dydt

= k · y, (1.3)

is an appropriate mathematical model . Equation (1.3) is called the lineargrowth equation. The word “linear” refers to the ODE, not its solution.

The linear growth equation can be used as a mathematical model for thebalance on a bank account with continuously compounded interest, as wellas many other phenomena, including population growth and radioactivedecay.

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Example 1.2.1 The family of solutions y = Cekt of equation (1.3) was found inexample 1.1.1. Verify that it is the general solution.

SOLUTION. We must show that any solution y = φ(t) of equation (1.3)belongs to the family y = Cekt, or, equivalently, that the product e−ktφ(t) isconstant.By the product rule for differentiation,

ddt[e−ktφ(t)] = −ke−ktφ(t) + e−ktφ′(t).

Since φ(t) is a solution of equation (1.3), we can replace φ′(t) with kφ(t),and then

ddt[e−ktφ(t)] = −ke−ktφ(t) + e−ktkφ(t) = 0.

By the equal derivatives theorem, e−ktφ(t) = C, where C is constant. Thusφ(t) = Cekt.

Every solution of equation (1.3) will display exponential growth if k > 0,and exponential decay if k < 0. Figures 1.2 and 1.3 show graphs of y = ekt fork > 0 and k < 0, respectively.

Compound interest is a familiar example of exponential growth.

Example 1.2.2 The IVP

y′ = 0.05 y; y(0) = 1000

is a mathematical model for the balance of a savings account that earns 5% interest—compounded continuously—where the initial principal is $1000. Solve this IVP tofind the balance as a function of time.

SOLUTION. The general solution of the ODE, y′ = k y is y = Cek t, where Cis constant. For this bank account, k = 0.05, so y = C e0.05 t. When wesubstitute t = 0 and y = 1000 to incorporate the initial condition, we findC = 1000. Hence y = 1000e0.05 t.

The following example shows how the relative growth rate, k, can becalculated from two data points, y(0) and y(t1) for some t1 > 0.

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Example 1.2.3 An investment of $10,000, made on June 30, 2005 in the Stan-dard and Poor’s 500 Stock Index would grow to $12,920 by December 31, 2007,assuming all dividends were reinvested when received. What rate of compoundinterest would be required to match this performance? If the investment continuedto increase at this rate, when would the investment be expected to double, thusreaching a value of $20,000?

SOLUTION. Let k be the interest rate. Then if y(t) denotes the balance attime t, the IVP y′ = ky; y(0) = 10,000 holds. The general solution of theODE is y = Cekt, and the initial condition specifies C = y(0) = 10,000. Thetime elapsed is t = 2.5 years. Hence 12,920 = 10, 000e2.5k, sok = 1

2.5 ln(1.292) = 0.1025 = 10.25%.Under our assumption that the investment will continue to grow at this rate,the date it would reach the value of $20,000 would be t years after June30, 2005, where 10,000e0.1025t = 20,000. Solving for t by taking logarithms,we have t = 1

0.1025 ln(2) = 6.76 years. Thus we would expect theinvestment to reach $20,000 at the end of March, 2012. (The rate ofgrowth of stock prices in not constant, so we can’t “bank” on thisprediction.)

The doubling time of a solution of (1.3) is the time it takesfor a solution whose initial value is y(0) to reach the value2y(0).

We saw in example 1.2.3 that the growth rate of the stock market forthe 18 months ending December 31, 2007 can be expressed as a doublingtime of about 6.75 years. The relation between the doubling time and therelative growth rate can be expressed as follows:

Doubling time =ln 2

relative growth rate(1.4)

To verify equation (1.4) just note that if k is the relative growth rate and Tthe doubling time, then

ekT = 2.

Taking logarithms, kT = ln 2, which is equivalent to (1.4).A solution of the linear growth equation (1.3) will double over any time

interval of length T, quadruple in 2T units of time, and so on. When nTunits of time have elapsed, the initial value will be multiplied by 2n.

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1.2. LINEAR GROWTH AND DECAY 15

Population growth. Let y(t) represent the population, at time t years af-ter a reference time. Although the population at any time is an integer,we will be using a continuous variable to represent it. This continuous ap-proximation of a discrete variable restricts this mathematical model to largepopulations. Under the assumption that the relative birth and death rates,b and d, respectively (expressed per thousand population), are constant, therelative rate of change for the population is b−d

1000

y′(t) =b− d1000

· y(t),

which is the linear growth equation (1.3), with k = b−d1000 . In the past, birth

and death rates for human populations have not remained constant forlong periods of time, so predictions based on the linear model are onlyreliable over short time spans.

Example 1.2.4 According to the 1980 census, the population of the United Stateswas then 227 million. The birth rate in 1980 was 15.9 per thousand, and the deathrate was 8.7 per thousand. Use these data with the linear growth model to estimatethe population in 2000.

SOLUTION. The relative growth rate is k = 15.9−8.71000 = 0.0072. Let y

represent the population, in millions. Our model takes the form of an IVP,y′ = 0.0072y; y(1980) = 227. The general solution of the ODE isy = Ce0.0072t. If we set t = 1980, and y = 227, we find

227 = Ce(0.0072)(1980)

and hence C = 227e−(0.0072)(1980). The solution of the IVP is therefore

y = 227e−(0.0072)(1980) e0.0072t = 227e0.0072(t−1980).

We therefore estimate that in 2000, the population will bey(2000) = 227e0.0072(20) = 262 million (The 2000 census recorded aconsiderably larger population, 281 million. Thediscrepancy—approximately the population of New York State—shows thelinear growth model is an oversimplification.)

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Radioactive decay. The linear model for the decay of a radioactive iso-tope agrees extremely well with experimental evidence. In this model, y(t)denotes the mass (in grams) of the isotope that has not yet decayed at timet. It is known that the rate at which atoms of the isotope disintegrate is pro-portional to the number of atoms present. Furthermore, all atoms of theisotope have the same mass, so the rate of disintegration is proportional toy(t). The situation is described by a differential equation, called the lineardecay model:

dydt

= −ky.

The parameter k is the decay constant. In this instance, the linear decaymodel is a special case of the linear growth model (1.3), where the growthrate is negative.

The general solution of the linear decay equation is y = C e−kt so wecan say that the mass of the isotope is undergoing exponential decay.

The analog of the doubling time in this case is the half life, which is thetime it takes for a solution of equation (1.3) whose initial value is y0 to reachthe value 1

2 y0.

Example 1.2.5 The half life of the isotope 131I of iodine is 8 days. A nuclearaccident releases 100 kilograms of 131I into the environment. How much 131I willremain in the environment after a period of one year? After two years?

SOLUTION. A rough estimate is made by noting that a year is made up ofslightly more that 45 eight day periods. At the end of each eight dayperiod, there is half as much 131I present as there was at the beginning.Thus, after the first eight days, there are 50 kg of the isotope left, after thefirst 16 days, 25 kg will be left, and so on. After a year, the amount will beabout 100 · 2−45 = 2.84× 10−12 kilograms left. Two years make up morethan 91 eight day periods, so the amount left after two years will be about100 · 2−91 = 4.04× 10−26 kilograms left.

It is prudent to view answers involving extremely large or extremelysmall numbers critically. An atom of 131I has mass 2.17× 10−25 kilograms,and it is impossible to have less than an atom of any substance. There-fore, the estimate we have made of the amount of 131I left after two yearsis wrong. This is not surprising since the model that we are using is notaccurate when only a few atoms of isotope are present.

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1.2. LINEAR GROWTH AND DECAY 17

Homogeneous linear equations

A first order linear ODE is an equation of the form

a1(t)y′ + a0(t)y = b(t).

The functions a1(t) and a0(t) are called the coefficient functions , and b(t)is the source term.

For example, equation (1.3), y′ − k y = 0 is linear, and the falling bodyequation with air drag, v′ + kv2 = g, is not linear because of the quadraticterm in the variable v.

A linear ODE is homogeneous if the source term is 0. Equation (1.3) ishomogeneous, and every homogeneous first order linear ODE can be putin the form

y′ = k(t) y, (1.5)

where k(t) = −a0(t)/a1(t). Thus a homogeneous linear equation with vari-able coefficients can be used as a model for growth or decay when the rel-ative growth rate is a function of time.

To solve a homogeneous first order linear ODE we can mimic the deriva-tion of the solution of equation (1.3) as follows: substitute y = CeK(t), whereK(t) is a function that we will attempt to determine and C is an arbitraryconstant. Using the chain rule, we find that

dydt

= C K′(t) eK(t)

Referring to equation (1.5),

C K′(t) eK(t) = k(t)C eK(t).

Hence K′(t) = k(t). In other words, if K(t) is an antiderivative of k(t) theny = CeK(t) is a family of solutions of equation (1.5). This generalizes whatwe found in the constant coefficient case.

Proposition 1.2.1 Let k(t) be a function that is continuous on an interval (c, d),and let K(t) be an antiderivative of k(t) on that interval. Then the family of solu-tions y = CeK(t) of y′ = k(t)y is the general solution.

PROOF. Proceed as in Example 1.2.1. Let y = φ(t) be any solution ofequation (1.5). By the product rule for differentiation,

ddt[e−K(t) φ(t)] = −K′(t) e−K(t) φ(t) + e−K(t) φ′(t)

= −k(t) e−K(t) φ(t) + e−K(t) k(t) φ(t)= 0.

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18 CHAPTER 1. FIRST-ORDER EQUATIONS

By the equal derivatives theorem, e−K(t) φ(t) is constant, and hence φ(t) =CeK(t).

The homogeneous case

An ODE that can be put into the form a1(t)y′ + a0(t)y = 0 iscalled a homogeneous first order linear equation. If k(t) =−a0(t)/a1(t) is continuous on an interval (c, d) then the gen-eral solution is y = CeK(t), where K(t) is an antiderivative ofk(t) and C is a constant. Thus, the following formula for thegeneral solution holds:

y = Ce−∫(a0(t)/a1(t)) dt

A linear ODE with variable coefficients may have singular points, wherek(t) = −a0(t)/a1(t) is discontinuous. The expression y = CeK(t) can rep-resent the general solution only on an interval that does not contain anysingular points. This comment is applicable to the next example, since theODE there has a singular point at 0. In this example we will substitutey = CeK(t) into the equation without first dividing through by the coeffi-cient of y′.

Example 1.2.6 Find the general solution of

ty′ + my = 0

where m is a constant, on the interval (0, ∞), and find the solution that satisfiesthe initial condition y(1) = −1.

SOLUTION. Substitute y = CeK(t), and y′ = CK′(t)eK(t) to obtain

tCK′(t)eK(t) + mCeK(t) = 0.

Solving for K′(t) we obtain K′(t) = −m/t. Hence

K(t) = −∫ m

tdt = −m ln |t|.

Because our solution is to be found on (0, ∞), |t| = t. Also note that theintegration constant does not appear, because we don’t need the most

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1.2. LINEAR GROWTH AND DECAY 19

general antiderivative of m/t.) It follows that y = Ce−m ln t = C t−m is thegeneral solution. You should check this result by differentiation.To satisfy the initial condition, set t = 1 and y = −1 to obtain −1 = C. Thesolution of the IVP is y = −tm.

Exercises

1. The doubling time for a certain population is 35 years. How longdoes it take for this population to triple?See answer

2. The U.S. national debt amounted to $1 trillion in 1980. By 1988 ithad grown to $2 trillion, and by 1992, $4 trillion. Does this informa-tion indicate that the debt was growing exponentially between 1980and 1992, or does it indicate otherwise?

3. Thomas Robert Malthus1 was concerned that the population willincrease exponentially if it is not controlled by limitations of resourcesor governmental regulation. In his words,

In the United States of America, where the means of sub-sistence have been more ample, the manners of the peoplemore pure, and consequently the checks to early marriagesfewer than in any of the modern states of Europe, the pop-ulation has been found to double itself in twenty-five years.

By how much did the birth rate (per thousand) exceed the death ratein the United States at the time of Malthus?See answer

4. If the birth and death rates of 1980 (see example 1.2.4) were sus-tained indefinitely, how long would it take for the population to dou-ble?Answer: 96 years.

5. Two countries, A and B, each had a population of 1 million in1900. The population of A obeys the exponential growth equation (1.3)

1Essay on the Principle of Population, 1798. The quotation was taken from page 22 ofNorton Critical Edition, edited by Philip Appleman, Norton, New York 1976

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20 CHAPTER 1. FIRST-ORDER EQUATIONS

with relative growth rate k = .04, and it is observed that the popula-tion of B, in millions, is always equal to the square root of the popu-lation of A, in millions. Thus, for example, when the population of Areached 4 million, the population of B was 2 million. Show that thepopulation of B also obeys equation (1.3), and find its relative growthrate.See answer

6. An investor sold her stock in two companies, A and B, for $44,200and $2800, respectively. Each stock had been bought for an initialpurchase price of $1000, with all dividends reinvested. The A stockhad been held for 13.26 years, while the B stock had been held for5.75 years. For each stock, compute the compound interest rate whichwould yield an equivalent return over the same time period. Whichwas the more lucrative investment?

7. A bond, dated 1700, promises to pay the bearer £1 sterling plusaccumulated interest, compounded continuously at 6% per annum.What is the bond worth in 2000?See answer

In Exercises 8–15, find the singular points (if any) and the general so-lution of the ODE on intervals not containing singular points. Sketch thegraph of the solution that satisfies the given initial condition.

8.

(a) y′ − 2y = 0; y(0) = 1

(b) y′ + 2y = 0; y(0) = 1.

9. ty′ − 12y = 0; y(2) = 1.See answer

10. y′ + sin(t)y = 0; y(0) = 1.

11. cos(t)y′ − sin(t)y = 0; y(0) = −1.See answer

12. y′ + ty = 0; y(1) = 1.

13. ty′ + y = 0; y(1) = 1.See answer

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1.2. LINEAR GROWTH AND DECAY 21

14.

(a) ty′ + |t|y = 0; y(0) = 1.

(b) ty′ − |t|y = 0; y(0) = 1.

15. (t2 + 1)y′ − ty = 0; y(3) = 0.See answer

Carbon dating

A radioactive isotope of carbon, 14C, has a half life of 5730 years. The frac-tion of 14C found in atmospheric carbon dioxide is constant at about onepart per billion (abbreviated 1 ppb), because the decay rate is matched bythe rate at which new 14C is created by the effect of cosmic radiation. Aliving plant acquires all of its carbon by respiration of atmospheric carbondioxide; hence the fraction of a living plant’s carbon which is the isotope14C is also 1 ppb. When the plant dies, decaying 14C is not replaced, so thetime elapsed since the death of the plant can be estimated by measuringthe fraction of 14C which remains. This technique, known as 14C radiomet-ric dating, was discovered in 1947 by Willard F. Libby. Before his work,no one had noticed that atmospheric carbon dioxide contained 14C. Libbyreceived the 1960 Nobel Prize for Chemistry as a result of this discovery.

16. Suppose that a tree dies at t = 0. Show that the fraction of 14C inwood from the tree, in ppb, is modelled by the IVP,

y′ = −0.000121y; y(0) = 1.

17. King Tutankhamen died in 1325 B.C.E., and his tomb was discov-ered in 1922 by Howard Carter. What would the proportion of 14C ina wood sample taken from the tomb? (Of course, this wouldn’t havebeen measured until after 1947).See answer

18. A sample of wood from an archaeological investigation contains0.7 ppb 14C. Assuming that the wood, when part of a living tree, had1 ppb 14C, estimate the age of the wood.

19. A sample of wood taken from a bristlecone pine tree on WhiteMountain is determined to be 4850 years old by counting growth

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22 CHAPTER 1. FIRST-ORDER EQUATIONS

rings. What would the expected proportion of 14C in the sample be?See answer

20. 14C radiometric dating is considered useful to determine the ageof objects between 500 and 50,000 years old. Use this information toestimate the following measures of precision of the device used todetermine the number of ppb 14C, on the basis of this statement.

Sensitivity The least concentration that can be detected.

Relative error The measurement error, expressed as a percentage ofthe true concentration.

Answer: sensitivity: .002 ppb, relative error, 6%.

1.3 Linear First Order Equations

A hot object in a cold environment will cool, and the temperature of a coldobject in a warm environment will increase. Newton’s Law of Cooling , isa simple model of this phenomenon. Expressed as an ODE it is

T′(t) = −k[T(t)− A(t)] (1.6)

where T(t) is the temperature of the object at time t, and A(t) is the ambi-ent temperature (the temperature of the environment). The coefficient k isthe transmission coefficient , which tells the rate of heat transfer betweenthe object and its environment.

Equation (1.6) can be written in the form

T′ + kT = kA(t),

and this reveals that it is linear, but inhomogeneous, because of the pres-ence of the source term, kA(t).

To solve an inhomogeneous equation, solve the associated homoge-neous equation—obtained by removing the source term—first. The asso-ciated homogeneous equation for

a1(t)y′ + a0(t)y = b(t), (1.7)

isa1(t)y′ + a0(t)y = 0. (1.8)

Denote the particular solution eK(t) of (1.8) by yh(t). The subscript hstands for “homogeneous;” thus the general solution of the homogeneous

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1.3. LINEAR FIRST ORDER EQUATIONS 23

equation (1.8) is y = C yh(t). The following theorem tells how to expressthe general solution of an inhomogeneous linear ODE.

Theorem 1.2 Suppose that the coefficient functions a0(t) and a1(t), and the sourceterm b(t) in the inhomogeneous linear ODE (1.7) are continuous on an interval(c, d), and that a1(t) 6= 0 on that interval. Let yp(t) be a particular solutionof (1.7), defined on (c, d), and let yh(t) be a particular solution of the associatedhomogeneous equation (1.8). Then the general solution of (1.7) on (c, d) is

y = yp(t) + C yh(t) (1.9)

PROOF. To verify that every function of the form (1.9) is a solution, justsubstitute y = yp(t) + C yh(t) in (1.7):

a1(t)[yp(t) + C yh(t)]′ + a0(t)[yp(t) + C yh(t)] =a1(t)y′p(t) + a0(t)yp(t)︸ ︷︷ ︸

=b(t)

+C[a1(t)y′h(t) + a0(t)yh(t)︸ ︷︷ ︸=0

]

= b(t).

Now let’s verify that every solution of (1.7) belongs to the family (1.9).Let y = φ(t) be any solution of (1.7) that is defined on (c, d), and put ψ(t) =φ(t)− yp(t). Then

a1(t)ψ′(t) + a0(t)ψ(t) =a1(t)φ′(t) + a0(t)φ(t)︸ ︷︷ ︸

=b(t)

−[a1(t)y′p(t) + a0(t)yp(t)︸ ︷︷ ︸=b(t)

]

= 0.

Therefore, ψ(t) is a solution of the associated homogeneous equation (1.8).By proposition 1.2.1, there is a constant C such that ψ(t) = C yh(t). It fol-lows that

φ(t) = yp(t) + C yh(t).

The technique that we will use to determine a particular solution yp(t)of an inhomogeneous linear ODE is known as the method of variation ofconstants 2. We replace the constant C that appears in (1.9) with a newdependent variable v, and substitute the resulting expression in the inho-mogeneous equation. Thus, substitute y = vyh(t) in (1.7) and replace y′

2Some texts refer to the method as ”variation of parameters.”

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24 CHAPTER 1. FIRST-ORDER EQUATIONS

with v′yh(t)+ vy′h(t) (obtained by the product rule for differentiation). Thisyields the equation

a1(t)(v′yh(t) + vy′h(t)) + a0(t)vyh(t) = b(t)

or, more simply,

a1(t)v′(t)yh(t) + v(t)[a1(t)y′h(t) + a0(t)yh(t)] = b(t). (1.10)

Since yh(t) is a solution of the homogeneous equation,

a1(t)y′h(t) + a0(t)yh(t) = 0,

so the expression in square brackets in equation (1.10) drops out. Now wehave

a1(t)v′(t)yh(t) = b(t).

which we solve for v′(t) to get

v′(t) =b(t)

a1(t)yh(t),

and hence

v(t) =∫ b(t)

a1(t)yh(t)dt + C,

where C is constant. Upon multiplying this v(t) by yh(t) we obtain the fol-lowing expression for the general solution of equation (1.7) on the interval(c, d):

y(t) = yh(t)∫ b(t)

a1(t)yh(t)dt + Cyh(t) (1.11)

whereyh = Ce−

∫(a0(t)/a1(t)) dt

is the solution of the homogeneous equation

a1(t)y′ + a0(t)y = 0.

We will revisit the method of variation of constants when solving higherorder linear equations and systems of linear equations. In these cases, morethan one constant will be “varied” — that is the reason for the plural “con-stants” in “variation of constants.”

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1.3. LINEAR FIRST ORDER EQUATIONS 25

Example 1.3.1 Find the general solution of

y′ + 3y = 2e−t

on the interval (−∞, ∞), and determine the solution that satisfies the initial con-dition y(0) = 0.

SOLUTION. The associated homogeneous equation is y′ + 3y = 0, and wecan take yh = e−3t.Substitute y = ve−3t and y′ = v′e−3t − 3ve−3t in the in the inhomogeneousODE to get

v′e−3t − 3ve−3t + 3ve−3t = 2e−t.

When simplified, this reduces to v′ = 2e2t. Thus

v =∫

2e2t dt = e2t + C

Since y = ve−3t, the general solution is the family y = e−t + Ce−3t.To satisfy the initial condition, set y = 0 and t = 0. This yields 0 = 1 + C soC = −1, and

y = e−t − e−3t

As expected, the general solution found in example 1.3.1 splits as thesum of a particular solution, yp = e−t, and the general homogenous solu-tion, Cyh = Ce−3t. Figure 1.5 displays the graphs of several solutions.

In the following example, we apply the method of variation of constantsto a cooling problem.

Example 1.3.2 The temperature in an oven is 200◦C when the oven is turnedoff. After 10 minutes, the temperature is 175◦C. The temperature in the kitchen is20◦C. Find an expression for the temperature of the oven as a function of time.

SOLUTION. We will use the IVP

dTdt

= −k(T − A(t)); T(0) = 200

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26 CHAPTER 1. FIRST-ORDER EQUATIONS

as our model, where T is the temperature in ◦C, and A(t) = 20 is theambient temperature. We can write the ODE simply as T′ + kT = 20k. Theassociated homogeneous equation is

T′ + kT = 0,

and hence Th = e−kt. Substitute T = ve−kt in the inhomogeneous equationand simplify to obtain v′ = 20kekt. Integrate and get

v = 20ekt + C.

The general solution is therefore T = e−ktv = 20 + Ce−kt. To evaluate theparameters C and k, we need to use the data. Substituting the initialcondition T(0) = 200, we get

200 = 20 + Ce0,

and C = 180. Using the second data point, t = 10, T = 175 we get

175 = 20 + 180e−10k,

so e−10k = 155/180, and, k = − 110 ln(155/180) ≈ 0.015. Thus

T(t) = 20 + 180e−0.015t.

The model predicts that the difference between the oven temperatureand the ambient temperature will decay exponentially.

If the object whose temperature is being modeled contains a source ofheat, equation (1.6) must be modified by inserting another source term. LetH(t) denote the rate that heat is generated within the object (H(t) wouldbe negative in some cases, such as air conditioning). Then

T′(t) = −k[T(t)− A(t)] + mH(t), (1.12)

where m is a positive constant, inversely proportional to the heat capacityof the object. For a small object, such as a toaster oven, m would be rel-atively large, while for a large object, such as a domed sports stadium, mwould be small. In practice, we do not need to know the individual valuesof m or H(t); all we need is the product mH(t).

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1.3. LINEAR FIRST ORDER EQUATIONS 27

Example 1.3.3 The insulation of a building has transmission coefficient k = 0.2hour−1. When the furnace of the building is operating continuously and the out-door temperature is 0◦C, the indoor temperature will be maintained at a constant20◦C. Find the temperature inside the building as a function of time, with thefurnace operating continuously, assuming that the outdoor temperature varies be-tween −7◦C and 13◦C each day, and that at time t = 0 the initial temperatureindoors is 20◦C, and outdoors it is 3◦C and getting warmer.

SOLUTION. We are given that when the ambient temperature is 0, thesolution is T(t) ≡ 20. This specifies a constant solution of equation (1.12),which can be used to determine the magnitude of the heat source term.Substitute T(t) = 20, T′(t) = 0, and A(t) = 0 to get

0 = −0.2(20− 0) + mH(t).

Hence mH(t) ≡ 4 degrees per hour.We assume the ambient temperature A(t) varies sinusoidally with aperiod of 24 hours, average temperature Ta =

−7+132 = 3 degrees

centigrade, and amplitude V = 13− Ta = 10 degrees. Hence

A(t) = 3 + 10 sin(

2πt24

),

where t = 0 is the time in the morning that A(t) = Ta. Hence thetemperature satisfies the IVP,

T′(t) = −0.2{

T(t)−[

3 + 10 sin(

πt12

)]}+ 4; T(0) = 20.

The ODE can be written more simply as

T′ + 0.2T = 4.6 + 2 sin(

πt12

).

The homogeneous solution is e−0.2t so we substitute T = e−0.2tv andsimplify to find

v′ = e0.2t [4.6 + 2 sin (.26t)]

( π12 ≈ 0.26). Now integrate (it is fair to use a table of integrals here) to find

v = 23e0.2t + 18.6e0.2t(0.2 sin(0.26t)− 0.26 cos(0.26t)) + C

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28 CHAPTER 1. FIRST-ORDER EQUATIONS

Thus

T(t) = e−0.2tv = 23 + 3.7 sin(0.26t)− 4.8 cos(0.26t) + Ce−0.2t.

When t = 0 and T = 20, and we have 20 = 23− 4.8 + C. It follows thatC ≈ 1.8.

Figure 1.4 displays a graph of the temperature function found in exam-ple 1.3.3. In our solution, we took the trouble to evaluate the constant C, us-ing the initial condition. It is clear from figure 1.4 that this was unnecessaryfor most purposes, because the temperature settles into a periodic regimeafter some time has elapsed. The term Ce−0.2t, which is the homogeneoussolution, is called a transient because it decays to 0 with increasing time.The periodic solution

T = 23 + 3.7 sin(0.26t)− 4.8 cos(0.26t))

of the inhomogeneous ODE is said to be stable because every solution isthe sum of this and a transient term. The keys to this example are that

• There is a stable periodic temperature over 24 hours.

• The mean temperature is 23◦C.

• The temperature ranges between 17◦ and 29◦C (the temperature rangewas determined from the graph).

• The initial condition is practically irrelevant; it determines only thetransient.

The occupants of the building (or a thermostat) will turn off the heatto prevent it from getting as warm as 29◦C, which invalidates our model,as the ODE would no longer be linear (see Exercise 22 at the end of thissection).

Example 1.3.4 Find the solution of the IVP,

ty′ − y = t2 ln(t); y(1) = 2. (1.13)

on the interval (0, ∞).

SOLUTION. The general solution y = Ct−m of the homogeneous equation,

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1.3. LINEAR FIRST ORDER EQUATIONS 29

ty′ + my = 0 was found in example 1.2.6. Here m = −1, so we will use thehomogeneous solution y = Ct.Thus, substitute y = v t, and y′ = v′ t + v in equation (1.13). Aftersimplifying, we have v′ = ln(t). Thus

v =∫

ln(t) dt = t ln(t)− t + C,

and the general solution of the ODE is

y = t2 ln(t)− t2 + Ct.

Now substitute initial conditions (t, y) = (1, 2) to obtain C = 3. The solutionof the IVP is

y = t2(ln(t)− 1) + 3t.

The general solution in example 1.3.4 again has the form yp +Cyh, whereyp = t2(ln(t)− 1) and yh = t.

Exercises

You may check your answers with a CAS, if you like; if an antiderivative ishard to figure out, go ahead and use the CAS for that, too.

In problems 1 – 12, find the general solutions of the ODEs on the inter-vals indicated (if no interval is indicated, the solution should be valid for allreal t). Then find the particular solutions that satisfy the initial conditions.In some of these problems, a pair of differential equations that differ onlyin the sign of the coefficient is given. Use a CAS (or graphing calculator) tocompare the graphs of several solutions of each of these paired equations.In cases where the solutions are defined for all real t, do the solutions seemto converge to some value as t→ ∞, or as t→ −∞?

1. y′ = 3t− 4y on (−∞, ∞); y(0) = 0.See Answer

2. ty′ − y = t3 − 2t on (0, ∞); y(1) = 0.

3. 2y′ + y = t−1e−t/2 on (0, ∞); y(2) = 0.See Answer

4. ty′ + y = 1 on (0, ∞); y(1) = 0.

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30 CHAPTER 1. FIRST-ORDER EQUATIONS

5. y′ + 2ty = e−t2on (−∞, ∞); y(0) = 0.

See Answer

6. y′ + ty = t on (−∞, ∞); y(0) = 0.

7.

(a) y′ + 10y = 1 on (−∞, ∞).

(b) y′ − 10y = 1 on (−∞, ∞).

Initial condition for both: y(0) = 1.See Answer

8.

(a) ty′ + 12y = t3 on (0, ∞).

(b) ty′ − 12y = t3 on (0, ∞).

Initial condition for both: y(1) = 0.

9.

(a) y′ + 4y = 2e−4t sin(2t) on (−∞, ∞).

(b) y′ − 4y = 2e−4t sin(2t) on (−∞, ∞).

Initial condition for both: y(0) = 0.See Answer

10.

(a) y′ + tan(t)y = sec3(t) on (−π/2, π/2).

(b) y′ − tan(t)y = sec3(t) on (−π/2, π/2).

Initial condition for both: y(0) = 1.

11. cos(t)y′ = (y− 1) sin(t); y(0) = 0 on (−π/2, π/2).See Answer

12. y′y−1 = t; y(0) = 1− 10−6 on (−∞, ∞).

13. A penny is heated to 800◦C and is then allowed to cool. Thetemperature after a minute is 600◦, and the room temperature is 20◦.When will it be safe to pocket the coin (the temperature should be lessthan 50◦C)?See Answer

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1.3. LINEAR FIRST ORDER EQUATIONS 31

14. A roast has an internal temperature of 15◦C when it is put in a175◦C oven. After 1 hour, a meat thermometer placed in the roastregisters 50◦C. How much longer will it take for the roast to reach thestate of medium rare (65◦C)?

15. A turkey is at room temperature (20◦C) when it is put into theoven, and it is removed after 5 hours, when its internal temperaturehas reached 85◦C. The oven temperature is 160◦C. The turkey is al-lowed to stand outside the oven for 30 minutes prior to carving. Esti-mate the internal temperature of the turkey when it is carved. Whatassumptions are necessary?

See Answer

16. The building described in example 1.3.3 is given additional insu-lation, reducing the transmission coefficient to k = 0.1 hour−1. Thefurnace output is reduced to half of its capacity to compensate for theadditional insulation. Find the new mean temperature of the buildingand the amplitude of its variation. Ignore transients.

17. At 3:00 am, the temperature inside a house is 15◦C, and the heatis turned off by a timer. When the heat is turned on again at 6:00am, the temperature has fallen to 10◦C. Throughout this period, thetemperature outdoors is−20◦C. Find the rate constant for heat loss inthe building.See Answer

18. (continuation of problem 17) The outdoor temperature remainsat −20◦C. Assume that the furnace produces heat at a constant rate,and remains on until 9:00 am, when the temperature inside the housereaches 20◦C. How long would it take to warm the house from 10◦Cto 20◦C if the temperature outdoors were 0◦C?Answer: 2 hours 17.5 minutes

19. A cabin has two identical wood stoves; only one is in operation.The temperature outside is −15◦C, and the indoor temperature hasstabilized at an uncomfortable 10◦. Therefore, the second stove willbe put into use. At what temperature will the temperature now sta-bilize?See Answer

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32 CHAPTER 1. FIRST-ORDER EQUATIONS

20. When the stove is lighted in a mountain cabin, the temperaturein the cabin is the same as the temperature outside: −20◦C. Afteran hour the temperature inside has reached 0◦C, and after 2 hours,the temperature is 10◦. Find the temperature as a function of time,assuming that the stove continues to operate. How warm will thecabin eventually get?Answer: 20◦C.

21. The insulation of a house has a transmission coefficient of k = .04hour−1. The furnace is capable of heating the house at 1◦C per hour(neglecting heat loss). Assume that the outdoor temperature is 0◦C,and that the heat is alternately turned on for 4 hours, then off for 1hour, by an automatic timer. What will be the average temperatureinside the house, if this continues indefinitely?See answer

22. The building described in example 1.3.3 is equipped with a ther-mostat that turns off the furnace when the interior temperature T >20◦C and turns it on again when T ≤ 20◦C.

(a) Let

H(T) ={

4 if T ≤ 200 if T > 20.

Show that the IVP T′ + 0.2T = 0.6 + 2 sin(

πt12

)+ H(T); T(0) =

20 is an appropriate model for the temperature in the building.

(b) Explain why the ODE in part (a) is not linear.

(c) Use an IVP solver to solve the IVP in part (a). Explain the dips inthe graph and the jagged appearance of the horizontal segments.

23. For each of the following ODEs, find the periodic solution, if thereis one, and decide whether or not it is stable. In other words, expressthe general solution as the sum of the periodic solution and a familyof exponential functions. If the exponential functions decay to 0 ast→ ∞, they represent transients and the periodic solution is stable.

(a) y′ + 5y = 5 cos 2t.

(b) y′ − y = 7 cos 4t.

(c) y′ + 2y = cos t− 3 sin t.

(d) y′ − 5y = 4 cos t + 3 sin t.

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1.3. LINEAR FIRST ORDER EQUATIONS 33

(e) y′ + y = e−t sin t.

See Answer

24. Find the periodic solution of each of the following ODEs andshow that it is stable.

(a) y′ + y = sin(2t)

(b) y′ + y = sin(2000t)

(c) y′ + 5y = cos(πt).

(d) y′ + 10000y = sin t.

(e) y′ + .0006y = sin(.0008t).

(f) y′ + py = cos(ωt), where p > 0 is a constant. What wouldhappen if p < 0?

The remaining problems in this section are designed to be done with aCAS.

25. Find the general solution of each of the following.

(a) y′ + 0.1y = sin(2t).

(b) y′ + 2ty = 1. Attempt to solve this equation first without thecomputer’s help. How does the computer get around the prob-lem of evaluating

∫et2

dt?

(c) ty′ + 12y = [ln(t)]3.

(d)√

ty′ + y = t.

See Answer

26. For each of the following linear differential equations, find thegeneral solution, and graph several solutions by substituting valuesfor the constant, using the domain −10 ≤ t ≤ 10, and range −10 ≤y ≤ 10. Make note of any properties that are common to all solutions.

(a) y′ + y = 1.

(b) y′ − y = 1.

(c) y′ + y = t.

(d) y′ − y = −t.

(e) y′ + y = sin(t).

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34 CHAPTER 1. FIRST-ORDER EQUATIONS

(f) y′ + 0.05y = sin(t).

(g) 1π sin(πt)y′ + cos(πt)y = 1

(h) cos(

π20 t)

y′ +(

π20

)sin(

π20 t)

y = 1.

27. Using the plot range −π ≤ t ≤ π, −10 ≤ y ≤ 10, graph thesolution of the differential equation

ty′ + t cot(t)y = 1

that has a finite value at t = 0. What is that value?See Answer

28. Let g(t) denote the solution of

ty′ − 2y = e−t

that has the property g(1) = g′(1). Draw the graph of g(t) on theinterval 0 ≤ t ≤ 2.

29. Plot the nonconstant solution of y′ + 2y = ty2 that has a relativeminimum at t = 1

4 . What is the value of y(1/4)?See Answer

30. Let φ(t) be the solution of y′ + 2ty = y2 that has a maximum att = 1. Plot the graph of φ(t) on the interval 0 ≤ t ≤ 2, and calculateφ(0).ANSWER: φ(0) = 1.074386372.

1.4 Mixture Problems

A typical mixture problem involves a tank that initially contains V0 litersof a salt solution, with concentration C0 grams per liter. A salt solutioncontaining K grams of salt per liter is being poured into the tank at J litersper second; simultaneously, solution is pumped out of the tank at L litersper second. The problem is to find the concentration of salt in the tank as afunction of time. See figure 1.6.

The mathematical model of the mixture problem depends on a simpli-fying assumption, the uniformity hypothesis: the concentration of the solutionthroughout the tank is uniform. Let C(t) denote the salt concentration in thetank, in grams per liter.

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1.4. *MIXTURE PROBLEMS 35

We will use the amount of salt in the tank, denoted x(t), rather than theconcentration C(t), as the dependent variable. We can determine the rate ofchange of x(t) by subtracting the rate at which the salt removed (the outputrate) from the rate that salt enters the tank (the input rate). The input rateis JK grams per minute, since J liters of solution, each containing K gramsof salt, enter the tank each minute. The output rate is LC(t), grams perminute, where C(t) is the concentration of salt in the tank. Thus

x′(t) = JK− LC(t).

To complete this model, we have to express the concentration in termsof x(t). Let V(t) be the volume of solution in the tank at that time. If V0 isthe initial volume in the tank, then

V(t) = V0 + (J − L)t

since the tank gains J − L liters per minute. Therefore,

C(t) =x(t)V(t)

=x(t)

V0 + (J − L)t.

It follows that x(t) satisfies the linear ODE

dxdt

= JK− Lx

V0 + (J − L)t, (1.14)

with the initial condition x(0) = C0V0.In our first example, the volume V is constant, because the rate J at

which fluid flows in is the same as the rate L at which fluid is pumped out.

Example 1.4.1 An accident has caused 10 kg of potassium permanganate to spillinto a mountain lake, coloring its water purple. The lake covers an area of 100square meters, and its average depth is 5 meters. A stream feeds the lake at 1000liters per minute, and another stream takes water from the lake at the same rate.

Find the concentration of potassium permanganate in the lake as a function oftime, assuming that the uniformity hypothesis holds in this situation. How muchof the substance will remain in the lake after 24 hours have elapsed?

SOLUTION. The volume V of the lake is

5 m. deep× 100 m2 × 1000 liters per m3 = 5× 105liters.

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36 CHAPTER 1. FIRST-ORDER EQUATIONS

Let x(t) denote the amount of potassium permanganate dissolved in the

lake; the concentration is C(t) =x(t) kg

5× 105 liters.

The input of potassium permanganate zero (the stream feeding the lakewas not affected by the spill), and the output is the product of theconcentration and the rate of flow: C(t) kg/liter × 1000 liters per minute.Thus

x′(t) = − x(t)5× 105 × 1000 = −0.002 x(t) kg/minute

Thus, the amount of potassium permanganate in the lake satisfies thelinear decay equation, x′ = −0.002 x. It follows that x(t) = Ae−0.002 t, whereA = 10, the initial value of x. The half-life of x(t) is (ln(2)/0.002) ≈ 347minutes, about 6 hours. Since 24 hours amounts 4 half-lives, we expectthat 10× 2−4 ≈ 0.6 kg will remain in the lake after a day.

When the volume of solution is not constant, we encounter variablecoefficients in the input-output equation, as in the following example.

Example 1.4.2 A tank initially contains 1000 liters of salt solution with 70 gramsof salt per liter. A solution containing 120 grams of salt per liter enters the tankat the rate of 9 liters per minute, and the well mixed solution is pumped out at 10liters per minute. Find the concentration of salt as a function of time.

SOLUTION. The salt input rate is

(120 grams/liter)× (9 liters/min.) = 1080 grams/min.

The volume of solution in the tank is decreasing at a rate of 1 liter/min., soV(t) = 1000− t liters. Thus, the output rate is(

x1000− t

grams/liter)×(

10 liters/min.)

The input-output model results in the ODE

dxdt

= 1080− 10(

x1000− t

),

or, in standard form,

x′ +(

101000− t

)x = 1080.

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1.4. *MIXTURE PROBLEMS 37

The solution of the associated homogeneous equation is

exp(−∫ 10

1000− tdt)= e10 ln(1000−t) = (1000− t)10.

Now we substitute x = v(1000− t)10 in the differential equation, andsimplify to get

v′ = 1080(1000− t)−10.

Integration yields v = 120(1000− t)−9 + A, where A is a constant. Sincex(t) = v(1000− t)10, it follows that

x(t) = 120(1000− t) + A(1000− t)10 (1.15)

The initial value is x(0) = (70 grams/liter)× (1000 liters) = 70, 000 grams;thus equation (1.15) with (t, x) = (0, 70,000) yields70, 000 = 120, 000 + A · 100010, or A = −50× 1000−9. If we substitute thisvalue of A into equation (1.15), and simplify a little, we will have

x(t) = 120(1000− t)− 50(1000− t)(

1− t1000

)9

.

To obtain the concentration C(t), divide by V(t) = 1000− t:

C(t) = 120− 50(1− .001t)9 grams per liter.

This formula is valid for 0 ≤ t < 1000. When t = 1000 the tank becomesempty, and the model is no longer applicable.

Input-output problems in personal finance

Many investment plans consider factors such as inflation and the effect ofregular withdrawals. Suppose, for example, that a person has a pensionaccount of P dollars upon retirement, and withdraws money at a rate ofE(t) dollars per year to meet living expenses. If y(t) denotes the accountbalance t years after retirement, then the rate of change of y(t) is the rate atwhich interest income is received minus E(t).

If r denotes the interest rate, then our model is an ODE,

y′ = ry− E(t).

Assuming that the expenses grow with inflation according to the expo-nential growth law E′ = k E, where k is the annual rate of inflation, theexpenses after t years will be E(t) = E0ekt, where E0 is the initial rate ofwithdrawal.

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38 CHAPTER 1. FIRST-ORDER EQUATIONS

Example 1.4.3 Ms. Doe retired yesterday at age 65. Her IRA account has aprincipal of $450,000, which invested with a guaranteed interest rate of 5.25%,compounded continuously. Her budget calls for annual expenses of $20,000, withprojected inflation of 2.5%. Determine Ms. Doe’s savings account balance t yearsafter her retirement. How long will her money last?

SOLUTION. Let y(t) denote the balance t years after retirement, in dollars.The input in this problem is interest amounting to 0.0525y(t), and theoutput is expenses of 20, 000e0.025t dollars per year. Thus

y′ = 0.0525y− 20000e0.025t.

This can be put into the standard form for a linear ODE:

y′ − 0.0525y = −20000e0.025t.

The solution of the associated homogeneous equation is y = Ce0.0525t, sowe substitute y = ve0.0525t and simplify to obtain v′ = −20, 000e−0.0275t.Therefore v = 20, 000e−0.017t/0.0275 + C, and since y = ve0.0525t,

y =200000.0275

e0.025t + Ce0.0525t

Since y(0) = 450, 000, and 20, 000/0.0275 ≈ 730, 000,450, 000 = 730, 000 + C, and thus C ≈ −280, 000. Ms. Doe’s accountbalance in t years will be about

y(t) = 730, 000e0.025t − 280, 000e0.0525t.

This function is decreasing, and y(t) = 0 when 730, 000 = 280, 000e0.0275t;that is, when e0.0275t = 2.6. This will be when Ms. Doe reaches the age of100.

Exercises

In Exercises 1–10, assume that the uniformity hypothesis holds.

1. A tank initially contains 100 liters of water. A 16% salt solutionenters the tank at 2 liters per second. Brine is pumped out of the tankat 2 liters per second. How long does it take for the concentration ofsalt in the tank to reach: 8%? 12%? 15%?See answer

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1.4. *MIXTURE PROBLEMS 39

2. A 4% salt solution is poured into a large tank at 2 liters per second.Brine is removed from the tank at 1.9 liters per second. Given thatthe tank initially contains 50 liters of 2% salt solution, what is theconcentration of salt in the tank after one minute has elapsed?Answer: 3.79%

3. A tank contains 16 liters of 100 parts per million (ppm) potassiumiodide (KI), in which the iodide is the unstable isotope 131I, with ahalf life of 8 days. Solution is drained from the tank at the rate of 1liter per day, and fresh 100 ppm solution is added at the rate of 1 literper day. Find the steady state concentration of 131I in the tank.See answer

4. (continuation of problem 3) Suppose fresh 100 ppm solution ofK 131I is added at the rate of 2 liters per day, while the well mixed so-lution drains out at 1 liter per day, as before. If the initial volume is 16liters and the initial concentration is 100 ppm, find the concentrationafter 8 days have elapsed.Answer: 63 ppm

5. A beverage bottling plant has had an accident. Someone pouredquinine (intended for the tonic) in the orange drink mixing tank, andthe resulting liquid is unpalatable. No one knows (or will say) whenthis happened, but we must find out, because the orange drink bot-tled since the mishap must not leave the plant. The concentration ofquinine in the tank is now 0.01%. The tank now holds 10000 liters,and orange drink is being transferred to the carbonation tank at 1000liters per hour. Fresh ingredients have been added to the tank at therate of 900 liters per hour. It is known that the amount of quinine thatwas put into the tank was not more than 4 kilograms, and that it wasintroduced into the tank within the last 24 hours. Estimate how longago, at worst, that the quinine was introduced.See answer

6. A tank contains 4 kg of salt dissolved in 100 liters of water. A saltsolution with concentration 0.01 kilograms per liter enters the tank atthe rate of 2 liters per day. Evaporation removes 1 liter of water perday (salt does not evaporate), and an additional liter of solution isdrained from the tank each day. Find the concentration of salt in thetank as a function of time.

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40 CHAPTER 1. FIRST-ORDER EQUATIONS

7. When brought to the emergency room, an accident victim has 3liters of blood, and is still losing blood at 0.25 liter per hour. He is im-mediately given continuous blood transfusions at 0.5 liter per hour,and an antibiotic drug is administered intravenously at 0.5 gram perhour. Four hours later, the bleeding is stopped, and the transfusionsstop in another two hours. Determine the concentration of the antibi-otic in the patient’s blood at the time that the transfusions stop.See answer

8. Two mixing tanks each initially contain 2 liters of pure water. A10% salt solution enters the first tank at 0.1 liter per minute. The wellmixed solution is pumped out of this tank at the same rate, and intothe second tank. Instantaneous and perfect mixing occurs in the sec-ond tank, and the resulting solution is removed from it, again at therate of 0.1 liters per minute. Determine the concentration of salt in thesecond tank as a function of time.

9. Consider two mixing tanks, labeled A and B. Initially, tank A con-tains 3 liters of 12% salt solution, and tank B contains 1 liter of water.Both tanks are stirred constantly, and solution is pumped from tankA to tank B at 0.2 liters per hour. Solution is also pumped from tankB to tank A at the same rate. Find the salt concentration of each tankas a function of time. Hint: Since this is a closed system, the totalamount of salt held by the tanks is constant.See answer

10. A 200 liter solar water heater absorbs heat at a rate proportional tothe sine of the angle of the sun over the horizon. The sun rises at 0600and sets at 1800 on the 24 hour scale. Assume that the heat absorptionrate is 1500 sin

(π12 (t− 6)

)kilocalories per hour, where t is hours after

midnight of the equinox. Of course, no heat is absorbed at night —the absorption rate is 0 for 0 < t < 6, 18 < t < 30, etc.

The water heater loses heat by conduction to the environment, and byhot water usage. The ambient temperature is 20◦C, and the rate con-stant for heat conduction to the rest of the house is k = 0.01 hour−1.Hot water is removed from the tank at an average rate of 5 liters perhour, and is replaced at the same rate by cold water with a tempera-ture of 10◦C with instant and perfect mixing. If the water in the tankis at 60◦C at 0600, find its temperature 24 hours later.

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1.5. *BERNOULLI EQUATIONS 41

Hint: treat this as a mixture problem, and keep track of heat inputand output. One kilocalorie of heat will raise the temperature of oneliter by one centigrade degree.

11. A retired citizen’s living expenses are $30,000 per year now, andhe wants to allow for inflation of 3% per year. How much moneyshould he invest, at 6% interest, so that he can meet his living ex-penses and not touch the principal, no matter how long he lives?See answer

12. An endowed chair in mathematics is to be funded with one mil-lion dollars. The endowment committee believes it can realize at leastan 8% return on the funds, and expects to offer the occupant of thechair a 5% salary increase each year to keep pace with inflation. Whatshould the initial salary be, if the principal of the endowment fund isnever to be drawn upon?

13. The age-specific death rate d(t) is the number of deaths of individu-als who are t years old per 1000 individuals of that age. Assume thatd(20) = 2.8, d(50) = 17.6, and that d(t) satisfies the linear growthequation d′(t) = kd(t). Of a sample of 100,000 twenty year olds, howmany will survive to age 50?See answer

1.5 Bernoulli Equations

It is sometimes possible to convert a nonlinear first order ODE to a linearone by a clever change of variables, For example, substitute v = ym in thenonlinear ODE,

ym−1y′ + p(t)ym = q(t) (1.16)

By the chain rule, dv/dt = mym−1y′. Multiply equation (1.16) by m andmake the substitution. The resulting ODE,

v′ + mp(t)v = mq(t),

is linear, and thus we can find its general solution. It is then a simple matterto replace v with ym and obtain the general solution of (1.16).

A Bernoulli equation is a first order nonlinear ODE of the form

y′ + p(t)y = q(t)yn, (1.17)

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42 CHAPTER 1. FIRST-ORDER EQUATIONS

where n 6= 0, 1. The power n to which y is raised on the right side is theexponent. Observe that by dividing equation (1.17) by yn, it can be madeto assume the form of equation (1.16), in which m = 1− n. Thus, the sub-stitution v = ym also converts Bernoulli equation to a linear equation.

The first Bernoulli equation was encountered by Johann Bernoulli in1695. Bernoulli was trying to determine the trajectory of a projectile subjectto the forces of gravity and air resistance. He knew that the force of air re-sistance would be proportional to the square of the speed v of the projectile,and derived the ODE,

dvdt

= −kv2 − g sin(θ), (1.18)

where g is the gravitational acceleration and θ denotes the inclination of thetrajectory from the horizontal. Of course, θ is not constant, and Bernoulliderived an ODE that it would satisfy:

dt= − g

vcos(θ) (1.19)

The two ODEs, (1.18) and (1.19), form a coupled system, a topic that wewill take up in chapter 2.

It may seem strange to do so, but Bernoulli decided to treat the speed vas a function of the inclination θ. Then, by the chain rule,

dvdt

=dvdθ

dt,

and hencedvdθ

=v′(t)θ′(t)

.

Bernoulli divided (1.18) by (1.19) to obtain

dvdθ

=k v3

g cos(θ)+ tan(θ) v. (1.20)

Example 1.5.1 Show that (1.20) is a Bernoulli equation and make the appropriatesubstitution to convert it to a linear ODE.

SOLUTION. Rearrange (1.20) in the form

dvdθ− tan(θ) v =

kg

sec(θ) v3 (1.21)

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1.5. *BERNOULLI EQUATIONS 43

This has the form of (1.17), where v is the dependent variable, θ is theindependent variable, p(θ) = − tan(θ), and q(θ) = (k/g) sec(θ). Thus it isa Bernoulli equation, with exponent n = 3. To linearize, define a newvariable u = v1−n = v−2. Then

dudθ

= −2v−3 dvdθ

;

ordvdθ

= −v3

2dudθ

.

Substitute this expression in (1.21) to obtain

−v3

2dudθ− tan(θ) v =

kg

sec(θ) v3.

Divide through by v3 to obtain

−12

dudθ− tan(θ) v−2 =

kg

sec(θ).

Finally, replace v−2 with u, and multiply through by −2 to obtaindudθ

+ 2 tan(θ) u. = −2kg

sec(θ).

Exercises

Find the general solution of the following equations:

1. ty2y′ + y3 = 1.See answer

2. y′ + (tan t)y = y2.

3. y′ + 3y = y3 sin t.See answer

4. e2yy′ + e2y = e−2t.

5. y′ + y2t+1 = 12[(2t + 1)/y]3.

See answer

6. y′ + (1− 3t )y = y4/3

t .

7. Can your CAS handle Bernoulli equations? Attempt to find thegeneral solution of each of these equations with your computer.

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44 CHAPTER 1. FIRST-ORDER EQUATIONS

1.6 Separable ODEs

A separable ODE has the form

dydt

= g(t) h(y), (1.22)

in which the right side can be expressed as the product of two single-variable functions. Here are some examples:

dydt

= t2 sin y

anddydt

= et+3y (= et e3y)

are separable. Homogeneous linear ODEs are separable because a1(t) y′ +a0(t)y = 0 is equivalent to

y′ = −(

a0(t)a1(t)

)y = g(t) h(y),

where g(t) = − a0(t)a1(t)

and h(y) = y. However, inhomogeneous linear ODEsare not separable (unless the coefficients and source term are constants).

For an example of a nonlinear ODE that is not separable, consider

dydt

= 1 + t y2.

The reason this equation is not separable is that the right side can’t be fac-tored as the product of a function of t and a function of y.

ODEs in terms of differentials Consider an ODE

dydt

= F(y, t). (1.23)

If y = φ(t) is a solution of (1.23) then the differentials of t and y are relatedby the equation dy = φ′(t) dt. Also, because φ is a solution,

φ′(t) = F(φ(t), t) = F(y, t).

It follows that the (1.23) is equivalent to

dy = F(y, t) dt. (1.24)

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1.6. SEPARABLE ODES 45

This form of an ODE is often convenient, and we will call (1.24) an ODE indifferential form.

Let’s put the generic separable ODE, (1.22), in differential form:

dy = g(t) h(y) dt, (1.25)

and see how to approach solving it. We will want to divide through byh(y), but we have to be careful, since this would be questionable if h(y) haszeros. Thus, at the outset, find the zeros of h(y). Suppose that h(y0) = 0and plug y = y0 (constant function) into either (1.22) or (1.25). Becauseboth sides (either equation) are 0, this y = y0 is a solution of the ODE.

To find the solutions that are not constant, divide (1.25) by h(y) to getthe following separated equation:

dyh(y)

= g(t) dt. (1.26)

Becauseφ′(t) dth(φ(t))

= g(t) dt (1.27)

implies that φ′(t) = g(t)h(φ(t)), every solution of (1.26) is also a solutionof the original ODE (1.22). It must be emphasized that constant solutionsof (1.22) are usually not solutions of the separated equation, (1.26).

Integrating the separated equation Let H(y) =∫ 1

h(y) dy be an antideriva-

tive of h(y)−1 and let G(t) =∫

g(t) dt. If y = φ(t) is a solution of theseparated equation, then we can make the following substitutions in equa-tion (1.26):

1/h(φ(t)) ; H′(φ(t))dy ; φ′(t) dt

g(t) ; G′(t).

This yieldsH′(φ(t)) φ′(t) dt = G′(t) dt. (1.28)

By the chain rule,ddt

H(φ(t)) = H′(φ(t)) φ′(t).

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46 CHAPTER 1. FIRST-ORDER EQUATIONS

Thus, (1.28) tells us that H(φ(t)) and G(t) have equal derivatives. It fol-lows from the equal derivatives theorem that on any interval upon whichH(φ(t)) and G(t) are both defined, there is a constant C such that

H(φ(t)) = G(t) + C. (1.29)

Equation (1.29) can be expressed as H(y) = G(t) + C. This is the solu-tion of (1.22) in implicit form.

If H(y) happens to be strictly increasing or strictly decreasing, there isa unique inverse function H−1, and

φ(t) = H−1(G(t) + C)

is the general solution of the separated equation (1.26).Here’s a summary of what we’ve learned so far:

Solving a separable ODE

To solve the ODE

dy = g(t) h(y) dt

start by finding the constant solutions y ≡ yi. This is doneby determining the roots y1, y2, . . . of h(y) = 0. Then findthe nonconstant solutions by integrating both sides of theseparated equation:∫ dy

h(y)=∫

g(t) dt.

As a first example, consider the ODE (1.1) that is a model for the velocityof a falling object with a drag force.

Example 1.6.1 Solve the IVP v′ = g− k v2; v(0) = 0.

SOLUTION. In differential form we have dv = (g− k v2) dt. The constant

solutions are v = ±v∞, where v∞ =

√gk

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1.6. SEPARABLE ODES 47

We have noted that the ODE, which—because g = kv2∞—we can write as

dv = k(v2∞ − v2) dt,

is separable. Of the two constant solutions, v = ±v∞, v = v∞ representsthe terminal velocity, while v = −v∞ is extraneous, as it indicates aconstant upward velocity. In fact, the ODE (1.1) is incorrect for v < 0,because the motion and air resistance cannot be in the same direction.See exercise 16. Divide through by (v∞ + v)(v∞ − v) to get the separatedequation:

1(v∞ + v)(v∞ − v)

dv = k dt

Expanding in partial fractions, we get

12v∞

(1

v∞ − v+

1v∞ + v

)dv = k dt

which we integrate and obtain

12v∞

(ln |v∞ + v| − ln |v∞ − v|

)= kt + C.

The value of C can be determined at this point by referring to the initialcondition, t = 0 and v = 0. We get C = 0, and after making thisreplacement and simplifying the logarithmic expression

ln∣∣∣∣v∞ + vv∞ − v

∣∣∣∣ = 2kv∞t,

To solve for v, first exponentiate

100 + v100− v

= ±e2kv∞t.

and choose the plus sign since it is in accord with the initial condition.Thus

v = v∞e2kv∞t − 1e2kv∞t + 1

= v∞ tanh (kv∞t)

The solution of example 1.6.1 has the expected properties: starting froma velocity of 0, the object will accelerate and its velocity will approach theterminal velocity v∞.

To explore further the properties of the linear and quadratic models, seeExercise 6.

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48 CHAPTER 1. FIRST-ORDER EQUATIONS

Example 1.6.2 Solve the initial value problem

y′ = t y−1; y(0) = −1

SOLUTION. Since h(y) = y−1 6= 0 for all y, there are no constant solutions.The separated equation is y dy = t dt, and integration of both sides yields12 y2 = 1

2 t2 + C. To determine the value of C, substitute t = 0 and y = −1:

12(−1)2 =

12(0)2 + C

so C = 12 . Thus 1

2 y2 = 12 t2 + 1

2 . Solving for y yields two solutions

y = ±√

t2 + 1

Only one of these solutions,

y = −√

t2 + 1,

is valid, since the initial condition specifies that y(0) is negative. Figure 1.8shows several solutions of the differential equation y′ = t y−1.

It is sometimes impractical to find a formula for y as an explicit func-tion of t and the constant C, and even if it is possible to do this, it may beunnecessary. We will say that a function F(t, y) is an integral of an ODEy′ = f (t, y) if, for all solutions y = φ(t) of the ODE, F(t, φ(t)) is a constantfunction of t. Technically speaking, a constant function F(t, y) ≡ C will bean integral for any ODE, but it does not define a solution implicitly. Inte-grals are usually obtained by integrating the ODE in some way, as we havedone in the process of solving separable equations.

If F(t, y) is a function of two variables, its level curves are the curvesdefined by the equations F(t, y) = C, where C is constant. If we have foundan integral F(t, y) of an ODE, then the graphs of all solutions will be subsetsof level curves of F. For example, in the process of working out the solutionof example 1.6.2, we found that

F(t, y) =12

y2 − 12

t2 is an integral of y′ = t y−1.

It follows that if y = φ(t) is a particular solution of y′ = t y−1, then thegraph of φ(t) is a subset of a hyperbola F(t, y) = C.

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1.6. SEPARABLE ODES 49

Example 1.6.3 Find the general solution of

y′ =2yt

. (1.30)

SOLUTION. Since h(y) = y has a zero at y = 0, there is a constantsolution, y ≡ 0. The separated equation is

dyy

=2 dt

t.

Integrating both sides, we have

ln |y| = 2 ln |t|+ C,

so F(t, y) = ln |y| − 2 ln |t| is an integral. Simplifying, we have

F(t, y) = ln∣∣∣ yt2

∣∣∣ .

Since F(t, y) is an integral, so is ±eF(t,y) = y/t2. Thus, solutions take theform y/t2 = C, or y = C t2, for some constant C. It happens that theconstant solution y ≡ 0 that we found at the outset belongs to this family(put C = 0).

In the following example we can again find a solution y ≡ 0 becausethe right side vanishes when y = 0. This time, the constant solution doesnot belong to the family obtained by integrating. When this happens, theconstant solution is called a singular solution.

Example 1.6.4 Find a family of solutions of the ODE

y′ = 3y2/3. (1.31)

SOLUTION. We see that there is the constant solution y ≡ 0. Let’s turn tothe separated equation

13

y−2/3 dy = dt.

for y 6= 0. Integrating, we have y1/3 = t + C; thus

y = (t + C)3,

which is not constant for any value of C. Thus our constant solution issingular. The graphs of several solutions from the family are shown infigure 1.9. Notice that the t-axis (the graph of the singular solution) istangent to the graph of each of these solutions.

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50 CHAPTER 1. FIRST-ORDER EQUATIONS

Exercises

1. Which of the following ODEs is separable?

(a) dydt = t2+5

ty

(b) y dt = t dy

(c) dydt = ety

(d) dydt = e−(t

2+y2)

(e) dydt = cot y

(f) y′ = t+yt−y

See answer

2. Find an integral for each of the following ODEs, if it is separable.The non-separable equations should be skipped!

(a) y′ = tan(ty).

(b) y′ = ln(ty).

(c) y′ = sec y.

(d) y′ = ln(t + y)

3. A ball whose mass is 1 kilogram rolls on a level surface, subjectonly to the force of friction. The initial velocity is 1 meter per second;after 1 second, the velocity is 0.8 meters per second. Assuming themagnitude of the frictional force is proportional to the velocity,

(a) What is the friction constant?

(b) When will the velocity be .5 meters per second?

(c) How far did the ball roll in the first second?

(d) Evaluate an improper integral to determine how far the ball willroll if given an infinite amount of time.

See answer

4. Repeat problem 3, assuming the magnitude of the frictional forceis proportional to the square of the velocity. The friction constant isnow the ratio between the magnitude of the friction force and v2.Answer: (a) 0.25 kg/m. (c) 0.893 m.

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1.6. SEPARABLE ODES 51

5. An object with a mass of 1 kg in free fall is subject to gravitationalforce of 9.8 newtons and a frictional force of −bv, where v is the ve-locity. Given that the terminal velocity is v∞ = 10 m/s, calculate b.See answer

6. An object falls from rest. The terminal velocity is observed to be98 meters per second. Assuming that the force of air resistance is

(a) proportional to the velocity,

(b) proportional to the square of the velocity,

how long does it take for the object to reach a speed of 49 meters persecond? Plot graphs of the velocity as a function of time to make acomparison of the two models.Answer: (a) 6.93 sec., (b) 5.49 sec.

In problems 7 – 15, find a family of solutions for the ODE. Then findthe particular solution that satisfies the given initial condition (be sure tospecify its domain).

7. dydt = (t− 1)y2; y(1) = 0.

See answer

8. y′ = 3t2y2; y(0) = −1.

9. dydt = y

2t ; y(1) = 2.See answer

10. tyy′ = 1; y(e) = 2.

11. y′ = yet; y(0) = 1.See answer

12. dy = (y2 + 1) dt; y(0) = 1.

13. dydt = y2; y(0) = 1.

See answer

14. dydt = 1+y√

t; y(4) = 0.

15. y′ = t(y2 − 1); y(0) = 0.See answer

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52 CHAPTER 1. FIRST-ORDER EQUATIONS

16. If an object is initially moving upward (we consider upward ve-locity to be negative) then the force of air resistance will be downward(positive), and the quadratic model for air resistance implies the fol-lowing ODE holds during the upward motion:

v′ = g + kv2.

When v > 0, the sign of the quadratic term should be changed tominus.

(a) Show that the ODEv′ = g− kv |v|

is applicable to both upward and downward motion.

(b) Suppose that the terminal velocity of the object is 100 m/s. Theobject is thrown upward with an initial velocity of 100 meters/second(v(0) = −100). Give its velocity as a function of time for the du-ration of its upward motion.

(c) How far upward does the object go?

(d) Find the velocity as a function of time for downward motion.

(e) Sketch the graph of v(t) for t > 0.

17. A basketball is dropped from a tower. There is no wind, and thedrag force is proportional to the square of the velocity. After 2 secondsof fall, its velocity is 14.7 meters per second. Calculate its terminalvelocity. It will be necessary to use Newton’s method.See answer

18. In this exercise, we must modify our model for the motion of afalling body to take into account a variable mass. Let m(t) denote themass of a falling body. Then the velocity of the body is a solution ofthe following ODE, derived from Newton’s second law of motion:

m(t)v′ = −m(t) g− k|v|v.

A rocket has a mass of 6 kilograms, including 2 kilograms of fuel.The fuel burns uniformly for 20 seconds, leaving no residue, and pro-duces a constant thrust of 120 newtons. After 20 seconds, the fuel isspent, and the only forces are gravity and air resistance. The force ofair resistance is proportional to the square of the velocity, with dragcoefficient k = 0.02 kilograms per meter. Write initial value problemsto determine the velocity of the rocket as a function of time for

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1.7. EXACT FORM 53

(a) 0 ≤ t ≤ 20,(b) 20 ≤ t ≤ T1, where T1 is the time that the rocket reaches its

maximum altitude, and for(c) T1 ≤ t ≤ T2, where T2 is the time that the rocket hits the ground.

Solving these ODEs is optional, and is recommended only for readerswho are using a CAS. Use the convention that upward motion haspositive velocity.

19. A ball is thrown upward with a velocity v0 (negative). Its terminalvelocity in free fall is v∞, and the drag force is proportional to thesquare of the velocity. Find formulas for the following as as functionsof v0 and v∞.

(a) the maximum height attained by the ball,(b) the time taken to reach that height,(c) the time taken to return to the ground, and(d) the velocity when the ball hits the ground.

Answer: |v0|v∞√v2

0+v2∞

What happens if the drag force is negligible (v∞ = ∞)?See answer

20. Does the ball in problem 19 spend more time on the way up or onthe way down?Answer: On the way down.

21. A ball falls from rest from a high tower, with drag force propor-tional to the square of its velocity. How long does it take for the ballto attain a speed of half its terminal velocity?See answer

22. A ball is thrown downward from a high tower, with v(0) = 4v∞.How long does it take for the ball to reach a speed of 2v∞? Assumethe drag force is proportional to the square of the velocity.Answer: t = (|v∞| ln(1.8))/(2g)

1.7 Exact Form and Integrating Factors

If we start with a function F(x, y) that has continuous partial derivatives,we can ask, is F an integral for some ODE? Suppose that there is such an

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54 CHAPTER 1. FIRST-ORDER EQUATIONS

ODE, and let y = φ(x) be a solution. Thus

F(x, y) = C

would be an implicit solution of that ODE.Let’s apply implicit differentiation with respect to x:

∂F∂x

+∂F∂y· dy

dx= 0. (1.32)

Equation (1.32) is an ODE that has F as an integral. The answer to ourquestion is yes.

The total differential of F(x, y) is defined as

dF =∂F∂x

dx +∂F∂y

dy.

The ODE (1.32) is equivalent to the equation dF = 0.One approach to integrating an ODE

P(x, y) dx + Q(x, y) dy = 0 (1.33)

would be to try to find a function F(x, y) whose total differential is

dF = P(x, y) dx + Q(x, y) dy.

This function F(x, y) would have to satisfy the requirements

∂F∂x (x, y) = P(x, y) and ∂F

∂y (x, y) = Q(x, y). (1.34)

Equations (1.34) overdetermine F—that is, usually there is no function thatsatisfies both equations. If function F(x, y) that satisfies (1.34) for all (x, y)in a domainD in the x, y-plane does exist, the ODE (1.33) is said to be exacton D. The remainder of this section will be devoted to explaining how torecognize exact ODEs, and how to find an integral for an exact ODE. Wewill also see that—in some circumstances—we can replace an ODE that isnot exact with an equivalent ODE that is exact.

Theorem 1.3 Assume that the first partial derivatives of P(x, y) and Q(x, y) arecontinuous on a rectangular domain D in the plane. Then the ODE (1.33) is exacton D if and only if

∂P∂y

(x, y) =∂Q∂x

(x, y) (1.35)

for all (x, y) ∈ D

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1.7. EXACT FORM 55

We will call condition (1.35) the exactness condition.PROOF. First we will prove that if the ODE (1.33) is exact, then the

exactness condition holds (this establishes that the exactness condition isnecessary). Suppose that there is a function F(x, y) such that P(x, y) =∂F∂x (x, y) and Q(x, y) = ∂F

∂y (x, y). Since P and Q have continuous first partialderivatives, F has continuous second partial derivatives on the domain D.

We can now refer to a theorem on partial derivatives that says if a func-tion F(x, y) has continuous second partial derivatives on a domain in theplane, then

∂2F∂x∂y

=∂2F

∂y∂x.

Thus∂P∂y

=∂

∂y

(∂F∂x

)=

∂2F∂y∂x

and∂Q∂x

=∂

∂x

(∂F∂y

)=

∂2F∂x∂y

are equal: the exactness condition holds.Now we will show that if P(x, y) and Q(x, y) satisfy the exactness con-

dition, then there is a differentiable function F(x, y) defined on D such thatdF = P(x, y) dx + Q(x, y) dy.

Let (x0, y0) be a point in D. Define a function

H(y) =∫ y

y0

Q(x0, s) ds.

andF(x, y) =

∫ x

x0

P(t, y) dt + H(y).

By the fundamental theorem of calculus,

∂F∂x

= P(x, y) +∂

∂xH(y) = P(x, y).

Since P has continuous partial derivatives, we can apply the Leibniz rulewhen differentiating F with respect to y:

∂F∂y

=∂

∂y

∫ x

x0

P(t, y) dt + H′(y)

=∫ x

x0

∂P∂y

(t, y) dt + H′(y).

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56 CHAPTER 1. FIRST-ORDER EQUATIONS

Now we will use the exactness condition to replace ∂P∂y (t, y) with ∂Q

∂t (t, y)in the integral. This enables us to use another form of the fundamentaltheorem of calculus:∫ x

x0

∂Q∂t

(t, y) dt = Q(x, y)−Q(x0, y).

Thus we have shown that

∂F∂y

= Q(x, y)−Q(x0, y) + H′(y). (1.36)

Since H(y) is by definition an antiderivative of Q(x0, y) the last two termsin (1.36) cancel, and the proof is complete.

Example 1.7.1 Which of the following ODEs are exact?

(i) (y2 + x2 − 2x + 3) dx + (2xy− y2 + 10) dy = 0.

(ii) x(y2 + x2 − 2x + 3) dx + x(2xy− y2 + 10) dy = 0.

SOLUTION.

(i) ∂P∂y (x, y) = 2y and ∂Q

∂x (x, y) = 2y so the equation is exact.

(ii) ∂P∂y (x, y) = 2xy and ∂Q

∂x (x, y) = 4xy− y2 + 10x; therefore theexactness condition does not hold.

Observe that equation (ii) in example 1.7.1 was obtained by multiplyingequation (i) by x, but only equation (i) is exact. This shows that even thoughtwo equations may be equivalent, one may be exact while the other is not.

To integrate an exact equation, we will use a simplified version of themethod used in the proof of theorem 1.3. Start by computing one of thefollowing indefinite integrals, whichever is easier:

M(x, y) =∫

P(x, y) dx or N(x, y) =∫

Q(x, y) dy.

When integrating with respect to x to find M(x, y), treat y as a constant,and when computing N(x, y), treat x as a constant. Let us assume that wehave computed M(x, y). Then any function of the form

F(x, y) = M(x, y) + H(y)

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1.7. EXACT FORM 57

will satisfy the requirement ∂F∂x = P(x, y). We can view the H(y) term here

as a “constant of integration,” which depends on y since y was a constantwhen the antiderivative M(x, y) was computed.

Now set ∂F∂y = Q(x, y). In other words, solve the equation

∂M∂y

+ H′(y) = Q(x, y)

to determine H(y). This yields

H(y) =∫ (

Q(x, y)− ∂M∂y

)dy.

Of course, H(y) is not allowed to depend on x. The exactness conditionensures that all terms involving x in the difference Q(x, y)− ∂M

∂y cancel out,since

∂x

(Q(x, y)− ∂M

∂y

)=

∂Q∂x− ∂2M

∂x∂y

=∂P∂y− ∂P

∂y= 0.

Therefore F(x, y) = M(x, y) + H(y) is an integral of the ODE (1.33).

Example 1.7.2 Find an integral for the ODE

(y2 + x2 − 2x + 3) dx + (2xy− y2 + 10) dy = 0.

SOLUTION. According to example 1.7.1, part (i), this equation is exact.The integral F(x, y) will have the form

F(x, y) =∫(y2 + x2 − 2x + 3) dx + H(y)

= xy2 +13

x3 − x2 + 3x + H(y).

To determine H(y), differentiate this expression with respect to y:

∂F∂y

= 2xy +dHdy

.

Since ∂F∂y = Q(x, y) = 2xy− y2 + 10, it follows that

2xy +dHdy

= 2xy− y2 + 10.

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58 CHAPTER 1. FIRST-ORDER EQUATIONS

After cancelling, we obtain dHdy = −y2 + 10. Therefore H(y) = − 1

3 y3 + 10y,and the integral is

F(x, y) = xy2 +13

x3 − x2 + 3x− 13

y3 + 10y.

To find an explicit solution, we would have to solve the cubic equationF(x, y) = C for y in terms of x.

Integrating factors

Equation (ii) of example 1.7.1 isn’t exact, but comparing it with the exactequation (i) of the example shows that it can be put into exact form by mul-tiplying by an appropriate function, namely x−1. This inspires the follow-ing definition: A function m(x, y) is an integrating factor for the equation

P(x, y) dx + Q(x, y) dy = 0. (1.37)

ifm(x, y)P(x, y) dx + m(x, y)Q(x, y) dy = 0

is exact.If we can determine an integrating factor for a given ODE, then we have

an equivalent ODE in exact form, and we can find an integral. In specialcircumstances, we can put this strategy into action.

One-variable integrating factors. Suppose that the ODE (1.37) has an in-tegrating factor m. Then

m P(x, y) dx + m Q(x, y) dy = 0

is exact, so by the exactness condition,

∂y[m P(x, y)] =

∂x[m Q(x, y)] (1.38)

Equation (1.38) is a PDE whose unknown function is the integratingfactor m. It is not always easy to solve for m, but if it happens that there isan integrating factor m(x) or m(y) that depends on only one of the variablesx or y, then equation (1.38) reduces to a separable ODE, and the integratingfactor can be found.

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1.7. EXACT FORM 59

Assume that there is an integrating factor of the form m(x). By applyingthe product rule for differentiation to equation (1.38), we obtain

m(x)∂P∂y

= m′(x) Q(x, y) + m(x)∂Q∂x

.

This equation can be rearranged as

m′(x)m(x)

=1

Q(x, y)

(∂P∂y− ∂Q

∂x

)(1.39)

Since its left side is independent of y, equation (1.39) only makes sense if itsright side is also independent of y. Thus, if the right side is not independentof y, there is no integrating factor m(x). If the right side is independent ofy it is straightforward to integrate both sides and calculate m(x).

The same reasoning shows that there is an integrating factor m(y) if andonly if

1P(x, y)

(∂Q∂x− ∂P

∂y

)is a function of y alone, and that in this case

m′(y)m(y)

=1

P(x, y)

(∂Q∂x− ∂P

∂y

)(see exercise 21).

Do not memorize the above material. It is easy to derive at a moment’snotice from the exactness condition (in the form of equation (1.38)) by mak-ing the assumption that m is a function of x alone or of y alone.

Example 1.7.3 If possible, find an integrating factor m(x) for the ODE

[2(x + y2) cos x + sin x] dx + 2y sin x dy = 0 (1.40)

and use it to determine an integral for the ODE.

SOLUTION. An integrating factor m(x) must satisfy

∂y[m(x)(2(x + y2) cos(x) + sin(x))] =

∂x[m(x)(2y sin(x))]

If you do the differentiation (remember to use the product rule on the rightside) you will obtain

m(x)(4y cos(x)) = m′(x)(2y sin(x)) + m(x)(2y cos(x)).

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60 CHAPTER 1. FIRST-ORDER EQUATIONS

We can cancel a factor of y and simplify this equation to the form

m′(x)m(x)

=cos(x)sin(x)

.

Now we can integrate: ln(|m(x)|) = ln(| sin(x)|) + C. Therefore,m = sin(x) is an integrating factor. After multiplying (1.40) by thisintegrating factor, we have the exact equation

[2(x + y2) sin(x) cos(x) + sin2(x)] dx + 2y sin2(x) dy = 0.

Since it’s easier to integrate Q(x, y) = 2y sin2(x) with respect to y than it isto integrate P(x, y) with respect to x, set

F(x, y) =∫

2y sin2(x) dy + K(x) = y2 sin2(x) + K(x).

Then ∂F∂y = 2y sin2(x). Since ∂F

∂x = 2y2 sin(x) cos(x) + K′(x),

2y2 sin(x) cos(x) + K′(x) = 2(x + y2) sin(x) cos(x) + sin2(x),

and K′(x) = 2x sin(x) cos(x) + sin2(x). A final integration shows thatK(x) = x sin2(x) and so F(x, y) = (x + y2) sin2(x) is the integral that wehave sought. Graphs of all solutions of the ODE are subsets of levelcurves

(x + y2) sin2(x) = C.

Example 1.7.4 Find, if possible, a one-variable integrating factor for the ODE(2xy− 2

x3

)dx +

(4x2 +

3x2y

)dy = 0, (1.41)

and use it to integrate the ODE.

SOLUTION.We’ll start by looking for an integrating factor m(x). If there is one, it mustsatisfy

∂y

[m(x)

(2xy− 2

x3

)]=

∂x

[m(x)

(4x2 +

3x2y

)].

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1.7. EXACT FORM 61

After you carry out the differentiation and simplify, you should obtain

m(x)(−6x +

6x3y

)= m′(x)

(4x2 +

3x2y

),

and thenm′(x)m(x)

=6− 6x4y4x5y + 3

Since the right side depends on y, this equation has no solution, andhence there is no integrating factor m(x).To find if there is an integrating factor m(y) we must find solutions (if any)of

∂y

[m(y)

(2xy− 2

x3

)]=

∂x

[m(y)

(4x2 +

3x2y

)].

Now we use the product rule in differentiating the left side of the equation:

m′(y)(

2xy− 2x3

)+ m(y)(2x) = m(y)

(8x− 6

x3y

).

You can simplify this equation and obtain

m′(y)m(y)

=3y

.

Integrate both sides to get ln |m(y)|) = 3 ln |y|+ C. Thus m(y) = y3 is anintegrating factor. Multiplying equation (1.41) by y3 results in the exactequation (

2xy4 − 2y3

x3

)dx +

(4x2y3 +

3y2

x2

)dy = 0.

Put

F(x, y) =∫(2xy4 − 2x−3y3) dx + H(y)

= x2y4 + x−2y3 + H(y).

Then ∂F∂y = 4x2y3 + 3x−2y2 + H′(y), and it follows that H′(y) ≡ 0. The

integral is F(x, y) = x2y4 + x−2y3.

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62 CHAPTER 1. FIRST-ORDER EQUATIONS

Exercises

General instruction for problems 1 – 14: decide if the ODE is exact; if it is,find an integral, and if it isn’t, do nothing. You are encouraged to use acomputer algebra system to help find antiderivatives.

1. (2x + 5y + 3) dx + (5x− 4y + 2) dy = 0.Answer

2. y dx + (x + y) dy = 0.

3. (y2 − 1) dx + (2xy− x) dy = 0.Answer

4. (y2 − y) dx + (2xy− x) dy = 0.

5. 3e3xy(ln y− 1) dx + ln y(e3x − y) dy = 0.Answer

6. (x2 + 2xy− y2) dx + (x2 − 2xy− y2) dy = 0.

7. x(x2 + y2 − 1) dx + y(x2 + y2 + 1) dy = 0.Answer

8. x(3√

x2 + y2 − 2) dx + y(3√

x2 + y2 + 2) dy = 0.

9.[

1 +(

yx+y

)2]

dx +

[(x

x+y

)2− 1]

dy = 0.

Answer

10. y dxx + ln x dy = 0.

11. (3x2 + 6xy + 9y2) dx + (3x2 + 18xy + 51y2) dy = 0.Answer

12. 2xy dx + y2−x2

y2 dy = 0.

13. 2xy dx + (y2 − x2) dy = 0.Answer

14. 2xy dx− (y2 − x2) dy = 0.

In problems 15 – 19, find an integrating factor for the given ODE, anduse it to determine an integral.

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1.7. EXACT FORM 63

15. (x2 + xy2 + 1) dx + 2y dy = 0.Answer

16. x dy− (y− x) dx = 0.

17. (x2 + 2x + 2xy + 2y + 3y2) dx + (2x + 6y) dy = 0.Answer

18. (5x + 20y + 28y3) dx + (5x + 21xy2) dy = 0.

19. 2xy dx + (3x2 + 2y) dy = 0.Answer

20. Show that the ODE (5xy− 2y2) dx + (5x2 − 3xy) dy = 0 does nothave an integrating factor m(x) or m(y).

21. Show that there is an integrating factor m(y) for the ODE P(x, y) dx+Q(x, y) dy = 0 if and only if the expression

1P(x, y)

(∂Q∂x− ∂P

∂y

)is independent of x.Answer

22. Show that if y′ = f (x, y) is a separable ODE, with f (x, y) =g(x) h(y) then [h(y)]−1 is an integrating factor for

dy− f (x, y) dx = 0.

Thus, the method for solving separable ODEs is a special case of theintegrating factor method.

23. Write the linear ODE, y′ + p(x)y = q(x), in the equivalent form

dy + [p(x)y− q(x)] dx = 0,

and find a one-variable integrating factor m(x). (This integrating fac-tor, discovered by Leibniz in 1692, has been memorized by ten gener-ations of students in ODE courses.)Answer

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64 CHAPTER 1. FIRST-ORDER EQUATIONS

1.8 Graphical Analysis of ODEs

Imagine a sheet of graph paper representing a section of the t, y-plane. Wewant to sketch some solutions of an ODE y′ = f (t, y) on this paper. Thereare vertical grid lines t = ti and horizontal grid lines y = yj, and we willcall the points (ti, yj) where the grid lines intersect grid points.

A direction field for the ODE y′ = f (t, y) can be drawn by placing ateach grid point (ti, yj) a short line segment, called an element. The elementis centered at the grid point, and its slope is equal to f (ti, yj). Figure 1.10shows a direction field for the ODE y′ = t − y2 on graph paper coveringthe region −1.5 ≤ t ≤ 1.5, −1 ≤ y ≤ 1.

The elements of the direction field can be drawn at only a finite setof points, but we should imagine an element located at each point of theplane. If y = φ(t) is a solution of the ODE, then at each point (t, φ(t)) onthe graph of φ the slope of the graph of φ is equal to f (t, φ(t)), the slope ofthe direction field element at that point. Thus at each point, the graph of asolution is tangent to the element at its location.

Figure 1.11 shows two typical direction fields, as drawn by a CAS. Theybelong to the ODEs,

v′ = 10− .001v2, (1.42)

andv′ = 10− .001e−.01tv2, (1.43)

respectively. Both equations represent the velocity of a falling body. In (1.42),the drag coefficient is constant, while in (1.43), it decays with time. The hor-izontal elements at v =

√1000 in the constant drag case indicate that solu-

tions of (1.42) tend to a terminal velocity while the direction field of (1.43)does not have this property, and hence there is no terminal velocity.

Graphing the solution of an IVP

To make a rough sketch of the solution of an IVP, y′ = f (t, y); y(t0) = y0,draw the direction field element centered at (t0, y0). Let (t1, y1) be the pointat the right end of that element, and draw the right half of the direction fieldelement at (t1, y1), extending to a point (t2, y2). Continuing this process, weobtain a polygonal curve (broken line graph) that approximates the graphof the solution of the IVP. The accuracy of the approximation depends onthe on the length of the direction field elements being used. With shorterelements the graph will be more accurate, but requires more effort to draw.

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1.8. GRAPHICAL ANALYSIS OF ODES 65

The endpoints (t0, y0), (t1, y1), (t2, y2), . . . of the direction field elementsyield two sequences tn and yn. We will stipulate that the direction fieldelements are drawn so that the tn are evenly spaced, with tn = t0 + nh,where h is a constant increment.

We will use “difference notation:” For any sequence u0, u1, u2, . . . ofnumbers, form a sequence of forward differences

∆u0 = u1 − u0, ∆u1 = u2 − u1, . . . , ∆un = un+1 − un, . . . .

The forward differences for the sequence tn are constant: ∆tn = h. Wecan calculate the forward differences for the yn because the slope of thedirection field element at the point (tn, yn) is

∆yn

∆tn= f (tn, yn).

Thus,∆yn = h f (tn, yn). (1.44)

Equation (1.44) is a difference equation. Difference equations are likeODEs, and we can exploit them to learn more about ODEs (and vice versa).

The analogy between difference equations and differential equations isillustrated by compound interest. Bank A pays r% interest, compoundedk times per year. If an initial deposit of $P is made, and the account isforgotten, let’s determine how the balance grows. Let yn denote the balanceat the end of the nth compounding period. The interest payment to theaccount will be

∆yn =r

100kyn

The account balance at any time is determined by solving this differenceequation with initial condition y0 = P.

Bank B uses continuous compounding, thus replacing the above differ-ence equation with an ODE,

y′ =r

100y

with initial condition y(0) = P to track the balance. Both methods give ap-proximately the same result, with the value due to discrete compoundingincreasing as k → ∞ and approaching the value due to continuous com-pounding in the limit.

In the context of difference equations, an IVP takes the form

∆ym = F(m, ym); y0 = A.

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66 CHAPTER 1. FIRST-ORDER EQUATIONS

Provided that F(m, y) is defined for all (m, y) such that m is a nonnegativeinteger, this IVP is guaranteed to have a unique solution. We don’t evenhave to assume F is continuous. To see why this is so, rewrite the differenceequation as

ym+1 = ym + F(m, ym). (1.45)

Equation (1.45) is recursive, in the sense it gives a formula for computing thenext term in the sequence. Since we have a starting value for y0 (the initialcondition), y1, y2, and so on can be computed, one after the other. Manywell known sequences are defined recursively—for example the factorialsequence n! is defined by 0! = 1 and n! = n · (n− 1)!.

Euler’s method

Leonhard Euler, one of the brightest stars in the history of mathematics,showed how to use a difference equation IVP to approximate the solutionof a differential equation IVP. Let y = φ(t) denote the solution of the IVP,

y′ = f (t, y); (1.46)y(t0) = A. (1.47)

Choose a time step h (h must be nonzero, but is allowed to be negative)and set tn = nh + t0. Euler’s method uses the difference equation (1.44)with initial condition y0 = A to determine a sequence y0, y1, y2, . . . . In thissequence, yn serves as the approximation of φ(tn).

We have pictured the approximate solution of an IVP given by Euler’smethod by plotting the sequence of points (tm, ym) and connecting adjacentpoints with straight line segments to form a polygonal curve.

Example 1.8.1 Use Euler’s method with time step h = ±0.25 to approximate thesolution of the IVP y′ = t− y2, y(0) = 0, for −1.5 ≤ t ≤ 1.5

SOLUTION. The solution must be propagated to the left and to the rightfrom the initial point (0, 0). To propagate leftward, we use h = −0.25; andto propagate rightward, h = 0.25. We will start with h = −0.25. Euler’smethod then generates a sequence ym defined by the initial conditiony(0) = 0 and the difference equation ∆ym = −0.25(tm − y2

m), where t0 = 0,t−1 = −0.25, t−2 = −0.5, and so on. Since t−6 = −1.5, we will need tocompute ym for −1 ≥ m ≥ −6.

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1.8. GRAPHICAL ANALYSIS OF ODES 67

Write the difference equation in recursive form as

ym−1 = ym − 0.25(tm − y2m).

Take a moment to calculate y−1 through y−6, starting with y0 = 0. Youshould get y−1 = 0, y−2 = 0.0625, y−3 ≈ 0.1885, and after three moresteps, y−6 ≈ 1.0972.This calculation approximates values of the solution y = φ(t) of the IVP atthe points t = −.25,−.5.− .75,−1,−1.25, and −1.5. Due to the large stepsize, these approximations poor. A CAS provides a solution of the IVP andevaluation yields φ(−0.5) = 0.1266 and φ(−1.5) = 1.7857 (both rounded to4 decimal places).To propagate to the right, start over with h = +0.25. The numberst0, t1, t2 . . . are 0, 0.25, 0.5 . . . , Again, calculate y1 through y6; you shouldget y1 = 0, y2 = 0.0625, and so on, with y6 = 0.8098. Again, theapproximations you will obtain are not accurate. The values of the solutionφ(t) computed by the CAS, rounded to 4 decimal places, areφ(0.25) = .0312, φ(0.5) = 0.1235, φ(0.75) = 0.2700, and φ(1.5) = 0.8574.Figure 1.12 shows the graph of the approximate solution that we haveproduced, and the exact solution produced by the CAS, with the directionfield for the ODE in the background.

To show that Euler’s method really does produce an approximation ofthe solution φ(t) of a given IVP, we need to refer to a special case of Taylor’stheorem, the second mean value theorem, The proof can be found in anycalculus text.

Theorem 1.4 (Second mean value theorem) Suppose that a function φ(t) istwice differentiable on an interval (A, B), and that t0 and h are numbers such thatt0, t0 + h ∈ (A, B). Then there is a number c1 between t0 and t0 + h, such that

φ(t0 + h) = φ(t0) + h φ′(t0) +12

h2 φ′′(c1). (1.48)

For any differentiable function φ(t),

φ(t0 + h) ≈ φ(t0) + hφ′(t0)

is the linear approximation of φ. The second mean value theorem gives a wayto see how precise it is.

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68 CHAPTER 1. FIRST-ORDER EQUATIONS

Euler’s method relies on linear approximation because each step uses it.If φ(t) is the solution of the IVP (1.46), (1.47) then we know that φ(t0) = y0and

h φ′(t0) = h f (t0, y0) = ∆y0

(∆y0 = y1 − y0 as in (1.44)). If φ is sufficiently differentiable then by theo-rem 1.4

φ(t0 + h) = y0 + ∆y0︸ ︷︷ ︸y1

+12

h2 φ′′(c1).

and hence φ(t0 + h) − y1 = 12 h2 φ′′(c1), where c1 is between t0 and t1. As

h→ 0, the error term 12 h2φ′′(c1)→ 0 at a faster rate, and that is why y1 is a

good approximation of φ(t0 + h) if h is sufficiently small.We still need to justify the Euler approximation as it proceeds through

many steps. In the second step, Euler’s method follows the direction fieldelement starting at (t1, y1). Let φ1(t) be the solution of the IVP

y′ = f (t, y) y(t1) = y1

Applying theorem 1.4 to φ1(t), we find that

φ1(t1 + h) = φ1(t2) = y1 + ∆y1︸ ︷︷ ︸y2

+12

h2 φ′′1 (c2).

with c2 between t1 and t2. The approximation error after the second step is

φ(t2)− y2 = [φ(t2)− φ1(t2)] + [φ1(t2)− y2]

= [φ(t2)− φ1(t2)] +12

h2 φ′′1 (c2).

The two terms of this expression are accumulated error, and the localerror, respectively. The local error is defined to be the error due to one stepof the approximation. In the step where y2 is computed, the local error,which we denote LE2, is equal to φ1(t2) − y2. The purpose of referring tothe second mean value theorem is to estimate local error, and it tells us thatLE2 = 1

2 h2 φ′′1 (c2).The accumulated error, denoted AEm, is the error due to all previous

steps taken in propagating the solution by Euler’s method. Thus the firstaccumulated error is equal to

AE1 = φ(t2)− φ1(t2).

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1.8. GRAPHICAL ANALYSIS OF ODES 69

This is the difference between two solutions of the same ODE at t = t2,given that the two solutions are close when t = t1.

As we continue, each step approximates the solution φm(t) of a newIVP, whose initial point is the point (tm, ym) reached in the previous step.There is a new local error LEm+1 = 1

2 φ′′(cm+1), and a new accumulatederror,

AEm = φ(tm+1)− φm(tm+1)

We can’t expect that AEm will be equal to the sum of the local errors LE1,LE2 . . . , LEm. Solutions of ODEs with nearby initial conditions may divergeaway from each other (if this happens, the accumulated error will be largerthan the sum of the previous local errors), or they may converge towardeach other, causing the accumulated error to be less than the sum of thelocal errors.

Example 1.8.2 Determine the solution of the initial value problem

y′ = y +14

; y(0) = 0.

Using a time step h = 0.5, use Euler’s method to calculate

(i) y1, y2, y3, and y4.

(ii) LE1, LE2, LE3 and LE4.

(iii) AE1, AE2 and AE3.

SOLUTION. We will start by computing y1 . . . y4. The large time step willmake the approximation unreliable, but we are studying errors.The difference equation prescribed by Euler’s method is∆ym = 0.5(ym + 0.25), with initial condition y0 = 0. Solving this differenceequation IVP, we obtain the following (See details).

i 0 1 2 3 4ti 0.0 0.5 1.0 1.5 2.0yi 0.0 0.125 0.3125 0.59375 1.015625

To determine the errors precisely, we need to have the general solution ofthe ODE (in practice, we will never have access to this, since the purposeof Euler’s method is to approximate a solution that can’t be obtainedanalytically). Since y′ = y + 1

4 is a linear ODE, you can derive the solution

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70 CHAPTER 1. FIRST-ORDER EQUATIONS

y = Cet − 14 . The solution that we are trying to approximate is satisfies the

initial condition φ(0) = 0, so C = 14 .

Each solution φm(t), satisfies initial conditions φ(tm) = ym, soφm(t) = Cmet − 1

4 where the constant Cm is determined by

ym = Cmetm − 14

.

Thus Cm = (ym + 0.25)e−tm and

φm(t) = (ym + 0.25)et−tm − 0.25.

In particular

φ1(t) = 0.375et−0.5 − 0.25φ2(t) = 0.5625et−1 − 0.25φ3(t) = 0.84375et−1.5 − 0.25

Figure 1.13 displays the graphs of φ(t), and φm(t) for m = 1, 2, 3. It alsoshows the broken line graph connecting (t0, y0), (t1, y1), (t2, y2), (t3, y3),and (t4, y4). Notice that the graph of φ(t) is tangent to the first segment ofthe broken line graph, φ1(t) is tangent to the second segment, and so on.The local and accumulated errors are shown in the following table (Seedetails), in which LEi = φi−1(ti)− yi and AEi = φ(ti)− φi−1(ti).

i 1 2 3 4LEi 0.037 0.056 0.084 0.125AEi — 0.061 0.193 0.456

In figure 1.13 we can picture LE4 as the vertical distance between thebottom two curves at t = 2 (at the right edge of the graph). AE3 is thedistance from the top curve to the second-to-the-bottom curve, andrepresents the consequences of the local errors LE1, LE2, and LE3.

Example 1.8.3 The velocity y(t) (in meters per second) of a certain rock sinkingin sea water is given by the differential equation

y′ = 8− 2y2. (1.49)

Given that y(0) = 0, use Euler’s method with h = 0.1 to approximate the solutionfor 0 ≤ t ≤ 2 seconds.

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SOLUTION. The difference equation given by Euler’s method with h = 0.1is

∆ym = 0.1(8− 2y2m).

The solution of this difference equation is the sequence beginning0,0.8,1.472,. . . . By connecting successive points (tm, ym), (tm+1, ym+1)with line segments we obtain a broken line graph that serves as anapproximation of the solution of our initial value problem. This graph isshown in figure 1.14. Although Euler’s method can’t be expected to give avery accurate approximation with such a large time step, it should be notedthat the graph shown levels off at a velocity of 2 meters per second, whichis the correct terminal velocity. This difference equation, like the differentialequation it approximates, has a constant solution, ym = 2, and our solutionconverges to it. In other words, the qualitative behavior of the Eulerdifference equation matches that of the differential equation. Unfortunately,this does not always happen—see problem 21 at the end of this section.

How to implement Euler’s method with a spreadsheet

Euler’s method uses four columns. Column A lists t0, t1, t2, t3, . . . ; ColumnB is for y0, y1, y2, y3, . . . ; column C is for ∆t (all entries will be the same);and we put ∆y0, ∆y1, ∆y2, ∆y3, . . . in column D. Enter the initial values of tand y in A1 and B1, respectively, and put ∆t in C1. In D1, type the formulafor ∆y. Thus, if the ODE is

y′ = y(1− y) +1

10sin2(t) (1.50)

the formula

=C1 * (B1 * (1 - B1) + (SIN(A1)ˆ2)/10)

would appear in cell D1.Cell A2 will contain the formula for t1, =A1+C1, and cell B2 will have

the formula for y1, =B1+D1 One could copy cell C1 to C2, but it is better toplace in cell C2 the formula =C1 because this makes it possible to changethe time step by altering cell C1. Copy D1 to the clipboard and paste to cellD2. Because copying is relative, if you are working with equation (1.50),the expression

=C2 * (B2 * (1 - B2) + (SIN(A2)ˆ2)/10)

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72 CHAPTER 1. FIRST-ORDER EQUATIONS

will appear in D2.Now copy row 2 of the spreadsheet to the clipboard, and paste to rows

3, . . . , n. Column B will contain the approximations of yi corresponding tothe ti in column A.

Spreadsheets are capable of producing a graph of the approximate so-lution. The menu of graph types contains many choices; the appropriateone is the “xy-graph,” with x-axis values from column A and “first series”values from column B. Although a polygonal graph is appropriate, com-mercial spreadsheets can draw smooth curves by employing sophisticatedinterpolation techniques. Smooth curves are no more accurate than thepolygonal graphs, but they look nicer.

Figure 1.15 displays the spreadsheet for the approximate solution ofequation (1.50), with initial condition y(0) = 0.5 and ∆t = 0.25.

Exercises

1. Exercise 16 in section 1.6 the ODE

v′ = g− kv |v|

introduced a model for the velocity of a ball that was initially thrownupward, subject to gravitational and drag forces (in the model, thepositive direction is downward). Draw a direction field for this ODE.Use as parameters g = 9.8 meters per second2 and k = 0.002 meters−1.Answer

Print exercises 2 – 7, and sketch graphs of solutions of the IVPs on thegiven direction fields.

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2. y′ = ty; initial conditions y(0) = 0, y(0) = 1, and y(0) = −1.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

3. y′ = 3t− 2y; initial conditions y(0) = −1, − 34 , and 0.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

Answer

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74 CHAPTER 1. FIRST-ORDER EQUATIONS

4. y′ = 2te−y; initial conditions y(0) = 0, y(1) = 0, and y(−1) = 0.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

5. y′ = 1t−3y ; y(0) = 1.

-2 -1 0 1 2 3

-1

-0.5

0

0.5

1

1.5

Answer

6. y′ = 2t; y(0) = 0.

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1.8. GRAPHICAL ANALYSIS OF ODES 75

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

7. y′ = −2y; y(0) = −1.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

Answer

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76 CHAPTER 1. FIRST-ORDER EQUATIONS

8. Match the ODEs with their direction fields.

(a) y′ = sin y

(b) y′ = t2 − y2

(c) y′ = y2 − 2ty + t2

(d) y′ = y− y2

(e) y′ = t2 − y

(I) -2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

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1.8. GRAPHICAL ANALYSIS OF ODES 77

(II) -2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

(III) -2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

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78 CHAPTER 1. FIRST-ORDER EQUATIONS

(IV) -2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

(V) -7.5 -5 -2.5 0 2.5 5 7.5

-4

-2

0

2

4

9. Show that the graph of the solution y(t) of the IVP y′ = t − 3y;y(0) = 1 crosses the line t− 3y = 0 exactly once, and has a relativeminimum at the crossing point. Do not use the explicit solution, y =19 (3t − 1 + 10e−3t). Instead, consider what the slope of the solutionwill be at a crossing point.Answer

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1.8. GRAPHICAL ANALYSIS OF ODES 79

10. Find the sequence of forward differences for each of the followingsequences.

(a) 1,2,4,8,16,32,. . . ,2m,. . .

(b) 1,3,9,27,81,243,. . . ,3m,. . .

(c) 0,1,8,27,64,125,. . . ,m3,. . .

(d) 1, 12 , 1

3 , 14 , 1

5 ,. . . , 1m ,. . .

11. For each difference equation, determine the first three or fourterms of the sequence ym and then find an expression for ym.

(a) ∆ym = 1, y1 = 0.

(b) ∆ym = ym, y1 = 1.

(c) ∆ym = kym, y1 = C, where k and C are constants.

(d) ∆ym = ym + 1, y1 = 0.

Answer

12. The bank offers you a choice of an account bearing an interestrate of 6%, compounded continuously, and an account with 6.25%interest, compounded annually. Both rates are guaranteed for fiveyears, and you don’t intend to make withdrawals before that time isup. Which account should you choose?

13. Let am and bm be sequences. The difference equation

∆ym = amym + bm

is called the linear difference equation with coefficients am and sourcebm. You will want to compare this definition with that of a linear ODE.If bm = 0 for all m, the difference equation is homogeneous.

Define a new sequence Am by A1 = 1, and

Am = (1 + a1)(1 + a2) · · · (1 + am−1) for m > 1.

(a) Show that every solution of a homogeneous difference equation∆ym = amym has the form ym = CAm, where C is constant.

(b) Explain how to solve an initial value problem involving an in-homogeneous linear difference equation. Hint: the procedureresembles the variation of constants method for solving an inho-mogeneous linear ODE.

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80 CHAPTER 1. FIRST-ORDER EQUATIONS

(c) Find the general solution of

∆ym =1m

ym.

(d) Use the method of part (b) to solve

∆ym =1m

ym + m + 1, y1 = 0.

Answer

14.

(a) Show that the solution of the IVP

y′ = f (x); y(a) = b, (1.51)

isy(x) = b +

∫ x

af (u) du. (1.52)

(b) Show that approximating the solution of the IVP 1.51 by Euler’smethod amounts to calculating a Riemann sum for the integralin equation (1.52).

Use Euler’s method and a spreadsheet or programmable calculator tofind approximate solutions of the differential equations in problems 15and 16, over the intervals indicated, and with the prescribed step sizes.Draw broken line graphs of the solutions.

15. y′ = y2; y(0) = 1. h = .05, 0 ≤ t ≤ 1.Answer

16.y′ = t√

t2+y2; y(0) = 1. h = ±0.1

for −1 ≤ t ≤ 1. Can you save time by exploiting the symmetry in thedifference equations for propagating to the left and to the right?

17. Let y = φ(t) denote the solution of the IVP,

y′ = t + y; y(0) = 0.

Since the differential equation is linear, you can solve this IVP andfind a formula for φ(t). This problem asks you to calculate φ(1) and

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1.8. GRAPHICAL ANALYSIS OF ODES 81

to approximate φ(1) by using Euler’s method. The purpose is to seehow fast the approximation improves when the time step is reduced.

Use Euler’s method with time steps h = 1, 0.5, 0.25, 0.1, 0.05, 0.02, and0.01 to approximate φ(1), and let E(h) = |φ(1) − yN | (where Nh =1) denote the approximation error obtained with time step h. Plot agraph of E(h) as a function of h.Answer

18. The purpose of this problem is to trace the consequence of onelocal error in Euler’s method. Calculate the approximation y1 of e−0.1

given by solving the IVP y′ = −y; y(0) = 1 by Euler’s method withh = 0.1. Let y = φ1(t) be the solution of y′ = −y with initial conditiony(0.1) = y1. Calculate e−t − φ1(t) for t = 10 and for t = 20.

Repeat this calculation for the IVP y′ = y, y(0) = 1 and its solution,y = et.

19. Solve the initial value problem y′ = y + sin(πt); y(0) = 0, andapproximate the solution by calculating y1, y2, y3, and y4 by Euler’smethod, using time step h = 0.25. Determine all of the local and ac-cumulated errors.Answer

20. Given that the general solution of the ODE y′ = 1 + y2 is y =tan(t − C) where C is an arbitrary constant, calculate the local andaccumulated errors in approximating the solution of the IVP withinitial condition y(0) = 0 with time step h = 0.5 for y1, y2, and y3.Remember to use radians, and draw a graph showing the solutionφ(t) = tan(t) as well as the functions φ1(t) and φ2(t).

Euler’s method can give an answer that is drastically wrong if too largea time step is used. Exercises 21–21b demonstrate this and give a modifica-tion of Euler’s method to avoid this problem.

21. The solution y = e−100t of the initial value problem

y′ = −100y; y(0) = 1 (1.53)

converges to 0 very rapidly. This is apparent even if we restrict ourattention to 0 ≤ t ≤ 1, since e−100×1 is of the order of magnitude of10−43.

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82 CHAPTER 1. FIRST-ORDER EQUATIONS

(a) Use Euler’s method with h = 0.1 to approximate the solutionof the initial value problem (1.53) for 0 ≤ t ≤ 1. Does yourcomputed solution appear to converge to 0?

(b) How small should h be to ensure that the solution converges to0?

Answer

22. The Backward Euler Method. One approach that avoids erraticbehavior observed in the solution computed in exercise 21 is to usebackward differences instead of forward differences with Euler’s Method.Thus, we use the difference equation3

∇ym+1 = h f (tm+1, ym+1) (m ≥ 0),

where ∇ym+1 = ym+1 − ym, as a model for the differential equationy′ = f (t, y). This difference scheme is said to be implicit because itdefines ym+1 implicitly in terms of ym by means of the relation

ym+1 − ym − h f (tm+1, ym+1) = 0.

Test the Backward Euler Method by approximating the solution of theinitial value problem (1.53), again with h = 0.1, for 0 ≤ t ≤ 1.

23. This problem uses the forward and backward versions of Euler’smethod with time step h = 0.1 to approximate the solution of y′ =√

1− y2 with y(0) = 0 for 0 ≤ t ≤ 2.

(a) Explain why the forward version fails.

(b) The difference equation for the backward version is

∇ym+1 = 0.1√

1− y2m+1.

Square both sides and use the quadratic formula to derive thefollowing recursive equation:

ym+1 =ym ± 0.1

√1.01− y2

m1.01

Decide which of the signs is correct, and which is extraneous.

3∇ is pronounced “nabla.”

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1.8. GRAPHICAL ANALYSIS OF ODES 83

(c) Calculate ym for 1 ≤ m ≤ 20.

(d) The solution of the IVP is

y =

{sin(t) if 0 ≤ t ≤ π/21 if t > π/2.

Determine the errors in both the forward and backward ver-sions. Which is the more accurate?

Answer

In problems 24 – 27, a separable equation and a corresponding directionfield are given. Find a family of solutions and note any singular solutions.Print the given direction field and sketch on it several solutions, includingany singular solutions.

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84 CHAPTER 1. FIRST-ORDER EQUATIONS

24. y′ =√

ty.

0 2 4 6 80

1

2

3

4

25. y′ =√

9− y2

-4 -2 0 2 4-3

-2

-1

0

1

2

3

Answer

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1.8. GRAPHICAL ANALYSIS OF ODES 85

26. y′ = cos(t) tan(y)

-7.5 -5 -2.5 0 2.5 5 7.5 10

-4

-2

0

2

4

6

27. (2− t) dy = (1− y) dt.

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

Answer

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86 CHAPTER 1. FIRST-ORDER EQUATIONS

1.9 Initial Value Problems

We can solve IVPs involving linear first order ODEs, and we can approx-imate solutions of IVP involving nonlinear first order ODE. Now we willsee why we should expect an IVP to have one and only one solution.

Let us consider an IVP

y′ = f (t, y); (1.54)y(t0) = y0. (1.55)

Let’s refer to the point in the t, y-plane with coordinates (t0, y0) as the initialpoint. For the IVP (1.54), (1.55) to be correctly posed, the initial point mustbe in the domain of f (t, y).

A solution φ(t) of (1.54) will be considered to be a solution of the IVP ifthe domain on which φ(t) is defined is an open interval that contains t0, andφ(t0) = y0. It is important to be aware that the domain of the solution of any IVPis an open interval. The following example is meant to illustrate this point.

Example 1.9.1 Solve the IVP

dydt

= − yt + 1

; y(0) = 1. (1.56)

SOLUTION. The ODE is linear and homogeneous, and so its generalsolution is of the form y = CeK(t). Substituting this and y′ = CK′(t)eK(t)

yields

CK′(t)eK(t) = −CeK(t)

t + 1,

so that K′(t) = −1/(t + 1). Hence we will put K(t) = − ln(t + 1) andeK(t) = 1/(t + 1). The general solution of the ODE is

y =C

t + 1.

Substituting t = 0 and y = 1 yields C = 1. The solution of the IVP is

y =1

t + 1,

defined on the largest interval containing t = 0 and not containing t = −1:(−1, ∞).

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1.9. INITIAL VALUE PROBLEMS 87

The graph of the solution of example 1.9.1 is the blue curve in fig-ure 1.16. In the figure, the portion of the graph shown as a red curve is notpart of the solution, because the solution must be defined on an interval.Intuitively speaking, it should be possible to trace the graph of a solutionof an IVP without lifting one’s pencil from the paper.

Existence and Uniqueness Theorems

Is there some property of a two-variable function f (t, y) that we can easilyverify and that guarantees the IVP (1.54), (1.55) has a solution? This ques-tion is answered by the following theorem, which is one of the foundationsof our subject. Briefly, it says that the property we are looking for is conti-nuity of f (as a function of two variables). Like most theorems, it must bestated carefully.

Theorem 1.5 (Existence theorem) Let f (t, y) be a function that is continuousat every point of some rectangle

D = {(t, y) : a < t < b, c < y < d},

and let (t0, y0) be a point4 in D.Then there is a function φ(t), defined on an open interval (h, k) that contains

t0, such that φ(t0) = y0 and for all t ∈ (h, k),

φ′(t) = f (t, φ(t)).

The existence theorem does not tell us how to find a solution of theIVP. Furthermore, it says nothing about the extent of the interval on whichthe solution is defined. The following example shows that even if f (t, y)is continuous on the entire t, y-plane, the solution may only exist on anextremely short interval.

Example 1.9.2 Show that

y =100

1− 100t2 . (1.57)

is a solution of the IVP y′ = 2ty2; y(0) = 100. What is the domain of thissolution?

4To ensure that (t0, y0) is not on an edge of the domain where f (t, y) is continuous, ourrectangle is open: D contains the interior points, but not the edges.

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SOLUTION. Setting t = 0 in 100/(1− 100t2) yields y(0) = 100. This is theinitial condition. Differentiating both sides of (1.57),

dydt

=−100(−200 t)(1− 100t2)2

=2t (100)2

(1− 100t2)2

= 2t y2.

Therefore (1.57) satisfies the ODE. This solution is undefined at t = ± 110 ,

and the domain is the largest interval containing t = 0 but not t = ± 110 :(

− 110 , 1

10

).

Figure 1.17 displays several graphs of solutions of the ODE that was thefocus of example 1.9.2.

There are many proofs of the existence theorem, the most direct beingthe proof given by the Italian mathematician Giuseppe Peano in 1890. Thisproof shows that a solution of the IVP can be obtained as a limit of approx-imations given by Euler’s method as the time step h→ 0. The details of theproof require a good dose of advanced calculus, and are therefore omitted.

Uniqueness of solutions

It is easy to verify that y ≡ 0 and y = t3 are solutions of the IVP

y′ = 3y2/3; y(0) = 0

(see figure 1.18). If y = t3, then y′ = 3t2 and y2/3 = t2; hence the firstsolution is valid. If y ≡ 0, then the ODE is satisfied because both sides areidentically 0.

IVPs often serve as mathematical models for physical, biological, so-cial, or engineering phenomena. Predictions will not be reliable unless weknow that the IVP doesn’t have multiple solutions. Fortunately, if the func-tion f (t, y) is nice enough, there will be only one solution. In rough terms,a nice function should not increase or decrease too rapidly. Requiring afunction to be continuous rules out functions that have sudden jumps (dis-continuities), but not functions that increase infinitely fast as the function

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f (y) = 3y2/3 does at y = 0: f ′(0) = 2(0−1/3) = ∞. We will show that mul-tiple solutions are impossible if the function f (t, y) satisfies the followingcondition, known as a Lipschitz condition :

Lipschitz condition

A function f (t, y) satisfies a Lipschitz condition with respectto the variable y in a rectangleD in the plane if there is a con-stant K (called a Lipschitz constant) such that for any (t, y1)and (t, y2) ∈ D,

| f (t, y2)− f (t, y1)| ≤ K |y2 − y1|.

Example 1.9.3 Show that f (t, y) = 3y2/3 does not satisfy a Lipschitz conditionon any domain that intersects the t-axis.

SOLUTION. If we take y1 = 0 then

| f (t, y2)− f (t, y1)| = 3 y2/32 ,

while K|y2 − y1| = K|y2|. If the Lipschitz condition holds, then for all y2

3 y2/32 ≤ K |y2|.

However, this can only hold for |y2| > (3/K)3. No matter how large wechoose K to be, there will be values of y2 close to 0 for which the inequalitydoes not hold.

The following theorem provides an easy way to check if a functionf (t, y) satisfies a Lipschitz condition.

Theorem 1.6 Suppose that the a function f (t, y) is defined of a closed rectangleD in the t, y-plane. If ∂ f

∂y is continuous at each point (t, y) of D then f satisfies aLipschitz condition on D.

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PROOF Since ∂ f∂y is continuous on the closed, bounded set D, there is a

finite maximum value of∣∣∣ ∂ f

∂y

∣∣∣ in D. Let K denote this maximum.Suppose that (t0, y1) and (t0, y2) are points in D, and for convenience,

assume y1 < y2. . Let g(y) = f (t0, y); then g is differentiable on [y1, y2] —in fact,

g′(y) =∂ f∂y

(t0, y).

Thus, by the mean value theorem, there is a number c ∈ (y1, y2) such that

g(y2)− g(y1) = g′(c)(y2 − y1).

Let’s take absolute values of both sides, and put this equation in terms off :

| f (t0, y2)− f (t0, y1)| =∣∣∣∣∂ f∂y

(t0, c)∣∣∣∣ |y2 − y1|.

Since∣∣∣ ∂ f

∂y (t0, c)∣∣∣ ≤ K, it follows that | f (t0, y2)− f (t0, y1)| ≤ K|y2 − y1|.

There are functions f (t, y) that satisfy Lipschitz conditions even thoughthey are not differentiable. See problem 22 for an example.

Theorem 1.7 (Uniqueness theorem) Suppose that y = φ1(t) and y = φ2(t)are two solutions of the IVP

y′ = f (t, y); y(t0) = y0.

If the function f (t, y) is continuous and satisfies a Lipschitz condition with respectto y on a rectangle

D = {(t, y) : a < t < b, c < y < d}

that contains the initial point (t0, y0), then there is an open interval (h, k) witht0 ∈ (h, k), such that φ1(t) = φ2(t) for all t ∈ (h, k).

Suppose that y = φ1(t) and y = φ2(t) are solutions of an ODE, y′ =f (t, y), that satisfy initial conditions φ1(t0) = y1 and φ2(t0) = y2, respec-tively. The proof of the uniqueness theorem is based on estimating how fastthe graphs of φ1(t) and φ2(t) can diverge from one another. This estimateis of practical importance, beyond the proof of the uniqueness theorem, be-cause initial conditions are often derived from measurements that can besubject to error. The true initial value might be y1 and the measured valuemight be y2. The estimate tells us how far the true solution can diverge from

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1.9. INITIAL VALUE PROBLEMS 91

the one based on a measurement error in the initial condition, in the worstcase.

In the case of a linear ODE,

y′ − k(t)y = q(t),

we can compute d(t) = φ1(t)− φ2(t) precisely — no estimation is needed.Since

d′(t) = φ′1(t)− φ′2(t)= (k(t)φ1(t) + q(t))− (k(t)φ2(t) + q(t))= k(t)d(t),

d(t) satisfies the homogeneous linear ODE y′ − k(t)y = 0. It follows thatd(t) = C eK(t), where C is a constant and K(t) is an antiderivative of k(t).Since d(t0) = y1 − y2, we can calculate that C = (y1 − y2)e−K(t0) and hence

d(t) = (y1 − y2) eK(t)−K(t0).

This proves the uniqueness theorem in the linear case, because if y1 =y2, then clearly d(t) ≡ 0. When y1 6= y2 we can also draw interesting con-clusions. For example, suppose it is known that K(t) → ∞ as t → ∞. Thensolutions with unequal initial conditions will diverge from one another. Onthe other hand, if K(t)→ −∞ as t→ ∞, solutions will approach each otherand become indistinguishable as t→ ∞.

Proof of the Uniqueness Theorem5

To see how fast solutions of a nonlinear ODE can diverge from one another,we need two lemmas. The first allows us to replace an IVP with an integralequation.

Lemma 1.9.1 The function φ(t) satisfies the IVP (1.54), (1.55) if and only if

φ(t) = y0 +∫ t

t0

f (u, φ(u)) du (1.58)

PROOF. By the fundamental theorem of calculus, for any differentiablefunction φ,

φ(t) = φ(t0) +∫ t

t0

φ′(u) du.

5This proof can be skipped if necessary.

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92 CHAPTER 1. FIRST-ORDER EQUATIONS

If φ(t) satisfies the IVP (1.54), (1.55) then we can replace φ(t0) with y0 andφ′(u) with f (u, φ(u)). This shows that φ(t) satisfies equation (1.58).

If φ(t) is a solution of equation (1.58) then φ(t0) = y0 because when t =t0 the integral vanishes. Furthermore, the rule for differentiating integralsshows that φ′(t) = f (t, φ(t)); that is, φ(t) satisfies the ODE (1.54).

The second lemma is due to a Swedish-American mathematician, ThomasGronwall.

Lemma 1.9.2 (Gronwall’s inequality) Suppose g(t) is a function that is con-tinuous on an interval [a, b) and that g(t) ≥ 0 for all t ∈ (a, b). Suppose also thatthere are constants M ≥ 0 and C ≥ 0 such that the inequality

g(t) ≤ C + M∫ t

ag(u) du (1.59)

holds for all t ∈ [a, b). Then g(t) ≤ C eM (t−a) for all t ∈ (a, b).

PROOF. Put

F(t) = e−M t∫ t

ag(u) du.

By the product rule for differentiation and the Leibniz rule for differentiat-ing integrals,

F′(t) = −M e−M t∫ t

ag(u) du + e−M tg(t)

= e−M t(

g(t)−M∫ t

ag(u) du

)Referring to the inequality (1.59), it follows that F′(t) ≤ C e−M t for all t ∈[a, b). Since F(a) = 0,

F(t) =∫ t

aF′(u) du

≤∫ t

aCe−M u du

= − CM

(e−M t − e−M a)

= e−M t CM

(eM(t−a) − 1)

Hence ∫ t

ag(u) du = eM t F(t)

≤ CM

(eM(t−a) − 1).

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1.9. INITIAL VALUE PROBLEMS 93

Referring to the inequality (1.59) again, it follows that

g(t) ≤ C + MCM

(eM(t−a) − 1) = C eM(t−a).

The uniqueness theorem now follows from the following proposition.

Proposition 1.9.3 Let f (t, y) be continuous and satisfy a Lipschitz conditionwith respect to y on a rectangular domain

D = {(t, y) : a < t < b, c < y < d}.

Let φ1(t) and φ2(t) be solutions of the ODE (1.54) that are defined on an interval(p, q) ⊂ (a, b), and let t0 ∈ (p, q). Put y1 = φ1(t0) and y2 = φ2(t0). If c <y1, y2 < d then there is an interval (h, k) such that for all t ∈ (h, k),

|φ1(t)− φ2(t)| ≤ |y1 − y2|eK|t−t0|,

where K is a Lipschitz constant for f on D.

PROOF. Since φ1 and φ2 are continuous and c < φ1(t0), φ2(t0) < dthere is an interval (h, k) ⊂ (a, b) such that for all t ∈ (h, k) the inequalitiesc < φ1(t), φ2(t) < d hold. By lemma 1.9.1,

φi(t) = yi +∫ t

t0

f (u, φi(u)) du

for i = 1, 2. By subtracting we find

φ1(t)− φ2(t) = y1 − y2 +∫ t

t0

[ f (u, φ1(u))− f (u, φ2(u))] du.

Assume t ∈ (t0, k). By the triangle inequality,

|φ1(t)− φ2(t)| ≤ |y1 − y2|+∫ t

t0

| f (u, φ1(u))− f (u, φ2(u))| du.

Since (u, φi(u)) ∈ D for u ∈ [t0, t] the Lipschitz condition allows us tosubstitute K|φ1(u)− φ2(u)| for | f (u, φ1(u))− f (u, φ2(u))| in the integral toget

|φ1(t)− φ2(t)| ≤ |y1 − y2|+ K∫ t

t0

|φ1(u)− φ2(u)| du.

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94 CHAPTER 1. FIRST-ORDER EQUATIONS

By Gronwall’s inequality (lemma 1.9.2) it follows that for t0 ≤ t < k,

|φ1(t)− φ2(t)| ≤ |y1 − y2| eK(t−t0).

For t < t0 a symmetrical argument completes the proof.To complete the proof of the uniqueness theorem we only have to apply

proposition 1.9.3 with y1 = y2 since the φi satisfy the same initial condition.The conclusion is that |φ1(t)− φ2(t)| = 0 for all t ∈ (h, k).

Proposition 1.9.3 also proves that the solution of an IVP depends con-tinuously on the initial conditions.

Theorem 1.8 Suppose that f (t, y) is continuous and satisfies a Lipschitz condi-tion with respect to y on a rectangular domain

D = {(t, y) : a < t < b, c < y < d}.

Let t0 ∈ (a, b), and denote by φ(t, v) the solution of the IVP

y′ = f (t, y); y(t0) = v.

Given an interval (C, D) with c < C < D < d, there is an interval (h, k)containing t0 such that φ(t, v) is defined and continuous on the rectangle

D′ = {(t, v) : h < t < k, C < y < D} ⊂ D.

While the proof of theorem 1.8 will not be given in its entirety it is worthnoting where proposition 1.9.3 fits in. To show φ(t, v) is continuous, weneed to show that the difference |φ(t, v)− φ(s, w)| can be made arbitrarilysmall by choosing (s, w) sufficiently close to (t, v). By the triangle inequal-ity,

|φ(t, v)− φ(s, w)| ≤ P + Q

where P = |φ(t, v) − φ(t, w)| and Q = |φ(t, w) − φ(s, w)|. We can makeQ small by choosing s close enough to t because φ(t, w), as a solution ofan IVP (with initial condition y(t0) = w) is continuous. Proposition 1.9.3is used to show that P can be made arbitrarily small by choosing w closeenough to v.

The Picard proof of existence and uniqueness

An alternate proof of the existence and uniqueness of solutions of an IVPy′ = f (t, y); y(t0 = y0 was published by Emile Picard in 1890. The proof

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1.9. INITIAL VALUE PROBLEMS 95

based on lemma 1.9.1, and it constructs a sequence {φn(t)}n≥0 of approxi-mate solutions that converge to a solution of the original IVP.

The first function in the sequence is constant: φ0(t) ≡ y0. For n > 0 thefunction φn(t) is defined recursively as

φn(t) = y0 +∫ t

t0

f (s, φn−1(s)) ds.

We’ll call this sequence the Picard sequence for the IVP. To summarize thedefinition of the Picard sequence, the following observation is useful. Let[a, b] be an interval such that a < t0 < b, and let X [a, b] denote the set of allfunctions φ(t) that are defined and continuous on [a, b]. For any φ ∈ X [a, b]let φ∗ = I(φ) ∈ X [a, b] be the function defined by

φ∗(t) = y0 +∫ t

t0

f (s, φ(s)) ds.

Then the recursive part of the definition of the sequence {φn} is just φn =I(φn−1).

One can view I as a mapping

I : X [a, b] −→ X [a, b];

then lemma 1.9.1 can be restated as follows:

φ ∈ X [a, b] has the property

I(φ) = φ (1.60)

if and only if φ is a solution of the IVP y′ = f (t, y), y(t0) = y0.

The property expressed in (1.60) may be stated verbally as “φ is fixed by I .”Thus the existence problem for IVP’s can be rephrased as “Find a functionthat is fixed by the transformation I .”

The following theorem is a tool from advanced calculus that allows usto find functions that are fixed by certain transformations. It is stated interms of Banach spaces; our space X [a, b] qualifies as a Banach space if,given φ, ψ ∈ X [a, b], we define

‖φ− ψ‖ = max{|φ(t)− ψ(t)| : t ∈ [a, b]

.A definition is necessary to proceed. Let X be a Banach space. (If you

haven’t heard of Banach spaces, please remember that the only Banach

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96 CHAPTER 1. FIRST-ORDER EQUATIONS

space that we have in mind is X [a, b].) A transformation T : X −→ Xis contractive if there is a constant C, with C < 1 such that for all φ, ψ ∈ Xthe following inequality holds:

‖T (φ)− T (ψ)‖ ≤ C‖φ− ψ‖.

Intuitively, the transformation T brings elements of X closer together.

Theorem 1.9 (Banach Fixed Point Theorem) Let X be a Banach space and letT : X −→ X be a contractive transformation. Then there is a unique φ ∈ X thatis fixed by T .

To apply the Banach Fixed Point Theorem, we just have to show thetransformation I is contractive; then it will have a unique fixed point—andas we have noted—fixed points of I and solutions of the IVP are the samethings.

The Picard proof works only if we assume that the function f (t, y) sat-isfies a Lipschitz condition with respect to y. In that case, let φ, ψ ∈ X [a, b]and denote I(φ), I(ψ) by φ∗ and ψ∗, respectively. By definition of I ,

φ∗(t)− ψ∗(t) =

(y0 +

∫ t

t0

f (s, φ(s)) ds)−(

y0 +∫ t

t0

f (s, ψ(s)) ds)

=∫ t

t0

( f (s, φ(s))− f (s, ψ(s))) ds

If L denotes the Lipschitz constant for f then for s ∈ [a, b],

| f (s, φ(s))− f (s, ψ(s))| ≤ L|φ(s)− ψ(s)|

In turn, for all s ∈ [a, b] we have |φ(s)− ψ(s)| ≤ ‖φ− ψ‖, because ‖φ− ψ‖is defined to be the maximum value of |φ(s)− ψ(s)| for s ∈ [a, b]. Thus, forall s ∈ [a, b],

| f (s, φ(s))− f (s, ψ(s))| ≤ L‖φ− ψ‖.

But

|φ∗(t)− ψ∗(t)| ≤∫ t

t0

| f (s, φ(s))− f (s, ψ(s))| ds.

We can put the above two inequalities together to get

|φ∗(t)− ψ∗(t)| ≤∣∣∣∣∫ t

t0

L‖φ− ψ‖ ds∣∣∣∣ = |t− t0|L‖φ− ψ‖.

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1.9. INITIAL VALUE PROBLEMS 97

Thus, if we choose the interval [a, b] to be of the form [t0− h, t0 + h], so that|t− t0| ≤ h for all t ∈ [a, b] then

‖φ∗ − ψ∗‖ ≤ hL‖φ− ψ‖.

Thus, provided h < L−1, the transformation I is contractive, and this com-pletes a combined proof of existence and uniqueness of solutions of theIVP y′ = f (t, y), y(t0) = y0 under the assumption that f (t, y) satisfies aLipschitz condition.

Because this proof depends critically on the Banach Fixed Point Theo-rem, a sketch of the proof is in order. See Exercise 27.

Exercises

In problems 1–8, solve the IVP, and find the largest interval on which thesolution is defined.

1. y′ = 0; y(1) = −2.Answer

2. y′ − 5y = 25t; y(0) = 10.

3. y′ + 2y = sin 5t; y(0) = 0.Answer

4. ty′ + y = et−1; y(1) = 1.

5. ty′ − 2y = t3; y(1) = 0.Answer

6.√

ty′ − y = −t; y(1) = 2

7. y′ − 2ty = t; y(0) = 12

Answer

8. y′ − (tan t)y = sec3 t; y(π6 ) = 5.

9. Suppose that f (t, y) is continuous and satisfies a Lipschitz con-dition with respect to y on a rectangular domain D. Show that ify = φ1(t) and y = φ2(t) are solutions of the ODE y′ = f (t, y), thentheir graphs do not intersect each other inD, unless they are identicalin D. Hint: use the uniqueness theorem.Answer

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98 CHAPTER 1. FIRST-ORDER EQUATIONS

10. Solve the the ODE cos(t)y′ + y = 1, with initial condition y(0) =1.

11. Let f (t) be a function that is continuous on (−∞, ∞), with f (t) 6=0 for all t. Solve the IVP f (t)y′ + y = 1; y(0) = 1. Hint: The IVP inexercise 10 is of this type.Answer

In problems 12 – 21, find the set of initial points (t0, y0) for which theODE, with initial value y(t0) = y0 satisfies the hypotheses of

(i) the existence theorem, and

(ii) the uniqueness theorem.

12. y′ = t−√y.

13. y′ = ty2+1 .

Answer

14. yy′ = t2 + y2.

15. sin(t)y′ + y = 0.Answer

16. (y′)3 = t + 2y.

17. y = ty′ + 1y′ .

Answer

18. y = ty′ + (y′)2.

19. y′ =3√

y−1√ty .

Answer

20. y′ = ln(t2 + y2 − 1) ln(9− t2 − y2).

21. y′ = 3√

t− yAnswer

22. Determine which of the following functions satisfy a Lipschitzcondition with respect to y on the domain

D = {(t, y) : −1 < t < 1,−1 < y < 1},

and find a Lipschitz constant for those that do.

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1.9. INITIAL VALUE PROBLEMS 99

(a) f (t, y) = t− y2.(b) f (t, y) = 4 t y.(c) f (t, y) = |y|.(d) f (t, y) = byc (byc is the greatest integer ≤ y).(e) f (t, y) = btc.

23. Show that f (t, y) =√

1− y2 satisfies a Lipschitz condition on thedomain

D = {(t, y) : a < t < b, c < y < d}

if −1 < c and d < 1, but not if c = −1 or d = 1Answer

24. Use the uniqueness theorem to prove the equal derivatives theo-rem. Hint: Consider the IVP satisfied by f1(t)− f2(t).

25. CAS exercise. Find an initial value y0 such that the solution of theIVP

cos(x)y′ + y = sin(x) + 2 sin(2x); y(0) = y0

satisfies y(1) = 1.Answer

26. CAS exercise. Find an initial value y0 such that the solution of theIVP

y′ + 2xy = 1, y(0) = y0

satisfies y(1) = 0.ANSWER: y0 = −1.462651746.

27. Let T : X −→ X be a contractive mapping with contraction con-stant C < 1.

(a) Show that T can leave at most one φ ∈ X fixed.(b) Starting with any ψ0 ∈ X , recursively define, for n ≥ 1, ψn =T (ψn−1). Show that for n, k > 0,

‖ψn+k − ψn‖ ≤Cn

1− C‖ψ1 − ψ0‖

(standard properties of Banach spaces show that this inequalityallows us to conclude that the sequence {ψn} converges, in thesense that there is an element ψ∞ ∈ X such that

limn→∞‖ψ∞ − ψn‖ = 0.

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100 CHAPTER 1. FIRST-ORDER EQUATIONS

(c) Show that if the limiting element ψ∞ exists, in the sense expressedby the comment in part (b), then ψ∞ is fixed by t.

(d) Parts (a)–(c) show that the solution of the IVP

y′ = f (t, y); y(t0) = y0

is the limit of the Picard sequence—provided, of course, that fsatisfies the Lipschitz condition. For each of the following IVP’s,construct the first four terms of the Picard sequence. Can youguess the limiting solution?

i. y′ = y, y(0) = 1ii. y′ = −t y, y(0) = 1.

iii. y′ = 3√

y, y(0) = 0

iv. y′ = 2yt , y(1) = 1

Answer

1.10 Nonlinear Growth Models

The linear model for population growth presented in section 1.2 was basedon an assumption that the relative growth rate of the population is constant.It leads to the linear homogeneous ODE,

y′ = r y,

in which r = 1y

dydt is a constant relative growth rate.

It is reasonable to expect the relative growth rate to decrease as re-sources become scarce. To include resource limits in our model we canput

1y

dydt

= r(y),

where r(y) is a decreasing function of the population. The model thus leadsto an ODE,

dydt

= y r(y). (1.61)

This ODE is not linear because r(y) is not a constant function.Both the linear and the nonlinear models can be generalized by allow-

ing the relative growth rate to depend on time. This is necessary to model

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1.10. *NONLINEAR GROWTH MODELS 101

a population whose reproduction is seasonal. In this section, we will onlyconsider models that are nonlinear and independent of time.

The simplest way to impose resource limitations on a growth model isto stipulate a carrying capacity M, representing the largest population thatthe environment can support. If for some reason the population exceedsthe carrying capacity, it will decrease; and as long as the population is lessthan the carrying capacity, it will increase.

In 1838, P. F. Verhulst presented such a model for population growth. Inhis model , the relative growth rate decreases linearly from a value k whenthe population is a negligible fraction of the carrying capacity to 0 whenthe carrying capacity M is reached. The relative growth rate in this modelis

r(y) = k(

1− yM

).

When we substitute this r(y) in equation (1.61), the following ODE, knownas the logistic equation results:

y′ = k y(

1− yM

). (1.62)

The logistic equation has two constant solutions, y = 0, and y = M.The graphs of these constant solutions divide the t, y-plane into three re-gions. Below the line y = 0 and above the line y = M the right side ofequation (1.62) is negative. Therefore, any solution of the logistic equationwith initial conditions in one of these two regions will be decreasing. Thesame reasoning shows that a solution with initial condition between thetwo constant solutions must be increasing. The red curve in figure 1.19 isone such solution. A nonconstant solution cannot cross either of the linesy = 0 or y = M, because at the crossing point, the uniqueness theorem forinitial value problems would be violated.

The population values y = 0 and y = M are called stationary points ofthe logistic equation, because when the initial population has either of thesevalues, it will remain constant (that is, stationary). The population y = Mis a stable stationary point because every solution whose initial value issufficiently close to M (any positive initial value will do) will converge toM as t → ∞. On the other hand, y = 0 is an unstable stationary point,because there are solutions with initial values y(0) arbitrarily close to 0 andwhich diverge away from 0 as t→ ∞.

In the following example, we will derive the general solution of thelogistic equation. The solution involves three parameters: M, k, and theinitial population; hence three data points are required to determine the

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102 CHAPTER 1. FIRST-ORDER EQUATIONS

solution. To make the calculation as easy as possible, the data have beentaken at evenly spaced time intervals.

Example 1.10.1 Find the general solution of the logistic equation (1.62), and findvalues of the parameters to fit the data y(0) = 1, y(1) = 2, y(2) = 3.

SOLUTION. The logistic equation is separable, and in separated form itappears as follows:

1y(M− y)

dy =kM

dt.

Integration by partial fractions yields

1M

∫ (1y+

1M− y

)dy =

1M

ln∣∣∣∣ y

M− y

∣∣∣∣ = kM

t + B,

where B denotes the integration constant. Thus

yM− y

= ±ekt+MB

Let’s assume our solution satisfies 0 < y < M; then the plus sign applies,and we can solve for y to obtain

y =M

1 + Ae−kt ,

where A = e−MB. To simplify further, set v = e−k; then

y =M

1 + Avt .

Since v < 1, we have, limt→∞ y(t) = M. This confirms that the populationwill converge to the carrying capacity as t→ ∞. The parameters to beevaluated are M, A, and v. The three data points yield the followingequations for these parameters:

M1 + A

= 1

M1 + Av

= 2

M1 + Av2 = 3

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By the first equation, M = 1 + A which we put in the second equation andsolve for A to get A = (1− 2v)−1. Now substitute this and

M = 1 + A = 1 +1

1− 2v

in the third equation. After simplifying, the result is a quadratic equation,

3v2 − 4v + 1 = 0,

whose solutions are v = 13 and v = 1. Since v < 1, the second solution is

extraneous and v = 13 . The remaining parameters are readily evaluated:

A = 3 and M = 4, and the solution is

y =4

1 + 31−t .

The graph of this solution is shown in figure 1.20.

The classic application of the logistic equation to demography is a 1920study by R. Pearl and L. Reed6, who used the censuses of 1790, 1850, and1910 to determine parameters for the logistic equation. They calculatedthat the carrying capacity of the United States was 197,274,000 (in fact, theactual population exceeded this value by 1970). Their solution of the lo-gistic equation did not display a significant deviation from the census dataof the years 1800 - 1840 and 1860 - 1900, and was considered to be a greatsuccess. Their predictions continued to be amazingly accurate until 1950.Figure 1.21 shows the Pearl and Reed logistic curve, and the census datacovering the years 1790 – 2000. It is interesting to consider the causes offailure for this model in the years after 1950.

General nonlinear growth models.

The key property of the growth models where the relative growth rate de-pends on the population is that they are based on an ODE

dydt

= y r(y)

6Raymond Pearl and Lowell Reed, “On the rate of growth of the population of theUnited States since 1790 and its mathematical representation,” Proceedings of the NationalAcademy of Sciences, Washington vol. 6 (1920), pp. 275 - 288, http://www.pnas.org/content/vol6/issue6/

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104 CHAPTER 1. FIRST-ORDER EQUATIONS

that is autonomous ; that is, the independent variable t does not appearon the right side. In general, an autonomous first order ODE will have theform

dydt

= g(y) (1.63)

Every autonomous ODE is separable, but we should not rush to inte-grate. Properties of solutions such as whether they are increasing or de-creasing functions, the location of any asymptotes, etc. can be determinedwithout calculating an integral. We will be assuming that

(i) g′(y) is continuous.

(ii) the zeros of g(y) are isolated. This means that if g(yj) = 0, then thereis an open interval (yj − h, yj + h) that contains no other zeros of g.

The purpose of assumption (i) is to enable us to use the Existence andUniqueness Theorems (recall that if g′(y) is continuous, then g(y) is alsocontinuous and satisfies a Lipschitz condition).

The zeros yi of g(y) are the stationary points of the ODE (1.63). Eachstationary point represents a constant solution y ≡ yi. It follows from as-sumption (ii) that the stationary points separate the real line into a collec-tion of open intervals of the form (−∞, y1), (yi, yi+1), or (yn, ∞), and oneach of these intervals, g is nonzero, with no sign changes. If g(y) has nozeros, there is just one interval, (−∞, ∞).

We will call an interval on which g(y) is positive an up interval, and aninterval where g(y) is negative is a down interval. It will be seen that therange of any solution of the ODE (1.63) will be one of these intervals. If therange is an up interval, the solution will be increasing—because in this casey′ = g(y) is positive—and if the range is a down interval, then the solutionis decreasing.

The stationary points of the logistic equation, g(y) = k y(1− y

M

), are

y1 = 0 and y2 = M. There is one up interval, (0, M), and two down in-tervals, (−∞, 0) and (M, ∞). Figure 1.19 displays typical solutions whoseranges are these intervals: note that the solution y = φ(t) with φ(0) ∈(0, M) is strictly increasing, with φ(t) → M as t → ∞ and φ(t) → 0 ast → −∞. If φ(0) lies in one of the two down intervals, then φ(t) is strictlydecreasing, but its graph has only one asymptote.

Proposition 1.10.1 Let y = φ(t) be a solution of an autonomous ODE of theform (1.63). Then φ(t) is either a strictly increasing function, a strictly decreasingfunction, or a constant function.

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PROOF. If φ(t) is not strictly increasing or decreasing, then it must havea relative minimum or maximum. Thus there is a number t0 such thatφ′(t0) = 0.

Set y0 = φ(t0). Since g(y0) = φ′(t0) = 0, y0 is a stationary point. Itfollows that y ≡ y0 and φ(t) are solutions of the same initial value prob-lem, y′(t) = g(y); y(t0) = y0. By the Uniqueness Theorem, φ(t) ≡ y0. Wehave thus shown that if φ is neither an increasing function nor a decreasingfunction then it must be constant.

To emphasize that proposition 1.10.1 is only applicable to autonomousequations whose right sides g(y) are nice, let us consider the followingexample.

Example 1.10.2 Explain why the following two examples are consistent with propo-sition 1.10.1.

(i) The function

y(t) ={

t2 for t > 0,0 for t ≤ 0.

is a solution of the ODE y′ = 2√

y, and it is not strictly monotone.

(ii) The function y(t) = et2is a solution of the ODE y′ = 2ty, and it is not

monotone.

SOLUTION.

(i) g(y) = 2√

y is not differentiable at y = 0.

(ii) The ODE y′ = 2ty is not autonomous.

Our next observation will be that if φ(t) is a nonconstant solution ofequation (1.63) then the range of φ(t) will be an entire up interval (if φ isincreasing) or an entire down interval (if φ is decreasing). We reason firstthat the range cannot overlap into two such intervals. If it did so, then thegraph of φ(t) would cross the graph of a constant solution correspondingto a stationary point that separates the two intervals—a violation of theuniqueness theorem. To see that the range of φ cannot be a proper subset ofan up or down interval, we will use the following proposition.

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106 CHAPTER 1. FIRST-ORDER EQUATIONS

Proposition 1.10.2 Let y(t) denote a solution of equation (1.63). If

limt→∞

y(t) = L,

where L is finite, then L is a stationary point of equation (1.63); and if N =limt→−∞ y(t) is finite, then N is a stationary point.

PROOF. If L is finite,

limt→∞

y′(t) = limt→∞

g(y(t))

= limy→L

g(y)

= g(L).

Thus, if g(L) 6= 0, then y′(t) would have a nonzero limit as t → ∞. Forexample, suppose g(L) = c > 0. Then for some number m, y′(t) > c/2for all t > m. Integrating, we would have y(t) > ct/2 + d (where d is anintegration constant) for t > m. This contradicts the fact that y(t) convergesto a finite number L as t → ∞. We can reach the same sort of contradictionif we assume that g(L) < 0; hence g(L) = 0.

The proof that N is a stationary point if it is finite is the same.Propositions 1.10.1 and 1.10.2 are the key ingredients of the proof of the

following theorem.

Theorem 1.10 Let y = φ(t) be a solution of the autonomous equation y′ = g(y),where g(y) has isolated zeros and g′(y) is continuous everywhere. Then either φ isconstant, or φ is a strictly increasing function whose range is an entire up interval,or φ is strictly decreasing, with range an entire down interval.

With the aid of theorem 1.10, we can sketch graphs of solutions of anautonomous ODE like equation (1.63) without integrating. We start bydrawing on the y-axis a phase diagram for the ODE, as follows. Locatethe stationary points yi by solving the equation g(y) = 0. These will be theendpoints of the up and down intervals. Mark the stationary points on they-axis and draw horizontal lines to show the constant solutions. Test thesign of g(y) by evaluating g at a convenient point in each interval (yi, yi+1)on the y-axis to see if it is an up or down interval. Finally, mark each upinterval with an arrow directed upward, and each down interval with adownward arrow.

When the phase diagram and stationary solutions have been drawn, itis easy to sketch graphs of the nonconstant solutions. All solutions with

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initial points in a given up or down interval will be strictly increasing ordecreasing, with ranges covering the entire interval. Stationary points atthe ends of the up and down intervals will be horizontal asymptotes of thesolution.

Example 1.10.3 Draw the phase diagram for the ODE

y′ = 4y− y3, (1.64)

and make a sketch of the graphs of the solutions of the four initial value problemswith y(0) = ±1, ±3.

SOLUTION. The stationary points are determined by solving 4y− y3 = 0 toobtain y1 = −2, y2 = 0, and y3 = 2. The up and down intervals aretherefore (−∞,−2), (−2, 0), (0, 2), and (2, ∞). To determine the directionof (−∞,−2), evaluate g(y) at an arbitrarily chosen point in the interval:f (−3) = 15. Since we have found a positive value, this is an up interval,and is to be marked accordingly. The other three intervals are marked asfollows. Since f (−1) < 0, (−2, 0) is a down interval; f (1) > 0 so (0, 2) isan up interval; and, f (3) < 0, indicating that (2, ∞) is a down interval. Thephase diagram is imposed on the y-axis of figure 1.22.The stationary solutions are indicated by the horizontal lines in figure 1.22.Each nonconstant solution is asymptotic to the constant solutionscorresponding to the boundary of its range. Hence the solution of theinitial value problem with y(0) = 1 is asymptotic to the t-axis as t→ −∞and to the line y = 2 as t→ ∞. In addition, the solution is strictlyincreasing and has a slope of 3 at its crossing of the y-axis. With thisinformation, it is not difficult to make a sketch of the solution.Similarly, if y(0) = −1, then y(t) is decreasing, asymptotic to the t-axis ast→ −∞, and to y = −2 as t→ ∞. If y(0) = 3, y(t) is again decreasing,and asymptotic to y = 2 as t→ ∞; if y(0) = −3, then y(t) is increasingand asymptotic to y = −2.

Stability

Consider an isolated stationary point y1 of an autonomous ODE y′ = f (y).Since y1 is isolated, there is an interval (a, b) containing y1 and no otherstationary points. We will say that y1 is a stable stationary point if all so-lutions φ(t) with initial values in (a, b) converge to y1 as t → ∞. If there

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108 CHAPTER 1. FIRST-ORDER EQUATIONS

are solutions with initial values in (a, b) that do not converge to y1, then wewould say that y1 is an unstable stationary point.

Figure 1.23 displays three phase diagrams in which the points yi are sta-tionary points of autonomous ODEs. The intervals immediately above andbelow the stationary point y1 are both directed toward it, so that stationarypoint is stable. In fact, the interval I consisting of the point y1, the up in-terval below it, and the down interval above it has the required propertythat every solution with initial point in I converges to y1 as t → ∞. If oneor both of the intervals adjacent to a stationary point yi are directed awayfrom yi, then yi is unstable. In these cases, if we only know that the initialpoint y0 is near to an unstable stationary point yi, but do not know whethery0 < y1 or y0 > y1, it is impossible to predict the limit of the function y(t).

The following proposition offers an alternative way to determine thestability of a stationary point.

Proposition 1.10.3 Let y1 be a stationary point of the autonomous ODE y′ =g(y) such that g′(y1) 6= 0. Then, y1 is stable if g′(y1) < 0, and is unstable ifg′(y1) > 0.

The proof of proposition 1.10.3 is left as an exercise. (See problem 25 atthe end of this section.) In the following example, we will use the proposi-tion to determine the stability of each stationary point of an ODE, and thenuse the stability information to draw a phase diagram.

Example 1.10.4 Determine the stability status of each stationary point of

y′ = sin(y)− 2π

y,

and draw the phase diagram.

SOLUTION. Figure 1.24 shows the graphs of u = sin(v) and u = 2v/π.They cross at v = 0, and v = ±π/2; therefore the set of stationary pointsis {−π/2, 0, π/2}. Since f ′(0) = cos(0)− 2/π is positive, 0 is unstable.f ′(±π/2) = cos(π/2)− 2/π < 0, so the other two stationary points arestable.It follows that the intervals (0, π/2) and (−π/2, 0) are directed away fromthe stationary point at 0: (0, π/2) is up and (−π/2, 0) is down. Since theequilibria at ±π/2 are stable, adjacent intervals are directed toward thesepoints. Therefore (−∞,−π/2) is an up interval and (π/2, ∞) is a down

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1.10. *NONLINEAR GROWTH MODELS 109

interval. The phase diagram is drawn on the y−axis of figure 1.24 (that isthe horizontal axis in this case). When sketching graphs of solutions ofy′ = sin(y)− 2y/π, the y-axis would be vertical, and the phase diagram infigure 1.24 would be rotated counterclockwise 90◦ before sketching.

Proposition 1.10.3 does not apply to any stationary point y1 where f ′(y1) =0. Exercise 26 shows that in this case, it is possible for a stationary point tobe either stable or unstable.

Dependence on Parameters

The population P of codfish in a certain marine fishery is modeled by amodified logistic equation,

P′ = kP(1− P/M)− H. (1.65)

The growth parameter k and the carrying capacity M are taken from thelogistic equation, while H is the rate at which fish are harvested. We areinterested in how the fate of the fish population depends on the parameterH.

Let f (P) = −kP2/M + kP− H be the quadratic expression on the rightside of equation (1.65). Figure 1.25 shows graphs of f (P) representing fourdifferent harvest rates.

Solving f (P) = 0 with the quadratic formula, we find that the station-ary points are

P1, P2 =12

M

[1±

√1− 4H

kC

], (1.66)

provided that these are real numbers. If there are no real stationary points,then f (P) is negative for all P, as in the graph corresponding to H = 12000in figure 1.25. Thus (−∞, ∞) is a down interval, and the codfish will beextinct when the population reaches 0. If there are two distinct stationarypoints, as in the graphs with H = 0 and H = 5000 of the figure, P1 will belocated between 0 and 1

2 M, and P2 will be between 12 M and M. The inter-

val (P1, P2) is an up interval, and (−∞, P1) and (P2, ∞) are down intervals.Hence P2 is stable, and will represent the limiting population.

We can identify a critical situation when P1 and P2 merge together as onestationary point, as in the graph corresponding to H = 10000 of figure 1.25.The intervals above and below are both down intervals, and the stationarypoint is unstable, with extinction looming. The quadratic equation f (P) =

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110 CHAPTER 1. FIRST-ORDER EQUATIONS

0 has a double root when1− 4H

kC= 0;

that is, when H = k4 M. This is the critical harvest rate.

Equation (1.65) is actually an infinite family of ODEs depending on aparameter H. The solutions undergo a fundamental change, called a bi-furcation at a critical value H0 of the parameter. It is best to describe thissituation with a two-dimensional diagram called a bifurcation diagram,as shown in figure 1.26. The horizontal axis represents the value of the H,which is called the bifurcation parameter in this context. The vertical axis rep-resents the population. The stationary points are considered as functionsP1(H) and P2(H) and plotted on the diagram. By convention, stable equi-libria are depicted by solid curves; unstable equilibria by dashed curves.Vertical arrows indicate whether the population is increasing or decreas-ing.

Exercises

1. A lake can support a population of 1000 fish. There are now 600fish in the lake, and on the same date last year there were 300. As-suming that the logistic model determines the fish population in thelake, how many fish will be in the lake a year from now?Answer

2. Suppose that the lake in problem 1 has been stocked, so that nowit contains 1200 fish. How many will be in the lake a year from now?

3. Solve the initial value problem:

y′ = .02y(200− y); y(0) = 10.

Answer

4. The populations of two communities are governed by the logisticequation,

u′ = k u(

1− uM

).

The second community has a carrying capacity twice as large as thefirst, and the growth constant k for the each community is the same.Show that if the initial population of the second community is twicethe initial population of the first, then the second community willalways have twice the population of the first.

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1.10. *NONLINEAR GROWTH MODELS 111

Let’s update Pearl and Reed, whose predictions of the US populationwere based on census data for the years 1790, 1850, and 1910. We willmove our data points to 1890, 1950, and 2010, respectively and answer thesame questions that they did a century ago.

Problems 5 – 8 are based on the following data concerning the popula-tion of the United States.

Year Population (millions)1890 63.01950 151.32010 309.4

5. Find a solution of the logistic equation to fit the data.Answer

6. What is the predicted carrying capacity, according to the logisticmodel?

7. When will the population reach a level of 500 million?Answer

8. Use the logistic model to predict the population in 2040.

9. A rumor that a top psychic has predicted the sky is falling isspreading in Gossipville. Each citizen of that town calls ten others perday at random to chat; if one of the parties has heard the rumor andthe other hasn’t, the rumor spreads. The population of Gossipvilleis 10,000, and at this point 1000 have already heard the rumor. Howlong will it be before 90% of the people in Gossipville have heard therumor?Hint: Let N(t) be the number of people who have been warned aboutthe sky. Show that N can be approximated by a continuous variablewhich satisfies the logistic equation. Of course, it is necessary to fig-ure out the values of M and k.Answer

10. The growth of a certain population is limited by two resources:food and water. The food resources will support a population of M,and there is enough water for a population of 2M. A researcher postu-lates the following nonlinear growth model, which accounts for bothlimitations

y′ = k(M− y)(2M− y)y.

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112 CHAPTER 1. FIRST-ORDER EQUATIONS

Draw a phase diagram and show that the population has three sta-tionary values, but that only one is stable. Do all solutions convergeto the same stable population? What happens if the population ini-tially exceeds 2M? Does this make sense?

Problems 11 – 14 refer to an alternative to the logistic model for popu-lation growth, the Gompertz model.

It is based on the ODE

y′ = ky ln(

My

),

where M is the carrying capacity and k is constant.

11. Show that the Gompertz model has, like the logistic model, twostationary populations, y ≡ 0 and y ≡ M, where y ≡ 0 is unstable,and that y ≡ M is stable.Answer

12. Show that the substitution v = ln(y) produces a linear ODE withv as the dependent variable.

13. Given the data y(0) = 1, y(1) = 2, and y(2) = 3, determine thecarrying capacity according to the Gompertz model. It is interestingto compare this result with that of example 1.10.1.Answer: 5.31261.Details

14. Use the following data to find the carrying capacity of the UnitedStates, according to the Gompertz model (These are the data used byPearl and Reed in the study cited above)Answer: 18.35 billion

Year Population (millions)1790 41850 231910 92

A chemical reaction proceeds at a rate determined by the concentrationof the reactants and catalysts. Specifically, if y(t) denotes the concentra-tion of one of the products of the reaction, and u(t), v(t), etc. denote the

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1.10. *NONLINEAR GROWTH MODELS 113

concentrations of the reactants and catalysts, then

dydt

= k[u(t)]n[v(t)]m · · ·

where the exponents n, m, etc. are called the orders of the reaction withrespect to the corresponding reactant or catalyst. The number k is calledthe rate constant.

15. In an autocatalytic reaction, the product acts as a catalyst to furtherthe reaction. Assume that an autocatalytic reaction has order 1 withrespect to the reactant and the catalyst, and that for each moleculeof the reactant that is consumed, one molecule of the catalyst is pro-duced. If the reaction is allowed to proceed in a closed system, showthat its rate is governed by the logistic equation.Answer

16. The reaction of gasses

2NOBr→ 2NO + Br2

is second order with respect to the concentration of NOBr. In anexperiment, a 1 liter flask initially contains 2.7 × 1022 molecules ofNOBr. After 2 minutes, 1022 molecules of the reactant are left. Deter-mine the rate constant for this reaction. When was exactly half of theNOBr that was originally present consumed?Answer: 1.2 minutes.

17. Which of the following equations are autonomous?

(a) y′ = 1− y.

(b) y′ = sin(t) + y2.

(c) y′ ={

y for 0 < t < 10, otherwise.

(d) y′ ={

y for 0 < y < 10, otherwise.

Answer

Find the stationary points of the ODEs in problems 18 – 22 and drawtheir phase diagrams.

18. y′ = 4 + 3y− y2.

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114 CHAPTER 1. FIRST-ORDER EQUATIONS

19. y′ = y2.Answer

20. y′ = sin(y)− 12 .

21. y′ = ey

Answer

22. y′ = y− cos(y)

23. Which of the following functions are solutions of first order au-tonomous ODEs? In the case of those that are solutions, where wouldthe stationary points be located?

(a) y = 1 + t. Domain, (−∞, ∞).

(b) y = 1 + t2. Domain, (−∞, ∞).

(c) y = e−2t. Domain, (−∞, ∞).

(d) y = tan(t). Domain, (−π2 , π

2 ).

(e) y = sin(t). Domain, (−∞, ∞).

Answer

24. The set of stationary points for an ODE y′ = f (y) is {0, 1, 2, 3, 4}.Suppose that f ′(0) = 2, f ′(1) = −5, f ′(2) = 0, f ′(3) = −1, andf ′(4) = 12. Determine the stability of each stationary point, and drawthe phase diagram.

25. Prove proposition 1.10.3. [Hint: Show that the hypothesis impliesg(y) changes sign at y1.]Answer

26. An stationary point y1 of the ODE y′ = g(y) is said to be degener-ate if f ′(y1) = 0. For example, every point is a degenerate stationarypoint of the equation y′ = 0. Give an example of a ODE with a de-generate, isolated, stable stationary point, and an example of a ODEwith a degenerate, isolated, unstable stationary point.

27. Show that an ODE whose phase diagram is identical to the phasediagram for y′ = y2 has a degenerate stationary point at 0.Answer

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1.10. *NONLINEAR GROWTH MODELS 115

28. Use proposition 1.10.3 to determine the stability of 0 as an station-ary point of

y′ = sin(y)− ky,

in the case where the parameter k > 1. Also determine the stabilitywhen k < 1. What can be said when k = 1? Draw phase diagramsk = 1, for k slightly less than 1, and for k slightly greater than 1.

29. For the growth equation with harvesting,

P′ = P(1− P)− H

(a) What is the critical harvest rate?(b) Suppose H = 0.1, and P(0) = 1. What will be the limiting pop-

ulation?(c) Draw the bifurcation diagram.

Answer

30. The following ODE depends on a parameter m:

y′ = y2 −m.

(a) For which values of m are there stationary points?(b) Draw the bifurcation diagram.

31. Fitting a logistic curve to data. In example 1.10.1 we found thatwe could find a logistic curve passing through three points in theplane. What if there are more than three points? In this case, wecannot expect the logistic curve to pass through all of the points, butwe can try to find parameters that will give the best logistic curve forthe data. The following method works well in cases where the logisticequation is an appropriate model for population growth, such as theU.S. population, 1790 – 1950.

Write the logistic equation P′ = kP(1− P/M) in the form

1P

dPdt

= mP + b,

where m = −k/M and b = k. If we know the population at timest0 − h, t0, and t0 + h, then we can approximate the relative growthrate as follows:

1P

dPdt

∣∣∣∣t=t0

≈ P(t0 + h)− P(t0 − h)2hP(t0)

. (1.67)

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116 CHAPTER 1. FIRST-ORDER EQUATIONS

(a) Let P(t) denote the U.S. population (see table 1.3) in year t. Tab-ulate, for t = 1800, 1810, 1820, . . . , 1940, P(t) and (P(t + 10)−P(t− 10))/(20P(t)).

(b) Using a statistical calculator or a spreadsheet7, find the leastsquares line that best fits the data, with the population as theindependent variable and the approximate relative growth rate,as calculated by equation (1.67), as the dependent variable.

(c) The result of part (b) will be the slope m and the intercept b of theleast squares line. Use these to determine the carrying capacity.

(d) Use a CAS (or a pencil and paper) to find the solution of

P′ = P(mP + b)

with initial condition P(1900) = 76.21 million. Plot a graph ofthe solution curve extending from the years 1790 to 2010. In-clude the census data, as was done in figure 1.21.Answer

1.11 Review Exercises

Find the general solution of each of the ODEs in problems 1 – 12.

1. y′ = ty + t + y + 1.See Answer

2. y′ = y− t2.See Answer

3. y′ = y + et.See Answer

4. y′ + tan(t)y = cos2 t.See Answer

5. y′ = et − y.See Answer

6. ty′ + y = et sin 2t.See Answer

7Statistical calculators and spreadsheets have least squares functions that you may use.

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1.11. REVIEW EXERCISES 117

7. ty′ + 12y = 5t2 + 3t− 2.See Answer

8. y′ + tan(t)y = 1.See Answer

9. y′ = ty + 1.See Answer

10. y′ = 5yt

See Answer

11. y′ + y = e−t

See Answer

12. y′ = y cos t.See Answer

In problems 13 – 6, solve the initial value problem.

13.

y′ = t+yt ; y(1) = 2.

See Answer

14. y′ + y = e−t; y(0) = 0.See Answer

15. y′ = 2t; y(2) = 4.See answer

16. y′ + 4y = 3e−4t sin 3t; y(0) = 0.See answer

17. ty′ − 3y = 5t3; y(1) = 1.See answer

18. y′ = 3√

y; y(1) = 1.See answer

19. y′ = (t + 3)y + 2t + 6; y(0) = 0.See answer

20. y′ = (t + 3)y + 2t + 6; y(0) = −2.See answer

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118 CHAPTER 1. FIRST-ORDER EQUATIONS

21. y′ − 2y4−t2 = 1√

2−t; y(0) = 3

√2.

See answer

22. y′ + ty = 2t + t3; y(0) = 1.See answer

23. A the balance of bank account bearing continuous compound in-terest at a constant rate triples in 30 years, with no deposits or with-drawals. How long did it take to double?See answer

24. If the world’s population follows the linear growth model with arelative growth rate of 1.3% (0.013), how long will it take the popula-tion to double?See answer

25. A piece of wood contains 0.8 ppb 14C, and when it was part of aliving tree, it contained 1 ppb. How old is it? The half-life of 14C is5730 years.See answer

26. Find the general solution of y′ + y = cos(4t). Is there a stableperiodic solution?See answer

27. A tank initially contains 1000 liters of brine with a concentrationof 100 grams per liter. Brine, containing 50 grams salt per liter, ispumped into the tank at 200 liters per hour, and the well-mixed so-lution is pumped out at 100 liters per hour. After 10 hours, the tankoverflows. What is the concentration of salt in the tank at that time?See answer

28. Print a copy of the direction field shown below, and sketch onit the graphs of the solutions of y′ = t2 − y2, with initial conditionsy(−2) = 0 and y(2) = 0, respectively.

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1.11. REVIEW EXERCISES 119

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

Answer

29. Use Euler’s method with time step h = 0.25 to compute an ap-proximate solution of the IVP, y′ = t2− y2; y(0) = 0. Draw the polyg-onal graph of the approximate solution on your copy of the directionfield accompanying exercise 28.Answer

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120 CHAPTER 1. FIRST-ORDER EQUATIONS

30.

Sketch the graph of the solution of the IVP, y′ = t−√y; y(0) = 2 ona hard copy of the direction field shown below.

-2 0 2 40

1

2

3

4

Answer

31. Use Euler’s method with time step h = 0.5 to approximate the so-lution of the IVP y′ = t−√y; y(−2) = 4. Draw the polygonal graphof this solution on your copy of the direction field accompanying ex-ercise 30.Answer

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1.11. REVIEW EXERCISES 121

32. Match the direction fields shown below with the ODEs.

(a) y′ = 0.03(5t− y2).

(b) y′ = sin(y).

(c) y′ = sin(t + y).

(d) y′ = −0.01y

(e) y′ = 0.05y(π − y).

(f) y′ = 0.02(t2 + y2).

IV V VI

I II III

Answer

In exercises 33 — 40, find a family of solutions and note any singularsolutions.

33. y′ = ty2.Answer

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122 CHAPTER 1. FIRST-ORDER EQUATIONS

34. y′ = ey.Answer

35. y′ = y(1− y).Answer

36. y′ =√

1−y2

1−t2 ; −1 < t < 1.Answer

37. y′ =√

y2−1t2−1 ; |t| > 1.

Answer

38. y′ = t(1 + y2)/(1 + t2).Answer

39. y′ =√

yt

Answer

40. y′ = e4y tan(3t).Answer

41. An object that is sinking in water is subject to forces due to gravity(downward), and buoyancy and drag (both upward). By the Princi-ple of Archimedes, the buoyancy force is equal to the weight of waterdisplaced by the object. Thus, if σ denotes the specific gravity of theobject, the buoyancy force is equal to mg

σ . The drag force is propor-tional to the square of the velocity. A diamond (σ = 3.5) has a ter-minal velocity of 21 meters per second when sinking in water. If youtoss it gently into the middle of Lake Tanganiyika (1463 meters deep),

(a) how far will the diamond travel in the first 10 seconds?

(b) Estimate how long it will take for the diamond to get to the bot-tom of the lake.

(c) Suppose that you throw the diamond vertically into the lake sothat its initial velocity after it is immersed in water is 42 metersper second. Answer questions (a) and (b).

Answer

42. A certain population is growing according to the logistic equation.In year A, the population is 100,000 and growing at 4500 per year, andin year B the population is 200,000, growing at 8000 per year.

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1.11. REVIEW EXERCISES 123

(a) Calculate the relative growth rate in each of the years.

(b) Find the carrying capacity.

(c) If year A is 1980, when was year B?

Answer

In problems 43 — 48, draw phase diagrams for the ODEs, identify allstable equilibrium points, and and describe the limiting behavior of thesolution of the IVP with increasing time. Do not solve these ODEs.

43. y′ = y (2y + 1); y(0) = −1.Answer

44. y′ = −y sin2(y); y(0) = 30.Answer

45. y′ = y2 + 2y + 2; y(0) = 0.Answer

46. y′ = 5− 3 y; y(0) = −2.Answer

47. y′ = y−1 − 1; y(0) = 3.Answer

48. y′ = y−1 − 1; y(0) = −3.Answer

In problems 49 – 51, it will be useful to differentiate both sides of theODE to obtain an expression for y′′. You don’t need formulas for the solu-tions to do these problems!

49. Show that the graph of any solution of y′ = 1y is either increasing

and concave down, or decreasing and concave up.Answer

50. Imagine that the set of points in the plane satisfying the inequalityt+ y+ 1 > 0 is colored red, and the set of points satisfying t+ y+ 1 <0 is colored blue. Show that if y(t) is a solution of the ODE y′ = t + y,then the graph of y(t) is concave up at red points, and concave downat blue points.Answer

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124 CHAPTER 1. FIRST-ORDER EQUATIONS

51. Show that the graph of any solution of y′ = ey is increasing andconcave up. What can be said about the graph of a solution of y′ =e−y?Answer

52. Richardson’s extrapolation. Let φ(t) be the solution of the IVPy′ = t− y2; y(0) = 0.

(a) Using a spreadsheet or other computer version of Euler’s methodwith h = 0.1, 0.05, and 0.025, compute approximate values ofφ(1).

(b) Richardson’s extrapolation works with the assumption that theaccumulated error in each of these approximations is approx-imately equal to Ch where C is an unknown constant. If Yh1

and Yh2 are approximations of φ(1) obtained using steps h1 andh2 = h1/2, respectively,

φ(1) ≈ Yh1 + Ch1 (1.68)

φ(1) ≈ Yh2 +12

Ch1. (1.69)

Eliminate the constant C from (1.68) and (1.69), to find a bet-ter approximation of φ(1). Let Zh1 denote the approximation ofφ(1) thus obtained. This Zh1 is called a first-order extrapolatedapproximation.

(c) Calculate Z0.1 and Z0.05.(d) Further analysis shows that there is a constant D such that

φ(1) ≈ Zh + Dh2.

Use the values of Z(0.1) and Z0.05 just calculated to eliminate theconstant D, thus obtaining a second-order extrapolated approx-imation.ANSWER SUMMARY: The second-order extrapolation yields φ(1) ≈ 0.4555383.This is considerably closer to the correct value (found by solving the initialvalue problem using a CAS), φ(1) = 0.4555445 than Y0.025 = 0.4477666.

Full Answer

53. For which values of y0 does the IVP

y′ = 3

√1 + y + t1− y + t

; y(0) = y0

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1.11. REVIEW EXERCISES 125

have a solution? A unique solution?Answer

54. The IVP y′ = y sin(t− y), y(0) = 0 has the solution y ≡ 0. Howcan you be sure that it’s the only solution?Answer

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126 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.1: Several solutions of the ODE v′ = g− kv2. Each graph represents the velocityof a falling object subject to air resistance. The values g = 9.8 meters/second2 and k =9.8× 10−4 kilogram /meter were used.

5 10 15 20

20

40

60

80

100

120

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FIGURES FOR CHAPTER 1 127

Figure 1.2: Exponential growth: y = ekt (k > 0). Vertical lines mark the doubling time.

2 4 6 8 10 12t

2

4

6

8

10

y

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128 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.3: Exponential decay: y = ekt (k < 0). Vertical lines are at intervals of one halflife.

2 4 6 8 10 12t

0.2

0.4

0.6

0.8

1

y

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FIGURES FOR CHAPTER 1 129

Figure 1.4: Temperature in a heated building (black curve), and the stable periodic tem-perature (blue curve). The outdoor temperature is indicated by the red curve. See exam-ple 1.3.3.

10 20 30 40

-5

5

10

15

20

25

30

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130 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.5: Solutions of y′ + 3y = 2e−t. See example 1.3.1.

-2 -1 1 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

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FIGURES FOR CHAPTER 1 131

Figure 1.6: A typical mixture problem.

INPUT:K grams solute per

liter

MIXING TANKx(t) grams solute in

V(t) liters

@@@@@@@R

J liters/second

OUTPUT

@@@@@@@R

L liters/second

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132 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.7: The velocity of a falling object.

5 10 15 20

10

20

30

40

50

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FIGURES FOR CHAPTER 1 133

Figure 1.8: Solutions of the differential equation y′ = t/y. The solutionwith initial value y(0) = −1 is shown in blue.

-4 -2 2 4

-2

-1.5

-1

-0.5

0.5

1

1.5

2

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134 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.9: Family of solutions of a separable equation y′ = 3y2/3. See ex-ample 1.6.4.

-2 -1 1 2

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

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FIGURES FOR CHAPTER 1 135

Figure 1.10: Direction field for y′ = t− y2

-1.5 -1 -0.5 0 0.5 1 1.5

-1

-0.5

0

0.5

1

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136 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.11: Direction fields corresponding to models for the velocity of afalling object. On the left, the drag coefficient is constant, and on the right,the drag coefficient decreases with time.

0 1 2 3 4 5 6

25

30

35

40

0 1 2 3 4 5 6

25

30

35

40

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FIGURES FOR CHAPTER 1 137

Figure 1.12: Approximate solution (in red) by Euler’s method, the exactsolution (in green), and the direction field for y′ = t− y2.

-1.5 -1 -0.5 0 0.5 1 1.5

-1

-0.5

0

0.5

1

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138 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.13: Approximate solution of y′ = y + 0.25; y(0) = 0 (in blue) byEuler’s method. The actual solution φ(t) (in red), and the graphs of φm(t)for m = 1, 2, 3 are also shown.

0.5 1 1.5 2

0.25

0.5

0.75

1

1.25

1.5

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FIGURES FOR CHAPTER 1 139

Figure 1.14: Velocity of a rock sinking in seawater, determined by Euler’smethod. See example 1.8.3.

0.25 0.5 0.75 1 1.25 1.5 1.75 2

0.5

1

1.5

2

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140 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.15: Spreadsheet solution of equation (1.50). Column A containsthe t values, column B, the y-values. ∆t appears in column C, and ∆y incolumn D. The graph is an XY-chart with data series from column B.

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FIGURES FOR CHAPTER 1 141

Figure 1.16: The blue curve is the solution of the IVP 1.56; the red curveis part of the graph of the equation y = 1/(t + 1) and satisfies the ODEy′ = −y/(t + 1), but is not considered to be part of the solution of the IVPwith y(0) = 1.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

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142 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.17: Solutions of y′ = ty2.

-0.6 -0.4 -0.2 0.2 0.4 0.6

25

50

75

100

125

150

175

200

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FIGURES FOR CHAPTER 1 143

Figure 1.18: Two solutions of the IVP y′ = 3y2/3; y(0) = 0.

-4 -2 0 2 4

-1.5

-1

-0.5

0

0.5

1

1.5

2

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144 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.19: Solutions of the logistic equation satisfying the initial condi-tions y(0) = 2, y(0) = 1

4 , and y(0) = −1. The parameters k and M were0.04 and 1, respectively.

-50 50 100 150t

-2

-1

1

2

3

y

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FIGURES FOR CHAPTER 1 145

Figure 1.20: Solution of the logistic equation with parameters chosen to fitthe data given in example 1.10.1.

-1 1 2 3 4 5 6t

1

2

3

4

y

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146 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.21: Population of the United States in millions, censuses of 1790 –2000. The logistic curve resulting from Pearl and Reed’s study is superim-posed, and the data points used in their study are shown with diamond-shaped marks. The dashed line represents the carrying capacity that Pearland Reed projected.

1850 1900 1950 2000

50

100

150

200

250

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FIGURES FOR CHAPTER 1 147

Figure 1.22: Solutions of the ODE y′ = 4y − y3. The phase diagram issuperimposed on the y-axis.

-1.5 -1 -0.5 0.5 1 1.5

-4

-2

2

4

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148 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.23: Stationary points: On the left, y1 is stable. In the middle andright diagrams, y2 and y3 are unstable.

s y1

6

? s y2

6

6 s y3

?

6

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FIGURES FOR CHAPTER 1 149

Figure 1.24: Graphs of u = sin(v) and u = 2v/π.

-4 -2 2 4

-3

-2

-1

1

2

3

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150 CHAPTER 1. FIRST-ORDER EQUATIONS

Figure 1.25: Population growth rates, as functions of the population, forvarious rates of harvest. In each graph, the growth parameter has been setat k = 0.4 per individual organism per year, and the carrying capacity atM = 105.

50000 100000

-5000

H=10000

50000 100000

-5000

H=12000

50000 100000

5000

10000

H=0

50000 100000

-5000

5000

H=5000

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FIGURES FOR CHAPTER 1 151

Table 1.1: Propagation of the solution of y′ = t− y2, y(0) = 0, to the left;see Example 1.8.1.

y0 = 0y−1 = 0− 0.25(0− 02) = 0y−2 = 0− 0.25(−0.25− 02) = 0.0625y−3 = 0.0625− 0.25(−0.5− 0.06252) ≈ 0.1885y−4 ≈ 0.1885− 0.25(−0.75− 0.18852) ≈ 0.3849y−5 ≈ 0.3849− 0.25(−1− 0.38492) ≈ 0.6719y−6 ≈ 0.6719− 0.25(−1.25− 0.67192) ≈ 1.0972

Propagation to the right:

y1 = 0 + 0.25(0− 02)

y2 = 0 + 0.25(0.25− 02) = 0.0625y3 = 0.0625 + 0.25(0.5− 0.06252) ≈ 0.1865y4 ≈ 0.1865 + 0.25(0.75− 0.18852) ≈ 0.3653y5 ≈ 0.3653 + 0.25(1− 0.36532) ≈ 0.5820y6 ≈ 0.5820 + 0.25(1.25− 0.58202) ≈ 0.8098

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152 CHAPTER 1. FIRST-ORDER EQUATIONS

Table 1.2: Euler method solution of the IVP in Example 1.8.2

y1 = y0 + ∆y0 = 0 + 0.5(0 + .25) = 0.125y2 = y1 + ∆y1 = 0.125 + 0.5(0.125 + 0.25) = 0.3125y3 = y2 + ∆y2 = 0.3125 + 0.5(0.3125 + .25) = 0.59375y4 = y3 + ∆y3 = 0.65625 + 0.5(0.65625 + .25) = 1.015625

Local errors:

LE1 = φ(0.5)− y1 = (0.25e0.5 − 0.25)− 0.125 ≈ 0.037LE2 = φ1(1)− y2 = (0.375e0.5 − 0.25)− 0.3125 ≈ 0.056LE3 = φ2(1.5)− y3 = (0.5625e0.5 − 0.25)− 0.59375 ≈ 0.084LE4 = φ3(2)− y4 = (0.84375e0.5 − 0.25)− 1.015625 ≈ 0.125

Accumulated errors

AE2 = φ(1)− φ1(1) = (0.25e1 − 0.25)− (0.375e0.5 − 0.25) ≈ 0.061AE3 = φ(1.5)− φ2(1.5) = (0.25e1.5 − 0.25)− (0.5625e0.5 − 0.25) ≈ 0.193AE4 = φ(2)− φ3(2) = (0.25e2 − 0.25)− (0.84375e0.5 − 0.25) ≈ 0.456

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FIGURES FOR CHAPTER 1 153

Figure 1.26: Bifurcation for the logistic equation with harvesting. The har-vesting rate is the bifurcation parameter; the growth parameter and thecarrying capacity are the same as in figure 1.25.

5000 10000H

50000

100000

P

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154 CHAPTER 1. FIRST-ORDER EQUATIONS

Table 1.3: U.S. population in millions, censuses of 1790 – 2010.

t P(t) t P(t) t P(t)1790 3.9 1870 38.6 1950 151.31800 5.3 1880 50.2 1960 179.31810 7.2 1890 63.0 1970 203.31820 9.6 1900 76.2 1980 225.61830 12.9 1910 92.2 1990 248.71840 17.1 1920 106.0 2000 281.41850 23.2 1930 123.2 2010 309.41860 31.4 1940 132.2

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Chapter 2

Systems of DifferentialEquations

155

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156 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

2.1 Introduction

A system of ODEs is a set of two or more ODEs to be treated simultane-ously. Each ODE involves the same independent variable (we will alwaysuse t, for elapsed time, as the independent variable), but there are as manydependent variables as there are equations. The systems that we will studyhave the form

x′ = f (t, x, y)y′ = g(t, x, y),

}where x′ and y′ denote the derivatives of x and y with respect to t.

Consider the velocity of a object in motion, subject to forces such asgravity and air resistance. In the previous chapters the object was assumedto fall in a vertical direction, and the velocity was represented by a sin-gle ODE. If we are determining the trajectory of a ball that has been hitor thrown, the motion will not be in a straight line. There will be two de-pendent variables, u(t) and v(t) representing the horizontal and verticalcomponents of the velocity, respectively. The drag force vector is pointedin the direction opposite to the velocity vector, and we will denote its hori-zontal and vertical components by−h(u, v) and−k(u, v), respectively. Thegravitational force, mg, is vertical. Our model is a system of ODEs,

m u′ = −h(u, v)m v′ = mg− k(u, v)

Just as single ODEs have families of solutions, so do systems. To finda particular solution of interest, one may specify initial values to createan IVP. The values of all dependent variables in an IVP must be specifiedat a common starting time t0. For example, suppose we are determiningthe motion of a ball thrown horizontally from a tower 100 m. high with avelocity of 30 m/s. Take the moment of release to be t = 0; then the initialconditions are u(0) = 30, and v(0) = 0. The above system of ODEs can beembellished by including the initial conditions:

m u′ = −h(u, v); u(0) = 30m v′ = mg− k(u, v); v(0) = 0

The height of the tower is not relevant in this model, because it ignores theposition of the ball. We could include the position by introducing two addi-tional dependent variables, x(t), the horizontal distance from the tower attime t, and y(t), the vertical distance from the ground. Noting that x′ = u

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2.1. INTRODUCTION 157

and y′ = v the following system is a model for the position and the velocityof the ball:

x′ = u; x(0) = 0y′ = v; y(0) = 100m u′ = −h(u, v); u(0) = 30m v′ = mg− k(u, v); v(0) = 0

While the system we just constructed is a nice model, it is not for begin-ners. Here is a simpler IVP.

Example 2.1.1 Show that x = cos(t), y = − sin(t) is a solution of the initialvalue problem

x′ = y; x(0) = 1y′ = −x; y(0) = 0

}(2.1)

SOLUTION. Substituting x = cos(t) and y = − sin(t) in the first equationyields the identity − sin(t) = − sin(t). The same substitution in the secondequation results in cos(t) = cos(t). Finally, cos(0) = 1 = x0 and− sin(0) = 0 = y0.

The system in the Example 2.1.1 occurs in the following application tomechanics. The dependent variables are the position x(t) and the velocityy(t) of an object in motion. (Air resistance is ignored.) The object is attachedto a spring that obeys Hooke’s law: when the spring is stretched x unitspast its equilibrium position (if the spring is compressed, x is negative), thenet force on the object, including the spring and gravity, is F = −k x. Asusual, the positive direction is downward. The variable y is the velocity, sothat y = dx

dt . By Newton’s second law of motion, the product of the massand the acceleration is equal to the net force. Since the acceleration is equalto dy

dt , it follows that y satisfies

mdydt

= −k x.

In (2.1), units were chosen so that the mass m = 1, and the spring constantk = 1 as well. The initial values represent initial position and velocity, andindicate that at t = 0, the object was pulled 1 unit below its equilibriumposition and released.

A second order ODE can also serve as a model for the motion of anobject suspended from a spring. The dependent variable x(t) represents

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158 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

position. The acceleration is the second derivative of the position with re-spect to time, and Newton’s second law of motion provides the ODE,

md2xdt2 = −kx.

By viewing this physical phenomenon first as a system we see that an IVPinvolving a second order ODE must be requires the initial values of theposition x and the velocity dx

dt . Second order ODEs have their purposes,but the for the present we will be content with learning how to get rid ofthem (and higher-order ODEs) by replacing them with equivalent systemsof first-order ODEs.

Given an explicit second order ODE,

d2xdt2 = f

(t, x,

dxdt

), (2.2)

put y = dxdt . The system of first-order ODEs,

x′ = yy′ = f (t, x, y).

}(2.3)

is equivalent to (2.2), because if x = φ(t), y = ψ(t) is a solution of thesystem (2.3) then φ′(t) = ψ(t) (by the first equation). Thus

φ′′(t) = ψ′(t) = f (t, φ(t), φ′(t));

Conversely, if x = φ(t) is a solution of equation (2.2), put ψ(t) = φ′(t).Then x = φ(t), y = ψ(t) is a solution of the system (2.3). We say that thesystem (2.3) replaces the ODE (2.2).

Example 2.1.2 Find a system of first order ODEs that replaces the ODE

y′′ + t y′ + sin(y) = 0

SOLUTION. The ODE can be written as

d2ydt2 = −

(tdydt

+ sin(y))

Set v = dydt and v′ = d2y

dt2 . Then the system

y′ = vv′ = −(t v + sin(y))

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2.1. INTRODUCTION 159

replaces the given ODE.

An explicit higher-order ODE can also be replaced by a system of first-order ODEs, with the number of equations in the system equal to the orderof the original equation. Thus, given an ODE

dnxdtn = f

(t, x,

dxdt

, . . . ,dn−1xdtn−1

),

define new dependent variables y1(t), y2(t), . . . , yn−1(t) by y1(t) = x′(t),y2(t) = x′′(t), . . . , yn−1(t) = x(n−1)(t). The nth order equation shown aboveis replaced by a system

x′ = y1y′1 = y2

...y′n−2 = yn−1y′n−1 = f (t, x, y1, y2, . . . , yn−1)

Uncoupled Systems

A system of two ODEs is uncoupled, if one of the differential equationsdoes not involve one of the dependent variables. Thus, an uncoupled sys-tem has the form

x′ = f (t, x)y′ = g(t, x, y)

To find the general solution of this system, we start by solving the firstequation. This gives a formula x = φ(t, C), where C is a constant. Substi-tuting this formula for x in the second equation yields another first orderequation with one dependent variable, y′ = g(t, φ(t, C), y). Thus, solvingan uncoupled system of two differential equations is accomplished by solv-ing two single first order equations.

Example 2.1.3 Find the general solution of the system

x′ = 2 t x2

y′ = 2 t x y

In addition, find the solutions that satisfy the following sets of initial conditions.

(a) x(0) = 1, y(0) = 1;

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160 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

(b) x(0) = 0, y(0) = 1;

(c) x(0) = 1, y(0) = 0.

SOLUTION. The equation x′ = 2x2t is separable, and can be written

dxx2 = 2t dt.

Integration of both sides yields −x−1 = t2 + C, or

x =−1

t2 + C. (2.4)

There is also a singular solution, x ≡ 0. We now substitute formula (2.4)for x in the second equation to get

y′ = 2t(−1

t2 + C

)y

which is also separable, and can be written in the equivalent form

dyy

= − 2t dtt2 + C

.

Integrating again, we have ln |y| = − ln |t2 + C|+ D, where D is a secondconstant. Taking the exponential of each side of this equation, we obtain

|y| = e− ln |t2+C|+D

=eD

t2 + C

Thusy =

Kt2 + C

where K = ±eD. It is also possible that K = 0 (why?).If we substitute the singular solution x ≡ 0 in the second equation (insteadof formula (2.4)), then y′ ≡ 0, so y would be constant.The solution of the system will thus be one of the following function pairs:{

x = − 1t2+C

y = Kt2+C

}or{

x = 0y = K

}where C and K are constants.We now solve the initial value problems.

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2.1. INTRODUCTION 161

(a) Set t = 0 and x = 1 in x = −1t2+C . This yields 1 = −1/C, so C = −1.

Now we set t = 0, y = 1, and C = −1 in y = Kt2+C , to get

1 = K/(−1). Hence K = −1 and the solution is

x =−1

t2 − 1

y =−1

t2 − 1.

(b) Since x(0) = 0, the solution of the first equation is the singularsolution x ≡ 0. It follows that y is constant, and must be equal to itsinitial value of 1. The solution is

x ≡ 0y ≡ 1.

(c) As in part (a), x = −1t2−1 . Since y(0) = 0, y ≡ 0. The solution is

x ≡ −1t2 − 1

y ≡ 0.

Example 2.1.4 Find the general solution of the second order ODE,

d2xdt2 =

dxdt

+ 1.

SOLUTION. Put x′ = y, so that the ODE is replaced by the system,

x′ = yy′ = y + 1

}(2.5)

The second equation does not involve the variable x, and thus (2.5) isuncoupled. That equation is linear, and its solution is y = Cet − 1.Referring to the first equation of (2.5), we have x′ = Cet − 1, which can besolved by antidifferentiating. The solution is

x = Cet − t + D,y = Cet − 1

where C and D are constants.

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162 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

Exercises

1. Show that for all real C, x = C et, y = C et is a solution of thesystem

x′ = yy′ = x,

and find all solutions of the form x = A e−t, y = B e−t, where A andB are constants.Answer

2. Show that x = C e−2t, y = −3C e−2t is a solution of

x′ = x + yy′ = 3x− y,

and find all solutions of the form x = A e2t, y = B e2t, where A and Bare constants.

3. Show that x = 2C cos(3t), y = C[cos(3t) + 3 sin(3t)] is a solutionof

x′ = x− 2yy′ = 5x− y,

and find all solutions where x = 2A sin(3t).Answer

4. Show that x = sin(2t) + et, y = 2 cos(2t)− 4et is a solution of

x′ = 5 et + yy′ = −4x,

and find a family of solutions of this system.

5. Show that x = e2t(t + 1), y = e2t(t− 1) is a solution of

x′ = x + y + 3e2t

y′ = 2x− 3e2t.

Furthermore, show that x = y = C e2t is a family of solutions of the“associated homogeneous system”

x′ = x + yy′ = 2x,

and find a family of solutions of the inhomogeneous system.Answer

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2.1. INTRODUCTION 163

In problems 6 – 9 find a system of two first order equations that replacesthe given second order ODE.

6. d2xdt2 = x dx

dt + 1.

7. y′′ + 3y′ + 4y = t2.Answer

8. z′′ + (z2 − 1) z′ + z = 0.

9. u′′u′u = 1 + t2

Answer

In problems 10 – 18 determine if the system is uncoupled. If the systemis uncoupled, find its general solution.

10.{

x′ = 4x + 1y′ = y

11.{

x′ = x2

y′ = 1/xAnswer

12.{

x′ = xyy′ = −y

13.{

x′ = y2

y′ = −x2

Answer

14.{

x′ = x− et

y′ = x + y

15.{

x′ = t/xy′ = t x y

Answer

16.{

x′ = x + y2

y′ = y/2

17.{

x′ = x + yy′ = x− y

Answer

18.{

x′ = etxy′ = e2tx + y

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164 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

2.2 The Phase Plane

There are several ways to present a solution x = φ(t), y = ψ(t) of a systemof two ODEs graphically, the simplest of which is to draw separate graphsof φ(t) and ψ(t). This is useful, but it does not show any relationship be-tween x and y. For example, it is easily verified that x = cos(t), y = sin(t)is a solution of the system

x′ = −yy′ = x,

but it is not obvious from the graphs of the sine and cosine functions thatx2 + y2 = 1.

The x, y-plane, whose coordinates represent the dependent variables ofthe system of ODEs, is called the phase plane. Let x = φ(t), y = ψ(t)be a solution of a system of two differential equations. The curve in thephase plane described by the parametric equations x = φ(t), y = ψ(t) iscalled the orbit of the solution. It is usually marked to show the directionof increasing t. For example, the orbit of the solution x = cos(t), y = sin(t)is the circle x2 + y2 = 1, marked with an arrow in the counterclockwisedirection, since as t increases from 0 to π

2 , (cos(t), sin(t)) moves along thecircle from (1, 0) to (0, 1).

While it is natural to think of an orbit as a curve in the phase plane,it is possible for an orbit to be just a point. If (x1, y1) is a point such thatf (x1, y1) = 0 and g(x1, y1) = 0 then (x, y) ≡ (x1, y1) is a solution of thesystem

x′ = f (x, y)y′ = g(x, y),

}(2.6)

corresponding to an orbit that is the single point, (x1, y1). In this case,(x1, y1) is a stationary point of the system (2.6).

Example 2.2.1 Describe the orbits of the system

x′ = xy′ = y

SOLUTION. This system is readily solved because it is uncoupled. Thegeneral solution of x′ = x is x = C et and the solution of y′ = y is y = D et.(The constants C and D are independent of each other.)If C = D = 0, the orbit is simply the origin itself (no arrow is attached tothis “stationary” orbit). If C = 0 the orbit is the positive y-axis, directed

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2.2. THE PHASE PLANE 165

upward, if D > 0 or the negative y-axis, directed downward, if D < 0.Assuming C 6= 0, we can substitute et = x/C in the formula for y to obtainy = (D/C) x. The orbit consists of all points on this straight line that lie onthe same side of the origin as (C, D), and is directed away from the origin.See figure 2.1.

Example 2.2.2 Find a system of ODEs that replaces

y′′ = y′ + 2y (2.7)

Given that y = e−t, y = e2t and y = e2t − e−t are solutions of equation (2.7) findthe corresponding solutions of the system and describe their orbits in the phaseplane.

SOLUTION. Let v = y′. Then the system

y′ = vv′ = v + 2y

replaces equation (2.7). The solution corresponding to y = e−t is y = e−t,v = −e−t. Notice that v = −y for all t, and that as t→ ∞, (y, v)→ (0, 0).Furthermore y is positive and v is negative, which puts the orbit in thefourth quadrant of the y, v plane. Therefore, the orbit is the part of thestraight line v = −y in the fourth quadrant, and is directed toward theorigin.The orbit of the solution corresponding to y = e2t is (y, v) = (e2t, 2 e2t).Here we have v = 2y, and as t→ ∞, both y and v→ +∞. This orbit istherefore the part of the line v = 2y in the first quadrant, directed awayfrom the origin.The parametric equation of the orbit of the solution corresponding toy = e2t − e−t is (y, v) = (e2t − e−t, 2 e2t + e−t). This will be left in parametricform, since it is not easy to eliminate t to obtain a relation between v andy. However, we can note that (y(0), v(0)) = (0, 3), and that as t→ −∞,(y(t), v(t))→ (−∞, ∞), while (y(t), v(t))→ (∞, ∞) as t→ ∞. The threeorbits are displayed in figure 2.2.

A systemx′ = f (x, y)y′ = g(x, y)

}(2.8)

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166 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

with the property that the independent variable does not appear on theright side of either equation is said to be autonomous. An autonomoussystem describes a vector field that assigns to each point (x, y) a vector

~v(x, y) = f (x, y)~i + g(x, y)~j.

(Here,~i and ~j denote the usual horizontal and vertical unit vectors in theplane.) A vector field can be represented graphically by drawing the vec-tors~v(xm, yn) at grid points (xm, yn) on graph paper. Such drawings resem-ble direction fields for ODEs, but there is a distinction: the vectors of thevector field have varying magnitude—displayed by giving them differentlengths—and each has a direction and so is drawn as an arrow. Figure 2.3shows the vector field ~v(x, y) = x~i− y~j that corresponds to the uncoupledsystem

x′ = xy′ = −y.

The orbits are not shown, but you should verify that they consist of the raysof the x-axis, directed away from the origin, the rays of the y-axis, directedtoward the origin, and components of the hyperbolas xy = C, directeddownward in the first and second quadrants, and directed upward in thethird and fourth quadrants.

We have seen that an autonomous system of two ODEs can be dis-played as a vector field in the plane. It is also useful to interpret a vectorfield

~v(x, y) = f (x, y)~i + g(x, y)~j

as a system of ODEs,

x′ = f (x, y)y′ = g(x, y)

If we think of ~v(x, y) as the velocity vector for a particle located at (x, y),then the particle will follow an orbit of this system.

An integral curve of a vector field ~v(x, y) is a curve with parametricequations (x, y) = (φ(t), ψ(t)) with tangent vector φ′(t)~i + ψ′(t)~j equal tothe field vector ~v(x, y) at each point of the curve; that is, for all t,

~v(φ(t), ψ(t)) = φ′(t)~i + ψ′(t)~j.

The term stationary point is used in the context of a vector field ~v toindicate a point (x1, y1) where~v(x1, y1) =~0. A stationary point of~v(x, y) =f (x, y)~i + g(x, y)~j is also a stationary point of the system x′ = f (x, y), y′ =g(x, y).

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2.2. THE PHASE PLANE 167

Displaying a solution of a system

In chapter 3, we will learn how to solve systems such as the one in thefollowing example.

Example 2.2.3 Verify that

x(t) = et cos(10t)y(t) = et sin(10t)

satisfies the systemx′ = x− 10yy′ = 10x + y,

(2.9)

and display the solution graphically.

SOLUTION. By the product rule for differentiation,

x′(t) = et cos(10t)− 10et sin(10t) = x(t)− 10y(t),

andy′(t) = et sin(10t) + 10et cos(10t) = y(t) + 10x(t).

There are several ways to display graphs of solutions of systems. Our firstplot, Figure 2.4, shows the components x(t) and y(t) of the solutionof (2.9) plotted together.We see that the solution functions are oscillating, with increasingamplitude. The orbit, displayed with the vector field representation of thesystem (2.9), is shown in figure 2.5. The orbit is an outward spiral, which iscertainly consistent with the graphs shown in figure 2.4.Using a three dimensional plot, we can display the graph of (x, y) as afunction of t. See figure 2.6, in which the t-axis, for the independentvariable, is vertical, and the x, y-plane is the “axis” for the dependentvariables.

Exercises

In exercises 1 – 8, show that the given pair of functions is a solution of thesystem of differential equations, and sketch the orbit determined by thesolution. If a second order ODE is given, replace it with a system of twofirst order equations and proceed.

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168 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

1.{

x′ = 2y2/ty′ = y2/x

x = t2, y = t.Answer

2. y′′ = y; draw the orbits corresponding to three solutions: y =−e−t, y = et, and y = et − e−t.

3.{

x′ = yy′ = x

x = e−t, y = −e−t.Answer

4.{

x′ = yy′ = x

x = cosh(t), y = sinh(t). Hint: Remember the identity cosh2(t) −sinh2(t) = 1.

5. y′′ = y−3; solution, y =√

t2 + 1Answer

6.{

x′ = xyy′ = x2

x = sec(t), y = tan(t).

7. y′′ = 12 t y′ − y; solution, y = t2 − 2.

Answer

8.{

x′ = x− 2yy′ = 2x + y

x = et cos(2t), y = et sin(2t).

In problems 9 – 12, find stationary points for the given vector fields.Draw some of the integral curves. (You may want to print copies of thevector fields and draw the integral curves on these copies.)

9. v(x, y) = [4yi− xj]/16.

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2.2. THE PHASE PLANE 169

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

Answer

10. v(x, y) = [(y2 − y)i + xj]/4.

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

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170 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

11. v(x, y) = [(x2 + 1)yi− 2xj]/16.

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

Answer

12. v(x, y) = [xi + 2yj]/8.

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2.2. THE PHASE PLANE 171

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

13. Use a CAS to plot the graphs of the component functions, the two-dimensional orbits and the three-dimensional graphs of the solutionsof the following systems.

(a) {x′ = 5yy′ = −5x.

with initial conditions x(0) = 0, y(0) = 1; 0 ≤ t ≤ 2π.(b) {

x′ = 2x− 6yy′ = x− 2y.

with initial conditions x(0) = 1, y(0) = 0; 0 ≤ t ≤ 4π.(c) {

x′ = 2y + sin(t)y′ = −2x− sin(2t).

with initial conditions x(0) = 0, y(0) = 0; −2π ≤ t ≤ 4π.

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172 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

(d) {x′ = x + yy′ = t− x− y.

with initial conditions x(0) = 0, y(0) = 0; −2 ≤ t ≤ 4.

Answer

14. Draw graphs of the component functions of the solution

(x, y) = (sinh(t), cosh(t))

ofx′ = y; x(0) = 0y′ = x; y(0) = 1,

and plot its orbit.

2.3 A User’s Guide to IVP Solvers

An IVP solver is a numerical method for computing solutions of IVPs.We already know of one: Euler’s method. Although it was presented in

the context of a single ODE, it is easily modified to approximate solutionsof IVPs involving systems of ODEs. Recall that Euler’s method works byreplacing the ODE by an easily solved difference equation. For a system,

x′ = f (t, x, y); x(t0) = Ay′ = g(t, x, y); y(t0) = B,

we use a system of difference equations,

∆xj = h f (ti, xi, yi); x0 = A∆yj = hg(ti, xi, yi); y0 = B,

where h is the time step, and ti is the time after i steps have been taken; thatis, ti = t0 + i · h.

Using this system, we obtain sequences {xi} and {yi}. If x = φ(t),y = ψ(t) denotes the solution of the IVP, then xi is an approximation ofφ(ti) and yi approximates ψ(ti).

Example 2.3.1 Plot the orbit of the system

x′ = 2x + 2yy′ = −x + 4y.

with initial conditions x(0) = 1, y(0) = 0, for t ranging from −1 to 1.

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2.3. A USER’S GUIDE TO IVP SOLVERS 173

SOLUTION. It is possible to solve this system analytically, or to use an IVPsolver incorporated in a CAS to produce a numerical solution. We will usea spreadsheet, with a moderately large stepsize, h = 0.1. Ourspreadsheet, shown in figure 2.7, has five columns, representing t, x, y,∆x, and ∆y — the latter two are labelled Dx and Dy. The calculations aredone in two blocks, the first for 0 ≤ t ≤ 1, and the second for 0 ≥ t ≥ −1.The formulas shown in figure 2.7 were entered once in each block, andthen copied. Thus, the entry in cell D2, +0.1*(2*B2+2*C2) was typed,copied onto the clipboard, and then pasted into the cells D3 through D12,and so on. This formula says to multiply the entry in cell B2 (the value ofx0) by 2 and add the result to 2× the entry in cell C2 (the value of y0) andthen multiply this sum by 0.1. The result is 0.1(2x0 + 2y0), which is equalto ∆x1. The entry in cell B3 adds the entries in cells B2 and D2 to formx1 = x0 + ∆x1.The spreadsheet in figure 2.7 shows the formulas instead of numericalvalues. The numerical values, and the orbit of the computed solution areshown in figure 2.8. The graph was made by selecting xy graph in thegraphics type menu, with x values from column B and series valuesfrom column C.

To assess the precision of the approximation just computed, see fig-ure 2.9.

The strength of Euler’s method is its simplicity. As we have seen, the ac-cumulated error varies proportionately with the step size. Since the amountof effort is inversely proportional to the time step, we can expect to use 10times as much computer time to improve accuracy by 1 decimal place. It isworthwhile to find a more efficient method.

A classical IVP solver is an IVP solver in which the user specifies atime step that will not be altered when the calculation is in progress. Theorder is a number n such that the accumulated error varies proportionatelywith |h|n, where h is the time step. Euler’s method, for example, is a classi-cal method of order 1. Classical solvers can be constructed with arbitrarilyhigh order. Runge-Kutta methods (the Euler method is one of these) oper-ate by setting

∆ym = h× ( a weighted average of values of f (t, y)), with tm ≤ t ≤ tm+1.

For example, the “improved Euler method” is typical. Given (tm, ym), set

ym+1 = ym + h f (tm, ym) and tm+1 = tm + h.

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174 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

Then set

∆ym =12( f (tm, ym) + f (tm+1, ym+1))

This method is second order. It requires two function evaluations per step,but the payoff is that the accumulated error is proportional to h2. Thus itwill take about

√10 times as much computer time to improve accuracy by

one decimal place.The most commonly used classical methods are of order 4. To obtain a

fourth order Runge-Kutta method, it turns out that four function evalua-tions are required per step. For the “RK4” method, we define

k1 = f (tm, ym)

k2 = f (tm + 0.5h, ym + 0.5k1h)k3 = f (tm + 0.5h, ym + 0.5k2h)k4 = f (tm + h, ym + k3h),

and then

∆ym = h(

16

k1 +13

k2 +13

k3 +16

k4

).

To obtain an additional decimal place of accuracy with the RK4 method,we expect to do 4

√10 ≈ 1.8 times as much work.

The predictor methods work a little differently. The Adams-Bashforthmethod (also a fourth order method) calculates ∆ym as follows: let p0 =f (tm, ym), p1 = f (tm−1, ym−1), p2 = f (tm−2, ym−2), p3 = f (tm−3, ym−3).Then

h× ( a weighted average of p0 . . . p3).

To be precise,

∆ym = h(

5524

p0 −5924

p1 +3724

p2 −924

p3

).

The advantage of this method is that it recycles previously computed data,and requires only one new function evaluation per step. This strength isalso a weakness, because the error in the computed value of ym is recycledand can cause accuracy to degenerate. To control error propagation, theAdams-Bashforth method can be coupled with a “correction stage.” Theresulting method is known as the Adams-Bashforth-Moulton method, andit works as follows. Numerical analysts refer to the sequence of calculationsby the acronym “PECE”.

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2.3. A USER’S GUIDE TO IVP SOLVERS 175

Predict Use the Adams-Bashforth method to find a tentative value for ym+1:

ym+1 = ym + h(

5524

p0 −5924

p1 +3724

p2 −9

24p3

)Evaluate Set p−1 = f (tm+1, ym+1).

Correct Recalculate ym+1, as

ym+1 = ym + h(

924

p−1 +1924

p0 −524

p1 +124

p2

)Evaluate Calculate f (tm+1, ym+1) to prepare for the next step.

The improved Euler method that was mentioned above is a predictor-cor-rector method as well as a Runge-Kutta method, since ym+1 is predictedby the Euler method, and used in a correction phase to get an improvedapproximation of the solution at t = tm+1.

The classical methods are not used for high precision work, becauseit is hard to assess the accuracy of a solution without recomputing thewhole thing with smaller time steps and making comparisons. Methodsthat change the stepsize as the computation progresses are called dynamicmethods. The Runge-Kutta-Fehlberg (RKF45) method is widely used. Withthis method, each value of ym+1 is calculated twice in each step, first usingthe RK4 method, followed by a Runge-Kutta method of order 5. When thetwo computed values differ by more than a specified tolerance, both arerejected, the time step is reduced, and the process is repeated. There is aprovision to stop when a minimum time step is reached: this avoids hav-ing the program running forever if, for example, the solution has a verticalasymptote.

Other dynamic methods are designed for either increased precision orto address problems encountered with certain types of higher order equa-tions.

It is not practical to use a spreadsheet with dynamic IVP solvers. Fortu-nately, the programs supplied with CAS systems are excellent and easy touse.

It may be convenient to use an IVP solver independently of a CAS, andseveral such programs are available. John Polking has made pair of Javaapplets, DFIELD and PPLANE, available on the web site

http://math.rice.edu/∼dfield/dfpp.html1

1You will have to type this URL, because the tilde symbol will not copy correctly in yourbrowser.

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176 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

These applets will draw direction fields for first order ODEs, and vectorfields representing systems of two autonomous first-order ODEs, respec-tively. By clicking any point in the field with the mouse pointer, one cansee a graph (or orbit) of the solution, with the initial condition specifiedby the point that was selected.There are several graphing calculators thatincorporate IVP solvers as well.

Exercises

1. The solution of the IVP

x′ = −y x(0) = 1y′ = x y(0) = 0

is (x, y) = (cos(t), sin(t)). Use the RKF45 algorithm with this IVP toestimate sin(1). Compare the answer you get to the value from yourcalculator (be sure it is set to radians!).Answer

2. The solution of the IVP

x′ = −4y x(0) = 1y′ = x y(0) = 0

is (x, y) =(cos(2t), 1

2 sin(2t))

. Thus, the identity x2 + 4y2 = 1 is satis-fied, and the orbit is an ellipse. Use an IVP solver to draw this ellipse.If you use a crude method like Euler’s, do you get a closed curve? Re-peat the experiment with a classical fourth order method, and againwith a dynamic method.

3. Consider the following system, which depends on a parameter ε:

x′ = −y + ε x(x2 + y2 − 1) x(0) = Ay′ = x + ε y(x2 + y2 − 1) y(0) = 0

When ε = 0 this system is the same as that of exercise 1, and if A =1, the solution of the IVP is the same as in exercise 1. This exerciseinvestigates what happens when ε 6= 0 and A 6= 1. Let r(x, y) denotethe distance of the point (x, y) to the origin. Differentiation of r2 =x2 + y2 with respect to t yields

r r′ = x x′ + y y′

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2.3. A USER’S GUIDE TO IVP SOLVERS 177

Furthermore, if θ = arg(x + i y) is the polar angle,

θ′ =x y′ − y x′

x2 + y2

Show that if (x(t), y(t)) is the solution of the IVP then

r′ = ε r(r2 − 1); r(0) = A

and draw a phase diagram for this IVP. Furthermore, show that θ′ =1. Thus orbits rotate counterclockwise about the origin, spiraling out-ward if r is increasing and inward if r is decreasing.

(a) Put ε = −.25 and A = .05. Use the RKF45 algorithm to draw theorbit of the solution for 0 ≤ 30. Explain how the orbit shown isconsistent with the conclusions reached above.

(b) Repeat the experiment with the following data.

i. (ε, A) = (−.25, 4).ii. (ε, A) = (.25, .95).

iii. (ε, A) = (.25, 1.05).

Answer

4. The solution φ(t) of the IVP

y′ = t2 + y2; y(0) = 0 (2.10)

has two vertical asymptotes, t = c and t = −c. Estimate the value ofc.

5. Produce an accurate plot of the solution φ(t) of the IVP (2.10), for−c < t < c. ranging from the negative asymptote t = −c to the pos-itive asymptote t = c. Change the time step to 0.05, and use Euler’smethod to generate an approximate solution. Reduce the time stepby a factor of 2 repeatedly. Describe precisely the effect that reducingthe time step has on the error. Repeat the experiment with a secondorder method, and the classical RK4 method.Answer

6. Draw the phase diagram for the ODE y′ = 4y− y3, and sketch thegraphs of four solutions with different asymptotic behavior. Use anIVP solver to graph the four solutions with initial conditions that youshould select so that the graphs resemble your sketches.

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7. Problem 6 suggests a way to find real roots of a polynomial f (y)by using an IVP solver. The roots will be horizontal asymptotes ofsolutions of the differential equation y′ = f (y), so by plotting numer-ous solutions, they can be located. Hint: if you try to solve y′ = f (y),where f (y) is a high degree polynomial, your computer will proba-bly complain of overflow problems. In this case, try replacing f (y)with f (y)/10N . Here you would choose N to be a positive integerlarge enough to eliminate the overflow, but small enough so that thesolution you are plotting “levels off” soon.

(a) Try the method on a polynomial with known roots: f (y) =(2y− 1)(2y+ 1)2(2y+ 3)3(2y+ 5). Can you locate all of the roots(including the multiple ones)?

(b) The polynomial

f (y) = y7− 49y6 + 882y5− 7350y4 + 29400y3− 52920y2 + 35280y− 5040

has seven real roots in the interval [0, 20]. Approximate them.

Answer

8. The differential equation

y′ + .05y = 2 sin(2t) + sin(3t)

has a periodic solution. Use an IVP solver to plot the graph of thissolution.

9. Use the computer to determine and graph several solutions of thefollowing differential equations, and note any periodic solutions.

(a) y′ + 0.1y = sin(2t).

(b) y′ − y = sin(20t).

(c) y′ + 10y = sin(100t).

Answer

10. Let S(t) = +1 if the integer part of t is even, (i.e. t ∈ [0, 1), [2, 3),etc., and S(t) = −1 otherwise. This function is sometimes called asquare wave. Graph the periodic solution of

y′ + y = S(t).

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To represent S(t) with a simple formula your CAS can handle, notethat S(t) = (−1)btcwhere btc is the “floor” function that rounds num-bers down to their integer parts. It is also possible to use S(t) =sign(sin t) if the floor function is not available.

2.4 Autonomous Systems

A phase portrait for an autonomous system of ODEs,

x′ = f (x, y)y′ = g(x, y)

}(2.11)

is a drawing of the phase plane, showing a representative collection of or-bits of the system.

Example 2.4.1 Draw a phase portrait of the system

x′ = 2xy′ = y,

SOLUTION. The system is uncoupled, and each equation in the system isa homogeneous linear ODE. The general solution of the first equation isx = Ce2t, and the general solution of the second is y = Det, where C andD are constants. An orbit in the phase plane is therefore described byparametric equations

x = Ce2t

y = Det,

}(2.12)

To eliminate t from these equations, note that y2 = D2e2t. Therefore, ifD 6= 0,

xy2 =

CD2 = constant.

Therefore, x = Ay2, where A is a constant. The orbit follows a parabolawith a horizontal axis of symmetry, opening to the right if A > 0, and to theleft if A < 0. If A = 0 the orbit lies on the y-axis, rising upward if D > 0,and downward if D < 0. Finally, if D = 0 then y = 0: the orbit follows thex-axis, headed away from the origin to the right if C > 0, and to the left ifC < 0. If C = D = 0 we have a stationary orbit.Thus, each parabola or axis actually consists of three orbits. The origin isa stationary point, and an orbit by itself. The other two orbits follow the

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180 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

parabola or axis, leading away from the origin. Figure 2.10 is a phaseportrait of the system (2.12). This figure includes the vector field, butnormally the vector field is not shown in a phase portrait.

Example 2.4.1 is unusual because we were able to determine the generalsolution of the system. Often, we must approach the problem of drawing aphase portrait by relying on qualitative analysis of the system and an IVPsolver. The qualitative analysis should be done first, as it will aid us inchoosing initial points for the IVP solver to use.

A set of points where f (x, y) = 0 is called an x-nullcline of the sys-tem (2.11). If (x, y) = (φ(t), ψ(t)) is a solution whose orbit crosses an x-nullcline at a point (x1, y1) = (φ(t1), ψ(t1)), then φ′(t1) = f (x1, y1) = 0. Ifψ′(t1) = g(x1, y1) 6= 0, the orbit must have a vertical direction at (x1, y1),which will be upward if g(x1, y1) > 0, and downward if g(x1, y1) < 0. Thex-nullcline should be marked with vertical arrows according to the signof g(x, y). Similarly, a y-nullcline is a set of points where g(x, y) = 0, andnon-constant orbits have horizontal tangents at points where they cross ay-nullcline. We mark the y-nullcline with horizontal arrows, pointed to theright where f (x, y) > 0, and to the left where f (x, y) < 0.

Stationary points are located at the intersections of the x- and y-nullclines.They can be found by solving the pair of equations f (x, y) = 0, g(x, y) = 0.

Example 2.4.2 Find the nullclines and stationary points of the system

x′ = x(x− 4y)y′ = y(y2 − 2x− 9)

SOLUTION. The x-nullclines are given by the equation

x(x− 4y) = 0.

Since this equation is satisfied if either x = 0 or if x = 4y, the x-nullclinesare the y-axis and the line x = 4y.The y-nullclines are determined by the equation

y(y2 − 2x− 9) = 0,

and hence they are the x-axis and the parabola x = (y2 − 9)/2.The stationary points lie at the intersection of x-nullclines with y-nullclines.The axes meet at (0, 0), and the y-nullcline x = (y2 − 9)/2 meets the

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y-axis at (0,±3). Finally, the x-nullcline x = 4y meets the y-nullclinex = (y2 − 9)/2 where y2 − 8y− 9 = 0. The solutions are y = −1 andy = 9, so (−4,−1) and (36, 9) are stationary points. Figure 2.11 shows thenullclines and stationary points.

Integrals

An integral of a system of autonomous ODEs is a function F(x, y) suchthat for every solution x = φ(t), y = ψ(t) of the system, F(φ(t), ψ(t)) isa constant function of t on the interval where (φ(t), ψ(t)) is defined. Thisdefinition can be compared to the definition of integral of a single ODE. Forexample, it is known that all solutions of the system

x′ = −yy′ = x

have the form x = R cos(t − C), y = R sin(t − C), where R and C arearbitrary constants. Hence the trigonometric identity cos2(θ) + sin2(θ) = 1implies that x2 + y2 = R2. In other words, F(x, y) = x2 + y2 is an integralof this system.

Finding an integral of a system by using the general solution of thesystem is not practical, and a fact about parametric equations leads to abetter approach. If x = φ(t), y = ψ(t) are parametric equations for a curve,then the slope of the tangent to the curve at a point (x0, y0) = (φ(t0), ψ(t0))is equal to

dydx

=dy/dtdx/dt

=ψ′(t0)

φ′(t0)

If x = φ(t), y = ψ(t) are solutions of the system, then φ′(t) = f (φ(t), ψ(t))and ψ′(t) = g(φ(t), ψ(t)). Thus, the slope of the orbit passing through apoint (x0, y0) is

dydx

=ψ′(t0)

φ′(t0)=

g(x1, y1)

f (x1, y1).

The orbit therefore satisfies the first order ODE,

dydx

=g(x, y)f (x, y)

This ODE is often written in differential form, as g(x, y) dx− f (x, y) dy = 0.

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Finding an Integral of an Autonomous System

Let F(x, y) be an integral for the ODE

g(x, y) dx− f (x, y) dy = 0. (2.13)

Then F(x, y) is also an integral of the autonomous system

x′ = f (x, y)y′ = g(x, y).

Example 2.4.3 Draw a phase portrait of the system

x′ = −p2 yy′ = q2 x

where p and q are nonzero constants.

SOLUTION. We will find an integral of the system by finding an integral forthe corresponding ODE

q2x dx + p2y dy = 0,

which is easily done since it is exact (and separable as well). The solutionis

q2x2 + p2y2 = C. (2.14)

We can assume C ≥ 0 since otherwise equation (2.14) represents theempty set. When C = 0 we have the origin, which is a stationary point. IfC > 0, put a2 = C/q2 and b2 = C/p2. Dividing equation (2.14) by C, wehave

x2

a2 +y2

b2 = 1

which is the equation of an ellipse, with principal axes of lengths 2a and 2bon the two coordinate axes. Since x′ = −p2y, x is decreasing (directed tothe left) on the upper half plane y > 0, and increasing (directed to theright) on the lower half-plane. Hence the orbit is directed counterclockwise.

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2.4. *AUTONOMOUS SYSTEMS 183

What do these orbits have in common? Each is an ellipse, for which theratio of length of the vertical axis to the length of the horizontal axis isequal to

ba=

√C/q√C/p

=pq

.

Thus the orbits are the origin, and a family of similar ellipses. The phaseportrait is shown in figure 2.12.

The second order ODE, y′′ + y2 = 0, can be replaced by the system

y′ = vv′ = −y2.

(2.15)

Example 2.4.4 Find an integral of the system (2.15) and draw its phase portrait.

SOLUTION. An integral F(y, v) of the ODE

y2 dy + v dv = 0

will also be an integral of the system (2.15). This ODE is exact; integratingyields

F(y, v) =13

y3 +12

v2.

The orbits lie on level curves of F(y, v). Before drawing the phase portrait,we note that the y-nullcline is the v-axis, and the v-nullcline is the y-axis.There is one stationary point, the origin. The phase portrait shown infigure 2.13, was drawn as a contour plot of the level curves of F(y, v). Thevector field

~v(y, v) = v~i− y2~j

was superimposed on the contour plot.

Can orbits intersect?

We will see that for autonomous systems, the answer is no. Is there an or-bit passing through each point of the plane? Here the answer is yes. Theproofs are based on the existence and uniqueness theorems for systems ofODEs. These theorems are modifications of the single-equation existenceand uniqueness theorems. As in the case of the single equation, there is a

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184 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

continuity requirement for f and g for the existence statement, and a Lips-chitz condition is also needed for uniqueness.

Consider an IVP (where the ODEs are not necessarily autonomous),

x′ = f (t, x, y); x(t0) = x0y′ = g(t, x, y); y(t0) = y0

}(2.16)

The functions f and g depend on three variables, and are required to becontinuous in a box in 3-dimensional space, R3, rather than a rectangle inthe t, y-plane. We will denote by B the box

B = {(t, x, y) : |t− t0| < k, |x− x0| < l, |y− y0| < m},

centered at the point (t0, x0, y0) ∈ R3. Here k, l, and m are positive numberswhich specify the size of the box.

To formulate the Lipschitz condition that we need for a system, it ishelpful to consider (2.16) as a single IVP where the dependent variable is avector. Let ~v denote the vector whose components are the dependent vari-ables x and y, and let ~F(t,~v) be the vector function whose components aref (t, x, y) and g(t, x, y). In vector terms, the IVP (2.16) is stated as follows:

d~vdt

= ~F(t,~v); ~v(t0) = ~v0

where ~v0 is the initial vector, with components (x0, y0).We will use absolute value signs to denote distance in the plane: thus if

~v1 = (x1, y1) and ~v2 = (x2, y2),

|~v1 −~v2| =√(x1 − x2)2 + (y1 − y2)2.

With these definitions in mind, the function ~F(t,~v) is said to satisfy aLipschitz condition with respect to ~v in the box B if there is a constant Ksuch that for all (t,~v1), (t,~v2) ∈ B,

|~F(t,~v1)− ~F(t,~v2)| ≤ K |~v1 −~v2|.

Our reason for using vector notation here is economy. The Lipschitzcondition can be expressed without it, but it looks more complicated thatway. See problem 18 at the end of this section.

Theorem 2.1

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2.4. *AUTONOMOUS SYSTEMS 185

Existence Suppose that f (t, x, y) and g(t, x, y) are each continuous at everypoint in the box B. Then there is a number h > 0 and there are func-tions φ(t) and ψ(t), defined on the interval t0 − h < t < t0 + h suchthat x = φ(t), y = ψ(t) is a solution of the IVP (2.16).

Uniqueness If the vector function ~F(t,~v) = ( f (t, x, y), g(t, x, y)) satisfies aLipschitz condition with respect to ~v in B, and x = φ(t), y = ψ(t) isanother solution of the IVP 2.16, then there is a positive number h1 suchthat for all t with t0 − h1 < t < t0 + h1, φ(t) = φ(t) and ψ(t) = ψ(t).

Continuous Dependence Assume that ~F(t,~v) satisfies a Lipschitz conditionwith respect to ~v ∈ B. If ~u = (u1, u2) is a vector such that (t0,~u) ∈ B, letφ(t,~u) denote the solution of

d~vdt

= ~F(t,~v); ~v(t0) = ~u.

Then there is a box B∗ ⊆ B such that φ(t,~u) is continuous as a function ofthe three variables (t, u1, u2) with domain the box B∗.

The proof of this theorem involves the same ideas as those of the single-equation theorems 1.5, 1.7 and 1.8, and is omitted.

Now let’s adapt the uniqueness theorem to the question of intersectingorbits. If a system is not autonomous, it is certainly possible for orbits tointersect. If (x, y) = (φ1(t), ψ1(t)) and (x, y) = (φ2(t), ψ2(t)) are solutionsof a system,

x′ = f (t, x, y)y′ = g(t, x, y)

their orbits could intersect at a point

(x∗, y∗) = (φ1(t1), ψ1(t1)) = (φ2(t2), ψ2(t2))

and as long as t1 6= t2 this would be consistent with the uniqueness the-orem. In the case of autonomous systems, we can show that the orbits donot intersect. To apply the uniqueness theorem, the following lemma isneeded.

Lemma 2.4.1 Let (x, y) = (φ(t), ψ(t)) be a solution of the autonomous system

x′ = f (x, y)y′ = g(x, y).

Then for any t0, (x, y) = (φ(t + t0), ψ(t + t0)) is also a solution.

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PROOF. Put φ(t) = φ(t + t0) and ψ(t) = ψ(t + t0). By the Chain Rule,dφdt = φ′(t + t0) and dψ

dt = ψ′(t + t0). Since for any u,

φ′(u) = f (φ(u), ψ(u)),

it follows that

dt= φ′(t + t0) = f (φ(t + t0), ψ(t + t0))

= f (φ(t), ψ(t)).

Similarly, dψdt = g(φ(t), ψ(t)), and therefore (x, y) = (φ(t), ψ(t)) is a solu-

tion.The proof of this lemma was rather simple, and you should consider

why it does not work for systems that are not autonomous.

Theorem 2.2 Suppose that the functions f (x, y) and g(x, y), are continuous andsatisfy a Lipschitz condition in a rectangleD ⊂ R2. Then for any point (x∗, y∗) ∈D, there is exactly one orbit of the system

x′ = f (x, y)y′ = g(x, y)

}(2.17)

that contains (x∗, y∗).

PROOF. The Existence and Uniqueness Theorems imply that there isexactly one solution, which we will denote (x, y) = (φ1(t), ψ1(t)), of thesystem (2.17) with initial values (x(0), y(0)) = (x∗, y∗). However, for anyt0 6= 0, there is also a solution, which we will denote (x, y) = (φ2(t), ψ2(t)),with initial values (x(t0), y(t0)) = (x∗, y∗). Both solutions will describe anorbit that passes through (x∗, y∗).

We have to show that the two solutions describe the same orbit. Putφ2(t) = φ2(t+ t0), ψ2(t) = ψ2(t+ t0). By Lemma 2.4.1, (x, y) = (φ2(t), ψ2(t))is a solution of the system (2.17), and it describes the same orbit that (φ2, ψ2)does. Since φ2(0) = φ2(t0) = x0, and ψ2(0) = y0, it follows that (φ2, ψ2)satisfies the same initial conditions as (φ1, ψ1). It follows from the Unique-ness Theorem that φ1(t) = φ2(t + t0) and ψ1(t) = ψ2(t + t0) for all t suchthat all four functions are defined.

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2.4. *AUTONOMOUS SYSTEMS 187

Closed orbits

Theorem 2.2 does not preclude an orbit from intersecting itself. This wouldhappen if the system (2.11) had a solution (x, y) = (φ(t), ψ(t)) such that forsome pair of numbers t1 6= t2,

(φ(t1), ψ(t1)) = (φ(t2), ψ(t2)). (2.18)

For example, if an orbit is simply an stationary point, the functions φ andψ would be constant; it therefore intersects itself infinitely often.

Theorem 2.3 Let (x, y) = (φ(t), ψ(t)) be a solution of the system (2.11) suchthat equation (2.18) holds for distinct numbers t1 < t2. Then φ and ψ are T-periodic, where T = t2 − t1, and the orbit described by (φ(t), ψ(t)) is a closedcurve.

PROOF. By lemma 2.4.1, (x, y) = (φ(t + T), ψ(t + T)) is also a solu-tion of the system (2.11), and since it satisfies the same initial conditions as(φ(t), ψ(t)) at t = t1, we can use the uniqueness theorem to conclude that(φ(t + T), ψ(t + T)) ≡ (φ(t), ψ(t)). Therefore the functions φ and ψ are T-periodic. A curve that is described by periodic parametric equations mustbe closed, so the corresponding orbit is closed.

Van der Pol’s equation.

To illustrate theorem 2.3, let’s consider the second order ODE,

x′′ + c(x2 − 1) x′ + x = 0.

which is known as van der Pol’s equation.The number c is a constant thatwe can vary to produce a variety of examples. Van der Pol’s equation wasdeveloped in the 1920’s as a model for the behavior of a triode vacuumtube. The dependent variable x represents the current, as a function oftime, in an electrical circuit that includes a triode tube. We can replace vander Pol’s equation with a system

x′ = yy′ = −x + c(1− x2) y.

}(2.19)

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188 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

The case c = 0. If c = 0, the system (2.19) is identical to the system thatwe examined in example 2.4.3, where both parameters p and q have thevalue 1. We found there that all orbits, except the stationary point at theorigin, were closed ellipses (in our case, since p = q, the orbits are actuallycircles). Therefore, theorem 2.3 tells us that all of the solutions except thestationary solution x ≡ 0, y ≡ 0 are periodic. In fact, you can easily verifythat the orbit passing through the point (a, 0) on the x-axis is traced by thesolution

x = sin(t)y = cos(t)

and is thus 2π-periodic.

The case c > 0. When c = 0, we are dismissing the important middleterm of van der Pol’s equation. We’ll use a positive value for c and seewhat happens.

Example 2.4.5 Draw the x- and y-nullclines of the van der Pol system (2.19),with c = 1

2 .

SOLUTION. The x-nullcline is given by the equation y = 0, and is thus thex-axis. This means that all orbits cross the x-axis in the vertical direction.Further inspection of the system (2.19) shows that when y = 0 we havey′ = −x; thus the orbits crossing the positive x-axis are directeddownward, and the orbits crossing the negative part of the x-axis aredirected upward.The y-nullcline has the equation

−x + c(1− x2) y = 0.

We will substitute c = 12 and solve for y to get

y =2x

1− x2 .

The orbits will be directed to the right as they cross the y-nullcline in theupper half-plane, and in the lower half plane they are directed to the left.Figure 2.14 shows the nullclines.

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2.4. *AUTONOMOUS SYSTEMS 189

Figure 2.14 indicates that orbits of the van der Pol system travel aroundthe origin in a counterclockwise direction, but they may not be closed orbits— it is possible that they are spirals. Figure 2.15 displays segments of twoorbits. I have chosen initial points (x0, y0) = (1, 0) and (x0, y0) = (3, 0)for these orbits, and followed each of them until it returned to the positivex-axis. You can see that neither orbit is closed, since neither returns to itsstarting point.

Letx = φ(t, u)y = ψ(t, u)

denote the solution of the van der Pol system (2.19) with c = 12 and initial

point (φ(0), ψ(0)) = (u, 0). By the continuous dependence statement in theo-rem 2.1, the functions φ and ψ are continuous functions of (t, u). Let g(u)be the function, defined for all u ∈ [1, 3], as follows: First let s(u) be theleast positive number such that ψ(s(u), u) = 0 and φ(s(u), u) > 0. Thus,the orbit starting at (u, 0) on the positive x-axis will return to the positivex-axis for the first time at t = s(u). Now let

g(u) = φ(s(u), u)

be the x-coordinate of the solution with initial point at (u, 0), when it firstreturns to the positive x-axis.

It follows from the continuity of φ and ψ (and the implicit function the-orem) that g is continuous. Now refer to figure 2.15. You will see that1 < g(1) < g(3) < 3. Thus the function h(u) = g(u) − u is continuous,with h(1) > 0, and h(3) < 0. By the intermediate value theorem, there isa number u∗ ∈ (1, 3) with h(u∗) = 0; that is, g(u∗) = u∗. The solution(x, y) = (φ(t, u∗), ψ(t, u∗)) thus satisfies

(φ(0, u∗), ψ(0, u∗)) = (u∗, 0) = (φ(s(u∗), u∗), ψ(s(u∗), u∗)).

Now it follows from theorem 2.3 that φ(t, u∗) and ψ(0, u∗)) are periodicfunctions of t, and the orbit that they describe is a closed curve. This orbitis shown in figure 2.16.

Exercises

In problems 1 – 7, sketch the nullclines and find the stationary points of thegiven system of ODEs. Then use an IVP solver to draw a few orbits.

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190 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

1.{

x′ = x2 + y2

y′ = x2 − y2

Answer

2.{

x′ = x(x + y)y′ = y(2x− y)

3.{

x′ = −xy′ = 4y− y2,

Answer

4.{

x′ = x(x− y2)y′ = y(x− 4)

5.{

x′ = x2

y′ = yAnswer

6.{

x′ = y + 1y′ = y

7.{

x′ = x + y− 2y′ = x− y

Answer

Find integrals for the systems of ODEs in problems 8 – 13.

8.{

x′ = xy′ = y + x2

9.{

x′ = 2xy′ = x + y

Answer

10.{

x′ = x(1− y)y′ = y(x− 1)

11.{

x′ = 3xy′ = 5y

Answer

12.{

x′ = 3yy′ = 5x

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2.4. *AUTONOMOUS SYSTEMS 191

13.{

x′ = x2 + y2

y′ = −2xyAnswer

14. Suppose that each orbit of an autonomous system of ODEs is anellipse of the form x2 + 4y2 = C, where C ≥ 0. Show that the origin isa stationary point.

15. Show that the graph of any solution of the non-autonomous firstorder ODE,

dydx

= f (x, y),

passing through the initial point (x0, y0), is the same as the orbit ofthe autonomous system

dxdt

= 1

dydt

= f (x, y)

passing through the same point.Answer

16. Problem 15 shows that a non-autonomous first order ODE canbe treated as a system of two autonomous ODEs. Find a way totreat a non-autonomous system of two ODEs as a system of threeautonomous ODEs.

17. Prove the existence and uniqueness theorem for second orderODEs: Suppose that f (t, x, y) is continuous in a box B in R3 and satis-fies a Lipschitz condition there with respect to the variables x, y (specifythis Lipschitz condition). Then, given (t0, x0, y0) ∈ B, there is a uniquesolution of the IVP

d2xdt2 = f

(t, x, dx

dt

)x(t0) = x0

dxdt

∣∣∣t=t0

= y0.

Base your proof on theorem 2.1.Answer

18. Write out the Lipschitz condition without using vector notation.

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192 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

19. Consider the system

ax′ + by′ = f (t, x, y)cx′ + dy′ = g(t, x, y)

with initial conditions x(t0) = x0, y(t0) = y0. Show that if the func-tions f (t, x, y), g(t, x, y) are continuous and satisfy a Lipschitz con-dition with respect to (x, y) in a box centered at (t0, x0, y0) and ifad− bc 6= 0 then this IVP has a unique solution.Answer

20. Here is a nice application of Green’s theorem from advanced cal-culus. (People who haven’t had advanced calculus are excused). Let

~v(x, y) =

[f (x, y)g(x, y)

]be a vector field defined on a rectangular do-

main R, and suppose that for all (x, y) ∈ R, div~v(x, y) > 0. Showthat the system

x′ = f (x, y)y′ = g(x, y)

has no closed orbits inR. This fact, first proved in 1900 by the Swedishmathematician Ivar O. Bendixson, is called Bendixson’s criterion.

2.5 Populations of Interacting Species

Suppose that two species, A and B, occupy the same environment. It mayhappen that A is a predator species and B is its prey, or that A and B com-pete for the same food supply. In either case, the population of one of thespecies cannot be successfully modeled without considering the influenceof the other. We will consider simple models involving systems of twoODEs.

Denote the populations of the two species by x and y. The derivativesx′(t) and y′(t) represent the absolute growth rates of these populations,and the relative growth rates are x′(t)/x(t) and y′(t)/y(t), respectively. Ourmodel posits that each relative growth rate is a function of x and y. It canthus be expressed as a system of two ODEs,

x′ = x f (x, y)y′ = y g(x, y).

}(2.20)

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2.5. *POPULATIONS OF INTERACTING SPECIES 193

The logistic equation

x′ = kx(1− x/C)

is a one-species model similar to the system (2.20). We used it in section 1.10to study the dynamics of the population x of a single species in an environ-ment with a carrying capacity of C individuals. In this model, the relativegrowth rate of x is a linear function of x, and in the two-species models thatwe will consider, the relative growth rate for each species will be a linearfunction of x and y.

The Lotka-Volterra Equations

The Lotka-Volterra equations,

x′ = x(a− by)y′ = cy(x− d),

}(2.21)

was the first differential equations model for the populations of more thanone species. It was developed in 1926 by the Italian mathematician VitoVolterra as a model to explain the fluctuating populations of predator andprey fish in the Adriatic Sea. The same system was proposed in 1920 byan American biophysicist, Alfred Lotka, to determine the rate of a hypo-thetical chemical reaction (See problem 2 at the end of this section). Lotka’sintention was to show that the concentrations of chemicals involved in areaction could vary periodically2. In Volterra’s model, x is the prey popu-lation, and y represents the population of a predator whose sole source offood is this prey. The relative growth rate of the prey species is set equalto a− by, indicating that in the absence of the predator, the number of preywill increase exponentially, but the relative rate of increase decreases lin-early with the predator population. Of course, even in the absence of thepredator, the prey population would be limited by the availability of re-sources, but the predator prevents the prey population from approachingthe carrying capacity. The predator keeps the prey from starvation. Thesecond equation of the system (2.21) indicates that the predator will starvewithout the prey: when x = 0 it reduces to y′ = −cdy; thus y = y0e−cdt

(where y0 is the initial population). The parameter d represents the mini-mum prey population necessary to support predators.

2Oscillating chemical reactions are no longer hypothetical. The first observations of anoscillating reaction were made in 1951 by B. P. Belousov, a chemist in the former SovietUnion.

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The system (2.21) has two stationary points. These are the solutions of

x(a− by) = 0cy(x− d) = 0,

(x, y) = (0, 0) and (x, y) = (d, b/a). The stationary point at the origin isnot unexpected: if both species are extinct, they will stay extinct. The otherstationary point represents a situation in which the appetite of the predatoris exactly matched by the reproductive rate of the prey. Our experiencewith the logistic equation suggests that the stationary point at the originis probably unstable, and the stationary point at (x, y) = (d, b/a) is stable.This is true, but we will see that it is an oversimplification.

We can apply the procedure introduced in section 2.4 to find an integralfor the system (2.21), by solving the ODE

cy(x− d) dx− x(a− by) dy = 0.

This ODE is separable, and upon dividing through by−xy we can integrateto obtain ∫

c(

dx− 1)

dx +∫ ( a

y− b)

dy = constant

and it follows that

F(x, y) = cd ln(x) + a ln(y)− cx− by

is an integral.To keep the notation simple, we will use the notation Fx, Fy, and so on,

for the partial derivatives ∂F∂x , ∂F

∂y , etc. The critical points of F(x, y) are foundby setting Fx and Fy equal to 0 and solving for x and y:

Fx(x, y) = c(d/x− 1) = 0Fy(x, y) = a/y− b = 0.

The stationary point (d, a/b) is the only critical point, and we will apply thesecond derivative test to it. Since Fxx(d, a/b) = −c/d, Fyy(d, a/b) = −b2/a,and Fxy ≡ 0,

Fxx(d, a/b)Fyy(d, a/b)− [Fxy(d, a/b)]2 =(− c

d

)(−b2

a

)> 0,

and we conclude that F(x, y) has an absolute maximum at (x, y) = (d, a/b).

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2.5. *POPULATIONS OF INTERACTING SPECIES 195

It follows that the level curves of F(x, y) in the first quadrant — whichare the orbits of the Lotka-Volterra equations—are closed curves (see fig-ures 2.17 and 2.18) surrounding the stationary point at (d, a/b). Orbits donot converge to this stationary point as t→ ∞, but they don’t diverge awayfrom it either. Since the orbits are closed curves, we conclude that the so-lutions of the Lotka-Volterra equations are periodic functions of time, andhence that the populations of predator and of prey will oscillate.

The Lotka-Volterra system is only a starting point in the study of popu-lations of predators and their prey.

Competing Species

We will now consider an ecosystem in which two species A and B, withpopulations x and y, compete for resources. The organisms may be plants,competing for nutrients and sunlight, grazing animals competing for for-age, or predators in competition for prey. The model is thus applicable to asingle level of the food chain.

If the species B were removed from the environment, the population ofthe species A would grow according to the logistic equation, which we willwrite in the form

dxdt

= ax(C− x),

where C is its carrying capacity, and a = k/C is the ratio of the initial rela-tive growth rate to the carrying capacity. The species B will consume someof the resources required by A. This is taken into account by subtractingfrom the carrying capacity an amount to reflect the rate at which the re-sources needed by species A are consumed by species B. Thus the relativegrowth rate for the species A is

a(C− ry− x),

where r is the rate at which species B consumes the resources.The relative growth rate of B is, by the same reasoning, proportional to

L− sx− y, where L denotes the carrying capacity for B and s is the rate atwhich A consumes the resources that B needs. Putting these together, wehave the system

1x

dxdt

= a(C− x− ry)

1y

dydt

= d(L− sx− y),

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196 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

We have thus derived the competing species equations,

x′ = x(k− ax− by)y′ = y(l − cx− dy)

}(2.22)

where k = aC, b = ar, l = dL, and c = ds.The system (2.22) has four stationary points. Three of these occur when

one or both of the species are extinct. For example, if A is extinct (x = 0)then the species B can have the stationary populations y = 0 (also extinct)or L (its carrying capacity). Furthermore, in this case the orbits would justbe those of the phase diagram of the logistic equation.

The fourth stationary point, (x1, y1) is located at the intersection of thelines

ax + by = k, and (2.23)cx + dy = l. (2.24)

The line (2.23) is an x-nullcline, and the line (2.24) is a y-nullcline. If thelines do not intersect in the first quadrant, this stationary point is not ofinterest, because neither population can be negative.

Figure 2.19 shows four possible configurations of the nullclines, de-pending on the relative size of the parameters. The intercepts of the line (2.23)are at (C, 0) and p = (0, k/b) , and since (C, 0) is a stationary point, it ismarked with a heavy dot. The intercepts of the line (2.24) are q = (l/c, 0)and the stationary point (0, L). In configurations (a) and (d) the stationarypoint E = (x1, y1) is also present. The arrows on the nullclines indicate thedirection orbits must take when they cross. Thus, the x-nullcline has ver-tical arrows. The part of the x-nullcline that lies above the y-nullcline is ina region where y′ < 0, so the arrows point down. When the x-nullcline isbelow the y-nullcline, the arrows point up, for y′ > 0. The directions of thearrows marking the y-nullclines are explained in the same way.

It is not feasible to find an integral for the system (2.22), and to un-derstand it we will need to use a different approach. A closed polygonT in the phase plane is called a trap if every orbit that crosses an edge orvertex of T is directed toward the interior. Thus, orbits can enter T , butonce they do so, they are trapped, since they cannot cross any edge fromthe inside outward. Identifying traps is an important technique for study-ing autonomous systems of ODEs. For the system (2.22) we will use trapswhose edges are nullclines.

Configurations (b) and (d) are the easiest to analyze, and we will startwith (b). The quadrilateral LqCP is a trap, because all orbits are directed

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2.5. *POPULATIONS OF INTERACTING SPECIES 197

downward as they cross the x-nullcline (the line pC), to the right as theycross the y-nullcline (the line Lq), and no orbits cross the axes. Furthermore,all orbits starting outside this quadrilateral must either enter it, or convergeto one of the stationary points C or L.

Inside the trap, x′ is positive (it doesn’t become negative until the otherside of the x-nullcline is reached), and, since the quadrilateral lies abovethe y-nullcline, Lq, y′ is negative. With orbits trapped inside, and directeddownward and to the right, it can be shown that they all converge to thestationary point (C, 0). The conclusion: in configuration (b), species A isdominant, and B will become extinct. The same reasoning shows that ex-actly the reverse is true in configuration (c): species B is dominant, and Adies out. The phase portrait corresponding to configuration (b) is shown infigure 2.20.

In configuration (a) of Figure 2.19 the quadrilateral OCEL (O stands forthe origin) is a trap. Within this quadrilateral, the triangles pEL and qCEare also traps. In triangle pEL, orbits are directed upward and to the left;and in qCE they are directed downward and to the right. Thus orbits start-ing near E will converge to L or C if they enter the interior of one of thesetriangles. In particular, consider the situation when the initial populationis inside the quadrilateral OqEp. Its orbit must either converge to E, or en-ter one of the above-mentioned triangles, and hence converge to C or L. Itcan be shown that there is one orbit in OqEp that does converge to E; it iscalled the separatrix. Orbits that start above the separatrix will cross theedge Ep, entering the triangle where all orbits converge to L; and orbitsstarting below the separatrix will enter the triangle qCE and converge toE. This situation is called competitive exclusion: the species with the ini-tial population advantage dominates; the other faces extinction. The phaseportrait in this case is shown in figure 2.21.

Now let us consider a happier situation: configuration (d). The quadri-lateral OqEp is a trap, and within it, the triangles LEp and CqE are alsotraps. The distinction between this and configuration (a) is that in this case,the orbits inside the triangles are directed toward E instead of away fromit. Since ever orbit either must converge directly to E or enter one of thesetriangles, we conclude that all orbits converge to E. Here we have a stableequilibrium in which the species share the resources. The phase portrait isdisplayed as figure 2.22.

It is interesting to speculate on which of the above cases is applicable infamiliar situations, such as crabgrass in the lawn, benign versus pathogenicbacteria in the body, introduction of new species of fish in a pond, and soon. Conclusions reached by using this model are not to be trusted without

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198 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

further study, because it is based on many simplifying assumptions.

Exercises

1. A population of insect pests is to be controlled by introducingpredator insects. The objective is to keep the maximum number ofpests present at any time as small as possible. Assuming that theLotka-Volterra model is valid in this situation,

(a) If the growth parameters a, b, c, d of the Lotka Volterra equa-tions on page 193 are known, what is the optimum number ofpredators to release?

(b) Evaluate this strategy: Apply a pesticide to reduce the numberof pests as much as possible; when the pesticide has degraded,introduce predators to keep the pests under control.

(c) If the number of pests in the environment is initially small, is itpreferable to introduce the predators immediately, or to wait fora while?

Answer

2. Given a chemical reaction Q + R → S, the Law of Mass Actionstates that the rate at which the product S is produced is proportionalto the product of the concentrations of Q and R. Thus, if x, y, and zdenote the concentrations of Q, R, and S, respectively, then

dzdt

= kxy,

where k is constant.

Lotka3 considered a hypothetical sequence of three reactions

P + Q→ 2Q, Q + R→ 2R, R→ S.

Suppose that the concentration of P is held fixed, by replenishing thatreactant at the same rate that it is consumed. Let x and y denote theconcentrations of Q and R, and derive the Lotka-Volterra equationsas a model of this chemical system.

3A. J. Lotka, “Undamped oscillations derived from the law of mass action,” J. Amer.Chem. Soc. 42 (1920) pp. 1595 – 1599

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2.5. *POPULATIONS OF INTERACTING SPECIES 199

3. Two species of fish inhabit a pond; neither is the prey of the other.The pond’s carrying capacity is 1000 fish of species A, or 1500 ofspecies B. Also, if there are 800 A-fish and 800 B-fish, the popula-tion will be stationary. Set up a system of ODEs to model the fishpopulations, and draw a sketch of the phase plane, identifying traps.How will the populations stabilize, assuming both populations areinitially positive?Answer

4. Repeat problem 3 under the assumption that the carrying capacityof the pond for each individual species is unchanged, but the combi-nation of A and B-fish in the pond for a stationary population is

(a) 800 A-fish and 250 B-fish.

(b) 600 fish of each species.

5. Symbiosis. If the relationship between two species is mutuallybeneficial, the species are said to be in symbiosis. Lichens are familiarexamples. While each lichen appears to be a single organism, in factit consists of two species, an alga and a fungus, in symbiosis. Draw athe nullclines for the two species model,

x′ = ax(K− x + By)y′ = dy(L + Cx− y).

and identify any traps. You will need to consider the cases whereBC ≥ 1 and BC < 1 separately.Answer

6. This problem is for readers who have had an advanced calculuscourse. Let (x, y) = (φ(t), ψ(t)) be a solution of the Lotka-Volterraequations, and let T denote its period. The average populations are

x =1T

∫ T

0φ(t) dt and y =

1T

∫ T

0ψ(t) dt.

Prove that x = d and y = a/b. Hint: One proof is based on Green’sTheorem.

7. This problem does not require advanced calculus, but readerswho have not had this course will have to take the result of problem 6on faith.

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200 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

Volterra’s predator-prey model4 was based on data from fish catchesin the Adriatic Sea. Volterra observed that the ratio of predators toprey varied periodically, and this is reflected by his mathematicalmodel. Let R denote the catch rate: that is, the fraction of the pop-ulation (of both species) that is caught. Taking fishing into account,the Lotka-Volterra equations become

x′ = x(a− by)− Rxy′ = cy(x− d)− Ry.

Assuming that a > R, show that as the catch rate increases, the av-erage population of prey increases, and the average predator popu-lation decreases. You may use the result of problem 6. Although theLotka-Volterra equations oversimplify, this effect is substantiated byVolterra’s data. For the same reason, it is unwise to apply a pesticideto “help” an insect predator control a pest species: the result will beto increase the average number of pests.Answer

2.6 Chapter Glossary

R3 Three-dimensional space.

Autonomous system A system of ODEs in which the independent vari-able does not appear on the right side.

Bendixson’s criterion If a vector field, defined of a rectangleR in the plane,has nonvanishing divergence then there are no closed integral curvesinR.

Classical IVP solver An IVP solver that has a fixed time step.

Competing species equations A system of ODEs that models the dynam-ics of the populations of two species that compete for the same re-sources. See page 196.

Competitive exclusion A situation in which a species with an initial pop-ulation advantage is able to overwhelm its competitor.

4V. Volterra, “Variazionie fluttuazioni del numero d’individui in specie animali con-viventi,” Mem. Acad. Lincei 2 (1926) pp. 31 – 113. English translation: “Variations andfluctuations in the population of animals living together,” in Animal Ecology, by R. N. Chap-man (New York: McGraw-Hill, 1931), pp. 409 – 448.

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2.6. CHAPTER GLOSSARY 201

Dynamic IVP solver An IVP solver that automatically adjusts its time stepto maintain precision.

IVP solver A numerical method for computing solutions of initial valueproblems.

Integral (of a system of ODEs) A function F(x, y) that is constant on allorbits of the system.

Integral curve (of a vector field v(x, y) = f (x, y)i+ g(x, y)j) A curve givenparametrically by x = p(t), y = q(t), where for all t, the tangentvector p′(t)i + q′(t)j is equal to the vector at (p(t), q(t)), that is,

v(p(t), q(t)) = p′(t)i + q′(t)j.

Lipschitz condition See page 184.

Lotka-Volterra equations A system of ODEs that is a model for the popu-lation dynamics of a predator and its prey. See page 193.

Nullcline (for a system x′ = f (x, y), y′ = g(x, y)) The set of points wheref (x, y) = 0 (the x-nullcline) of g(x, y) = 0 (the y-nullcline).

Order (of an IVP solver) A number n such that the accumulated errorvaries proportionately with the nth power of the time step.

Orbit The graph in the phase plane of parametric equations given by asolution of a system of two ODEs. It is customary to mark orbits toindicate the direction of increasing t.

Phase plane The plane, in which the coordinates correspond to the valuesof the dependent variables of a system of two ODEs.

Phase portrait A drawing of the phase plane, showing a representativecollection of orbits of a given system of two ODEs.

Replace an ODE The system

dydt = vdvdt = f (t, y, v)

}(2.25)

is said to replace the second order ODE

d2ydt2 = f

(t, y,

dydt

), (2.26)

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202 CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS

since (y, v) = (φ(t), ψ(t)) is a solution of (2.25) if and only if y = φ(t)satisfies (2.26) and ψ = φ′.

Separatrix An orbit of an autonomous system of two ODEs that separatesorbits with different limiting behavior.

Stationary point (of a vector field v = f (x, y)i + g(x, y)j) A point (x1, y1)such that v(x1, y1) = 0. An integral curve containing a stationarypoint consists of the stationary point alone.

(of a system of ODEs, x′ = f (x, y), y′ = g(x, y)) A point (x1, y1) suchthat f (x1, y1) = g(x1, y1) = 0

System of ODEs Two or more ODEs to be treated simultaneously. To bewell-posed, the system must have one equation for each dependentvariable, and only one independent variable.

Trap A closed polygon in the phase plane such that all orbits that cross itsedges or vertices are pointed inward.

Uncoupled system A system of two ODEs in which one of the dependentvariables does not appear in one of the equations.

Vector field A function, defined on a subset of the xy plane, that assigns toeach point (x, y) in its domain a vector v(x, y) = f (x, y)i + g(x, y)j inthe plane.

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2.7. REVIEW EXERCISES 203

2.7 Review Exercises

1. Solve the initial value problem

x′ = x + et; x(0) = 0y′ = xy2; y(0) = 1

Answer

2. Find the general solution of

x′ = −xy′ = y.

Also find an integral for this system and draw its phase portrait.Answer

3. Repeat Exercise 2 for the system

x′ = xy′ = 2y

Answer

4. Suppose that F′(y) = f (y). Find an integral for the system thatreplaces the ODE y′′ + f (y) = 0. Apply the result to the followingODEs, in which α denotes a constant parameter:

(a) y′′ = −α2 sin(y) (the pendulum equation).

(b) y′′ = −α2y (the linearized pendulum).

(c) y′′ = −α2y−2 (a falling body).

Answer

5. Find systems of first order ODEs to replace the given second orderODEs.

(a) y′′ + y = 0.

(b) y′′ = y′ − t2 sin(y)

Answer

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204 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

6. Given that y = et cos(t) is a solution of the ODE

y′′ = 2y′ − 2y

sketch the corresponding orbit of the system that replaces this ODE.Answer

7. Find an integral for the system

x′ = −x(x + 4y)y′ = 2y(x + y)

Answer

8. Find the stationary points of the following vector fields:

(a) ~v = (x + y− 2) i + (x− 3y + 2) j.

(b) ~v = (y− x) i + (y− x3) j.

(c) ~v = x(x + y + 4) i + y(x + 5y) j.

Answer

9. Write out the systems of ODEs corresponding to the vector fieldsin Exercise 3 and sketch their nullclines.Answer

10. Draw the nullclines of the system

x′ = (x + y)(y2 − 1)y′ = (y− x)(x2 − 1)

}and identify a trap.Answer

11. Use an IVP solver to draw several integral curves for each of thevector fields in Exercise 3. Choose initial points near (but not at!) thestationary points, and let the time variable range from −5 to 5.Answer

12. Use an IVP solver to plot the graph of the solution of the dampedpendulum equation, y′′ = −.05y′− sin(y) with initial conditions y(0) =0.25, y′(0) = 0. Repeat with the “linearized version” of the equation,y′′ = −.05y′ − y with the same initial conditions.Answer

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FIGURES FOR CHAPTER 3 205

Figure 2.1: Orbits of the system x′ = x, y′ = y.

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206 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.2: Three orbits of the system that replaces the ODE y′′ = y′ + 2y. The horizontalaxis is the y-axis; the vertical axis represents y′.

-7.5 -5 -2.5 2.5 5 7.5 10

-5

5

10

15

20

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FIGURES FOR CHAPTER 3 207

Figure 2.3: The vector field xi− yj

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

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208 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.4: Component graphs of a solution (x(t), y(t)) of the system (2.9).

-1 -0.5 0.5 1t

-2

-1

1

2

y

x

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FIGURES FOR CHAPTER 3 209

Figure 2.5: Orbit of a solution (x(t), y(t)) of the system (2.9). The vector field represen-tation of the system of ODEs is also displayed.

-3 -2 -1 1 2

-1.5

-1

-0.5

0.5

1

1.5

2

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210 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.6: Graph of a solution (x(t), y(t)) of the system (2.9).

-2

-1

0

1

-1

0

1

2

-2

-1

0

1

-2

-1

0

1

-1

0

1

2

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FIGURES FOR CHAPTER 3 211

Figure 2.7: Spreadsheet for approximating the solution of x′ = 2x + 2y, y′ = −x + 4y,with initial conditions (x(0), y(0)) = (1, 0), by Euler’s method with stepsize h = 0.1.

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212 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.8: Spreadsheet calculation and orbit plot for the IVP, x′ = 2x+ 2y, y′ = −x+ 4y;(x(0), y(0) = (1, 0),. The formulas in the spreadsheet are shown in figure 2.7.

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FIGURES FOR CHAPTER 3 213

Figure 2.9: Orbit of x′ = 2x + 2y, y′ = −x + 4y, with initial condition x(0) = 1, y(0) = 0.The upper curve was drawn from the analytic solution, and the lower one was drawn fromthe approximation obtained by Euler’s method in example 2.3.1.

-6 -4 -2 2

-15

-12.5

-10

-7.5

-5

-2.5

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214 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.10: : Vector field and phase portrait for the system of ODEs in example 2.12.

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FIGURES FOR CHAPTER 3 215

Figure 2.11: : Stationary points and nullclines for the system of ODEs in example 2.4.2.The x-nullcline is marked with vertical arrows indicating the direction of orbital crossing;the y-nullcline is marked with horizontal arrows. Stationary points are located where thetwo nullclines intersect.

-10 10 20 30 40

-5

-2.5

2.5

5

7.5

10

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216 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.12: Orbits of x′ = −p2y, y′ = q2x.

-6 -4 -2 2 4 6

-3

-2

-1

1

2

3

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FIGURES FOR CHAPTER 3 217

Figure 2.13: Phase portrait of y′′ = −y2: see example 2.4.4.

-3 -2 -1 1 2 3y

-2

-1.5

-1

-0.5

0.5

1

1.5

2v

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218 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.14: Nullclines of the van der Pol system (2.19), where c = 1/2.

-4 -2 2 4

-4

-2

2

4

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FIGURES FOR CHAPTER 3 219

Figure 2.15: Segments of two orbits of the van der Pol system (2.19), with c = 1/2.

-2 -1 1 2 3

-2

-1

1

2

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220 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.16: A closed orbit of the van der Pol system (2.19), with c = 1/2.

-2 -1 1 2

-2

-1

1

2

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FIGURES FOR CHAPTER 3 221

Figure 2.17: An integral for the Lotka-Volterra equations. The orbits are level curves.The parameters are a = 0.1, b = 0.01, c = 0.0005, and d = 2000. Thus, the stationary pointis located at (x, y) = (2000, 10).

10002000

3000

4000

5000

10

20

30

40

5.5

6

6.5

10002000

3000

4000

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222 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.18: Integral curves of the Lotka-Volterra equations.

500 1000 1500 2000 2500 3000

10

20

30

40

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FIGURES FOR CHAPTER 3 223

Figure 2.19: Four possible configurations for the nullclines of the system (2.22) The sta-tionary points are marked with heavy dots; the x-nullcline is marked with vertical arrows,and the y-nullcline is marked with horizontal arrows.

6 6

- -

6 6

- -

s s

s s

s

sss

ss

ssss

(a) (b)

(c) (d)

L

L

L

Lp

p

p

p

q q

q qC C

C C

E

E

CCCCCCCC

CCCCCCCC

PPPPPPPPP

PPPPPPPPP

@@@@@@@@@

@@@@@@@@@

JJJJ

JJJJ

-

6

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6

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6

-

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224 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.20: In this phase portrait, the parameters correspond to the configuration shownin figure 2.19(b). The species whose population is represented by x dominates the speciesrepresented by y regardless of the initial conditions. The nullclines are represented bydashed lines, and the parameters are (k, a, b, l, c, d) = (30, 5, 6, 6, 2, 3).

2 4 6 8

2

4

6

8

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FIGURES FOR CHAPTER 3 225

Figure 2.21: Competitive exclusion: the phase portrait of the system 2.22, when thenullclines (shown as dashed lines) are configured as in figure 2.19(a). The parameter valuesare: k = 3, a = 1, b = 2, l = 5, c = 2, and d = 3.

0.5 1 1.5 2 2.5 3

0.5

1

1.5

2

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226 CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS

Figure 2.22: When the population growth parameters correspond to the configurationshown in figure 2.19(d), the two species will coexist with populations represented by thestationary point in the first quadrant. The initial conditions have no effect on the outcome,as long as both species are present. Nullclines are represented by dashed lines, and theparameters are (k, a, b, c, l, c, d) = (15, 5, 3, 12, 2, 6).

1 2 3 4 5 6

1

2

3

4

5

6

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Chapter 3

Linear Systems

227

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228 CHAPTER 3. LINEAR SYSTEMS

3.1 The Initial Value Problem

A system of linear, first order ODEs,

x′ = p1(t) x + q1(t) y + r1(t)y′ = p2(t) x + q2(t) y + r2(t),

}(3.1)

expresses the rate of change of each dependent variable as a function of theindependent variable and both dependent variables. Thus, although eachequation in the system (3.1) is first order, we cannot directly solve for xwithout knowing y, or for y without having first determined x. The prob-lem is that the two equations are coupled. This kind of coupling appearsin many applications, including mechanical systems, parallel electrical cir-cuits, and in economic models involving supplies of various commodities.

This chapter is an introduction to the general properties of linear sys-tems. If the coefficients pi(t) and qi(t) are constants, we will learn how todecouple the two equations of the system (3.1) and thus to solve it. Systemswith variable coefficients rarely decouple.

Systems in matrix form.

We will use matrix notation for systems. A matrix is a rectangular arrayof numbers or functions. If there are m rows and n columns, the matrixis called an m× n matrix. For example, a 2× 2 matrix would be a squarearray [

a11 a12a21 a22

].

This matrix uses the standard notation: the number or function that is lo-cated in the horizontal row i and vertical column j is aij. These aij’s arecalled the entries of the matrix.

A matrix with just one column is called a column vector. We’ll usearrow notation~v,~a, etc. for column vectors, and capital letters A, B, etc. willdenote more general matrices. Scalars (that is, constants and real-valuedfunctions) will be denoted with lower case italic letters x, a, t, etc.

It is possible to add matrices and to multiply them, provided that thedimensions of the matrices are compatible. Any two matrices of the samedimensions can be added; each entry of C = A + B is the sum of the corre-sponding entries of A and B. Thus cij = aij + bij.

The matrix product P = AB of an m× n matrix A and a p× q matrix Bcan be formed if and only if n = p; that is, the number of columns of A is

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3.1. THE INITIAL VALUE PROBLEM 229

equal to the number of rows of B. When this compatibility condition holds,P will be a matrix of dimension m × q, with as many rows as A has, andas many columns as B. For example, if A is a 2 × 2 matrix and B is a 2-dimensional column vector (a 2× 1 matrix), AB would be a 2-dimensionalcolumn vector. The formula for the product is[

a11 a12a21 a22

] [b1b2

]=

[a11b1 + a12b2a21b1 + a22b2

].

More generally, if A is an m × k matrix and B is a k × n matrix, theentries of the product C = AB are computed as follows:

cij = ai1b1j + ai2b2j + · · ·+ aikbkj.

If you are familiar with the dot-product of vectors, the i, jth entry of C is thedot-product of row i of A and column j of B. Matrix multiplication satisfiesthe associative law: Given three matrices A, B, and C, of dimensions suchthat the products P = AB and Q = BC are defined, it always holds thatPC = AQ; in other words,

(AB)C = A(BC).

Matrix multiplication also satisfies the distributive laws: assume that A andA′ are matrices of the same dimensions, and that B is a matrix of dimen-sions such that AB is defined. Let B′ be a matrix of the same dimensions asB. Then

(A + A′)B = AB + A′B and A(B + B′) = AB + AB′.

Matrix algebra shares these properties with the arithmetic of real numbers.However, the commutative law does not hold. If A is a p × q matrix andB is q× p, then unless p = q, AB and BA are of different sizes (p× p andq× q, respectively); thus AB 6= BA. Even it A and B are square matrices,so that p = q, usually AB 6= BA.

To convert the system (3.1) to matrix form, let the column vector

~v =

[xy

]represent the dependent variables, and the matrix

A(t) =[

p1(t) q1(t)p2(t) q2(t)

]

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230 CHAPTER 3. LINEAR SYSTEMS

be the coefficient matrix. The vector function

~r(t) =[

r1(t)r2(t)

]is the source vector. To see that the matrix ODE

~v′ = A(t)~v +~r(t)

is equivalent to the system (3.1), we have only to multiply the matrix A andthe column vector ~v to obtain the column vector

A(t)~v =

[p1(t) x + q1(t) yp2(t) x + q2(t) y

],

add the column vector~r(t),[p1(t) x + q1(t) yp2(t) x + q2(t) y

]+

[r1(t)r2(t)

]=

[p1(t) x + q1(t) y + r1(t)p2(t) x + q2(t) y + r2(t)

]and write out the matrix equation as[

x′

y′

]=

[p1(t) x + q1(t) y + r1(t)p2(t) x + q2(t) y + r2(t)

]Example 3.1.1 Put the system

x′ = yy′ = 4 x

in matrix form.

SOLUTION.We need to find a coefficient matrix

A =

[a11 a12a21 a22

]such that

A[

xy

]=

[a11 x + a12 ya21 x + a22 y

]=

[y

4 x

].

It follows that a12 = 1, a21 = 4, and the other entries are 0. Hence

A =

[0 14 0

],

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3.1. THE INITIAL VALUE PROBLEM 231

and the matrix form of the system is ~v′ = A~v.

It is not difficult to check that x = e2t, y = 2 e2t is a solution of thesystem that we considered in example 3.1.1. We can put this solution invector notation as

~v =

[e2t

2 e2t

].

It is possible to multiply any matrix by a scalar constant or function:just multiply each entry by the scalar. Thus we can write the above vectorsolution as

~v = e2t[

12

]Example 3.1.2 Show that

~w = e−2t[

1−2

]is also a solution of the system in example 3.1.1.

SOLUTION. Let

~c =[

1−2

],

so that ~w = e−2t~c. We need to show that

~w′ = A · (e−2t~c) (3.2)

By matrix multiplication,

A~c =[

0 14 0

] [1−2

]=

[(0)(1) + (1)(−2)(4)(1) + (0)(−2)

]=

[−2

4

]= −2~c.

Thus~w′ =

ddt(e−2t~c) = e−2t(−2)~c = e−2t A~c.

We can now conclude that equation (3.2) holds, sincee−2t A~c = A · (e−2t~c).

With vector notation, the existence and uniqueness theorem 2.1 for sys-tems looks the same as the existence and uniqueness theorems for first or-der ODEs that we encountered in chapter 1. The following theorem is for

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232 CHAPTER 3. LINEAR SYSTEMS

linear systems only; it differs from theorem 2.1 in that the domain of thesolution of an IVP involving a linear system can be specified. We will referto this theorem frequently.

Theorem 3.1 (Existence & Uniqueness) Suppose that the entries of the coeffi-cient matrix A(t) and the components of the source vector ~f (t) are continuous onan interval (a, b) containing the initial point t0. Then the IVP

~v′ = A(t)~v + ~f (t); ~v(t0) = ~v0

has a solution ~v(t) that is defined for all t ∈ (a, b). If ~u(t) is another solution ofthe same IVP, then ~u(t) = ~v(t) for all t ∈ (a, b).

Homogeneous systems

A system of ODEs~v′ = A(t)~v

with a zero source vector is said to be homogeneous . The following corol-lary of theorem 3.1 is a useful property of homogeneous systems.

Corollary 3.1.1 Let A(t) be an n× n matrix whose entries are continuous on theinterval (a, b), and let ~v(t) be a solution of the homogeneous system ~v′ = A(t)~v.If, for some t0 ∈ (a, b), ~v(t0) =~0, then ~v(t) ≡~0 on (a, b).

PROOF. Notice that ~v(t) and~z(t) ≡~0 are solutions of the IVP

~v′ = A(t)~v; ~v(t0) =~0.

The uniqueness assertion of theorem 3.1 implies that ~v(t) = ~z(t) for allt ∈ (a, b)

Definition: If ~v1(t) and ~v2(t) are vector functions that are defined on forall t in an interval (a, b) If c1 and c2 are scalar constants, the vector function

~v3(t) = c1~v1(t) + c2~v2(t) (3.3)

is called a linear combination of ~v1 and ~v2. (It is possible to form linearcombinations of three or more vector functions in the same way.)

Now suppose that ~v1(t) and ~v2(t) in (3.3) are solutions of a homoge-neous linear system of ODEs, ~v′ = A~v. Then

~v′3 = c1~v′1 + c2~v′2= c1 A~v1 + c2 A~v2

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3.1. THE INITIAL VALUE PROBLEM 233

Therefore, by the distributive law for matrix multiplication,

~v′3 = A (c1~v1 + c2~v2)

= A(t)~v3.

The result of this calculation can be summarized as the following theorem.

Theorem 3.2 Let ~v1(t), ~v2(t), . . . , ~vn(t) be solutions of a linear homogeneoussystem, ~v′ = A(t)~v. Then every linear combination

c1~v1 + c2~v2 + · · ·+ cn~vn

of these solutions is also a solution.

If we have just one solution ~v1(t), then theorem 3.2 tells us that the familyof functions c1~v1(t) is an infinite family of solutions (unless v1(t) ≡ 0, forthen the family would not be infinite). If there are two solutions available,c1~v1(t) + c2~v2(t) is a larger family of solutions, unless ~v2(t) happens to bea constant multiple of ~v1(t).

Definition: The linear span of a set of solutions S = {~v1(t),~v2(t), . . . ,~vn(t)}of a system of ODEs is the set of all vector functions that can be expressedas linear combinations of the functions in S . We can also consider the linearspan of a set T of vectors {~b1, . . . ,~bk} ⊂ Rn: it is the set of all possible linearcombinations of the vectors in T .

Theorem 3.2 can be rephrased as follows: Every vector function in thelinear span of any set of solutions of a homogeneous linear system of ODEsis also a solution.

In example 3.1.2 we found that

~v1 = e2t[

12

]and ~v2 = e−2t

[1−2

]are solutions of the system

~v′ =[

0 14 0

]~v.

The linear span of these solutions is the family

~v = c1 e2t[

12

]+ c2 e−2t

[1−2

].

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234 CHAPTER 3. LINEAR SYSTEMS

Definition: The vector functions~v1(t) and~v2(t) are linearly independentif neither function is equal to a constant scalar multiple of the other. Moregenerally, we will say that a set of vector functions

S = {~v1(t),~v2(t), . . . ,~vm(t)}

is linearly independent if none of the ~vi(t) belongs to the linear span of theother (m− 1) vector functions in S.

A similar definition applies to vectors in the plane (or n-dimensionalspace Rn). We say that a set of constant vectors, S = {~a1,~a2, . . . ,~am} islinearly independent if none of the vectors in S belongs to the linear spanof the other vectors in S.

Proposition 3.1.2 The set S (of vectors in Rn or vector functions) is linearlyindependent if and only if the only linear combination of the elements of S that isequal to~0 is the combination in which all of the coefficients are zero.

PROOF: (We will refer to the case of vector functions; the proof for thecase of constant vectors is the same.) Suppose that S is linearly indepen-dent, and that

c1~v1(t) + c2~v2(t) + · · ·+ cn~vn(t) ≡ 0.

We have to show that c1 = c2 = · · · cn = 0. We will do so by supposingthat this is not the case and reaching a contradiction. Thus suppose that atleast one of the coefficients ci is nonzero. We can select this ci to be the lastcoefficient that is nonzero, so that cj = 0 for j > i. Thus

c1~v1(t) + c2~v2(t) + · · ·+ ci−1~vi−1 + ci~vi(t) ≡ 0.

Since ci 6= 0 we can divide through by ci and solve for vi(t), thus expressingit as a linear combination of {~v1(t), . . . ,~vi−1(t)}:

~vi(t) =−1ci

(c1~v1(t) + c2~v2(t) + · · ·+ ci−1~vi−1).

This shows that ~vi(t) belongs to the linear span of {~v1(t),~v2(t), . . . ,~vi−1},contradicting our hypothesis that S is linearly independent.

To prove the converse, assume that the only linear combination of Sthat is zero is the combination where all coefficients are zero. Now we willassume that one of the elements of S,~vi(t) lies in the span of the other n− 1elements of S. This means there are scalar constants c1, . . . , ci−1, ci+1, . . . , cnsuch that

~vi(t) = c1~v1(t) + c2~v2(t) + · · ·+ ci−1~vi−1 + ci+1~vi+1 + · · ·+ cn~vn(t).

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3.1. THE INITIAL VALUE PROBLEM 235

Put ci = −1. Then

c1~v1(t) + c2~v2(t) + · · ·+ ci−1~vi−1 + ci~vi(t) + ci+1~vi+1 + · · ·+ cn~vn(t) = 0

is a linear combination of S in which not all of the coefficients are zero,contradicting our hypothesis.

Theorem 3.3 Let A(t) be an n×n matrix whose entries are continuous functionson an interval (a, b), and let t0 be any point in (a, b). Then solutions ~v1(t), ~v2(t). . . , ~vn(t) of the linear homogeneous system ~v′ = A(t)~v are linearly independentif and only if, as vectors in Rn,

~v1(t0),~v2(t0), . . . ,~vn(t0)

are linearly independent.

PROOF. For simplicity, let n = 2. (The proof can be used for generaln with only slight modification.) If ~v1(t0) and ~v2(t0) are not linearly inde-pendent, then we can assume that there is a number c such that ~v2(t0) =c~v1(t0).

~v3(t) = c~v1(t)−~v2(t)

is a linear combination of solutions of ~v′ = A(t)~v, and is therefore also asolution. Since ~v3(t0) = ~0, it follows from corollary 3.1.1 that ~v3(t) = ~0 forall t, and hence ~v2(t) = c~v1(t) for all t ∈ (a, b).

Thus, if ~v1 and ~v2 are linearly independent as vector functions on (a, b),the initial vectors ~v1(t0) and ~v2(t0) must also be linearly independent.

To complete the proof we have to show that if ~v1(t0) and ~v2(t0) arelinearly independent, then the vector functions ~v1 and ~v2 are linearly inde-pendent. If there were a scalars c such that

~v2(t) = c~v1(t)

or all t ∈ (a, b) then~v2(t0) = c~v2(t0).

This contradicts our assumption that~v1(t0) and~v2(t0) are linearly indepen-dent.

Theorem 3.3 implies that the solution vectors ~v1(t), ~v2(t) . . . , ~vn(t) areeither linearly independent for all t in the domain (a, b) where A(t) is con-tinuous, or they are linearly dependent for all t ∈ (a, b). They cannot belinearly independent at some points of the domain (a, b), and linearly de-pendent at others.

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236 CHAPTER 3. LINEAR SYSTEMS

Example 3.1.3 Show that the solutions

~v1 = e2t[

12

]and ~v2 = e−2t

[1−2

]of the system

~v′ =[

0 14 0

]~v.

are linearly independent.

SOLUTION. Since ~v2(0) =[

1−2

]is not a scalar multiple of

~v2(0) =[

12

], the two solutions are independent.

We are now almost ready for a full description of the general solutionof a homogeneous linear system of ODEs. The following definition is fromlinear algebra:

Definition: A basis for Rn is a set S of vectors that is linearly indepen-dent, and whose linear span is all of Rn.

There is a parallel definition in the realm of systems of ODEs:

Definition: Let A(t) be an n× n matrix whose entries are functions of t. Afundamental set of solutions of the homogeneous linear system ~v′ = A(t)~vis a set of vector functions that is linearly independent, and whose linearspan is the set of all solutions of the system.

Just one more fact from linear algebra is needed: Let S = {~a1,~a2, . . . ,~an}be a set of n vectors in Rn. Then the following are equivalent:

• S is a basis for Rn.

• S is linearly independent.

• The linear span of S is equal to Rn.

Theorem 3.4 Let A(t) be an n × n matrix of functions, all continuous on aninterval (a, b), let t0 be any point in (a, b), and let S = {~a1,~a2 . . . ,~an} be a basisof Rn.

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3.1. THE INITIAL VALUE PROBLEM 237

Then the solutions ~v1(t), ~v2(t), . . . , ~vn(t) of the system

~v′ = A(t)~v, (3.4)

satisfying the initial conditions ~vi(t0) = ~ai, respectively, form a fundamental setof solutions of (3.4).

PROOF. To keep the notation as simple as possible, we take n = 2. (Theproof works in the same way for any n.)

We’ll start by showing that the linear span of T = {~v1(t),~v2(t)} con-tains an arbitrary solution ~w(t) of (3.4). Because {~a1,~a2} is a basis for R2

and ~w(t0) ∈ R2 there are scalars c1, c2 such that ~w(t0) = c1~a1 + c2~a2.Define the vector function ~u(t) = ~w(t) − (c1~v1(t) + c2~v2(t)). As it is

a linear combination of solutions of (3.4), u(t) is also a solution. Because~vi(t0) = ~ai, it follows that ~u(t0) = ~0. Then, by corollary 3.1.1, ~u = ~0 identi-cally. Thus,

~w(t) = c1~v1(t) + c2~v2(t)

for all t ∈ (a, b). This establishes that the linear span of T comprises allsolutions of (3.4).

To complete the proof, we need to show that T is linearly independent.This follows from theorem 3.3, because S = {~a1, ~a2} is a basis for R2, thuslinearly independent.

The general solution of a system of ODEs is a family of solutions thatencompasses all solutions of the system.

Corollary 3.1.3 Let A(t) be an n× n matrix of functions that are continuous onthe interval (a, b). Let t0 ∈ (a, b), and let ~v1(t), ~v2(t) . . . , ~vn(t) be solutions of~v′ = A(t)~v such that the set of vectors

S = {~v1(t0),~v2(t0), . . . ,~vn(t0)}

is a basis of Rn. Then the general solution of ~v′ = A(t)~v is

~v = c1~v1 + c2~v2 + · · ·+ cn~vn.

Example 3.1.4 Solve the IVP,

x′ = y; x(0) = 1y′ = 4x; y(0) = 0

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238 CHAPTER 3. LINEAR SYSTEMS

SOLUTION. In matrix form, this system is[x′

y′

]=

[0 14 0

] [xy

].

In example 3.1.3 we identified linearly independent solutions

~v1(t) = e2t[

12

]and ~v2(t) = e−2t

[1−2

].

Therefore, the general solution is ~v = c1~v1 + c2~v2, and we need todetermine values for the coefficients so that

~v(0) =[

10

].

Thus we need to solve

c1

[12

]+ c2

[1−2

]=

[10

],

which is equivalent to the pair of equations

c1 + c2 = 12c1 − 2c2 = 0

Thus c1 = c2 = 12 . The solution is x = 1

2 e2t + 12 e−2t = cosh(2t) and

y = e2t − e−2t = 2 sinh(2t).

Vector Spaces

Definition: A vector space is a set V of objects, together with an additivestructure, so that if ~v, ~w ∈ V then there is a “vector” ~u = ~v + ~w ∈ V . It isalso required that if r is a scalar (that is, a real or complex constant) and~v ∈ V then there is a “vector” ~x = r~v ∈ V . It is customary to call V a realvector space if the scalars are real, or a complex vector space if complexscalars are used.

We can summarize this definition as follows: a vector space is a setof objects that can be combined by taking linear combinations. To avoidambiguities, some rules have to be imposed. These are as follows (~u,~v, ~wrepresent arbitrary elements of V , and r, s are arbitrary scalars):

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3.1. THE INITIAL VALUE PROBLEM 239

• Addition is associative: If ~u = ~v + ~w and ~w +~x = ~y then

~u +~x = ~v +~y.

(This can also be expressed as (~v + ~w)︸ ︷︷ ︸~u

+~x = ~v + (~w +~x)︸ ︷︷ ︸~y

.)

• Addition is commutative: ~v + ~w = ~w +~v.

• There is an additive identity, denoted~0, such that~0 +~v = ~v for eachvector ~v.

• Each vector ~v has a unique additive inverse −~v such that ~v + (−~v) =~0.

• Multiplication by the scalar 1 sends each vector to itself: 1~v = ~v.

• Two distributive laws that govern the interaction of scalar multipli-cation and addition:

Left: (r + s)~v = r~v + s~v

Right: r(~v + ~w) = r~v + r~w

Here are some exercises:

1. Show that 0~v = ~0 by using a distributive law and the familiar fact0 + 0 = 0 for scalars.

2. Show that (−1)~v = −~v

In addition to Rn and Cn, there are other vector spaces that will attractout attention. If (a, b) is an interval, Cn(a, b) will denote the collection of

all functions f (t) with the property thatdn fdtn (t) is continuous on (a, b). In

particular C0(a, b) is the collection of continuous functions on (a, b). If thefunctions are real valued, Cn(a, b) is a real vector space, because the func-tions (which play the role of “vectors”) can be added and multiplied byscalars. When working with complex-valued functions, Cn(a, b) becomes acomplex vector space.

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240 CHAPTER 3. LINEAR SYSTEMS

Definition: If V is a vector space andW is a nonempty subset of V thenwe will say W is a subspace of V if W is a vector space in its own right.To see if W is a subspace, you have to show that if ~v, ~w ∈ W and r, s arescalars, then r~v + s~w ∈ W too. If r = s = 0 this would mean~0 ∈ W , so thefirst thing to check when verifying thatW is a subspace is that~0 ∈ W .

Vector spaces are the environment for taking linear combinations, sogiven S = {~v1, . . . ,~vn} ⊂ V we can form the span of S: the set of all possiblelinear combinations of the vectors in S. If S is nonempty then span S is asubspace of V . It also makes sense to ask if the set S is linearly independent.If that is the case then S is a basis of the subspaceW = span S.

Definition: LetW be a vector space that has a finite basis S. The numberof vectors that comprise S is the dimension ofW . You will wonder: “Whatif S and T are separate bases ofW . Must they have the same cardinality?”The answer, proved in all linear algebra courses, is yes.

A vector space may not have a finite basis. For example, let P be the setof all polynomials with real coefficients. Addition and scalar multiplicationare included in the rules for manipulating polynomials, so P is a real vectorspace. Let S = {1, x, x2, x3, . . .}. Every polynomial is a linear combinationof a finite number of elements of S, so P = span S. Furthermore, we canshow S is linearly independent by noting that if the polynomial

f (x) = a0 + a1x + · · ·+ anxn

is identically zero, then it has more than n zeros, which is all a nontrivialpolynomial of degree n can have—therefore a0 = a1 · · · = an = 0. It followsthat S is a basis for P , and is not finite. A vector space that does not havea finite basis is infinite dimensional. Besides P , the spaces Cn[a, b] that wejust introduced are infinite dimensional vector spaces—indeed, all containP as a subspace.

The vector space concept is important in the context of differential equa-tions. In fact, let A(t) be an n× n matrix whose entries are continuous onan interval (a, b). Solutions of the homogeneous system ~v′ = A(t)~v willbe vector functions that are differentiable on (a, b). Each such function hasn components, each belonging to C1(a, b); thus all solutions will belong toV = (C1(a, b))n. The principal results of this section can be written com-pactly in terms of vector spaces as follows:

• The solutions of ~v′ = A(t)~v form a subspace of V (Theorem 3.2). LetW denote that subspace.

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3.1. THE INITIAL VALUE PROBLEM 241

• If S = {~v1(t), ~v2(t) . . . ,~vn(t)} ⊂ W and t0 ∈ (a, b) then S is linearlyindependent if and only if the set of vectors

S(t0) = {~v1(t0),~v2(t0) . . . ,~vn(t0)}

is a linearly independent subset of Rn (Theorem 3.3).

• Let S = {~a1,~a2 . . . ,~an} be a basis of Rn. Then there exist solutions~vi(t) ∈ W such that ~vi(t0) =~ai, for i = 1, 2, . . . , n (Theorem 3.1).

• The set of solutions ~vi(t), whose existence was just asserted, is a fun-damental set of solutions of ~v′ = A(t)~v (Theorem 3.4).

• The definition of a fundamental set of solutions is the same as thedefinition of a basis forW . Thus S is a basis forW .

• The vector space W of solutions of the homogeneous system ~v′ =A(t)~v is n-dimensional, where n is the size of the square matrix A(t).

Exercises

1. Let A be a 2× 3 matrix, B be a 3× 2 matrix, ~v ∈ R3, and ~w ∈ R2.Make a list of six products that can be formed with A, B,~v, and ~wusing matrix multiplication.Answer

2. Let A =

(1 4 72 0 5

), B =

1 −12 3−5 4

, and ~v =

(87

). Form

the following products: ~w = B~v, C = AB, ~x = A~w, and ~y = C~v. Ifyou multiplied correctly, you should get ~x = ~y. Why?

3. Let E =

(1 00 0

), and A =

(a bc d

).

(a) Calculate EA.

(b) Calculate AE.

(c) What property must the matrix A have in order to commutewith E (that is,what conditions must the entries a, b, c, d satisfyto ensure that AE = EA)?

Answer

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242 CHAPTER 3. LINEAR SYSTEMS

4. Let F =

(1 11 0

), and let A be the same as in Exercise 3.

(a) Calculate FA.(b) Calculate AF.(c) What property must the matrix A have in order to commute

with F?

5. Put the system

x′ = 3x− 2yy′ = x + 7y

in matrix form.Answer

6. Put the system that replaces the second order ODE

y′′ + 2y′ + y = 0

in matrix form.

7. Find the matrix form of the system that replaces the inhomoge-neous third order ODE

x′′′ = (sin t)x′′ − tx′ + 2etx + tan t

Answer

8. Put the inhomogeneous system

x′ = x + t y + t2

y′ = t x− y + 1

in matrix form.

9. Show that {e3t[

11

], e−t

[1−1

]}is a fundamental set of solutions of the system[

x′

y′

]=

[1 22 1

] [xy

].

Find the general solution.Answer

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3.1. THE INITIAL VALUE PROBLEM 243

10. Show that [x(t)y(t)

]=

[3t− 5−6t + 4

]is a solution of the system[

x′

y′

]=

[1 22 1

] [xy

]+

[9t0

].

In Exercises 11 – 15, solve the IVP.

11.[

x′

y′

]=

[2 00 3

] [xy

];[

x(0)y(0)

]=

[12

].

Answer

12.[

x′

y′

]=

[−1 0

0 1

] [xy

];[

x(0)y(0)

]=

[20

].

13. ddt

[xy

]=

[1 22 1

] [xy

];[

x(0)y(0)

]=

[20

]. (See Exercise 9.)

Answer

14. ddt

[xy

]=

[1 22 1

] [xy

];[

x(0)y(0)

]=

[05

]. (See Exercise 9.)

15.[

x′

y′

]=

[0 0−1 1

] [xy

];[

x(0)y(0)

]=

[11

].

Answer

In Exercises 16 – 19, write the IVP in matrix form. For each system, usecorollary 3.1.3 to verify that the given family of solutions is actually thegeneral solution. Finally, use the general solution to solve the IVP.

16.x′ = 4x + 3y; x(0) = 2y′ = 3x− 4y; y(0) = 4.

General solution:[

xy

]= c1e5t

[31

]+ c2e−5t

[1−3

].

17.{

x′ = 2x + 3y; x(0) = −1y′ = −x− 2y; y(0) = 1.

General solution:[

xy

]= c1et

[3−1

]+ c2e−t

[1−1

].

Answer

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244 CHAPTER 3. LINEAR SYSTEMS

18.x′ = x− 2y; x(0) = 1y′ = 2x− 4y; y(0) = −1

General solution[

xy

]= e−t

(c1

[1 + 2t

2t

]+ c2

[−2t

1− 2t

]).

19.x′ = 4x + 3y; x(0) = 7y′ = −2x− y; y(0) = −2

General solution[

xy

]= c1et

[1−1

]+ c2e2t

[3−2

].

Answer

20. Show that{

e5t(

3−1

), e−t

(1−3

)}is a fundamental set of so-

lutions of the system in Exercise 16.

21. Show that{

et(

3−1

), e−t

(1−1

)}is a fundamental set of so-

lutions of the system in Exercise 17.Answer

22. The system

x′ = x + yy′ = y

is uncoupled. Write the system in vector form and find a fundamentalset of solutions.

In Exercises 23–26 solve the IVP. The systems are not homogeneous,but their general solutions are given.

23.{

x′ = 2x− y; x(0) = 1y′ = 4x− 2y + 2; y(0) = −1.

General solution:[

xy

]=

[−t2

2t− 2t2

]+ c1

[12

]+ c2

[2t + 1

4t

].

Answer

24.{

x′ = x + 2y + 2et; x(0) = 0y′ = −x− y; y(0) = 0.

General solution:[xy

]= et

[2−1

]+ c1

[2 cos(t)

− cos(t)− sin(t)

]+ c2

[2 sin(t)

− sin(t) + cos(t)

].

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 245

25.x′ = y + cosh t; x(0) = 2y′ = x + sinh t; y(0) = 2

General solution:[

xy

]= (c1 + t)

[cosh tsinh t

]+ c2

[sinh tcosh t

]Answer

26.x′ = 2x + 2y + et; x(0) = 0y′ = 3x + y; y(0) = −2

.

General solution:[

xy

]= c1e−t

[2−3

]+ c2e4t

[11

]+ et

[−5

3

].

27. A linear second-order ODE is an ODE that can be put in the form

y′′ + p(t)y′ + q(t)y = r(t). (3.5)

The functions p(t) and q(t) are the coefficients of the ODE, and r(t) isthe source. Show that a linear second-order ODE can be replaced witha linear system of first-order ODEs, and write the system correspond-ing to equation (3.5) in matrix form.Answer

28. State and prove, as a corollary to theorem 3.1, an existence anduniqueness theorem for linear second-order ODEs.

3.2 Systems with Constant Coefficients

The general solution of the linear homogeneous ODE,

dydt

= a(t) y,

is found by substituting y = C e f (t). This yields

C e f (t) f ′(t) = a(t)(C e f (t)).

After dividing through by Ce f (t), we obtain f ′(t) = a(t). It follows that

y = C e∫

a(t) dt

is a family of solutions. By proposition 1.2.1, it is the general solution.This idea can be used to solve a system of linear homogeneous ODEs

if the coefficient matrix of the system is a constant matrix. Consider a ho-mogeneous system with n dependent variables x1, . . . xn forming a vector~v,

~v′ = A~v, (3.6)

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246 CHAPTER 3. LINEAR SYSTEMS

where A is an n × n constant matrix. Substitute ~v = est~b, where ~b is aconstant vector, and s is a constant that will be determined. Then~v′ = sest~b,and A~v = est A~b, so with this substitution, (3.6) becomes

sest~b = est A~b.

We may divide by the nonzero factor est to obtain

s~b = A~b. (3.7)

Equation (3.7) will hold only if the vector A~b is a scalar multiple of ~b(and s is the scalar). A vector~b 6=~0 with this property is called a eigenvec-tor of the matrix A, and s is a eigenvalue of that matrix. Since there is anassociation between s and~b, we say that~b belongs to s, and s belongs to~b.

Theorem 3.5 If ~b is an eigenvector of the matrix A, and s is an eigenvalue be-longing to~b, then

~v(t) = est~b

is a solution of the system (3.6).

Theorem 3.5 is motivation to find eigenvalues and eigenvectors of squarematrices—then we can solve systems of ODEs with constant coefficients.The method presented here is in many linear algebra texts. It is placed inthe context of the 2× 2 matrix, and although it works in principle for n× nmatrices, the effort required makes it impractical for n ≥ 3; then one shouldresort to a CAS to find eigenvalues and eigenvectors.

The matrix

I =[

1 00 1

]is called the 2× 2 identity matrix. There is an n× n identity matrix for anyn, with the same pattern of zeros and ones, and you can readily check thatmatrix I has the property that I~b =~b for any vector~b.

For any scalar r, the matrix

rI =[

r 00 r

]is called a scalar matrix. Now, if r is an eigenvalue of an n × n matrix Aand~b is an eigenvector belonging to r, then

A~b = r I~b,

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 247

or, in more compact form,

(A− r I)~b =~0. (3.8)

A matrix C with the property that there is a nonzero vector ~d such thatC~d = ~0 is said to be singular. Since ~b is a nonzero vector (Remember!~0 is not permitted to be an eigenvector), equation (3.8) implies that that(A− r I) is singular.

Conversely, if r is given so that (A − r I) is singular, then there mustexist a nonzero vector~b such that (A− r I)~b =~0, and thus r is an eigenvalueof A belonging to the eigenvector~b.

We have proved the following result:

Proposition 3.2.1 Let A be a square matrix. A scalar r is an eigenvalue of A ifand only if (A− r I) is singular matrix.

To find eigenvalues, find all values of r such that (A− r I) is a singularmatrix. Let

B =

[a cb d

].

be a 2 × 2 matrix, whose entries can be constants or functions. The de-terminant of B is defined to be a scalar that is calculated by the formuladet(B) = a d− b c.

Example 3.2.1 Find the determinant of[9 73 5

]

SOLUTION.

det[

9 73 5

]= 9 · 5− 3 · 7 = 24.

The determinant can be defined for square matrices of any size, but inthis text we will not consider determinants of matrices larger than 2× 2.The following proposition will be familiar to all who have studied linearalgebra, where it is shown to be valid for n× n matrices.

Proposition 3.2.2 Let C =

[m np q

]be a 2× 2 matrix. Then C is singular if

and only if det(C) = 0.

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248 CHAPTER 3. LINEAR SYSTEMS

PROOF Assume that C is singular. Then there is a vector ~b =

[hk

],

(where h and k are not both 0), such that C~b =~0. By matrix multiplication,

C~b =

[mh + nkph + qk

]=

[00

]Assuming for definiteness that h 6= 0,

m = − kh

n and p = − kh

q.

It follows that

det(C) = mq− pn =kh

nq− kh

qn = 0.

Conversely, assume that det(C) = 0. Since mq = np, you can easilyshow that

C[

n−m

]= C

[q−p

]=~0

If C has any nonzero entries, then at least one of the vectors[

n−m

]and[

q−p

]is nonzero, and hence C is singular.

If all of the entries of C are equal to 0, then C~b = ~0 for every vector~b,and hence C is singular.

By propositions 3.2.1 and 3.2.2, r is an eigenvalue of A if and only ifwhen s = r, the determinant of the matrix (entries are functions of s)

A− sI =[

a− s bc d− s

]is equal to 0. Let’s calculate this determinant, as a function of s!

det([

a− s bc d− s

])= (a− s)(d− s)− bc.

Thus det(A− s I) is a quadratic polynomial in the variable s, which can besimplified as

det(A− sI) = s2 − (a + d)s + (ad− bc).

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 249

The coefficients of this equation are significant. You will recognize theconstant term as det(A). The coefficient of s is equal to the sum of the diag-onal entries of A, and is called the trace of A. (Notation: tr (A).)

We have thus proved the following result:

Proposition 3.2.3 Let A be a 2× 2 matrix. Then eigenvalues of A are the rootsof the quadratic equation

s2 − tr (A) s + det(A) = 0. (3.9)

Equation (3.9) is called the characteristic equation of A.Once an eigenvalue of a matrix A is known, the eigenvectors belonging

to it are found by solving a system of linear equations.

Example 3.2.2 Find the eigenvalues and eigenvectors of

A =

[1 23 2

]. (3.10)

SOLUTION. Since tr (A) = 1 + 2 = 3 and det(A) = (1)(2)− (3)(2) = −4,the characteristic equation of A is

s2 − 3s− 4 = 0 or (s + 1)(s− 4) = 0.

It follows that the eigenvalues are s = −1 and s = 4.The eigenvectors belonging to s = −1 satisfy the equation A~b = −~b. Set

~b =

[hk

].

Then

A~b =

[h + 2k

3h + 2k

]=

[−h−k

]Hence h + 2k = −h and 3h + 2k = −k. Both of these equations reduce tok = −h, so any nonzero vector that is a scalar multiple of

~b1 =

[1−1

]is an eigenvector belonging to −1.To find an eigenvector belonging to s = 4, we have to solve

A~b =

[1 23 2

] [hk

]=

[4h4k

].

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250 CHAPTER 3. LINEAR SYSTEMS

This is equivalent to solving the equations

h + 2k = 4h3h + 2k = 4k

These equations both reduce to 3h = 2k, so it follows that any nonzeroscalar multiple of

~b2 =

[23

]is an eigenvector belonging to 4.

To determine the general solution of a system of ODEs, it is helpfulto know that solutions corresponding to different eigenvalues are linearlyindependent. The following theorem from linear algebra can be applied forthis purpose.

Theorem 3.6 Let s1, s2, . . . sm be distinct eigenvalues of a matrix A, and let~b1,~b2 . . . ,~bm be eigenvectors belonging to them. Then{~b1,~b2, . . . ,~bm} is linearly independent.

PROOF FOR m = 2. Suppose that~b2 = c~b1. Then A~b2 = cA~b1 = cs1~b1 =

s1~b2. However, ~b2 is an eigenvector belonging to s2. Hence A~b2 = s2~b2 Itfollows that s1 = s2, a contradiction.

Corollary 3.2.4 Let A be a 2× 2 matrix that has two distinct eigenvalues s1 ands2. and let~b1 and~b2 be eigenvectors belonging to them. Then the general solutionof ~v′ = A~v is

~v(t) = c1es1t~b1 + c2es2t~b2. (3.11)

PROOF. Let ~v1(t) = es1t~b1 and ~v2(t) = es2t~b2. Then by theorem 3.6~v1(0) = ~b1 and ~v2(0) = ~b2 are linearly independent. By corollary 3.1.3 onpage 237, it follows that the general solution can be expressed as in (3.11).

Example 3.2.3 Find the general solution of the system

x′ = 4x− 4yy′ = 3x− 3y

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 251

SOLUTION. Let

~v =

[xy

]and A =

[4 −43 −3

]so that our system is equivalent to ~v′ = A~v. Since tr (A) = 1 anddet(A) = 0, the characteristic equation is s2 − s = 0 and the eigenvaluesare 0 and 1. Let

~b1 =

[hk

]be an eigenvector belonging to 0. Then

4h− 4k = 03h− 3k = 0.

This reduces to h = k, so we can take

~b1 =

[11

]Now suppose that

~b2 =

[hk

]is an eigenvector belonging to 1. Then

4h− 4k = h3h− 3k = k,

so that 3h = 4k. Hence we can set

~b2 =

[43

].

The general solution to the matrix equation is

~v = c1e0t[

11

]+ c2e1t

[43

].

Thus x = c1 + 4c2et and y = c1 + 3c2et.

Example 3.2.4 Solve the IVP

~v′ =[

2 43 −2

]~v; ~v(0) =

[10

].

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252 CHAPTER 3. LINEAR SYSTEMS

SOLUTION. The trace of the coefficient matrix is 0, the determinant is −16,and the characteristic equation is s2 − 16 = 0. The eigenvalues are ±4. Tofind an eigenvector belonging to 4, we solve[

2 43 −2

] [hk

]=

[4h4k

],

or,

2h + 4k = 4h3h− 2k = 4k.

Both of these equations reduce to 2k = h, so

~b1 =

[21

]is an eigenvector belonging to 4. Turning to −4,

~b2 =

[hk

]is an eigenvector if

2h + 4k = −4h3h− 2k = −4k.

These equations reduce to 2k = −3h so we will put h = 2 and k = −3.The general solution is

~v(t) = c1 e4t[

21

]+ c2 e−4t

[2−3

]Thus

~v(0) =[

2c1 + 2c2c1 − 3c2

]Since the initial condition specifies that ~v(0) =

[10

], we can find c1 and

c2 by solving

2c1 + 2c2 = 1c1 − 3c2 = 0.

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 253

The solutions are c1 = 3/8, c2 = 1/8. We can now assemble the solutionof the IVP:

~v =

[ 34 e4t + 1

4 e−4t

38 e4t − 3

8 e−4t

].

Double eigenvalues

If the characteristic equation of a matrix A has a double root r, and~b1 is aneigenvector, then the system~v′ = A~v has the family of solutions~v = c ert~b1.If there is a second, independent eigenvector~b2 belonging to r, then everyvector in the plane can be expressed as a linear combination of~b1 and~b2.Since

A(c1~b1 + c2~b2) = c1A~b1 + c2A~b2

= r(c1~b1 + c2~b2),

every vector in the plane is an eigenvector belonging to r; that is, A~v = rI~vfor all ~v. Hence A = rI is a scalar matrix, and the general solution of~v′ = A~v is ~v = ert~c where~c is an arbitrary constant vector.

Not all 2× 2 matrices with double eigenvalues are scalar matrices, andhere is an example of such a matrix. The characteristic polynomial of

A =

[r 01 r

]is s2 − 2rs + r2 = (s − r)2, so r is a double eigenvalue. A is not a scalarmatrix, so it cannot have two independent eigenvectors. To see how to findthe general solution of ~v′ = A~v, let us put

~v =

[xy

]and write the matrix equation as the system

x′ = rxy′ = x + ry.

This system is uncoupled. The solution of the first equation is x = c ert, andwe substitute this into the second equation to obtain

y′ = ry + c ert

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254 CHAPTER 3. LINEAR SYSTEMS

We will solve this inhomogeneous first order equation by the method ofvariation of constants, starting with the homogeneous solution yh(t) = ert.Set y = w ert, where w is a new dependent variable, to obtain

w′ ert + rw ert = rw ert + c ert

which can be simplified as w′ = c. Thus w = ct + d and we have y =(ct + d) ert. In vector form,

~v = ert(

c[

1t

]+ d

[01

]).

This solution is a linear combination of two solutions,

~v1(t) = ert[

1t

]= ert(~i + t~j)

and

~v2(t) = ert[

01

]= ert~j,

where~i and~j are the standard basis vectors,

~i =[

10

]and~j =

[01

].

The solution ~v2 derives from the eigenvalue r and its eigenvector,~j.Let us now turn to an arbitrary matrix A that has a double eigenvalue

r. We will assume that A is not a scalar matrix, since scalar matrices arehandled differently.

Following the example computed above, we will try to derive a solutionof the form ~v(t) = ert(t~b +~c). By the product rule for differentiation,

~v′ = ert[r(t~b +~c) +~b].

Since ~v′ = A~v andA~v = ert(tA~b + A~c),

it follows that ~v is a solution if and only if for all t,

r(t~b +~c) +~b = tA~b + A~c. (3.12)

Setting t = 0 in (3.12), we have

~b = (A− rI)~c. (3.13)

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 255

On the other hand, if we differentiate (3.12) with respect to t, we have

A~b = r~b. (3.14)

Equations (3.13) and (3.14) must both be satisfied. The following the-orem from linear algebra tells us that since r is a double eigenvalue of A,equation (3.14) will be satisfied automatically if (3.13) holds.

Theorem 3.7 (Cayley - Hamilton Theorem) Let B be a 2× 2 matrix, and let

s2 − tr (B) s + det(B) = 0

be its characteristic equation. Let B2 be the matrix formed by multiplying thematrix B by itself. Then

B2 − tr (B) B + det(B)I = 0I

Although stated here in terms of 2 × 2 matrices, the Cayley - Hamiltontheorem holds for square matrices of any size. It can be summarized to saythat “every square matrix satisfies its own characteristic equation.”

PROOF OF THE CAYLEY - HAMILTON THEOREM. If B =

[a bc d

], then

tr (B) = a + d and det(B) = ad− bc. By matrix multiplication,

B2 =

[a2 + bc ab + bdac + cd bc + d2

]and

tr (B) B =

[a2 + ad b(a + d)c(a + d) ad + d2

].

It follows that

B2 − tr (B) B =

[bc− ad 0

0 bc− ad

]= −det(B) I.

Because r is a double eigenvalue, the characteristic equation of A is (s−r)2 = 0. By the Cayley - Hamilton Theorem, (A− rI)2 = 0I. Since A 6= rI,we can choose a vector ~c 6= ~0 that is not an eigenvector: A~c 6= r~c. Put~b = (A− rI)~c; then (3.13) holds. Furthermore,

(A− rI)~b = (A− rI)2~c = 0I~c =~0.

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256 CHAPTER 3. LINEAR SYSTEMS

It follows that A~b = r~b: thus (3.14) also holds. Hence ~v(t) = es0t(t~b +~c) is asolution of the system ~v′ = A~v.

Matrices with Double Eigenvalues

Let A 6= rI be a 2× 2 matrix that has a double eigenvalue, r.Choose a vector~c such that A~c 6= r~c and define~b = A~c− r~c.Then ~v1(t) = ert~b and ~v2(t) = ert(t~b +~c) are linearly inde-pendent solutions of the system

~v′ = A~v.

Example 3.2.5 Find the general solution of the system ~v′ = A~v, where A =[1 −22 −3

].

SOLUTION. The characteristic equation of A is s2 + 2s + 1 = (s + 1)2 = 0,which has the double root −1. Since A 6= (−1)I, we need a vector~c that isnot an eigenvector. For simplicity, take one of the standard basis vectors.

(They cannot both be eigenvectors; if they were, A = −I.) Put~c =[

10

];

~c is not an eigenvector, because A~c is the first column of A, not a scalarmultiple of~c. Set

~b = A~c− (−1)~c =[

12

]+

[10

]=

[22

].

Then ~v1(t) = e−t~b and ~v2(t) = e−t(t~b +~c) are independent solutions. Thegeneral solution is

~v = c1

[2e−t

2e−t

]+ c2

[(1 + 2t)e−t

2te−t

]

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3.2. SYSTEMS WITH CONSTANT COEFFICIENTS 257

Exercises

In Exercises 1– 5, find all real eigenvalues and eigenvectors of the givenmatrix.

1.[

1 00 −4

]Answer

2.[

2 31 −2

]

3.[

1 10 −4

]Answer

4.[

3 3−3 −3

]

5.[

2 3−3 4

]Answer

6. Show that the matrix[

1 −22 1

]has no real eigenvalues.

7. Let s = r be an eigenvalue of A =

[a bc d

].

(a) Show that[

br− a

]is an eigenvector belonging to r unless b = 0

and r = a.

(b) Show that if b = 0 then s = a is an eigenvalue, and[

a− dc

]is

an eigenvector belonging to it, unless a = d and c = 0 as well.

(c) How do you find the eigenvectors if a = d and b = c = 0?

Answer

In Exercises 8 – 13, find the general solution for each system.

8. ~v′ =[

3 4−2 −3

]~v

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258 CHAPTER 3. LINEAR SYSTEMS

9. ~v′ =[

3 21 2

]~v

Answer

10.{

x′ = −3x− yy′ = x− y

11. ~v′ =[

0 10 0

]~v

Answer

12.{

x′ = x− yy′ = x− y

13.{

x′ = x + 2yy′ = 2x + y

Answer

In Exercises 14 – 18 solve the IVP.

14. ~v′ =[

1 2−1 4

]~v; ~v(0) =

[10

]

15. ~v′ =[−1 1−1 −3

]~v; ~v(0) =

[01

]Answer

16.{

x′ = x + y; x(0) = 4y′ = x + 2y; y(0) = 2

17.{

x′ = 5x− 2y x(0) = −1y′ = 2x + y y(0) = −2

Answer

18.{

x′ = x + y; x(0) = 3y′ = x + y; y(0) = 1

19. Show that the characteristic equation of the system that replacesthe second order linear ODE

y′′ + py′ + qy = 0

is s2 + ps + q = 0Answer

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3.3. SYSTEMS WITH OSCILLATING SOLUTIONS 259

20. Let A be a 2× 2 matrix with distinct eigenvalues s1 and s2, andlet ~c be a vector that is not characteristic. Show that (A − s2 I)~c isan eigenvector belonging to s1, and that (A− s1 I)~c is an eigenvectorbelonging to s2.

3.3 Systems with Oscillating Solutions

We have already observed that x = cos(t), y = sin(t) is a solution of thelinear system

x′ = −yy′ = x.

}(3.15)

This system can be converted to vector form as ~v′ = A~v, where

A =

[0 −11 0

].

Since tr (A) = 0 and det(A) = 1, the characteristic equation of A, s2 + 1 =0, has no real roots. It does have two imaginary roots, ±i, where i =

√−1.

The system (3.15) is typical of the case where there are no real eigenvalues:all such systems have solutions that oscillate.

Euler’s Formula

To solve a systems that has no real eigenvalues, we need to work with com-plex numbers. The following review is for the convenience of readers wholack experience with the complex number system.

The complex number system is an extension of the real number sys-tem formed by including “imaginary numbers” to serve as square roots ofnegative real numbers. Every complex number z can be expressed as

z = x + iy

where x and y are real numbers. The operations of arithmetic, +,−,×,÷are all defined for complex numbers, and they satisfy the same commu-tative, associative, distributive, and existence of inverse laws that the realnumbers do. In addition, complex arithmetic has one operation that realarithmetic does not have: conjugation. The conjugate of z = x + iy isz = x − iy. (It is standard practice to use z to denote the conjugate of z.)For example, 3 + 4i = 3− 4i. Here is how to do complex arithmetic:

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Addition: Use vector addition:

(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)

Thus, (3 + 4i) + (5− i)− 6 = 2 + 3i.

Multiplication: Use binomial multiplication, and i2 = −1. Thus

(x1 + iy1)(x2 + iy2) = x1x2 + i(x1y2 + x2y1) + y1y2i2

= (x1x2 − y1y2) + i(x1y2 + x2y1).

Thus, (12− i)× (1 + i) = (12− i2) + 12i− i = 13 + 11i. Notice thatzz = x2 + y2 is real and positive (unless z = 0). If z = 3 + 4i, thenzz = 32 + 42 = 25.

Division: Multiply the numerator and the denominator by the conjugateof the denominator. The resulting quotient will be the same, but witha real denominator.

x1 + iy1

x2 + iy2=

(x1 + iy1

x2 + iy2

)(x2 − iy2

x2 − iy2

)=

(x1x2 + y1y2

x22 + y2

2

)+ i(−x1y2 + x2y1

x22 + y2

2

)Thus,

2 + i3 + 4i

=(2 + i)(3− 4i)

25=

10− 5i25

= 0.4− 0.2i

Just as it is customary to visualize real numbers as points on a line,complex numbers are visualized as points in a plane, called the complexplane. The horizontal axis of the complex plane is identified with the realline, and is called the real axis. The vertical axis is the imaginary axis.Figure 3.1 displays the complex plane.

The distance from z = x + iy to 0 in the complex plane is, by the usualdistance formula from analytic geometry,

√x2 + y2. This distance is the ab-

solute value of z, and is denoted |z|. Notice that |z| =√

z z. All of the usualproperties of absolute value for real numbers are still true. For example thetriangle inequality,

|z1 + z2| ≤ |z1|+ |z2|can be interpreted as saying that one side of a triangle must be shorter thanthe sum of the lengths of the other two sides: see figure 3.2.

The components of a complex number z = x + iy are the real and imag-inary parts of z, Re(z) = x and Im(z) = y, respectively. By convention,the imaginary part of a complex number is a real number. For example,Im(3 + 4i) = 4, not 4i.

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3.3. SYSTEMS WITH OSCILLATING SOLUTIONS 261

Polar representation of complex numbers. The polar representation of acomplex number is useful for multiplying and dividing complex numbers,and for raising them to powers. It is not useful for complex addition. Thepolar representation of z = x + iy is simply a set of polar coordinates (r, θ)for the point (x, y), in which r ≥ 0. Thus, r =

√x2 + y2 = |z|. The angle

θ is called the argument of z, and is denoted arg(z). The representationz = x + iy can be recovered from the polar representation by the formulasx = r cos(θ) and y = r sin(θ); hence

z = r(cos(θ) + i sin(θ)).

To compute arg(z), use the inverse cosine function:

arg(z) =

cos−1(

Re(z)|z|

)if Im(z) ≥ 0

− cos−1(

Re(z)|z|

)if Im(z) < 0

The range of the argument as defined here is−π < arg(z) ≤ π, but we willfollow the custom of considering arg(z) as a function with values takenmod 2π. For example, if I say arg(1 −

√3i) = −π

3 and you say arg(1 −√3i) = 5π

3 then we agree. Figure 3.3 illustrates the polar representation(r, θ) of −1 + 2i, where r = | − 1 + 2i| =

√5 and θ = arg(−1 + 2i) =

cos−1(−1/√

5), approximately 117◦.

Euler’s formula (Leonhard Euler)

eiθ = cos θ + i sin θ

We will take Euler’s formula as the definition of the exponential of animaginary number. It can be justified by substituting x = iθ in the Maclau-rin series for ex :

eiθ =∞

∑n=0

(iθ)n

n!.

In this series, the even terms are real and the odd terms are imaginary,because (iθ)2m = (−1)mθ2m and (iθ)2m+1 = (−1)miθ2m+1. Therefore,

eiθ =∞

∑m=0

(−1)m θ2m

(2m)!+ i

∑m=0

(−1)m θ2m+1

(2m + 1)!. (3.16)

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262 CHAPTER 3. LINEAR SYSTEMS

The first sum on the right side of equation (3.16) is the Maclaurin series forthe cosine function, and the second is i times the Maclaurin series for thesine.

The identity eu+v = eu ev extends the definition of the exponential to allcomplex numbers. Thus

eλ+iω = eλeiω

= eλ(cos(ω) + i sin(ω)).

To make use of the complex exponential, we need to be able to differ-entiate it. The derivative of a complex-valued function f (t) = p(t) + iq(t)is

f ′(t) = p′(t) + i q′(t).

Proposition 3.3.1 Let a = λ + iω be a complex constant. Then

ddt

eat = a eat.

PROOF.

ddt

eat =ddt{eλt[cos(ωt) + i sin(ωt)]}

=ddt[eλt cos(ωt)] + i

ddt[eλt sin(ωt)]

= [λeλt cos(ωt)− eλtω sin(ωt)]+ i[λeλt sin(ωt) + ωeλt cos(ωt)]

= eλt[λ cos(ωt) + iω cos(ωt)−ω sin(ωt) + iλ sin(ωt)]= eλt[(λ + iω) cos(ωt) + (iω + λ)i sin(ωt)]= a eλt[cos(ωt) + i sin(ωt)] = a eat.

Let A be a square matrix with real entries. The characteristic polynomialof A may have some complex roots, and these are considered to be complexeigenvalues. For example, the characteristic polynomial of the matrix

A =

[1 −11 1

]is s2− 2s+ 2. The eigenvalues of A can be calculated by using the quadraticformula:

s1, s2 =2±

√(−2)2 − 8

2= 1± i.

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3.3. SYSTEMS WITH OSCILLATING SOLUTIONS 263

Since complex arithmetic follows the same rules as real arithmetic, we canwork with matrices and vectors that have complex entries just as we do

when the entries are real. Thus, to find an eigenvector~b1 =

[hk

]belong-

ing to s1 = 1 + i, we solve (A− s1 I)~b1 =~0:([1 −11 1

]−[

1 + i 00 1 + i

]) [hk

]=

[00

]After simplifying, this matrix equation reduces to two complex equations,

−i h− k = 0h− i k = 0

Observe that the second of these equations can be obtained from the firstby multiplying through by i. Thus for any h 6= 0, k = −i h will satisfy bothequations. Let’s put h = 1; then

~b1 =

[1−i

]is an eigenvector belonging to 1 + i.

Suppose that s is a complex eigenvalue of a square matrix A, and ~bis an eigenvector belonging to s. Put ~v(t) = est~v. By proposition 3.3.1,~v′(t) = sest~b = s~v. Since A~v(t) = est A~b = ests~b = s~v, it follows that ~v(t) is a

solution of ~v′ = A~v. Returning to our matrix A =

[1 −11 1

], we see that

~v(t) = e(1+i)t[

1−i

]=

[et(cos(t) + i sin(t))et(sin(t)− i cos(t))

]is a solution of ~v′ = A~v.

This solution is complex since~b is a complex vector, but we require areal solution, since ~v′ = A~v is a real system of ODEs. By the followingtheorem, we can exchange a complex solution of a system of linear, homo-geneous ODEs with real coefficients for two real solutions.

Theorem 3.8 Let~z(t) be a complex solution of a homogeneous linear ODE or sys-tem of ODEs with real coefficients. Then ~v1(t) = Re[~z(t)] and ~v2(t) = Im[~z(t)]are solutions as well.

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264 CHAPTER 3. LINEAR SYSTEMS

PROOF. The vector form of the system is ~v′ = A(t)~v, where the entriesof the matrix A(t) are real-valued. It is given that~z(t) = ~v1(t) + i~v2(t) is acomplex-valued solution. Expand~z′ = A(t)~z as

~v′1 + i~v′2 = A(t)~v1 + iA(t)~v2. (3.17)

Equating real and imaginary parts of both sides of (3.17), we see that ~v′1 =A(t)~v1, and ~v′2 = A(t)~v2. It follows that ~v1(t) and ~v2(t) are solutions.

By theorem 3.8 the complex solution ~v(t) =[

et(cos(t) + i sin(t))et(sin(t)− i cos(t))

]of

~v′ =[

1 −11 1

]~v, yields two real solutions,

~v1(t) = et[

cos(t)sin(t)

]and ~v2(t) = et

[sin(t)

− cos(t)

]

Since ~v1(0) =[

10

]and ~v2(0) =

[0−1

]are linearly independent, the

solutions ~v1(t) and ~v2(t) are linearly independent.

Example 3.3.1 Find the general solution of the system ~v′ = A~v, where

A =

[−5 −45 −1

].

SOLUTION. Since tr (A) = −6 and det(A) = 25, the characteristicequation of A is s2 + 6s + 25 = 0. By the quadratic formula, theeigenvalues are

s =−6±

√62 − 4× 252

=−6±

√−64

2= −3± 4i.

To find a complex eigenvector

~b =

[hk

](h and k will be complex numbers) belonging to s = −3 + 4i, we need tosolve

(A− (−3 + 4i)I)~b =~0.

This matrix equation is([−5 −45 −1

]−[(−3 + 4i) 0

0 (−3 + 4i)

] [hk

])=

[00

],

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3.3. SYSTEMS WITH OSCILLATING SOLUTIONS 265

or [(−2− 4i) −4

5 (2− 4i)

] [hk

]=

[00

].

We must choose h and k so that (−2 + 4i)h− 4k = 0 and5h + (2− 4i)k = 0. These equations are redundant: any (h, k) thatsatisfies one will also satisfy the other. Thus we will put

~b =

[−2

1 + 2i

].

By theorem 3.8, the real and the imaginary parts of

e(−3+4i)t~b = e−3t[

−2(cos(4t) + i sin(4t))cos(4t) + i sin(4t) + 2i(cos(4t) + i sin(4t))

]= e−3t

([−2 cos(4t)

cos(4t)− 2 sin(4t)

]+ i[

2 sin(4t)2 cos(4t) + sin(4t)

])

are solutions. These solutions are ~v1(t) = e−3t[

−2 cos(4t)cos(4t)− 2 sin(4t)

]and

~v2(t) = e−3t[

−2 sin(4t)2 cos(4t) + sin(4t)

]. Since

~v1(0),~v2(0) =[−21

],[

02

]are linearly independent, corollary 3.1.3 tells us that the general solution is

~v(t) = c1~v1(t) + c2~v2(t)

If we exchange a complex solution of a linear system for two real so-lutions, are the real solutions always linearly independent? The answer isyes:

Theorem 3.9 Let s = λ + iω, where ω 6= 0, be a complex eigenvalue of a realmatrix A, and ~b = ~h + i~k be an eigenvector belonging to s. Then ~h and ~k arelinearly independent.

For a proof, see Exercise 11.

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266 CHAPTER 3. LINEAR SYSTEMS

Corollary 3.3.2 Let A be a 2 × 2 real, constant matrix that has an eigenvalueλ + iω with ω 6= 0, and let~h + i~k be an eigenvector belonging to that eigenvalue.Then the general solution of

~v′ = A~v

is

~v(t) = c1eλt(

cos(ωt)~h− sin(ωt)~k)+ c2eλt

(sin(ωt)~h + cos(ωt)~k

),

(3.18)where c1 and c2 are constants.

Exercises

1. Find the reciprocal of 2 + i.Answer

2. Locate each of the complex numbers on the complex plane, calcu-late arg(z), |z|, and find λ + ωi such that z = eλ+ωi.

(a) z = 2.

(b) z = −2.

(c) z = 1− i.(d) z = −1 + i

√3.

(e) z = −2i.

3. Find the sixth roots of 1 (there are 6 of them), and locate them onthe complex plane. Hint: 1 = e2nπi.Answer

4. Show that es = es.

5. Show that |eλ+iω| = eλ (λ and ω are real).Answer

6. Find all solutions of

e2s + 2es + 2 = 0.

In Exercises 7 – 10, find the general solution

7.{

x′ = − 12 x− 1

2 yy′ = 1

2 x− 12 y

Answer

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3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS 267

8.{

x′ = 2x + 5yy′ = −x

9.{

x′ = 4x− 8yy′ = 4x− 4y

Answer

10.{

x′ = −4x− 41yy′ = 1

2 x− 3y

11. Let s = λ + iω, where ω 6= 0, be a complex eigenvalue of a realmatrix A, and~b =~h + i~k be a eigenvector belonging to s.

(a) Show that the conjugate of~b is an eigenvector belonging to theconjugate of s.

(b) Using theorem 3.6 , show that~b and ~b are linearly independent.

(c) Prove theorem 3.9.

(d) Prove corollary 3.3.2.

Answer

12. Let A be a 2× 2 matrix such that tr A = 0 and det A < 0. Showthat the orbits of the system~v′ = A~v are hyperbolas, their asymptotes,and the origin.

Hint: Imitate the case where det A > 0.

3.4 Matrix Solutions and Variation of Constants

Let A(t) be an n× n matrix whose entries are either constants or continuousfunctions of t, all defined of an interval (a, b). A homogeneous system of nlinear ODEs

~v′ = A(t)~v, (3.19)

admits a family of solutions on (a, b), which can be described completelyif a linearly independent set {~v1(t),~v2(t), . . . ,~vn(t)} of solutions is known.By Theorem 3.4, every solution is then uniquely expressed as a linear com-bination

c1~v1(t) + c2~v2(t) + · · ·+ cn~vn(t).

It will be convenient to use matrix notation to simplify this statement.

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268 CHAPTER 3. LINEAR SYSTEMS

Given an n× n matrix X (t) whose entries are functions of t, let X ′ bethe matrix formed by differentiating each entry of X . If the equation

X ′ = A(t)X (3.20)

holds on the interval (a, b), then X (t) is called a matrix solution of equa-tion (3.19).

Proposition 3.4.1 Let the columns of the matrix X (t) be denoted ~v1(t), ~v2(t),. . . , ~vn(t). Then X (t) is a matrix solution of the homogeneous system (3.19) ifand only if the ~vi(t) are solutions of (3.19) as vector functions.

PROOF. The columns of X ′ are ~v′1, ~v′2, . . . , ~v′n. Furthermore, the columnsof A(t)X are A(t)~vi, for i = 1, 2, . . . , n. It follows each column of a matrixsolution of (3.19) is itself a (vector) solution, and conversely that if each~vi(t) is a solution, then X (t) is a matrix solution.

We can use a matrix solution to generate a family of solutions of (3.19)as follows.

Proposition 3.4.2 Suppose that X (t) is a matrix solution of (3.19), and let ~c ∈Rn be a constant vector. Then the vector function

~v(t) = X (t)~c

is a solution of the system (3.19).If, in addition, the columns of X (t) are linearly independent, then given any

solution ~w(t) of the system (3.19), there is a unique~c ∈ Rn such that ~v = X (t)~c.

PROOF. Let ~v1, ~v2, . . . , ~vn denote the columns of X . By the definition ofmatrix multiplication,

X~c = c1~v1 + c2~v2 + · · ·+ cn~vn

is a linear combination of solutions of the system (3.19) and thus itself asolution of the system.

If {~v1, ~v2, . . . , ~vn} is linearly independent, then by theorem 3.4 there areunique constants c1, . . . , cn such that

~w = c1~v1 + c2~v2 + · · ·+ cn~vn.

The constants ci are the components of the vector~c such that ~w = X~c.DEFINITION A matrix solution X (t) of equation (3.19) is called a funda-

mental matrix solution if there is a point t0 ∈ (a, b) such that the columnsof X (t0) are linearly independent.

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3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS 269

An n × n matrix P is said to be nonsingular if it is not singular. InExercise 13 at the end of this section, you are asked to prove that a matrixP is nonsingular if and only if, as vectors, the columns of P are linearlyindependent.

Proposition 3.4.3 Let X (t) be a matrix solution of (3.19), defined on an inter-val (a, b). Then X (t) is a fundamental matrix solution if and only if one of thefollowing equivalent conditions holds.

• X (t) is nonsingular for all ordinary points t ∈ (a, b).

• X (t0) is nonsingular for some ordinary point t0 ∈ (a, b).

• det(X (t)) 6= 0 for all ordinary points t ∈ (a, b).

• det(X (t0)) 6= 0 for some ordinary point t0 ∈ (a, b).

This proposition implies that a matrix solution that is nonsingular atone point t0 ∈ (a, b) is automatically nonsingular for all t ∈ (a, b). This isa special property of matrix solutions of homogeneous linear systems ofODEs: it is possible for matrix functions in general to be nonsingular insome places, and not in others.

PROOF OF PROPOSITION 3.4.3: The first statement follows from Exer-cise 13. For the second, we will apply theorem 3.4. Let~a1 = ~v1(t0), . . . ~an =~vn(t0) be the columns of X (t0), which is then nonsingular if and only if{~a1, . . . , ~an} is linearly independent. By theorem 3.4, the columns~v1(t), . . . ,~v(t)of X (t) are then also linearly independent—that is, X (t) is nonsingular—for all t ∈ (a, b).

The statements about determinants follow from proposition 3.2.2.

Example 3.4.1 Find a fundamental matrix solution of

~v′ =[

0 11 0

]~v. (3.21)

SOLUTION. The characteristic equation of

A =

[0 11 0

]is s2 − 1 = 0; hence the eigenvalues are ±1. We can use eigenvectors

~e1 =

[11

]and~e2 =

[1−1

]

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270 CHAPTER 3. LINEAR SYSTEMS

belonging to 1 and −1, respectively, to obtain the solutions ~v1(t) = et~e1and ~v2(t) = e−t~e2. Since

~v1(t) =[

et

et

]and ~v2(t) =

[e−t

−e−t

],

X (t) =[

et e−t

et −e−t

].

is a matrix solution.

Then X (0) =[

1 11 −1

]; since detX (0) = −2 if follows from

proposition 3.4.3 that X (t) is a fundamental matrix solution.

Example 3.4.2 Find the general solution of the system 3.21, and also the solution

with initial value ~v(0) =[

35

].

SOLUTION. In example 3.4.1, we found a fundamental matrix solution,

X =

[et e−t

et −e−t

]The general solution is

~v = X~c

=

[et e−t

et −e−t

] [c1c2

]=

[c1et + c2e−t

c1et − c2e−t

]= c1et

[11

]+ c2e−t

[1−1

].

To find the solution of the IVP, we must find a vector~c such that X (0)~c isequal to the specified value of ~v(0). Thus we must solve the equation[

1 11 −1

] [c1c2

]=

[35

].

or, equivalently, (c1, c2) must satisfy the system

c1 + c2 = 3c1 − c2 = 5

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3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS 271

which yields (c1, c2) = (4,−1). The solution of the IVP is therefore

~v(t) = 4et[

11

]− e−t

[1−1

]

Inhomogeneous systems

The method for solving an inhomogeneous system of linear ODEs,

~v′ = A(t)~v + ~f (t), (3.22)

where A(t) is an n× n matrix, is basically the same as the method we usedin section 1.3 to solve a single first order linear ODE. We start with thegeneral solution of the associated homogeneous system,

~v′ = A(t)~v,

which we can express in terms of a fundamental matrix solution,

~vh(t) = X (t)~c

and add a particular solution ~vp(t) of the inhomogeneous equation.

Theorem 3.10 Suppose that all entries of the coefficient matrix A(t) and sourcevector ~f (t) in (3.22) are continuous on an interval (a, b). Let~vp(t) be a particularsolution of the system (3.22), and let X (t) be a fundamental matrix solution of theassociated homogeneous system. Then the general solution of the system (3.22) on(a, b) is

~v(t) = ~vp(t) +X (t)~c.

PROOF. Let ~v(t) be an arbitrary solution of the inhomogeneous sys-tem (3.22), and put ~y(t) = ~v(t)−~vp(t). Then

~y′ = ~v′ −~v′p= [A(t)~v + ~f (t)]− [A(t)~vp + ~f (t)]= A(t)~v− A(t)~vp = A~y.

It follows that ~y(t) is a solution of the associated homogeneous system.Therefore there is a unique constant vector~c such that

~y(t) = X (t)~c;

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272 CHAPTER 3. LINEAR SYSTEMS

and it follows that ~v(t) = ~vp(t) +X (t)~c.Conversely, suppose that ~v(t) = ~vp(t) + X (t)~c, where ~c is a constant

vector. Thenddt(~v(t)) = ~v′p(t) +X ′(t)~c

= A(t)~vp(t) + ~f (t) + A(t)X (t)~c

= A(t)(~vp(t) +X (t)~c) + ~f (t)

= A(t)~v(t) + ~f (t);

in other words, ~v(t) is a solution of (3.22).Theorem 3.10 can best be deployed in conjunction with a method for

finding particular solutions of systems.

Variation of constants

The method of variation of constants is used to find a particular solution ofan inhomogeneous system. Recall from section 1.3, where the method wasintroduced for the scalar case, that the general solution of the associatedhomogeneous equation was a required input. For systems, it is convenientto start with a fundamental matrix solution of the associated homogeneoussystem.

The key to adapting the method for systems of linear ODEs is inversionof a matrix solution. This replaces division by yh(t) in the scalar case (as inequation (1.11)).

Let P be a square matrix. A square matrix Q is called the inverse matrixof P if P Q = Q P = I, where I is the identity matrix. We will use thecustomary notation P−1 for the inverse matrix of P. The following theoremis from linear algebra. Although it is true for square matrices of any size,our proof only works for 2× 2 matrices.

Theorem 3.11 Let

P =

[a cb d

]be a nonsingular matrix. Then P has an inverse matrix.

PROOF By proposition 3.2.2, det(P) = ad− bc is equal to 0 if and only ifP is singular.

Thus, if P is nonsingular, det(P) 6= 0. Define a matrix Q as

Q =1

det(P)

[d −b−c a

].

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3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS 273

You can verify that P Q = I and Q P = I by matrix multiplication.

We will now see how to find a particular solution of an inhomogeneoussystem of ODEs. Let X (t) be a fundamental matrix solution of a homo-geneous system ~v′ = A(t)~v. To find a particular solution of an inhomoge-neous system

~v′ = A(t)~v + ~f (t) (3.23)

set

~vp(t) = X (t)~w(t),

where ~w(t) is a vector function that will be determined. By the product rulefor differentiation,

~v′p = X ′(t)~w(t) +X (t)~w′(t)

= A(t)X (t)~w(t) +X (t)~w′(t)= A(t)~vp(t) +X (t)~w′(t).

It follows that ~vp(t) is a particular solution of the system (3.23) if and onlyif

X (t)~w′(t) = ~f (t). (3.24)

Since X (t) is nonsingular, it has an inverse X−1(t). Multiplying (3.24) bythis, we have

~w′(t) = X−1(t)~f (t).

If t0 ∈ (a, b), we can put

~w(t) =∫ t

t0

X−1(s)~f (s) ds.

Now that ~w(t) has been determined we can find ~vp(t) by multiplying byX (t). This can be summarized as follows:

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274 CHAPTER 3. LINEAR SYSTEMS

Variation of Constants Formula

Let X (t) be a fundamental matrix solution of ~v′ = A(t)~v, whereA(t) is a matrix whose entries are all continuous on an interval(a, b). If ~f (t) is a vector function, also continuous on (a, b), andt0 ∈ (a, b), then

~vp(t) = X (t)∫ t

t0

X−1(s)~f (s) ds

is a particular solution of the inhomogeneous equation

~v′ = A(t)~v + ~f (t).

It is not necessary to memorize this formula; you just need to rememberto substitute ~v = X (t)~w in (3.23).

Example 3.4.3 Find a particular solution of

x′1 = x2 − e−t

x′2 = x1 + e−t

SOLUTION. In example 3.4.1, it was shown that X (t) =[

et e−t

et −e−t

]is a

fundamental matrix solution of the associated homogeneoussystem (3.21). Set ~vp(t) = X (t)~w(t). Thus[

x1x2

]=

[et e−t

et −e−t

] [w1w2

]=

[etw1 + e−tw2etw1 − e−tw2

]Differentiating, and using the product rule we have[

x′1x′2

]=

[etw′1 + e−tw′2etw′1 − e−tw′2

]+

[etw1 − e−tw2etw1 + e−tw2

]and the right side of our system is[

x2 − e−t

x1 + e−t

]=

[etw1 − e−tw2etw1 + e−tw2

]+

[−e−t

e−t

]

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3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS 275

After canceling, we have

etw′1 + e−tw′2 = −e−t

etw′1 − e−tw′2 = e−t.

The solution of this equation is w′1 = 0, w′2 = −1, or

~w′(t) =[

0−1

].

Thus ~w(t) =[

0−t

], and we have the particular solution

~vp(t) =[

et e−t

et −e−t

] [0−t

]=

[−te−t

te−t

].

The general solution is ~vp(t) +~vh(t), where ~vh(t) = X (t)~c denotes thegeneral solution of the associated homogeneous equation.

Exercises

In Exercises 1 – 4, find a fundamental matrix solution of the system.

1.{

x′ = 32 x + 1

2 yy′ = 1

2 x + 32 y;

Answer

2.{

x′ = −3x + yy′ = −5x− y;

3.{

x′ = 2x− 4yy′ = 5x− 2y;

Answer

4.{

x′ = −x− yy′ = x− 3y;

In each of Exercises 5 – 8, use the method of variation of constants tofind a particular solution of the system of ODEs, and then write down thegeneral solution. Fundamental matrix solutions of the associated homoge-neous systems were found in Exercises 1 – 4.

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276 CHAPTER 3. LINEAR SYSTEMS

5.{

x′ = 32 x + 1

2 y + b1(t)y′ = 1

2 x + 32 y + b2(t),

where

(a)[

b1b2

]=

[et

et

].

(b)[

b1b2

]=

[et

−et

].

(c)[

b1b2

]=

[tet

0

].

Answer

6.{

x′ + 3x− y = −3e−2t

y′ + 5x + y = e−2t

7.[

x′

y′

]=

[2 −45 −2

] [xy

]+

[−32 cos(4t)

0

].

Answer

8.[

x′

y′

]+

[1 1−1 3

] [xy

]=

[b1(t)b2(t)

], where

(a)[

b1(t)b2(t)

]= e−2t

[13

].

(b)[

b1(t)b2(t)

]=

[2e−2t

0

].

9. Show that

X =

[2t2 t3

−t2 −t3

]is a fundamental matrix solution of

tx′ = x− 2yty′ = x + 4y.

Using variation of constants, find a particular solution of

tx′ = x− 2y + b1(t)ty′ = x + 4y + b2(t),

where

(a)[

b1(t)b2(t)

]=

[t

2t

].

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 277

(b)[

b1(t)b2(t)

]=

[t3

−t3

].

Warning: the variation of constants formula is based on the assump-tion that the system has the form~v′ = A(t)~v +~b(t); the system in thisExercise must be written in that form before the formula can be used.However, if you use the method of variation of constants there are noworries.Answer

10. Find the general solution of each of the following inhomogeneoussystems.

(a){

x′ = x + y + t−1

y′ = −x− y.

(b){

x′ = x + y + t−1

y′ = −(x + y + t−1).

(c){

x′ = 4x + 5y + 5 sin(3t)y′ = −5x− 4y + 3 cos(3t)− 4 sin(3t)

(d){

x′ = sec3(t)− yy′ = x− sec3(t)

11. Let C be a constant matrix, and suppose that X (t) is a matrixsolution of ~v′ = A(t)v. Show that XC is also a matrix solution.Answer

12. Let X1(t) and X2(t) be fundamental matrix solutions of the linearsystem of differential equations ~v′ = A(t)~v, defined on an interval I .Show that X−1

1 · X2 is a constant matrix.

13. Let Q be an n× n matrix. Show that Q is nonsingular if and onlyif the columns of q are linearly independent.Answer

3.5 A Fundamental Matrix Solution of Interest

The Maclaurin series expansion for the function eat is

eat = 1 + a t +a2

2!t2 +

a3

3!t3 + · · ·+ an

n!tn + · · · ,

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278 CHAPTER 3. LINEAR SYSTEMS

and it converges for all values of t.Now consider the series with the scalar constant a replaced by a con-

stant matrix A. By convention, A0 = I is the identity matrix, A1 = A,A2 = A× A, etc.

The Matrix Exponential Function

Let A be an m×m matrix. Then let

eAt = I + A t +A2

2!t2 + · · ·+ An

n!tn + · · · . (3.25)

The first thing to consider is convergence. Let pij,k denote the i,jth entryof Ak. Since A0 = I, and A1 = A,

pij,0 =

{1 if i = j0 if i 6= j ,

pij,1 = aij, and so on. Equation (3.25) defines eAt to be the m × m matrixwhose i,jth entry is

pij(t) =∞

∑k=0

1k!

pij,k tk, (3.26)

provided that this series is convergent for each i, j.

Theorem 3.12 For any m × m matrix A, the series (3.26) converges for all i, jand for all values of t.

PROOF. We will show that for any i, j,

∑k=0

∣∣∣∣ 1k!

pij,k tk∣∣∣∣ (3.27)

is convergent. Choose a number B such that for each entry aij of A, |aij| ≤ B.We’ll see that

|pij,k| ≤ mk−1Bk for k ≥ 1. (3.28)

The proof of (3.28) is by mathematical induction. If k = 1, (3.28) holdsbecause |pij,1| = |aij| ≤ B. Let k ≥ 2, and assume that (3.28) holds for thepower k− 1; that is

|pij,k−1| ≤ mk−2Bk−1.

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 279

Since Ak = A · Ak−1,

pij,k = ai1 p1j,k−1 + ai2 p2j,k−1 + · · ·+ aim pmj,k−1.

Starting with the triangle inequality, we have

|pij,k| ≤ |ai1||p1j,k−1|+ |ai2||p2j,k−1|+ · · ·+ |aim||pmj,k−1|≤ B ·mk−2Bk−1 + B ·mk−2Bk−1 + · · ·+ B ·mk−2Bk−1︸ ︷︷ ︸

m terms= mk−1Bk.

Therefore, if (3.28) holds for the power k− 1, then it holds for the kth poweras well. By the principle of mathematical induction, (3.28) is valid for allk ≥ 1. Then

1k!|pij,ktk| ≤ mk−1Bk

k!|tk|,

which implies that the series (3.27) is dominated by the series

∑k=0

mk−1Bk

k!|t|k.

The latter series converges since it is the Maclaurin series expansion off (t) = 1

m emB|t|. By the comparison test, the series (3.27) also converges.This means that the series (3.26) defining pij converges absolutely. Sinceevery absolutely convergent series is convergent, the proof is complete.

As a simple example, suppose that Z denotes the zero matrix (all entriesequal 0). Then the only nonzero term in the series

I + Zt +12

Z2t2 + · · ·

defining eZt is I. It follows that eZt = I. The next example is also fairlysimple, because the series has only a finite number of nonzero terms.

Example 3.5.1 Let

A =

0 1 00 0 10 0 0

.

Calculate eAt.

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280 CHAPTER 3. LINEAR SYSTEMS

SOLUTION. By matrix multiplication,

A2 =

0 0 10 0 00 0 0

,

and Ak = 0 for k ≥ 3. Hence

eAt = I + At +12(A)2t2

=

1 t 12 t2

0 1 t0 0 1

Example 3.5.2 Calculate eAt, where

A =

[1 00 2

].

SOLUTION. If

B =

[a 00 b

]and C =

[c 00 d

]are diagonal matrices, you can readily verify that

BC =

[ac 00 bd

].

Thus, the product of two diagonal matrices is also a diagonal matrix. Thediagonal entries of the product matrix are obtained simply by multiplyingthe respective diagonal entries of B and C. It follows that we can find thenth power of a diagonal matrix by raising each diagonal entry to the nthpower:

An =

[1n 00 2n

].

Thus

eAt =∞

∑k=0

1k!

Aktk

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 281

=

[∑∞

k=0tk

k! 0

0 ∑∞k=0

(2t)k

k!

]

=

[et 00 e2t

].

The matrix exponential shares many properties with the ordinary expo-nential function, but there are limitations. The proof that er+s = eres doesnot emphasize the commutative law, rs = sr, but it uses that fact. Multipli-cation of square matrices is not always commutative, so the matrix versionof that identity is as follows.

Theorem 3.13 Let A and B be n× n matrices that commute: AB = BA. Then

e(A+B)t = eAteBt.

The proof is an exercise.

Corollary 3.5.1 If A is a square matrix, then for all t ∈ R, the matrix eAt isnonsingular.

PROOF: The matrices A and−A commute; therefore eAte−At = e(A−A)t =eZt, where Z is the zero matrix. Since eZt = I it follows that e−At is the in-verse matrix of eAt. A matrix is nonsingular iff it is invertible; hence eAt isnonsingular.

Theorem 3.14 Let A be a square matrix whose entries are constants. Then eAt isa fundamental matrix solution of ~x′ = A~x.

The proof depends on the chain rule applied to matrix polynomials. Iff (x) = c0 + c1x + c2x2 + · · · ckxk is an ordinary polynomial, f (At) denotesthe matrix

f (At) = c0 I + c1 A t + c2 A2 t2 + · · ·+ ck Aktk.

Lemma 3.5.2 Let f ′(x) denote the derivative of f with respect to x of a polynomialf (x). If A is a square matrix with constant entries, then

d[ f (At)]/dt = A · f ′(At).

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282 CHAPTER 3. LINEAR SYSTEMS

PROOF. We only need to consider the special case where f (x) = c xk isa monomial, because a general polynomial is a sum of monomials. f ′(x) =c k xk−1, so

A · f ′(At) = A · (c k Ak−1 tk−1) = c k Ak tk−1.

On the other hand,

d[ f (At)]/dt =ddt(c Ak tk) = c Ak (k tk−1).

Since these are equal, the proof is complete

PROOF OF THEOREM 3.14. By Corollary 3.5.1 eAt is nonsingular, so we onlyneed to show that deAt/dt = AeAt. Let Fm(At) denote the partial sum

Fm(At) =m

∑k=0

1k!(At)k.

By Lemma 3.5.2, dFm(At)/dt = A · F′m(At). Since

F′m(At) =m

∑k=0

1k!

k(At)k−1

= AFm−1(At),

it follows that limm→∞ F′m(At) = limm→∞ Fm−1(At) = eAt.It is the definition of the sum of an infinite series that

limm→∞

Fm(At) =∞

∑k=0

1k!(At)k = eAt.

Hencelim

m→∞d[Fm(At)]/dt = A · eAt,

and it might appear that the proof is complete. However, it still should bechecked that

limm→∞

(ddt[Fm(At)]

)=

ddt[ limm→∞

Fm(t)];

that is, that the limit of the derivatives is the derivative of the limit. Theproof of this fact is omitted.

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 283

Calculation

The calculation of eAt, where A is a constant square matrix, involves devel-opments that are found in a full course in linear algebra. While all of thenecessary ingredients are available in any linear algebra text, it is not un-usual for 14-week linear algebra courses to omit some of this material. Inthis text, the procedure for calculating eAt, where A is an arbitrary constantmatrix with real or complex entries, will be described in detail, but we willdepend on linear algebra texts for proofs.

There is a simple algorithm to calculate eAt when A is a 2× 2 matrix.Although the algorithm does not extend to n > 2, it is a good warm up forthe more general case.

Suppose that B is a 2× 2 matrix with the special property that

tr B = 0.

The characteristic equation of B is then s2 + d = 0, where d = det B. Hereis the key formula:

B2 = −dI (3.29)

The proof is just a reference to the Cayley-Hamilton theorem!It follows that B2k = (−d)k I, and B2k+1 = B B2k = (−d)kB. Now that

we know the powers of B we can figure out the exponential:

eBt =∞

∑k=0

1(2k)!

B2kt2k +∞

∑k=0

1(2k + 1)!

B2k+1t2k+1

=∞

∑k=0

(−d)k

(2k)!t2k I +

∑k=0

(−d)k

(2k + 1)!t2k+1B (3.30)

We can simplify this formula if we let |d| = ω2. If d < 0, so that −d =ω2 then (3.30) becomes

eBt =∞

∑k=0

(ωt)2k

(2k)!I +

∑k=0

(ωt)2k+1

(2k + 1)!B

= cosh(ωt)I +1ω

sinh(ωt)B (3.31)

As an exercise, you should prove that when d > 0,

eBt = cos(ωt)I +1ω

sin(ωt)B (3.32)

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284 CHAPTER 3. LINEAR SYSTEMS

Now for a matrix A with tr A = s 6= 0, put B = A− s2 I, and note that

tr B = tr A− tr s2 I = s− s

2 (1 + 1) = 0. Furthermore, because the scalar ma-trix s

2 I commutes with any matrix, by Theorem 3.13 we have eAt = est/2eBt.

Example 3.5.3 Find eAt, where (a) A =

[1 24 3

], (b) A =

[2 −24 6

], and

(c) A =

[2 10 2

].

SOLUTION. In each case, we will put B = A− ( 12 tr A)I.

(a) Since tr A = 4 we get B = A− 2I =[−1 24 1

]. Thus d = det B = −9,

and ω =√| − 9| = 3. Hence

eBt = cosh(3t)I +13

sinh(3t)B =

[cosh(3t)− 1

3 sinh(3t) 23 sinh(3t)

43 sinh(3t) cosh(3t) + 1

3 sinh(3t)

],

and

eAt = e2tB = e2t[

cosh(3t)− 13 sinh(3t) 2

3 sinh(3t)43 sinh(3t) cosh(3t) + 1

3 sinh(3t)

]=

[ 13 e5t + 2

3 e−t 13 (e

5t − e−t)23 (e

5t − e−t) 23 e5t + 1

3 e−t

].

(b) Since tr A = 8 we get B = A− 4I =[−2 −24 2

], and d = det B = 4.

Thus

eBt = cos(2t)I +12

sin(2t)B =

[cos(2t)− sin(2t) − sin(2t)

2 sin(2t) cos(2t) + sin(2t)

],

and

eAt = e4teBt = e4t[

cos(2t)− sin(2t) − sin(2t)2 sin(2t) cos(2t) + sin(2t)

].

(c) Here tr A = 4 so B = A− 2I =[

0 10 0

]. Because d = 0, we can’t

plug into (3.31) or (3.32). However, we see that B2 = 0 so

eBt = I + tB =

[1 t0 1

]It follows that eAt = e2teBt =

[e2t te2t

0 e2t

].

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 285

Inhomogeneous equations.

When applying the variation of constants formula to a system

~v′ = A~v + ~f (t), (3.33)

where A is an n× n constant matrix, and ~v(t) and ~f (t) are n-dimensionalvector functions, we can use the matrix exponential eAt as the fundamentalmatrix solution of the associated homogeneous system, ~v′ = A~v.

Thus, set ~v(t) = eAt~z(t). By the product rule for differentiation,

~v′(t) = A eAt~z(t) + eAt~z′(t)= A~v(t) + eAt~z′(t).

Substituting this expression for ~v′(t) in (3.33),

A~v(t) + eAt~z′(t) = A~v(t) + ~f (t)

and hence eAt~z′(t) = ~f (t). The inverse matrix of eAt is simply e−At, so

~z′(t) = e−At~f (t).

Any antiderivative of e−At~f (t) can be used as~z(t); we will take

~z(t) =∫ t

0e−As~f (s) ds.

Then ~v(t) = eAt~z(t), or

~v(t) = eAt∫ t

0e−As~f (s) ds

Since eAt does not depend on the variable s of integration,

~v(t) =∫ t

0eAte−As~f (s) ds.

Noting that At and As are commuting matrices, we can combine exponen-tials to obtain a result:

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286 CHAPTER 3. LINEAR SYSTEMS

Variation of Constants FormulaSystems with constant coefficients

The general solution of

~v′ = A~v + ~f (t),

where ~v and ~f are n× 1 matrices and A is an n× n constantmatrix, is

~v(t) =∫ t

0eA(t−s)~f (s) ds + eAt~c.

Example 3.4.3 was to illustrate the variation of constants method forfinding a particular solution of an inhomogeneous system. We will returnto the system of that example, to see how the variation of constants formulaworks.

Example 3.5.4 Use the variation of constants formula to find a particular solu-tion of ~v′ = A~v + ~f (t), where

A =

[0 11 0

]and ~f = e−t

[−11

].

SOLUTION. The first step is to calculate eAt. Since tr (A) = 0 anddet(A) = −1, the characteristic equation of A is s2 − 1 = 0, and therefore,by the Cayley-Hamilton theorem, A2 − I = 0. Even powers of A are thusequal to I, and odd powers of A are equal to A. Thus

eAt =

(∞

∑k=0

1(2k)t2k!

)I +

(∞

∑k=0

1(2k + 1)!

t2k+1

)A

= (cosh t)I + (sinh t)A

=

(cosh t sinh tsinh t cosh t

)We can now apply the variation of constants formula. A particular solutionof ~v′ = A~v + ~f (t), will be given by

~v(t) =∫ t

0

[cosh(t− s) sinh(t− s)sinh(t− s) cosh(t− s)

] [−11

]e−s ds

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 287

=∫ t

0

[−es−t−s

es−t−s

]ds

= s[−e−t

e−t

]∣∣∣∣ts=0

= te−t[−11

].

Exercises

1. Show that if A =

0 1 0 00 0 1 00 0 0 10 0 0 0

, then A4 = 0. Hence calculate

eAt.Answer

2. Show that exp[

t 10 t

]=

[et et

0 et

].

3. Prove theorem 3.13. Hint: review the proof that es+t = eset, basedon the formula et = ∑∞

k=01k! t

k. Where is the commutative law used?Answer

4. Let A(t) be a square matrix whose entries are functions of t, andlet B(t) = dA/dt. Show that if A(t) and B(t) commute, then

eA(t) =∞

∑k=0

1k![A(t)]k

is a fundamental matrix solution of the system ~x′ = B(t)~x.

5. Show that if A(t) =

[t 10 2t

], then A and dA/dt do not com-

mute.Answer

6. Let A =

0 1 −10 0 20 0 0

, and B =

1 1 00 1 20 0 1

. Show that eA =

B.

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288 CHAPTER 3. LINEAR SYSTEMS

7. Compute all powers of

A =

0 a b0 0 c0 0 0

,

and hence compute

(a) eAt.

(b) e(A+λI)t, where λ is a scalar constant.

Use the result of part (b) to solve the following IVP:

x′ = −2x + 3y + z; x(0) = 3y′ = −2y− 4z; y(0) = 0z′ = −2z; z(0) = 1

Answer

8. Let K =

[0 −11 0

]. Determine all powers of K and hence com-

pute

(a) eKt.

(b) e(K+λI)t, where λ is a scalar constant.

Use the result of part (b) to solve the following IVP:

x′ = −x− y; x(0) = 1y′ = x− y; y(0) = 0

}

In Exercises 9 – 15, calculate the matrix eAt.

9. A =

[1 1−2 −1

].

Answer

10. A =

[0 42 2

].

11. A =

[−1 2−3 3

].

Answer

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3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST 289

12. A =

[−3 4−4 −1

].

13. A =

[−3 −2

2 1

].

Answer

14. A =

[2 −54 −2

].

15. A =

0 1 20 0 10 0 0

.

Answer

In Exercises 16– 19, find the general solution.

16.

x′ = x− 4y + 2e−t

y′ = 2x− 3y + e−t

17.

x′ = x + et

y′ = x− y + cosh t

Answer

18.

x′ = x− y + t−1

y′ = −x + y + t−1

19.

x′ = 2x− y + e2t

y′ = 4x− 2y + 2e2t

Answer

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290 CHAPTER 3. LINEAR SYSTEMS

The following two Exercises are intended to be challenging.

20. Let A(t) be an n× n matrix whose entries are functions of t. Showthat if the matrices A(t) and A(u) commute for all t and u, then A(t)and A′(t) = dA/dt commute. Hence find a fundamental matrix so-lution for ~x′ = A′(t)~x.

21. Let P be the vector space of all polynomials, and let D : P −→ Pdenote the differentiation operator. Show that if f (x) ∈ P , then

(etD f )(x) = f (x + t)

Answer

3.6 Exponentiating a Matrix

Every CAS has the matrix exponential function: if a square matrix M isentered, one issues the command Exp(M) and the CAS returns eM. To per-form manual calculations, there are two steps. First, we’ll develop a set ofmatrices that we can handle easily. We will call these matrices E-T-E (standsfor Easy-To-Exponentiate). That done, we will see how to adapt any n× nmatrix to fit our scheme.

Here is our first E-T-E:

Definition: A matrix N is nilpotent if there is an integer p > 0 such thatNp = 0.

We have seen some nilpotent matrices already (Examples 3.5.1 and 3.5.3(c)).If Np = 0 then

eNt = I + t N +12

t2 N2 + · · ·+ 1(p− 1)!

tp−1Np−1.

There is a particular sequence of nilpotent matrices of interest:

N2 =

(0 10 0

)

N3 =

0 1 00 0 10 0 0

...

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3.6. EXPONENTIATING A MATRIX 291

Np =

0 1 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 10 0 0 · · · 0

.

Lemma 3.6.1 If k < p the matrix Nkp can be obtained from Ip−k by adding k

columns of zeros on the left, and k rows of zeros on the bottom. For k ≥ p,Nk

p = Zp.

Lemma 3.6.1 leads to a formula for eNpt:

eNpt = I + t Np +12

t2 (Np)2 + · · ·+ 1

(p− 1)!tp−1(Np)

p−1

=

1 t 1

2 t2 · · · 1(p−1)! t

p−1

0 1 t · · · 1(p−2)! t

p−2

......

.... . .

...0 0 0 · · · t0 0 0 · · · 1

. (3.34)

The proofs of lemma 3.6.1 and of formula (3.34) are asked for in Exer-cise 3 at the end of this section.

We can build more E-T-E’s by adding scalar matrices and nilpotent ma-trices. For example, if A = kIp + Np then—because the matrices kIp and Npcommute—Theorem 3.13 tells us that

eAt = ekIpt eNpt = ekteNp

By (3.34), then

e(k Ip+Np)t = ekt

1 t 1

2 t2 · · · 1(p−1)! t

p−1

0 1 t · · · 1(p−2)! t

p−2

......

.... . .

...0 0 0 · · · t0 0 0 · · · 1

. (3.35)

Example 3.6.1 Display the matrix A = −3I4 + N4 and calculate eAt.

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292 CHAPTER 3. LINEAR SYSTEMS

SOLUTION. −3I4 + N4 =

−3 0 0 00 −3 0 00 0 −3 00 0 0 −3

+

0 1 0 00 0 1 00 0 0 10 0 0 0

=

−3 1 0 00 −3 1 00 0 −3 10 0 0 −3

. By formula (3.35),

e(−3I4+N4)t = e−3t(

I4 + tN4 +12

t2N24 +

13!

t3N3k

)

= e−3t

1 t 1

2 t2 16 t3

0 1 t 12 t2

0 0 1 t0 0 0 1

Diagonal matrices. In a square matrix A, with entries denoted aij, the di-agonal entries are the entries aii. These run down the diagonal extendingfrom the upper left corner to the lower right corner. The off-diagonal en-tries are the aij with i 6= j. If all of the off-diagonal entries of a matrix arezeros, the matrix is called a diagonal matrix. Examples include scalar ma-trices, where all of the diagonal entries are equal. The diagonal matrix withdiagonal entries d1, d2, . . . , dn is denoted diag (d1, d2, . . . dn). Unlike scalarmatrices, an arbitrary diagonal matrix does not commute with all othermatrices, although it does commute with all other diagonal matrices.

Diagonal matrices qualify as E-T-E’s. Following Example 3.5.2, you canderive this formula:

ediag (d1,d2,...,dn)t = diag (ed1t, ed2t, . . . , ednt). (3.36)

Direct sums To build still more E-T-E’s, we will use a new matrix opera-tion.

Let P be a p× p matrix, and let Q be a q× q matrix, and let n = p + q.The direct sum of these matrices is an n × n matrix P ⊕ Q that looks likethis: [

P 0p×q0q×p Q

],

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3.6. EXPONENTIATING A MATRIX 293

where 0p×q denotes a p × q matrix of zeros. Note that if p = q = 1 thenP⊕ Q is just a diagonal 2× 2 matrix. Similarly, an n× n diagonal matrixcan be expressed as the direct sum of n 1 × 1 matrices. Here is anotherexample:

[9]⊕[

3 −12 1

]⊕ (−I2 + N2) =

9 0 0 0 00 3 −1 0 00 2 1 0 00 0 0 −1 10 0 0 0 −1

(3.37)

In this example we have used the fact, which you can prove, that the ⊕operation is associative.

Direct sums are easily added and multiplied, provided that the sum-mands are of the same size. Let P and P′ be p× p matrices, and Q, Q′ beq× q matrices. Then (see exercise 5)

(P⊕Q)(P′⊕Q′) = PP′⊕QQ′ and (P⊕Q)+ (P′⊕Q′) = (P+ P′)⊕ (Q+Q′).(3.38)

We can use formula (3.38) to raise direct sums to powers, (P⊕Q)n = Pn ⊕Qn, and to take exponentials:

e(P⊕Q)t = ePt ⊕ eQt. (3.39)

Formula (3.39) provides a way to find the exponential of direct sums ofE-T-E’s; thus these too are E-T-E’s.

Example 3.6.2 Find eAt where A is the matrix in equation (3.37).

SOLUTION. Let U = [9], V =

[3 −12 1

], and W =

[−1 10 −1

], so that

A = U ⊕V ⊕W. Then eAt = eUt ⊕ eVt ⊕ eWt.First, eUt = [e9t]. The other two matrices are 2× 2. Noting that tr V = 4, we

have V = 2I + B, where B =

[1 −12 −1

]. Since tr B = 0 and det B = 1, the

Cayley-Hamilton Theorem implies B2 = −I. Hence B2n = (−1)n I andB2n+1 = (−1)nB. Reasoning as we have before, we geteBt = (cos t)I2 + (sin t)B, and

eVt = e2teBt = e2t[

cos t + sin t − sin t2 sin t cos t− sin t

]=

[e2t(cos t + sin t) −e2t sin t

2e2t sin t e2t(cos t− sin t)

]

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294 CHAPTER 3. LINEAR SYSTEMS

Finally, W = −I + N2 so

eWt = e−t(I + tN2) =

[e−t te−t

0 e−t

].

It follows that

eAt =

e9t 0 0 0 00 e2t(cos t + sin t) −e2t sin t 0 00 2e2t sin t e2t(cos t− sin t) 0 00 0 0 e−t te−t

0 0 0 0 e−t

.

Matrix conjugation

Proposition 3.6.2 If B = P−1AP, where A, B, and P are n× n matrices and Pis invertible, then

eBt = P−1eAtP.

PROOF: Notice that for any power p,

Bp = (P−1AP)(P−1AP) · · · (P−1AP)︸ ︷︷ ︸p factors

= P−1ApP

Therefore

eBt = I + Bt +12

B2t2 + · · ·+ 1m!

Bmtm + · · ·

= I + P−1APt +12

P−1A2Pt2 + · · ·

= P−1(

I + At +12

A2t2 + · · ·)

P

= P−1eAtP

Definition: Let A and P be n × n matrices, P being invertible. Thematrix B = P−1AP is called the conjugate of A by P.

Definition. An n× n matrix A is semisimple if A has n linearly inde-pendent eigenvectors.

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3.6. EXPONENTIATING A MATRIX 295

Proposition 3.6.3 Let A be a semisimple n × n matrix. Then A is conjugateto a diagonal matrix D = diag (s1, s2, . . . , sn), where the s1, s2, . . . , sn are theeigenvalues of A (multiple eigenvalues are repeated in the list).

There is a converse to proposition 3.6.3.Proof: Let ~b1,~b2, . . . ,~bn be n linearly independent eigenvectors of A,

with eigenvalues s1, s2, . . . , sn. We will take P to be the matrix whose columnsare the vectors~bi. Because the columns of P are linearly independent, P isnonsingular, and thus invertible. Set Q = P−1.

The columns of AP are the vectors A~b1, . . . A~bn. Since the vectors~bi areeigenvectors, the ith column of AP is si~bi.

Now consider the matrix PD. When any matrix M on the right by adiagonal matrix, the ith column of M is multiplied by the ith diagonal en-try. Thus the columns of PD are also s1~b1, s2~b2, . . . , sn~bn, and we have theequality

AP = PD, and thus AQ−1 = Q−1D.

Multiply on the right by Q to obtain

A = Q−1DQ.

Proposition 3.6.3 shows how our program is to work: since all diagonalmatrices are E-T-E’s and all semisimple matrices are conjugate to diagonalmatrices, we have a method to compute eAt—and thus to find a fundamen-tal matrix solution of ~v′ = A~v—for any semisimple constant matrix A.

Example 3.6.3 Compute eAt where A =

2 −3 −31 −6 −3−1 9 4

. The characteris-

tic polynomial of A is t3 − t.

SOLUTION. The eigenvalues are the roots of the characteristic polynomial,−1, 0, 1. The eigenvectors satisfy equations (A− si I)~bi =~0. Thus~b1 isdetermined by

(A− (−1)I)~b1 =

3 −3 −31 −5 −3−1 9 5

xyz

=

000

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296 CHAPTER 3. LINEAR SYSTEMS

The matrix A + I is row equivalent to

2 0 −10 2 10 0 0

. We can therefore

take~b1 =

1−12

. By similar means, eigenvectors belonging to 0 and 1

are determined to be~b2 =

3−13

and~b3 =

301

. Thus

P−1AP = diag (−1, 0, 1), where P is the matrix with columns~b1,~b2,~b3.Exponentiating and referring to proposition 3.6.2 yields

P−1eAtP = diag (e−t, 1, et)

Multiplying on the left by P and on the right by P−1 we have

eAt = P (e−t, 1, et)P−1

=

1 3 3−1 0 −12 1 3

e−t 0 00 1 00 0 et

1 −6 −31 −3 −2−1 5 3

=

e−t + 3− 3et −6e−t − 9 + 15et −3e−t − 6 + 9et

−e−t + et 6e−t − 5et 3e−t − 3et

2e−t + 1− et −12e−t − 3 + 15et −6e−t − 2 + 9et

If a matrix is not semisimple, there is a theorem from linear algebra thatwe can use. It refers to the following definition:

Definition: An elementary Jordan matrix is a an m×m matrix E(s, m) =sIm + Nm, where Nm is the standard m × m nilpotent matrix. (Note thatE(s, 1) = [s], because N1 = [0].)

We have encountered elementary Jordan matrices already; Equation (3.35)shows that they are E-T-E’s. See also Example 3.6.1. The following theoremincludes the result of Proposition 3.6.3 as the special case when each mi = 1.It can be paraphrased as “Every square matrix is conjugate to a direct sumof elementary Jordan matrices.” We have seen (equation (3.39)) that a ma-trix that can be expressed as a direct sum of E-T-E’s is itself an E-T-E; henceevery square matrix is conjugate to an E-T-E.

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3.6. EXPONENTIATING A MATRIX 297

Theorem 3.15 Let A be an n× n matrix. There exist elementary Jordan matricesEi = E(si, mi) (i = 1, . . . , p), such that A is conjugate to the direct sum of the Ei.In other words, there exists an invertible n× n matrix P such that

A = P−1

( p⊕i=1

Ei

)P.

Furthermore, the dimensions mi of the matrices Ei satisfy m1 + · · ·+ mp = n andthe si are the eigenvalues of A.

We will omit the proof of Theorem 3.15. It is a corollary of the existenceof a matrix called the Jordan canonical form of the matrix A, which is provedin all worthy linear algebra texts. It is worth reminding you that some ofthe eigenvalues si of A may be complex, even if all entries of A are real.Then the corresponding matrices Ei, and the matrix P, would have somecomplex entries.

The procedure for calculating eAt is to find an E-T-E B that is conjugateto A (theorem 3.15 guarantees that there is one); of course one also needs amatrix P such that P−1AP = B; then A = PBP−1 and by proposition 3.6.2,

eAt = PeBtP−1.

This reduces finding a fundamental matrix solution of a system of linear,constant coefficient ODE’s to a routine problem in linear algebra. Becausethe matrix computations can be strenuous, they are often left to a computerrunning a CAS, or specialized software.

In Example 3.6.4, we will find the exponential of a matrix that is notsemisimple. Before proceeding, we will need to introduce some helpfuldefinitions from linear algebra.

Definitions: Let s be an eigenvalue of a matrix A. The set of all eigenvec-tors of A that belong to s is called the eigenspace of A belonging to s. Thegeneralized eigenspace belonging to s is the set of all vectors~a ∈ Rn suchthat for some m,

(A− sI)m~a =~0

You can verify that the eigenspace, and the generalized eigenspace belong-ing to an eigenvalue of A are subspaces of Rn.

The algebraic multiplicity of an eigenvalue s of a matrix A is the mul-tiplicity of s as a root of the characteristic polynomial of A. Contrast this

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298 CHAPTER 3. LINEAR SYSTEMS

with the geometric multiplicity, which is the dimension of the eigenspacebelonging to s.

If ~a is an eigenvector of A belonging to an eigenvalue of s then (A −sI)1 = ~0; thus ~a is also a generalized eigenvector. It is shown in linearalgebra courses that the generalized eigenspace of an eigenvalue is a sub-space of Rn of dimension equal to its algebraic multiplicity. The set ofeigenvectors belonging to the same eigenvalue also form a subspace of thegeneralized eigenspace. If a matrix is semisimple, the the geometric andalgebraic multiplicities of each eigenvalue are the same. If a matrix is notsemisimple, then for at least one eigenvalue, the algebraic multiplicity isgreater than the geometric multiplicity.

Example 3.6.4 Find eAt, where

A =

−19 22 21−9 11 9−9 10 11

,

given that the characteristic polynomial of A is p(s) = (s + 1)(s− 2)2.

This example could be restated as “Find a fundamental matrix solutionX (t) of the following system such that X (0) = I.”

x′ = −19x + 22y + 21zy′ = −9x + 11y + 9zz′ = −9x + 10y + 11z

SOLUTION. The eigenvalues of A are 2 (double) and −1. The solutions of

(A− 2I)~b =~0 are all multiples of

101

. Therefore the geometric

multiplicity of the eigenvalue 2 is 1, while the algebraic multiplicity of thesame eigenvalue is 2. It follows that A is not semisimple. We calculate

(A− 2I)2 =

54 −54 −5427 −27 −2718 −18 −18

which is row equivalent to the matrix

C =

1 −1 −10 0 00 0 0

.

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3.6. EXPONENTIATING A MATRIX 299

The solutions of (A− 2I)2~a =~0 are the same as the solutions of C~a =~0,and one can easily see that the solutions of the latter equation are the

linear span of

1

10

,

101

. These two vectors will form the first

two columns of the matrix P; the third will be the eigenvector

232

belonging to the eigenvalue −1. Thus

P =

1 1 21 0 30 1 2

,

and by matrix multiplication we find that

P−1AP =

2 0 01 2 00 0 −1

,

which is E-T-E, as the direct sum of 2I2 + N, where N =

(0 01 0

)is

nilpotent, and a 1× 1 matrix. We have e(2I2+N)t = e2teNt = e2t(I + Nt).Thus, P−1eAtP = (e2t(I + Nt)⊕ [e−t]F, and so

eAt = P

e2t 0 0te2t e2t 00 0 e−t

P−1

=13

e2t(−3t− 5) + 2e−t 2(e2t − e−t) e2t(3t + 2)− 2e−t

3(−e2t + e−t) −3e−t 3(e2t − e−t)e2t(−3t− 2) + 2e−2t 2(e2t − e−t) e2t(3t− 1)− 2e−t

Clearly, A is not an E-T-E.

Exercises

1. Calculate eAt, where A =

2 1 0−1 0 0

0 0 −1

.

Answer

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300 CHAPTER 3. LINEAR SYSTEMS

2. Let M be a p× n matrix, and let Np be the standard p× p nilpotentmatrix. Then M′ = Np M is also a p× n matrix. Show that the bottomrow of M′ is a zero row, and for i < p the ith row of M′ is equal to the(i + 1)st row of M.

3. Prove lemma 3.6.1, and formula (3.34).Answer

4. Prove that the direct sum operation of matrices, (P, Q) 7→ P⊕ Q,is associative. That is, show that for any 3 matrices P, Q, and R, if weput A = P⊕Q and B = Q⊕ R then A⊕ R = P⊕ B.

5. Use the definitions of matrix addition and multiplication to verifythe formula (3.38) for the product of direct sums of matrices. Thenprove formula 3.39) for the matrix exponential of direct sums of ma-trices.Answer

6. Let the n × n matrix A be conjugate to a diagonal matrix D; inother words, we are assuming that there exists an invertible matrix Psuch that P−1AP = D. Prove that A is semisimple.

7. If B a conjugate of A by some matrix P, write B ∼ A. Prove that∼ is an equivalence relation: you need to prove the following:

• A ∼ A (reflexive law),

• A ∼ B⇒ B ∼ A (symmetric law), and

• (A ∼ B & B ∼ C)⇒ A ∼ C (transitive law).

Answer

In Exercises 8–13, calculate eAt. As a practical matter, you will prob-ably want to use a CAS for routine matrix calculations. Please limityour automated calculations to the following operations:

• Matrix multiplication,

• Matrix inversion,

• Row reduction.

8. A =

−2 1 0 01 −2 0 00 0 −5 40 0 −2 1

.

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3.6. EXPONENTIATING A MATRIX 301

9. A =

−2 0 0 00 0 −4 00 1 0 00 0 0 1

.

Answer

10. A =

−2 −4 −82 2 20 1 3

. The characteristic polynomial of A is

f (s) = s3 − 3s2 + 2s.

11. A =

−1 1 21 −1 −2−2 2 4

. The characteristic polynomial of A is

f (s) = s3 − 2s2.Answer

12. A =

7 −4 −111 −6 −26 −4 0

. The characteristic polynomial of A is

f (s) = s3 − s2.

13. A =

2 −8 −2 52 −4 −6 62 −2 −10 82 −2 −8 6

. The characteristic polynomial of A is

f (s) = (s2 + 2s + 2)(s + 2)2.Answer

14. Use the result of Exercise 8 to determine the solution of the IVP

x′ = −2x + y, x(0) = 1y′ = x− 2y, y(0) = −1

z′ = −5z + 4w, z(0) = 0w′ = −2z + w, w(0) = 3

15. Use the result of Exercise 11 and the method of variation of con-stants to determine the solution of the IVP

x′ = −x + y + 2z + 1, x(0) = 0y′ = x− y− 2z− 1, y(0) = 0

z′ = −2x + 2y + 4y + 1 z(0) = 0

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302 CHAPTER 3. LINEAR SYSTEMS

Answer

3.7 Chapter Glossary

eAt The matrix exponential.

I The identity matrix.

Im(z) The imaginary part of z = x + iy: Im(x + iy) = y.

Re(z) The real part of z = x + iy: Re(x + iy) = x.

Associated homogeneous system (of a system of ODEs, v′ = A(t)v +b(t)) The system obtained by removing the source term: v′ = A(t)v.

Absolute value The distance of a complex number z = x + iy to 0. Theformula is |z| =

√x2 + y2

Argument (of a complex number z) The angle made by a line connecting zto the origin with the positive real axis in the complex plane.

arg(z) =

cos−1(

Re(z)|z|

)if Im(z) ≥ 0

− cos−1(

Re(z)|z|

)if Im(z) < 0

Belongs to Corresponding characteristic roots and characteristic vectors ofa matrix are said to belong to each other.

Cayley-Hamilton theorem Let f (s) = 0 be the characteristic equation of amatrix A. Then f (A) is equal to the zero matrix.

Characteristic equation A polynomial equation used to determine the char-acteristic roots of a matrix A. For an n× n matrix A, this is

det(A− sI) = 0.

For a 2× 2 matrix

A =

[a cb d

],

the characteristic equation is s2 − p s + q = 0, where p = trace (A) =a + d, and q = det(A).

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3.7. CHAPTER GLOSSARY 303

Coefficient matrix The matrix A(t) in the system

v′ = A(t)v + b(t).

Column vector A matrix that has only one column. We use boldface lettersfor column vectors.

Complex plane The ordinary xy-plane, in which the point (x, y) is madeto correspond the complex number z = x + iy.

Determinant A scalar-valued function of square matrices. For a 2× 2 ma-trix

A =

[a cb d

],

det(A) = ad − bc. Determinants are defined for n × n matrices inlinear algebra courses.

Eigenvalue (of a matrix A) A scalar s such that for some characteristic vec-tor b, Ab = sb.

Eigenvector (of a matrix A). A nonzero vector b such that Ab = sb forsome scalar s.

Entry A scalar that appears at a specified location in a matrix.

Exponential (of a matrix A) A matrix function, denoted eAt, that is the sumof the power series

eAt = I + At +12

A2t2 + · · ·+ 1n!

Antn + · · · .

If A is a constant matrix, eAt is the fundamental matrix solution ofv′ = Av having initial value X (0) = I.

Fundamental matrix solution A matrix solutionX (t) that is a nonsingularsquare matrix.

Generalized eigenspace belonging to an eigenvalue of a matrix A. The setof vectors~a with the property that (A− sL)p~a =~0 for some p.

Homogeneous system A system of linear ODEs that is satisfied by the vec-tor function v(t) ≡ 0. A homogeneous system always has a zerosource vector.

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304 CHAPTER 3. LINEAR SYSTEMS

Imaginary axis The vertical axis of the complex plane, which correspondsto complex numbers z = 0 + iy.

Imaginary part (of a complex number z = x + iy) Im(z) = y.

Identity matrix The matrix

I =

1 0 0 · · · 00 1 0 · · · 00 0 1 · · · 0

0 0 0. . . 0

0 0 0 · · · 1

.

I has the property that for all b ∈ Rn, Ib = b.

Inverse matrix (of a square matrix A) A matrix B such that AB = BA = I.

Matrix A rectangular array of scalars.

Matrix product The result of multiplying an m× n matrix A with entriesaij and a p× q matrix B with entries bkl is defined if and only if n = p;in this case it is an m× q matrix C whose entries are

crs = ar1b1s + ar2b2s + · · ·+ arnbns.

Matrix solution (of a homogeneous system of ODEs, v′ = A(t)v) A matrixfunction X (t) such that dX (t)/dt = A(t)X (t).

Nonsingular A matrix that is not singular: A is nonsingular if and only ifAb 6= 0 for all b 6= 0.

Nilpotent matrix A square matrix N with the property that some powerNp is equal to the zero matrix.

Real axis The horizontal axis of the complex plane, which corresponds tonumbers z = x + 0i.

Real part (of a complex number z = x + iy) Re(z) = x.

Scalar A real or complex number, or a real or complex-valued expression.We use lower case italic letters for scalars.

Singular A matrix A such that Ab = 0 for some nonzero vector b.

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3.7. CHAPTER GLOSSARY 305

Source vector The vector function b(t) in the system

v′ = A(t)v + b(t).

Trace The sum of the diagonal entries of a square matrix.

Variation of constants A method for finding a particular solution of an in-homogeneous system. Given a fundamental matrix solution X ofv′ = Av, substitute v = Xw in

v′ = Av + f(t),

where w is to be determined. The resulting equation is readily sim-plified and integrated to find w. If A is a constant matrix then we cantake X(t) = eAt. Then the following formula for the solution results:

v(t) =∫ t

0eA(t−s) f (s) ds + eAtc.

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306 CHAPTER 3. LINEAR SYSTEMS

3.8 Review Exercises

1. Solve the initial value problem

x′ = 2x + y; x(0) = 1y′ = x + 2y; y(0) = 0.

Answer

2. Find a fundamental matrix solution for each differential equation.

(a){

x′ = x− 2yy′ = 5x− y

(b){

x′ = x− yy′ = x + y

(c)[

xy

]′=

[1 20 1

] [xy

].

(d) ~x′ = A~x, where

A =

1 2 30 2 30 0 3

.

Answer

3. Find the general solution of

x′ = x + 5y + tan 2ty′ = −x− y,

given that X (t) =

[2 sin 2t− cos 2t sin 2t + 2 cos 2t

cos 2t − sin 2t

]is a funda-

mental matrix solution of the associated homogeneous equation.Answer

4. Suppose A is an n× n matrix with independent characteristic vec-tors ~b1, . . . , ~bn. Denote the corresponding characteristic roots by r1,. . . , rn. Let E(t) denote the diagonal matrix whose diagonal entriesare er1t, . . . , ernt and let B denote the matrix whose columns are thecharacteristic vectors~b1, . . . , ~bn. Show that B · E(t) is a fundamentalmatrix solution of the system ~x′ = A~x. Find an expression for eAt.Answer

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3.8. REVIEW EXERCISES 307

5. Show that a 2× 2 matrix is nilpotent if and only if its trace anddeterminant are both zero.Answer

6. Let A =

(a bc d

)and let B = A− 1

2 tr (A)I. Find a formula for

det(B).Answer

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308 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS

Figure 3.1: The complex plane.

-

6

r r r r r

r r

r r

r r

r r r

−1 + 2i 2i

i 2 + i

−2 −1 0 1 2

−(2 + i) −i 2 + i

−1 + 2i −2i

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FIGURES FOR CHAPTER 4 309

Figure 3.2: The triangle inequality: |z1 + z2| ≤ |z1|+ |z2|

0

z1 z1 + z2

rr r

|z1|

|z2|

|z1 + z2|����

���

���

((((((((

(((((((

(((((((

(((((((

(

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310 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS

Figure 3.3: Polar representation of a complex number. The length of the heavy line to thepoint representing the complex number z = −1 + 2i is r = |z|, and the radian measure ofthe directed arc shown is θ = arg(z).

-1 -0.5 0.5 1 1.5 2

0.5

1

1.5

2z

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Chapter 4

Stability Theory

311

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312 CHAPTER 4. STABILITY THEORY

4.1 Phase Portraits of Linear Systems

Let A be a 2× 2 constant matrix. We will see how to draw a phase portraitthe system of ODEs,

~v′ = A~v. (4.1)

All linear systems have a stationary point at the origin, since~0′ = ~0 = A~0.If A is singular then there is a vector~b 6= ~0, such that A~b = ~0. In this case,all scalar multiples of~b are stationary points, and we say that the system isdegenerate. If A is nonsingular, then~0 is the only stationary point, and wesay that (4.1) is nondegenerate.

Nodes

The simplest case is that of a scalar matrix. Let A = rI, where r 6= 0 isconstant. Then eAt = ert I so the general solution of the system (4.1) is ~v =

ert~c, where ~c =

(c1c2

). Thus every solution has the equations x = c1er t,

y = c2er t. Since y/x = c2/c1 is constant, the orbits are half-lines directedtoward the origin (if r < 0) or away from the origin (if r > 0). The phaseportrait, shown as figure 4.1, is called a proper node. We say that the nodeis stable if r < 0, because the solutions approach the origin with increasingtime. If r > 0, the node is unstable.

Now let’s consider the case of a matrix A that has unequal real eigen-values. In this case, A is conjugate to a diagonal matrix

D =

(r 00 s

),

and the phase portrait of (4.1) can be obtained by distorting the phase por-trait of

~w′ = D~w (4.2)

in a very specific way.

To determine the general solution of (4.2) we note that eDt =

(er t 00 es t

),

so ~w = eDt~c =(

c1er t

c2es t

).

To be definite, let’s suppose that r = −2 and s = −1. Orbits wherec2 = 0 follow the x-axis toward the origin, and when c1 = 0 instead, the

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 313

orbit follows the y-axis toward the origin. If c1 and c2 are both nonzero,then the orbit is given by the parametric equations

x = c1 e−2t

y = c2 e−t.

Since (e−t)2 = e−2t we can eliminate t from the parametric equations andobtain

xc1

=

(yc2

)2

,

or x = a y2 where a = c1/c22. The orbits that do not follow either coordi-

nate axis follow parabolas toward the origin. The phase portrait, shownin figure 4.3, displays the orbits on the x and y-axes, and four other or-bits. The reason that not all orbits are straight lines is that as t increases,x = c1 e−2t decreases in magnitude faster than y = c2e−t. This causes theorbits to become squashed against the y-axis, unless c2 = 0. This phase por-trait is called an improper node. It is stable, because all orbits converge tothe origin.

In general, the phase portrait of a linear system of ODEs is called astable node if every orbit is directed toward the origin as t→ ∞. If all orbitsapproach the origin as t → −∞, the phase portrait is an unstable node.Stable and unstable nodes can be recognized by calculating the eigenvaluesof the coefficient matrix.

Proposition 4.1.1 Let A be a 2 × 2 matrix. The phase portrait of the system~v′ = A~v is a stable node if and only if both eigenvalues of A are negative, or arecomplex numbers with negative real parts, and it is an unstable node if and only ifboth eigenvalues are positive, or are complex numbers that have positive real parts.

The proof is left to you; see exercise 29 at the end of this section.The following example is to illustrate the way to draw a phase portrait

of an improper node.

Example 4.1.1 Draw a phase portrait of the system

x′ = 2x− yy′ = −x + 2y.

SOLUTION. The eigenvalues of the coefficient matrix,(2 −1−1 2

),

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314 CHAPTER 4. STABILITY THEORY

are 1 and 3, so the phase portrait is an unstable node. The vectors[

11

]and

[1−1

], respectively, are eigenvectors belonging to these

eigenvalues. The slopes of these eigenvectors are 1 and −1, respectively.The general solution of the system is therefore[

xy

]= c1et

[11

]+ c2e3t

[1−1

](4.3)

It follows that the half lines with slopes ±1, directed away from the origin,are orbits of the system. The other orbits are not straight. As t→ −∞, thesecond term of (4.3) is negligible; the orbit therefore recedes from theorigin with slope 1. As t becomes large, the first term of (4.3) becomesnegligible, so the slope approaches −1, the slope of the other eigenvector.See figure 4.4.

If a 2× 2 matrix A is not a scalar matrix and has a double eigenvaluer 6= 0, then the phase portrait of[

x′

y′

]= A

[xy

](4.4)

is an improper node, stable if r < 0, and unstable if r > 0. Here is anexample:

Example 4.1.2 Draw the phase portrait of

x′ = r xy′ = k x + r y

SOLUTION. The system has a double eigenvalue of r, and all eigenvectorsare vertical. Assuming that r > 0, the phase portrait is an unstable node,with straight orbits emanating from the origin and following the y-axis.The general solution of this system,

x = c ert (4.5)y = (kct + d)ert (4.6)

was derived on page 254 (for k = 1). We can eliminate t by solving (4.5) toget

t =1r

ln(x/c) =1r(ln |x| − ln |c|) (4.7)

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 315

(the absolute value signs are necessary because x may be negative — weonly know that x and c have the same sign). Now divide (4.6) by (4.5) toobtain y/x = (k t + (d/c)), and substitute (4.7) for t. This yields

yx=

kr

ln |x| − kr

ln |c|) + dc

Let b = − kr ln |c|) + d

c . Then

y = x(

kr

ln |x|+ b)

. (4.8)

Each orbit (except for the stationary point at the origin and the orbits onthe positive and negative y-axis) follows the graph of (4.8) for someparticular value of b. The phase portrait can be drawn by sketching graphsof equation (4.8) for various b. Figure 4.5 shows the phase portrait fork = 2 and r = 1.

Saddles

Let A be a 2 × 2 matrix whose eigenvalues r and s have opposite signs:r < 0 < s. Let ~e and ~f be eigenvectors belonging to r and s, respectively.Five orbits of ~v′ = A~v are readily identified: the stationary point at theorigin; the pair of half-lines, directed toward the origin and parallel to ~e;and the two half-lines directed away from the origin and parallel to ~f . Thelines through the origin in the ~e and ~f directions are called the stable lineand the unstable line, respectively.

The remaining nonstationary orbits resemble hyperbolas, with the sta-ble and unstable lines as asymptotes. As t → ∞ the orbits approach theunstable line, and as t→ −∞ they approach the stable line.

Example 4.1.3 Draw a phase portrait of the system

x′ = −xy′ = x + y.

SOLUTION. The eigenvalues of A =

[−1 01 1

]are −1 and 1,

corresponding to eigenvectors~e =[

2−1

]and ~f =

[01

]These vectors

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316 CHAPTER 4. STABILITY THEORY

determine the half-line orbits, which are drawn first. The remaining orbitsare asymptotic to these. Figure 4.6 displays the phase portrait and thevector field.

The phase portrait of a system having eigenvalues of opposite sign iscalled a saddle. To see why, let us find an integral for the system presentedin example 4.1.3. An integral for this system is also an integral for the ODE

(x + y) dx + x dy = 0,

which happens to be in exact form. Knowing this, It is straightforwardto obtain the integral F(x, y) = 1

2 x2 + xy. The orbits follow level curvesF(x, y) = C. The origin is a critical point of F (that is, both partial deriva-tives of F vanish there), and by the second derivative test,

∂2F∂x2

∂2F∂y2 −

(∂2F

∂x∂y

)= (1)(0)− (1)2 < 0,

this critical point is a saddle point of the function F(x, y).You can classify a linear system with constant coefficients as a saddle,

degenerate, or a node with very little effort. The following lemma fromlinear algebra is the key. Although it is stated in terms of on 2× 2 matricesit holds for all square matrices.

Lemma 4.1.2 Let A be an 2× 2 matrix, with eigenvalues r1 and r2. Then thetrace of A is equal to the sum of the eigenvalues:

tr(A) = r1 + r2;

and the determinant is equal to the product of the eigenvalues:

det(A) = r1 · r2.

PROOF. The characteristic polynomial of A can be factored as (s −r1)(s− r2). Multiply this out to obtain

s2 − (r1 + r2)s + r1r2 = s2 − tr(A) s + det(A).

It follows that tr (A) = r1 + r2 and det(A) = r1r2.

By the lemma, the eigenvalues of a 2× 2 matrix A are of opposite signif and only if det(A) < 0, and 0 is a eigenvalue if and only if det(A) = 0.

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 317

Thus, the phase portrait of the system ~v′ = A~v is a saddle if and only ifdet(A) < 0, and it is degenerate if and only if det(A) = 0.

If det(A) > 0, it is possible that A has no real eigenvalues at all. If weassume that A has real eigenvalues, then the eigenvalues must be of thesame sign, and hence the phase portrait is a node. If the eigenvalues arepositive, the node is unstable, and if they are negative, the node is stable.Since the trace of A is equal to the sum of the eigenvalues, it must be of thesame sign as the roots, and thus serves as a stability indicator.

Classifying phase portraits of 2× 2 Linear Systems:Real eigenvalues.

Let A be a 2× 2 matrix with real orbits of~v′ = A~v. The phaseportrait of

~v′ = A~v

is

• degenerate if det(A) = 0,

• A saddle if det(A) < 0,

• A stable node if tr (A) < 0 and det(A) > 0, and

• An unstable node if tr (A) > 0 and det(A) > 0.

Complex eigenvalues

Consider a system ~v′ = A~v, where A is a 2× 2 matrix with real entries andcomplex eigenvalues. We will denote the eigenvalues, which are conjugateto each other, by λ ± iω, and the eigenvectors will be ~h ± i~k. The generalsolution is

~v = eλt[(c1 cos(ωt) + c2 sin(t))~h + (−c1 sin(ωt) + c2 cos(ωt))~k],

where c1 and c2 are constants (see equation 3.18 on page 266). If λ > 0,the exponential factor will cause all solutions to converge to 0 as t → −∞,and the phase portrait will be an unstable node. Similarly, if λ < 0, the

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318 CHAPTER 4. STABILITY THEORY

phase portrait is a stable node. These are called spiral nodes, since thephase portraits turn out to be spirals, expanding outward in the unstablecase and inward in the stable case.

If λ = 0, then~v(t) is periodic (the period is 2π/ω), and hence the orbitsare closed. A phase portrait in which all orbits are closed is called a center.

Referring to lemma 4.1.2, we see that the trace of A is the sum of theeigenvalues:

tr A = (λ + iω) + λ− iω) = 2λ.

Therefore, if A has complex eigenvalues tr A = 0 if and only if the eigenval-ues of A are pure imaginary, and if the eigenvalues are not pure imaginary,their real part has the same sign as tr A.

These observations can be summarized as follows:

Classifying phase portraits of 2× 2 Linear Systems:Complex eigenvalues.

Let A be a 2 × 2 real matrix that has no real eigenvalues.Then the phase portrait of ~v′ = A~v is

• a center if tr A = 0;

• a stable node if tr A < 0; and

• an unstable node if tr A > 0.

Centers

Let

A =

[a bc −a

](4.9)

be a matrix whose trace is 0. An integral for the system

x′ = ax + byy′ = cx− ay

}(4.10)

can be found by solving the differential equation

(cx− ay) dx− (ax + by) dy = 0. (4.11)

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 319

This equation is in exact form:

−a =∂

∂y(cx− ay) =

∂x[−(ax + by)].

Integrating equation (4.11) yields F(x, y) = C, where

F(x, y) =12

c x2 − a xy− 12

b y2.

Recall from analytic geometry that an equation

px2 + qxy + ry2 = C

represents an ellipse if its discriminant d = 4pr − q2 is positive, and a hy-perbola if d < 0. The discriminant of F(x, y) = C is equal to −bc − a2 =det(A). Thus, if det A > 0, the level curves of F(x, y) are ellipses, andif det(A) < 0, the level curves are hyperbolas. Since the orbits of (4.10)must follow the level curves, they are ellipses when det(A) > 0. We wereaware that when det(A) < 0 the phase portrait is a saddle. Now we knowthat when tr A = 0 and det A < 0, the orbits follow hyperbolas and theirasymptotes.

To determine the direction of the orbits of a center, consider a point(0, y0) on the positive y-axis. At that point dx

dt = by0. Thus the orbit will bedirected to the right at (0, y0) if b > 0; this indicates the clockwise direction.Similarly, if b < 0, the direction is counterclockwise. (b = 0 is impossiblebecause then det A = −a2 indicates the phase portrait would be a saddlerather than a center.)

Example 4.1.4 Draw a phase portrait of the system

x′ = 2x + 8yy′ = −5x− 2y.

SOLUTION. The matrix A =

[2 8−5 −2

]has trace 0 and determinant 36.

Therefore,

eAt = (cos 6t)I +16(sin 6t)A =

[cos 6t + 1

3 sin 6t 43 sin 6t

− 56 sin 6t cos 6t− 1

3 sin 6t

].

It follows that the orbits rotate clockwise(since b = 8 > 0) around the originwith angular frequency is 6 radians per unit time.

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320 CHAPTER 4. STABILITY THEORY

The phase portrait is a center, and it can be drawn with the aid of theintegral found by integrating the exact equation

(−5x− 2y) dx− (2x + 8y) dy = 0.

Thus, the orbits are the ellipses

52

x2 + 2xy + 4y2 = c. (4.12)

See figure 4.7.Details related to how the ellipses are drawn are optional.

Spiral nodes

Suppose that the eigenvalues of the system

x′ = ax + byy′ = cx + dy

}(4.13)

are complex numbers λ± iω with λ, ω 6= 0. Let A =

[a bc d

]be the coef-

ficient matrix. Then λ = 12 tr A = 1

2 (a + d) and

ω =12

√−(a + d)2 + 4(ad− bc) =

12

√−(a− d)2 − 4bc

Let B = A− λI. A quick calculation shows tr B = 0 and det B = ω2 Hence

eBt = (cos ωt)I +1ω(sin ωt)B

and eAt = eλteBt.The phase portrait of

~w′ = B~w (4.14)

is a center. Orbits cycle around the origin with angular frequency ω, clock-wise if b > 0 and counterclockwise if b < 0. Again, b = 0 is impossible,for in that case the eigenvalues of A would be a and d, which are real; ourassumption is that the eigenvalues of A are not real.

If we start at some point other than the origin, the orbit of (4.14) followsan ellipse. An orbit of the system (4.13) starting at the same point willrevolve around the origin with the same angular frequency, but will moveoutward (if λ > 0) or inward (if λ < 0). In either case, the orbit willbe a spiral. Thus this phase portrait is called a spiral node. Focus is analternative term used instead of spiral node by some authors.

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 321

Example 4.1.5 Draw a phase portrait for the system

x′ = x− yy′ = x + y.

}(4.15)

SOLUTION. Let A =

[1 −11 1

]be the coefficient matrix of the system.

Noting that tr A = 2, we let B = A− 12 (tr A) =

[0 −11 0

]; then tr B = 0

and det B = 1 soeBt = (cos t)I + (sin t)B,

and

eAt = eteBt = et[

cos t − sin tsin t cos t

]The orbits are oriented counterclockwise and expand outward, with anangular frequency of 1. Thus the orbit that starts at (1,0) will cross thenegative x-axis at (−eπ, 0), then the positive x-axis at (e2π, 0), and so on.These crossing points form a geometric sequences on the positive andnegative x-axis, with ratio e2π ≈ 535. Unless a very large or very detaileddrawing is made, it is unlikely that more than one crossing will be seen.The phase portrait in figure 4.9 shows four orbits of the system.

Example 4.1.6 Draw a phase portrait of the system

x′ = x + 8yy′ = −5x− 3y

}(4.16)

SOLUTION. Again, let A =

[1 8−5 −3

]be the coefficient matrix. We

calculate tr A = −2 and put B = A− 12 (tr A)I =

[2 8−5 −2

]. Then

tr B = 0 and det B = 62, so

eBt = (cos 6t)I +16(sin 6t)B

The system ~w′ = B~w was the subject of example 4.1.6. Its orbits areclockwise ellipses (see figure 4.7), revolving around the origin with angularfrequency 6.

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322 CHAPTER 4. STABILITY THEORY

The orbits of system (4.16) can be obtained by following the elliptical orbitsof ~w′ = B~w and contracting inward at the same time at a rate e−t.The angular frequency of 6 indicates that each orbit, each orbit crossesthe positive x-axis once every 2π/6 = π/3 units of time. The exponentialfactor in the solution is e−t. It follows from this that the crossing points forma geometric sequence with ratio e−π/3 ≈ 0.351. Figure 4.10 displays oneorbit of the system (4.16), and the elliptical orbit of ~w′ = B~w with the samestarting point.

Degenerate systems

If A is a 2× 2 matrix that is singular, then the system~v′ = A~v is degenerate.All points of the phase plane are stationary if A is the zero matrix, and ifA 6= 0 we still know that 0 is an eigenvalue since there must be a nonzerovector~b with

A~b =~0 = 0 ·~b;

thus ~b is an eigenvector belonging to 0. The line through the origin withdirection~b consists of stationary points. Here are two examples of degen-erate systems: in the first, there is a nonzero eigenvalue; and in the second,0 is a double root.

Example 4.1.7 Draw a phase portrait of the system

x′ = −xy′ = 0

SOLUTION. Let A =

[−1 00 0

]be the coefficient matrix. The eigenvalues

are −1 and 0, and~i and~j are corresponding eigenvectors. The stationaryline is therefore the y-axis. The general solution is

~v = c1e−t~i + c2~j =[

c1e−t

c2

]or x = c1e−t, y = c2. The eigenvectors indicate the direction of the orbits, inthis case horizontal, pointed toward the stationary line. See figure 4.11.

Example 4.1.7 is typical of a degenerate system ~v′ = A~v, where A isa matrix whose eigenvalues are 0 and r 6= 0. Let ~e and ~f be eigenvectors

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 323

belonging to 0 and r, respectively. The stationary line has the directionof ~e, and the nonstationary orbits will be half lines parallel to ~f , directedtoward or away from the stationary line, depending on the sign of r. Inour example, the nonstationary orbits were perpendicular to the stationaryline, but in many cases the angle is oblique.

Example 4.1.8 Draw a phase portrait of the system

x′ = 0y′ = −x

}(4.17)

SOLUTION. The coefficient matrix A =

[0 0−1 0

]is nilpotent. Thus

eAt = I + At =[

1 0−t 1

]The solutions of the system (4.17) are therefore

eAt[

xy

]=

[x

y− tx

]

Thus all points[

0y

]on they-axis are stationary. Orbits not on the y-axis

follow vertical lines with speed |x|, directed downward if x > 0 and upwardif x < 0. See figure 4.12.

Notice that in this example, the nonstationary orbits are lines parallelto the stationary line, with opposite directions on either side of the station-ary line, and with speed → 0 as we approach the stationary line. This istypical of degenerate systems of two linear equations when 0 is a doubleeigenvalue. The only special property of the system in example 4.1.8 is thatthe stationary line is vertical — it could have been any direction.

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324 CHAPTER 4. STABILITY THEORY

Exercises

In Exercises 1 – 12 select the phase portrait on the next page that mostclosely corresponds to the system ~v′ = A~v, where A is the given matrix.

1. A =

[−1 1−1 1

]Answer

2. A =

[−3 11 −1

]

3. A =

[4 −36 −5

]Answer

4. A =

[2 00 1

]

5. A =

[1 32 2

]Answer

6. A =

[2 11 2

]

7. A =

[0 −1−1 0

]Answer

8. A =

[1 00 −1

]

9. A =

[2 00 2

]Answer

10. A =

[−4 3−3 2

]

11. A =

[−1 11 −1

]Answer

12. A =

[3 −24 −3

]

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 325

D EF

A BC

J

KL

G HI

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326 CHAPTER 4. STABILITY THEORY

13. Draw a phase portrait for the system x′ = y′ = x + y.Answer

14. Draw a phase portrait for the system{

x′ = x + yy′ = −x− y.

15. Show that the each orbit of the system in example 4.1.3 on page 315is a branch of a hyperbola x2 + 2xy = c, a half-line with slope − 1

2 , ora vertical half-line.Answer

16. Draw a phase portrait of the system{

x′ = xy′ = −3y.

Show that

the orbits are horizontal or vertical half lines, and curves (not hyper-bolas) satisfying the equation x3y = constant.

In problems 17 – 26, draw the phase portrait of the given system, andclassify it as degenerate, a saddle, a center, or a node (stable or unstable).Drawings of saddles must clearly indicate the stable and unstable lines.

17.{

x′ = 2x + yy′ = 3x

Answer

18.{

x′ = 5x + 5yy′ = 3x + 7y

19.{

x′ = −3x + 5yy′ = − 5

2 x + 2yAnswer

20.{

x′ = x + yy′ = −5x− y

21.{

x′ = x− yy′ = 0

Answer

22.{

x′ = 2x + yy′ = −x

23.{

x′ = 2x− 5yy′ = 17x− 2y

Answer

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4.1. PHASE PORTRAITS OF LINEAR SYSTEMS 327

24.{

x′ = 3x− 10yy′ = 8x− 5y

25.{

x′ = 3x + yy′ = −5x− 3y

Answer

26.{

x′ = 8x + 16yy′ = 4x + 8y

27. Show that if the phase portrait of example 4.1.1 is rotated 45◦

clockwise, the nonvertical orbits will lie on the graphs of functionsy = kx3.Answer

28. Let A denote a 2 × 2 matrix such that tr A = 0 and det A < 0.Show that the orbits of the system ~v′ = A~v are hyperbolas, theirasymptotes, and the origin. Hint: imitate the case where det A > 0.

29. Prove proposition 4.1.1.Answer

30. Construct if possible, linear systems of two ODEs that fulfill thespecifications. (Answers are not unique).

(a) A degenerate system; stationary line: y = 2x.

(b) A nondegenerate system; the positive and negative half lines onthe x-axis are orbits directed away from the origin; all other or-bits converge to the origin.

(c) A system in which all orbits except the stationary point at theorigin approach ∞ as t→ ∞.

(d) A system with exactly one nonstationary orbit that converges tothe origin.

(e) A system in which there are exactly two nonstationary orbitsthat converge to the origin.

(f) A system in which there are exactly four nonstationary orbitsthat converge to the origin.

31. Consider a homogeneous linear system

x′ = ax + byy′ = cx + dy

}(4.18)

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328 CHAPTER 4. STABILITY THEORY

(a) Show that if we substitute y = vx, where v is a new independentvariable, the result is the following system:

x′ = x(a + bv)xv′ + vx′ = x(c + dv)

}(4.19)

(b) Multiply the first equation in the system (4.19) by v and subtractit from the second. After dividing through by x you should ob-tain

v′ = −bv2 + (d− a)v + c. (4.20)

(c) Draw a phase diagram for equation (4.20). You will have to con-sider three cases, depending on the number of stationary points:0, 1, or 2.

(d) Show that the stationary points of equation (4.20) are the slopesof the eigenvectors of the system (4.18).

(e) Give an interpretation of the direction of the orbits in the phasediagram for equation (4.20), and explain how it is related to thephase portrait of the system (4.18).

(f) What does the phase diagram indicate in the case where thequadratic equation −bv2 + (d− a)v + c = 0 has no real roots?

Answer

Match the systems in problems 32– 37 with the phase portraits showbelow.

D E F

A B C

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4.2. NONLINEAR SYSTEMS 329

32.{

x′ = x + 3yy′ = x− y

33.{

x′ = 4yy′ = −x

Answer

34.{

x′ = 2x− 4yy′ = 2x− 2y

35.{

x′ = −3x + 10yy′ = −2x + y

Answer

36.{

x′ = x + yy′ = −x + y

37.{

x′ = −4x− 4yy′ = 2x

Answer

4.2 Stationary Points of Nonlinear Systems

Let us consider an autonomous system of ODEs

x′ = f (x, y)y′ = g(x, y),

}(4.21)

The system is nonlinear unless the functions f and g have the form f (x, y) =ax+ by and g(x, y) = cx+ dy (where a, b, c, d are constants). We will alwaysassume that the functions f and g have continuous first partial derivativeson an open domain D ⊂ R2. By a two-dimensional analogue of Theo-rem 1.6, this will ensure that the vector field

~v(x, y) =[

f (x, y)g(x, y)

]satisfies the Lipschitz condition. Therefore the existence and uniqueness,and continuous dependence theorem 2.1 for solutions of IVPs applies.

A stationary point (x1, y1) of ~v(x, y) is isolated if a circle can be drawnaround it that contains no other stationary points. For example, the originis an isolated stationary point of a linear system if and only if the system isnondegenerate, because a degenerate system would have a stationary line

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330 CHAPTER 4. STABILITY THEORY

through the origin, in the direction of eigenvectors belonging to the eigen-value 0. Stationary points arbitrarily close to the origin can be found on thestationary line.

A nonlinear system cannot be classified as stable or unstable as a whole;it must be analyzed locally. The one-dimensional case was covered in sec-tion 1.10. Each stationary point could be individually classified as stableor unstable. If the two adjacent orbits were directed toward it, the station-ary point was stable; and it was unstable if at least one of these orbits wasdirected away from it. In dimensions ≥ 2, we will need three categories:asymptotically stable, neutrally stable, and unstable.

Consider a nonstationary point (x0, y0), and let ~φ(t) be the solutionof (4.21) such that ~φ(0) = (x0, y0)

Definitions: If limt→∞ ~φ(t) = (x1, y1) then we will say (x0, y0) is attractedto (x1, y1), and if limt→−∞ ~φ(t) = (x2, y2) then we will say (x0, y0) is re-pelled from (x2, y2).

It is an Exercise to show that if a point (x0, y0) is either attracted to orrepelled from a point (x1, y1) then (x1, y1) is a stationary point. Perhaps itis surprising, but we will soon see an example where a point is simultane-ously attracted to and repelled from the same stationary point!

Definition: A stationary point ~p = (x1, y1) ∈ D of the system (4.21) isasymptotically stable if there is a circle C ⊂ D with center ~p such that everypoint inside C is attracted to ~p.

If A is a 2× 2 constant matrix, the phase portrait of

~v′ = A~v

was defined to be a stable node if every orbit is attracted to origin. Thus,the origin is an asymptotically stable stationary point of a stable node.

Before turning to a detailed analysis of stability for stationary points ofnonlinear systems, let’s look at a couple of pictures.

Example 4.2.1 Figure 4.13 is the phase portrait of

x′ = −4y + x(1−√

x2 + y2)

y′ = 4x + y(1−√

x2 + y2)

}(4.22)

Find a stationary point of the system, and draw a conclusion about its stability. Inwhat sense does the stability (or lack thereof) appear to be local?

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SOLUTION. You can see that the origin is a stationary point, and the unitcircle x2 + y2 = 1 is an orbit (in fact, it is easy to verify that (x, y) = (0, 0),and (x, y) = (cos(4t), sin(4t)) are solutions). Since (4.22) is nonlinear, theformula [

xy

]= C

[cos(4t)sin(4t)

]only yields a solution when |C| = 1. The orbits inside the unit circle appearto spiral away from the origin, and the orbits outside the circle spiral inwardtoward it. Therefore, the phase portrait indicates that the origin is notstable, but the circular orbit is stable. Of course, these statements arespeculative: we can’t use a computer-generated sketch as a proof, and wehaven’t defined the notion of a stable orbit. We can look at the figure as ascientist would view an experiment, and seek an explanation in the form ofa rigorous discussion.

Example 4.2.1 provides a preview of a topic we will explore in moredepth in section 4.5: a limit cycle. The system looks as if it might be simplerin terms of polar coordinates, r =

√x2 + y2 and θ = arctan

( yx

). Differenti-

ater2 = x2 + y2, tan θ =

yx

to get r r′ = x x′ + y y′ and sec2 θθ′ = r2

x2 θ′ = xy′−x′yx2 . We can simplify the

latter equation to get r2θ′ = xy′− x′y. Now substitute the expressions for x′

and y′ in the system (4.2.1) to get an equivalent system in polar coordinates(the details are for you to fill in):

r′ = r(1− r)θ′ = 4.

}(4.23)

This system is uncoupled. It says that orbits circulate counterclockwisearound the origin with an angular4 velocity of 4. The variable r satisfiesthe logistic equation, which has a stable stationary point at r = 1. Thusorbits all converge (in a sense we will be able to make precise later) to theperiodic orbit in the unit circle.

Example 4.2.2 Locate all stationary points of the system

x′ = yy′ = 2x− 3x2/2,

}(4.24)

whose phase portrait is shown in figure 4.14, and discuss their stability.

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332 CHAPTER 4. STABILITY THEORY

SOLUTION. If (x1, y1) is astationary point, then x′ = 0 implies y1 = 0. Thusall stationary points are located on the x-axis. Also, y′ = 0 implies

2x1 −32

x21 = 0.

This quadratic has two solutions: x1 = 0 and x1 = 43 . We conclude that

there are two stationary points: (0,0) and( 4

3 , 0)

. The origin looks like asaddle point of a linear system, and is not stable. The orbits appear tocirculate around

( 43 , 0)

as they do in a linear system that is acenter(eigenvalues are located on the imaginary axis). We infer that it toois not asymptotically stable, although here we are stretching. Thecomputer-drawn orbits circulating around

( 43 , 0)

appear to be closed ovals,but possibly they spiral inward (or outward) very gradually.

Linearly Stable Equilibria

Near the stationary points in the examples that we just considered, thephase portraits all appear to resemble phase portraits of linear systems. Torelate a stationary point (x1, y1) of (4.21) to a specific linear system, we usethe matrix of partial derivatives of f and g, all evaluated at (x1, y1). We willbe generalizing the stability test (proposition 1.10.3) for stationary points ofa one-dimensional system. The following definition may be familiar if youhave taken an advanced calculus course:

Definition: Let fx(x1, y1), fy(x1, y1), gx(x1, y1), and gy(x1, y1) denote thepartial derivatives of f and g. The derivative of the vector function~v(x, y) =[

f (x, y)g(x, y)

]at the point (x1, y1) is defined to be the matrix

∂( f , g)∂(x, y)

(x1, y1) =

[fx(x1, y1) fy(x1, y1)gx(x1, y1) gy(x1, y1)

].

The following proposition indicates that the derivative of a vector functionis the best linear approximation of the function. For simplicity, we willassume that ~v(x1, y1) =~0; that is, (x1, y1) is a stationary point.

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4.2. NONLINEAR SYSTEMS 333

Proposition 4.2.1 Let (x1, y1) be a point such that f (x1, y1) = g(x1, y1) = 0.Suppose that the second partial derivatives of f (x, y) and g(x, y) are continuouson D. Then

~v(x, y) = A(x1, y1)

[x− x1y− y1

]+ O((x− x1)

2 + (y− y1)2). (4.25)

as (x, y)→ (x1, y1).

Equation (4.25) uses “Big O” notation, which can be explained as fol-lows: let ~v(x, y) and ~w(x, y) be vector functions, and let h(x, y) be a real-valued function. The statement

~v(x, y) = ~w(x, y) + O(h(x, y)) as (x, y)→ (x1, y1)

means there is a circle C centered at (x1, y1) and a constant K such that

|~v(x, y)− ~w(x, y)| ≤ K h(x, y)

for all (x, y in the interior of C. Thus we can interpret (4.25) to say that thereis a constant K and a circle C centered at (x1, y1) such that∣∣∣∣~v(x, y)− A(x1, y1)

[x− x1y− y1

]∣∣∣∣ ≤ K((x− x1)2 + (y− y1)

2) (4.26)

for all (x, y) within C.PROOF OF PROPOSITION 4.2.1. Put x = x − x1 and y = y − y1, and

define a functionp(t) = f (x1 + tx, y1 + ty).

Since f has continuous second partial derivatives, it is easy to see that p hascontinuous second derivatives on the interval [0, 1]. By Taylor’s theoremthere is a number c1 ∈ (0, 1) such that

p(1) = p(0) + p′(0)(1− 0) +12

p′′(c1)(1− 0)2.

But, p(1) = f (x1 + x, y1 + y) = f (x, y), and p(0) = f (x1, y1) = 0. Further-more, by the chain rule for partial derivatives,

p′(0) =∂ f∂x

(x1, y1)x +∂ f∂y

(x1, y1)y

andp′′(c1) = B11 x2 + 2B12 xy + B13y2,

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334 CHAPTER 4. STABILITY THEORY

where

B11 =∂2 f∂x2 (x1 + c1 x, y1 + c1y),

B12 =∂2 f

∂x∂y(x1 + c1 x, y1 + c1y), and

B13 =∂2 f∂y2 (x1 + c1 x, y1 + c1y).

Thus

f (x, y)− [ fx(x1, y1)x + fy(x1, y1)y] = B11 x2 + 2B12 xy + B13y2. (4.27)

Similarly,

g(x, y)− [gx(x1, y1)x + gy(x1, y1)y] = B21 x2 + 2B22 xy + B23y2, (4.28)

where B21, B22, and B23 represent the second partial derivatives of g, evalu-ated at a point (x1 + c2 x, y1 + c2y). The Bij are not constants: all depend on(x, y). Since any function that is continuous on D must be bounded withinany circle C ⊂ D, there is a constant M such that each second partial deriva-tive of either f or g has absolute value less than M inside C. Since each Bijis equal to a second partial derivative of either f or g, the bounds |Bij| < Mhold.

By combining (4.27) and (4.28), we obtain∣∣∣∣~v(x, y)− A(x1, y1)

[xy

]∣∣∣∣ = ∣∣∣∣[ B11 x2 + 2B12 xy + B13y2

B21 x2 + 2B22 xy + B23y2

]∣∣∣∣ (4.29)

Using the fact that (x± y)2 ≥ 0, you can show that

2|xy| ≤ (x2 + y2).

You can use this inequality, the bounds |Bij| < M, and (4.29) to show that∣∣∣∣~v(x, y)− A(x1, y1)

[xy

]∣∣∣∣ < 2M(x2 + y2)

∣∣∣∣[ 11

]∣∣∣∣ = 2√

2M(x2 + y2).

Thus (4.26) holds, with K = 2√

2M.Proposition 4.2.1 provides the motivation to explore similarities of the

nonlinear system [x′

y′

]=

[f (x, y)g(x, y)

](4.30)

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4.2. NONLINEAR SYSTEMS 335

and the linear system [x′

y′

]= A(x1, y1) ·

[xy

]. (4.31)

Any comparison would only be valid when (x, y) is close to the stationarypoint (x1, y1). It turns out that the two systems have much in common,unless the matrix A(x1, y1) has an eigenvalue that is either zero or pureimaginary.

The system (4.30) is said to be linearly stable at the stationary point(x1, y1) if all eigenvalues of A(x1, y1) lie to the left of the imaginary axis ofthe complex plane. The following theorem, which was published in 1929 bythe German mathematician Oskar Perron, is the promised two-dimensionalversion of proposition 1.10.3. (It is actually valid in n dimensions.)

Theorem 4.1 Let (x1, y1) be a linearly stable stationary point of the system (4.21)of differential equations, where f and g are assumed to have continuous secondpartial derivatives in D. Then (x1, y1) is asymptotically stable.

You will be asked to prove this theorem in exercise 24 in section 4.5,using a technique that will be developed there.

The set of all points that are attracted to a stationary point ~p is calledthe attracted set of ~p. For example, consider a linear system~v′ = A~v, whereA is a constant matrix. The attracted set of the origin is

• the origin alone if A has no eigenvalues in the left half-plane,

• the entire plane if both eigenvalues of A are in the left half-plane, and

• the stable line if the eigenvalues of A are real and of opposite sign.

Similarly, the repelled set of ~p defined to be the set of points repelled by~p. For a linear system ~v′ = A~v, the repelled set of the origin is

• the origin alone if A has no eigenvalues in the right half-plane,

• the entire plane if both eigenvaluesof A are in the right half-plane,and

• the unstable line if the eigenvalues of A are real and of opposite sign.

For another example, consider an asymptotically stable stationary point~p of a nonlinear system. By definition, there is a circle C centered at ~v1, (C

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336 CHAPTER 4. STABILITY THEORY

may be very small), such that all points within C are attracted to ~p. Thuswe can say that the attracted set of ~p contains the points inside some circlecentered at ~p.

Advanced texts go beyond theorem 4.1 and show that the phase portraitof the system (4.30) and its “linearization” at ~p, the system (4.31), share thefollowing properties:

(i) If the phase portrait of the linear system (4.31) is an unstable node,then the repelled set of ~p in the nonlinear system (4.30) contains thethe set of points inside a circle C centered at ~p .

(ii) If the phase portrait of the linear system (4.31) is a saddle, then, in thenonlinear system (4.30), the repelled set of ~p is a curve, whose tan-gent direction at ~p is parallel to the eigenvector of A(x1, y1) belongingto the positive eigenvalue. The attracted set is a curve with tangentdirection at ~p parallel to the eigenvector belonging to the negativeeigenvalue. In figure 4.14, the origin is a saddle for the linearized sys-tem. The nonlinear system is more complicated (there is a center onthe positive x-axis), but the stationary point at the origin is of the sad-dle type. The stable and repelled sets are no longer straight lines; theyconsist of the orbits directed toward and away from the origin. Noticethat in the right half-plane, the attracted and repelled sets coincide.

(iii) If the phase portrait of the linear system (4.31) is a spiral node, therewill be a spiral node near at the stationary point ~p of the nonlinearsystem (4.30). This is illustrated by figures 4.13, in which the node isunstable, and by figure 4.15, where the node is stable, although orbitsnot starting inside the circular orbit will not converge to the origin.

(iv) If the phase portrait of the linear system (4.31) is a center, the orbitsnear ~v1 of the system (4.30) will swirl around ~v1, but they may not beclosed. The second order term that was dropped when passing fromthe nonlinear system to the linear system is small, but not too smallto disrupt a family of closed orbits. This is illustrated in figure 4.16.

Example 4.2.3 Find the stationary points of the system

x′ = x(2x + 3y− 7)y′ = y(3x− 4y− 2)

}(4.32)

and determine which are stable.

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4.2. NONLINEAR SYSTEMS 337

SOLUTION. The stationary points are the solutions of the equations

x(2x + 3y− 7) = 0y(3x− 4y− 2) = 0.

The graphs of these equations are the nullclines of the system. Whendrawing a phase portrait, it is often useful to include them. The nullclinesintersect at the points (0, 0),

(0,− 1

2

),( 7

2 , 0)

, and (2, 1). These are thestationary points.The derivative matrix is

A(x, y) =

[∂

∂x x(2x + 3y− 7) ∂∂y x(2x + 3y− 7)

∂∂x y(3x− 4y− 2) ∂

∂y y(3x− 4y− 2)

]

=

[4x + 3y− 7 3x

3y 3x− 8y− 2

].

This matrix is only of interest when (x, y) is one of the four stationarypoints of the system (4.32).At the stationary point (0, 0)

A(0, 0) =[−7 00 −2

],

which has eigenvalues −2 and −7. Since both are negative, (0, 0) is alinearly stable stationary point, and by theorem 4.1, asymptotically stable.The eigenvalues of

A(

0,−12

)=

[− 17

2 0− 3

2 2

]are − 17

2 and 2. Since these have opposite signs, there is a saddle point at(0,− 1

2

), with attracted set tangent to the eigenvector

[71

]belonging to

− 172 and with repelled set tangent to the eigenvector

[01

]belonging to 2.

The eigenvalues of A( 7

2 , 0)

are both positive, and it follows that thestationary point

( 72 , 0)

is unstable. Finally,

A(2, 1) =[

4 63 −4

],

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338 CHAPTER 4. STABILITY THEORY

with eigenvalues ±√

34 and corresponding eigenvectors[

6±√

34− 4

].

This indicates that there is a saddle point at (2, 1); the repelled set hasslope 1

6 (√

34− 4) ≈ 0.3 as it crosses the stationary point, and theattracted set has slope − 1

6 (√

34 + 4) ≈ −1.6 there. The phase portrait isshown in figure 4.17.

Invariant Sets

A subset S of the x, y-plane is an invariant set for a system (4.21) of differ-ential equations if every orbit of the system that contains a point of S is asubset of S . An equivalent way of stating this definition is that S is invari-ant if and only if S is the union of a collection of orbits of the system.Question: It is obvious that the union of two or more invariant sets is also an invariant set.What about the intersection?

A typical example of an invariant set would be a level curve of an inte-gral F(x, y) of the system (4.21).

Example 4.2.4 Show that the x- and y-axes, and the four quadrants of the coor-dinate plane are invariant sets for the system in example 4.2.3.

SOLUTION. The points on the y-axis have x-coordinate equal to 0. If weset x = 0, the system reduces to

x′ = 0y′ = y(−4y− 2).

In particular, x′ = 0 assures x is constant. Thus, if an orbit meets they-axis, then x ≡ 0 on that orbit, so that it is a subset of the y-axis.Furthermore, on the y-axis, the phase portrait is just the phase diagram ofy′ = −y(4y + 2). Similarly, the x-axis is invariant, and the the phaseportrait reduces to the phase diagram of x′ = x(2x− 7). No orbit cancross either of the coordinate axis, and thus every orbit is confined to thequadrant of its initial point. Thus, each quadrant is invariant.

When drawing phase portraits, it is important to draw the attracted andrepelled sets of each saddle point. Each of these will be a smooth curve,called a separatrix, passing through the saddle point, and they divide the

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4.2. NONLINEAR SYSTEMS 339

region around the saddle point into four quadrants. In each of the quad-rants, the behavior of the orbits is distinctive. To use an ODE solver to drawa separatrix, follow these steps:

1. Find the eigenvectors of the linearized system at the saddle point.Draw axes through the saddle point with these characteristic direc-tions. These axes are tangent to the separatrices.

2. Select four points ~v1, . . . , ~v4 on the axes, on opposite sides of the sad-dle point. These points should be close to the saddle point.

3. Use the ODE solver to draw the orbits passing through the points ~v1,. . . , ~v4. Each solution ~vi(t) should have initial point ~vi(0) = ~vi, andshould be drawn for a ≤ t ≤ b where a < 0 and b > 0.

4. If the above steps have been done correctly, the four orbits drawnwill seem to merge to form the two separatrices. This method wasemployed in drawing the separatrices for the two saddle points infigure 4.17.

Example 4.2.5 Find the attracted set of the origin for the system in example 4.2.3.

SOLUTION. In example 4.2.4 it was shown that each quadrant of the planeis an invariant set. We will determine the portion to the attracted set of theorigin in each quadrant, starting with the second quadrant and proceedingcounterclockwise.It is easy to see that at every point in the second quadrant, y′ < 0. Thus allorbits will eventually cross the x-nullcline, y = (7− 2x)/3, and after thatcrossing, x′ > 0. It follows that all orbits in the second quadrant areattracted toward the origin. In other words, the attracted set of the origincontains the second quadrant.In the third quadrant, x′ > 0, so all orbits in that quadrant are directed tothe right. One of these orbits is part of a separatrix of the stationary point(0,− 1

2

). It separates the other orbits in that quadrant into two classes:

those above it are attracted to the origin, and those below it are asymptoticto the negative y-axis. The portion of the third quadrant that lies above thisseparatrix is also in the attracted set of the origin,The entire fourth quadrant is contained in the repelled set of the stationarypoint

( 72 , 0)

on the positive x-axis. It also contains part of the separatrix ofthe stationary point

(0,− 1

2

). This is an orbit that goes from

( 72 , 0)

to

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340 CHAPTER 4. STABILITY THEORY

(0,− 1

2

), and the wedge-shaped region in the fourth quadrant that lies

above this orbit is the portion of the attracted set of the origin in the fourthquadrant. Outside this wedge orbits are asymptotic to the negative y-axis.Finally, in the first quadrant, the attracted set of the saddle at (2, 1) forms aboundary of the attracted set of the origin.Figure 4.18 shows the attracted set.

Neutrally Stable stationary points

If the phase portrait of a linear system is a center, then the origin is notasymptotically stable, because no orbit, other than the origin itself, actuallyconverges to the origin. In fact, the attracted and repelled sets of the originconsist of the origin alone. However, it is desirable to have a definition ofstability that encompasses centers. For example, it is in the context of sucha definition that we can say that the solar system, as described by Newton’slaws, is stable1.

Definitions: A stationary point ~p ∈ D of the system (4.21) of differentialequations is said to be neutrally stable if, given any circle C1 with center ~pthere is a (perhaps smaller) circle C2, also centered at ~p, such that for anysolution ~v(t) of (4.21) such that ~v(0) is inside C2, ~v(t) lies inside C1 for allt ∈ R.

On the other hand, the stationary point ~p is unstable if there exists acircle C centered at ~p such that there are points ~q arbitrarily close to ~p suchthat the solution ~φ(t) with φ(0) = ~q eventually gets outside C; that is, forsome t1 > 0, the point φ(t1) is outside C.

A stationary point that is not unstable is, naturally, said to be stableExamples of neutrally stable stationary points that are not asymptot-

ically stable are not hard to find: see example 4.2.6 below. Surprisingly,there are also examples of unstable stationary points that are asymptoticallystable. (See Exercise 17 at the end of this section.)

Example 4.2.6 Show that the stationary point (0, 0) of the system

x′ = 36yy′ = −x,

is neutrally stable.

1This has never been proved, but we have several billion years of experimental evidence.

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4.2. NONLINEAR SYSTEMS 341

SOLUTION.The phase portrait of this system is a center; the orbits are ellipses; in fact,the reader can show that each orbit has equation 36x2 + y2 = C, byfollowing the procedure established in example 4.1.4 on page 319. Themajor axis of each ellipse is horizontal, and 6 times as long as the minoraxis. Figure 4.19 displays one of the orbits.Let R > 0 be given, and put r = 1

6 R. Any orbit starting inside the circle ofradius r will be an ellipse whose minor semiaxis has length less than r; itsmajor semiaxis is therefore less than R units long. Hence the entire orbit isinside the circle of radius R.

Example 4.2.7 Show that (0, 0) is not a stable stationary point of the system

x′ = yy′ = x

SOLUTION. We will show that no matter how small r is taken to be, thereare soluitons with initial points inside the circle of radius r that leave thecircle of radius 1. In fact, the phase portrait is a saddle, so all solutionsstarting at a point inside the circle of radius r, except for the orbits lying onthe attracted set (the line y = −x), are unbounded as t→ ∞, andeventually leave the circle of radius 1.

Exercises

1. Find the repelled set to the stationary point of( 7

2 , 0)

of the systemin example 4.2.3.Answer

In problems 2 – 13, find the stationary points, and sketch a phase por-trait. In each case, is the stationary point stable? Linearly stable? asymp-totically stable? Draw the separatrices of each saddle point.

2.{

x′ = x− yy′ = 3x− 2y

3.{

x′ = x− yy′ = 5x− 3y

Answer

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342 CHAPTER 4. STABILITY THEORY

4.{

x′ = x(1− x)y′ = y(2− y)

5.{

x′ = x + 4yy′ = −9x− y

Answer

6.{

x′ = x + 2yy′ = 2x− y

7.{

x′ = y(1− x2 − y2)y′ = −x(1− x2 − y2)

Answer

8.{

x′ = y(1− x2)y′ = −x(1− x2)

9.{

x′ = x(2− x− y)y′ = y(x− y)

Answer

10.{

x′ = x(1 + x2 + y2)y′ = y(2 + x− y)

11.{

x′ = x(x2 + y2 − 10)y′ = y(xy− 3)

Answer

12.{

x′ = x(x + y + 6)y′ = y(5x + y− 10)

13.{

x′ = x(1− y)y′ = y(x− 1)

Answer

14. Show that the origin is the only stationary point of (4.22).

15. Analyze the system

x′ = −y + kx(1− x2 − y2)y′ = x + ky(1− x2 − y2).

(a) Show that the circle x2 + y2 = 1 is always a closed orbit.

(b) Show that the stability of the stationary point at the origin de-pends on k.

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4.3. *COMPETING SPECIES 343

(c) Show that if we set r2 = x2 + y2 then r satisfies the differentialequation r′ = kr(1− r2).

(d) Conclude that the phase portrait is a stable spiral node if k < 0,a center if k = 0, and an unstable spiral node if k > 0.

(e) Draw typical phase portraits for the system corresponding tok > 0, k = 0, and k < 0.

Answer

16. Consider the system{

x′ = −y + kx(x2 + y2)y′ = x + ky(x2 + y2),

where k is a

constant (see figure 4.16).

(a) Show that theorem 4.1 does not determine whether or not theorigin is a stable stationary point, and that the stability of thestationary point at the origin depends on k.

(b) Show that if we set r2 = x2 + y2 then r satisfies the differentialequation r′ = kr3.

(c) Conclude that the phase portrait is a stable spiral node if k < 0,a center if k = 0, and an unstable spiral node if k > 0.

17. Challenging problem: Asymptotic stability does not imply sta-bility. Show that all nonconstant orbits of

x′ = −y + x(1− x2 − y2) +xy√

x2 + y2

y′ = x + y(1− x2 − y2)− x2√x2 + y2

converge to the stationary point (1, 0) as t → ∞. Furthermore, showthat (1, 0) is unstable (see figure 4.20). Hint: try polar coordinates!Answer

18. Modify the proof of proposition 1.10.2 to show that in the phasespace of the system (4.21), if ~p = (x1, y1) is either repels or attracts apoint (x0, y0) then ~p is a stationary point.

4.3 Competing Species

When we considered the population dynamics of two-species systems insection 2.5, we had not yet developed the concept of stability. With the

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344 CHAPTER 4. STABILITY THEORY

definitions of neutrally and asymptotically stable stationary points in place,we will reexamine the competing species equations

x′ = x(k− ax− by)y′ = y(l − cx− dy)

}(4.33)

that were derived in section 2.5.The four stationary points of (4.33) are the origin, where both species

are extinct, (k/a, 0), where the species represented by y is extinct, (0, l/d),where the x-species is extinct, and a point (x1, y1) at the intersection of thenullclines ax + by = k and cx + dy = l. The latter stationary point has nobiological significance unless it is located in the first quadrant, but then itrepresents a situation where the two species share the resources. Since nei-ther nullcline enters the third quadrant, (x1, y1) must lie in the first, second,or fourth quadrant. Figure 2.19 displays the possible configurations of thenullclines. As in figure 2.19, let L = l/d be the coordinate of the stationarypoint on the y-axis, and C = k/a be the coordinate of the stationary pointon the x-axis. Also put p = k/b; this is the point where the x-nullcline in-tersects the y-axis, and let q = l/c be the point where the y-nullcline crossesthe y-axis. The points p and q are not stationary points.

The derivative matrix for the system (4.33) is

A(x, y) =[

k− 2ax− by −bx−cy l − cx− 2dy

].

Thus, A(0, 0) =[

k 00 l

]has two positive eigenvalues; as expected, there

is an unstable node at the origin.At the stationary point on the y-axis,

A(0, L) =[

k− bL 0−cL −l

].

The eigenvalues of A(0, L) are −l and k − bL = b(p − L). The second ofthese eigenvalues will be negative if p < L, and positive if L < p. It followsthat the stationary point at (0, L) is asymptotically stable if p < L, and isa saddle when L < p. Similar reasoning shows that the stationary point(K, 0) is stable if q < C and a saddle if C < q.

The fourth stationary point (x1, y1) lies in the first quadrant if and onlyif either

Case I. p < L and q < C, or

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4.3. *COMPETING SPECIES 345

Case II. L < p and C < q.

Furthermore, you can calculate that

A(x1, y1) =

[−ax1 −bx1−cy1 −dy1

].

Since det A(x1, y1) = x1y1(ad− bc), the eigenvalues are real with oppositesign if ad < bc and (x1, y1) is a saddle. In Case I,

ld>

kb

andka>

lc

.

Since each of the parameters a, b, c, d, k, and l is positive, we can multiplythe inequalities to obtain

klad

>klbc

(4.34)

from which it follows that ad− bc < 0. Hence, in Case I the stationary pointat (x1, y1) is a saddle.

In Case II the inequality in (4.34) is reversed, so det(A(x1, y1)) > 0.Hence the eigenvalues are either real with the same sign, or complex con-jugate. The trace of A(x1, y1) is negative, and thus the roots, if real, are neg-ative, and if they are complex, the real parts are negative. The system (4.33)is therefore linearly stable in Case II. By the theorem of Perron, the (x1, y1)is an asymptotically stable stationary point in Case II.

What are the biological implications? We will start with the situationin Case I, where there is a saddle at (x1, y1). If the two populations startedat their exact equilibrium values of x1 and x2 respectively, that equilibriumwould be maintained (unless disturbed by external influences). Since thestationary point at (x1, y1) is unstable, coexistence is extremely improba-ble in this situation. Usually competitive exclusion occurs: the species withthe initial population advantage overwhelms the other. The separatrix of(x1, y1) serves as a boundary between the attracted sets of (0, L) and (C, 0),and the destiny of the two competing species is determined by which sideof the separatrix the initial population vector lies. Figure 4.21 is a phaseportrait displaying the phenomenon of competitive exclusion.

When the stationary point (x1, y1) does not lie in the first quadrant, thetwo species cannot coexist: one species will dominate and eventually ex-clude the other without regard to the initial condition, unless all membersof the dominant species are removed from the environment. To see whythis is so, refer to figure 4.22, which is a phase portrait of the system for

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346 CHAPTER 4. STABILITY THEORY

case where C > q and L < p. We have noted that when the parameters areconfigured in this way, (0, L) is a saddle and (C, 0) is asymptotically stable.Furthermore, the attracted set for (0, L) is the positive y-axis, and we caninfer that the repelled set is a curve extending into the first quadrant. Wesaw in section 2.5 that the quadrilateral LqCp is a trap, in the sense thatevery orbit in the first quadrant enters LqCp unless it converges directlyto (C, 0), and no orbit can leave the quadrilateral. An orbit that enters thequadrilateral is easily seen to converge to (C, 0). Since every orbit in thefirst quadrant must enter LqCp, the attracted set of (C, 0) contains the en-tire first quadrant.

When L < p and C < q, both stationary points (C, 0) and (0, L) aresaddles. As long as both species are represented in the initial populationvector, the populations will approach the stationary point (x1, y1), whichis asymptotically stable: see figure 4.23. In this case, the attracted set of(x1, y1) is the first quadrant.

Exercises

1. In Exercise 5 in Section 2.5, the system

x′ = ax(K− x + By)y′ = dy(L + Cx− y).

was presented as model for the population dynamics of two speciesin symbiosis. Draw a phase portrait for the system, considering thecases where BC > 1 and BC < 1 separately.Answer

2. Suppose that the differential equations model of a predator-preysystem has a limit cycle. If the prey is a pest, how would you devisea pest control strategy?

3. Draw a phase portrait for the following differential equation modelfor a predator-prey system, in which the prey population would havelogistic growth in the absence of the predator.

x′ = x[A(C− x)− by]y′ = cy(x− d)

}(4.35)

Consider the following three cases, and draw conclusions about thefate of the predator in each.

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4.4. ENERGY INTEGRALS 347

(a) C− d < 0.

(b) 0 < C− d <Ad4c

(c) C− d > Ad4c

Answer

4. In the system (4.35) (see problem 3), let a = AC. Then, as C → ∞,with A = aC−1 so that a is constant, the right side of the system (4.35)approaches the right side of the Lotka-Volterra equations (2.21). How-ever, show that for any finite C > d, the stationary point (x1, y1) =(

d, a−Adb

)of the system (4.35) is asymptotically stable, even though

the corresponding stationary point of the Lotka-Volterra equationsis neutrally stable. We express this failure of thephase portrait ofthe Lotka-Volterra system to resemble the phase portrait of the sys-tem (4.35), no matter how large the carrying capacity is taken to be,by saying that the Lotka-Volterra system is not structurally stable.

4.4 Energy Integrals

Suppose that a particle with unit mass is in motion with one degree of free-dom. This is the case if the particle is constrained to move on an axis. Ifthe force on the particle is dependent only on its position, not on its veloc-ity (this rules out friction), then Newton’s Second Law of Motion takes theform

d2xdt2 = g(x), (4.36)

where g(x) denotes the force at position x. For example, the particle mightbe attached to a nonlinear spring with restoring force g(x).

Solutions of equation (4.36) can be represented on the phase plane asorbits of the system

x′ = yy′ = g(x).

}(4.37)

Proposition 4.4.1 Orbits of the system (4.37) are directed to the right at pointsin the first and second quadrants, and to the left in the third and fourth quadrants.An orbit has a vertical tangent only at points where it crosses the x-axis.

PROOF. In the first and second quadrants, y > 0. Since y = x′, it fol-lows that x is increasing in these quadrants; in other words, every orbit is

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348 CHAPTER 4. STABILITY THEORY

directed to the right as it passes through them. Similarly, in the third andfourth quadrants, x′ < 0, and it follows that an orbit will be directed to theleft while in these quadrants. A vertical tangent can only appear where theorbit crosses the x-nullcline, which is the x-axis.

We can find an integral for the system (4.37) by solving the differentialequation

y dy− g(x) dx = 0.

The result is12

y2 − G(x) = C, (4.38)

where G(x) denotes an antiderivative of g(x). It follows that

F(x, y) =12

y2 − G(x)

is an integral of the system (4.37). We call F an energy integral.The variable y the system (4.37) in represents the velocity of the object,

and 12 y2 is the kinetic energy. The potential energy of the particle is defined to

be U(x) = −G(x). Since G(x) is an antiderivative, the potential energy isdetermined only up to the addition of a constant. The integral F(x, y) rep-resents the total mechanical energy of the particle. In deriving this integralwe have established a special case of the Law of Conservation of Energy.

To draw a phase portrait of the system (4.37), it is helpful to sketch thegraph of the potential energy function first. Since on any orbit, the sumof the kinetic and potential energy is constant, when the potential energyincreases, the kinetic energy must decrease, and vice versa. All stationarypoints of the system lie at points (x1, 0) on the x-axis where g(x1) = 0; thatis, U′(x1) = 0, so that x1 is a critical point of the potential energy. We willsay that x1 is an isolated critical point if there is an interval (x1 − r, x1 + s)containing no other critical points.

Proposition 4.4.2 Suppose that the potential energy U(x) has an isolated criticalpoint at x = x1. If U(x1) is a relative minimum, then (x1, 0) is a neutrally stablestationary point of the system (4.37). Furthermore, there are numbers r, s > 0such that every orbit that crosses the x-axis in the interval (x1 − r, x1 + s) is aclosed orbit, traversed clockwise.

PROOF. For convenience, assume that x1 = 0. Let r > 0 and s > 0 bechosen so that U(x) has no other critical points in the interval I = (−r, s).The potential energy is then decreasing on I− = (−r, 0) and increasing onI+ = (0, s). We will assume that r and s are chosen so that U(−r) = U(s).

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4.4. ENERGY INTEGRALS 349

Suppose that an orbit crosses the x-axis at a point x0 ∈ I−. Since the kineticenergy at the crossing is 0, the total energy of this orbit is then U(x0). Asshown in figure 4.24, the orbit cannot proceed to the left, because then boththe kinetic and potential energy would increase, and the total energy wouldnot be constant.

The orbit must therefore proceed to the right. Since a right directed orbitmust go above the x-axis, our orbit enters the upper half-plane. It continuesto move to the right, reaching a maximum kinetic energy as it crosses they-axis, where the potential energy is minimum. Then the kinetic energy(and y) will decrease, until the x-axis is crossed again. This crossing willoccur at the point x ∈ I+ where U(x) = U(x0). Now the orbit cannot gofarther to the right, because the potential energy cannot exceed the valueU(x0). In going to the left, it enters the lower half-plane. To describe the restof the orbit we can appeal to symmetry. The orbit is the locus of solutionsof (4.38), which is obviously symmetric with respect to the x-axis. Thereforethe orbit in the lower half-plane is just the reflection of the portion alreadytraversed in the upper half-plane, and is therefore a closed curve.

We have verified the second assertion of the proposition. It still must beproved that the stationary point at the origin is neutrally stable. This meansthat for any number R > 0, it must be possible to find a number ε > 0 suchthat every orbit that starts inside the circle of radius ε centered at the originremains inside the circle of radius R.

We can assume that R < r, s. (If this is not the case, replace R by asmaller number that satisfies the assumption.) Let CR denote the circle ofradius R, centered at the origin. Put m = min{F(x, y) : (x, y) ∈ CR},the minimum value of the energy integral on CR. (This is the minimum onthe perimeter of the circle.) If m = F(x, y), where (x, y) ∈ CR then m =12 y2 + U(x). Our choice of r ensures that x ∈ I \ {0}, so F(x, y) ≥ U(x) >U(0) = f (0, 0). In short, m > F(0, 0). Appealing to the continuity of F,there is a number ε > 0 such that if (x, y) is inside the circle Sε with radiusε, then F(x, y) < m. Any orbit which starts inside Sε has total energy lessthan m, and since every point of CR has total energy at least m, the orbitcannot cross CR.

Proposition 4.4.3 Let x1 be an isolated critical point of the potential U(x). If thepotential does not have a relative minimum at x1, then (x1, 0) is a not a stablestationary point of the system (4.37).

PROOF. As in the proof of proposition 4.4.2, assume that x1 = 0, and letI = (−r, s) denote an open interval containing no other critical points of

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350 CHAPTER 4. STABILITY THEORY

U(x). Either U′(x) > 0 on I− or U′(x) < 0 on I+ since otherwise, by thefirst derivative test, U(0) would be a relative minimum. To be definite, wewill suppose that U′(x) > 0 on I−.

Let Cr denote the circle with center at the origin and radius r. We willshow that for any x0 ∈ (−r, 0), the orbit of the solution (x, y) = (φ(t), ψ(t))of the system (4.37) with initial condition (φ(0), ψ(0)) = (x0, 0) crosses thecircle Cr. As shown in figure 4.25, there will be two crossing points, sincethe orbit is symmetric with respect to the x-axis; the one in the lower half-plane is directed to the left and thus outward from the circle.

Since the initial point (x0, 0) can be made arbitrarily close to the origin,this will demonstrate instability.

The orbit through the point (x0, 0) is on the level curve of the total en-ergy given by the equation

U(x) +12

y2 = U(x0). (4.39)

The circle Sr has equationx2 + y2 = r2. (4.40)

We will show that there is a point (x∗, y∗), with −r < x∗ < 0, satisfyingequations (4.39) and (4.40) simultaneously. Multiply equation (4.39) by 2and subtract it from equation (4.40) to eliminate the variable y and get

x2 − 2U(x) = r2 − 2U(x0). (4.41)

Let P(x) denote the expression on the left side of equation (4.41). Noticethat

P(x0) = x20 − 2U(x0) < r2 − 2U(x0).

Also, since U(−r) < U(x0),

P(−r) = r2 − 2U(−r) > r2 − 2U(x0)

Appealing to the continuity of P(x) and the intermediate value theorem,there is a solution x∗ ∈ (−r, x0) of equation (4.41); the orbit through thepoint (x0, 0) crosses Sr at (x∗,±y∗), where y∗ =

√2[U(x0)−U(x∗)].

If x1 is an isolated critical point of U(x), not a relative minimum, then(x1, 0) lies on a level energy curve

12

y2 + U(x) = U(x1).

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4.4. ENERGY INTEGRALS 351

This level curve is the separatrix of the stationary point (x1, 0). If U(x) hasa relative maximum at x1, then (x1, 0), is a saddle point, as shown in fig-ure 4.26. If x1 is an inflection point of U(x), the separatrix has only onebranch; this is illustrated by figure 4.25.

Example 4.4.1 Draw a phase portrait for the differential equation

x′′ = 6x(1− x).

SOLUTION. We will draw the phase portrait of the system

x′ = yy′ = 6x(1− x).

The potential is U(x) = −∫ x

0 6z(1− z) dz = 2x3− 3x2, which has a relativemaximum at x = 0 and a minimum at x = 1. Hence the stationary point at(0, 0) is unstable, while the stationary point (1, 0) is stable. The separatrixcorresponding to the stationary point at the origin is the cubic curve

12

y2 + 2x3 − 3x2 = 0.

This curve consists of three orbits in addition to the origin itself. Theportion of the curve in the second quadrant is an orbit directed toward theorigin, and forms part of the attracted set. The intersection of the curvewith the third quadrant is an orbit directed away from the origin: it is part ofthe repelled set. The situation in the right half-plane seems paradoxical,but actually represents a relatively common occurrence: the portion of theseparatrix in the right half-plane is a single orbit, which converges to theorigin as t→ ∞ and as t→ −∞; it therefore lies in the intersection of theattracted and repelled sets.The orbits inside the separatrix in the right half-plane are closed; all otherorbits were drawn by reference to the potential function. The phaseportrait is shown in figure 4.14 on page 387.

The Pendulum

Figure 4.27 shows a pendulum. L denotes the length of the pendulum arm,θ the angle made by the arm with the vertical axis, and m the mass.

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352 CHAPTER 4. STABILITY THEORY

The force of gravity on the bob is F = mg, directed downward. We willuse a moving coordinate system in which the origin is the center of the bob,with the y-axis is aligned along the pendulum arm, and the x-axis tangentto the trajectory of the pendulum. In this coordinate system,

F = −mg(sin(θ)i + cos(θ)j).

The net force on the pendulum bob is F + Tj, where Tj, the support pro-vided by the pendulum arm, is exactly cancelled by the j-component of F.Thus, the net force is equal to −mg sin(θ)i.

By Newton’s Second Law of Motion, the net force on the pendulumbob is equal to its mass multiplied by its acceleration. The acceleration isthe product Lθ′′ of the angular acceleration θ′′ of the pendulum arm andits length L. Therefore, the motion of the pendulum is governed by thedifferential equation mLθ′′ = −mg sin θ, or

θ′′ = − gL

sin θ, (4.42)

The potential energy of the pendulum, normalized so that U(0) = 0, is

U(θ) =∫ θ

0

gL

sin(z) dz =gL(1− cos θ).

As we would expect, the stationary points at θ = 2nπ, when the pendulumis at the bottom of its trajectory, correspond to minima of the potential en-ergy, and are stable. The stationary points at odd multiples of π (where thependulum is inverted) correspond to maxima of the potential energy andare unstable. The separatrix is the curve

12

ω2 − gL

cos θ =gL

,

in the phase plane (the variable ω = θ′ represents angular velocity). In thiscase, the separatrix separates the closed (periodic) orbits from those thatare not periodic. See figure 4.28.

Exercises

1. Let x1 denote an isolated critical point of U(x). Is it possible forthe stationary point (x1, 0) of the system (4.37) to be asymptoticallystable?Answer

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4.4. ENERGY INTEGRALS 353

2. Suppose that U(x) has a non-degenerate critical point at (x1, 0); thatis, U′(x1) = 0 and U′′(x1) 6= 0. Let A denote the derivative matrix ofthe system (4.37) at (x1, 0). Show that if U(x) has a relative maximumat x1, then the eigenvalues of A have opposite sign, while if U(x) hasa relative minimum at x1, then the eigenvalues are pure imaginary.

3. Show that if g(x1) = 0 and g′(x1) < 0, then (x1, 0) is a neutrallystable stationary point of the system (4.37).Answer

In problems 4 – 19, find a potential energy function and sketch its graph.Then draw a phase portrait for the differential equation.

4. x′′ = ax, where a is a positive constant.

5. x′′ = ax, where a is a negative constant.Answer

6. x′′ = 0.

7. x′′ = 1.Answer

8. x′′ = x2.

9. x′′ = −x2.Answer

10. x′′ = 4x3.

11. x′′ = −4x3.Answer

12. x′′ = 4x − x3. Show that the attracted and repelled sets of theorigin are the same.

13. x′′ = 3x|x|.Answer

14. x′′ = e−x(cos x + sin x).

15. x′′ = x2−1(x2+1)2 . To save time in computing the potential energy, note

that x2−1(x2+1)2 = Re

[1

(x+i)2

].

Answer

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354 CHAPTER 4. STABILITY THEORY

16. x′′ = x2 sin x− 2x cos x.

17. x′′ = sin x− 12 .

Answer

18. x′′ = sin x− 1.

19. x′′ = sin x−( 2

π

)x.

Answer

20. Suppose that the potential energy function U(x) for the differen-tial equation (4.36) has a relative maximum at x = x1. Show that thewhen the two branches of the separatrix cross the x-axis of the phaseplane at x = x1, their slopes are ±

√g′(x1).

4.5 The Lyapunov Stability Tests

Let θ denote the angular displacement of a pendulum arm from the verticalposition. If the pendulum is subject to linear friction, then θ satisfies thedifferential equation

d2θ

dt2 = −bdθ

dt− k sin θ, (4.43)

where b is the friction constant, and k = g/L.Let ω denote the angular velocity of the pendulum. The phase portrait

of the differential equation (4.43) is that of the system

dθdt = ω

dωdt = −bω− k sin θ

}(4.44)

Mechanical energy is not conserved when this pendulum is in motion,because the friction dissipates kinetic energy as heat. The total energy,

F(θ, ω) =12

ω2 + k(1− cos θ),

is an integral for the motion of the frictionless pendulum, but it cannot beexpected to serve as an integral when there is friction.

We can validate this physical intuition by differentiating F(θ, ω).

ddt

F(θ, ω) = ωdω

dt+ k sin θ

dt= ω[−bω− k sin θ] + k(sin θ)ω

= −bω2. (4.45)

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4.5. LYAPUNOV STABILITY TESTS 355

It follows that F(θ, ω) decreases at a rate proportional to ω2 and is not anintegral.

When (θ, ω) = (2nπ, 0) the pendulum is motionless at the bottom ofits trajectory. Physical intuition suggests that if the pendulum starts witha modest initial velocity near this equilibrium position, it will eventuallystop there. In other words, we expect the stationary point (2nπ, 0) to beasymptotically stable. It is possible to prove this by using theorem 4.1, butwe will explore a second method, due to Lyapunov, which focuses on themechanical energy F(θ, ω).

Lyapunov Functions

Let L(x, y) be a function with L(0, 0) = 0. If there is a number R > 0with L(x, y) > 0. for all points (x, y) except the origin inside the circlex2 + y2 = R2, then L(x, y) is said to be positive definite near the origin.If the inequality is not strict; and we only have L(x, y) ≥ 0 for x2 + y2 < R2,then L(x, y) is positive semidefinite near the origin.

We also define a function N(x, y), with N(0, 0) = 0, to be negativedefinite or semidefinite if −N(x, y) is positive definite of semidefinite, re-spectively. A function L(x, y) with L(x1, y1) = 0 is positive or negative(semi)definite near (x1, y1) if the function F(x, y) = L(x + x1, y + y1) hasthe corresponding property near the origin.

Consider an autonomous system of differential equations,

x′ = f (x, y)y′ = g(x, y).

}(4.46)

Let L(x, y) be a function whose partial derivatives ∂L∂x and ∂L

∂y are continu-ous, and define the function

L′(x, y) =∂L∂x

f (x, y) +∂L∂y

g(x, y).

Proposition 4.5.1 Let (x, y) = (φ(t), ψ(t)) be a solution of (4.46). Then

ddt

L(φ(t), ψ(t)) = L′(φ(t), ψ(t)).

PROOF. By the Chain Rule,

dLdt

(x, y) =∂L∂x

dxdt

+∂L∂y

dydt

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356 CHAPTER 4. STABILITY THEORY

=∂L∂x

f (x, y) +∂L∂y

g(x, y)

= L′(x, y).

Now set x = φ(t), y = ψ(t), and the proof is complete.

Definition: A Lyapunov function for a stationary point (x1, y1) of (4.46)is a function L(x, y) that is positive definite near (x1, y1), while L′(x, y) iseither positive or negative semidefinite.

Example 4.5.1 The total energy, F(θ, ω) = 12 ω2 + k(1− cos θ), is a Lyapunov

function for the pendulum system (4.44), for each of the stationary points (2nπ, 0).

SOLUTION. We will confine our attention to the stationary point at theorigin. Let R = 2π. Then for 0 < |θ| < R, we have 1− cos θ > 0, andhence F(θ, ω) > 0 for 0 < θ2 + ω2 < R2.Since ∂F

∂θ = k sin(θ), and ∂F∂ω = ω, we have

F′(θ, ω) = (k sin θ)(ω) + (ω)(−bω− k sin θ)

= −bω2.

Thus F′ is negative semidefinite. Since it vanishes on the θ-axis, F′ is notnegative definite.

The first stability test of Lyapunov is as follows.

Proposition 4.5.2 Suppose that L(x, y) is a Lyapunov function for a stationarypoint (x1, y1) of the system (4.46), where L′(x, y) is negative semidefinite near(x1, y1). Then (x1, y1) is stable.

For simplicity, we will assume in the proof of this proposition, and ofsubsequent results in this section, that the point (x1, y1) is the origin. Theset of points in the phase plane that lie inside the circle with radius R, cen-tered at the origin, will be denoted BR. The boundary circle is CR.

PROOF OF PROPOSITION 4.5.2. Choose R > 0 so that for all (x, y) ∈BR ∪ CR with (x, y) 6= (0, 0), L(x, y) > 0 and L′(x, y) ≤ 0. Let m denotethe minimum value of L(x, y) for (x, y) ∈ CR. Since the origin does notlie on CR, this number m is positive. Furthermore, since L is continuousand L(0, 0) = 0, there is a number r, with 0 < r ≤ R, such that for all

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4.5. LYAPUNOV STABILITY TESTS 357

points (x, y) ∈ Br, L(x, y) < m. Let (x, y) = (φ(t), ψ(t)) be a solution of thesystem (4.46) with (φ(0), ψ(0)) ∈ Br. By proposition 4.5.1,

ddt

L(φ(t), ψ(t)) = L′(φ(t), ψ(t)) ≤ 0,

provided that (φ(t), ψ(t)) ∈ BR. Therefore2 L(φ(t), ψ(t)) cannot increaseunless the orbit of (φ(t), ψ(t)) crosses CR, and the circle CR cannot be reachedunless L(φ(t), ψ(t)) increases to at least m.

Proposition 4.5.2 and example 4.5.1 together imply that the lower equi-librium position of the pendulum is stable. Since we expect asymptotic sta-bility, this is disappointing, but we will see that the method of Lyapunovcan be refined to obtain the desired result.

Example 4.5.2 Show that L(x, y) = x2 + y2 is a Lyapunov function for the ori-gin as a stationary point of the system

x′ = yy′ = −x

and conclude that the origin is stable.

SOLUTION. Since L(x, y) > 0 for all (x, y) 6= 0, L is positive definite.Further, L′(x, y) = (2x) · y + (2y) · (−x) ≡ 0. Therefore, L′(x, y) ≤ 0 for all(x, y), and, by proposition 4.5.2, the origin is stable.

The result of example 4.5.2 is not surprising, since the phase portrait isa center, and the orbits are circles — level curves of L(x, y).

Limit Sets

A point (x, y) is said to be a forward limit point of a solution (x, y) =(φ(t), ψ(t)) of the system (4.46) if there is a sequence tn → ∞ such that

limn→∞

(φ(tn), ψ(tn)) = (x, y).

The set of all forward limit points is denoted lim+(φ, ψ).Here are some examples to consider. Proofs are omitted.

2Readers of Catch-22, by Joseph Heller, will recognize this line of thought.

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358 CHAPTER 4. STABILITY THEORY

• If a stationary point is asymptotically stable, it is the only forwardlimit point of each orbit whose initial point is sufficiently near to it.

• A saddle point of a linear system is the forward limit point of anysolution in its attracted set. Orbits which do not lie on the attractedset may not have any forward limit points.

• Some solutions of nonlinear systems are are only defined on an inter-val (a, b) with b < ∞; these automatically have empty forward limitsets.

• It is possible for the forward limit set of an orbit to contain more thanone point. For example, the forward limit set of a periodic orbit is theentire orbit.

• If an orbit converges to a limit cycle, as in figure 4.13, each point ofthe limit cycle is a forward limit point.

The forward limit set of a solution

(x, y) = (φ(t), ψ(t))

of the system (4.46) is the set of all its forward limit points. Our notationfor this set is

lim +(φ, ψ)

The following two propositions are presented without proof. The firstsays that an orbit passing close enough to a stable stationary point musthave forward limit points. These forward limit points are not necessarilystationary points themselves. For example, consider the case where theequilibrium is neutrally stable: the limit points may lie in a periodic orbit.Of course, if the stationary point is asymptotically stable, all nearby orbitsconverge to it.

Proposition 4.5.3 Suppose that (x1, y1) is a stable stationary point of the sys-tem (4.46). Then there is a circle C centered at (x1, y1) such that if (x, y) =(φ(t), ψ(t)) is a solution of the system (4.46) starting inside C then

lim +(φ, ψ) 6= ∅.

The forward limit set of an orbit may contain more than one point, butaccording to the next proposition, it must be an invariant set. In otherwords, if a point x1 belongs to the forward limit set of an orbit, then the

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4.5. LYAPUNOV STABILITY TESTS 359

entire orbit determined by x1 is either equal to or a subset of the forwardlimit set. The best example to visualize for this proposition is the limit cy-cle.

Proposition 4.5.4 Assume that the functions f (x, y) and g(x, y) on the rightside of the system (4.46) have continuous partial derivatives everywhere in thex, y-plane. Then the forward limit set lim+(φ, ψ) of any solution (φ(t), ψ(t)) isan invariant set.

The Test for Asymptotic Stability

We can now formulate a criterion for asymptotic stability by combining theconcepts of limit sets and Lyapunov functions.

Proposition 4.5.5 Let (x1, y1) be a stationary point of the system (4.46). Supposethat there is a Lyapunov function L(x, y) for (x1, y1) such that L′(x, y) is negativesemidefinite near (x1, y1), and that there is a radius R > 0 such that the onlynonempty invariant subset of

AR = BR ∩ {(x, y) : L′(x, y) = 0}

is {(x1, y1)} itself (BR denotes the set of points inside the circle with center (x1, y1)and radius R). Then (x1, y1) is asymptotically stable.

The proof depends on the following lemma.

Lemma 4.5.6 Let g(t) be a continuous, nonincreasing function, defined for allt > 0. Suppose that g(t) ≥ 0 for all t > 0. Then limt→∞ g(t) exists.

The proof of lemma 4.5.6 is omitted. It is based on a fundamental prop-erty of the real number system: every set that has a lower bound has agreatest lower bound.

PROOF OF PROPOSITION 4.5.5. Again, assume that (x1, y1) = (0, 0). LetR be a radius such that AR, the set of points (x, y) ∈ BR where L′(x, y) = 0,contains no invariant set except (0, 0). By proposition 4.5.2, this stationarypoint is stable; hence there is an r > 0 such that if (x, y) = (φ(t), ψ(t)) is asolution of the system (4.46) with

(φ(0), ψ(0)) ∈ Br (4.47)

then(φ(t), ψ(t)) ∈ BR (4.48)

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360 CHAPTER 4. STABILITY THEORY

for all t > 0.We will show that any solution (x, y) = (φ(t), ψ(t)) satisfying the con-

dition (4.47) converges to the origin. Since condition (4.48) then holds for allt > 0, L(φ(t), ψ(t)) ≥ 0 and L′(φ(t), ψ(t)) ≤ 0. It follows that L(φ(t), ψ(t))is a non-increasing non-negative function. Therefore limt→∞ L(φ(t), ψ(t))exists, by lemma 4.5.6. Denote this limit by l.

If (x, y) is a forward limit point of the solution (φ(t), ψ(t)) then thereis a sequence tn → ∞ such that (φ(tn), ψ(tn)) → (x, y). Therefore, for any(x, y) ∈ lim+(φ, ψ)

L(x, y) = limn→∞

L(φ(tn), ψ(tn)) = l.

Let (x, y) = (φ(t), ψ(t)) be the solution of the system (4.46) with initialcondition (φ(0), ψ(0)) = (x, y). Since, by proposition 4.5.4, lim+(φ, ψ) isinvariant, (φ(t), ψ(t)) ∈ lim+(φ, ψ) for all t. It follows that L(φ(t), ψ(t)) ≡ lfor all t; differentiating, we have L′(φ(t), ψ(t)) ≡ 0. It follows that L′(x, y) =0; thus (x, y) ∈ AR.

We have shown that the forward limit set of any orbit starting in Bris an invariant subset of AR. By proposition (4.5.3), this forward limit setcannot be empty; therefore it must be {(0, 0)}. It follows that the origin isasymptotically stable.

Example 4.5.3 Show that the stationary points (θ, ω) = (2nπ, 0) of the sys-tem (4.44), which describes the motion of a linearly damped pendulum, are asymp-totically stable.

SOLUTION. We have seen in example 4.5.1 that the total energyF(θ, ω) = 1

2 ω2 + k(1− cos θ) is a Lyapunov function for each of thesestationary points, and F′(θ, ω) = −bω2 is negative semidefinite. To applyproposition 4.5.5, it is necessary to find a number R such that the onlynonempty invariant subset of AR is the stationary point (2nπ, 0).The function F′ vanishes only on the θ-axis. The invariant subsets of thisaxis are composed of the equilibria (nπ, 0). Hence, if R < π, the set ARwill only contain one stationary point, and thus only one nonemptyinvariant subset. The equilibrium point, (2nπ, 0) is therefore asymptoticallystable. Figure 4.29 is a phase portrait for equation (4.44).

There is a Lyapunov test for instability:

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4.5. LYAPUNOV STABILITY TESTS 361

Proposition 4.5.7 Suppose that L(x, y) is a Lyapunov function for the station-ary point (x1, y1) of the system (4.46), such that L′(x, y) is positive definite near(x1, y1). Then (x1, y1) is unstable.

PROOF. As usual, we will take (x1, y1) to be the origin. Choose R > 0such that L(x, y) > 0 and L′(x, y) > 0 for all (x, y) ∈ BR or on its boundarycircle CR (except for (x, y) = (0, 0)). We will show that every orbit thatstarts at a point inside BR other than the origin eventually crosses the circleCR.

Let M be the maximum value of L(x, y) for (x, y) ∈ BR ∪ CR. Supposethat (x, y) = (φ(t), ψ(t)) is a solution of the system (4.46) with (φ(0), ψ(0)) =(x0, y0) ∈ BR. Put m = L(x0, y0); assuming that (x0, y0) 6= (0, 0), m > 0.Since L is continuous and L(0, 0) = 0, a number r > 0 can be chosen suchthat for all (x, y) ∈ Br, L(x, y) < m

2 . Since

ddt

L(φ(t), ψ(t)) = L′(φ(t), ψ(t)) > 0

as long as L(φ(t), ψ(t)) ∈ BR, L(φ(t), ψ(t)) is increasing. Therefore

L(φ(t), ψ(t)) ≥ m

for all t ≥ 0, unless (φ(t), ψ(t)) escapes from BR.Let K denote the annulus {(x, y) : r ≤

√x2 + y2 ≤ R}. We will show

that the orbit of (φ(t), ψ(t)) must leave K. Since it cannot cross the innerboundary circle, it must cross the outer one, and the proof will be complete(see figure 4.30).

Let k = L′(x∗, y∗) be the minimum value of L′(x, y) for all (x, y) ∈ K.Since

L′(x, y) > 0

for all (x, y) ∈ K, L(x∗, y∗) > 0, and thus k > 0. Put T = M−m+1k . If we

assume that (φ(t), ψ(t)) ∈ K for all t > 0, then

L(φ(T), ψ(T)) = m +∫ T

0L′(φ(t), ψ(t)) dt

≥ m + kT = M + 1.

This is impossible, since L(x, y) ≤ M for all (x, y) ∈ K.

Example 4.5.4 Let q(x, y) be a function with continuous partial derivatives. Showthat L(x, y) = x2 + y2 is a Lyapunov function for the origin as a stationary pointof the system

x′ = −y + xq(x, y)y′ = x + yq(x, y)

}(4.49)

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362 CHAPTER 4. STABILITY THEORY

if either

(a) q is positive definite, or

(b) q is negative definite.

Use this Lyapunov function to determine whether or not the origin is stable, asymp-totically stable, or unstable.

SOLUTION. The function L(x, y) = x2 + y2 is positive definite, and

L′(x, y) = 2x(y + xq(x, y)) + 2y(−x + yq(x, y))= 2(x2 + y2)q(x, y)

is positive (negative) definite if q(x, y) is positive (negative) definite.Therefore, L(x, y) is a Lyapunov function in either case. By Lyapunov’sstability tests, the origin is an unstable stationary point if q is positivedefinite, and it is asymptotically stable if q is negative definite.

The origin is not a linearly stable stationary point of the system (4.49). A

brief calculation shows that the derivative of F(x, y) =

[−y + xq(x, y)x + yq(x, y)

]is

A(x, y) =[

xqx(x, y) + q(x, y) 1 + xqy(x, y)−1 + yqx(x, y) yqy(x, y) + q(x, y)

]where qx and qy denote the partial derivatives of q. Since q(0, 0) = 0,

A(0, 0) =[

0 1−1 0

], with eigenvalues±i. Since the eigenvalues are not in

the left half-plane this does not imply linear stability. Indeed, A(0, 0) doesnot involve q at all, whereas we have seen that the stability of the origindepends on q.

Example 4.5.5 Show that L(x, y) = x2 + y2 is a Lyapunov function for the sys-tem

x′ = −3kx + y + kx3

y′ = −x

}(4.50)

and use it to draw conclusions about the stability of the equilibrium at the origin.

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4.5. LYAPUNOV STABILITY TESTS 363

SOLUTION. L(x, y) is obviously positive definite, and

L′(x, y) = 2x(−3kx + y + kx3)− 2xy= 2kx2(−3 + x2).

If k > 0, L′(x, y) ≤ 0 for |x| <√

3, with equality only if x = 0. Thus theorigin is stable by proposition 4.5.2. To apply proposition 4.5.5, theinvariant subsets of the y-axis, where L′(x, y) vanishes, must bedetermined. If x = 0, the first equation reduces to x′ = y. If y 6= 0, an orbitstarting at (0, y) cannot be a subset of the y-axis, since any orbit on they-axis must have x′ = 0. Therefore, the origin is the only invariant subsetof the y-axis, and we conclude that if k > 0, then the origin isasymptotically stable.If k < 0, the origin is not stable. Unfortunately, L′ is not positive definitenear the origin (it is only positive semidefinite), so we cannot applyproposition 4.5.7 to reach this conclusion. Instead, we will assume stabilityand reach a contradiction. By proposition 4.5.4, there is a number r > 0such that if (x, y) = (φ(t), ψ(t)) is a solution of the system with(φ(0), ψ(0)) ∈ Br(0, 0), then lim+(φ, ψ) 6= ∅. Reasoning similar to theproof of proposition 4.5.5 shows that lim+(φ, ψ) must be an invariantsubset of the y-axis. The only such subset is the origin. However, if(φ(0), ψ(0)) 6= (0, 0) then L(φ(t), ψ(t)) ≥ L(φ(0), ψ(0)) > 0, and it followsthat lim+(φ, ψ) 6= {(0, 0)}, a contradiction.

Notice that if k = 0, the system (4.50) is the linear system x′ = y, y′ =−x, whose orbits are circles. This system is neutrally stable. The derivativematrix of the system (4.50) is

A(x, y) =[

3k(x2 − 1) 1−1 0

].

The characteristic polynomial of A(0, 0) is p(s) = s2 + 3ks + 1. The eigen-values are complex for |k| < 2

3 , double for k = ± 23 , and a pair of real num-

bers for |k| > 23 . The roots have negative real parts for k > 0 and positive

real parts for k < 0. Therefore, theorem 4.1 is applicable in this case, andthe phase portrait will be an unstable node if k < 0 and a stable node ifk > 0, of spiral type if |k| < 2

3 . See figure 4.31.

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364 CHAPTER 4. STABILITY THEORY

Finding Lyapunov Functions

The principal difficulty in using the Lyapunov tests for stability is in find-ing a Lyapunov function. There is no general procedure, but systems whicharise in applications are frequently accompanied by natural Lyapunov func-tions. For example, the total mechanical energy of the damped pendulumis a Lyapunov function.

If the functions f (x, y) and g(x, y) on the right side of the system (4.46)are polynomials in x and y, there may be a Lyapunov function of the formL(x, y) = a(x − x1)

2 + b(x − x1)(y − y1) + c(y − y1)2. You can show that

this function is definite near (x1, y1) if and only if b2 − 4ac < 0. Of course,L′ will not be quadratic (unless we are dealing with a linear system). Thefollowing proposition makes it possible to use this test for definiteness forsome non-quadratic functions. A function P(x, y) is said to vanish to order 2at the origin if P(0, 0) = 0, and

∂P∂x

(0, 0) =∂P∂y

(0, 0) =∂2P∂2x

(0, 0) =∂2P

∂x∂y(0, 0) =

∂2P∂2x

(0, 0) = 0.

Proposition 4.5.8 Let L2(x, y) = ax2 + bxy + cy2 and let

L(x, y) = L2(x, y) + P(x, y),

where P(x, y) vanishes to order 2 at the origin. Then

• If L2 is positive or negative definite, then L has the same property near theorigin.

• If L2 is positive semidefinite, L is not necessarily positive semidefinite nearthe origin.

The proof of the proposition is left to you; see problem 26 at the endof this section for hints. To see that there can be no conclusion when L2is only semidefinite, consider L(x, y) = x3 + y3, which is neither positivenor negative semidefinite. If we put L2(x, y) = 0 (this is both positive andnegative semidefinite, then L(x, y) = L2(x, y) + P(x, y) where P(x, y) =x3 + y3 vanishes to order 2 at the origin.

Example 4.5.6 Find a Lyapunov function of the form L(x, y) = ax2 + cy2 forthe system

x′ = −x + 4y + x3 + xy2

y′ = −x− 2y + 2y3 + 2x2y

and determine the stability of the origin.

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4.5. LYAPUNOV STABILITY TESTS 365

SOLUTION. Since the function L(x, y) = ax2 + cy2 is required to bepositive definite, a and c must be positive. We will now calculate L′.

L′(x, y) = 2ax(−x + 4y + x3 + xy2) + 2by(−x− 2y + 2y3 + 2x2y)= −2ax2 + (8a− 2c)xy− 4cy2 + 2ax4 + (2a + 4c)x2y2 + 4cy4.

By proposition 4.5.8, L′(x, y) will be positive or negative definite if theexpression obtained by deleting the third and higher order terms is alsopositive or negative definite. We will therefore try to adjust the parametersa and c so that the expression

L′2(x, y) = −2ax2 + (8a− 2c)xy− 4cy2

is negative definite (positive definite is out of the question, since thecoefficients of x2 and y2 are negative). Observe that if we put a = 1 andc = 4, the coefficient of xy will vanish, giving the expressionL′2(x, y) = −2x2 − 16y2, which is negative definite. Therefore L′(x, y) isalso negative definite, and hence L(x, y) = x2 + 4y2 is a Lyapunovfunction that shows the origin to be asymptotically stable.

Exercises

1. Show that L(x, y) = ax2 + bxy + cy2 is positive definite near theorigin if and only if b2 − 4ac < 0 and a > 0.Answer

2. Suppose that L(x, y) is a Lyapunov function for a stationary point(x1, y1) of a system of differential equations. Show that if L′ ≡ 0then L is an integral, and that (x1, y1) is stable but not asymptoticallystable.

In each of problems 3 – 8, show that L(x, y) = x2 + y2 is a Lyapunovfunction for the given system at the origin, and draw a conclusion aboutstability.

3.{

x′ = −y + x3

y′ = x + y3.Answer

4.{

x′ = −5x + 2yy′ = 4x− 3y.

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366 CHAPTER 4. STABILITY THEORY

5.{

x′ = y− sin xy′ = −x− sin y

Answer

6.{

x′ = 2xy2 − x3 + 4y3

y′ = 4x3 − 8x2y− y3.

7.{

x′ = x3 + x + yy′ = −2x + y

Answer

8.{

x′ = −2y + y2

y′ = 2x + y− xy− y3

In problems 9 – 17, try to find a Lyapunov function of the form L(x, y) =ax2 + cy2 and determine the stability of the origin as a stationary point. Itwill occasionally be necessary to take L(x, y) = ax2 + bxy + cy2.

9.{

x′ = −x + yy′ = −x− y

Answer

10.{

x′ = 100yy′ = −x

11.{

x′ = 2x− yy′ = 40x + 2y

Answer

12.{

x′ = −x− 5yy′ = 5x + 5y.

13.{

x′ = x− yy′ = 50x− y

Answer

14.{

x′ = −x3 − yy′ = x− 4y3

15.{

x′ = 2x + 5(y2 − y)y′ = 5(x− x2)− 4y

Answer

16.{

x′ = 4y− xy2

y′ = −x− x2y

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4.5. LYAPUNOV STABILITY TESTS 367

17.{

x′ = 3y− x3 − xy2

y′ = −4x− x2y− y3

Answer

18. Prove Lyapunov’s Test for Asymptotic Stability: a stationary point isasymptotically stable if there is a Lyapunov function L for it with theproperty that L′ is negative definite.

19. What happens if, in the system 4.49 in example 4.5.4, q(x, y) isnegative semidefinite?Answer

20. The Lotka-Volterra equations

x′ = x(a− by)y′ = cy(x− d),

}are a model for describing the dynamics of a predator-prey relation-ship, with x representing the prey population, and y the predatorpopulation. The system was explored in some detail in section 2.5,where we found that there are two stationary points: the origin ~0,and ~p = (d, a

b ).

Show that ~p is stable, but not asymptotically stable.

21. Put q(x, y) = (x2 + y2)(1− x2 − y2) in the system (4.49) in exam-ple 4.5.4

(a) Show that the origin is unstable.

(b) Show that x = cos t, y = sin t is a solution of the system (4.49),whose orbit is the unit circle, traversed counterclockwise.

(c) Show that the orbit found in part (b) is a limit cycle.

(d) Computer laboratory problem. Draw the phase portrait of thissystem.

Answer

22. A certain nonlinear, damped mass spring apparatus is governedby the differential equation

x′′ + b|x′|x′ + k(x− x3) = 0.

Show that the total mechanical energy is a Lyapunov function for thisspring, and that the solution x = x′ ≡ 0 is asymptotically stable.

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368 CHAPTER 4. STABILITY THEORY

23. Computer laboratory problem. Figure 4.31 indicates that the sys-tem 4.50 has a limit cycle. Investigate what happens for other valuesof k. Does the limit cycle persist?Answer

24. The following is a proof, in outline form, of theorem 4.1 on page 335.Suppose that ~x1 is a linearly stable stationary point of the system

~x′ = ~f (~x); (4.51)

that is, ~f (~x1) =~0, and all eigenvalues of A(~x1), the derivative matrixof ~f at ~x1, are either negative real numbers or complex numbers withnegative real parts. We will refer to this matrix simply as A.

A symmetric matrix S will be called positive definite if the functionL(~x) = ~xT · S · ~x is positive definite. Here, we view ~x as a columnmatrix; ~xT denotes its transpose, a row matrix, and L(~x), which is theproduct of the three matrices, is a 1× 1 matrix.

(a) Show that if there is a nonsingular matrix P such that S = P · PT

then S is positive definite.

(b) Show that L(~x) = ~xT · S · x is a Lyapunov function for the originas a stationary point of

~x′ = A~x

if S is positive definite, and

AT · S + S · A is negative definite. (4.52)

(c) Use problem 18 to conclude that if there exists a positive definitesymmetric matrix S that satisfies condition (4.52), then ~x1 is anasymptotically stable stationary point of the system (4.51).

To prove theorem 4.1, we will show how to find a positive definitesymmetric matrix S that satisfies condition (4.52), provided that alleigenvalues of A have negative real parts.

(d) Let Y(t) = eAT t · (eAT t)T. Show that if the eigenvalues of A areeither negative, or have negative real parts, then∫ ∞

0Y(t) dt

converges, and let S denote the value of this integral.

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4.6. CHAPTER GLOSSARY 369

(e) Show that S, as defined in part (d), is positive definite.

(f) Show that AT · S+ S · A = −Y(0), and complete the proof. Hint:differentiate Y(t), and note that

−Y(0) =∫ ∞

0[dY(t)/dt] dt.

25. Take A =

[−1 10 −1

]. Calculate S by the method outlined in

problem 24 and show that S is positive definite and satisfies condi-tion (4.52).Answer

26. Prove proposition 4.5.8. Hints: (1) The origin is a critical point ofL. (2) Use the second derivative test.

4.6 Chapter Glossary

In this glossary, definitions apply to a system of n ODEs except when statedto the contrary. Thus we refer to phase space rather than the phase plane. Tofix notation, the system will be denoted

v′ = F(v),

if it is autonomous, and v′ = F(v, t) if it is not. Here,

v =

x1x2...

xn

,

is the vector of dependent variables, and

F(v, t) =

f1(v, t)f2(v, t),

...fn(v, t)

denotes a vector function defined on a subset of Rn ×R. Finally, if v∗ ∈ Rn

and r > 0, let Br(v∗) denote the set of points in Rn at a distance less than rfrom v∗. The set Br(v∗) is called the ball of radius r centered at v∗.

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370 CHAPTER 4. STABILITY THEORY

Asymptotically stable (stationary point) A stationary point v∗ such thatthere is a ball Br(v∗) and every solution v(t) with v(t1) ∈ Br(v∗) atsome time t1 has the property

limt→∞

v(t) = v∗

Attracted A point v1 is attracted to a stationary point v∗ if any solutionwith v(t1) = v1 for some t1 converges to v∗. The set of points that areattracted to v∗ is called the attracted set of v∗. More generally, if S isan invariant set, then v1 is attracted to S if the forward limit set ofevery solution with v(t1) = v1 for some t1 is contained in S .

Center A system of two linear ODEs in which the characteristic roots areimaginary numbers. The orbits are ellipses, centered at the origin.

Degenerate system A system of ODEs v′ = Av in which the matrix A issingular.

Forward limit point See Limit point.

Improper node A system of linear ODEs in which not all orbits are half-lines, but all orbits are directed toward the origin as t→ ∞ (the stablecase), or all orbits are directed toward the origin as t → −∞ (theunstable case).

Invariant set (of an autonomous system) A subset S of phase space withthe property that every orbit of the system that contains a point of Sis a subset of S .

lim+ See Limit point.

Limit cycle A closed orbit that is the forward limit set of all nearby orbits.

Limit point (of a solution v(t)). There are forward and backward limitpoints. v∗ is a forward limit point if there is an increasing, unboundedsequence of real numbers (we use the suggestive notation tm ↑ ∞)such that

limm→∞

v(tm) = v∗. (4.53)

Similarly, v∗ is a backward limit point if there is a sequence tm ↓ −∞such that (4.53) holds. Limit points can also be ascribed to the orbitrepresented by v(t). The set of forward limit points of v(t) is denotedlim+(v) and the set of backward limit points is lim−(v). These sets

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4.6. CHAPTER GLOSSARY 371

can be empty, consist of one point to which the orbit converges, orcan be infinite. Some texts refer to the backward and forward limitsets as the alpha and omega sets, respectively, of the orbit. In thesetexts, the notations α(v) and ω(v) are used in place of lim−(v) andlim+(v), respectively

Limit set See Limit point.

Linearly stable (stationary point of an autonomous system) A stationarypoint v∗ such that all of the characteristic roots of the derivative A(v∗)of F at v∗ are situated to the left of the imaginary axis of the complexplane. By Perron’s theorem (page 335), linearly stable implies asymp-totically stable, provided that the second partial derivatives of thecomponents of F are continuous.

Node A system of linear ODEs in which all orbits are directed toward theorigin as t → ∞ (the stable case), or all orbits are directed toward theorigin as t→ −∞ (the unstable case).

Positive definite A function L(x1, x2, . . . , xn) such that L(0, 0, . . . , 0) = 0and L(x1, x2, . . . , xn) > 0 for (x1, x2, . . . , xn) 6= L(0, 0, . . . , 0). The func-tion L is said to be positive semidefinite. if L(x1, x2, . . . , xn) ≥ 0. Ifone of the above inequalities holds only for all (x1, x2, . . . , xn) ∈ Br(0)for some r > 0, then we say that L is positive definite or semidefinitenear the origin.

Proper node A linear system of ODEs in which the orbits are all half-lines.The half-lines will all be directed toward the origin (the stable case)or all directed away from the origin (the unstable case).

Repelled A point v1 is repelled by stationary point v∗ if any solution withv(t1) = v1 for some t1 converges to v∗ as t → −∞. The set of pointsthat are repelled by v∗ is called the repelled set of v∗. More generally,if S is an invariant set, then v1 is repelled by S if the backward limitset of every solution with v(t1) = v1 for some t1 is contained in S .

Saddle A linear system of two ODEs that has two real characteristic rootsof opposite sign, or a stationary point v∗ of a nonlinear autonomoussystem of two equations v′ = F(v) such that the characteristic rootsof the derivative A(v∗) of F at v∗ are real and of opposite sign. Nearthe stationary point, the phase portrait looks like a linear saddle.

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372 CHAPTER 4. STABILITY THEORY

Semisimple characteristic root A characteristic root of a matrix with mul-tiplicity (as a root of the characteristic equation) equal to the maxi-mum number of linearly independent characteristic vectors belong-ing to it.

Separatrix (applies to an autonomous system of two equations) An orbitthat converges to a saddle. Orbits on either side of a separatrix havedifferent limiting behavior.

Spiral node A linear system of two ODEs in which the characteristic rootsare neither real nor imaginary. The orbits swirl around the origin inspirals.

Stable (stationary point v∗ of an autonomous system) For any number R >0, there is a number r > 0 such that every orbit that passes within adistance r from v∗ will be entirely within the set of points at distanceless than R from v∗.

Stable line A straight line through the origin that is the union of the origin,and two orbits directed toward the origin. A linear system of twoODEs that is a saddle has a unique stable line. In the case of a stableproper node, all lines through the origin are stable.

Stable node A system of linear ODEs in which all orbits approach the ori-gin as t→ ∞.

Stationary point (of an autonomous system) A point v∗ such that F(v∗) =0. A stationary point represents a constant solution of the system.

Unstable line A straight line through the origin that is the union of theorigin, and two orbits directed away from the origin. A linear systemof two ODEs that is a saddle has a unique unstable line. In the case ofan unstable proper node, all lines through the origin are unstable.

Suggestions for Further Reading

Many advanced texts give complete proofs of the relation between thephase portrait of a system of differential equations near an equilibriumpoint and the phase portrait of the linearized system. The books OrdinaryDifferential Equations, by Philip Hartman3 and Ordinary Differential Equa-

3Second edition, New York: Wiley, 1973.

Page 374: Ordinary Differential Equations: A Systems Approach

4.6. CHAPTER GLOSSARY 373

tions, by Jack Hale4 are graduate level texts which contain rather completetreatments of this topic.

Applications of differential equations to biology is the subject of a greatdeal of current research. An outstanding text in this area is MathematicalBiology, by J. D. Murray5. Murray’s book is comprehensive, has few pre-requisites, and is well written. It has the additional advantage, unusual fora mathematics text, of being suitable for browsing.

4Second edition, Malabar, Florida: Krieger, 19805New York: Springer-Verlag, 1989.

Page 375: Ordinary Differential Equations: A Systems Approach

374 CHAPTER 4. STABILITY THEORY

Figure 4.1: A proper node: the phase portrait of v′ = Iv.

Page 376: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 375

Figure 4.2: Obtaining the phase portrait of a system by distortion of a thephase portrait of a conjugate system with a diagonal coefficient matrix.

Let A be a 2× 2 matrix with distinct real eigenvalues r and s, and let ~e and ~f beeigenvectors belonging to these eigenvalues. Let P be the matrix with columns~e, ~f , and let D be the diagonal matrix diag (r, s). Then A and D are conjugate:A = PDP−1.The general solution of (4.2) is

~w(t) = eDt~c = αer t~i + βes t~j,

where ~c =

[αβ

]. Therefore the phase portrait of (4.2) is made up of orbits with

parametric equationsx = αert, y = βest.

These will be half-lines on the x- and y-axes (directed toward the origin if the eigen-values are positive, and away from it if the eigenvalues are negative) and curves

|y| = constant · |x|s/r

We want to see how to distort this phase portrait to obtain the phase portraitof (4.1).The general solution of (4.1) is

~v(t) = eAt~k = αer t~e + βes t~f ,

where we have expressed the vector~k as a linear combination of~e = P~i and ~f = P~j:

~k = α~e + β~f = P(α~i + β~j).

Because eAt = PeDtP−1, then it follows that

~v(t) = PeDtP−1~k= PeDtP−1[P(α~i + β~j)]

= PeDt(α~i + β~j) = P~w(t)

Therefore the transformation ~w 7→ ~v = P~w distorts the phase portrait of (4.2) toobtain the phase portrait of (4.1).

Page 377: Ordinary Differential Equations: A Systems Approach

376 CHAPTER 4. STABILITY THEORY

Figure 4.3: A stable improper node: phase portrait of x′ = −2x, y′ = −y.

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FIGURES FOR CHAPTER 4 377

Figure 4.4: A node with two distinct, positive characteristic roots: see example 4.1.1. Thearrows in this figure show the vector field corresponding to the system of ODEs.

Page 379: Ordinary Differential Equations: A Systems Approach

378 CHAPTER 4. STABILITY THEORY

Figure 4.5: An unstable improper node corresponding to a double characteristic root:the vector field and phase portrait of x′ = x, y′ = 2x + y. See example 4.1.2.

Page 380: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 379

Figure 4.6: A saddle: see example 4.1.3.

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380 CHAPTER 4. STABILITY THEORY

Figure 4.7: A center: see example 4.1.4.

Page 382: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 381

Figure 4.8: Optional: Drawing the elliptical orbits in example 4.1.4

Let R(t) = [x(t)]2 + [y(t)]2. Then

dRdt

= 2x · x′ + 2y · y′.

Substituting x′ = 2x + 8y and y′ = −5x− 2y (from the differentialequations), we find that

dRdt

= 2x(2x + 8y) + 2y(−5x− 2y)

= 4x2 + 6xy− 4y2

= 2(2x− y)(x + 2y)

At the extreme points of the ellipse, dRdt = 0; this will occur when

y = 2x and when y = − 12 x. These lines contain the axes of each

orbit. You may wish to show that the longer axis is on the line y =2x and that the ratio of the lengths of the axes is 3.

Page 383: Ordinary Differential Equations: A Systems Approach

382 CHAPTER 4. STABILITY THEORY

Figure 4.9: A spiral node: see example 4.1.5.

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FIGURES FOR CHAPTER 4 383

Figure 4.10: A spiral node. See example 4.1.6. The dashed ellipse is an orbit of the relatedsystem that was presented in Example 4.1.4.

Page 385: Ordinary Differential Equations: A Systems Approach

384 CHAPTER 4. STABILITY THEORY

Figure 4.11: A degenerate system: see example 4.1.7

- �

-�

-

-

Page 386: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 385

Figure 4.12: A degenerate system: 0 is a double characteristic root. See example 4.1.8

6 6

? ?

Page 387: Ordinary Differential Equations: A Systems Approach

386 CHAPTER 4. STABILITY THEORY

Figure 4.13: ). Phase portrait of the nonlinear system (4.22 The unit circle is a limit cycle.(See example 4.2.1.)

-1.5 -1 -0.5 0.5 1 1.5

-1.5

-1

-0.5

0.5

1

1.5

Page 388: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 387

Figure 4.14: The origin is a saddle point of the system 4.24

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Page 389: Ordinary Differential Equations: A Systems Approach

388 CHAPTER 4. STABILITY THEORY

Figure 4.15: The origin is a stable stationary point of the systemx′ = −y− x(1−

√x2 + y2), y′ = x− y(1−

√x2 + y2).

-2 -1.5 -1 -0.5 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

Page 390: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 389

Figure 4.16: The system x′ = −y− x(x2 + y2), y′ = x− y(x2 + y2) has an asymptoticallystable stationary point at the origin. The characteristic roots of the linearization at the originare ±i, so the linearization is a center. One orbit is shown.

-1 1

-0.5

0.5

Page 391: Ordinary Differential Equations: A Systems Approach

390 CHAPTER 4. STABILITY THEORY

Figure 4.17: Phase portrait of the nonlinear system in example 4.2.3. The dashed linesare nullclines.

-4 -2 2 4

-4

-2

2

4

Page 392: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 391

Figure 4.18: Attracted set of the origin for the system in examples 4.2.3 – 4.2.5.

-4 -2 2 4

-4

-2

2

4

Page 393: Ordinary Differential Equations: A Systems Approach

392 CHAPTER 4. STABILITY THEORY

Figure 4.19: A stable stationary point. See example 4.2.6.

-1 -0.5 0.5 1

-1

-0.5

0.5

1

R

r

Page 394: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 393

Figure 4.20: The stationary point at (1, 0) is unstable, although all orbits except x ≡0, y ≡ 0 converge to it. See problem 17.

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

Page 395: Ordinary Differential Equations: A Systems Approach

394 CHAPTER 4. STABILITY THEORY

Figure 4.21: Competitive exclusion The attracted set of the stationary point (x1, y1) is theseparatrix. If the initial population vector lies below the separatrix, the x-species dominatesand the y-species becomes extinct. If the initial population is above the separatrix, the y-species is dominant. The parameter configuration is p < L and q < C. The nullclines areshown as dashed lines. In drawing this phase portrait, the parameter values (k, a, b, l, c, d) =(3, 1, 2, 5, 2, 3) were used.

0.5 1 1.5 2 2.5 3

0.5

1

1.5

2

Page 396: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 395

Figure 4.22: Dominance. The species with population x will outcompete its competitorunless it is entirely removed from the environment. The parameter configuration is C > qand L < p. The nullclines are represented by dashed lines. In drawing the phase portrait,the parameter values (k, a, b, l, c, d) = (30, 5, 6, 6, 2, 3) were used.

2 4 6 8

2

4

6

8

Page 397: Ordinary Differential Equations: A Systems Approach

396 CHAPTER 4. STABILITY THEORY

Figure 4.23: Coexistence.When L < p and C < q the species will coexist at the stationarypoint in the first quadrant. The initial population vector has no effect on the outcome, aslong as both species are represented. The nullclines are represented by dashed lines. Indrawing this phase portrait, the parameter values (k, a, b, c, l, c, d) = (15, 5, 3, 12, 2, 6) wereused.

1 2 3 4 5 6

1

2

3

4

5

6

Page 398: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 397

Figure 4.24: A potential well: the orbit cannot escape.

-4 -2 2 4

-1.5

-1

-0.5

0.5

1

Page 399: Ordinary Differential Equations: A Systems Approach

398 CHAPTER 4. STABILITY THEORY

Figure 4.25: An unstable stationary point. The critical point of U(x) at the origin is not arelative minimum.

Page 400: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 399

Figure 4.26: A separatrix: U(x) has a relative maximum at x1.

Page 401: Ordinary Differential Equations: A Systems Approach

400 CHAPTER 4. STABILITY THEORY

Figure 4.27: A simple pendulum

r

q

�����

������������

������������

����HHHHHH

HHj x

θ

���������

y

,

?

mg

Page 402: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 401

Figure 4.28: Phase portrait of the pendulum equation. The variables θ and ω representangular displacement and angular velocity, respectively.

Page 403: Ordinary Differential Equations: A Systems Approach

402 CHAPTER 4. STABILITY THEORY

Figure 4.29: A phase portrait of the linearly damped pendulum equation (4.44).

Page 404: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 403

Figure 4.30: L and L′ are both positive definite. In the annulus K, L′(x, y) ≥ k, andL(x, y) ≤ M.

x

y

K

Page 405: Ordinary Differential Equations: A Systems Approach

404 CHAPTER 4. STABILITY THEORY

Figure 4.31: Phase portrait of the system (4.50). The parameter k = −0.2.

Page 406: Ordinary Differential Equations: A Systems Approach

FIGURES FOR CHAPTER 4 405

Answers to Selected Exercises

1.

(i) order 2, (ii) ODE.Return

Page 407: Ordinary Differential Equations: A Systems Approach

406 ORDINARY DIFFERENTIAL EQUATIONS

3.

(i) order 1, (ii) ODE.Return

Page 408: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 407

5.

(i) order 1, (ii) ODE.Return

Page 409: Ordinary Differential Equations: A Systems Approach

408 ORDINARY DIFFERENTIAL EQUATIONS

7.

(i) order 1, (ii) ODE.Return

Page 410: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 409

9.

(i) order 1, (ii) ODE.Return

Page 411: Ordinary Differential Equations: A Systems Approach

410 ORDINARY DIFFERENTIAL EQUATIONS

11.

(i) order 2, (ii) PDE.Return

Page 412: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 411

13.

(i) order 3, (ii) ODE.Return

Page 413: Ordinary Differential Equations: A Systems Approach

412 ORDINARY DIFFERENTIAL EQUATIONS

15.

(a) y = C sin(t)

(b) y = C sin(t)

(c) y = 2 sin(t)

(d) y = − sin(t) + cos(t)

ReturnDetails

Page 414: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 413

17.

y = 3et2.

ReturnDetails

Page 415: Ordinary Differential Equations: A Systems Approach

414 ORDINARY DIFFERENTIAL EQUATIONS

19.

Use implicit differentiation.Return

Details

Page 416: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 415

21.

ddt

( yt

)= 1

t . Use the equal derivatives theorem.ReturnDetails

Page 417: Ordinary Differential Equations: A Systems Approach

416 ORDINARY DIFFERENTIAL EQUATIONS

1.

55.47 years.ReturnDetails

Page 418: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 417

3.

By 27.726 per thousand.DetailsReturn

Page 419: Ordinary Differential Equations: A Systems Approach

418 ORDINARY DIFFERENTIAL EQUATIONS

5.

The relative growth rate of B is 2% per year.DetailsReturn

Page 420: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 419

7.

£65,659,969.14DetailsReturn

Page 421: Ordinary Differential Equations: A Systems Approach

420 ORDINARY DIFFERENTIAL EQUATIONS

9.

y = 1212 t12.

0.5 1 1.5 2 2.5

2

4

6

8

10

12

14

DetailsReturn

Page 422: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 421

11.

y = − 1cos(t) for −π

2 < t < π2

-1.5 -1 -0.5 0.5 1 1.5

-14

-12

-10

-8

-6

-4

-2

DetailsReturn

Page 423: Ordinary Differential Equations: A Systems Approach

422 ORDINARY DIFFERENTIAL EQUATIONS

13.

y = 1t for t > 0.

2 4 6

1

2

3

4

DetailsReturn

Page 424: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 423

15.

y = 0.DetailsReturn

Page 425: Ordinary Differential Equations: A Systems Approach

424 ORDINARY DIFFERENTIAL EQUATIONS

17.

0.675 ppb.DetailsReturn

Page 426: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 425

19.

0.556 ppb.DetailsReturn

Page 427: Ordinary Differential Equations: A Systems Approach

426 ORDINARY DIFFERENTIAL EQUATIONS

1.

y = 34 t− 3

16 +3

16 e−4t.DetailsReturn

Page 428: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 427

3.

y = 12 ln(t)e−t/2 − 1

2 ln(2)e−t/2.DetailsReturn

Page 429: Ordinary Differential Equations: A Systems Approach

428 ORDINARY DIFFERENTIAL EQUATIONS

5.

y = te−t2.

DetailsReturn

Page 430: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 429

7.

(a) y = 110 +

910 e−10t.

(b) y = − 110 +

1110 e10t.

DetailsReturn

-1 -0.5 0.5 1

0.25

0.5

0.75

1

1.25

1.5

Page 431: Ordinary Differential Equations: A Systems Approach

430 ORDINARY DIFFERENTIAL EQUATIONS

9.

(a) y = − cos(2t)e−4t + e−4t.

(b) y = − 117 e−4t(4 sin(2t) + cos(2t)) + 1

17 e4t.

-2 -1 1 2

-40

-20

20

40

60

80

DetailsReturn

Page 432: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 431

11.

y = 1− sec(t).DetailsReturn

Page 433: Ordinary Differential Equations: A Systems Approach

432 ORDINARY DIFFERENTIAL EQUATIONS

13.

Wait 11 minutes.DetailsReturn

Page 434: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 433

15.

81◦C. The assumptions are that the transmission coefficient in theoven is the same as that outside the oven, and that there are no othersources of heat.DetailsReturn

Page 435: Ordinary Differential Equations: A Systems Approach

434 ORDINARY DIFFERENTIAL EQUATIONS

17.

The heat loss rate is 0.0514 hour−1

DetailsReturn

Page 436: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 435

19.

35◦C.DetailsReturn

Page 437: Ordinary Differential Equations: A Systems Approach

436 ORDINARY DIFFERENTIAL EQUATIONS

21.

There are two approaches, with slightly different answers. For a quickapproximation, just assume the heat source mH(t) = 0.8, because theheat is on four-fifths of the time. This yields an average temperatureof 20◦C.

A more ambitious project is to determine the stable periodic temper-ature, which is not constant—the heat does cycle on and off. Takingthe average value of that function over a full period, [0, 5] one obtainsan average temperature of about 19.7◦C.DetailsReturn

Page 438: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 437

23.

(a) 529 (5 cos 2t + 2 sin 2t); stable.

(b) y = − 717 (cos 4t− 4 sin 4t) ; not stable.

(c) y = cos t− sin t; stable.

(d) y = − 126 (23 cos t + 11 sin t) ; not stable.

(e) There is no periodic solution.

DetailsReturn

Page 439: Ordinary Differential Equations: A Systems Approach

438 ORDINARY DIFFERENTIAL EQUATIONS

25.

(a)C

e(0.1) t +(−0.498753) e0.1 t cos(2. t) + (0.0249377) e0.1 t sin(2. t)

e(0.1) t

(b) The solution is

y = e−t2∫

et2dt + C e−t2

The CAS foundCet2 +

√π Erfi(t)

2 et2 ,

and seems to be using a special function of the form

Erfi(t) =2√π

∫et2

dt.

(c)Ct12 +

−t12

3456 +t12 ln(t)

288 − t12 ln(t)2

48 + t12 ln(t)3

12t12 .

(d)12−√

t + t +C

e2√

t

DetailsReturn

Page 440: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 439

27.

y(0) = 1.DetailsReturn

Page 441: Ordinary Differential Equations: A Systems Approach

440 ORDINARY DIFFERENTIAL EQUATIONS

29.

y(1/4) = 8.DetailsReturn

Page 442: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 441

In Exercises 1–10, assume that the uniformity hypothesis holds.

1.

35, 69, and 138 seconds, respectively. Notice that the time doubles foreach increment — can you say why?DetailsReturn

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442 ORDINARY DIFFERENTIAL EQUATIONS

3.

42 ppmDetailsReturn

Page 444: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 443

5.

No more than 13 hours ago.DetailsReturn

Page 445: Ordinary Differential Equations: A Systems Approach

444 ORDINARY DIFFERENTIAL EQUATIONS

7.

0.55 grams per liter.DetailsReturn

Page 446: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 445

9.

The concentration in tank A is 0.09 + 0.03e−4

15 t, and the concentrationin tank B is 0.09− 0.09e−

415 t.

DetailsReturn

Page 447: Ordinary Differential Equations: A Systems Approach

446 ORDINARY DIFFERENTIAL EQUATIONS

11.

$1 million.DetailsReturn

Page 448: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 447

13.

About 79,000DetailsReturn

Page 449: Ordinary Differential Equations: A Systems Approach

448 ORDINARY DIFFERENTIAL EQUATIONS

1.

y = (1 + Ct−3)1/3.DetailsReturn

Page 450: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 449

3.

y = ±√

37√2 cos t + 12 sin t + Ce6t

.

DetailsReturn

Page 451: Ordinary Differential Equations: A Systems Approach

450 ORDINARY DIFFERENTIAL EQUATIONS

5.

y = 4√

4(2t + 1)4 + C(2t + 1)−2.DetailsReturn

Page 452: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 451

1.

(a) Separable.

(b) Separable.

(c) Not separable.

(d) Separable.

(e) Separable.

(f) Not separable.

DetailsReturn

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452 ORDINARY DIFFERENTIAL EQUATIONS

3.

(a) − ln(0.8) = 0.223.

(b)ln(0.5)ln(0.8)

= 3.106 seconds.

(c) − 0.2ln(0.8)

= 0.8963 meters.

(d) − 1ln(0.8)

= 4.4814 meters.

DetailsReturn

Page 454: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 453

5.

b = 0.98 kg/s.DetailsReturn

Page 455: Ordinary Differential Equations: A Systems Approach

454 ORDINARY DIFFERENTIAL EQUATIONS

7.

y = −2( 1

2 t2 − t + C)−1

; y ≡ 0.DetailsReturn

Page 456: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 455

9.

y = ±C√

t; y = 2√

t.DetailsReturn

Page 457: Ordinary Differential Equations: A Systems Approach

456 ORDINARY DIFFERENTIAL EQUATIONS

11.

y = ±Ceet;y = eet−1.

DetailsReturn

Page 458: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 457

13.

y = − 1t+C ; y = − 1

t−1 .DetailsReturn

Page 459: Ordinary Differential Equations: A Systems Approach

458 ORDINARY DIFFERENTIAL EQUATIONS

15.

y−1y+1 = ±Cet2

; y = −et2+11+et2

.DetailsReturn

Page 460: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 459

17.

About 19 meters per second.DetailsReturn

Page 461: Ordinary Differential Equations: A Systems Approach

460 ORDINARY DIFFERENTIAL EQUATIONS

19.

(a)v2

∞g

ln

√v2

0 + v2∞

v∞.

(b) T1 =v∞

garctan

|v0|v∞

.

(c) Let T2 be the time taken to fall from the maximum height to the

ground. Then T2 = v∞g ln√

v20+v2

∞+|v0|v∞

.

If the drag force is negligible then

(a) the maximum height attained by the ball is v20

2g

(b) the time taken to reach that height is v0g

(c) the time taken to fall from the maximum height to the ground isv0g

(d) the velocity when the ball hits the ground is |v0|

DetailsReturn

Page 462: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 461

21.

t = (|v∞| ln(3))/(2g)DetailsReturn

Page 463: Ordinary Differential Equations: A Systems Approach

462 ORDINARY DIFFERENTIAL EQUATIONS

1.

F(x, y) = x2 + 5xy + 3x− 2y2 + 2y.DetailsReturn

Page 464: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 463

3.

The exactness condition does not hold.DetailsReturn

Page 465: Ordinary Differential Equations: A Systems Approach

464 ORDINARY DIFFERENTIAL EQUATIONS

5.

F(x, y) = e3xy(ln y− 1) + y2

4 −y2 ln y

2 .DetailsReturn

Page 466: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 465

7.

F(x, y) = x4

4 + x2y2

2 −x2

2 + y4

4 + y2

2 .DetailsReturn

Page 467: Ordinary Differential Equations: A Systems Approach

466 ORDINARY DIFFERENTIAL EQUATIONS

9.

F(x, y) = x− y2

x+y .DetailsReturn

Page 468: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 467

11.

F(x, y) = x3 + 3yx2 + 9y2x + 17y3.DetailsReturn

Page 469: Ordinary Differential Equations: A Systems Approach

468 ORDINARY DIFFERENTIAL EQUATIONS

13.

The exactness condition does not hold.DetailsReturn

Page 470: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 469

15.

m = ex22 is an integrating factor, and F(x, y) = y2e

x22 + xe

x22 is an

integral.DetailsReturn

Page 471: Ordinary Differential Equations: A Systems Approach

470 ORDINARY DIFFERENTIAL EQUATIONS

17.

m = ex is an integrating factor, and F(x, y) = 2xyex + 3y2ex + x2ex isan integral.DetailsReturn

Page 472: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 471

19.

m = y2 is an integrating factor, and F(x, y) = x2y3 + y4

2 is an integral.DetailsReturn

Page 473: Ordinary Differential Equations: A Systems Approach

472 ORDINARY DIFFERENTIAL EQUATIONS

23.

m = e∫

p(x) dx is an integrating factor.DetailsReturn

Page 474: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 473

1.

0 2 4 6 8 10

-100

-50

0

50

100

Return

Page 475: Ordinary Differential Equations: A Systems Approach

474 ORDINARY DIFFERENTIAL EQUATIONS

3.

-2 -1 0 1 2 3

-1

-0.5

0

0.5

1

1.5

Return

Page 476: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 475

5.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

ReturnDetails

Page 477: Ordinary Differential Equations: A Systems Approach

476 ORDINARY DIFFERENTIAL EQUATIONS

7.

-2 -1 0 1 2 3

-1

-0.5

0

0.5

1

1.5

Return

Page 478: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 477

8.

-2 -1 0 1 2 3

-1

-0.5

0

0.5

1

1.5

Return

Page 479: Ordinary Differential Equations: A Systems Approach

478 ORDINARY DIFFERENTIAL EQUATIONS

9.

At a crossing point, y′ = 0 and y′′ = 1 > 0; hence there is a relativeminimum by the second derivative test.DetailsReturn

Page 480: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 479

11.

(a) 0, 1, 2, 3, 4, . . .; ym = m− 1.

(b) 1, 2, 4, 8, . . . ; ym = 2m−1.

(c) C, Ck + 1, C(k + 1)2, C(k + 1)3, . . .; ym = C(k + 1)m−1.

(d) 0, 1, 3, 7, . . .; ym = 2m−1 − 1.

DetailsReturn

Page 481: Ordinary Differential Equations: A Systems Approach

480 ORDINARY DIFFERENTIAL EQUATIONS

13.

(c) ym = C m, where C is a constant.

(d) ym = m(m− 1).

DetailsReturn

Page 482: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 481

15.

y20 = 9.55. In this case, the final local error (LE20) is infinite, since theanalytic solution, y = 1/(1− t), has a singularity at t = 1.

0.2 0.4 0.6 0.8 1

2

4

6

8

10

DetailsReturn

Page 483: Ordinary Differential Equations: A Systems Approach

482 ORDINARY DIFFERENTIAL EQUATIONS

17.

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

DetailsReturn

Page 484: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 483

19.

Local errors: 0.101691, 0.0771727, 0.013463, -0.0506145Accumulated errors: 0.130573, 0.266751, 0.359802.DetailsReturn

Page 485: Ordinary Differential Equations: A Systems Approach

484 ORDINARY DIFFERENTIAL EQUATIONS

21.

(a) No.

(b) 0 < h < 0.02

DetailsReturn

Page 486: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 485

23.

(a) In the forward version,using ym+1 = ym +√

1− y2m, ym even-

tually increases to a value bigger than 1, and then√

1− y2m is

undefined.

(b) The plus sign should be used.

(c) Forward version: -0.000166583, 0.199499, 0.297489, 0.392961, 0.484917,0.572373, 0.654372, 0.729989, 0.798335, 0.858556, 0.909828, 0.951327,0.982145, 1.00096 (terminates)Backward version: 0.0995037, 0.00113599, 0.29314, 0.385415, 0.473494,0.55657, 0.633914, 0.70485, 0.768799, 0.825273, 0.873886, 0.914373,0.946611, 0.970657, 0.986832, 0.99589, 0.999388, 0.999982, 1, 1, . . .

(d) The backward version is the more accurate.

DetailsReturn

Page 487: Ordinary Differential Equations: A Systems Approach

486 ORDINARY DIFFERENTIAL EQUATIONS

25.

y = ±3 (singular solutions); y = 3 sin(t + C), where C is a constant,is a solution for all values of t such that y′ = 3 cos(t + C) is positive.

-4 -2 0 2 4-3

-2

-1

0

1

2

3

DetailsReturn

Page 488: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 487

27.

|1− y| = C|2− t|, where C is a constant.

1 2 3 4 5 6

0.5

1

1.5

2

2.5

3

3.5

DetailsReturn

Page 489: Ordinary Differential Equations: A Systems Approach

488 ORDINARY DIFFERENTIAL EQUATIONS

1.

y = −2, defined on (−∞,+∞).DetailsReturn

Page 490: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 489

3.

y = − 129

(5 cos(5t)− 2 sin(5t)− 5e−2t), defined on (−∞,+∞).DetailsReturn

Page 491: Ordinary Differential Equations: A Systems Approach

490 ORDINARY DIFFERENTIAL EQUATIONS

5.

y = t3 − t2, defined on (−∞,+∞).DetailsReturn

Page 492: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 491

7.

y = −12+ et2

defined on (−∞,+∞).DetailsReturn

Page 493: Ordinary Differential Equations: A Systems Approach

492 ORDINARY DIFFERENTIAL EQUATIONS

9.

If the graphs intersect, then there would be two solutions to the IVPwith initial condition at the intersection point.DetailsReturn

Page 494: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 493

11.

y ≡ 1.DetailsReturn

Page 495: Ordinary Differential Equations: A Systems Approach

494 ORDINARY DIFFERENTIAL EQUATIONS

13.

(i) t0, y0 could be any real numbers.

(ii) t0, y0 could be any real numbers.DetailsReturn

Page 496: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 495

15.

(i) t0 6= kπ, where k is an integer.

(ii) t0 6= kπ, where k is an integer.DetailsReturn

Page 497: Ordinary Differential Equations: A Systems Approach

496 ORDINARY DIFFERENTIAL EQUATIONS

17.

(i) y20 > 4t0, and t0 6= 0.

(ii) ∅.DetailsReturn

Page 498: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 497

19.

(i) t0y0 > 0.

(ii) t0y0 > 0 and y0 6= 1.DetailsReturn

Page 499: Ordinary Differential Equations: A Systems Approach

498 ORDINARY DIFFERENTIAL EQUATIONS

21.

(i) t0, y0 could be any real numbers.

(ii) t0 − y0 6= 0.DetailsReturn

Page 500: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 499

23.

∂ f∂y (t, y) = − y√

1−y2continuous if |y| < 1, but not continuous on |y| =

1 This implies that f (t, y) =√

1− y2 satisfies a Lipschitz conditionon the domain D if −1 < c and d < 1, but there is no implication ifeither c = −1 or d = 1.

In the case d = 1 let y2 = 1 and y1 < 1. Then

| f (t, y2)− f (t, y1)|y2 − y1

=

√1− y2

1

1− y1=

√1 + y1

1− y1.

It follows from this calculation that if there is a Lipschitz constant

K, then

√1 + y1

1− y1≤ K for all y1 ∈ (−1, 1), which is not possible:

limy1→1−

√1 + y1

1− y1= ∞. Therefore f does not satisfy a Lipschitz condi-

tion when d = 1. A similar argument works when c = −1.Return

Page 501: Ordinary Differential Equations: A Systems Approach

500 ORDINARY DIFFERENTIAL EQUATIONS

25.

y0 = −2.001386626.DetailsReturn

Page 502: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 501

27.

(a) If φ and ψ are fixed by T . Then

‖φ− ψ‖ ≤ C‖φ− ψ‖and hence ‖φ− ψ‖ = 0.

(b) By the triangle inequality,

‖ψn+k − ψn‖ ≤n+k−1

∑j=n‖ψj+1 − ψj‖ <

∑j=n‖ψj+1 − ψj‖

By the contractive property of T ,

‖ψj+1 − ψj‖ ≤ Cj‖ψ1 − ψ0‖.Using the formula for the sum of a geometric series, the formulafollows.

(c) Because T is contractive,

‖T (ψ∞)− ψn‖ = ‖T (ψ∞)− T (ψn−1)‖ ≤ C‖ψ∞ − ψn−1‖. Thus

limn→∞‖T (ψ∞)− ψn‖ = 0

It follows that the sequence {ψn} converges to both ψ∞ and toT (ψ∞) and hence the two are equal.

i. φ0 = 1, φ1(t) = 1 + t, φ2(t) = 1 + t + 12 t2, and φ3(t)1 + t +

12 t2 + 1

6 t3. Thus φn(t) = ∑nm=0

1m! t

m and φ∞(t) = et.ii. φ0 = 1, φ1(t) = 1− 1

2 t2, φ2(t) = 1− 12 t2 + 1

8 t4, and φ3(t) =1− 1

2 t2 + 18 t4− 1

48 t6 Thus φn(t) = ∑nm=0

12mm! t

2m and φ∞(t) =et2/2.

iii. We have φn ≡ 0 for 0 ≤ n ≤ ∞.iv. φ0 = 1, φ1(t) = 1 + 2 ln t, φ2(t) = 1 + 2 ln t + 2(ln t)2, and

φ3(t) = 1+ 2 ln t+ 2(ln t)2 + 43 (ln t)3. Thus φn(t) = ∑n

m=01

m! (2 ln t)m

and φ∞(t) = t2.

ReturnDetails

Page 503: Ordinary Differential Equations: A Systems Approach

502 ORDINARY DIFFERENTIAL EQUATIONS

1.

840DetailsReturn

Page 504: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 503

3.

y = 200/(19e−4t + 1).DetailsReturn

Page 505: Ordinary Differential Equations: A Systems Approach

504 ORDINARY DIFFERENTIAL EQUATIONS

5.

Let p(t) denote the population, in millions, 60t years after 1890. Then

p(t) =772.6

1 + (11.264) (0.3246t)million.

DetailsReturn

Page 506: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 505

7.

In 2070.DetailsReturn

Page 507: Ordinary Differential Equations: A Systems Approach

506 ORDINARY DIFFERENTIAL EQUATIONS

9.

About 10 12 hours, assuming they chatter 24 hours per day.

DetailsReturn

Page 508: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 507

11.

The function f (y) = ky ln(

My

)is defined for y > 0 only, but limy→0 f (y) =

0. Thus we can say 0 is stationary. The other stationary point at M isverified because ln(1) = 0. If 0 < y < M then ln(M/y) > 0 indicat-ing the population is increasing; for y > M we have ln(M/y) < 0and the population is decreasing. Thus M is a stable stationary pointand 0 is unstable.Return

Page 509: Ordinary Differential Equations: A Systems Approach

508 ORDINARY DIFFERENTIAL EQUATIONS

14.

dydt = ku(t)y(t). In a closed system, the number of total molecules is aconstant C. Therefore u(t) + y(t) = C, and dy

dt = ky(t)(C− y(t)). It’sa logistic equation.DetailsReturn

Page 510: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 509

16.

(a) and (d)DetailsReturn

Page 511: Ordinary Differential Equations: A Systems Approach

510 ORDINARY DIFFERENTIAL EQUATIONS

18.

Stationary point: 0.

- s -

0

DetailsReturn

Page 512: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 511

20.

No stationary points.

-

DetailsReturn

Page 513: Ordinary Differential Equations: A Systems Approach

512 ORDINARY DIFFERENTIAL EQUATIONS

22.

(a) This is the solution of first order autonomous ODE with no sta-tionary points.

(b) This is not a solution of a first order autonomous ODE.

(c) This is the solution of first order autonomous ODE with a sta-tionary point y = 0.

(d) This is the solution of first order autonomous ODE with no sta-tionary points.

(e) This is not a solution of a first order autonomous ODE.

DetailsReturn

Page 514: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 513

24.

If g′(y1) < 0, there exists ε > 0 such that g′(y) < 0 in (y1 − ε, y1 + ε),because g′(y) is continuous. That means g(y) is strictly decreasing in(y1 − ε, y1 + ε). Because g(y1) = 0, it follows that g(y) changes signfrom positive to negative at y = y1 Therefore (y1 − ε, y1) is an upinterval, and (y1, y + ε) is a down interval; the phase diagram for theODE looks like this near y1:s- � .Thus y1 is stable.

If g′(y1) > 0, the same reasoning shows that near y1, the phase dia-gram looks like this:s� - . Thus y1 is an unstable stationary point.DetailsReturn

Page 515: Ordinary Differential Equations: A Systems Approach

514 ORDINARY DIFFERENTIAL EQUATIONS

26.

Suppose the ODE is y′ = f (y). Then f (y) has a same stationary pointas y′ = y2 at y ≡ 0, thus f (0) = 0. The two up intervals (−∞, 0),(0,+∞) for y′ = y2 is also the up intervals for f (y). therefore f ′(0) =limh→0

f (0+h)− f (0)h = limh→0

f (h)h ≥ 0, and f ′(0) = limh→0

f (0−h)− f (0)−h =

limh→0f (−h)−h ≤ 0, We obtain f ′(0) = 0 and y ≡ 0 is a degenerate sta-

tionary point.Return

Page 516: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 515

28.

(a) H = 0.25

(b) P2 = 0.816

(c)0.25

H

1

P

DetailsReturn

Page 517: Ordinary Differential Equations: A Systems Approach

516 ORDINARY DIFFERENTIAL EQUATIONS

30.

1850 1900 1950 2000

50

100

150

200

250

DetailsReturn

Page 518: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 517

Find the general solution of each of the ODEs in problems 1 – 12.

1.

y = −1 + Cet2/2+t.DetailsReturn

Page 519: Ordinary Differential Equations: A Systems Approach

518 ORDINARY DIFFERENTIAL EQUATIONS

2.

y = (2 + 2t + t2) + Cet.DetailsReturn

Page 520: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 519

3.

y = (t + C)et.DetailsReturn

Page 521: Ordinary Differential Equations: A Systems Approach

520 ORDINARY DIFFERENTIAL EQUATIONS

4.

y = (sin t + C) cos t.DetailsReturn

Page 522: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 521

5.

y = 12 et + Ce−t.

DetailsReturn

Page 523: Ordinary Differential Equations: A Systems Approach

522 ORDINARY DIFFERENTIAL EQUATIONS

6.

y = − 15t et(2 cos(2t)− sin(2t)) + C/t.

DetailsReturn

Page 524: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 523

7.

y = 514 t2 + 3

13 t− 16 + Ct−12.

DetailsReturn

Page 525: Ordinary Differential Equations: A Systems Approach

524 ORDINARY DIFFERENTIAL EQUATIONS

8.

y = (ln | sec(t) + tan(t)|+ C) cos t.DetailsReturn

Page 526: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 525

9.

y = (∫

e−t2/2 dt + C)et2/2.DetailsReturn

Page 527: Ordinary Differential Equations: A Systems Approach

526 ORDINARY DIFFERENTIAL EQUATIONS

10.

y = Ct5

DetailsReturn

Page 528: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 527

11.

y = (t + C)e−t.DetailsReturn

Page 529: Ordinary Differential Equations: A Systems Approach

528 ORDINARY DIFFERENTIAL EQUATIONS

12.

y = Cesin t

DetailsReturn

Page 530: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 529

13.

y = t ln(t) + 2t.DetailsReturn

Page 531: Ordinary Differential Equations: A Systems Approach

530 ORDINARY DIFFERENTIAL EQUATIONS

14.

y = te−t.DetailsReturn

Page 532: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 531

15.

y = t2.ReturnDetails

Page 533: Ordinary Differential Equations: A Systems Approach

532 ORDINARY DIFFERENTIAL EQUATIONS

16.

y = − cos(3t)e−4t + e−4t.ReturnDetails

Page 534: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 533

17.

y = 5t3 ln t + t3.ReturnDetails

Page 535: Ordinary Differential Equations: A Systems Approach

534 ORDINARY DIFFERENTIAL EQUATIONS

18.

y = (2/3t + 1/3)3/2.ReturnDetails

Page 536: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 535

19.

y = −2 + 2et2/2+3t.ReturnDetails

Page 537: Ordinary Differential Equations: A Systems Approach

536 ORDINARY DIFFERENTIAL EQUATIONS

20.

y = −2.ReturnDetails

Page 538: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 537

21.

y = 2 2+t√(2−t)

+√

2 2+t2−t .

ReturnDetails

Page 539: Ordinary Differential Equations: A Systems Approach

538 ORDINARY DIFFERENTIAL EQUATIONS

22.

y = t2 + e−t22 .

ReturnDetails

Page 540: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 539

23.

30 ln 2ln 3 years.

ReturnDetails

Page 541: Ordinary Differential Equations: A Systems Approach

540 ORDINARY DIFFERENTIAL EQUATIONS

24.

53 years.ReturnDetails

Page 542: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 541

25.

1844 years.ReturnDetails

Page 543: Ordinary Differential Equations: A Systems Approach

542 ORDINARY DIFFERENTIAL EQUATIONS

26.

y = 117 (cos(4t) + 4 sin(4t)) + Ce−t; the solution with C = 0 is the

stable periodic solution.ReturnDetails

Page 544: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 543

27.

62.5 grams per liter.ReturnDetails

Page 545: Ordinary Differential Equations: A Systems Approach

544 ORDINARY DIFFERENTIAL EQUATIONS

28.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

Return

Page 546: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 545

29.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

DetailsReturn

Page 547: Ordinary Differential Equations: A Systems Approach

546 ORDINARY DIFFERENTIAL EQUATIONS

30.

-3 -2 -1 0 1 2 3 4

1

2

3

4

5

Return

Page 548: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 547

31.

-3 -2 -1 0 1 2 3 4

0

1

2

3

4

5

DetailsReturn

Page 549: Ordinary Differential Equations: A Systems Approach

548 ORDINARY DIFFERENTIAL EQUATIONS

32.

(a) y′ = 0.03(5t− y2) : Graph IV.

(b) y′ = sin(y) : Graph III.

(c) y′ = sin(t + y) : Graph II.

(d) y′ = −0.01y : Graph VI.

(e) y′ = 0.05y(π − y) : Graph V.

(f) y′ = 0.02(t2 + y2) : Graph I.

DetailsReturn

Page 550: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 549

33.

y = 2C−t2 , with a singular solution, y ≡ 0.

DetailsReturn

Page 551: Ordinary Differential Equations: A Systems Approach

550 ORDINARY DIFFERENTIAL EQUATIONS

34.

y = − ln(−t− C).DetailsReturn

Page 552: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 551

35.

y = Cet

1+Cet , with a singular solution, y ≡ 1.DetailsReturn

Page 553: Ordinary Differential Equations: A Systems Approach

552 ORDINARY DIFFERENTIAL EQUATIONS

36.

y = sin(arcsin t + C), with two singular solutions: y = ±1.DetailsReturn

Page 554: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 553

37.

y = cosh(cosh−1(t) + C), with two singular solutions: y = ±1.DetailsReturn

Page 555: Ordinary Differential Equations: A Systems Approach

554 ORDINARY DIFFERENTIAL EQUATIONS

38.

y = tan(ln√

1 + t2 + C).DetailsReturn

Page 556: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 555

39.

y = (√

t + C)2, with a singular solution Y ≡ 0.DetailsReturn

Page 557: Ordinary Differential Equations: A Systems Approach

556 ORDINARY DIFFERENTIAL EQUATIONS

40.

y = − 14 ln(ln(C(cos(3t))4/3)).

DetailsReturn

Page 558: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 557

41.

(a) 166.4 meters.

(b) 72 seconds.

(c) 236 meters; 68.5 seconds.

DetailsReturn

Page 559: Ordinary Differential Equations: A Systems Approach

558 ORDINARY DIFFERENTIAL EQUATIONS

42.

(a) For year A, 4.5%. For year B, 4%

(b) 1,000,000

(c) 1996.

DetailsReturn

Page 560: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 559

43.

- r� r -

− 12 0

y = − 12 is stable. y(t)→ − 1

2 .DetailsReturn

Page 561: Ordinary Differential Equations: A Systems Approach

560 ORDINARY DIFFERENTIAL EQUATIONS

44.

- r−π

- r0� r

π

� r2π

� r3π

� r4π

� r5π

� r6π

� r7π

� r8π

� r9π

� r10π

y = 0 is stable. If y(0) = 30 then y(t)→ 9π.DetailsReturn

Page 562: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 561

45.

-

No stationary points; y(t)→ +∞.Return

Page 563: Ordinary Differential Equations: A Systems Approach

562 ORDINARY DIFFERENTIAL EQUATIONS

46.

- r�53

y ≡ 53 is stable, and y(t)→ 5

3 .DetailsReturn

Page 564: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 563

47.

� r- �0 1

y ≡ 1 is stable and y(t)→ 1.DetailsReturn

Page 565: Ordinary Differential Equations: A Systems Approach

564 ORDINARY DIFFERENTIAL EQUATIONS

48.

� r- �0 1

There are no stationary points in (−∞, 0); y(t)→ −∞.DetailsReturn

Page 566: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 565

49.

If y′ = 1y then y′′ = − y′

y2 . Thus the signs of y′ and y′′ are opposite.DetailsReturn

Page 567: Ordinary Differential Equations: A Systems Approach

566 ORDINARY DIFFERENTIAL EQUATIONS

50.

Notice that y′′ = 1 + y′ = 1 + t + y, and use the second derivative todetermine concavity.DetailsReturn

Page 568: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 567

51.

For y′ = ey, we have y′′ = e2y. Both are positive. Solutions of y′ = e−y

are increasing and concave down.DetailsReturn

Page 569: Ordinary Differential Equations: A Systems Approach

568 ORDINARY DIFFERENTIAL EQUATIONS

52.

(a)

h Y(h)0.1 0.4230.05 0.4400.025 0.448

(b) Z(h1) = 2Yh2 −Yh1 .

(c) Z0.1 = 0.456638, and Z0.05 = 0.455814.

DetailsReturn

Page 570: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 569

53.

There will be a solution unless y0 = 1, and a unique solution unlessy0 = ±1.DetailsReturn

Page 571: Ordinary Differential Equations: A Systems Approach

570 ORDINARY DIFFERENTIAL EQUATIONS

54.

The right side of the differential equation satisfies a Lipschitz condi-tion; with respect to y with Lipschitz constant equal to 2.DetailsReturn

Page 572: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 571

55.

x = A e−t, y = −A e−t

ReturnDetails

Page 573: Ordinary Differential Equations: A Systems Approach

572 ORDINARY DIFFERENTIAL EQUATIONS

57.

x = 2A sin(3t), y = A sin(3t)− 3A cos(3t).ReturnDetails

Page 574: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 573

59.

x = e2t(t + 1 + C), y = e2t(t− 1 + C).ReturnDetails

Page 575: Ordinary Differential Equations: A Systems Approach

574 ORDINARY DIFFERENTIAL EQUATIONS

61.

y′ = vv′ = −3v− 4y + t2

ReturnDetails

Page 576: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 575

63.

u′ = vv′ = 1+t2

uv

ReturnDetails

Page 577: Ordinary Differential Equations: A Systems Approach

576 ORDINARY DIFFERENTIAL EQUATIONS

65.

x =−1

t + C, y = −1

2t2 − Ct + D, where C, D are constants.

ReturnDetails

Page 578: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 577

67.

This system is not uncoupled.Return

Page 579: Ordinary Differential Equations: A Systems Approach

578 ORDINARY DIFFERENTIAL EQUATIONS

69.

x = ±√

t2 + C, y = De±13 (t

2+C)3/2, where C and D are constants.

ReturnDetails

Page 580: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 579

71.

This system is not uncoupled.Return

Page 581: Ordinary Differential Equations: A Systems Approach

580 ORDINARY DIFFERENTIAL EQUATIONS

73.

1 2 3 4

-2

-1

1

2

ReturnDetails

Page 582: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 581

75.

1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

ReturnDetails

Page 583: Ordinary Differential Equations: A Systems Approach

582 ORDINARY DIFFERENTIAL EQUATIONS

77.

1 2 3 4

-1

-0.5

0.5

1

ReturnDetails

Page 584: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 583

79.

-2 -1 1 2

-4

-2

2

4

ReturnDetails

Page 585: Ordinary Differential Equations: A Systems Approach

584 ORDINARY DIFFERENTIAL EQUATIONS

81.

Stationary point: (0, 0).

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

ReturnDetails

Page 586: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 585

83.

Stationary point (0, 0).

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

ReturnDetails

Page 587: Ordinary Differential Equations: A Systems Approach

586 ORDINARY DIFFERENTIAL EQUATIONS

85.

(a)

1 2 3 4 5 6

-1

-0.5

0.5

1

-1 -0.5 0.5 1

-1

-0.5

0.5

1 -1-0.500.51-1-0.500.51

0

2

4

6-1-0.500.51

Page 588: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 587

(b)

2 4 6 8 10 12

-6

-4

-2

2

4

6

-6 -4 -2 2 4 6

-2-1

12

-50

5-202

0

5

10

-202

0

5

10

Page 589: Ordinary Differential Equations: A Systems Approach

588 ORDINARY DIFFERENTIAL EQUATIONS

(c)

-5 -2.5 2.5 5 7.5 10 12.5

-4

-2

2

4

6

-4 -2 2 4 6

-4

-2

2

4

6 -5-2.502.55-2.502.55

-5

0

5

10

-2.502.55

Page 590: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 589

(d)

-2 -1 1 2 3 4

-1

-0.5

0.5

1

1.5

-1 1 2 3

-1

-0.5

0.5

1

1.5 05

10

-101-2

0

2

4

-2

0

2

4

ReturnDetails

Page 591: Ordinary Differential Equations: A Systems Approach

590 ORDINARY DIFFERENTIAL EQUATIONS

1.

y(1) ≈ 0.841471076.ReturnDetails

Page 592: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 591

3.

The phase diagram depends on ε:- r−1� r

0- r

1�

Phase diagram for ε < 0

� r−1

- r0� r

1-

Phase diagram for ε > 0

(a)

-1 -0.5 0.5 1

-1

-0.5

0.5

1

(b) i.

-1 1 2 3 4

-1

-0.5

0.5

1

1.5

Page 593: Ordinary Differential Equations: A Systems Approach

592 ORDINARY DIFFERENTIAL EQUATIONS

ii.

-1 -0.5 0.5 1

-1

-0.5

0.5

1

iii.

-1.5 -1 -0.5 0.5 1

-4

-3

-2

-1

1

ReturnDetails

Page 594: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 593

5.

The blue curve in the following graph is the actual solution, and theblack curve is the Euler approximation using step size 0.05.

-2 -1 1 2

-300

-200

-100

100

200

300

Return

Page 595: Ordinary Differential Equations: A Systems Approach

594 ORDINARY DIFFERENTIAL EQUATIONS

7.

(a) Solve the ODE y′ = f (y)/1000 with initial conditions y(0) =−3,−2,−1, and 0, over the interval [−5, 5]. The zeros of f willbe the ordinates of the horizontal asymptotes of these solutions.

(b) The roots are approximately 0.193, 1.027, 2.568, 4.900, 8.182, 12.734,and 19.396.

ReturnDetails

Page 596: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 595

9.

(a)

92 94 96 98 100

-0.4

-0.2

0.2

0.4

(b)

-14.5 -14 -13.5 -13

-0.04

-0.02

0.02

0.04

Page 597: Ordinary Differential Equations: A Systems Approach

596 ORDINARY DIFFERENTIAL EQUATIONS

(c)

0.8 0.85 0.9 0.95

-0.01

-0.005

0.005

0.01

Return

Page 598: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 597

1.

x-nullcline: (0.0).

y-nullclines: y = ±x.

Stationary point: (0, 0).

-10 -7.5 -5 -2.5 2.5 5 7.5 10

-10

-7.5

-5

-2.5

2.5

5

7.5

10

ReturnDetails

Page 599: Ordinary Differential Equations: A Systems Approach

598 ORDINARY DIFFERENTIAL EQUATIONS

3.

x-nullcline: the y-axis.

y-nullclines: y = 4 and y = 0.

Stationary points: (0, 0) and (0, 4).

-3 -2 -1 1 2 3

-2

2

4

6

ReturnDetails

Page 600: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 599

5.

x-nullcline: the y-axis.

y-nullcline: the x-axis.

Stationary point: (0, 0).

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

ReturnDetails

Page 601: Ordinary Differential Equations: A Systems Approach

600 ORDINARY DIFFERENTIAL EQUATIONS

7.

x-nullcline: y = 2− x.

y-nullcline: y = x.

Stationary point: (1, 1).

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

ReturnDetails

Page 602: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 601

9.

F(x, y) = −2x−1/2y + 2x1/2.ReturnDetails

Page 603: Ordinary Differential Equations: A Systems Approach

602 ORDINARY DIFFERENTIAL EQUATIONS

11.

F(x, y) = 5 ln x− 3 ln y.ReturnDetails

Page 604: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 603

13.

F(x, y) = x2y + y3

3 .ReturnDetails

Page 605: Ordinary Differential Equations: A Systems Approach

604 ORDINARY DIFFERENTIAL EQUATIONS

15.

The system is uncoupled. The solution of the first equation, x′ = 1,with x(0) = x0 is x = t + x0. Thus the second equation can be writtendydx = f (x, y); the solution of the system that passes through (x0, y0)when t = 0 would satisfy the initial condition y(0) = y0, and sincex = t + x0 the solution follows the graph of the solution of the IVP

dydx

= f (x, y); y(x0) = y0.

Return

Page 606: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 605

17.

Set y = dxdt and y′ = d2x

dt2 . Then the system

x′ = y; y(t0) = y0

y′ = f (t, x, y); x(t0) = x0

replaces the given ODE. Let g(t, x, y) = y, then f (t, x, y) and g(t, x, y)satisfy Lipschitz conditions. By the existence and uniqueness theo-rems there is a unique solution of the system. Thus there is a uniquesolution of the IVP.Return

Page 607: Ordinary Differential Equations: A Systems Approach

606 ORDINARY DIFFERENTIAL EQUATIONS

19.

The system is equivalent to

x′ =1

ad− bc(a g(t, x, y)− c f (t, x, y))

y′ =1

ad− bc(d f (t, x, y)− b g(t, x, y))

If the functions f (t, x, y), g(t, x, y) are continuous and satisfy a Lips-chitz condition then 1

ad−bc (ag(t, x, y)− c f (t, x, y)) and 1ad−bc (d f (t, x, y)−

bg(t, x, y)) are continuous and satisfy a Lipschitz condition. Thus bythe existence and uniqueness theorems the system has a unique solu-tion with the initial conditions.Return

Page 608: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 607

1.

(a) ab .

(b) The strategy is ineffective.

(c) Wait until there are ≈ d pests.

ReturnDetails

Page 609: Ordinary Differential Equations: A Systems Approach

608 ORDINARY DIFFERENTIAL EQUATIONS

3.

The fish populations are governed by the system

x′ = k1x(4000− 4x− y)y′ = k2y(12000− 7x− 8y)

The phase portrait has the configuration (d) of Figure 2.19. The tri-angles ABC and CDE in the figure below are traps, and orbits withinthese triangles converge to the point C = (800, 800).

500 1000 1500 2000

1000

2000

3000

4000 A

B

C

D E

ReturnDetails

Page 610: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 609

5.

The x-nullclines are the lines x = 0 and K − x + By = 0. The y-nullclines are the lines y = 0 and L + Cx − y = 0. The system hasat most four stationary points. Three of these occur when one or bothof the species are extinct. The fourth stationary point, (x1, y1) exists ifand only if BC 6= 1. Then the stationary point is

(x1, y1) =

(BL + K1− BC

,CK + L1− BC

).

If BC > 1 the fourth stationary point is in the third quadrant. Thereis no trap, and the species will increase in numbers without bound. IfBC < 1 we have the fourth stationary point in the first quadrant, andit is stable. All orbits in the first quadrant converge to this point.

The drawing shows the configuration when BC < 1. The quadrilat-eral whose vertices are the four stationary points is a trap

500 1000 1500 2000

500

1000

1500

2000

ReturnDetails

Page 611: Ordinary Differential Equations: A Systems Approach

610 ORDINARY DIFFERENTIAL EQUATIONS

7.

The system with fishing can be written as Lotka-Volterra equations:

x′ = x[(a− R)− by]

y′ = cy[

x−(

d +Rc

)].

By the result of problem 6, the average populations will be x = d + Rc

and y = a−Rb . Thus the average prey population increases proportion-

ally to the catch rate, and the average predator population decreases.ReturnDetails

Page 612: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 611

1.

x = tet, y = 1et(1−t) .

ReturnDetails

Page 613: Ordinary Differential Equations: A Systems Approach

612 ORDINARY DIFFERENTIAL EQUATIONS

2.

General solution: x = Ce−t, y = Det.

Integral: F(x, y) = xy.

Phase Portrait:

ReturnDetails

Page 614: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 613

3.

General solution: x = Cet, y = De2t.

Integral: F(x, y) = x2/y.

Phase portrait:

ReturnDetails

Page 615: Ordinary Differential Equations: A Systems Approach

614 ORDINARY DIFFERENTIAL EQUATIONS

4.

F(y, v) = v2

2 +∫

f (y) dy is an integral for the system y′ = v, v′ = f (y).

(a) F(y, v) = v2

2 − α2 cos(y).

(b) F(y, v) = v2

2 + α2y2

2 .

(c) F(y, v) = v2

2 − α2y−1.

ReturnDetails

Page 616: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 615

5.

(a) y′ = v, v′ = −y

(b) y′ = v, v′ = v− t2 sin(y)

ReturnDetails

Page 617: Ordinary Differential Equations: A Systems Approach

616 ORDINARY DIFFERENTIAL EQUATIONS

6.

ReturnDetails

Page 618: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 617

7.

F(x, y) = yx2 + 2y2xReturnDetails

Page 619: Ordinary Differential Equations: A Systems Approach

618 ORDINARY DIFFERENTIAL EQUATIONS

8.

(a) (1, 1).

(b) (0, 0), (−1,−1), (1, 1).

(c) (0, 0), (−4, 0), (−5, 1).

Return

Details

Page 620: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 619

9.

(a) x′ = x + y− 2, y′ = x− 3y + 2.

(b) x′ = y− x, y′ = y− x3.

(c) x′ = x(x + y + 4), y′ = y(x + 5y).

Page 621: Ordinary Differential Equations: A Systems Approach

620 ORDINARY DIFFERENTIAL EQUATIONS

ReturnDetails

Page 622: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 621

10.

In the drawing, the trap is shown in blue.

ReturnDetails

Page 623: Ordinary Differential Equations: A Systems Approach

622 ORDINARY DIFFERENTIAL EQUATIONS

11.

(a)

(b)

(c)

Page 624: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 623

Return

Page 625: Ordinary Differential Equations: A Systems Approach

624 ORDINARY DIFFERENTIAL EQUATIONS

12.

Both the linear and nonlinear versions are shown: the linear versionis in black; the nonlinear version in blue. The bottom figure is anenlargement of the region inside the rectangle in the top figure, andshows more detail.

5 10 15 20

-0.2

-0.1

0.1

0.2

Return

Page 626: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 625

1.

AB, BA, A~v, B~w, AB~w, BA~vReturnDetails

Page 627: Ordinary Differential Equations: A Systems Approach

626 ORDINARY DIFFERENTIAL EQUATIONS

3.

(a) EA =

(a b0 0

).

(b) AE =

(a 0c 0

).

(c) b = c = 0. In other words, A must be a diagonal matrix.

ReturnDetails

Page 628: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 627

5.

Let A =

(3 −21 7

). Then the matrix equation ~v′ = A~v is equivalent

to the given system.ReturnDetails

Page 629: Ordinary Differential Equations: A Systems Approach

628 ORDINARY DIFFERENTIAL EQUATIONS

7.

~v′ = A~v +~b where ~v =

xx′

x′′

, A =

0 1 00 0 1

2et −t sin t

, and~b = 00

tan t

.

ReturnDetails

Page 630: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 629

9.

[xy

]= C1e3t

[11

]+ C2e−t

[1−1

], where C1, C2 are constants.

ReturnDetails

Page 631: Ordinary Differential Equations: A Systems Approach

630 ORDINARY DIFFERENTIAL EQUATIONS

11.

x = e2t, t = 2e3t

ReturnDetails

Page 632: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 631

13.

x = e3t + e−t, y = e3t − e−t.ReturnDetails

Page 633: Ordinary Differential Equations: A Systems Approach

632 ORDINARY DIFFERENTIAL EQUATIONS

15.

x ≡ 1, y ≡ 1.ReturnDetails

Page 634: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 633

17.

e−t[

1−1

]ReturnDetails

Page 635: Ordinary Differential Equations: A Systems Approach

634 ORDINARY DIFFERENTIAL EQUATIONS

19.

[xy

]= −8et

[1−1

]+ 5e2t

[3−2

].

ReturnDetails

Page 636: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 635

21.

Put t = 0 and see that the vectors ~v1(0) and ~v2(0) are linearly inde-pendent.ReturnDetails

Page 637: Ordinary Differential Equations: A Systems Approach

636 ORDINARY DIFFERENTIAL EQUATIONS

23.

[−t2 + 3t + 1−2t2 + 8t + 1

].

ReturnDetails

Page 638: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 637

25.

t[

cosh(t)sinh(t)

]+ 2

[et

et

]ReturnDetails

Page 639: Ordinary Differential Equations: A Systems Approach

638 ORDINARY DIFFERENTIAL EQUATIONS

27.

[y′

v′

]=

[0 1−q(t) −p(t)

] [yv

]+

[0

r(t)

]ReturnDetails

Page 640: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 639

1.

Eigenvalues: s1 = −4 and s2 = 1. Eigenvalues:~b1 =

[01

]and

[10

]ReturnDetails

Page 641: Ordinary Differential Equations: A Systems Approach

640 ORDINARY DIFFERENTIAL EQUATIONS

3.

Eigenvalues: s1 = −4 and s2 = 1; eigenvectors: ~b1 =

[1−5

]and

~b2 =

[10

].

ReturnDetails

Page 642: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 641

5.

This matrix has no real eigenvalues.ReturnDetails

Page 643: Ordinary Differential Equations: A Systems Approach

642 ORDINARY DIFFERENTIAL EQUATIONS

7.

(c) Every nonzero vector is an eigenvector in this case.ReturnDetails

Page 644: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 643

9.

c1et[−11

]+ c2e4t

[21

]ReturnDetails

Page 645: Ordinary Differential Equations: A Systems Approach

644 ORDINARY DIFFERENTIAL EQUATIONS

11.

c1

[10

]+ c2

[t1

]ReturnDetails

Page 646: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 645

13.

c1e3t[

11

]+ c2e−t

[1−1

]ReturnDetails

Page 647: Ordinary Differential Equations: A Systems Approach

646 ORDINARY DIFFERENTIAL EQUATIONS

15.

~v = e−2t[

t1− t

]ReturnDetails

Page 648: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 647

17.

x = (2t− 1)e3t, y = (2t− 2)e3t.ReturnDetails

Page 649: Ordinary Differential Equations: A Systems Approach

648 ORDINARY DIFFERENTIAL EQUATIONS

19.

The coefficient matrix of the system is A =

[0 1−q −p

]; its charac-

teristic equation is s2 + ps + q = 0.ReturnDetails

Page 650: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 649

1.

15(2− i)

DetailsReturn

Page 651: Ordinary Differential Equations: A Systems Approach

650 ORDINARY DIFFERENTIAL EQUATIONS

3.

1,12+ i√

32

, −12+ i√

32

, −1, −12− i√

32

, and12+ i√

32

.

DetailsReturn

Page 652: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 651

5.

|eλ+iω| = eλ|eiω| = eλ.ReturnDetails

Page 653: Ordinary Differential Equations: A Systems Approach

652 ORDINARY DIFFERENTIAL EQUATIONS

7.

e−t/2[−c1 sin(t/2) + c2 cos(t/2)c1 cos(t/2) + c2 sin(t/2)

]Details

Return

Page 654: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 653

9.

[(c1 + c2) cos(4t) + (c2 − c1) sin(4t)

c1 cos(4t) + c2 sin(4t)

]DetailsReturn

Page 655: Ordinary Differential Equations: A Systems Approach

654 ORDINARY DIFFERENTIAL EQUATIONS

11.

(a) Verify that the conjugate of a product (or sum) of complex num-bers is the product (or sum) of the conjugates of these numbers.Apply this when conjugating the equation A~b = s~b, noting thatsince A has real entries, A = A.

(b) Because s is complex, s 6= s.

(c) Since~b and ~b are linearly independent,~h and~k must also be lin-early independent.

(d) The real part and the imaginary part of a complex-valued solu-tion of v′ = Av are themselves solutions of the system when Ais a matrix with real entries.

DetailsReturn

Page 656: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 655

1.

X (t) =[

et e2t

−et e2t

]ReturnDetails

Page 657: Ordinary Differential Equations: A Systems Approach

656 ORDINARY DIFFERENTIAL EQUATIONS

3.

X (t) =[

2 cos(4t) 2 sin(4t)cos(4t) + 2 sin(4t) sin(4t)− 2 cos(4t)

]ReturnDetails

Page 658: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 657

5.

(a) ~v(t) =[

et e2t

−et e2t

]~c +

[−et

−et

](b) ~v(t) =

[et e2t

−et e2t

]~c +

[tet

−tet

](c) ~v(t) =

[et e2t

−et e2t

]~c +

[(t2/4− t/2− 1/2)et

(−t2/4− t/2− 1/2)et

]ReturnDetails

Page 659: Ordinary Differential Equations: A Systems Approach

658 ORDINARY DIFFERENTIAL EQUATIONS

7.

~v(t) =[

2 cos(4t) 2 sin(4t)cos(4t) + 2 sin(4t) sin(4t)− 2 cos(4t)

]~c

+

[−16t cos(4t)− 8t sin(4t)− cos(4t)− 2 sin(4t)

−20t sin(4t)− 52 cos(4t)

]ReturnDetails

Page 660: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 659

9.

(a)[ −7

2 t12 t

].

(b)[

t3 ln |t|−t3 ln |t|

].

ReturnDetails

Page 661: Ordinary Differential Equations: A Systems Approach

660 ORDINARY DIFFERENTIAL EQUATIONS

11.

(XC)′ = X ′C = A(t)(XC).Return

Page 662: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 661

13.

Suppose Q is singular; then there is a nonzero vector d such that Q~d =~0. Let~qi denote the i th column of Q. You can show that

Q~d = d1~q1 + · · ·+ dn~qn

and hence that the columns of Q are linearly dependent.

The proof that a matrix with linearly dependent columns is singularalso follows from the above observation.ReturnDetails

Page 663: Ordinary Differential Equations: A Systems Approach

662 ORDINARY DIFFERENTIAL EQUATIONS

1.

eAt =

1 t 1

2 t2 16 t3

0 1 t 12 t2

0 0 1 t0 0 0 1

ReturnDetails

Page 664: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 663

3.

eAeB = (∞

∑i=0

1i!

Ai)(∞

∑j=0

1j!

Bj)

=∞

∑i=0

∑j=0

1i!

Ai 1j!

Bj

=∞

∑i=0

∑j=0

1i!j!

AiBj

=∞

∑k=0

k

∑i=0

1i!(k− i)!

AiBk−i

=∞

∑k=0

1k!

k

∑i=0

k!i!(k− i)!

AiBk−i.

By the binomial theorem, if A and B commute, then

(A + B)k =k

∑i=0

k!i!(k− i)!

AiBk−i.

If A and B do not commute, this won’t work: for example (A + B)2 =A(A + B) + B(A + B) = A2 + AB + BA + B2. If AB 6= BA, it wouldnot be possible to combine the two middle terms to get (A + B)2 =A2 + 2AB + B2.

It follows that for commuting square matrices A and B,

eAeB =∞

∑k=0

1k!(A + B)k = eA+B.

Return

Page 665: Ordinary Differential Equations: A Systems Approach

664 ORDINARY DIFFERENTIAL EQUATIONS

5.

A(t) =[

t 10 2t

], dA/dt =

[1 00 2

]. Thus A(t) · dA/dt =

[t 20 4t

],

but dA/dt · A(t) =[

t 10 4t

].

Return

Page 666: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 665

7.

A2 =

0 0 ac0 0 00 0 0

, and Ak = 0 for k ≥ 3.

(a) eAt =

1 a t b t + 12 ac t2

0 1 c t0 0 1

(b) e(A+λI)t =

eλt ateλt (bt + 12 act2)eλt

0 eλt cteλt

0 0 eλt

.

x = e−2t(3 + t + 6t2), y = −4te−2t, and z = e−2t. ReturnDetails

Page 667: Ordinary Differential Equations: A Systems Approach

666 ORDINARY DIFFERENTIAL EQUATIONS

9.

[cos(t) + sin(t) sin(t)−2 sin(t) cos(t)− sin(t)

]Return

Details

Page 668: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 667

11.

eAt = et

(cos√

2t−√

2 sin√

2t√

2 sin√

2t− 3√

2sin√

2t cos√

2t +√

2 sin√

2t

)Return

Details

Page 669: Ordinary Differential Equations: A Systems Approach

668 ORDINARY DIFFERENTIAL EQUATIONS

13.

[1− 2t −2t

2t 1 + 2t

]ReturnDetails

Page 670: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 669

15.

1 t 2t + 12 t2

0 1 t0 0 1

ReturnDetails

Page 671: Ordinary Differential Equations: A Systems Approach

670 ORDINARY DIFFERENTIAL EQUATIONS

17.

x = (cosh t + sinh t)(t + c1)

y = sinh t(t + c1) + (cosh t− sinh t)(t + c2)

ReturnDetails

Page 672: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 671

19.

x =12

e2t + c1(1 + 2t)− c2t

y = e2t + 4c1 + c2(1− 2t).

ReturnDetails

Page 673: Ordinary Differential Equations: A Systems Approach

672 ORDINARY DIFFERENTIAL EQUATIONS

21.

Let f (t) be a polynomial of degree N. Then for n > N, Dn( f ) ≡ 0.Hence

(etD f )(x) =

{[I + tD +

t2

2!D2 + · · ·+ tn

n!Dn + · · ·] f

}(x)

= f (x) + t f ′(x) +t2

2!f ′′(x) + · · ·+ tN

N!f (N)(x).

By Taylor’s theorem, for any function g that is (N + 1) times differen-tiable on R, there is a number ξ between x and t such that

g(x+ t) = g(x)+ t g′(x)+t2

2!g′′(x)+ · · ·+ tN

N!g(N)(x)+

tN+1

(N + 1)!g(N+1)(ξ).

Since f (N+1)(ξ) = 0 it follows that

(etD f )(x) = f (x + t).

Return

Page 674: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 673

1.

(t + 1)et tet 0−tet et − tet 00 0 e−t

ReturnDetails

Page 675: Ordinary Differential Equations: A Systems Approach

674 ORDINARY DIFFERENTIAL EQUATIONS

3.

We can refer to the result of the previous exercise.

The formula asserted by lemma 3.6.1 is true if k = 1, because itmatches the actual definition of Np. the Assume it holds for pow-ers of Np less than some number r. Then the rows of Nr

p = Np Nr−1p

are obtained by moving the rows of Nr−1p up one level and adding

a row of zeros at the bottom. This results in a matrix an additionalrow of zeros at the bottom, and lower p− 1 rows of Nr−1

p above it. Itfollows that Nr

p has r rows of zeros at the bottom, and r columns ofzeros at the left. The complement of these rows and columns is Ip−r.By the principle of mathematical induction, the proof of lemma 3.6.1is complete.

The proof of (3.34) now follows directly from the lemma. We can seethat Nr

p is to be multiplied by tr/r!; the results are added to get eNpt.Thus when r = 0 we have the identity matrix; as we move up fromthe diagonal we encounter entries of t, 1

2 t2, and so on, until we get tothe first row.Return

Page 676: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 675

5.

Let P and P′ be n× n matrices, and let Q, Q′ be m×m matrices, andlet A, A′ denote P⊕Q and P′⊕Q′, respectively. Also let R = PP′ andS = QQ′ We will use the notation pij, p′ij, qij, q′ij, aij, a′ij for the entriesof these matrices.

Thus if B = AA′ then bij =n+m

∑k=1

aika′kj. If 1 ≤ i, j ≤ n then aika′kj =

pik p′kj when k ≤ n; and aika′kj = 0 for k > n. Hence when 1 ≤ i, j ≤n, we have bij = rij. Also when n + 1 ≤ i, j, k ≤ n + m, aika′kj =q(i−n)(k−n)p′(k− n)(j− n), and when n + 1 ≤ i, j ≤ n + m and k ≤ nboth aik = 0 and a′kj = 0. It follows that bij = s(i−n)(j−n) as well.

Now suppose i ≤ n < j Choose any k between 1 and n + m. If k ≤ nthen a′kj = 0, while if k > n we have aik = 0. Thus in the sum bij =

∑n+mk=1 aika′kj one factor of each term is equal to zero, and it follows

that bij = 0. Identical reasoning shows that bij = 0 when j ≤ n < i..We have shown that B = R⊕ S, thus verifying (3.38).

It now follows that we can raise direct sums to powers by the rule(P⊕Q)n = Pn ⊕Qn. By definition,

e(P⊕Q)t =∞

∑n=0

1n!(P⊕Q)ntn

=∞

∑n=0

1n!(Pn ⊕Qn)tn

=∞

∑n=0

1n!

Pntn ⊕∞

∑n=0

1n!

Qntn

= ePt ⊕ eQt,

and thus we have verified 3.39).Return

Page 677: Ordinary Differential Equations: A Systems Approach

676 ORDINARY DIFFERENTIAL EQUATIONS

7.

Recall that B ∼ A means that there exists an invertible matrix P suchthat B = P−1AP.

Reflexive: To prove A ∼ A let P = I (the identity matrix): A is theconjugate of itself by I.

Symmetric: If A ∼ B then A is the conjugate of B by some invertiblematrix P. It follows that B is the conjugate of A by P−1; thusB ∼ A.

Transitive: If (A ∼ B and B ∼ C) then there are invertible matricesP, Q such that A is conjugate to B by P and B is conjugate to Cby Q. Thus A = P−1BP and B = Q−1CQ. Substitute for B in thefirst equation to get

A = P−1(Q−1CQ)P = (QP)−1 C(QP);

in other words A ∼ C because A is conjugate to C by the invert-ible matrix QP.

Return

Page 678: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 677

9.

eAt =

e−2t 0 0 0

0 cos 2t −2 sin 2t 00 1

2 sin 2t cos 2t 00 0 0 et

.

ReturnDetails

Page 679: Ordinary Differential Equations: A Systems Approach

678 ORDINARY DIFFERENTIAL EQUATIONS

11.

eAt =

12 (3− e2t) 1

2 (e2t − 1) e2t − 1

12 (e

2t − 1) 12 (3− e2t) 1− e2t

1− e2t e2t − 1 2e2t − 1

.

ReturnDetails

Page 680: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 679

13.

eAt =

e−t

cos t + 3 sin t 3 cos t− 3e−t − 11 sin t

2 sin t 4 cos t− 3e−t − 6 sin t2 sin t 4 cos t− 4e−t − 6 sin t2 sin t 4 cos t− 4e−t − 6 sin t

4(e−t − cos t) + 8 sin t− 6te−t cos t− e−t + 6te−t

4(e−t − cos t)− 4 sin t− 6te−t cos t− e−t + 6te−t

5e−t − 4 cos t + 4 sin t− 8te−t cos t− e−t + 8te−t

4(e−t − cos t) + 4 sin t− 8te−t cos t + 8te−t

ReturnDetails

Page 681: Ordinary Differential Equations: A Systems Approach

680 ORDINARY DIFFERENTIAL EQUATIONS

15.

~v = t

1−11

.

ReturnDetails

Page 682: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 681

1.

x =12

e3t +12

et, y =12

e3t − 12

et.DetailsReturn

Page 683: Ordinary Differential Equations: A Systems Approach

682 ORDINARY DIFFERENTIAL EQUATIONS

2.

Please notice that answers are not unique; in fact none of the answersbelow is equal to the matrix exponential. (It would be correct—andeasier—to compute eAt in each case.)

(a)[

cos(3t)− 3 sin(3t) sin(3t) + 3 cos(3t)5 cos(3t) 5 sin(3t)

](b)

[−et sin(t) et cos(t)et cos(t) et sin(t)

](c)

[2et 2tet

0 et

]

(d)

et 2e2t 92 e3t

0 e2t 3e3t

0 0 e3t

DetailsReturn

Page 684: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 683

3.

x = −12+

(12

sin(2t)− 14

cos(2t))

ln(

1 + sin(2t)cos(2t)

)+ C1[2 sin(2t)− cos(2t)] + C2[sin(2t) + 2 cos(2t)]

y =14

cos(2t) ln(

1 + sin(2t)cos(2t)

)+ C1 cos(2t) + C2 sin(2t)

ReturnDetails

Page 685: Ordinary Differential Equations: A Systems Approach

684 ORDINARY DIFFERENTIAL EQUATIONS

4.

eAt = V · E(t) ·V−1.ReturnDetails

Page 686: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 685

5.

Let r be an eigenvalue of a nilpotent matrix M, and let ~v be corre-sponding eigenvector. Since M is nilpotent, Mn is the zero matrixfor some n. Therefore Mn~v = ~0. But Mn~v = rn~v so it follows thatr = 0. Since all eigenvalues of M are equal to zero, and since the traceof any matrix is the sum of its eigenvalues (counted with algebraicmultiplicity), and the determinant of any matrix is the product of itseigenvalues, it follows that the determinant of M is also zero.

Conversely, if the trace and determinant of M are equal to zero, thenthe characteristic equation of M is s2 = 0. The Cayley-Hamilton theo-rem says that every matrix a satisfies its own characteristic equation;hence M2 is is equal to the zero matrix.Return

Page 687: Ordinary Differential Equations: A Systems Approach

686 ORDINARY DIFFERENTIAL EQUATIONS

6.

det(B) = det(A)− 14 tr (A)2.

ReturnDetails

Page 688: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 687

1.

E.ReturnDetails

Page 689: Ordinary Differential Equations: A Systems Approach

688 ORDINARY DIFFERENTIAL EQUATIONS

3.

I.ReturnDetails

Page 690: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 689

5.

F.ReturnDetails

Page 691: Ordinary Differential Equations: A Systems Approach

690 ORDINARY DIFFERENTIAL EQUATIONS

7.

L.ReturnDetails

Page 692: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 691

9.

A.ReturnDetails

Page 693: Ordinary Differential Equations: A Systems Approach

692 ORDINARY DIFFERENTIAL EQUATIONS

11.

D.ReturnDetails

Page 694: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 693

D EF

A BC

J

KL

G HI

Page 695: Ordinary Differential Equations: A Systems Approach

694 ORDINARY DIFFERENTIAL EQUATIONS

13.

Degenerate system:

-4 -2 2 4

-4

-2

2

4

ReturnDetails

Page 696: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 695

15.

F(x, y) = x2 + 2xy is an integral of this system. Thus the orbits fol-low level curves F(x, y) = c, which are hyperbolas (for c 6= 0). Theasymptotes x = 0 and x + 2y = 0 are invariant sets consisting of theorigin and two half-line orbits, directed respectively away from andtoward the origin.ReturnDetails

Page 697: Ordinary Differential Equations: A Systems Approach

696 ORDINARY DIFFERENTIAL EQUATIONS

17.

The phase portrait is a saddle:

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

ReturnDetails

Page 698: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 697

19.

The phase portrait is a stable spiral node:

-3 -2 -1 1 2 3

-4

-2

2

4

ReturnDetails

Page 699: Ordinary Differential Equations: A Systems Approach

698 ORDINARY DIFFERENTIAL EQUATIONS

21.

This degenerate system has y = x as the stationary line:

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

ReturnDetails

Page 700: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 699

23.

The phase portrait is a center, with orbits directed counterclockwiseon ellipses 5y2 − 4xy + 17x2 = C

-4 -2 2 4

-7.5

-5

-2.5

2.5

5

7.5

ReturnDetails

Page 701: Ordinary Differential Equations: A Systems Approach

700 ORDINARY DIFFERENTIAL EQUATIONS

25.

This phase portrait is a saddle; the stable line is y = −5x, and theunstable line is y = −x.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

ReturnDetails

Page 702: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 701

27.

The rotated solution is[cos(π/4) sin(π/4)− sin(π/4) cos(π/4)

] [c1et + c2e3t

c1et − c2e3t

]=√

2[

c1et

−c2e3t

].

ReturnDetails

Page 703: Ordinary Differential Equations: A Systems Approach

702 ORDINARY DIFFERENTIAL EQUATIONS

29.

If both eigenvalues are negative −r,−s, then the general solution is~v(t) = c1e−rt~b1 + c2e−st~b2, where ~b1,~b2 are corresponding eigenvec-tors, and c1, c2 are constants. As t → ∞, ~v(t) → ~0. Thus all orbitsconverge toward the origin; the phase portrait is therefore a stablenode.

Now suppose that the eigenvalues are complex numbers −r ± is,with negative real parts. Let ~h + i~k be an eigenvector belonging to−r+ is. Then the general solution of the system is~v(t) = c1e−rt(cos(st)h−sin(st)~k)+ c2e−rt(sin(st)~h+ cos(st)~k). Again,~v(t) approaches the ori-gin as t→ ∞; thus the phase portrait is a stable node.

If the eigenvalues are positive real numbers r, s, then the general so-lution is~v(t) = c1ert~b1 + c2est~b2, where~b1,~b2 are corresponding eigen-vectors. Since limt→−∞ ekt = 0 when k > 0, it follows that as t→ −∞,~v(t) approaches the origin. Hence the phase portrait is an unstablenode.

In the case where the eigenvalues are complex numbers r ± is, withpositive real parts, let~h + i~k be a eigenvector belonging to r + is. Thegeneral solution of the system is ~v(t) = ert[c1(cos(st)~h− sin(st)k) +c2(sin(st)~h + cos(st)~k)]. As t → −∞, ~v approach the origin. Thus thephase portrait is an unstable node.

If one of the eigenvalues is 0, the system is degenerate and hencethere is a stationary line through the origin. Each point (other thanthe origin) on the stationary line is an orbit that does not approachthe origin. If the eigenvalues are complex numbers with 0 as real part,the phase portrait is a center. Each orbit is an ellipse, which does notconverge to the origin. If one of the eigenvalues is positive and theother is negative, the phase portrait is a saddle. The stable line willconsist of orbits directed toward the origin as t → ∞; the unstableline consists of orbits that converge to the origin as t → −∞, and allother orbits go to ∞ as t→ ±∞.Return

Page 704: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 703

Details

Page 705: Ordinary Differential Equations: A Systems Approach

704 ORDINARY DIFFERENTIAL EQUATIONS

31.

(a) Use the product rule to differentiate y = vx and simplify.

(c) If there are 2 stationary points and b > 0, the phase diagram thatlooks like this:� �r r-

Thus the greater of the two stationary points is stable; the lesseris unstable. If b < 0 all arrows in the phase diagram reverse, andthe lesser of the two stationary points is stable.If there is only one stationary point then the phase portrait is asfollows if b > 0 (arrows reversed if b < 0):� �rFinally, if there are no stationary points, the phase diagram is adown arrow if b > 0, or an up arrow if b < 0.

(d) The eigenvalues of the coefficient matrix are 12 (a+ d±

√(a− d)2 + 4bc).

The slopes of the eigenvectors are

−a + d±√(a− d)2 + 4bc2b

,

the same as the the stationary points of the ODE (4.20).

(e) Assume b > 0. If the equation (4.20) has two stationary points,s1 < s2 then the two lines l1 : cx + dy = s1(ax + by) and l2 :cx + dy = s2(ax + by) are parallel to eigenvectors. The half-plane on one side of the line ax + by = 0 is divided into threeregions by l1 and l2. The region of the phase plane where

s1 <cx + dyax + by

< s2

corresponds to the middle section of the first phase diagram inpart (c). The slope v of an orbit in this region increases as t in-creases, converging to s2 as t→ ∞ and to s1 as t→ −∞.

Page 706: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 705

In the region wherecx + dyax + by

< s1 the slope v → −∞ when the

orbit crosses the line ax+ by = 0 with a vertical tangent; then thesign of v changes and the slope will decrease from ∞, convergingto s2 as t→ ∞.If the system (4.20) has just one stationary point s1, then the sys-tem (4.19) has a double eigenvalue. The slope of each orbit isdecreasing, but changes sign as it crosses the line ax + by = 0with a vertical tangent. Hence the slope converges to s1 both ast→ −∞ and as t→ +∞!

(f) Again we will assume that b > 0. If the equation (4.20) has nostationary points, then the system (4.18) has no real eigenvalues.The orbits will be either clockwise ellipses (if the eigenvalues arepurely imaginary) or clockwise spirals. In either case, the slopewill be decreasing until the orbit assumes a vertical direction,and then, as the orbit turns farther, the slope will decrease againuntil a vertical direction is assumed again. The slope changesfrom +∞ to −∞ twice in each revolution about the origin.

ReturnDetails

Page 707: Ordinary Differential Equations: A Systems Approach

706 ORDINARY DIFFERENTIAL EQUATIONS

33.

D.ReturnDetails

Page 708: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 707

35.

B.ReturnDetails

Page 709: Ordinary Differential Equations: A Systems Approach

708 ORDINARY DIFFERENTIAL EQUATIONS

37.

A.ReturnDetails

Page 710: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 709

1.

Stable, but not asymptotically stable: degenerate system.

Page 711: Ordinary Differential Equations: A Systems Approach

710 ORDINARY DIFFERENTIAL EQUATIONS

3.

Unstable spiral: every nonstationary orbit is unbounded as t → ∞.

Example of an unbounded solution:[

xy

]= et

[cos(t)− sin(t)

]

Page 712: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 711

5.

Unstable. Here is an unbounded solution: v = e3t

0011

.

Page 713: Ordinary Differential Equations: A Systems Approach

712 ORDINARY DIFFERENTIAL EQUATIONS

7.

Unstable. Here is an unbounded solution: v = et

0011

.

Page 714: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 713

9.

Unstable. Here is an unbounded solution: v = Rev1(t) =

t cos(t)

cos(t) + t sin(t)00

.

Page 715: Ordinary Differential Equations: A Systems Approach

714 ORDINARY DIFFERENTIAL EQUATIONS

13.

The probability is equal to 0.

Page 716: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 715

1.

The repelled set consists of the fourth quadrant, and the points of thefirst quadrant that lie below repelled set (part of the separatrix) of thestationary point (2, 1).ReturnDetails

Page 717: Ordinary Differential Equations: A Systems Approach

716 ORDINARY DIFFERENTIAL EQUATIONS

3.

The origin is the only stationary point. It is asymptotically stable, andthe attracted set is the entire plane.

-4 -3 -2 -1 1 2 3 4

-4

-2

2

4

ReturnDetails

Page 718: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 717

5.

The origin is the only stationary point, and is neutrally stable.

-4 -3 -2 -1 1 2 3 4

-4

-2

2

4

ReturnDetails

Page 719: Ordinary Differential Equations: A Systems Approach

718 ORDINARY DIFFERENTIAL EQUATIONS

7.

The stationary points are the origin, and all points on the circle ofradius 1, centered at the origin. The origin is neutrally stable. Thepoints on the circle are unstable.

-1.5 -1 -0.5 0.5 1 1.5

-1.5

-1

-0.5

0.5

1

1.5

ReturnDetails

Page 720: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 719

9.

The stationary points (0, 0), which is linearly degenerate (eigenval-ues, 0 and −2) and unstable, and its attracted set is the positive y-axis; (2, 0), a saddle point with the x-axis as the stable separatrix; and(1, 1), a stable spiral node. Its attracted set is the first quadrant.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

ReturnDetails

Page 721: Ordinary Differential Equations: A Systems Approach

720 ORDINARY DIFFERENTIAL EQUATIONS

11.

The seven stationary points are (0, 0) (asymptotically stable), (±√

10, 0)(saddle points with the x-axis as unstable separatrices), (±3,±1) (un-stable), and (±1,±3) (saddle points).

-4 -2 2 4

-4

-3

-2

-1

1

2

3

4

ReturnDetails

Page 722: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 721

13.

There are two stationary points, (0, 0) (a saddle with stable separatrixon the y-axis and unstable separatrix on the y-axis) and (1, 1) (eigen-values are ±i). There is an integral: F(x, y) = x + y− ln |x y| definedexcept on the axes; F has a local minimum at (1,1), confirming thatthis stationary point is a center.

-1 -0.5 0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

ReturnDetails

Page 723: Ordinary Differential Equations: A Systems Approach

722 ORDINARY DIFFERENTIAL EQUATIONS

15.

The origin is asymptotically stable if k < 0 (a counterclockwise in-ward spiral), stable if k = 0 (a linear center, oriented counterclock-wise), and unstable if k > 0 (a counterclockwise outward spiral).

-2 2

-2

2k=-0.05 k=0

-2 2

-2

2k=0.05

ReturnDetails

Page 724: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 723

17.

The unit circle minus the point (1, 0) is an orbit contained in both theattracted and repelled set of (1, 0). Because the repelled set containspoints near (1, 0), this stationary point is unstable. The attracted setcontains the entire plane, except the origin. Thus all points inside acircle of radius 1

2 centered at (1, 0) are attracted to (1, 0); it is thereforeasymptotically stable.ReturnDetails

Page 725: Ordinary Differential Equations: A Systems Approach

724 ORDINARY DIFFERENTIAL EQUATIONS

1.

There are four stationary points, including the origin, (K, 0) on the x-axis, and (0, L) on the y-axis. If BC < 1 the fourth stationary point isa stable node in the first quadrant, and if BC > 1 the fourth stationarypoint is in the third quadrant, where it is not of interest—populationscan’t be negative.

In the phase portraits shown below, (B, C) =

(2,

13

)and (B, C) =(

2,23

), respectively. The other parameters are the same for both:

(a, d, K, L) = (0.05, 0.05, 2, 1).

2 4 6 8 10 12 14

2

4

6

8

2 4 6 8 10 12 14

2

4

6

8

ReturnDetails

Page 726: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 725

3.

(a) (C− d < 0) The small carrying capacity for the prey means thatthere will never be enough prey in this environment to supportthe predators. In the phase portrait that follows, the parametershave been set as to be (A, C, b, c, d) = (0.5, 1500, 1, 0.125, 2000)

500 1000 1500 2000

100

200

300

400

500

(b) ( Ad4c > C − d > 0) There is an asymptotically stable stationary

point in the first quadrant, representing equilibrium populationsof each species. Their populations will converge directly to thesenumbers. In the phase portrait that follows, the parameters havebeen set to be (A, C, b, c, d) = (0.5, 1500, 1, 0.125, 1000)

500 1000 1500 2000

100

200

300

400

500

(c) When C− d > Ad4c the stationary point in the first quadrant is a

Page 727: Ordinary Differential Equations: A Systems Approach

726 ORDINARY DIFFERENTIAL EQUATIONS

stable spiral node, because the eigenvalues have nonzero imag-inary parts, and their real part is negative. Orbits are directedcounterclockwise, indicating fluctuations as they converge to theequilibrium. In the following phase portrait, the parametershave been set to be (A, C, b, c, d) = (0.5, 1500, 2, 0.125, 500).

500 1000 1500 2000

100

200

300

400

500

ReturnDetails

Page 728: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 727

1.

No.ReturnDetails

Page 729: Ordinary Differential Equations: A Systems Approach

728 ORDINARY DIFFERENTIAL EQUATIONS

3.

Use the second derivative test to detect a relative minimum of thepotential at x1. Apply proposition 4.4.2.ReturnDetails

Page 730: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 729

5.

U(x) = − a2

x2.

Potential Phase Portrait

ReturnDetails

Page 731: Ordinary Differential Equations: A Systems Approach

730 ORDINARY DIFFERENTIAL EQUATIONS

7.

U(x) = −x

Potential Phase Portrait

ReturnDetails

Page 732: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 731

9.

U(x) =13

x3

Potential Phase Portrait

ReturnDetails

Page 733: Ordinary Differential Equations: A Systems Approach

732 ORDINARY DIFFERENTIAL EQUATIONS

11.

U(x) = x4

Potential Phase Portrait

ReturnDetails

Page 734: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 733

13.

U(x) = −|x3|Potential Phase Portrait

ReturnDetails

Page 735: Ordinary Differential Equations: A Systems Approach

734 ORDINARY DIFFERENTIAL EQUATIONS

15.

U(x) =x

x2 + 1Potential Phase Portrait

ReturnDetails

Page 736: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 735

17.

U(x) = cos(x) +12

x

Potential Phase Portrait

ReturnDetails

Page 737: Ordinary Differential Equations: A Systems Approach

736 ORDINARY DIFFERENTIAL EQUATIONS

19.

U(x) = cos(x) +(

)x2

Potential Phase Portrait

ReturnDetails

Page 738: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 737

1.

Complete the square to get

L(x, y) = a(x +b

2ay)2 +

14a

(4ac− b2)y2

If b2 − 4ac < 0, and a > 0, then L(x, y) > 0 for (x, y) 6= (0, 0). There-fore L(x, y) is positive definite.

Converse: Assume L(x, y) is positive definite. Then L(ε, 0) = aε2

must be positive; hence a > 0. Aslo

L(bε,−2aε) = a(4ac− b2)ε2,

so 4ac− b2 > 0.ReturnDetails

Page 739: Ordinary Differential Equations: A Systems Approach

738 ORDINARY DIFFERENTIAL EQUATIONS

3.

The origin is unstable.ReturnDetails

Page 740: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 739

5.

The origin is asymptotically stable.ReturnDetails

Page 741: Ordinary Differential Equations: A Systems Approach

740 ORDINARY DIFFERENTIAL EQUATIONS

7.

The origin is unstable.ReturnDetails

Page 742: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 741

9.

The origin is asymptotically stable. Use L(x, y) = x2 + y2 as the Lya-punov function.ReturnDetails

Page 743: Ordinary Differential Equations: A Systems Approach

742 ORDINARY DIFFERENTIAL EQUATIONS

11.

Using the Lyapunov function L(x, y) = 40x2 + y2, the origin can beshown to be unstable.ReturnDetails

Page 744: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 743

13.

L(x, y) = 50x2 − 2xy + y2 is a Lyapunov function that shows the ori-gin to be neutrally stable but not asymptotically stable.ReturnDetails

Page 745: Ordinary Differential Equations: A Systems Approach

744 ORDINARY DIFFERENTIAL EQUATIONS

15.

L(x, y) = 5x2 − 5xy + 4y2 is a Lyapunov function that shows the ori-gin to be asymptotically stable.ReturnDetails

Page 746: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 745

17.

L(x, y) = 4x2 + 3y2 is a Lyapunov function that shows the origin tobe asymptotically stable.ReturnDetails

Page 747: Ordinary Differential Equations: A Systems Approach

746 ORDINARY DIFFERENTIAL EQUATIONS

19.

If q(x, y) is negative semidefinite, L(x, y) is still a Lyapunov functionfor the system, and the origin is a stable stationary point that may ormay not be asymptotically stable.ReturnDetails

Page 748: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 747

21.

(a) L′(x, y) = 2(x2 + y2)2(1− x2 − y2) is positive definite for on theset x2 + y2 < 1. Therefore, the origin is unstable.

(b) Since q(cos t, sin t) = 0, this is obvious.

(c) In polar coordinates, this system decouples as

r′ = r3(1− r2)

θ′ = 1

The first equation has a stable stationary point at r = 1, whichshows solutions approach the unit circle as t → ∞. The secondequation says that every solution (except the stationary solutionat the origin) has angular velocity of 1.

(d)

-2 -1.5 -1 -0.5 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

ReturnDetails

Page 749: Ordinary Differential Equations: A Systems Approach

748 ORDINARY DIFFERENTIAL EQUATIONS

23.

The system has a limit cycle for any negative value of k.

-2 -1 1 2

-4

-2

2

4k=-1

-2 -1 1 2

-20

-10

10

20

k=-10

ReturnDetails

Page 750: Ordinary Differential Equations: A Systems Approach

ANSWERS TO SELECTED EXERCISES 749

25.

S =

[ 12

14

14

34

]ReturnDetails

Page 751: Ordinary Differential Equations: A Systems Approach

0 ORDINARY DIFFERENTIAL EQUATIONS

2.

(i) order 1

(ii) This equation is an ODE

Page 752: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 1

4.

(i) order 1

(ii) This equation is an ODE

Page 753: Ordinary Differential Equations: A Systems Approach

2 ORDINARY DIFFERENTIAL EQUATIONS

6.

(i) order 1

(ii) This equation is an ODE

Page 754: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 3

8.

(i) order 1

(ii) This equation is an ODE

Page 755: Ordinary Differential Equations: A Systems Approach

4 ORDINARY DIFFERENTIAL EQUATIONS

10.

(i) order 2

(ii) This equation is a PDE for there are two independent variables:x and y.

Page 756: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 5

12.

(i) order 3

(ii) This equation is an ODE

Page 757: Ordinary Differential Equations: A Systems Approach

6 ORDINARY DIFFERENTIAL EQUATIONS

14. If y = C sin(t) + D cos(t) then y′ = C cos(t) − D sin(t) andy′′ = −C sin(t) − D cos(t), therefore y′′ + y = 0. So y = C sin(t) +D cos(t)is a family of solutions of y′′ + y = 0.

(a) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D = 0,therefore the solutions are y = C sin(t).

(b) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D = 0.Substitute y = 0, t = π in y = C sin(t), we get C could be anynumber. Therefore the solutions are y = C sin(t).

(c) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D =0. Substitute y = 1, t = π/6 in y = C sin(t),we get C = 2.Therefore the solution is y = 2 sin(t).

(d) Substitute y = 1, t = 0 in y = C sin(t) + D cos(t),we get D = 1.Substitute y′ = −1, t = 0 in y′ = C cos(t) − D sin(t) we getC = −1. So the solution is y = − sin(t) + cos(t).

Return

Page 758: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 7

16.

If y = Cet2then y′ = Cet2 · (2t) = 2t · Cet2

= 2ty, therefore y = Cet2

is a solution of the ODE. Substitute t = 0, y = 3 in y = Cet2: we get

C = 3,therefore the solution of the IVP is y = 3et2.

Return

Page 759: Ordinary Differential Equations: A Systems Approach

8 ORDINARY DIFFERENTIAL EQUATIONS

18. Use implicit differentiation: if t2 + y2 = 0, then

2t + 2ydydt

= 0.

Divide by 2 to obtain yy′ + t = 0. Return

Page 760: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 9

20. If y > 0 and y = t ln t + Ct for some constant C, then y/t =ln(t) + C, so

ty′ − yt2 =

ddt

(yt

)=

1t

.

Thus, y = t ln t + Ct satisfies the ODE for any constant C. Now wewill show that all solutions have this form. Suppose that y is a solu-tion; that is,

ty′ − yt2 =

1t

.

Since(ty′ − y)/t2 =

ddt

(yt

)and

ddt(ln t) =

1t

,

we haveddt

yt=

ddt

ln(t).

Therefore, by the equal derivatives theorem, there is a constant C suchthat y

t= ln t + C

and it follows that y = t ln t + Ct. Therefore y = t ln t + Ct is thegeneral solution of the ODE ty′−y

t2 = 1t on the interval (0, ∞). Return

Page 761: Ordinary Differential Equations: A Systems Approach

10 ORDINARY DIFFERENTIAL EQUATIONS

1. Substitute t = 0 in the formula y = Cekt: y(0) = Ce0 = C. Nowsubstitute t = D, where D denotes the doubling time, and y = 2C.We find that 2C = CekD, and hence ekD = 2. Taking logarithms, kD =

ln(2), and therefore k = ln(2)D . Now substitute t = T, in y = Cekt where

T denotes the tripling time, and y = 3C. We find that 3C = CekT,and hence ekT = 3. Taking logarithms, kT = ln(3), and thereforeT = ln(3)

k . = 35 · ln(3)ln 2) = 55.47 years.

Return

Page 762: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 11

3. Let c denote the relative growth rate (the birth rate minus thedeath rate). Our model is y′ = c

1000 y; y(t + 25) = 2y(t), where t ≥ 0.The general solution of the ODE is y = Ce

c1000 t, so y(0) = C, 2C =

2y(0) = y(25) = Cec

1000 ·25. Therefore, 2 = ec

40 so c = 40 ln 2 = 27.726.Return

Page 763: Ordinary Differential Equations: A Systems Approach

12 ORDINARY DIFFERENTIAL EQUATIONS

5. For population A our model is y′1 = 0.04y1; y1(1900) = 1 mil-lion. The general solution for this ODE is y1 = Ce0.04t. Substitut-ing t = 1900, y1 = 1, we get C = e−1900×0.04, so y1 = e0.04(t−1900).Thus the population B is y2 =

√y1 = e0.02(t−1900), therefore y′2 =

0.02e0.02(t−1900) = 0.02y2 obeys equation (1.3), and its relative growthrate is 2% per year.Return

Page 764: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 13

7. The solution for the IVP y′ = 0.06y; y(1700) = 1 is y = e0.06(t−1700),so y(2000) = e0.06(2000−1700) = e18 = 65, 659, 969.1Return

Page 765: Ordinary Differential Equations: A Systems Approach

14 ORDINARY DIFFERENTIAL EQUATIONS

9. Substitute y = CeK(t), and y′ = CK′(t)eK(t) to obtain CtK′(t)eK(t)−12CeK(t) = 0. Solving for K′(t) we obtain K′(t) = 12

t . Hence K(t) =∫ 12t dt = 12 ln t. It follows that the general solution of this ODE is

y = Ce12 ln t = Ct12. To satisfy the initial condition, set t = 2 and y = 1to obtain C = 1

212 . Therefore the solution of the IVP is y = 1212 t12.

0.5 1 1.5 2 2.5

2

4

6

8

10

12

14

Return

Page 766: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 15

11. Substitute y = CeK(t), and y′ = CK′(t)eK(t) to obtain C cos(t)K′(t)eK(t)−C sin(t)eK(t) = 0. Solving for K′(t) we obtain K′(t) = tan(t). HenceK(t) =

∫tan(t) dt = − ln cos(t). It follows that the general solution

of this ODE is y = Ce− ln cos(t) = Celn 1cos(t) = C

cos(t) . To satisfy the initialcondition, set t = 0 and y = −1 to obtain C = −1. Therefore thesolution of the IVP is y = − 1

cos(t) for −π2 < t < π

2

-1.5 -1 -0.5 0.5 1 1.5

-14

-12

-10

-8

-6

-4

-2

Return

Page 767: Ordinary Differential Equations: A Systems Approach

16 ORDINARY DIFFERENTIAL EQUATIONS

13. Substitute y = CeK(t), and y′ = CK′(t)eK(t) to obtain tCK′(t)eK(t)+CeK(t) = 0. Solving for K′(t) we obtain K′(t) = − 1

t . Hence K(t) =

−∫ 1

t dt = − ln t. It follows that the general solution of this ODE isy = Ce− ln t = Celn 1

t = C 1t . To satisfy the initial condition, set t = 1

and y = 1 to obtain C = 1. Therefore the solution of the IVP is y = 1t

for t > 0.

2 4 6

1

2

3

4

Return

Page 768: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 17

15. Substitute y = CeK(t), and y′ = CK′(t)eK(t) to obtain (t2 + 1)CK′(t)eK(t)−tCeK(t) = 0. Solving for K′(t) we obtain K′(t) = t

t2+1 . Hence K(t) =∫ tt2+1 dt = 1

2

∫ 1t2+1 d(t2 + 1) = 1

2 ln(t2 + 1). It follows that the general

solution of this ODE is y = Ce12 ln(t2+1) = Celn

√(t2+1) = C

√(t2 + 1).

To satisfy the initial condition, set t = 3 and y = 0 to obtain C = 0.Therefore the solution of the IVP is y = 0.Return

Page 769: Ordinary Differential Equations: A Systems Approach

18 ORDINARY DIFFERENTIAL EQUATIONS

17. y = e−0.000121t = e−0.000121(1922+1325) = e−0.3929 = 0.675 ppb.Return

Page 770: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 19

19. y = e−0.000121t = e−0.000121×4850 = e−0.58685 = 0.556 ppb.Return

Page 771: Ordinary Differential Equations: A Systems Approach

20 ORDINARY DIFFERENTIAL EQUATIONS

1. The associated homogeneous equation is y′ + 4y = 0, and we cantake yh = e−4t. Substitute y = ve−4t and y′ = v′e−4t − 4ve−4t in theinhomogeneous ODE to get v′e−4t − 4ve−4t + 4ve−4t = 3t. When sim-plified, this reduces to v′e−4t = 3t. Multiplying through by e4t yieldsv′ = 3te4t. Thus v =

∫3te4t dt = 3

4

∫t d(e4t) = 3

4 te4t − 34

∫e4t dt =

34 te4t − 3

16 e4t + C Since y = ve−4t, the general solution is the familyy = 3

4 t − 316 + Ce−4t. To satisfy the initial condition, set y = 0 and

t = 0. This yields 0 = − 316 + C so C = 3

16 , and y = 34 t− 3

16 +3

16 e−4t.Return

Page 772: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 21

3. The associated homogeneous equation is 2y′ + y = 0, and we cantake yh = e−t/2. Substitute y = ve−t/2 and y′ = v′e−t/2 − 1

2 ve−t/2 inthe inhomogeneous ODE to get 2v′e−t/2 − ve−t/2 + ve−t/2 = t−1e−t/2

When simplified, this reduces to v′ = 12 t−1 Thus v = 1

2

∫t−1 dt =

12 ln t + C Since y = ve−t/2, the general solution is the family y =12 ln te−t/2 + Ce−t/2. To satisfy the initial condition, set y = 0 andt = 2. This yields 0 = 1

2 ln 2e−1 + Ce−1 so C = − 12 ln 2, and y =

12 ln(t)e−t/2 − 1

2 ln(2)e−t/2.Return

Page 773: Ordinary Differential Equations: A Systems Approach

22 ORDINARY DIFFERENTIAL EQUATIONS

5. The associated homogeneous equation is y′ + 2ty = 0, and wecan take yh = e−t2

. Substitute y = ve−t2and y′ = v′e−t2 − 2tve−t2

in the inhomogeneous ODE to get v′e−t2 − 2tve−t2+ 2tve−t2

= e−t2.

When simplified, this reduces to v′e−t2= e−t2

. Multiplying throughby et2

yields v′ = 1. Thus v =∫

dt = t + C Since y = ve−t2, the

general solution is the family y = te−t2+ Ce−t2

. To satisfy the initialcondition, set y = 0 and t = 0. This yields C = 0 so y = te−t2

.Return

Page 774: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 23

7.

(a) The associated homogeneous equation is y′ + 10y = 0, and wecan take yh = e−10t. Substitute y = ve−10t and y′ = v′e−10t −10ve−10t in the inhomogeneous ODE to get v′e−10t − 10ve−10t +10ve−10t = 1. When simplified, this reduces to v′e−10t = 1. Mul-tiplying through by e10t yields v′ = e10t. Thus v =

∫e10t dt =

110 e10t + C Since y = ve−10t, the general solution is the familyy = 1

10 + Ce−10t. To satisfy the initial condition, set y = 1 andt = 0. This yields 1 = 1

10 + C so C = 910 , and y = 1

10 +9

10 e−10t.

(b) The associated homogeneous equation is y′ − 10y = 0, and wecan take yh = e10t. Substitute y = ve10t and y′ = v′e10t + 10ve10t

in the inhomogeneous ODE to get v′e10t + 10ve10t − 10ve10t = 1.When simplified, this reduces to v′e10t = 1. Multiplying throughby e−10t yields v′ = e−10t. Thus v =

∫e−10t dt = − 1

10 e−10t + CSince y = ve10t, the general solution is the family y = − 1

10 +Ce10t. To satisfy the initial condition, set y = 1 and t = 0. Thisyields 1 = − 1

10 + C so C = 1110 , and y = − 1

10 +1110 e10t.

-1 -0.5 0.5 1

0.25

0.5

0.75

1

1.25

1.5

Return

Page 775: Ordinary Differential Equations: A Systems Approach

24 ORDINARY DIFFERENTIAL EQUATIONS

9.

(a) The associated homogeneous equation is y′ + 4y = 0, and wecan take yh = e−4t. Substitute y = ve−4t and y′ = v′e−4t − 4ve−4t

in the inhomogeneous ODE to get v′e−4t − 4ve−4t + 4ve−4t =2e−4t sin(2t). When simplified, this reduces to v′e−4t = 2e−4t sin(2t).Multiplying through by e4t yields v′ = 2 sin(2t). Thus v = 2

∫sin(2t) dt =

− cos(2t) + C Since y = ve−4t, the general solution is the fam-ily y = − cos(2t)e−4t + Ce−4t. To satisfy the initial condition,set y = 0 and t = 0. This yields 0 = −1 + C so C = 1, andy = − cos(2t)e−4t + e−4t.

(b) The associated homogeneous equation is y′ − 4y = 0, and wecan take yh = e4t. Substitute y = ve4t and y′ = v′e4t + 4ve4t in theinhomogeneous ODE to get v′e4t + 4ve4t − 4ve4t = 2e−4t sin(2t).When simplified, this reduces to v′e4t = 2e−4t sin(2t). Multiply-ing through by e−4t yields v′ = 2e−8t sin(2t). Thus v = 2

∫e−8t sin(2t) dt =

− 28

∫sin(2t) de−8t = − 1

4 e−8t sin(2t)+ 14

∫e−8t cos(2t)(2) dt = − 1

4 e−8t sin(2t)+12

∫e−8t cos(2t) dt = − 1

4 e−8t sin(2t)− 116

∫cos(2t) de−8t = − 1

4 e−8t sin(2t)−1

16 e−8t cos(2t)− 216

∫e−8t sin(2t) dt = − 1

4 e−8t sin(2t)− 116 e−8t cos(2t)−

116 v, so v = 16

17 (−14 e−8t sin(2t)− 1

16 e−8t cos(2t))+C = − 117 (4e−8t sin(2t)+

e−8t cos(2t)) + C. Since y = ve4t, the general solution is the fam-ily y = − 1

17 (4e−4t sin(2t) + e−4t cos(2t)) + Ce4t. To satisfy theinitial condition, set y = 0 and t = 0. This yields 0 = − 1

17 + C soC = 1

17 , and y = − 117 e−4t(4 sin(2t) + cos(2t)) + 1

17 e4t.

-2 -1 1 2

-40

-20

20

40

60

80

Return

Page 776: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 25

11. The associated homogeneous equation is y′ = tan(t)y, and wecan take yh = e− ln cos(t) = sec(t). Substitute y = v sec(t) and y′ =v′ sec(t) + v sec(t) tan(t) in the inhomogeneous ODE to get

v′ sec(t) cos(t) + v sec(t) tan(t) cos(t) = (v sec(t)− 1) sin(t).

When simplified, this reduces to v′ = − sin(t). Thus v = −∫

sin(t) dt =cos(t)+C, the general solution is the family y = cos(t) sec(t)+C sec(t) =1 + C sec(t). To satisfy the initial condition, set y = 0 and t = 0. Thisyields 0 = 1 + C so C = −1, and y = 1− sec(t).Return

Page 777: Ordinary Differential Equations: A Systems Approach

26 ORDINARY DIFFERENTIAL EQUATIONS

13. We will use the IVP dTdt = −k(T − A(t)); T(0) = 800 as our

model, where T is the temperature in ◦C, and A(t) = 20 is the am-bient temperature. The first step is to find the general solution of theODE, dT

dt = −k(T − 20), or T′ + kT = 20k. The homogeneous solu-tion is Th = e−kt. Substitute T = ve−kt; after cancelling this resultsin e−ktv′ = 20k, or v′ = 20kekt. Integration yields v = 20ekt + C, andhence T = e−ktv = 20 + Ce−kt. To evaluate the parameters C and k,we need to use the data. Substituting the initial condition T(0) = 800,800 = 20 + Ce0, we find that C = 780. Now substitute the other con-straint that was given: t = 1, and T = 600: 600 = 20 + 780e−k.It follows that e−k = 0.7436. Taking logarithms, k ≈ 0.2963. ThusT(t) = 20 + 780e−0.2963t. Let 50 = 20 + 780e−0.2963t, we get t = 11minutes.Return

Page 778: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 27

15. While the turkey is in the oven, we will use the IVP dTdt = −k(T−

A(t)); T(0) = 20 as our model, where T is the temperature in ◦C, andA(t) = 160 is the ambient temperature. The first step is to find thegeneral solution of the ODE, dT

dt = −k(T − 160), or T′ + kT = 160k.The homogeneous solution is Th = e−kt. Substitute T = ve−kt; af-ter cancelling this results in e−ktv′ = 160k, or v′ = 160kekt. Integra-tion yields v = 160ekt + C, and hence T = e−ktv = 160 + Ce−kt. Toevaluate the parameters C and k, we need to use the data. Substi-tuting the initial condition T(0) = 20, 20 = 160 + Ce0, we find thatC = −140. Now substitute the other constraint that was given: t = 5,and T = 85: 85 = 160− 140e−5k. It follows that e−5k = 0.62415. Takinglogarithms, k = − ln(0.62415) ≈ 0.12483.

Once the turkey is removed from the oven, we will use the IVP dTdt =

−k(T− A(t)) = −0.12483(T− A(t)); T(0) = 85 as our model, whereT is the temperature in ◦C, and A(t) = 20 is the ambient tempera-ture. The first step is to find the general solution of the ODE, dT

dt =−0.12483(T − 20), or T′ + 0.12483T = 20k. The homogeneous solu-tion is Th = e−0.12483t. Substitute T = ve−0.12483t; after cancelling thisresults in e−0.12483tv′ = 20× 0.12483, or v′ = 20× 0.12483× e0.12483t.Integration yields v = 20e0.12483t + C, and hence T = e−0.12483tv =20 + Ce−0.12483t. To evaluate the parameters C, we need to use thedata. Substituting the initial condition T(0) = 85, 85 = 20 + Ce0,we find that C = 65. Thus T(t) = 20 + 65e−0.12483t. So T(0.5) =20 + 65e−0.12483×0.5 = 81◦C. The assumptions are that the transmis-sion coefficient in the oven is the same as that outside the oven, andthat there are no other sources of heat.Return

Page 779: Ordinary Differential Equations: A Systems Approach

28 ORDINARY DIFFERENTIAL EQUATIONS

17. We will use the IVP dTdt = −k(T− A(t)); T(0) = 15 as our model,

where T is the temperature in ◦C, and A(t) = −20 is the ambienttemperature. The first step is to find the general solution of the ODE,dTdt = −k(T + 20), or T′ + kT = −20k. The homogeneous solutionis Th = e−kt. Substitute T = ve−kt; after cancelling this results ine−ktv′ = −20k, or v′ = −20kekt. Integration yields v = −20ekt + C,and hence T = e−ktv = −20 + Ce−kt. To evaluate the parametersC and k, we need to use the data. Substituting the initial conditionT(0) = 15, 15 = −20 + Ce0, we find that C = 35. Now substitutethe other constraint that was given: t = 6 − 3 = 3, and T = 10:10 = −20 + 35e−3k. It follows that e−3k = 0.857143. Taking loga-rithms, k = ln(0.857143)/3 ≈ 0.0514 hour−1.Return

Page 780: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 29

19. We use the model: T′(t) = −k[T(t) − A(t)] + mN, where N isthe number of stoves in use. We are given that when the ambienttemperature is −15 and N = 1, the solution is T(t) ≡ 10. SubstituteT(t) = 10, T′(t) = 0, A(t) = −15, and N = 1 to get 0 = −k(10 +15) + m. Hence m ≡ 25k. Then when N = 2 and the temperature hasstabilized at T = T∞, we have

0 = −k(T∞ + 15) + 2m = −k(T∞ + 15) + 50k = −k(T∞ − 35)

Therefore the temperature will stabilize at 35◦C.Return

Page 781: Ordinary Differential Equations: A Systems Approach

30 ORDINARY DIFFERENTIAL EQUATIONS

21. We use the model: T′(t) = −k[T(t)− A(t)] + mH(t), with A(t) =0, k = 0.04, and

mH(t) =

1 if 0 ≤ t < 40 if 4 ≤ t < 51 if 5 ≤ t < 9

...

In simpler terms, T′+ 0.04T = mH(t). Since the furnace is on 45 of the

time, an intuitive approach would be to use a different model, wheremH(t) = 4

5 = 0.8.

The homogeneous solution is Th = e−0.04t. Substitute T = ve−0.04t;after canceling this results in e−0.04tv′ = mH(t), or v′ = mH(t)e0.04t.With mH(t) = 0.8, we can integrate to get v = 20e0.04t + C. HenceT = e−0.04tv = 20 + Ce−0.04t. As t → ∞, this solution converges to20◦C.

The periodic solution Tp(t) of the ODE

T′ + 0.04T = mH(t)

is not constant, because the furnace does cycle on and off. First notethat Tp is stable, because if T(t) is any other solution then y = T− Tpsatisfies the ODE y′ = −.04y; hence y = Ce−0.04t → 0.

Let f1(t) be the solution of the IVP

y′ = −.04y + 1; y(0) = a, (54)

where a is a parameter that we will determine. Let f2(t) be the solu-tion of the IVP

y′ = −.04y; y(4) = f1(4).

If Tp(0) = a, then

Tp(t) ={

f1(t) if 0 ≤ t < 4f2(t) if 4 ≤ t ≤ 5,

and, because Tp is 5-periodic, Tp(5) = Tp(0) = a. Since Tp(5) = f2(5),we must find the value of A such that f2(5) = a.

Solving equation 54, we have

f1(t) = 25[(.04a− 1)e−.04t + 1]

Page 782: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 31

. Thus, f1(4) = 25(1− e−.16) + a e−.16 = 3.696 + 0.8521 a, where weare using four-digit precision. Thus f2(t) = (3.696+ 0.8521 a)e−.04(t−1)

and hencef2(5) = (3.686 + 0.8521 a)e−.04 = a

Using the approximation e−.04 = 0.9608, we can solve for a to geta = 19.59◦C. This is the temperature when the furnace cycles on.

Using our value of a, we have, for 0 ≤ t ≤ 5,

f1(t) = 25− 5.411 e−.04t

andf2(t) = 20.39 e−.04(t−1)

The average temperature is

15

∫ 5

0Tp(t) dt =

15

[∫ 4

0f1(t) dt +

∫ 5

4f2(t) dt

]=

15

[100 +

1.04

5.411(e(−.04)4 − 1)− 1.04

20.39(e−.04 − 1)]

= 19.72◦C.

The drawing shows the graph of the periodic temperature.

2 4 6 8 10 12 14

5

10

15

20

Return

Page 783: Ordinary Differential Equations: A Systems Approach

32 ORDINARY DIFFERENTIAL EQUATIONS

23.

(a) The associated homogeneous equation is y′+ 5y = 0, and we cantake yh = e−5t. Substitute y = ve−5t yields v′ = 5e5t cos 2t. Thusv = 5

29 e5t (5 cos 2t + 2 sin 2t) + C Since y = ve−5t, the generalsolution is the family y = 5

29 (5 cos 2t + 2 sin 2t) + Ce−5t. The pe-riodic solution is 5

29 (5 cos 2t + 2 sin 2t) when t → ∞, Ce−5t → 0,therefore it is stable.

(b) The associated homogeneous equation is y′ − y = 0, and we cantake yh = et. Substitute y = vet and y′ = v′et + vet in the inho-mogeneous ODE to get v′et = 7 cos 4t. So v′ = 7e−t cos 4t. Thusv = − 7

17 e−t (cos 4t− 4 sin 4t) + C and the general solution is thefamily y = − 7

17 (cos 4t− 4 sin 4t) + Cet. The periodic solution isy = − 7

17 (cos 4t− 4 sin 4t) . When t → ∞, Cet → ∞, therefore itis not stable.

(c) The associated homogeneous equation is y′ + 2y = 0, and wecan take yh = e−2t. Substitute y = ve−2t yields v′ = e2t(cos t−3 sin t). Thus v = e2t(cos t− sin t) + C and the general solutionis the family y = cos t − sin t + Ce−2t. The periodic solution isy = cos t− sin t when t→ ∞, Ce−2t → 0, therefore it is stable.

(d) The associated homogeneous equation is y′ − 5y = 0, and wecan take yh = e5t. Substitute y = ve5t and y′ = v′e5t + 5ve5t in theinhomogeneous ODE to get v′ = e−5t(4 cos t + 3 sin t) Thus v =− 1

26 e−5t (23 cos t + 11 sin t) + C and the general solution is thefamily y = − 1

26 (23 cos t + 11 sin t) + Ce5t.The periodic solutionis y = − 1

26 (23 cos t + 11 sin t) . When t→ ∞, Ce5t → ∞,thereforeit is not stable.

(e) The associated homogeneous equation is y′ + y = 0, and we cantake yh = e−t. Substitute y = ve−t yields v′ = ete−t sin t = sin t.Thus v = − cos t + C and the general solution is the family y =−e−t cos t + Ce−t, and there is no periodic solution.

Return

Page 784: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 33

25.

(a)C

e(0.1) t +(−0.498753) e0.1 t cos(2. t) + (0.0249377) e0.1 t sin(2. t)

e(0.1) t

(b) Substitute y = eK(t) in the homogeneous equation y′ + 2ty = 0,to get K′(t)eK(t) + 2teK(t) = 0. Cancel eK(t) to obtain k′(t) = −2t,so K(t) = −t2. Thus y = e−t2

is a solution of the associatedhomogeneous equation. Put y = ve−t2

in the original equationto obtain

ddt(v e−t2

) + 2t ve−t2= 1

Use the product rule and simplify to obtain v′ e−t2= 1, so v′ =

et2. It follows that

v =∫

et2dt.

Hence the solution is

y = e−t2∫

et2dt + C e−t2

The CAS foundCet2 +

√π Erfi(t)

2 et2 ,

and seems to be using a special function of the form

Erfi(t) =2√π

∫et2

dt.

(c)Ct12 +

−t12

3456 +t12 ln(t)

288 − t12 ln(t)2

48 + t12 ln(t)3

12t12 .

(d)12−√

t + t +C

e2√

t

Return

Page 785: Ordinary Differential Equations: A Systems Approach

34 ORDINARY DIFFERENTIAL EQUATIONS

27. Put t = 0 in the ODE; noting that limt→0 t cot(t) = 1, we obtain0 · y′(0)+ 1 · y(0) = 1; thus y(0) = 1. Our CAS tells us that the generalsolution of the ODE is

(Si(t) + C) csc(t)

where Si(t) is a special function. You can see by solving this ODEthat Si(t) =

∫ 1t sin(t) dt. If C 6= 0, this solution would be undefined

at t = 0.

-3 -2 -1 1 2 3

-10

-7.5

-5

-2.5

2.5

5

7.5

10

Return

Page 786: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 35

29. If there is a critical point at t = 1/4 then y′(1/4) = 0. Put t = 1/4in the ODE; we obtain

0 + 2 y(

14

)=

14

y(

14

)2

which we solve to find y(1/4) is either 0 or 8. By differentiating theODE, we find

y′′ + 2y′ = y2 + 2t y y′.

Set t = 1/4. Then y′ = 0 so y′′ = y2. If y(1/4) = 8 then there is arelative minimum by the second derivative test. The test is indecisivewhen y(1/4) = 0, but the solution with initial value y(1/4) = 0 is theconstant solution y ≡ 0.

-0.2 0.2 0.4 0.6

2.5

5

7.5

10

12.5

15

17.5

20

Return

Page 787: Ordinary Differential Equations: A Systems Approach

36 ORDINARY DIFFERENTIAL EQUATIONS

In Exercises 1–10, assume that the uniformity hypothesis holds.

1. Let x be the amount of salt in the tank, in kilograms. We use theODE x′(t) = JK − LC(t). Substitute J = 2, K = 16%, L = 2, C(t) =x(t)/100, x(0) = 0. We get x′(t) + 0.02x(t) = 0.32 The homogeneoussolution is e−0.02t. Now we substitute x = ve−0.02t in the differentialequation, and simplify to get v′ = 0.32e0.02t. Integration yields v =16e0.02t +C, where C is a constant. Since x(t) = ve−0.02t, it follows thatx(t) = 16 + Ce−0.02t. Substitute x(0) = 0, we get C = −16. Thereforex(t) = 16− 16e−0.02t. Let x(t) = 8%× 100 = 8. Solving 8 = 16−16e−0.02t, we find it takes t = 35 seconds to reach a concentration of8%; solving 12 = 16− 16e−0.02t yields t = 69 seconds to reach 12%;and solving 15 = 16− 16e−0.02t yields t = 138 seconds. to reach aconcentration of 15%.Return

Page 788: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 37

3. Let y(t) denote the amount, in milligrams, of KI131 in the tank attime t. (If the concentration is 100 ppm, then there are 100 milligramsper liter.) The input rate is thus 100 milligrams per day; the outputrate is ky (due to radioactive decay, where k is the decay rate of I131)plus 1

16 y for the solution that is drained from the tank. To find k,solve y′ = −ky with initial condition y(0) = 1. Thus y = e−kt. Sincethe half-life is 8 days, y(8) = 0.5 = e−8k, which can be solved for k toobtain k = ln 2

8 .

Taking into account the decay of the radioactive iodide and the re-moval and replacement of solution, we will use the ODE

y′ = − ln 28

y− 116

y + 100 = 100− p y,

where p = 2 ln 2+116 . The homogeneous solution is y = e−p t. Substi-

tuting y = ve−p t results in v′ = 100ep t. Hence v = 100p ep t + C, and

y = 100p + Ce−p t. The limit, as t→ ∞, is y = 100

p milligrams. Dividingby the volume, 16 liters, the limiting concentration is

10016 p

=100

2 ln 2 + 1≈ 42 ppm.

Return

Page 789: Ordinary Differential Equations: A Systems Approach

38 ORDINARY DIFFERENTIAL EQUATIONS

5. Let y(t) denote the amount of quinine in the tank with time t.Our model will be the ODE y′ = − 1000

10000+(1000−900)t y = − 10100+t y. The

initial value will be the amount of quinine now in the tank: y(0) =10000 × 0.01% = 1. The ODE is homogeneous and linear, so y =

Ce−∫( 10

100+t dt) = Ce−10 ln(100+t) = C(100 + t)−10. Substituting y(0) = 1we find C = 10010. Therefore y(t) = 10010(100+ t)−10 = (1+ t

100 )−10.

To find when the accident occurred, set y(t) = 4. We get t = 100(4−0.1−1) = −12.9449, so at worst the quinine was introduced about 13 hoursago.Return

Page 790: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 39

7. Let y(t) denote the amount of antibiotic drug at time t. Our modelis the ODE y′ = 0.5 − 0.25

3+(0.5−0.25)t y. The homogeneous solution is

y = e−∫ 0.25

3+0.25t dt = 13+0.25t . Substitute y = v 1

3+0.25t ; after canceling thisresults in 1

3+0.25t v′ = 0.5, or v′ = 0.5(3+ 0.25t) = 1.5+ 0.125t. Integra-

tion yields v = 1.5t+ 116 t2 +C, and hence y =

1.5t+ 116 t2+C

3+0.25t . Substitutingy(0) = 0, we find C = 0, so y(4) = 7

4 , and after the transfusions stopthe amount of drug is 7

4 + 0.5× 2 = 114 . The concentration of drug is

114

3+4×(0.5−0.25)+0.5×2 = 1120 = 0.55 grams per liter.

Return

Page 791: Ordinary Differential Equations: A Systems Approach

40 ORDINARY DIFFERENTIAL EQUATIONS

9. let x(t) and y(t) denote the amounts of salt in tanks A and B,respectively, at time t. Then x′ = − 0.2

3 x + 0.21 y. Since x(t) + y(t) =

x()) = 0.36,, we can replace y with .36− x. Then, after simplifying,x′ = − 4

15 x+ 0.072. The homogeneous solution is x = e−4

15 t. Substitutex = ve−

415 t; after cancelling this results in e−

415 tv′ = 0.072, or v′ =

0.072e4

15 t. Integration yields v = 0.072 × 154 e

415 t + C = 0.27e

415 t + C,

and hence x(t) = 0.27 + Ce−415 t. Substitute x(0) = 3.6, get C = 0.09.

Then y(t) = 0.36− x(t) = 0.09− 0.09e−415 t. The concentration in tank

A is x(t)/3 = 0.09 + 0.03e−415 t, and the concentration in tank B is

y(t) = 0.09− 0.09e−415 t.

Return

Page 792: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 41

11. Let y(t) denote the funds available in the retirement accountat time t. Then y′ = 6%y − 30000e3%t, The homogeneous solutionis yh = e0.06t. Substitute y = ve0.06t; after canceling this results ine0.06tv′ = −30000e0.03t, or v′ = −30000e−0.03t. Integration yields v =1000000e−0.03t + C, and hence y = e0.06tv = 1000000e0.03t + Ce0.06t. IfC < 0, as t → ∞, y → −∞, so C ≥ 0, and y(0) = 1000000 + C ≥1,000,000.Return

Page 793: Ordinary Differential Equations: A Systems Approach

42 ORDINARY DIFFERENTIAL EQUATIONS

13. The solution of d′(t) = kd(t), is d(t) = Cekt. To determine thevalues of k and C, substitute d(20) = 2.8 and d(50) = 17.6. ThusCe20k = 2.8, and Ce50k = 17.6; dividing these equations results ine30k = 17.6

2.8 . Thus k = 130 ln

( 17.62.8

)≈ 0.061, and C = 2.8/e20k = 0.82.

Therefore d(t) = 0.82e0.061t.

Let N(t) be the number of survivors in the sample t years later. De-termining this function is an input-output problem where there is noinput and the output is by death. Thus, N(0) = 100, 000, and we areto determine N(30). We will use the ODE

N′(t) = −d(t + 20)1000

N(t)

since at time t everyone in the sample is t + 20 years old.

Substitute N(t) = C eK(t) and d(t + 20) = 0.82e0.061(t+20) in the ODEand simplify to obtain

K′(t) = −0.00082e0.061(t+20).

Let’s choose the constant of integration so that K(0) = 0; thus K(t) =−.013(e0.061(t+20)− e0.061(20)). Then N(0) = C = 100, 000, and we havethe formula

N(t) = 100, 000e−.013e0.061(t+20)

e−.013e0.061(20)

The number who survive to age 50 is N(30) ≈ 79,000.Return

Page 794: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 43

1. Divide through by ty2 to obtain y′+ t−1y = t−1y−2. To linearize,definea new variable v = y3. Then dv

dt = 3y2 dydt ; or dy

dt = 13 y−2 dv

dt . Substitutethis expression in the equation to obtain v′ + 3

t v = 3t . The associ-

ated homogeneous equation is v′ + 3t v = 0, and we can take vh =

e−∫ 3

t dt = e−3 ln t = t−3. Substitute v = ut−3 yields u′ = t3 3t = 3t2.

Thus u = t3 + C Since v = ut−3, the general solution is the familyv = 1 + Ct−3. Therefore y = (1 + Ct−3)1/3.Return

Page 795: Ordinary Differential Equations: A Systems Approach

44 ORDINARY DIFFERENTIAL EQUATIONS

3. To linearize, define a new variable v = y−2. Then v′ = −2y−3y′.Divide the ODE through by y−3; we obtain

y−3 y′ + 3y−2 = sin(t),

or,

−12

v′ + 3 v = sin(t)

The associated homogeneous equation is v′− 6v = 0, and we can takevh = e6t. Substituting v = ue6t yields

−12

u′e6t = sin(t),

so

u = −2∫

e−6t sin(t) dt =137

e−6t[2 cos(t) + 12 sin(t)] + C

Since v = ue6t, the general solution is the family v = 137 (2 cos t +

12 sin t) + Ce6t. Therefore

y = ±( 137

(2 cos t + 12 sin t + Ce6t))−1/2

= ±√

37√2 cos t + 12 sin t + Ce6t

Return

Page 796: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 45

5. To linearize,define a new variable v = y4. Multiply through by y3

to obtain

y3 y′ +y4

2t + 1= 12(2t + 1)3,

or,14

v′ +v

2t + 1= 12(2t + 1)3,

because v′ = 4y3 y′. The associated homogeneous equation is v′ +4v

2t+1 = 0, and we can take vh = e−∫ 4

2t+1 dt = e−2 ln(2t+1) = (2t +1)−2. Substitute v = u(2t + 1)−2 yields u′ = 48[(2t + 1)]5. Thus u =4(2t+ 1)6 +C Since v = u(2t+ 1)−2, the general solution is the familyv = 4(2t + 1)4 + C(2t + 1)−2. Therefore y = ±(4(2t + 1)4 + C(2t +1)−2)1/4.Return

Page 797: Ordinary Differential Equations: A Systems Approach

46 ORDINARY DIFFERENTIAL EQUATIONS

1.

(a) The right side of dydt = t2+5

ty = t2+5t ×

1y can be expressed as the

product of two single-variable functions, therefore the ODE isseparable.

(b) The right side of dydt = y

t = y× 1t can be expressed as the product

of two single-variable functions,therefore the ODE is separable.

(c) The ODE dydt = ety, is not separable, because ety can’t be factored

as the product of a function of t and a function of y.

(d) The right side of dydt = e−(t

2+y2) = e−t2 × e−y2can be expressed as

the product of two single-variable functions, therefore the ODEis separable.

(e) The right side of dydt = cot y depends only on y, therefore the

ODE is separable.

(f) The ODE y′ = t+yt−y is not separable, because t+y

t−y can’t be factoredas the product of a function of t and a function of y.

Return

Page 798: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 47

3. According to Newton’s second law of motion, we have mv′ =−b v, m = 1, v(0) = 1, v(1) = 0.8, where b is the friction constant.So dv

dt = −b v, v 6= 0, and the separated equation is dvv = −b dt, Inte-

grating both sides, we have ln |v| = −bt + C1, so v = Ce−bt, whereC = eC1 is a constant. Substituting v(0) = 1, we obtain 1 = C × e0.Therefore C = 1.

(a) Substitute v(1) = 0.8 in v = e−bt, and solve for b. Thus, 0.8 =e−b, and therefore b = − ln(0.8) = 0.223 kg/sec.

(b) Substituting v = 0.5 in v = eln(0.8)t, we obtain t = ln(0.5)ln(0.8) = 3.106

seconds.

(c) To obtain the distance s, integrate the velocity. Thus,

s =∫ 1

0eln(0.8)t dt = − 0.2

ln(0.8)= 0.8963 meters.

(d) In an infinite amount of time, the ball will roll∫ ∞

0eln(0.8)t dt = − 1

ln(0.8)= 4.4814 meters.

Return

Page 799: Ordinary Differential Equations: A Systems Approach

48 ORDINARY DIFFERENTIAL EQUATIONS

5. According to Newton’s second law of motion, mv′ = −b v + mg.Put m− 9.8 kg and g = 9.8 m/s2. Now put v = 10 m/s and v′ = 0.(The object will move at constant speed when v = 10.) This yields0 = −b(10) + 9.8 and hence b = 0.98 kg/s.Return

Page 800: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 49

7. Since h(y) = y2 has a zero at y = 0, there is a constant solution,y ≡ 0. The separated equation is dy

y2 = (t − 1) dt. Integrating both

sides, we get a family of solutions in implicit form, −y−1 = 12 t2 − t +

C. Solving for y,

y =−2

t2 − 2t + 2C.

The singular solution satisfies the initial conditionReturn

Page 801: Ordinary Differential Equations: A Systems Approach

50 ORDINARY DIFFERENTIAL EQUATIONS

9. Since h(y) = y has a zero at y = 0, there is a constant solution,y ≡ 0. The separated equation is dy

y = dt2t . Integrating both sides, we

get a family of solutions in implicit form, ln |y| = 12 ln |t|+C1. Solving

for y, y = ±C√

t, where C = eC1 is a constant. Substituting y(1) = 2,we obtain 2 = C

√1. Therefore C = 2 and the particular solution is

y = 2√

t.Return

Page 802: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 51

11. Since h(y) = y has a zero at y = 0, there is a constant solution,y ≡ 0. The separated equation is dy

y = et dt. Integrating both sides,we get a family of implicit solutions, ln |y| = et + C1. Solving fory, y = ±Ceet

, C = eC1 is a constant, substitute y(0) = 1 we obtain1 = Ce, therefore C = e−1 and the particular solution is y = eet−1.Return

Page 803: Ordinary Differential Equations: A Systems Approach

52 ORDINARY DIFFERENTIAL EQUATIONS

13. Since h(y) = y2 has a zero at y = 0, there is a constant solution,y ≡ 0. The separated equation is dy

y2 = dt. Integrating both sides,

we get a family of implicit solutions, − 1y = t + C. Solving for y, we

obtain y = − 1t+C , where C is a constant. Substitute y(0) = 1 to obtain

1 = − 1C ; therefore C = −1 and the particular solution is y = − 1

t−1 .Return

Page 804: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 53

15. Since h(y) = y2− 1 has two zeros, y = ±1, there are two constantsolutions y ≡ 1, and y ≡ −1. The separated equation is dy

y2−1 = t dt.

Integrating both sides, we get a family of implicit solutions, 12 ln

∣∣∣ y−1y+1

∣∣∣ =12 t2 + C1. Take the exponential of both sides to obtain y−1

y+1 = ±Cet2,

where C is a constant, and choose the plus sign since it is in accordwith the initial condition. Substitute y(0) = 0 we obtain −1 = Ce0;therefore C = −1 and the particular solution is y−1

y+1 = −et2. Hence

y = −et2+11+et2

.Return

Page 805: Ordinary Differential Equations: A Systems Approach

54 ORDINARY DIFFERENTIAL EQUATIONS

17. According to Newton’s second law of motion, mv′ = −k v2 + mg,

Set v′ = 0 to obtain v∞ =√

mgk . Thus the separated equation can be

writtendv

g− kv2/m= dt,

ordv

1− v2

v2∞

= g dt.

Integrating by the method of partial fractions, we obtain

v∞

2ln(

v∞ + vv∞ − v

)= gt + C.

By substituting v(0) = 0, we find C = 0. Therefore

v∞

2ln(

v∞ + vv∞ − v

)= gt.

Now substitute v(2) = 14.7 to obtain the equation

v∞ ln(v∞ + 14.7v∞ − 14.7

) = 39.2.

The following graph of f (u) = 39.2− u (ln(u + 14.7)− ln(u− 14.7))shows that f has a zero at approximately u = 19. This zero representsthe terminal velocity.

17 18 19 20 21 22

-10

-8

-6

-4

-2

2

4

Return

Page 806: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 55

19. According to Newton’s second law of motion, mv′ = kv2 + mg,when the object is moving upward, and mv′ = −kv2 + mg, whenthe object falls downward. Substitute v = v∞ and v′ = 0 in mv′ =−kv2 + mg, and obtain k = mg

v2∞

. When the object is moving upward,

v′ =g

v2∞

v2 + g,

and the separated equation is

dv1 + v2

v2∞

= g dt,

Integrating both sides, we have v∞ arctan(

vv∞

)= gt + C. Substitut-

ing v(0) = v0, we find

C = v∞ arctan(

v0

v∞

),

and therefore v∞ arctan vv∞

= gt + v∞ arctan v0v∞

. To find the time T1 toreach the maximum height, set v = 0, and obtain gT1 + v∞ arctan v0

v∞=

0. Thus

T1 =v∞

garctan

|v0|v∞

For 0 ≤ t ≤ T1 the velocity is negative and is given by the formula

v = v∞ tan((g/v∞)t + arctan(v0/v∞))

= −v∞

(v0 + v∞ tan(g t/v∞)

v0 tan(g t/v∞)− v∞

)(a) the maximum height attained by the ball is

H = v∞

∫ T1

0tan((g/v∞)t + arctan(v0/v∞)) dt

Notice that since the upward direction is negative, v0 < 0, and Hwill be negative as well. Substitute s = gt

v∞+ arctan v0

v∞and ds =

(g/v∞) dt. The limits of integration for s corresponding to t− 0and t = T1 are arctan v0

v∞(which is negative), and 0, respectively.

Page 807: Ordinary Differential Equations: A Systems Approach

56 ORDINARY DIFFERENTIAL EQUATIONS

Thus

H = (v2∞/g)

∫ 0

arctan( v0v∞ )

tan(s) ds

= −v2∞g

ln(sec(arctan(v0/v∞)))

= −v2∞g

ln

v20 + v2

v∞

.

(b) We have seen that the time taken to reach that height is

T1 =v∞

garctan

|v0|v∞

.

(c) When the ball falls downward, we have the initial value problem

mv′ = −mgv2

∞v2 + mg; v(0) = 0.

Dividing through by m, we obtain

v′ = g(

1− v2

v2∞

).

The separated equation is

dv1− v2

v2∞

= g dt,

Integrating both sides, we have

v∞

2ln(

v∞ + vv∞ − v

)= gt + C.

Substituting v(T1) = 0, we obtain C = 0, and then

v = v∞

(e

2gtv∞ − 1

e2gtv∞ + 1

)= v∞ tanh(gt/v∞).

Let T2 be the time taken for the ball to return to the ground fromthe maximum height. The distance fallen, which is positive now,is equal to −H. Thus

v∞

∫ T2

0tanh(gt/v∞) dt = −H =

v2∞g

ln

v20 + v2

v∞

,

Page 808: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 57

or

v2∞g

ln cosh(gT2/v∞)) =v2

∞g

ln

v20 + v2

v∞

It follows that

T2 =v∞

gcosh−1

v20 + v2

v∞

=

v∞

gcosh−1

√( v0

v∞

)2

+ 1

The identity cosh(x) =

√sinh2(x) + 1 can now be used to see

that

T2 =v∞

gsinh−1

(|v0|v∞

)The time taken to return the ground is T1 + T2

(d) The velocity when the ball hits the ground is

V = v∞ tanh(gT2/v∞) = v∞ tanh(

sinh−1(|v0|v∞

))Now refer to the hyperbolic identity

tanh(u) =sinh(u)cosh(u)

=sinh(u)√

sinh2(u) + 1.

It follows that

V = v∞|v0|/v∞√

(|v0|/v∞)2 + 1

If the drag force is negligible (v∞ = ∞)

(a) The maximum height attained by the ball equal limv∞→∞(v2

∞g ln√

v20+v2

∞v∞

) =v2

02g

(b) the time taken to reach that height equal limv∞→∞(v∞g arctan |v0|

v∞) =

v0g

Page 809: Ordinary Differential Equations: A Systems Approach

58 ORDINARY DIFFERENTIAL EQUATIONS

(c) the time taken to return to the ground equal limv∞→∞(T1 +T2) =2v0g

(d) the velocity when the ball hits the ground is limv∞→∞(|v0|v∞√

v20+v2

∞) =

|v0|

Return

Page 810: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 59

21. According to Newton’s second law of motion, we have mv′ =−kv2 + mg. To determine k, substitute v = v∞, v′ = 0. Thus k =mgv2

∞. It follows that v′ = − g

v2∞

v2 + g, and the separated equation isdv

1− v2

v2∞

= g dt. By integrating both sides, we find v∞2 ln

∣∣∣ v+v∞v−v∞

∣∣∣ = gt + C.

Substitute v(0) = 0, we find C = 0. Thus v∞2 ln

∣∣∣ v+v∞v−v∞

∣∣∣ = gt, and t =v∞2g (ln

∣∣∣ v+v∞v−v∞

∣∣∣). Finally, with v = v∞2 we obtain t = v∞

2g (ln(3)) =v∞ ln(3)

2g .Return

Page 811: Ordinary Differential Equations: A Systems Approach

60 ORDINARY DIFFERENTIAL EQUATIONS

1. ∂P∂y (x, y) = 5 and ∂Q

∂x (x, y) = 5 so the equation is exact. The inte-gral F(x, y) will have the form F(x, y) =

∫(2x + 5y + 3) dx + H(y) =

x2 + 5xy + 3x + H(y). To determine H(y), differentiate this expres-sion with respect to y: ∂F

∂y = 5x + dHdy . Since ∂F

∂y = Q(x, y) = 5x −4y + 2, it follows that 5x + dH

dy = 5x − 4y + 2. After cancelling, we

obtain dHdy = −4y + 2. Therefore H(y) = −2y2 + 2y, and the integral

is F(x, y) = x2 + 5xy + 3x− 2y2 + 2y.Return

Page 812: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 61

3. ∂P∂y (x, y) = 2y and ∂Q

∂x (x, y) = 2y− 1; therefore the exactness con-dition does not hold.Return

Page 813: Ordinary Differential Equations: A Systems Approach

62 ORDINARY DIFFERENTIAL EQUATIONS

5. ∂P∂y (x, y) = 3e3x(ln y− 1)+ 3e3x = 3e3x ln y and ∂Q

∂x (x, y) = 3e3x ln yso the equation is exact. The integral F(x, y) will have the form F(x, y) =∫

3e3xy(ln y − 1) dx + H(y) = e3xy(ln y − 1) + H(y). To determineH(y), differentiate this expression with respect to y: ∂F

∂y = e3x(ln y−1) + e3x + dH

dy = ln y(e3x) + dHdy . Since ∂F

∂y = Q(x, y) = ln y(e3x − y),

it follows that ln y(e3x) + dHdy = ln y(e3x − y). After cancelling, we ob-

tain dHdy = −y ln y. Therefore H(y) = y2

4 −y2 ln y

2 , and the integral is

F(x, y) = e3xy(ln y− 1) + y2

4 −y2 ln y

2 .Return

Page 814: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 63

7. ∂P∂y (x, y) = 2xy and ∂Q

∂x (x, y) = 2xy so the equation is exact. Theintegral F(x, y) will have the form F(x, y) =

∫x(x2 + y2 − 1) dx +

H(y) = x4

4 + x2y2

2 −x2

2 + H(y). To determine H(y), differentiate thisexpression with respect to y: ∂F

∂y = yx2 + dHdy . Since ∂F

∂y = Q(x, y) =

y(x2 + y2 + 1), it follows that yx2 + dHdy = y(x2 + y2 + 1). After can-

celling, we obtain dHdy = y3 + y. Therefore H(y) = y4

4 + y2

2 , and the

integral is F(x, y) = x4

4 + x2y2

2 −x2

2 + y4

4 + y2

2 .Return

Page 815: Ordinary Differential Equations: A Systems Approach

64 ORDINARY DIFFERENTIAL EQUATIONS

9. ∂P∂y (x, y) = 2 y

x+y ×x+y−y(x+y)2 = 2xy

(x+y)3 and ∂Q∂x (x, y) = 2 x

x+y ×x+y−x(x+y)2 =

2xy(x+y)3 ; so the equation is exact. The integral F(x, y) will have the form

F(x, y) =∫ [

1 +(

yx+y

)2]

dx+ H(y) = x− y2

x+y + H(y). To determine

H(y), differentiate this expression with respect to y: ∂F∂y = − y2+2xy

(x+y)2 +

dHdy =

(x

x+y

)2− 1 + dH

dy . Since ∂F∂y = Q(x, y) =

(x

x+y

)2− 1, it follows

that(

xx+y

)2− 1+ dH

dy =(

xx+y

)2− 1. After cancelling, we obtain dH

dy =

0. Therefore H(y) = 0, and the integral is F(x, y) = x− y2

x+y .Return

Page 816: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 65

11.∂P∂y (x, y) = 6x + 18y and ∂Q

∂x (x, y) = 6x + 18y; so the equation is ex-act. The integral F(x, y) will have the form F(x, y) =

∫(3x2 + 6xy +

9y2) dx+ H(y) = x3 + 3yx2 + 9y2x+ H(y). To determine H(y), differ-entiate this expression with respect to y: ∂F

∂y = 3x2 + 18xy + dHdy . Since

∂F∂y = Q(x, y) = 3x2 + 18xy + 51y2, it follows that 3x2 + 18xy + dH

dy =

3x2 + 18xy + 51y2. After cancelling, we obtain dHdy = 51y2. Therefore

H(y) = 17y3, and the integral is F(x, y) = x3 + 3yx2 + 9y2x + 17y3.Return

Page 817: Ordinary Differential Equations: A Systems Approach

66 ORDINARY DIFFERENTIAL EQUATIONS

13. ∂P∂y (x, y) = 2x and ∂Q

∂x (x, y) = −2x; therefore the exactness condi-tion does not hold.Return

Page 818: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 67

15. Let’s seek an integrating factor m(x). This must satisfy

∂y[m(x)(x2 + xy2 + 1)] =

∂x[m(x)(2y)]

Do the differentiation to obtain

m(x)(2xy) = m′(x)(2y).

Simplify this equation to the form

m′(x)m(x)

= x.

Now we can integrate: ln(|m(x)|) = x2

2 + C. Therefore m = ex22 is an

integrating factor. After multiplying the equation by this integratingfactor, we have the exact equation

ex22 (x2 + xy2 + 1) dx + 2e

x22 y dy = 0.

Since it’s easier to integrate Q(x, y) = 2ex22 y with respect to y than it

is to integrate P(x, y) with respect to x, set

F(x, y) =∫

2ex22 y dy + K(x) = y2e

x22 + K(x).

Then ∂F∂y = 2e

x22 y. Since ∂F

∂x = xy2ex22 + K′(x),

xy2ex22 + K′(x) = e

x22 (x2 + xy2 + 1),

and K′(x) = ex22 (x2 + 1). A final integration shows that K(x) = xe

x22

and so F(x, y) = y2ex22 + xe

x22 .

Return

Page 819: Ordinary Differential Equations: A Systems Approach

68 ORDINARY DIFFERENTIAL EQUATIONS

17. Let’s seek an integrating factor m(x). This must satisfy

∂y[m(x)(x2 + 2x + 2xy + 2y + 3y2)] =

∂x[m(x)(2x + 6y)]

Do the differentiation to obtain

m(x)(2x + 2 + 6y) = m′(x)(2x + 6y) + 2m(x).

Simplify this equation to the form

m′(x)m(x)

=2x + 6y2x + 6y

= 1.

Now we can integrate: ln(|m(x)|) = x + C. Therefore m = ex is anintegrating factor. After multiplying the equation by this integratingfactor, we have the exact equation

(x2 + 2x + 2xy + 2y + 3y2)ex dx + (2x + 6y)ex dy = 0.

Since it’s easier to integrate Q(x, y) = (2x + 6y)ex with respect to ythan it is to integrate P(x, y) with respect to x, set

F(x, y) =∫(2x + 6y)ex dy + K(x) = 2xyex + 3y2ex + K(x).

Then ∂F∂y = (2x + 6y)ex. Since ∂F

∂x = 2xyex + 2yex + 3y2ex + K′(x),

2xyex + 2yex + 3y2ex + K′(x) = (x2 + 2x + 2xy + 2y + 3y2)ex,

and K′(x) = (x2 + 2x)ex. A final integration shows that K(x) = x2ex

and so F(x, y) = 2xyex + 3y2ex + x2ex.Return

Page 820: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 69

19. Let’s seek an integrating factor m(x). This must satisfy

∂y[2xym(x)] =

∂x[m(x)(3x2 + 2y)]

Do the differentiation to obtain

2xm(x) = m′(x)(3x2 + 2y) + 6xm(x).

Simplify this equation to the form

m′(x)m(x)

= − 4x3x2 + 2y

.

The right side of this equation is not a function of x, so there is no inte-grating factor m(x). It is still possible that there will be an integratingfactor m(y). This must satisfy

∂y[m(y)2xy] =

∂x[m(y)(3x2 + 2y)]

Do the differentiation to obtain

2xm(y) + 2xym′(y) = 6xm(y).

Simplify this equation to the form

m′(y)m(y)

=4x

2xy=

2y

.

Now we can integrate: ln |m(y)| = 2 ln |y| = ln(y2); thus we can mul-tiply by the integrating factor m(y) = y2 to obtain the exact equation

2xy3 dx + (3x2 + 2y)y2 dy = 0.

Since it’s easier to integrate P(x, y) = 2xy3 with respect to x than it isto integrate Q(x, y) with respect to y, set

F(x, y) =∫

2xy3 dx + K(y) = x2y3 + K(y).

Then ∂F∂x = 2xy3. Since ∂F

∂y = 3x2y2 + K′(y),

3x2y2 + K′(y) = (3x2 + 2y)y2,

and K′(y) = 2y3. A final integration shows that K(y) = y4

2 and so

F(x, y) = x2y3 + y4

2 .Return

Page 821: Ordinary Differential Equations: A Systems Approach

70 ORDINARY DIFFERENTIAL EQUATIONS

21. (i) An integrating factor m(y) must satisfy

∂y[m(y)P(x, y)] =

∂x[m(y)Q(x, y)]

Do the differentiation to obtain

m(y)∂P∂y

+ m′(y)P(x, y) = m(y)∂Q∂x

.

Simplify this equation to the form

m′(y)m(y)

=1

P(x, y)

(∂Q∂x− ∂P

∂y

)Therefore 1

P(x,y)

(∂Q∂x −

∂P∂y

)is independent of x.

(ii) If R(y) = 1P(x,y)

(∂Q∂x −

∂P∂y

)is independent of x, let

m(y) = e∫

R(y) dy = e∫ 1

P(x,y)

(∂Q∂x −

∂P∂y

)dy

Then m(y) is an integrating factor for the ODE.Return

Page 822: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 71

24. Let’s seek the integrating factor m(x). This must satisfy

∂y[m(x)[p(x)y− q(x)]] =

∂xm(x)

Do the differentiation: we will obtain

m(x)p(x) = m′(x).

Simplify this equation to the form

m′(x)m(x)

= p(x).

Now we can integrate: ln(|m(x)|) =∫

p(x) dx + C. Therefore m =

e∫

p(x) dx is an integrating factor.Return

Page 823: Ordinary Differential Equations: A Systems Approach

72 ORDINARY DIFFERENTIAL EQUATIONS

1.

Page 824: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 73

7. Let f (t) denote the solution of y′ = t − 3y; y(0) = 1, and letg(t) = t/3. Then f (0) = 1 > g(0) = 0. Above the graph of g(t),f ′(t) = t− 3y < 0, so f (t) is decreasing above the line t− 3y = 0. Butg(t) is increasing, therefore f (t) and g(t) should intersect at somepoint (t0, f (t0))(existence). If the graphs of f (t) and g(t) intersectat another point (t1, f (t1)),then f ′(t1) > g′(t1) > 0, On the otherhand (t1, f (t1)) is a point on the graph of g(t), so f ′(t1) = 0. This is acontradiction, showing that the point of intersection is unique. At thecrossing, f ′(t0) = 0. By differentiating both sides of the ODE, we findy′′ = 1− 3y′. Hence f ′′(t0) = 1− 3 f ′(t0) = 1 > 0, and by the secondderivative test (t0, f (t0)) is a relative minimum point for f (t).Return

Page 825: Ordinary Differential Equations: A Systems Approach

74 ORDINARY DIFFERENTIAL EQUATIONS

9.

(a) y2 = ∆y1 + y1 = 1, y3 = ∆y2 + y2 = 2, y4 = ∆y3 + y3 = 3, andym = m− 1.

(b) y2 = ∆y1 + y1 = 2, y3 = ∆y2 + y2 = 4, y4 = ∆y3 + y3 = 8, andym = 2m−1.

(c) y2 = ∆y1 + y1 = C + kC = (k + 1)C, y3 = ∆y2 + y2 = (k + 1)2C,y4 = ∆y3 + y3 = (k + 1)3C, and ym = (k + 1)m−1C.

(d) y2 = ∆y1 + y1 = 0 + 0 + 1 = 2 − 1, y3 = ∆y2 + y2 = 4 − 1,y4 = ∆y3 + y3 = 8− 1, and ym = 2ym−1 + 1 = 2m−1 − 1

Return

Page 826: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 75

11.

(a) ∆ym = amym can be written ym − ym−1 = am−1ym−1. There-fore ym = (1 + am−1)ym−1, and similarly we have ym−1 = (1 +am−2)ym−2, ym−2 = (1 + am−3)ym−3, . . . , y2 = (1 + a1)y1, soym = (1 + a1)(1 + a2) · · · (1 + am−1)y1 = CAm, where C = y1is a constant.

(b) Substitute ym = vm Am, where vm is a sequence to be deter-mined, to obtain vm+1Am+1 = (1+ am)vm Am + bm. Since Am+1 =(1 + am)Am, we get: vm+1 = vm + bm

Am+1. It follows that vm =

m−1

∑k=1

bk

Ak+1.

(c)

Am =

(1 +

11

)(1 +

12

)(1 +

13

)· · ·(

1 +1

m− 1

)=

1 + 11· 2 + 1

2· 3 + 1

3· · · (m− 1) + 1

m− 1= m

Therefore ym = Cm, where C is a constant.

(d) Put ym = mvm, where vm is a sequence to be determined. Thenym+1 = (m + 1)vm+1, Substitute these in ∆ym = 1

m ym + m + 1 toobtain (m+ 1)vm+1−mvm = vm +m+ 1, which we can simplifyas vm+1 − vm = 1. Because v1 = y1 = 0, vm = m− 1, and ym =m(m− 1).

Return

Page 827: Ordinary Differential Equations: A Systems Approach

76 ORDINARY DIFFERENTIAL EQUATIONS

13. Write the difference equation in recursive form as ym+1 = ym +0.05y2

m. Starting with y0 = 1, we havey1 = 1 + 0.05(1)2 = 1.05y2 = 1.05 + 0.05(1.05)2 = 1.10513y3 = 1.10513 + 0.05(1.10513)2 = 1.16619y4 = 1.16619 + 0.05(1.16619)2 = 1.23419y5 = 1.23419 + 0.05(1.23419)2 = 1.31035y6 = 1.31035 + 0.05(1.31035)2 = 1.3962y7 = 1.3962 + 0.05(1.3962)2 = 1.49367y8 = 1.49367 + 0.05(1.49367)2 = 1.60522y9 = 1.60522 + 0.05(1.60522)2 = 1.73406y10 = 1.73406 + 0.05(1.73406)2 = 1.88441y11 = 1.88441 + 0.05(1.88441)2 = 2.06196y12 = 2.06196 + 0.05(2.06196)2 = 2.27454y13 = 2.27454 + 0.05(2.27454)2 = 2.53322y14 = 2.53322 + 0.05(2.53322)2 = 2.85408y15 = 2.85408 + 0.05(2.85408)2 = 3.26137y16 = 3.26137 + 0.05(3.26137)2 = 3.7932y17 = 3.7932 + 0.05(3.7932)2 = 4.51261y18 = 4.51261 + 0.05(4.51261)2 = 5.5308y19 = 5.5308 + 0.05(5.5308)2 = 7.06029y20 = 7.06029 + 0.05(7.06029)2 = 9.55267

0.2 0.4 0.6 0.8 1

2

4

6

8

10

Return

Page 828: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 77

15. The associated homogeneous equation is y′ − y = 0, and we cantake yh = et. Substitute y = vet to get v′ = te−t. Thus v = −(1 +t)e−t + C, the general solution is the family y = −(1 + t) + Cet. Tosatisfy the initial condition, set y = 0 and t = 0. This yields 1 = C soφ(t) = et − t− 1. and φ(1) = e− 2.

Write the difference equation in recursive form as ym+1 = ym + h(tm +ym) with initial condition y0 = 0.

For h = 1,y1 = 0 + 1× (0 + 0) = 0E(1) = |φ(1)− y1| = 0.71828

For h = 0.5,y1 = 0 + 0.5× (0 + 0) = 0y2 = 0 + 0.5× (0.5 + 0) = 0.25E(0.5) = |φ(1)− y2| = 0.46828

For h = 0.25,y1 = 0 + 0.25× (0 + 0) = 0y2 = 0 + 0.25× (0.25 + 0) = 0.0625y3 = 0.0625 + 0.25× (0.5 + 0.0625) = 0.203125y4 = 0.203125 + 0.25× (0.75 + 0.203125) = 0.441406E(0.25) = |φ(1)− y4| = 0.276874.

Similarly, if h = 0.1, then the approximation of φ(1) is y10 = 0.457948+0.1× (0.9 + 0.457948) = 0.593742, and hence E(0.1) = |φ(1)− y10| =0.124538.

If h = 0.05 then the approximation of φ(1) is y20 = 0.57695 + 0.05×(0.95 + 0.57695) = 0.653298, and E(0.05) = |φ(1)− y20| = 0.0649823.

if h = 0.02 then the approximation of φ(1) is y50 = 0.691588, andE(0.02) = |φ(1)− y50| = 0.026692.

Finally, if h = 0.01 then the approximation of φ(1) is y100 = 0.704814,and E(0.01) = |φ(1)− y100| = 0.0134662.

In the following graph, the horizontal axis represents the time step hand the vertical axis gives the approximation error. It shows that theerror decreases as h decreases; in fact the error seems to be decreasing

Page 829: Ordinary Differential Equations: A Systems Approach

78 ORDINARY DIFFERENTIAL EQUATIONS

almost as a linear function of h (follow the graph downward and tothe left as h decreases).

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Return

Page 830: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 79

17. The associated homogeneous equation is y′ − y = 0, and wecan take yh = et. Substitute y = vet to get v′ = e−t sin(πt). Thusv = − e−t(π cos(πt)+sin(πt))

π2+1 + C , the general solution is the family y =

−π cos(πt)+sin(πt)π2+1 + Cet. To satisfy the initial condition, set y = 0 and

t = 0. This yields C = ππ2+1 and y = −π cos(πt)−sin(πt)+etπ

π2+1 . Sub-stitute 0.25, 0.5, 0.75, 1 in we obtain:y(0) = 0, y(0.25) = 0.101691,y(0.5) = 0.384523, y(0.75) = 0.751185, y(1) = 1.07468.

Write the difference equation in recursive form as ym+1 = ym + 0.25(sin(π×tm) + ym). Starting with y0 = 0 ,we havey1 = 0 + 0.25× (sin(π × 0) + 0) = 0y2 = 0 + 0.25× (sin(π × 0.25) + 0) = 0.176777y3 = 0.176777 + 0.25× (sin(π × 0.5) + 0.176777) = 0.470971y4 = 0.470971 + 0.25× (sin(π × 0.75) + 0.470971) = 0.76549

Each solution φm(t), which we will use to quantify the errors, satisfiesinitial conditions φ(tm) = ym, so φm(t) = −π cos(πt)+sin(πt)

π2+1 + Cmet

where the constant Cm is determined by

ym = −π cos(πtm) + sin(πtm)

π2 + 1+ Cmetm

Thus Cm = (ym + π cos(πtm)+sin(πtm)π2+1 )e−tm and

φm(t) = −π cos(πt) + sin(πt)

π2 + 1+ (ym +

π cos(πtm) + sin(πtm)

π2 + 1)et−tm

In particularφ1(t) = −π cos(πt)+sin(πt)

π2+1 + 0.269425et−0.25,

φ2(t) = −π cos(πt)+sin(πt)π2+1 + 0.268776et−0.5,

φ3(t) = −π cos(πt)+sin(πt)π2+1 + 0.331653et−0.75,

Therefore we obtain:LE1 = φ(0.25)− y1 = 0.101691LE2 = φ1(0.5)− y2 = 0.0771727LE3 = φ2(0.75)− y3 = 0.013463LE4 = φ3(1)− y4 = −0.0506145and:AE1 = φ(0.5)− φ1(1) = 0.130573AE2 = φ(0.75)− φ2(1.5) = 0.266751AE3 = φ(1)− φ3(2) = 0.359802Return

Page 831: Ordinary Differential Equations: A Systems Approach

80 ORDINARY DIFFERENTIAL EQUATIONS

19.

(a) Write the difference equation in recursive form as ym+1 = ym +0.1(−100ym) = −9ym. Starting with y0 = 1 ,we have

y1 = −9× 1 = −9y2 = −9× (−9) = 81y3 = −9× (81) = −729y4 = −9× (−729) = 6561y5 = −9× 6561 = −59049y6 = −9× (−59049) = 531441y7 = −9× 531441 = −4.78297e× 106

y8 = −9× (−4.78297× 106) = 4.30467× 107

y9 = −9× 4.30467× 107 = −3.8742× 108

y10 = −9× (−3.8742× 108) = 3.48678× 109

Obviously it diverges.

(b) Write the difference equation in recursive form as ym+1 = ym +h(−100ym) = (1− 100h)ym. When |1− 100h| < 1, ym → 0. Thus,the solution of the difference equation converges for 0 < h <0.02.

Return

Page 832: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 81

21.

(a) In the forward version, using ym+1 = ym +√

1− y2m, ym even-

tually increases to a value bigger than 1, and then√

1− y2m is

undefined which causes the calculation to terminate.

(b) y′ =√

1− y2 > 0 therefore y is increasing on (0,2),and ym+1 >ym. therefore the ”+” sign is correct,and the ”-” sign is extrane-ous.

(c) Forward version: Write the difference equation in recursive formas ym+1 = ym + 0.1

√1− y2

m with initial condition y0 = 0.n yn Error: sin(tn)− yn

1 0 + 0.1×√

1− 02 = 0.1 −0.0001665832 0.1 + 0.1×

√1− 0.12 = 0.199499 −0.000829413

3 0.199499 + 0.1×√

1− 0.1994992 = 0.297489 −0.001968354 0.297489 + 0.1×

√1− 0.2974892 = 0.392961 −0.00354275

5 0.392961 + 0.1×√

1− 0.3929612 = 0.484917 −0.005491066 0.484917 + 0.1×

√1− 0.4849172 = 0.572373 −0.00773017

7 0.572373 + 0.1×√

1− 0.5723732 = 0.654372 −0.01015438 0.654372 + 0.1×

√1− 0.6543722 = 0.729989 −0.0126332

9 0.729989 + 0.1×√

1− 0.7299892 = 0.798335 −0.015008210 0.798335 + 0.1×

√1− 0.7983352 = 0.858556 −0.0170855

11 0.858556 + 0.1×√

1− 0.8585562 = 0.909828 −0.01862112 0.909828 + 0.1×

√1− 0.9098282 = 0.951327 −0.0192878

13 0.951327 + 0.1×√

1− 0.9513272 = 0.982145 −0.01858714 0.982145 + 0.1×

√1− 0.9821452 = 1.00096 −0.0155079

The calculation ends here: y15 is undefined because y14 > 1.

Backward version: Write the difference equation in recursive

form as ym+1 =ym±0.1

√1.01−y2

m1.01 with initial condition y0 = 0.

Page 833: Ordinary Differential Equations: A Systems Approach

82 ORDINARY DIFFERENTIAL EQUATIONS

n yn Error: sin(tn)− yn

1 0+0.1√

1.01−02

1.01 = 0.0995037 0.000329698

2 0.0995037+0.1√

1.01−0.09950372

1.01 = 0.197533 0.00113599

3 0.197533+0.1√

1.01−0.1975332

1.01 = 0.29314 0.00237993

4 0.29314+0.1√

1.01−0.293142

1.01 = 0.385415 0.00400371

5 0.385415+0.1√

1.01−0.3854152

1.01 = 0.473494 0.00593122

6 0.473494+0.1√

1.01−0.4734942

1.01 = 0.556574 0.00806836

7 0.556574+0.1√

1.01−0.5565742

1.01 = 0.633914 0.0103033

8 0.633914+0.1√

1.01−0.6339142

1.01 = 0.70485 0.012506

9 0.70485+0.1√

1.01−0.704852

1.01 = 0.768799 0.0145278

10 0.768799+0.1√

1.01−0.7687992

1.01 = 0.825273 0.0161984

11 0.825273+0.1√

1.01−0.8252732

1.01 = 0.873886 0.0173217

12 0.873886+0.1√

1.01−0.8738862

1.01 = 0.914373 0.0176661

13 0.914373+0.1√

1.01−0.9143732

1.01 = 0.946611 0.0169473

14 0.946611+0.1√

1.01−0.9466112

1.01 = 0.970657 0.0147923

15 0.970657+0.1√

1.01−0.9706572

1.01 = 0.986832 0.0106628

16 0.986832+0.1√

1.01−0.9868322

1.01 = 0.99589 0.00411033

17 0.99589+0.1√

1.01−0.995892

1.01 = 0.999388 0.000612076

18 0.999388+0.1√

1.01−0.9993882

1.01 = 0.999982 1.76663× 10−5

19 0.999982+0.1√

1.01−0.9999822

1.01 = 1 0

20 1+0.1√

1.01−12

1.01 = 1 0

Return

Page 834: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 83

23. The constant solutions are y = ±3. Notice that for −3 < y <

3, the sign of y′ is positive. The separated equation is dy√9−y2

= dt.

Integrating both sides we obtain arcsin( y3 ) = t + C. Therefore y =

3 sin(t + C), where C is a constant is a solution for all values of t suchthat y′ = 3 cos(t+C) is positive. The solutions y = ±3, not belongingto this family, are singular solutions.

-4 -2 0 2 4-3

-2

-1

0

1

2

3

Return

Page 835: Ordinary Differential Equations: A Systems Approach

84 ORDINARY DIFFERENTIAL EQUATIONS

25. The constant solution is y = 1. The separated equation is dy1−y =

dt2−t . Integrating both sides we obtain: − ln |1 − y| = − ln |2 − t| +C1, Therefore |1− y| = C|2− t|, where C = e−C1 is a constant. Thesolution y = 1, belongs to this family (when C = 0). There are nosingular solutions.

1 2 3 4 5 6

0.5

1

1.5

2

2.5

3

3.5

Return

Page 836: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 85

1. The general solution of y′ = 0 is y = C, where C is a constant.Substitute y(1) = −2 to obtain C = −2. Therefore the solution of theIVP is y = −2, defined on (−∞,+∞).Return

Page 837: Ordinary Differential Equations: A Systems Approach

86 ORDINARY DIFFERENTIAL EQUATIONS

3. The solution of associated homogeneous equation is yh = e−2t.Substitute y = ve−2t to get v′ = sin(5t)e2t. Thus v = − 1

29 (5 cos(5t)−2 sin(5t))e2t +C, and the general solution is the family y = − 1

29 (5 cos(5t)−2 sin(5t)) + Ce−2t. Substituting y(0) = 0 yields C = 5

29 . Therefore thesolution of the IVP is y = − 1

29 (5 cos(5t)− 2 sin(5t)− 5e−2t) definedon (−∞,+∞).Return

Page 838: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 87

5. The solution of associated homogeneous equation is yh = t2.Substitute y = vt2 to get v′ = 1. Thus v = t + C, and the generalsolution is the family y = t3 + Ct2. Substituting y(1) = 0 yieldsC = −1. Therefore the solution of the IVP is y = t3 − t2 defined onthe largest interval containing t = 1 : (−∞,+∞).Return

Page 839: Ordinary Differential Equations: A Systems Approach

88 ORDINARY DIFFERENTIAL EQUATIONS

7. The solution of associated homogeneous equation is yh = et2.

Substitute y = vet2to get v′ = te−t2

. Thus v = − 12 e−t2

+ C, and thegeneral solution is the family y = − 1

2 + Cet2. Substituting y(0) = 1

2yields C = 1. Therefore the solution of the IVP is y = − 1

2 + et2defined

on the largest interval containing t = 0 : (−∞,+∞).Return

Page 840: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 89

9. If y = φ1(t) and y = φ2(t) intersect at (t0, y0), then y = φ1(t)and y = φ2(t) are two solutions of the IVP y′ = f (t, y); y(t0) = y0.Because f (t, y) is continuous and satisfies a Lipschitz condition withrespect to y on a rectangular domain D, they are identical in D by theuniqueness theorem.Return

Page 841: Ordinary Differential Equations: A Systems Approach

90 ORDINARY DIFFERENTIAL EQUATIONS

11. The given IVP has a constant solution y = 1, just as in the previ-ous Exercise. We have to prove that it is unique.

Write the ODE in the form y′ = 1−yf (t) . Let ε > 0. On the interval

[−ε,+ε], f (t) 6= 0. Let g(t, y) = 1−yf (t) , then |g(t, y1)− g(t, y2)| = |y1−y2|

| f (t)| ≤K|y1 − y2|, K = max{| 1

f (t) |, t ∈ [−ε,+ε]} Therefore g(t, y) is contin-uous and satisfies Lipschitz condition with respect to y on {(t, y) :−ε < t < +ε,−∞ < y < +∞}. By the uniqueness theorem the IVPhas a unique solution.Return

Page 842: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 91

13.

(i) f (t, y) = ty2+1 is continuous in the plane. Thus we obtain t0, y0

could be any pair of real numbers.

(ii) ∂ f∂y (t, y) = − 2ty

(1+y2)2 is continuous in the plane. Thus we obtaint0, y0 could be any pair of real numbers.Return

Page 843: Ordinary Differential Equations: A Systems Approach

92 ORDINARY DIFFERENTIAL EQUATIONS

15.

(i) f (t, y) = − ysin t . is continuous if t 6= kπ. Thus we obtain t0 6= kπ,

where k is an integer.

(ii) ∂ f∂y (t, y) = − 1

sin t is continuous if t 6= kπ. Thus we obtain t0 6= kπ.k is an integer.Return

Page 844: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 93

17.

(i) Write the equation in the form: ty′2 − yy′ + 1 = 0, We obtain

y′ = y±√

y2−4t2t Let f (t, y) =

y±√

y2−4t2t , f (t, y) is continuous if

y2 > 4t, and t 6= 0 Thus we obtain y20 > 4t0, and t0 6= 0

(ii) There are two functions for y′, therefore the set of initial points(t0, y0) for the uniqueness theorem is ∅.Return

Page 845: Ordinary Differential Equations: A Systems Approach

94 ORDINARY DIFFERENTIAL EQUATIONS

19.

(i) f (t, y) =3√

y−1√ty . is continuous if ty > 0. Thus we obtain t0y0 > 0.

(ii) ∂ f∂y (t, y) = − t(y−1)1/3

2(ty)3/2 + 13(y−1)2/3√ty is continuous if ty > 0 and

y 6= 1 Thus we obtain t0y0 > 0 and y0 6= 1.Return

Page 846: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 95

21.

(i) f (t, y) = (t− y)1/3. is continuous in the plane. Thus we obtaint0, y0 could be any real numbers.

(ii) ∂ f∂y (t, y) = − 1

3(t−y)2/3 is continuous if t − y 6= 0 Thus we obtaint0 − y0 6= 0.Return

Page 847: Ordinary Differential Equations: A Systems Approach

96 ORDINARY DIFFERENTIAL EQUATIONS

23. ∂ f∂y (t, y) = − y√

1−y2continuous if |y| < 1, but not continuous

on |y| = 1 This implies that f (t, y) =√

1− y2 satisfies a Lipschitzcondition on the domain D if −1 < c and d < 1, but there is noimplication if either c = −1 or d = 1.

In the case d = 1 let y2 = 1 and y1 < 1. Then

| f (t, y2)− f (t, y1)|y2 − y1

=

√1− y2

1

1− y1=

√1 + y1

1− y1.

It follows from this calculation that if there is a Lipschitz constant

K, then

√1 + y1

1− y1≤ K for all y1 ∈ (−1, 1), which is not possible:

limy1→1−

√1 + y1

1− y1= ∞. Therefore f does not satisfy a Lipschitz condi-

tion when d = 1. A similar argument works when c = −1.Return

Page 848: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 97

25. Our CAS found the following expression for the general solu-tion, where a denotes the initial value y(0), and w(x) = sin(x/2)−cos(x/2):

y =w(x)[8 ln |w(x)|+ 4 sin(x) + x− a] + 2 sin(x/2)

sin(x/2) + cos(x/2)

Solving y(1) = 0 for a yields a = −2.001386626.Return

Page 849: Ordinary Differential Equations: A Systems Approach

98 ORDINARY DIFFERENTIAL EQUATIONS

27.

(a) Suppose φ and ψ are fixed by T . Then

‖φ− ψ‖ = ‖T (φ)− T (ψ)‖ ≤ C‖φ− ψ‖

and because 0 < C < 1, this inequality is impossible unless‖φ− ψ‖ = 0.

(b) By the triangle inequality,

‖ψn+k − ψn‖ ≤ ‖ψn+k − ψn+k−1‖+ · · ·+ ‖ψn+1 − ψn‖.

Therefore, by the contractive property of T ,

‖ψn+k − ψn‖ ≤ (Cn+k + · · ·+ Cn){ψ1 − ψ0‖.

Using the formula for the sum of a geometric series,

Cn+k + · · ·+ Cn <∞

∑m=n

Cm =Cn

1− C

and the formula follows.

(c) For n > 0, note that ‖T (ψ∞)−T (ψn−1)‖ ≤ C‖ψ∞−ψn−1‖. Alsonote that T (ψn−1) = ψn. Since |ψ∞ − ψn−1‖ → 0 as n → ∞, itfollows that

limn→∞‖T (ψ∞)− ψn‖ = 0

It follows that the sequence {ψn} converges to both ψ∞ and toT (ψ∞) and hence the two are equal: ψ∞ is fixed by T .

i. The initial condition tells us φ0 = 1. Then

φ1(t) = 1 +∫ t

01 ds = 1 + t

φ2(t) = 1 +∫ t

0(1 + s) ds = 1 + t +

12

t2

φ3(t) = 1 +∫ t

0(1 + s +

12

s2) ds = 1 + t +12

t2 +16

t3

One can deduce that

φn(t) =n

∑m=0

1m!

tm

and that φ∞(t) = et.

Page 850: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 99

ii. The initial condition yields φ0 = 1. Then

φ1(t) = 1 +∫ t

0−s ds = 1− 1

2t2

φ2(t) = 1 +∫ t

0−s(

1− 12

s2)

ds = 1− 12

t2 +18

t4

φ3(t) = 1+∫ t

0−s(

1− 12

s2 +18

s4)

ds = 1− 12

t2 +18

t4− 148

t6

One can deduce that

φn(t) =n

∑m=0

12mm!

t2m

and that φ∞(t) = et2/2.iii. We have φ0 = 0. Note that the sequence is thus given re-

cursively by φn+1 =∫ t

03√

φn(s) ds. Thus, if φn = 0, thenφn+1 = 0, too. It follows that φ∞ = 0.

iv. The initial condition yields φ0 = 1. Then

φ1(t) = 1 +∫ t

1

2s

ds = 1 + 2 ln t

φ2(t) = 1 +∫ t

1

2 + 4 ln ss

ds = 1 + 2 ln t + 2(ln t)2

φ3(t) = 1+∫ t

1

2 + 4 ln s + 2(ln s)2

sds = 1+ 2 ln t+ 2(ln t)2 +

43(ln t)3.

One can deduce that

φn(t) =n

∑m=0

1m!

(2 ln t)m

and that φ∞(t) = e2 ln t = t2.

Return

Page 851: Ordinary Differential Equations: A Systems Approach

100 ORDINARY DIFFERENTIAL EQUATIONS

1. We use the equation: y′ = k y(1− y

M

), where M = 1000. Write

it in separated form:1

y(1000− y)dy =

k1000

dt. Integration by partial

fractions yields1

1000ln(

y1000− y

)=

k1000

t + B, where B denotes

the integration constant. This equation, solved for y, is y = 10001+Ae−kt ,

where A = e−1000B is a constant. Substituting y(0) = 300 for last yearand y(1) = 600 this year, we get 1 + A = 1000/300 and 1 + Ae−k =1000/600. Solving these equations, we obtain A = 7/3, k = ln(3.5).Thus y(t) = 1000

1+ 73 (3.5)−t , and y(2) = 1000

1+ 73 (3.5)−2 = 840

Return

Page 852: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 101

3.

Write the ODE in separated form: 1y(200−y) dy = 0.02 dt. Integration by

partial fractions yields 1200 ln

(y

200−y

)= 0.02t + B, where B denotes

the integration constant. This equation, solved for y, is y = 2001+Ae−4t ,

where A = e−200B is a constant. Substituting y(0) = 10, we get: 1 +A = 200/10. Thus A = 19 and y(t) = 200

1+19e−4t .Return

Page 853: Ordinary Differential Equations: A Systems Approach

102 ORDINARY DIFFERENTIAL EQUATIONS

5. Let p0 = 63.0, p1 = 151.3, and p2 = 309.4 be the three populationdata. As these measurements were taken at 60-year intervals, we willuse 60 years as our unit of time.

Following the calculations in Example 1.10.1, the population can beexpressed as

p(t) =M

1 + Avt

where M is the carrying capacity, A is a constant we’ll need to deter-mine, and v = e−k, (where k is the ratio of p′(t) to p(t)(1− p(t)/M)and t is measured in 60-year increments after 1890, so when t = 1 theyear is 1850, etc. Thus

p0 =M

1 + A,

so M = p0(1 + A). Substitute for M in

p1 =M

1 + Av

to get

A =p1 − p0

p0 − p1vand M = p0 p1

(1− v

p0 − p1v

).

Substitute these inp2 =

M1 + Av2

and simplify to get

p2 = p0 p1

(1− v

(p1 − p0)v2 − p1v + p0

)=

p0 p1

p0 − (p1 − p0)v.

(The second equality is due to the factorization of the denominator as(1− v)(p0 − (p1 − p0)v).) Now solve for v:

v =p0

p2

(p2 − p1

p1 − p0

)Substituting our data, we get v = 0.3246, A = 11.264, and M = 772.6million. Thus, 60t years after 1890, the population is

p(t) =772.6

1 + (11.264) (0.3246t)million.

Return

Page 854: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 103

7. Let P = 500 million. Solve M1+Avt = P to obtain

t =ln(M− P)− ln(PA)

ln v,

and plug in the numerical values to get t = 3.00. Remembering tomultiply by 60, we see that the logistic model predicts that the popu-lation will reach 500 million in the year 1890 + 180 = 2070.Return

Page 855: Ordinary Differential Equations: A Systems Approach

104 ORDINARY DIFFERENTIAL EQUATIONS

9. Let N(t) be the number of people who have been warned aboutthe sky. Then the number of people who have not heard the rumoris 10000− N(t). The rumor will spread to another person in a givencall if one of the parties calling has not heard the rumor, and the otherhas. The probability that caller A will spread the rumor to caller Bis equal to the product of the probability that caller A has heard therumor, N(t)

10000 , and that caller B hasn’t, 1− N(t)10000 . There are 100000 calls

per day; hence

N′ = 100000N

10000(10000− N)

10000= 10N(1− N

10000).

Thus, the model that we will use is the logistic equation.

Solving we get N = 100001+Ae−10t , where A is a constant. Substituting

N(0) = 1000, we obtain A = 9. Let N(t) = 9000, we have 1+ 9e−10t =109 , 10t = − ln(1/81) = 4.39. Thus it takes 0.439 day or = 10.536

hours to warn 90% of the population.Return

Page 856: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 105

11. The function f (y) = ky ln(

My

)is defined for y > 0 only, but

limy→0 f (y) = 0. Thus we can say 0 is stationary. The other station-ary point at M is verified because ln(1) = 0. If 0 < y < M thenln(M/y) > 0 indicating the population is increasing; for y > M wehave ln(M/y) < 0 and the population is decreasing. Thus M is a sta-ble stationary point and 0 is unstable.Return

Page 857: Ordinary Differential Equations: A Systems Approach

106 ORDINARY DIFFERENTIAL EQUATIONS

13. Let v = ln(y). In terms of v the data are v(0) = 0, v(1) = ln 2, andv(2) = ln 3. The general solution the ODE v′ = k(ln(M)− v) is v =ln M +Ce−kt. From the data we get 0 = ln M +C, ln 2 = ln M +Ce−k,and ln 3 = ln M + Ce−2k. By the first equation, C = − ln(M); hencethe second and third equations are

ln(M)(1− e−k) = ln(2)ln(M)(1− e−2k) = ln(3).

Divide the second of these equations by the first (cancelling ln(M)) toobtain

1− p2

1− p= α

where p = e−k and α = ln(3)ln(2) . Since 1− p2 = (1− p)(1 + p) we have

1 + p = α,

or, p = α − 1. Now we can find M: Since ln(2) = ln(M)(1− ek) =ln(M)(1− p),

ln(M) =ln(2)2− α

and

M = exp(

ln(2)2

2 ln(2)− ln(3)

)≈ 5.31261.

Return

Page 858: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 107

15. dydt = ku(t)y(t). In a closed system, the number of total molecules

is a constant C. Therefore u(t) + y(t) = C, and dydt = ky(t)(C− y(t)).

It’s a logistic equation.Return

Page 859: Ordinary Differential Equations: A Systems Approach

108 ORDINARY DIFFERENTIAL EQUATIONS

17.

(a) This ODE is autonomous because the right side is independentof t.

(b) This ODE is not autonomous because the right side is dependson t.

(c) Again, the right side depends on t, so the ODE isn’t autonomous.

(d) The right side of this ODE is independent of t, so it is autonomous.

Return

Page 860: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 109

19. Solving f (y) = y2 = 0 we get one stationary point y1 = 0,f ′(y) = 2y. Thus f ′(0) = 0 if y > 0 or y < 0, we all get f (y) > 0.Therefore (−∞, 0) and (0,+∞) are up intervals, and the stationarypoint is unstable.

- s -

0

Return

Page 861: Ordinary Differential Equations: A Systems Approach

110 ORDINARY DIFFERENTIAL EQUATIONS

21. There is no solution for f (y) = ey = 0. Therefore there are nostationary points. Because f (y) = ey > 0, (−∞,+∞) is up interval.

-

Return

Page 862: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 111

23.

(a) This is the solution of first order autonomous ODE: y′ = 1. Thereare no stationary points.

(b) y′ = 2t, y′ > 0 if t > 0; y′ < 0 if t < 0. Therefore y is not strictlyincreasing, strictly decreasing or a constant. It is therefore not asolution of a first order autonomous ODE.

(c) This is the solution of first order autonomous ODEs: y′ = −2e−2t =−2y. There is a stationary point y = 0.

(d) This is the solution of first order autonomous ODEs: y′ = sec2(t) =1 + y2 There are no stationary points.

(e) y′ = cos(t), Thus y is not strictly increasing, strictly decreas-ing or constant. It is therefore not a solution of a first order au-tonomous ODE.

Return

Page 863: Ordinary Differential Equations: A Systems Approach

112 ORDINARY DIFFERENTIAL EQUATIONS

25. If g′(y1) < 0, there exists ε > 0 such that g′(y) < 0 in (y1− ε, y1 +ε), because g′(y) is continuous. That means g(y) is strictly decreasingin (y1− ε, y1 + ε). Because g(y1) = 0, it follows that g(y) changes signfrom positive to negative at y = y1 Therefore (y1 − ε, y1) is an upinterval, and (y1, y + ε) is a down interval; the phase diagram for theODE looks like this near y1:s- � .Thus y1 is stable.

If g′(y1) > 0, the same reasoning shows that near y1, the phase dia-gram looks like this:s� - . Thus y1 is an unstable stationary point.Return

Page 864: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 113

27. Suppose the ODE is y′ = f (y). Then f (y) has a same station-ary point as y′ = y2 at y ≡ 0, thus f (0) = 0. The two up inter-vals (−∞, 0), (0,+∞) for y′ = y2 is also the up intervals for f (y).therefore f ′(0) = limh→0

f (0+h)− f (0)h = limh→0

f (h)h ≥ 0, and f ′(0) =

limh→0f (0−h)− f (0)

−h = limh→0f (−h)−h ≤ 0, We obtain f ′(0) = 0 and y ≡ 0

is a degenerate stationary point.Return

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114 ORDINARY DIFFERENTIAL EQUATIONS

29.

(a) The critical harvest rate is the maximum harvest rate that thepopulation can sustain without becoming extinct. Let f (P) =P(1− P)−H = −P2 + P−H be the growth rate. The stationarypoints are the roots of f (P) = 0. By the quadratic formula, forH < 0.25 there are two stationary points, P1, P2 = 1±

√1−4H2 . The

phase diagram for this case is

s sP1 P2

� - �

As the harvest rate increases, the stationary point P1 moves tothe right, and P2 moves left, until H = 0.25, when there is onlyone stationary point, P = 0.5, and the phase diagram looks likethis:

s0.5

� �

When H > 0.25 there is no stationary point, and the populationbecomes extinct. Thus the critical harvest rate is 0.25.

Notice that 0 is not a stationary point. The model predicts thatpopulations will become negative if P(0) < P1 (in the case H <0.25), and for P(0) < 0.5 in the critical case. If the harvest rateexceeds this critical value, negative populations are predictedfor any starting population. Of course the model will cease toapply when the resource is exhausted; the model actually tell usthat that the state of extinction will not be approached asymp-totically, it will occur in a finite span of time.

(b) For H = 0.1 we have P1, P2 = 1±√

1−0.62 = 0.5 ± 0.316. P1 =

0.184, P2 = 0.816. Referring to the first phase diagram in part (a),(P1, P2) is an up interval, and (−∞, P1) and (P2,+∞) are downintervals. Therefore y = P2 is stable. Because P(0) = 1 > P2, thelimiting population will be P2 = 0.816

(c) The bifurcation diagram merges the phase diagrams (presentedvertically) in a graph, where the horizontal axis gives the harvestrate.

Page 866: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 115

0.25H

1

P

Return

Page 867: Ordinary Differential Equations: A Systems Approach

116 ORDINARY DIFFERENTIAL EQUATIONS

31.

(a) P(1800) = 5.3, (P(1810)− P(1790))/(20P(1800)) = 0.0311321P(1810) = 7.2, (P(1820)− P(1800))/(20P(1810)) = 0.0298611P(1820) = 9.6, (P(1830)− P(1810))/(20P(1820)) = 0.0296875P(1830) = 12.9, (P(1840)− P(1820))/(20P(1830)) = 0.0290698P(1840) = 17.1, (P(1850)− P(1830))/(20P(1840)) = 0.030117P(1850) = 23.2, (P(1860)− P(1840))/(20P(1850)) = 0.030819P(1860) = 31.4, (P(1870)− P(1850))/(20P(1860)) = 0.0245223P(1870) = 38.6, (P(1880)− P(1860))/(20P(1870)) = 0.0243523P(1880) = 50.2, (P(1890)− P(1870))/(20P(1880)) = 0.0243028P(1890) = 63, (P(1900)− P(1880))/(20P(1890)) = 0.0206349P(1900) = 76.2, (P(1910)− P(1890))/(20P(1900)) = 0.0191601P(1910) = 92.2, (P(1920)− P(1900))/(20P(1910)) = 0.0161605P(1920) = 106, (P(1930)− P(1910))/(20P(1920)) = 0.0146226P(1930) = 123.2, (P(1940)− P(1920))/(20P(1930)) = 0.0106331P(1940) = 132.2, (P(1950)− P(1930))/(20P(1940)) = 0.0106278

(b) m = −0.000165504 b = 0.0317447

(c) M = −b/m = − 0.0317447−0.000165504 = 191.806 million.

(d) P′ = P(mP + b) The separated form is dPP(mP+b) = dt, Integrating

both sides, we have 1b ln( P

mP+b ) = t + C1 Simplifying we obtainP = b

Ce−bt−m , where C is a constant. Substitute the initial condi-tion, P(0) = 76.21 to get C = 0.000251038. Thus

p(t) =0.0317447

0.000251038e−0.0317447t + 0.000165504.

Page 868: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 117

1850 1900 1950 2000

50

100

150

200

250

Return

Page 869: Ordinary Differential Equations: A Systems Approach

118 ORDINARY DIFFERENTIAL EQUATIONS

Find the general solution of each of the ODEs in problems 1 – 12.

1. The associated homogeneous equation is y′ − (t + 1)y = 0, andwe can take yh = e

∫(t+1) dt = et2/2+t. Substitute y = vet2/2+t yields

v′ = (t + 1)e−t2/2−t. Thus v = −e−t2/2−t + C. Since y = vet2/2+t, thegeneral solution is the family y = −1 + Cet2/2+t.Return

Page 870: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 119

2. The associated homogeneous equation is y′ − y = 0, and we cantake yh = e

∫dt = et. Substitute y = vet yields v′ = −t2e−t. Thus

v = e−t(2 + 2t + t2) + C. Since y = vet, the general solution is thefamily y = (2 + 2t + t2) + Cet.Return

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120 ORDINARY DIFFERENTIAL EQUATIONS

3. The associated homogeneous equation is y′ − y = 0, and we cantake yh = e

∫dt = et. Substitute y = vet yields v′ = 1 .general solution

Return

Page 872: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 121

4. The associated homogeneous equation is y′ + tan(t)y = 0, andwe can take yh = e−

∫tan(t) dt = eln cos t = cos t. Substitute y = v cos t

yields v′ = cos t. Thus v = sin t + C. Since y = v cos t, the generalsolution is the family y = (sin t + C) cos t.Return

Page 873: Ordinary Differential Equations: A Systems Approach

122 ORDINARY DIFFERENTIAL EQUATIONS

5. The associated homogeneous equation is y′ + y = 0, and we cantake yh = e−

∫dt = e−t. Substitute y = ve−t yields v′ = e2t. Thus

v = 12 e2t + C. Since y = ve−t, the general solution is the family y =

12 et + Ce−t.Return

Page 874: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 123

6. The associated homogeneous equation is ty′ + y = 0, and we cantake yh = e− ln t = 1/t. Substitute y = v/t and y′ = v′/t− v/t2 in theinhomogeneous ODE to get v′ = et sin 2t. Thus v =

∫(et sin 2t) dt =

− 15 et(2 cos(2t)− sin(2t)) + C. Since y = v/t, the general solution is

the family y = − 15t et(2 cos(2t)− sin(2t)) + C/t.

Return

Page 875: Ordinary Differential Equations: A Systems Approach

124 ORDINARY DIFFERENTIAL EQUATIONS

7. The associated homogeneous equation is ty′ + 12y = 0, and wecan take yh = e−12 ln t = t−12. Substitute y = vt−12 and y′ = v′t−12 −12vt−13 in the inhomogeneous ODE to get v′ = t11(5t2 + 3t − 2) =5t13 + 3t12 − 2t11. Thus v = 5

14 t14 + 313 t13 − 1

6 t12 + C. Since y = vt−12,the general solution is the family y = 5

14 t2 + 313 t− 1

6 + Ct−12.Return

Page 876: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 125

8. The associated homogeneous equation is y′ + tan(t)y = 0, andwe can take yh = e−

∫tan(t) dt = eln cos t = cos t. Substituting y = v cos t

yields v′ cos t = 1. Thus v =∫

sec(t) dt = ln | sec(t) + tan(t)| + C.Since y = v cos t, the general solution is the family y = (ln | sec(t) +tan(t)|+ C) cos t.Return

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126 ORDINARY DIFFERENTIAL EQUATIONS

9. The associated homogeneous equation is y′ − ty = 0, and we cantake yh = et2/2. Substitute y = vet2/2 in the inhomogeneous ODE toget v′ = e−t2/2. Thus v =

∫e−t2/2 dt + C. Since y = vet2/2, the general

solution is the family y = (∫

e−t2/2 dt + C)et2/2.Return

Page 878: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 127

10. y = Ce∫ 5

t dt = Ce5 ln t = Ct5

Return

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128 ORDINARY DIFFERENTIAL EQUATIONS

11. The associated homogeneous equation is y′ + y = 0, and we cantake yh = e−t. Substitute y = ve−t in the inhomogeneous ODE to getv′ = 1. Thus v = t + C Since y = ve−t, the general solution is thefamily y = (t + C)e−t.Return

Page 880: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 129

12. y = Ce∫

cos t dt = Cesin t

Return

Page 881: Ordinary Differential Equations: A Systems Approach

130 ORDINARY DIFFERENTIAL EQUATIONS

13.

The associated homogeneous equation is ty′ − y = 0, and we cantake yh = eln t = t. Substitute y = vt and y′ = v′t + v in the inho-mogeneous ODE to get v′ = t−1. Thus v = ln t + C Since y = vt, thegeneral solution is the family y = t ln t + Ct. To satisfy the initial con-dition, set y = 2 and t = 1. This yields C = 2, and y = t ln t + 2t.Return

Page 882: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 131

14.

The associated homogeneous equation is y′ + y = 0, and we can takeyh = e−t. Substitute y = ve−t in the inhomogeneous ODE to get v′ =1. Thus v = t + C Since y = ve−t, the general solution is the familyy = (t + C)e−t. To satisfy the initial condition, set y = 0 and t = 0.This yields C = 0, and y = te−t.Return

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132 ORDINARY DIFFERENTIAL EQUATIONS

15. y = t2 + C, To satisfy the initial condition, set y = 4 and t = 2.This yields C = 0, and y = t2.Return

Page 884: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 133

16. The associated homogeneous equation is y′ + 4y = 0, and we cantake yh = e−4t. Substitute y = ve−4t in the inhomogeneous ODE toget v′ = 3 sin(3t). Thus − cos(3t) + C Since y = ve−4t, the generalsolution is the family y = − cos(3t)e−4t + Ce−4t. To satisfy the initialcondition, set y = 0 and t = 0. This yields 0 = −1 + C so C = 1, andy = − cos(3t)e−4t + e−4t.Return

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134 ORDINARY DIFFERENTIAL EQUATIONS

17. The associated homogeneous equation is ty′− 3y = 0, and we cantake yh = e3 ln t = t3. Substitute y = vt3 to get v′t4 = 5t3, or v′ = 5

t .Thus v = 5 ln t + C Since y = vt3, the general solution is the familyy = 5t3 ln t + Ct3. To satisfy the initial condition, set y = 1 and t = 1.This yields 1 = C, so y = 5t3 ln t + t3.Return

Page 886: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 135

18. To linearize,define a new variable v = y1−1/3 = y2/3. Theny = v3/2, and y′ = 3

2 v1/2v′. Substitute this expression in the equa-tion to obtain 3

2 v1/2v′ = v1/2; or v′ = 2/3 the general solution is thefamily v = 2/3t + C. Therefore y = (2/3t + C)3/2. To satisfy the ini-tial condition, set y = 1 and t = 1. This yields 1 = (2/3 + C)3/2, soc = 1/3 and y = (2/3t + 1/3)3/2.Return

Page 887: Ordinary Differential Equations: A Systems Approach

136 ORDINARY DIFFERENTIAL EQUATIONS

19. The associated homogeneous equation is y′ − (t + 3)y = 0, andwe can take yh = e

∫(t+3) dt = et2/2+3t. Substituting y = vet2/2+3t yields

v′ = (2t + 6)e−t2/2−3t. Thus v = −2e−t2/2−3t + C, and the generalsolution is y = −2 + Cet2/2+3t. To satisfy the initial condition, set y =

0 and t = 0. This yields 0 = −2 + C, so C = 2 and y = −2 + 2et2/2+3t.Return

Page 888: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 137

20. The ODE is the same as in Exercise 19, so the general solution isagain y = −2 + Cet2/2+3t. To satisfy the initial condition, set y = −2and t = 0. This yields −2 = −2 + C, so C = 0 and y = −2.Return

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138 ORDINARY DIFFERENTIAL EQUATIONS

21. The associated homogeneous equation is y′ − 2y4−t2 = 0, and we

can take yh = e∫ 2

4−t2dt

= e1/2 ln 2+t2−t =

√2+t2−t . Substituting y = v

√2+t2−t

yields v′√

2+t2−t =

1√2−t

, and thus v′ = 1√2+t

. Integrating, v = 2√

2 + t+

C. Since y = v√

2+t2−t , the general solution is the family y = 2 2+t√

(2−t)+

C√

2+t2−t . To satisfy the initial condition, set y = 3

√2 and t = 0. This

yields 3√

2 = 2√

2 + C, so C =√

2 and y = 2 2+t√(2−t)

+√

2 2+t2−t .

Return

Page 890: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 139

22.

The associated homogeneous equation is y′ + ty = 0, and we can

take yh = e−t22 . Substitute y = ve−

t22 and y′ = v′e−

t22 − tve−

t22 in

the inhomogeneous ODE to get v′ = (2t + t3)et22 . Thus v =

∫(2t +

t3)et22 dt = t2e

t22 +C Since y = ve−

t22 , the general solution is the family

y = t2 + Ce−t22 . To satisfy the initial condition, set y = 1 and t = 0.

This yields 1 = C, so y = t2 + e−t22 .

Return

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140 ORDINARY DIFFERENTIAL EQUATIONS

23. Let y(t) denote the balance of the bank at time t, then y′(t) = ky,and y = Cekt (C is the balance at time 0). Substituting y(30) = 3C,we obtain e30k = 3, and k = 1

30 ln 3. Thus y = C 3t/30. To find thedoubling time, solve y(t) = 2C, or 3t/30 = 2. Thus t = 30 ln 2

ln 3 years.Return

Page 892: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 141

24. Let y(t) denote the population at time t, We use the ODE, y′ =0.013y, so y = Ce0.013t (C is the population at time 0) Solving y(t) =Ce0.013t = 2C, we obtain t = ln 2

0.013 = 53 years.Return

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142 ORDINARY DIFFERENTIAL EQUATIONS

25. The fraction of C14 in wood from the tree, in ppb,is modelledby the IVP y′ = −ky; y(0) = 1. The general solution for the ODE isy = Ce−kt. Substituting t = 0, y = 1 in y = Ce−kt, we get C = 1.Because C14 has a half life of 5730 years, e−5730k = 0.5, and thereforek = ln(0.5)

−5730 = 0.000121. Solving y = e−0.000121t = 0.8 for t, we obtaint = ln 0.8

−0.000121 = 1844 years.Return

Page 894: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 143

26. The associated homogeneous equation is y′ + y = 0, and we cantake yh = e−t. Substitute y = ve−t yields v′ = et cos(4t). Thus v =117 et(cos(4t) + 4 sin(4t)) + C. Since y = ve−t, the general solution isthe family y = 1

17 (cos(4t) + 4 sin(4t)) + Ce−t. The periodic solutionis y = 1

17 (cos(4t) + 4 sin(4t)), and when t → ∞, Ce−t → 0. Thereforeit is stable.Return

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144 ORDINARY DIFFERENTIAL EQUATIONS

27. Let x(t) denote the amount (in kilograms) of salt in the tank. Thevolume of brine in the tank starts at 1000 liters and increases at therate of 100 liters per hour. Thus the concentration is C(t) = x(t)

1000+100t .We will use the ODE, x′(t) = JK − LC(t) as our model. Substitut-ing J = 200, K = 0.050, L = 100, and using the initial conditionx(0) = 100, we get x′(t) + 1

10+t x(t) = 10, The homogeneous solu-

tion is e−∫ 1

10+t dt = (10 + t)−1. Now we substitute x = v(10 + t)−1

in the differential equation, and simplify to get v′ = 10(10 + t). In-tegration yields v = 100t + 5t2 + C, where C is a constant. Sincex(t) = v(10 + t)−1, it follows that x(t) = 100t+5t2+C

10+t . Substituting

x(0) = 100, we get C = 1000. Therefore x(t) = 100t+5t2+100010+t . Then

x(10) = 100×10+5(10)2+100010+10 = 125 kilograms, and the concentration of

the salt in the tank is 1252000 = .0625 kilograms per liter, or 62.5 grams

per liter.Return

Page 896: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 145

28.

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

Page 897: Ordinary Differential Equations: A Systems Approach

146 ORDINARY DIFFERENTIAL EQUATIONS

29. Write the difference equation in recursive form as ym+1 = ym +h× (t2

m − y2m) with y0 = 0, and h = −0.25 we have

y(−0.25) = 0 + (−0.25)× ((0)2 − (0)2) = 0y(−0.5) = 0 + (−0.25)× ((−0.25)2 − (0)2) = −0.015625y(−0.75) = −0.015625+(−0.25)× ((−0.5)2− (−0.015625)2) = −0.078064y(−1) = −0.078064+(−0.25)× ((−0.75)2− (−0.078064)2) = −0.217165y(−1.25) = −0.217165+(−0.25)× ((−1)2− (−0.217165)2) = −0.455375y(−1.5) = −0.455375+(−0.25)× ((−1.25)2− (−0.455375)2) = −0.794159y(−1.75) = −0.794159+(−0.25)× ((−1.5)2− (−0.794159)2) = −1.19899y(−2) = −1.19899+(−0.25)× ((−1.75)2− (−1.19899)2) = −1.60522y(−2.25) = −1.60522+(−0.25)× ((−2)2− (−1.60522)2) = −1.96104y(−2.5) = −1.96104+(−0.25)× ((−2.25)2− (−1.96104)2) = −2.26525y(−2.75) = −2.26525+(−0.25)× ((−2.5)2− (−2.26525)2) = −2.54491y(−3) = −2.54491+(−0.25)× ((−2.75)2− (−2.54491)2) = −2.81639with y0 = 0, and h = 0.25 we havey(0.25) = 0 + (0.25)× ((0)2 − (0)2) = 0y(0.5) = 0 + (0.25)× ((0.25)2 − (0)2) = 0.015625y(0.75) = 0.015625 + (0.25)× ((0.5)2 − (0.015625)2) = 0.078064y(1) = 0.078064 + (0.25)× ((0.75)2 − (0.078064)2) = 0.217165y(1.25) = 0.217165 + (0.25)× ((1)2 − (0.217165)2) = 0.455375y(1.5) = 0.455375 + (0.25)× ((1.25)2 − (0.455375)2) = 0.794159y(1.75) = 0.794159 + (0.25)× ((1.5)2 − (0.794159)2) = 1.19899y(2) = 1.19899 + (0.25)× ((1.75)2 − (1.19899)2) = 1.60522y(2.25) = 1.60522 + (0.25)× ((2)2 − (1.60522)2) = 1.96104y(2.5) = 1.96104 + (0.25)× ((2.25)2 − (1.96104)2) = 2.26525y(2.75) = 2.26525 + (0.25)× ((2.5)2 − (2.26525)2) = 2.54491y(3) = 2.54491 + (0.25)× ((2.75)2 − (2.54491)2) = 2.81639

Page 898: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 147

-2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

Return

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148 ORDINARY DIFFERENTIAL EQUATIONS

30.

-3 -2 -1 0 1 2 3 4

1

2

3

4

5

Page 900: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 149

31. Write the difference equation in recursive form as ym+1 = ym +h× (t−√y) with y0 = 4, and h = 0.5 we havey(−1.5) = 4 + (0.5)× (−2−

√4) = 2

y(−1) = 2 + (0.5)× (−1.5−√

2) = 0.542893y(−0.5) = 0.542893 + (0.5)× (−1−

√0.542893) = −0.325513

y(0) = −0.325513 + (0.5)× (−0.5−√−0.325513) = ERROR

with y0 = 4, and h = −0.5 we have

y(−2.5) = 4 + (−0.5)× (−2−√

4) = 6y(−3) = 6 + (−0.5)× (−2.5−

√6) = 8.47474

y(−3.5) = 8.47474 + (−0.5)× (−3−√

8.47474) = 11.4303y(−4) = 11.4303 + (−0.5)× (−3.5−

√11.4303) = 14.8708

-3 -2 -1 0 1 2 3 4

0

1

2

3

4

5

Return

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150 ORDINARY DIFFERENTIAL EQUATIONS

32.

(a) The direction field will have positive slope only when 5t > y2.Direction field IV has this property.

(b) This ODE has constant solutions y = kπ, for integers k. Directionfield III is the only one with this property.

(c) The direction field for this ODE must be periodic in t when yis constant, and periodic in y when t is constant. It is thereforedirection field II.

(d) This autonomous ODE has the constant solution 0, which is sta-ble. It therefore corresponds to direction field VI.

(e) This autonomous ODE has constant solutions y = 0 (unstable)and y = π (stable). Therefore, the direction field is V.

(f) All solutions of this ODE are increasing; therefore all directionfield elements have nonnegative slope. Direction field I is theonly one with this property.

Return

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SOLUTIONS MANUAL 151

33. Since h(y) = y2 has a zero at y = 0, there is a singular solution,y ≡ 0. The separated equation is dy

y2 = tdt. Integrating both sides, we

get the family of solutions satisfy − 1y = t2

2 + C1, that is y = − 1t22 +C1

=

2C−t2 , where C = −2C1 is a constant.Return

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152 ORDINARY DIFFERENTIAL EQUATIONS

34. The separated equation is e−y dy = dt. Integrating both sides, weget the family of solutions satisfy−e−y = t+C, that is y = − ln(−t−C), where C is a constant.Return

Page 904: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 153

35. Since h(y) = y(1 − y) has two zeros y = 0, y = 1, there aretwo singular solutions: y ≡ 0 and y ≡ 1. The separated equationis dy

y(1−y) = dt. Integrating both sides, we get the family of solutions

in implicit form, ln∣∣∣ y

y−1

∣∣∣ = t + C1. Solving for y yields the explicit

formula, y = Cet

1+Cet where C = −eC1 is a constant.Return

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154 ORDINARY DIFFERENTIAL EQUATIONS

36. Since h(y) =√

1− y2 has two zeros y = ±1, there are two sin-gular solutions: y = ±1. The separated equation is dy√

1−y2= dt√

1−t2 .

Integrating both sides, we get the family of solutions in implicit form,arcsin y = arcsin t + C. Solving for y yields the explicit formula, y =sin(arcsin t + C) where C is a constant.Return

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SOLUTIONS MANUAL 155

37. Since h(y) =√

y2 − 1 has two zeros y = ±1, there are two sin-gular solutions: y = ±1. The separated equation is dy√

y2−1= dt√

t2−1.

Integrating both sides, we get the family of solutions in implicit form,ln |y +

√y2 − 1| = ln |t +

√t2 − 1|+ C1. Thus y +

√y2 − 1 = C(t +√

t2 − 1) where C = eC1 is a constant.Return

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156 ORDINARY DIFFERENTIAL EQUATIONS

38. Since h(y) = (1 + y2) has no zeros. Thus there are no singu-lar solutions. The separated equation is dy

1+y2 = tdt1+t2 . Integrating

both sides, we get the family of solutions in implicit form, arctan y =ln√

1 + t2 + C. Thus y = tan(ln√

1 + t2 + C), where C is a constant.Return

Page 908: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 157

39. Since h(y) =√

y has a zero y = 0, there is a singular solutiony ≡ 0. The separated equation is dy√

y = dt√t. Integrating both sides, we

get the family of solutions in implicit form, 2√

y = 2√

t + C1. Thusy = (

√t + C)2, where C = C1/2 is a constant.

Return

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158 ORDINARY DIFFERENTIAL EQUATIONS

40. Since h(y) = e4y has no zero, there are no singular solutions. Theseparated equation is e−4y dy = tan(3t)dt. Integrating both sides, weget the family of solutions in implicit form,− e−4y

4 = − 13 ln | cos(3t)|+

C1. Thus e−4y = 43 ln | cos(3t)|+ C where C = −4C1 is a constant.

Return

Page 910: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 159

41. Our model is the ODE mv′ = mg− mg3.5− kv2. Divide through by

m and simplify to obtain v′ =5g7− kv2

m. With g = 9.8 m/s we have

57 g = 7, v′ = 7− kv2

m. so By solving v′ = 0 for v we obtain the terminal

velocity, v =

√7mk

= 21. Thusmk= 63.

With this parameter, our model is v′ = 7(

1− v2

212

). The separated

equation isdv

1− v2

212

= 7 dt. Integrating both sides, we have ln∣∣∣∣21 + v21− v

∣∣∣∣ =23

t + C. To evaluate C, set v(0) = 0: we find C = 0. In explicit form,

we have v = 21e2t/3 − 1e2t/3 + 1

.

(a) 21∫ 10

0

e2t/3 − 11 + e2t/3 dt = 210.08− 43.6683 = 166.4.meters.

(b) v(10) ≈ 21, thus T = 10 + (1463 − 166)/21 = 10 + 62 = 72seconds.

(c) Suppose that you throw the diamond vertically into the lake sothat its initial velocity after it is immersed in water is 42 me-ters per second. Answer questions (a) and (b). We know that

ln |21 + v21− v

| = 23

t + C. Thus21 + v21− v

= C1e2t/3, where C1 = ±eC.

Substituting v(0) = 42, we obtain C1 = −3. Thus v = 21(3e2t/3 + 1)3e2t/3 − 1

,

and the distance travelled in 10 seconds is 21∫ 10

0

(3e2t/3 + 1)3e2t/3 − 1

dt =

236 meters. Also v(10) ≈ 21; thus T = 10 + (1463− 236)/21 =10 + 58.4 = 68.5 seconds.Return

Page 911: Ordinary Differential Equations: A Systems Approach

160 ORDINARY DIFFERENTIAL EQUATIONS

42.

(a) In year A the relative growth rate wasy′

y= 4500/100,000 =

4.5%, and in year By′

y= 8000/200,000 = 4%

(b) According to the logistic equation,y′

y= k

(1− y

M

). Substitute

y′

y= .045 and y = 100,000 to obtain .045 = k(1− 100,000

M). With

y′

y= 0.04 and y = 200,000 we get 0.04 = k(1− 200,000

M). Solving

these two equations, we obtain M = 1,000,000 and k = 0.05.

(c) The general solution of the logistic equation y′ = k y(

1− yM

)is y =

M1 + Ae−kt . Thus y =

1,000,0001 + Ae−0.05t . Substituting y(0) =

100,000, we find A = 9. We can then solve 200,000 =1,000,000

1 + 9e−0.05t

to obtain t = ln(9/4)/0.05 ≈ 16. Thus year B is 1980+16=1996.

Return

Page 912: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 161

43. The stationary points are y ≡ 0, y ≡ − 12 . The up intervals are

(−∞,− 12 ) and (0,+∞), and (− 1

2 , 0) is a down interval. Thus y = − 12

is stable. and y(t)→ − 12 .

- r� r -

− 12 0

Return

Page 913: Ordinary Differential Equations: A Systems Approach

162 ORDINARY DIFFERENTIAL EQUATIONS

44. The stationary points are y = kπ, where k is an integer. The upintervals are (−(k + 1)π,−kπ) for k ≥ 0, and the down intervals are(kπ, (k + 1)π). Thus y = 0 is stable. If y(0) = 30 then y(t) → 9πbecause 30 belongs to the down interval (9π, 10π).

- r−π

- r0� r

π

� r2π

� r3π

� r4π

� r5π

� r6π

� r7π

� r8π

� r9π

� r10π

Return

Page 914: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 163

45. y′ = y2 + 2y+ 2 = (y+ 1)2 + 1 > 0 Thus (−∞,+∞) is up intervaland y(t)→ +∞. There are no stationary points.

-

Return

Page 915: Ordinary Differential Equations: A Systems Approach

164 ORDINARY DIFFERENTIAL EQUATIONS

46. There is one stationary point, y ≡ 53 . (−∞, 5

3 ) is up interval, and( 5

3 ,+∞) is down interval. Thus y ≡ 53 is stable, and y(t)→ 5

3 .

- r�53

Return

Page 916: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 165

47. There is one stationary point, y ≡ 1. (0, 1) is up interval, and(−∞, 0) and (1,+∞) are down intervals. Thus y ≡ 1 is stable, andy(t)→ 1.

� r- �0 1

Return

Page 917: Ordinary Differential Equations: A Systems Approach

166 ORDINARY DIFFERENTIAL EQUATIONS

48. The phase diagram is the same as in the previous exercise, butnow the initial point is in the down interval (−∞, 0). Therefore y(t)→−∞.Return

Page 918: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 167

49. Differentiate both sides of y′ = 1y to obtain y′′ = − y′

y2 . Now replace

y′ with y−1 to obtain the equation y′′ = −y−3. Thus when y < 0 thesolution graphs are decreasing (y′ < 0) and concave up (y′′ > 0), andwhen y > 0 the solution graphs are increasing (y′ > 0) and concavedown (y′′ < 0).Return

Page 919: Ordinary Differential Equations: A Systems Approach

168 ORDINARY DIFFERENTIAL EQUATIONS

50. Differentiate both sides of y′ = t + y with respect to t to obtainy′′ = 1 + y′. Replace y′ with t + y; thus y′′ = 1 + t + y. In red region,y′′ > 0, and thus the solution graphs are concave up. In the blueregion: y′′ < 0, and thus the solution graphs are concave down.Return

Page 920: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 169

51. Since y′ = ey > 0 the graph of any solution is increasing. Bydifferentiation, we find y′′ = y′ey; the substitution y′ = ey shows thaty′′ = e2y > 0. Thus the graph of any solution is concave up.

Similarly, for the ODE y′ = e−y we have y′ > 0 and y′′ = −e−yy′ =−e−2y < 0. Thus all solution graphs are increasing and concave down.Return

Page 921: Ordinary Differential Equations: A Systems Approach

170 ORDINARY DIFFERENTIAL EQUATIONS

52.

(a) Write the difference equation in recursive form as ym+1 = ym +h× (tm − y2

m). Thus, with y0 = 0, and h = 0.1 we have

y0.1(0.1) = 0 + (0.1)× (0− (0)2) = 0y0.1(0.2) = 0 + (0.1)× (0.1− (0)2) = 0.01y0.1(0.3) = 0.01 + (0.1)× (0.2− (0.01)2) = 0.02999y0.1(0.4) = 0.02999 + (0.1)× (0.3− (0.02999)2) = 0.0599001y0.1(0.5) = 0.0599001+ (0.1)× (0.4− (0.0599001)2) = 0.0995413y0.1(0.6) = 0.0995413 + (0.1)× (0.5− (0.0995413)2) = 0.14855y0.1(0.7) = 0.14855 + (0.1)× (0.6− (0.14855)2) = 0.206344y0.1(0.8) = 0.206344 + (0.1)× (0.7− (0.206344)2) = 0.272086y0.1(0.9) = 0.272086 + (0.1)× (0.8− (0.272086)2) = 0.344683y0.1(1) = 0.344683 + (0.1)× (0.9− (0.344683)2) = 0.422802

Similarly, with y0 = 0, and h = 0.05 we have

y0.05(0.05) = 0 + (0.05)× (0− (0)2) = 0,y0.05(0.1) = 0 + (0.05)× (0.05− (0)2) = 0.0025,

...

y0.05(1) = 0.400229 + (0.05)× (0.95− (0.400229)2) = 0.43972,

and with y0 = 0, and h = 0.025 we havey0.025(0.025) = 0 + (0.025)× (0− (0)2) = 0y0.025(0.05) = 0 + (0.025)× (0.025− (0)2) = 0.000625y0.025(0.075) = 0.000625+(0.025)× (0.05− (0.000625)2) = 0.00187499y0.025(0.1) = 0.00187499 + (0.025) × (0.075− (0.00187499)2) =0.0037499

...

y0.025(1) = 0.427971+(0.025)× (0.975− (0.427971)2) = 0.447767.

(b) Multiply (2.46) by 2 and subtract from (2.45). Thus φ(1) ≈ 2Yh2−Yh1 . Thus Zh1 = 2Yh2 −Yh1 .

Page 922: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 171

(c) Calculate Z0.1 and Z0.05. Z0.1 = 2Y0.05 − Y0.1 = 2 × 0.43972 −0.422802 = 0.456638.Z0.05 = 2Y0.025 −Y0.05 = 2× 0.447767− 0.43972 = 0.455814.

(d) φ(1) ≈ Z0.1 + 0.01D, and φ(1) ≈ Z0.05 + 0.0025 D. Thus,

4(Z0.05 + 0.0025 D)− (Z0.1 + 0.01 D) ≈ 3φ(1)

. Since the D terms cancel, we may divide through by 3 andobtain

φ(1) ≈ 4Z0.05 − Z0.1

3=

4× 0.455814− 0.4566383

= 0.455539

Return

Page 923: Ordinary Differential Equations: A Systems Approach

172 ORDINARY DIFFERENTIAL EQUATIONS

53. There will be a solution unless y0 = 1, since that is the only pointon the y-axis where the right side is not continuous. To apply theuniqueness theorem, we need a Lipschitz condition; this precludesy0 = −1.Return

Page 924: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 173

54. By the uniqueness theorem, we only need to see that the rightside of the differential equation satisfies a Lipschitz condition; this inturn can be shown by demonstrating that the partial derivative withrespect to y of the right side is continuous.Return

Page 925: Ordinary Differential Equations: A Systems Approach

174 ORDINARY DIFFERENTIAL EQUATIONS

55. Substituting x = C et and y = C et in the first equation yieldsthe identity C et = C et. The same substitution in the second equationresults in C et = C et.

Substituting x = A e−t and y = B e−t in the first equation yields−A e−t = B e−t. The same substitution in the second equation re-sults in −B e−t = A e−t. Solve these equations for A and B to obtainA = −B.Return

Page 926: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 175

57. Substituting x = 2C cos(3t) and y = C[cos(3t) + 3 sin(3t)] in thefirst equation yields the identity−6C sin(3t) = 2C cos(3t)− 2C[cos(3t)+3 sin(3t)]. The same substitution in the second equation results inC[−3 sin(3t) + 9 cos(3t)] = 10C cos(3t)− C[cos(3t) + 3 sin(3t)].

Substituting x = 2A sin(3t) in the first equation yields 6A cos(3t) =2A sin(3t) − 2y. Thus if y = A sin(3t) − 3A cos(3t), the first equa-tion is satisfied. Now substitute x = 2A sin(3t) and y = A sin(3t)−3A cos(3t) in the second equation. This results in an identity, 3A cos(3t)+9A sin(3t) = 10A sin(3t)− A sin(3t) + 3A cos(3t). Thus the solutionshave the form x = 2A sin(3t), y = A sin(3t)− 3A cos(3t).Return

Page 927: Ordinary Differential Equations: A Systems Approach

176 ORDINARY DIFFERENTIAL EQUATIONS

59. Substituting x = e2t(t + 1) and y = e2t(t− 1) in the first equationyields the identity 2e2t(t + 1) + e2t = e2t(t + 1) + e2t(t− 1) + 3e2t. Thesame substitution in the second equation results in 2e2t(t− 1) + e2t =2e2t(t + 1)− 3e2t.

Substituting x = C e2t and y = C e2t in the first equation of the “as-sociated homogeneous system” yields the identity 2C e2t = C e2t +C e2t. The same substitution in the second equation results in 2C e2t =2C e2t.

A family of solutions of the inhomogeneous system is the sum of theassociated homogeneous solutions plus a particular solution. Thuswe obtain a family of solutions:5 x = e2t(t+ 1+C), y = e2t(t− 1+C)Return

Page 928: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 177

61. The ODE can be written as d2ydt2 = −3 dy

dt − 4y + t2 Set v = dydt and

v′ = d2ydt2 . Then the system

y′ = vv′ = −3v− 4y + t2

replaces the given ODE.Return

Page 929: Ordinary Differential Equations: A Systems Approach

178 ORDINARY DIFFERENTIAL EQUATIONS

63. The ODE can be written as d2udt2 = 1+t2

u dudt

Set v = dudt and v′ = d2u

dt2 .

Then the systemu′ = vv′ = 1+t2

uv

replaces the given ODE.Return

Page 930: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 179

65. The first equation, x′ = x2, has a singular solution x ≡ 0, thatisn’t consistent with the second equation; thus it is extraneous.

Therefore we can separate the first equation and integrate to obtain−x−1 = t + C, or explicitly, x = −1

t+C . Substitute this in the sec-ond equation and obtain y′ = −t − C, which we integrate to gety = − 1

2 t2 − Ct + D, where C, D are constants.Return

Page 931: Ordinary Differential Equations: A Systems Approach

180 ORDINARY DIFFERENTIAL EQUATIONS

67. This system is not uncoupled.

Page 932: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 181

69. The first equation can be put in separated form as x dx = t dt.Thus x2 = t2 + C, or x = ±

√t2 + C. Substitute this in the second

equation to obtain y′ = ±y t√

t2 + C. This has a singular solution

y ≡ 0, and the separated equation isdyy

= ±t√

t2 + C. Integrating

both sides we obtain ln y = ± 13 (t

2 + C)3/2 + D1, or y = De±13 (t

2+C)3/2,

where C and D = eD1 are constants. (The singular solution is in thisfamily.)Return

Page 933: Ordinary Differential Equations: A Systems Approach

182 ORDINARY DIFFERENTIAL EQUATIONS

71. This system is not uncoupled.

Page 934: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 183

73. Substituting x = t2 and y = t in the first equation yields the iden-tity 2t = 2t2/t. The same substitution in the second equation resultsin 1 = t2/t2.

Eliminate t in the solution x = t2, y = t. we have x = y2, and ast → ∞, both x, y → ∞. This orbit is therefore the parabola x = y2

directed upward.

1 2 3 4

-2

-1

1

2

Return

Page 935: Ordinary Differential Equations: A Systems Approach

184 ORDINARY DIFFERENTIAL EQUATIONS

75. Substituting x = e−t and y = −e−t in the first equation yields theidentity −e−t = −e−t. The same substitution in the second equationresults in e−t = e−t.

Eliminate t in the solution x = e−t, y = −e−t to obtain y = −x. Ast → ∞, both x and y → 0. This orbit is therefore part of the liney = −x in the fourth quadrant directed toward the origin.

1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

Return

Page 936: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 185

77. The ODE can be written as d2ydt2 = y−3 Set v = dy

dt and v′ = y−3.Then the system

y′ = vv′ = y−3

replaces the given ODE.

If y =√

t2 + 1, then y′ =t√

t2 + 1and y′′ =

√t2 + 1− t2(t2 + 1)−1/2

t2 + 1=

1(t2 + 1)3/2 = y−3

The orbit of the solution corresponding to y =√

t2 + 1 is

(y, v) =(√

t2 + 1,t√

t2 + 1

).

To eliminate t, notice that v2 =t2

t2 + 1=

y2 − 1y2 . Hence v = ±

√1− 1

y2 .

As t → ∞, y → ∞ and v → 1, and when t → −∞, we have y → ∞and v → −1. The orbit in the y, v-plane is therefore directed upwardand to the right in the first quadrant, and asymptotic to v = ±1.

1 2 3 4

-1

-0.5

0.5

1

Return

Page 937: Ordinary Differential Equations: A Systems Approach

186 ORDINARY DIFFERENTIAL EQUATIONS

79.

The ODE can be written as d2ydt2 = t dy

2dt y′ − y Set v = dydt and v′ = y′′.

Then the systemy′ = vv′ = 1

2 tv− y

replaces the given ODE.

If y = t2 − 2, then y′ = 2t, and y′′ = 2 = t2 − (t2 − 2) = 12 t y′ − y.

The orbit of the solution corresponding to y = t2 − 2 is (y, v) = (t2 −2, 2t). Here we have v2 = 4y + 8, and as t → ∞, both y and v → ∞.This orbit is therefore the parabola v2 = 4y + 8 directed upward.

-2 -1 1 2

-4

-2

2

4

Return

Page 938: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 187

81. By solving the equations 4y/16 = 0, −x/16 = 0 we get the sta-tionary point (0, 0).

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

Return

Page 939: Ordinary Differential Equations: A Systems Approach

188 ORDINARY DIFFERENTIAL EQUATIONS

83. By solving the equations (x2 + 1)y/16 = 0, −2x/16 = 0 we getthe stationary point (0, 0).

-2 -1 0 1 2 3

-1.5

-1

-0.5

0

0.5

1

1.5

2

Return

Page 940: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 189

85.

Page 941: Ordinary Differential Equations: A Systems Approach

190 ORDINARY DIFFERENTIAL EQUATIONS

(a)

1 2 3 4 5 6

-1

-0.5

0.5

1

Graphs of x(t) = sin(5t) and y(t) = cos(5t).

-1 -0.5 0.5 1

-1

-0.5

0.5

1

Phase portrait

-1-0.50 0.5 1

-1-0.5

00.5

1

0

2

4

6

-0.500.5

1

Three-dimensional graph.

Page 942: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 191

(b)

2 4 6 8 10 12

-6

-4

-2

2

4

6

Graphs of x(t) = 4(cos(√

2t)+√

2 sin(√

2t)) and y(t) = 2√

2 sin[(√

2t).

-6 -4 -2 2 4 6

-2

-1

1

2

Phase portrait-5

0

5-2

02

0

5

10

-2

02

0

5

10

Three-dimensional graph.

Page 943: Ordinary Differential Equations: A Systems Approach

192 ORDINARY DIFFERENTIAL EQUATIONS

(c)

-5 -2.5 2.5 5 7.5 10 12.5

-4

-2

2

4

6

Graphs of x(t) = 112 (4 cos(t)+ (−4+ 6t) cos(2t)− 3 sin(2t)) and

y(t) = − 13 ((2 + (−2 + 3 t) cos(t)) sin(t)).

-4 -2 2 4 6

-4

-2

2

4

6

Phase portrait-5

-2.50

2.55

-2.502.5

5

-5

0

5

10

-2.502.5

5

Three-dimensional graph.

Page 944: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 193

(d)

-2 -1 1 2 3 4

-1

-0.5

0.5

1

1.5

Graphs of x(t) = 16 t3 and y(t) = 1

6 (3t2 − t3).

-1 1 2 3

-1

-0.5

0.5

1

1.5

Phase portrait0

5

10

-101-2

0

2

4

-2

0

2

Three-dimensional graph.

Return

Page 945: Ordinary Differential Equations: A Systems Approach

194 ORDINARY DIFFERENTIAL EQUATIONS

1. Using MDEP with an initial stepsize of 0.1, I found that y(1) ≈0.841471076. Your answer may be different, since other renditions ofthe RKF45 algorithm may not have the same default error tolerancesettings that MDEP has. The correct value of sin(1), to nine decimalplaces, is 0.8414709848.Return

Page 946: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 195

3.

(a) If we multiply the first equation by x, multiply the second by yand add, we have

xx′ + yy′ = ε(x2 + y2)(x2 + y2 − 1).

With r2 = x2 + y2 and rr′ = xx′ + yy′ we have

rr′ = εr2(r2 − 1),

or r′ = εr(r2 − 1). This differential equation has three stationarypoints, at 0 and ±1. The phase diagram depends on ε:

- r−1� r

0- r

1�

Phase diagram for ε < 0

� r−1

- r0� r

1-

Phase diagram for ε > 0

The stationary point 0 corresponds to the stationary point of thesystem at the origin and the stationary point 1 corresponds tothe circular orbit r = 1. Since r ≥ 0, the stationary point −1 hasno significance. If ε < 0, 0 is an unstable stationary point ofr′ = εr(r2− 1) and 1 is stable; then the origin is unstable and thethe orbit converges to the unit circle as shown in the followingphase portrait.

-1 -0.5 0.5 1

-1

-0.5

0.5

1

If ε > 0, the stability is reversed: the origin is stable, and thecircle is the limiting orbit for all non-constant orbits as t → −∞.Now the two orbits look like this:

Page 947: Ordinary Differential Equations: A Systems Approach

196 ORDINARY DIFFERENTIAL EQUATIONS

-2 -1.5 -1 -0.5 0.5 1 1.5

-1.5

-1

-0.5

0.5

1

1.5

(b) i. With (ε, A) = (−.25, 4), the orbit will spiral inward towardand converge to the unit circle, which is stable because ε <0..

-1 1 2 3 4

-1

-0.5

0.5

1

1.5

ii. With (ε, A) = (.25, .95) the orbit will spiral inward towardand converge to the origin, which is stable, and away fromthe circular orbit, which is unstable because ε > 0.

Page 948: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 197

-1 -0.5 0.5 1

-1

-0.5

0.5

1

iii. With (ε, A) = (.25, 1.05) the orbit will spiral outward to in-finity, and away from the circular orbit, which is unstablebecause ε > 0.

-1.5 -1 -0.5 0.5 1

-4

-3

-2

-1

1

Return

Page 949: Ordinary Differential Equations: A Systems Approach

198 ORDINARY DIFFERENTIAL EQUATIONS

5.

-2 -1 1 2

-300

-200

-100

100

200

300

Return

Page 950: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 199

7.

(a) Solve the ODE y′ = f (y)/100 with initial conditions y(0) =−3,−2,−1, and 0, over the interval [−5, 5]. The zeros of f willbe the ordinates of the horizontal asymptotes of these solutions.Notice that the convergence to the multiple roots is slow — canyou explain why?

-100 -50 50 100

-4

-3

-2

-1

(b) The roots are approximately 0.193, 1.027, 2.568, 4.900, 8.182, 12.734,and 19.396. To locate them, start with a numerical solution ofy′ = f (y)× 10−4 with initial condition y(0) = 10. You will findthat the graph of the solution rapidly converges to two asymp-totes: y → 8.18215 as t → −∞, and y → 12.7342 as t → +∞.Now try the same experiment with initial condition y(0) = 4.This locates two more asymptotes: y → 4.90035 as t → −∞,and y → 2.56788 as t → +∞. Are there zeros between 4.90035and 8.18215? To find out, find a solution with initial conditionbetween these values of y: we will take y(0) = 6. The solutionis a decreasing function converging to 8.19215 to the left, and4.90035 to the right. Thus, there can be no zeroes between thetwo values. Now try an initial condition y(0) = 1. This solutionconverges to 1.02666 to the right, and 0.193044 to the left. Nowtry a solution with y(0) = 0. The result is a decreasing func-tion, converging to 0.193044 as t → −∞, and diverging to −∞as t → ∞. We have located 6 of the seven zeros. To locate thelast one, try a solution with y(0) = 15. This solution will be de-creasing, with convergence to 19.3957 to the left and 12.7342 to

Page 951: Ordinary Differential Equations: A Systems Approach

200 ORDINARY DIFFERENTIAL EQUATIONS

the right. We have thus located the seventh zero.The followinggraph shows the solutions just computed.

-4 -2 2 4

5

10

15

Return

Page 952: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 201

9. The graphs shown in (a) and (c) are stable periodic solutions; thesolution shown was drawn by an IVP solver using arbitrary initialconditions y(0) = 0, and then displaying a segment of the graph afterthe transients have decayed. There is a periodic solution in (b), butit is not stable, so we have to go backwards: The solution shown ispart of the graph of a solution calculated on the interval−15 ≤ t ≤ 0,with initial condition y(0) = 0. Only the portion for −15 ≤ t ≤ −13is displayed.

(a)

92 94 96 98 100

-0.4

-0.2

0.2

0.4

(b)

-14.5 -14 -13.5 -13

-0.04

-0.02

0.02

0.04

(c)

Page 953: Ordinary Differential Equations: A Systems Approach

202 ORDINARY DIFFERENTIAL EQUATIONS

0.8 0.85 0.9 0.95

-0.01

-0.005

0.005

0.01

Return

Page 954: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 203

1. The x-nullcline is given by the equation x2 + y2 = 0. Thus thex-nullcline are the point (0, 0).

The y-nullclines are determined by the equation x2 − y2 = 0, andhence they are the lines y = ±x.

The stationary points lie at the intersection of x-nullclines with y-nullclines. Thus there is only one stationary point, (0, 0).

-10 -7.5 -5 -2.5 2.5 5 7.5 10

-10

-7.5

-5

-2.5

2.5

5

7.5

10

Return

Page 955: Ordinary Differential Equations: A Systems Approach

204 ORDINARY DIFFERENTIAL EQUATIONS

3. The x-nullcline is given by the equation −x = 0, hence it is they-axis.

The y-nullclines are determined by the equation 4y − y2 = 0, andhence they are the line y = 4 and the x-axis.

The stationary points lie at the intersection of x-nullclines with y-nullclines. Thus there are two stationary points, (0, 0) and (0, 4).

-3 -2 -1 1 2 3

-2

2

4

6

Return

Page 956: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 205

5. The x-nullcline is given by the equation x2 = 0; hence it is they-axis.

The y-nullcline is determined by the equation y = 0 and hence it isthe x-axis.

The stationary points lie at the intersection of x-nullclines with y-nullclines. Thus the stationary points are: (0, 0).

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Return

Page 957: Ordinary Differential Equations: A Systems Approach

206 ORDINARY DIFFERENTIAL EQUATIONS

7. The x-nullcline is given by the equation x + y− 2 = 0, hence it isthe line y = 2− x.

The y-nullcline is determined by the equation x− y = 0 and hence itis the line y = x.

The stationary points lie at the intersection of x-nullclines with y-nullclines. Thus there is one stationary point, (1, 1).

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

Return

Page 958: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 207

9. We will find an integral of the system by finding an integral forthe corresponding ODE (x + y) dx− 2x dy = 0,

This ODE can be written as

2xdydx− y = x,

and hence is linear. The general solution of the homogeneous equa-

tion 2xdydx− y = 0 will have the form y = C eK(x), where 2x K′(x) =

1. It is easily seen that K(x) = 12 ln(x) so y = C eln(x)/2 = C x1/2

satisfies the homogeneous equation. Let’s substitute y = x1/2 v(x)

in the inhomogeneous equation; we obtain 2x3/2 dvdx

= x, and thus

v(x) = x1/2 + C. It follows that the general solution of out ODE isy = x + C x1/2. Divide through by x1/2 to obtain an integral:

F(x, y) = x−1/2(y− x).

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208 ORDINARY DIFFERENTIAL EQUATIONS

11. We will find an integral of the system by finding an integral forthe corresponding ODE 5y dx − 3x dy = 0, which is separable. Di-vide through by xy to obtain (5/x) dx− (3/y) dy = 0. The separatedequation is thus (5/x) dx = (3/y) dy. Integrating both sides we get5 ln x = 3 ln y + C, Thus F(x, y) = 5 ln x− 3 ln y is an integral.Return

Page 960: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 209

13. We will find an integral of the system by finding an integral forthe corresponding ODE 2xy dx + (x2 + y2) dy = 0,

Because ∂P∂y (x, y) = 2x = ∂Q

∂x (x, y), the equation is exact. Set

F(x, y) =∫

2xy dx + K(y) = x2y + K(y).

The function K(y) satisfies ∂F∂y = x2 + K′(y),

Thus K′(y) = y2. A final integration shows that K(y) = y3

3 and so

F(x, y) = x2y + y3

3 .Return

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210 ORDINARY DIFFERENTIAL EQUATIONS

15. The system is uncoupled. The solution of the first equation, x′ =1, with x(0) = x0 is x = t + x0. Thus the second equation can bewritten dy

dx = f (x, y); the solution of the system that passes through(x0, y0) when t = 0 would satisfy the initial condition y(0) = y0, andsince x = t + x0 the solution follows the graph of the solution of theIVP

dydx

= f (x, y); y(x0) = y0.

Return

Page 962: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 211

17. Set y = dxdt and y′ = d2x

dt2 . Then the system

x′ = y; y(t0) = y0

y′ = f (t, x, y); x(t0) = x0

replaces the given ODE. Let g(t, x, y) = y, then f (t, x, y) and g(t, x, y)satisfy Lipschitz conditions. By the existence and uniqueness theo-rems there is a unique solution of the system. Thus there is a uniquesolution of the IVP.Return

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212 ORDINARY DIFFERENTIAL EQUATIONS

19. The system is equivalent to

x′ =1

ad− bc(ag(t, x, y)− c f (t, x, y))

y′ =1

ad− bc(d f (t, x, y)− bg(t, x, y))

If the functions f (t, x, y), g(t, x, y) are continuous and satisfy a Lip-schitz condition then, provided that ad − bc 6= 0, 1

ad−bc (ag(t, x, y) −c f (t, x, y)) and 1

ad−bc (d f (t, x, y)− bg(t, x, y)) are continuous and sat-isfy a Lipschitz condition. Thus by the existence and uniqueness the-orems the system has a unique solution with the initial conditions.Return

Page 964: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 213

1.

(a) For the Lotka Volterra equation the stable point is (d, ab ), Thus

the optimum number of predators to release is ab .

(b) The strategy is ineffective. For the solutions of the Lotka Volterraequation are periodic functions of time with the same averagepopulation d for pests and a

b for predators. If the number of preyis initially small, it will become even bigger at some other time.

(c) Wait until there are ≈ d pests.

Return

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214 ORDINARY DIFFERENTIAL EQUATIONS

3. Let x and y denote the populations of species A and B, respec-tively. The x-nullclines are given by the line x = 0, and line deter-mined by the points (1000, 0) and (800, 800). The equation of thisx-nullcline is therefore 4x + y = 4000. It follows that x′ = k1x(4000−4x− y).

The y-nullclines are given by the line y = 0, and line determined bythe points (0, 1500) and (800, 800), which has the equation 7x + 8y =12000. Hence y′ = k2y(12000− 7x− 8y).

Thus the fish populations are governed by the system

x′ = k1x(4000− 4x− y)y′ = k2y(12000− 7x− 8y)

The phase portrait has the configuration (d) of Figure 3.19. The tri-angles ABC and CDE in the figure below are traps, and orbits withinthese triangles converge to the point C = (800, 800).

500 1000 1500 2000

1000

2000

3000

4000 A

B

C

D E

Return

Page 966: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 215

5. The x-nullclines are the lines x = 0 and K − x + By = 0. The y-nullclines are the lines y = 0 and L + Cx− y = 0. The system has atmost four stationary points. Three of these occur when one or both ofthe species are extinct. The fourth stationary point, (x1, y1) if it exists,is located at the intersection of the lines K − x + By = 0, and L +Cx− y = 0. If BC = 1 then the two lines parallel, and there are onlythree stationary points. If BC 6= 1 we solve the equations to obtain afourth intersection point (x1, y1), where x1 = BL+K

1−BC and y1 = CK+L1−BC .

If BC > 1 (x1, y1) is in the third quadrant, and thus this stationarypoint is extraneous. If BC < 1 we have the fourth stationary point inthe first quadrant. This configuration is shown in the drawing. Thequadrilateral whose vertices are the four stationary points is a trap,and the dihedral with vertex at the stationary point interior to thefirst quadrant, and which is opposite to the quadrilateral, is also atrap. Orbits in these traps are directed toward that stationary point,which is therefore stable.

Thus. if BC < 1, the model indicates a stable population for eachspecies. In the case BC ≥ 1 the populations will increase withoutbound.

500 1000 1500 2000

500

1000

1500

2000

Return

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216 ORDINARY DIFFERENTIAL EQUATIONS

7. The system with fishing can be written as Lotka-Volterra equa-tions:

x′ = x[(a− R)− by]

y′ = cy[

x−(

d +Rc

)].

By the result of Exercise 6, the average populations will be x = d + Rc

and y = a−Rb . Thus the average prey population increases proportion-

ally to the catch rate, and the average predator population decreases.Return

Page 968: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 217

1. This system is uncoupled, because the first equation does notinvolve y. This equation s linear, and the general solution of its asso-ciated homogenous equation, x′ = x, is x = Aet. Substitute x = vet

in the inhomogeneous equation to get v′ = 1. Thus v = t + C andthe general solution of the first equation is x = (t + C)et. To satisfythe initial condition, substitute x(0) = 0. Solving, we get C = 0. Thusx = tet. Substitute this expression for x in the second equation toobtain y′ = tety2. This equation is separable. There is a singular solu-tion y = 0, which does not satisfy the initial condition, y(0) = 1. Theseparated equation is y−2 dy = tet dt. Integrating both sides we have−y−1 = et(t− 1) + C. Substituting y(0) = 1, we get C = 0. Thereforewe get

x = tet, y =1

et(1− t).

Return

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218 ORDINARY DIFFERENTIAL EQUATIONS

2. This system is uncoupled. The general solution of first equationis x = Ce−t, and the general solution of the second is y = Det, whereC and D are constants.

We will find an integral of the system by finding an integral for thecorresponding ODE y dx+ x dy = 0, which is separable. Divide through

by xy to getdxx

= −dyy

. Integrating both sides we have ln |x| =

− ln |y|+C1. An equivalent form of this equation is ln |xy| = C1. Thisshows that ln |xy| is an integral, but we can apply the exponential andobtain a simpler integral, F(x, y) = xy.

The orbits of this system will be hyperbolas asymptotic to the x- andy-axes, directed so as to approach the y-axis as t → ∞, and to ap-proach the X-axis as t→ −∞.

Return

Page 970: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 219

3. This equation is also uncoupled. The general solution of the firstequation is x = Cet, and the general solution of the second is y =De2t, where C, D are constants.

We will find an integral of the system by finding an integral for thecorresponding ODE 2y dx − x dy = 0, which is separable. Dividethrough by xy to get 2dx

x = dyy , and integrate both sides. Thus 2 ln |x| =

ln |y|+ C1, or ln |x2/y| = C1. Apply the exponential to obtain a sim-ple integral, F(x, y) = x2/y.

orbits will be halves of parabolas y = Ax2, directed away from thestationary point at the origin.

Return

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220 ORDINARY DIFFERENTIAL EQUATIONS

4. Let v = y′; then the system y′ = v, v′ = − f (y) replaces the givenODE. We will find an integral of the system by finding an integral forthe corresponding ODE, v dv + f (y) dy = 0, which is separable. We

obtain F(y, v) =v2

2+ F(y).

(a) Here f (y) = α2 sin(y), and we could take F(y) = −α2 cos(y).Thus the integral is F(y, v) = v2

2 − α2 cos(y).

(b) Here f (y) = α2y), and we could take F(y) = α2y2

2 . Thus the

integral is F(y, v) = v2

2 + α2y2

2 .

(c) Here f (y) = α2y−2, and we could take F(y) = −α2y−1. Thus theintegral is F(y, v) = v2

2 − α2y−1.

Return

Page 972: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 221

5.

(a) Let y′ = v; then the system y′ = v, v′ = −y replaces the givenODE.

(b) Let y′ = v; then the system y′ = v, v′ = v− t2 sin(y) replaces thegiven ODE.

Return

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222 ORDINARY DIFFERENTIAL EQUATIONS

6. Let y′ = v then the system y′ = v, v′ = 2v− 2y replace the givenODE. The corresponding orbit is (y, v) = (et cos(t), et cos t− et sin t).

Return

Page 974: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 223

7. We will find an integral of the system by finding an integral forthe corresponding ODE 2y(x + y) dx + x(x + 4y) dy = 0,∂P∂y (x, y) = 2x + 4y and ∂Q

∂x (x, y) = 2x + 4y so the equation is exact.The integral F(x, y) will have the form F(x, y) =

∫2y(x + y) dx +

H(y) = yx2 + 2y2x + H(y). To determine H(y), differentiate this ex-pression with respect to y: ∂F

∂y = x2 + 4yx + dHdy . Since ∂F

∂y = Q(x, y) =

x2 + 4yx, it follows that dHdy = 0. and the integral is F(x, y) = yx2 +

2y2x.Return

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224 ORDINARY DIFFERENTIAL EQUATIONS

8.

(a) Solve the equations: x + y − 2 = 0, x − 3y + 2 = 0 we get thestationary point (1, 1).

(b) Solve the equations: y− x = 0, y− x3 = 0 we get the stationarypoints (0, 0), (−1,−1), (1, 1).

(c) Solve the equations: x(x + y + 4) = 0, y(x + 5y) = 0 we get thestationary points (0, 0), (−4, 0), (−5, 1).

Return

Page 976: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 225

9.

(a) The system of ODEs corresponding to the vector fields is x′ =x + y− 2, y′ = x − 3y + 2. The x-nullclines are the line x + y−2 = 0, and the y-nullclines are the line x− 3y + 2 = 0,

(b) ~v = (y− x) i + (y− x3) j. The system of ODEs corresponding tothe vector fields is x′ = y− x, y′ = y− x3. The x-nullclines arethe line y− x = 0, and the y-nullclines are the line y− x3 = 0,

(c) ~v = x(x+ y+ 4) i+ y(x+ 5y) j. The system of ODEs correspond-ing to the vector fields is x′ = x(x + y + 4), y′ = y(x + 5y). Thex-nullclines are the lines x = 0, (x + y + 4) = 0. The y-nullclinesare the line y = 0, (x + 5y) = 0.

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226 ORDINARY DIFFERENTIAL EQUATIONS

Return

Page 978: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 227

10. The x-nullclines are the lines x + y = 0 and y = ±1. The y-nullclines are the line x− y = 0, and x = ±1.

Return

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228 ORDINARY DIFFERENTIAL EQUATIONS

11.

(a)

(b)

(c)

Page 980: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 229

Return

Page 981: Ordinary Differential Equations: A Systems Approach

230 ORDINARY DIFFERENTIAL EQUATIONS

12. Both the linear and nonlinear versions are shown: the linear ver-sion is in black; the nonlinear version in blue. The second figure isan enlargement of the region inside the rectangle on left, and showsmore detail.

5 10 15 20

-0.2

-0.1

0.1

0.2

Return

Page 982: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 231

1. If P is a p× q matrix and M is an m× n matrix then PM is definediff q = m; then it is a p × n matrix. Thus BA is a 2× 2 matrix, andAB is a 3× 3 matrix. A vector ~u ∈ Rd is a d× 1 matrix A~v is a 2× 1matrix (that is, a vector in R2; similarly B~w ∈ R3. As AB is a 3× 3matrix and BA is 2× 2, the products AB~w ∈ R2, and BA~v ∈ R3.Return

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232 ORDINARY DIFFERENTIAL EQUATIONS

3.

(a) The matrix E, when multiplied by A on the right, operates on Aby multiplying the first row of A by 1 and the second row by 0.

(b) When E is multiplied by A on the left, it results in multiplyingcolumn 1 of A by 1, and column 2 of A by 0.

(c) If EA = AE then the corresponding entries of each matrix mustbe equal. In row 2, column 1, the entry of EA is 0; the corre-sponding entry in AE is c. Thus c = 0. The entries of row 1, col-umn 2 of EA and AE are b and 0, respectively. Therefore b = 0as well. The only matrices that commute with E are the diagonal

matrices(

a 00 d

).

Return

Page 984: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 233

5. Let ~v =

(xy

). Then A~v =

(3x− 2yx + 7y

). Thus the matrix equa-

tion(

x′

y′

)= A~v is equivalent to the given system.

Return

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234 ORDINARY DIFFERENTIAL EQUATIONS

7. The system that replaces the given ODE is

x′ = yy′ = zz′ = 2etx− ty + (sin t)z + tan t,

or x′

y′

z′

=

yz

2etx− ty + (sin t)z + tan t

.

We can recast this as~v′ = A~v+~b where~v =

xx′

x′′

, A =

0 1 00 0 1

2et −t sin t

,

and~b =

00

tan t

.

Return

Page 986: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 235

9.

~v′1(t) = 3e3t[

11

]= e3t

[33

], and

[1 22 1

]~v1 = e3t

[1 22 1

] [11

]=

e3t[

1 + 22 + 1

]; thus ~v1(t) is a solution.

~v2(t) is also a solution because ~v′2(t) = −e−t[

1−1

], which is equal

to[

1 22 1

]~v2(t).

We must verify that {~v1(t),~v2(t)} is linearly independent. This is

done by noticing that~v1(0) =[

11

]is not a scalar multiple of~v2(0) =[

1−1

]. Therefore, the general solution of the system is of the form:

[xy

]= C1e3t

[11

]+ C2e−t

[1−1

]where C1, C2 are constants.Return

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236 ORDINARY DIFFERENTIAL EQUATIONS

11. x′ = 2x, thus x = Ce2t. Substituting x(0) = 1 in we have C = 1.y′ = 3y, thus y = De3t. Substituting y(0) = 2 in we have D = 2.

Therefore,[

xy

]=

[e2t

2e3t

].

Return

Page 988: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 237

13.

From Exercise 1 we know the general solution is~v(t) = C1e3t[

11

]+

C2e−t[

1−1

], and we need to determine values for the coefficients so

that ~v(0) =[

20

]. We need to solve C1

[11

]+ C2

[1−1

]=

[20

],

which is equivalent to the pair of equations

C1 + C2 = 2C1 − C2 = 0

Thus C1 = C2 = 1, and[

x(t)y(t)

]=

[e3t + e−t

e3t − e−t

]Return

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238 ORDINARY DIFFERENTIAL EQUATIONS

15. It follows from the first equation, x′ = 0, that x = C. Since x(0) =1, we have C = 1. Thus the second equation can be recast as y′ =−1+ y. The solution of the associated homogeneous equation is yh =Cet. Substituting y = vet in the inhomogeneous equation and usingvariation of constants, we have v′ = −e−t. Thus v = e−t + D, and y =1 + Det. The initial condition y(0) = 1 implies that D = 0. Therefore

the solution of the IVP is the constant,[

x(t)y(t)

]=

[11

].

Return

Page 990: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 239

17. In matrix form,this IVP is:[x′

y′

]=

[2 3−1 −2

] [xy

];[

x(0)y(0)

]=

[−11

]. Let~v1(t) = et

[3−1

]and ~v2(t) =

e−t[

1−1

]then~v′1(t) = et

[3−1

]= et

[2 3−1 −2

] [3−1

]=

[2 3−1 −2

]~v1

~v′2(t) = −e−t[

1−1

]= e−t

[2 3−1 −2

] [1−1

]=

[2 3−1 −2

]~v2

Since ~v1(0) =

[3−1

]is not a scalar multiple of ~v2(0) =

[1−1

], it

follows that ~v1,~v2 are independent solutions of the system and thegiven family of solutions in the general solution.

For the IVP we need to solve

c1

[3−1

]+ c2

[1−1

]=

[−11

],

which is equivalent to the pair of equations

3c1 + c2 = −1−c1 − c2 = 1

Thus c1 = 0, c2 = −1. The solution is[

xy

]= −e−t

[1−1

].

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240 ORDINARY DIFFERENTIAL EQUATIONS

19. In matrix form,this IVP is:[x′

y′

]=

[4 3−2 −1

] [xy

];[

x(0)y(0)

]=

[7−2

]. Let~v1(t) = et

[1−1

]and ~v2(t) =

e2t[

3−2

]then~v′1(t) = et

[1−1

]= et

[4 3−2 −1

] [1−1

]=

[4 3−2 −1

]~v1

~v′2(t) = 2e2t[

3−2

]= e−5t

[4 3−2 −1

] [3−2

]=

[4 3−2 −1

]~v2

Since ~v1(0) =

[1−1

]is not a scalar multiple of ~v2(0) =

[3−2

], it

follows that ~v1,~v2 are independent solutions of the system and thegiven family of solutions in the general solution.

For the IVP we need to solve

c1

[1−1

]+ c2

[3−2

]=

[7−2

],

which is equivalent to the pair of equations

c1 + 3c2 = 7−c1 − 2c2 = −2

Thus c1 = −8, c2 = 5. The solution is[

xy

]= −8et

[1−1

]+ 5e2t

[3−2

].

Return

Page 992: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 241

21. ~v1(0) =(

3−1

)and ~v2(0) =

(1−1

)are linearly independent;

thus the solutions are linearly independent.Return

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242 ORDINARY DIFFERENTIAL EQUATIONS

23. In matrix form,this IVP is:[x′

y′

]=

[2 −14 −2

] [xy

]+

[02

];[

x(0)y(0)

]=

[1−1

].

For the IVP we need to solve[00

]+ c1

[12

]+ c2

[10

]=

[1−1

],

which is equivalent to the pair of equations

c1 + c2 = 12c1 = −1

Thus c1 = − 12 , c2 = 3

2 . The solution is[−t2

2t− 2t2

]− 1

2

[12

]+ 3

2

[2t + 1

4t

]=[

−t2 + 3t + 1−2t2 + 8t + 1

].

Return

Page 994: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 243

25. In matrix form,this IVP is:[x′

y′

]=

[0 11 0

] [xy

]+

[cosh(t)sinh(t)

];[

x(0)y(0)

]=

[22

].

For the IVP we need to solve

(c1 + 0)[

10

]+ c2

[01

]=

[22

],

which is equivalent to the pair of equations

c1 = 2c2 = 2

Thus c1 = 2, c2 = 2. The solution is[

xy

]= (2 + t)

[cosh tsinh t

]+

2[

sinh tcosh t

]= t

[cosh(t)sinh(t)

]+ 2

[et

et

].

Return

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244 ORDINARY DIFFERENTIAL EQUATIONS

27. Let v = y′. Then the linear second-order ODE (4.6) can be re-placed with a linear system of first-order ODEs: y′ = v, v′ = −q(t)y−p(t)v + r(t). The system corresponding to (4.6) is[

y′

v′

]=

[0 1−q(t) −p(t)

] [yv

]+

[0

r(t)

].

Return

Page 996: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 245

1. Since tr (A) = 1 + (−4) = −3 and det(A) = (1)(−4)− (0)(0) =−4, the characteristic equation of A is s2 + 3s− 4 = 0 and it followsthat the eigenvalues are s = −4 and s = 1.

The eigenvectors belonging to s = −4 satisfy the equation A~b = −4~b.

Set~b =

[hk

]. Then A~b =

[h−4k

]=

[−4h−4k

]. Hence h = −4h and

−4k = −4k. these equations reduce to h = 0, so any vector that is a

scalar multiple of~b1 =

[01

]is an eigenvector belonging to −4.

To find an eigenvector belonging to s = 1, we have to solve A~b =[1 00 −4

] [hk

]=

[hk

]. This is equivalent to solving the equations

h = h,−4k = k These equations reduce to k = 0, so it follows that any

scalar multiple of~b2 =

[10

]is an eigenvector belonging to 1.

Return

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246 ORDINARY DIFFERENTIAL EQUATIONS

3. Since tr (A) = 1 + (−4) = −3 and det(A) = (1)(−4)− (0)(1) =−4, the characteristic equation of A is s2 + 3s− 4 = 0 and it followsthat the eigenvalues are s = −4 and s = 1.

The eigenvectors belonging to s = −4 satisfy the equation A~b = −4~b.

Set~b =

[hk

]. Then A~b =

[h + k−4k

]=

[−4h−4k

]. Hence 5h + k = 0

and −4k = −4k. these equations reduce to 5h + k = 0, so any vector

that is a scalar multiple of~b1 =

[1−5

]is a eigenvector belonging to

−4.

To find an eigenvector belonging to s = 1, we have to solve A~b =[1 10 −4

] [hk

]=

[hk

]. This is equivalent to solving the equations

h + k = h,−4k = k These equations reduce to k = 0, so it follows that

any scalar multiple of~b2 =

[10

]is an eigenvector belonging to 1.

Return

Page 998: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 247

5. Since tr (A) = 1 + (−4) = 6 and det(A) = (2)(4)− (−3)(3) =17, the characteristic equation of A is s2 − 6s + 17 = 0. The discrimi-nant of this equation is equal to (−6)2 − 4× 17 = −32 < 0, and thusit has no real roots. The matrix has no real eigenvalues or vectors.Return

Page 999: Ordinary Differential Equations: A Systems Approach

248 ORDINARY DIFFERENTIAL EQUATIONS

7.

(a) If either b 6= 0 or r 6= a, then~k =

[b

r− a

]is a nonzero vector,

and

A~k =

[a bc d

] [b

r− a

]=

[ab + rb− abbc + dr− ad

]=

[rb

bc + dr− ad

]Since r is an eigenvalue of A, we have r2− (a+ d)r+ ad− bc = 0.Thus bc + dr− ad = r2 − ar, and it follows that

A~k =

[rb

r2 − ra

]= r

[b

r− a

]= r~k.

Thus~k is an eigenvector belonging to r.

(b) If b = 0 the characteristic equation is s2 − (a + d)s + ad = 0,and thus s = a, d are eigenvalues. If either c 6= 0 or a 6= d, then

~k =

[a− d

c

]is a nonzero vector, and

A~k =

[a 0c d

] [a− d

c

]=

[a2 − ad

ac− dc + dc

]=

[a2 − ad

ac

]= a

[a− d

c

]= a~k.

Thus~k is an eigenvector belonging to a.

(c) If a = d and b = c = 0, then A = aI is a scalar matrix. It has oneeigenvalue, s = a, and every nonzero vector is an eigenvector.

Return

Page 1000: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 249

9. Let A =

[3 21 2

]so that our system is equivalent to ~v′ = A~v.

Since tr (A) = 5 and det(A) = 4, the characteristic equation is s2 −

5s + 4 = 0 and the eigenvalues are 1 and 4. Let ~b1 =

[hk

]be an

eigenvector belonging to 1. Then 3h+ 2k = h, h+ 2k = k. These equa-

tions reduce to h = −k, so we can take~b1 =

[−11

]Now suppose

that ~b2 =

[hk

]is an eigenvector belonging to 4. Then 3h + 2k =

4h, h + 2k = 4k, so that h = 2k. Hence we can set ~b2 =

[21

]. The

general solution to the matrix equation is

~v = c1et[−11

]+ c2e4t

[21

].

Return

Page 1001: Ordinary Differential Equations: A Systems Approach

250 ORDINARY DIFFERENTIAL EQUATIONS

11. Let A =

[0 10 0

]so that our system is equivalent to ~v′ = A~v.

Since tr (A) = 0 and det(A) = 0, the characteristic equation is s2 = 0,which has the double root 0.

Since A 6= 0I, the matrix A does not have two independent eigenvec-

tors. . Put~c =

[01

]. It is easy to see that A~c 6= 0c; that is,~c is not an

eigenvector. Set

~b = A~c− (0)~c =[

10

].

Then ~v1(t) = e0t~b and ~v2(t) = e0t(t~b +~c) are independent solutions.The general solution is

~v = c1

[10

]+ c2

[t1

]Return

Page 1002: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 251

13. Let ~v =

[xy

]and A =

[1 22 1

]so that our system is equivalent

to v′ = A~v. Since tr (A) = 2 and det(A) = −3, the characteristicequation is s2 − 2s − 3 = 0 and the eigenvalues are 3 and −1. Let

~b1 =

[hk

]be an eigenvector belonging to 3. Then h + 2k = 3h, 2h +

k = 3k. These equations reduce to h = k, so we can take~b1 =

[11

]Now suppose that ~b2 =

[hk

]is an eigenvector belonging to −1.

Then h + 2k = −h, 2h + k = −k, so that h = −k. Hence we can set~b2 =

[1−1

]. The general solution to the matrix equation is ~v =

c1e3t[

11

]+ c2e−t

[1−1

]. or x = c1e3t + c2e−t, y = c1e3t − c2e−t.

Return

Page 1003: Ordinary Differential Equations: A Systems Approach

252 ORDINARY DIFFERENTIAL EQUATIONS

15. Let A =

[−1 1−1 −3

]so that our system is equivalent to ~v′ =

A~v. Since tr (A) = −4 and det(A) = 4, the characteristic equation iss2 + 4s + 4 = 0 and which has the double root −2.

Since A 6= −2I, the matrix A does not have two independent eigen-

vectors. Put~c =[

10

]. It is easy to see that A~c 6= −2c; that is,~c is not

an eigenvector. Set

~b = A~c− (−2)~c =[−1−1

]+

[20

]=

[1−1

].

Then ~v1(t) = e−2t~b and ~v2(t) = e−2t(t~b +~c) are independent solu-

tions. The general solution is ~v = c1e−2t[

1−1

]+ c2e−2t

[t + 1−t

].

Substituting ~v(0) =

[01

], we have c1 + c2 = 0,−c1 = 1. Thus

c1 = −1, c2 = 1. Therefore the solution of the IVP is

~v = −e−2t[

1−1

]+ e−2t

[t + 1−t

]= e−2t

[t

1− t

].

Return

Page 1004: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 253

17. Let A =

[5 −22 1

]so that our system is equivalent to ~v′ =

A~v. Since tr (A) = 6 and det(A) = 9, the characteristic equation iss2 − 6s + 9 = 0 and which has the double root 3.

Since A 6= 3I, the matrix A does not have two independent eigenvec-

tors. Put ~c =

[10

]. It is easy to see that A~c 6= 3c; thus ~c is not an

eigenvector. Set

~b = A~c− (3)~c =[

52

]−[

30

]=

[22

].

Then ~v1(t) = e3t~b and ~v2(t) = e3t(t~b +~c) are independent solu-

tions. The general solution is ~v = c1e3t[

22

]+ c2e3t

[2t + 1

2t

]. Sub-

stituting ~v(0) =

[−1−2

], we have 2c1 + c2 = −1, 2c1 = −2. Thus

c1 = −1, c2 = 1. Therefore the solution of the IVP is ~v = −e3t[

22

]+

e3t[

2t + 12t

]= e3t

[2t− 12t− 2

]. Thus the solution of the IVP is x =

(2t− 1)e3t, y = (2t− 2)e3t.Return

Page 1005: Ordinary Differential Equations: A Systems Approach

254 ORDINARY DIFFERENTIAL EQUATIONS

19. Let y′ = v. The system that replaces the ODE is

y′ = vv′ = −q y− p v

The coefficient matrix of the system is A =

[0 1−q −p

]. Since the

trace of A is equal to −p and the determinant is equal to q, thus thecharacteristic equation is s2 + ps + q = 0.Return

Page 1006: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 255

1. 12+i =

2−i(2+i)(2−i) =

2−i5

Return

Page 1007: Ordinary Differential Equations: A Systems Approach

256 ORDINARY DIFFERENTIAL EQUATIONS

3. Suppose α6 = e2nπi, then α = e2nπi

6 = enπi

3 = cos( nπi

3

)+ i sin

( nπi3

),

for n = 0, 1, 2, 3, 4, 5. The six sixth roots are 1,12+ i√

32

,−12+ i√

32

,−1,

−12− i√

32

, and12+ i√

32

.

Return

Page 1008: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 257

5.

|eλ+iω| = eλ|eiω| = eλ| cos(ω) + i sin(ω)|

= eλ√

cos2(ω) + sin2(ω) = eλ.

Return

Page 1009: Ordinary Differential Equations: A Systems Approach

258 ORDINARY DIFFERENTIAL EQUATIONS

7. Let ~v =

[xy

]and A =

[− 1

2 − 12

12 − 1

2

]so that our system is equiv-

alent to~v′ = A~v. Since tr (A) = −1 and det(A) = 12 , the characteristic

equation is s2 + s + 12 = 0 and the eigenvalues are s =

−1± i2

.

To find a eigenvector~b =

[hk

](h and k will be complex numbers)

belonging to s = −1/2 + i/2, we need to solve[−1/2 −1/21/2 −1/2

] [hk

]= (−1/2 + i/2)

[hk

],

or −h − k = (−1 + i)h, h − k = (−1 + i)k. Each of these equations

reduces to h = ik. Thus we will put~b =

[i1

]. The corresponding

complex-valued solution is

e(−1/2+i/2)t~b = e−t/2[

i(cos(t/2) + i sin(t/2))cos(t/2) + i sin(t/2)

]= e−t/2

([− sin(t/2)cos(t/2)

]+ i[

cos(t/2)sin(t/2)

])The real part and the imaginary part of this solution are themselvessolutions of the system, and in fact form a fundamental set of solu-tions. Therefore, the general solution is

~v = c1e−t/2[− sin(t/2)cos(t/2)

]+ c2e−t/2

[cos(t/2)sin(t/2)

]= e−t/2

[−c1 sin(t/2) + c2 cos(t/2)c1 cos(t/2) + c2 sin(t/2)

].

Return

Page 1010: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 259

9. Let ~v =

[xy

]and A =

[4 −84 −4

]so that our system is equiva-

lent to ~v′ = A~v. Since tr (A) = 0 and det(A) = 16, the characteristicequation is s2 + 16 = 0 and the eigenvalues are ±4i.

To find a eigenvector~b =

[hk

](h and k will be complex numbers)

belonging to s = 4i, we need to solve[

4 −84 −4

] [hk

]= (4i)

[hk

],

or 4h− 8k = (4i)h, 4h− 4k = (4i)k. Each of these equations reduces

to h = (1 + i)k. Thus we will put~b =

[1 + i

1

]. The corresponding

complex-valued solution is

e(4i)t~b =

[(1 + i)(cos(4t) + i sin(4t))

cos(4t) + i sin(4t)

]=

[cos(4t)− sin(4t)

cos(4t)

]+ i[

sin(4t) + cos(4t)sin(4t)

]The real part and the imaginary part of this solution are themselvessolutions of the system, and in fact form a fundamental set of solu-tions. Therefore, the general solution is

~v = c1

[cos(4t)− sin(4t)

cos(4t)

]+ c2

[sin(4t) + cos(4t)

sin(4t)

]=

[(c1 + c2) cos(4t) + (c2 − c1) sin(4t)

c1 cos(4t) + c2 sin(4t)

].

Return

Page 1011: Ordinary Differential Equations: A Systems Approach

260 ORDINARY DIFFERENTIAL EQUATIONS

11.

(a) A~b = A~b = s~b = s~b. Thus ~b is a eigenvector corresponding to s.

(b) ~b and~b are eigenvectors corresponding to two distinct eigenval-ues s and s respectively, thus they are linearly independent.

(c) Suppose c1~h + c2~k = 0. Because~h = 12 (~b +~b) and~k = −i

2 (~b−~b),it follows that

c1 − ic2

2~b +

c1 + ic2

2~b = 0.

We know that~b and ~b are independent; thus c1 − ic2 = 0, c1 +ic2 = 0. Solving these equations, we have c1 = 0, c2 = 0. There-fore~h and~k are linearly independent.

(d) The complex-valued solution corresponding to the eigenvalueλ + iω and eigenvector~h + i~k is

e(λ+iω)t(~h + i~k) = eλt(cos(ωt) + i sin(ωt))(~h− i~k)

= eλt(cos(ωt)~h− sin(ωt)~k + i(sin(ωt)~h + cos(ωt)~k)).

The real part and the imaginary part of this solution are them-selves solutions of the system, and in fact form a fundamentalset of solutions. Therefore, the general solution is

~v(t) = c1eλt(

cos(ωt)~h− sin(ωt)~k)+ c2eλt

(sin(ωt)~h + cos(ωt)~k

),

where c1 and c2 are constants.

Return

Page 1012: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 261

1. The characteristic equation of A =

[ 32

12

12

32

]is s2 − 3s + 2 = 0;

hence the characteristic roots are 1, 2. We can use characteristic vec-

tors ~e1 =

[1−1

]and~e2 =

[11

]belonging to 1 and 2, respectively,

to obtain independent solutions ~v1(t) = et~e1 and ~v2(t) = e2t~e2. Thus

X (t) =[

et e2t

−et e2t

]is a fundamental matrix solution.

Return

Page 1013: Ordinary Differential Equations: A Systems Approach

262 ORDINARY DIFFERENTIAL EQUATIONS

3. The characteristic equation of A =

[2 −45 −2

]is s2 + 16 = 0;

hence the characteristic roots are ±4i. A characteristic vector corre-

sponding to 4i is ~v =

[2

1− 2i

]. The real and imaginary parts of

e4itv =

[2 cos(4t) + 2i sin(4t)

(1− 2i)(cos(4t) + i sin(4t))

]=

[2 cos(4t)

cos(4t) + 2 sin(4t)

]+ i[

2 sin(4t)sin(4t)− 2 cos(4t)

]are two independent solutions. Thus we can take

X (t) =[

2 cos(4t) 2 sin(4t)cos(4t) + 2 sin(4t) sin(4t)− 2 cos(4t)

]as a fundamental matrix solution.Return

Page 1014: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 263

5. X (t) =[

et e2t

−et e2t

]is a fundamental matrix solution of the as-

sociated homogeneous system.

(a) Set ~vp(t) = X (t)~w(t). It follows that X (t)~w′(t) = et[

11

], or

etw′1 + e2tw′2 = et

−etw′1 + e2tw′2 = et.

The solution of this equation is w′1 = 0, w′2 = e−t, or ~w′(t) =[0

e−t

]. Thus ~w(t) =

[0−e−t

], and we have the particular so-

lution

~vp(t) =[

et e2t

−et e2t

] [0−e−t

]=

[−et

−et

].

The general solution is ~vp(t) + ~vh(t), where vh(t) = X (t)~c de-notes the general solution of the associated homogeneous equa-tion.

(b) Set ~vp(t) = X (t)~w(t). It follows that X (t)~w′(t) = et[

1−1

], or

etw′1 + e2tw′2 = et

−etw′1 + e2tw′2 = −et.

The solution of this equation is w′1 = 1, w′2 = 0, or w′(t) =[10

]. Thus ~w(t) =

[t0

], and we have the particular solution

~vp(t) =[

et e2t

−et e2t

] [t0

]=

[tet

−tet

].

The general solution is ~vp(t) + ~vh(t), where vh(t) = X (t)~c de-notes the general solution of the associated homogeneous equa-tion.

(c)[

b1b2

]=

[tet

0

].

Set ~vp(t) = X (t)~w(t). It follows that X (t)~w′(t) = tet[

10

], or

etw′1 + e2tw′2 = tet

−etw′1 + e2tw′2 = 0.

Page 1015: Ordinary Differential Equations: A Systems Approach

264 ORDINARY DIFFERENTIAL EQUATIONS

The solution of this equation is w′1 = t/2, w′2 = te−t/2, or ~w′(t) =[t/2

te−t/2

]. Thus ~w(t) =

[t2/4

− 12 e−t(1 + t)

], and we have the

particular solution

~vp(t) =[

et e2t

−et e2t

] [t2/4

− 12 e−t(1 + t)

]=

[(t2/4− t/2− 1/2)et

(−t2/4− t/2− 1/2)et

].

The general solution is ~vp(t) + ~vh(t), where vh(t) = X (t)~c de-notes the general solution of the associated homogeneous equa-tion.

Return

Page 1016: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 265

7. X (t) =

[2 cos(4t) 2 sin(4t)

cos(4t) + 2 sin(4t) sin(4t)− 2 cos(4t)

]is a funda-

mental matrix solution of the associated homogeneous system. Set

~vp(t) = X (t)~w(t). It follows that X (t)~w′(t) =[−32 cos(4t)

0

], or

2 cos(4t)w′1 + 2 sin(4t)w′2 = −32 cos(4t)(cos(4t) + 2 sin(4t))w′1 + (sin(4t)− 2 cos(4t))w′2 = 0.

The solution of this equation is w′1 = −16 cos2(4t) + 8 sin(4t) cos(4t),w′2 = 8 sin2(4t)− 16 cos(4t) sin(4t)− 8, or

~w′(t) =[−16 cos2(4t) + 8 sin(4t) cos(4t)

8 sin2(4t)− 16 cos(4t) sin(4t)− 8

].

Thus ~w(t) =[− 1

2 (16t + cos(8t) + 2 sin(8t))12 (−8t + 2 cos(8t)− sin(8t))

], and we have the par-

ticular solution

~vp(t) = X (t)~w(t) =[−16t cos(4t)− 8t sin(4t)− cos(4t)− 2 sin(4t)

−20t sin(4t)− 52 cos(4t)

].

The general solution is~vp(t)+~vh(t), where~vh(t) = X (t)~c denotes thegeneral solution of the associated homogeneous equation.Return

Page 1017: Ordinary Differential Equations: A Systems Approach

266 ORDINARY DIFFERENTIAL EQUATIONS

9. Since det(X (1)) = (2)(−1)− (−1)(1) = −1 6= 0, X (1) is nonsin-

gular. Furthermore, dX/dt =[

4t 3t2

−2t −3t2

]= 1

t

[1 −21 4

] [2t2 t3

−t2 −t3

].

Thus X (t) is a matrix solution. A nonsingular matrix solution is afundamental matrix solution.

(a) Set ~vp(t) = X (t)~w(t). It follows that X (t)~w′(t)t =[

t2t

], or

2t2w′1 + t3w′2 = 1−t2w′1 − t3w′2 = 2.

The solution of this equation is w′1 = 3t−2, w′2 = −5t−3, or

~w′(t) =[

3t−2

−5t−3

].

Thus ~w(t) =[−3t−1

5t−2/2

], and we have the particular solution

~vp(t) =[

2t2 t3

−t2 −t3

] [−3t−1

5t−2/2

]=

[ −72 t12 t

].

(b) Set ~vp(t) = X (t)~w(t). It follows that X (t)~w′(t)t =[

t3

−t3

], or

2t2w′1 + t3w′2 = t2

−t2w′1 − t3w′2 = −t2.

The solution of this equation is w′1 = 0, w′2 = t−1, or

~w′(t) =[

0t−1

].

Thus ~w(t) =[

0ln |t|

], and we have the particular solution

~vp(t) =[

2t2 t3

−t2 −t3

] [0

ln |t|

]=

[t3 ln |t|−t3 ln |t|

].

Return

Page 1018: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 267

12. Suppose Q = [~q1,~q2, · · · , qn] and ~d =

d1d2...dn

.

Then Qd = [~q1, q2, · · · ,~qn]

d1d2...dn

= d1~q1 + d2q2 + · · ·+ dn~qn. Thus

any linear combination of columns of Q can be expresses as Q~d, wherethe components of the vector ~d are the coefficients of the linear com-bination.

If Q is nonsingular, then Q~d =~0 if and only if ~d =~0; in other words, ifand only if the only linear combination of columns that is equal to thezero vector is the zero combination. Thus, the columns of a nonsin-gular matrix are linearly independent. Conversely, if the columns ofQ are linearly independent, and Q~d = ~0, then the vector d must rep-resent the coefficients of the zero combination; in other words, ~d =~0.Hence Q is nonsingular.Return

Page 1019: Ordinary Differential Equations: A Systems Approach

268 ORDINARY DIFFERENTIAL EQUATIONS

1. By matrix multiplication, A2 =

0 0 1 00 0 0 10 0 0 00 0 0 0

, A3 =

0 0 0 10 0 0 00 0 0 00 0 0 0

,

and Ak = 0 for k ≥ 4. Hence

eAt = I + At +12(A)2t2 +

16(A)3t3

=

1 t 1

2 t2 16 t3

0 1 t 12 t2

0 0 1 t0 0 0 1

Return

Page 1020: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 269

3.

eAeB = (∞

∑i=0

1i!

Ai)(∞

∑j=0

1j!

Bj)

=∞

∑i=0

∑j=0

1i!

Ai 1j!

Bj

=∞

∑i=0

∑j=0

1i!j!

AiBj

=∞

∑k=0

k

∑i=0

1i!(k− i)!

AiBk−i

=∞

∑k=0

1k!

k

∑i=0

k!i!(k− i)!

AiBk−i.

By the binomial theorem, if A and B commute, then

(A + B)k =k

∑i=0

k!i!(k− i)!

AiBk−i.

If A and B do not commute, this won’t work: for example (A + B)2 =A(A + B) + B(A + B) = A2 + AB + BA + B2. If AB 6= BA, it wouldnot be possible to combine the two middle terms to get (A + B)2 =A2 + 2AB + B2.

It follows that for commuting square matrices A and B,

eAeB =∞

∑k=0

1k!(A + B)k = eA+B.

Page 1021: Ordinary Differential Equations: A Systems Approach

270 ORDINARY DIFFERENTIAL EQUATIONS

5. A(t) =

[t 10 2t

], dA/dt =

[1 00 2

]. Thus A(t) · dA/dt =[

t 20 4t

], but dA/dt · A(t) =

[t 10 4t

].

Page 1022: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 271

7. A2 =

0 0 ac0 0 00 0 0

,and Ak = 0 for k ≥ 3.

(a)

eAt = I + At +12(A)2t2

=

1 at bt + 12 act2

0 1 ct0 0 1

(b) e(A+λI)t = eλIteAt = eλteAt =

eλt ateλt (bt + 12 act2)eλt

0 eλt cteλt

0 0 eλt

.

Now let A =

0 3 10 0 −40 0 0.

Then the ODE is the same as ~v = (A−

2I)~v The solution is e(A−2I)t~c. Substitute the initial value we have~c = 301

. Thus

xyz

= e(A−2I)t

301

.

= e−2t

1 3t t− 6t2

0 1 −4t0 0 1

301

= e−2t

3 + t− 6t2

−4t1

Hence x = e−2t(3 + t− 6t2), y = −4te−2t, and z = e−2t.Return

Page 1023: Ordinary Differential Equations: A Systems Approach

272 ORDINARY DIFFERENTIAL EQUATIONS

9. Since tr A = 0 and det(A) = 1, the characteristic equation is s2 +1 = 0 and by the Cayley-Hamilton Theorem it follows that A2 + I =0. Therefore A2n = (A2)n = (−I)n = (−1)n I. Multiply this by A toget the odd powers: A2n+1 = (−1)n A.

To calculate eAt separate into two sums:

eAt =

(∞

∑n=0

(−1)n

(2n)!t2n

)I +

(∞

∑n=0

(−1)n

(2n + 1)!t2n+1

)A

= cos tI + sin tA =

(cos t + sin t sin t−2 sin t cos t− sin t

)Return

Page 1024: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 273

11. Since tr A = 2 we’ll set B = A− I =

(−2 2−3 2

), and note that

det A = 2. By the Cayley Hamilton theorem, B2 = −2I, and hence

eBt =∞

∑n=0

1(2n)!

B2n +∞

∑n=0

1(2n + 1)!

B2n+1

=∞

∑n=0

(−2)n

(2n)!I +

∑n=0

(−2)n

(2n + 1)!B

=∞

∑n=0

(−1)n(√

2)2n

(2n)!I +

1√2

∑n=0

(−1)n(√

2)2n+1

(2n + 1)!B

= (cos√

2t)I +1√2(sin√

2t)B

=

(cos√

2t−√

2 sin√

2t√

2 sin√

2t− 3√

2sin√

2t cos√

2t +√

2 sin√

2t

)

Noting that A = I + B, and that I and B commute, we have

eAt = eteBt = et

(cos√

2t−√

2 sin√

2t√

2 sin√

2t− 3√

2sin√

2t cos√

2t +√

2 sin√

2t

)

Return

Page 1025: Ordinary Differential Equations: A Systems Approach

274 ORDINARY DIFFERENTIAL EQUATIONS

13. Since tr A = −2, we’ll put

B = A + I =[−2 −2

2 2

].

Then tr B = 0 and det B = 0. It follows (from the Cayley-Hamiltontheorem) that B2 = Z, the zero matrix and hence

eBt = I + Bt =[

1− 2t −2t2t 1 + 2t

].

Therefore,

eAt = e−teBt = e−t[

1− 2t −2t2t 1 + 2t

].

Return

Page 1026: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 275

15. A2 =

0 0 10 0 00 0 0

,and Ak = 0 for k ≥ 3. Hence

eAt = I + At +12(A)2t2

=

1 t 2t + 12 t2

0 1 t0 0 1

Return

Page 1027: Ordinary Differential Equations: A Systems Approach

276 ORDINARY DIFFERENTIAL EQUATIONS

17. Write the system in the form ~v′ = A~v + ~f (t), where

~v =

(xy

), A =

(1 01 −1

), and ~f (t) =

(et

cosh t

).

Noting that tr A = 0 and det A = −1, we have A2 = (det A)I = I soA2n = I and A2n+1 = A. Thus

eAt =

(∞

∑n=0

t2n

(2n)!

)I +

(∞

∑n=0

t2n+1

(2n + 1)!

)A

= (cosh t)I + (sinh t)A

=

(cosh t + sinh t 0

sinh t cosh t− sinh t

).

Applying variation of constants, put ~v = eAt~w. We get ~v′ = AeAt~w +

eAt~w′ so it follows that eAt~w′ = ~f . To find ~w, multiply ~f by (eAt)−1 =e−At and integrate:

~w =∫ (

cosh t− sinh t 0− sinh t cosh t + sinh t

)(et

cosh t

)dt

=∫ ( et(cosh t− sinh t)

−et sinh t + cosh2 t + cosh t sinh t

)dt =

∫ (11

)dt

= t(

11

)+~c

(To get the second-to-last equality, substitute sinh t = 12 (e

t − e−t) andcosh t = 1

2 (et + e−t).) Finally, we have the general solution:

~v = eAt~w =

((cosh t + sinh t)(t + c1)

sinh t(t + c1) + (cosh t− sinh t)(t + c2)

).

Return

Page 1028: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 277

19. Let A =

[2 −14 −2

]and note that A2 = 0. (The trace and the

determinant of A are both 0.) Hence eAt = I + At. Turning to themethod of variation of constants, set ~v = eAt~w. Our system, which isequivalent to

~v′ = A~v + e2t[

12

],

reduces to ~w′ = e−Ate2t[

12

], or

~w′ = e2t(I − At)[

12

]= e2t

[1− 2t t−4t 1 + 2t

] [12

]= e2t

[12

].

Integrating, we get

~w =12

e2t[

12

]+

[c1c2

]The general solution is

~v = eAt~w = (I + At)~w

=

[1 + 2t −t

4t 1− 2t

] [ 12 e2t + c1e2t + c2

]=

12

e2t[

12

]+ c1

[1 + 2t

4t

]+ c2

[−t

1− 2t

]Return

Page 1029: Ordinary Differential Equations: A Systems Approach

278 ORDINARY DIFFERENTIAL EQUATIONS

21.

Page 1030: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 279

1. A = C⊕ [−1], where C =

(2 1−1 0

). Thus eAt = eCt ⊕ [e−t]. To

determine eCt put B = C − 12 (tr C)I = C − I =

(1 1−1 −1

). Since

tr B = det B = 0, it follows from the Cayley-Hamilton theorem thatB2 = 0 and hence eBt = I + Bt. Therefore,

eCt = et(I + Bt) =(

et(1 + t) tet

−tet et(1− t)

).

and

eAt =

et(1 + t) tet 0−tet et(1− t) 0

0 0 e−t

Return

Page 1031: Ordinary Differential Equations: A Systems Approach

280 ORDINARY DIFFERENTIAL EQUATIONS

3.

Page 1032: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 281

5.

Page 1033: Ordinary Differential Equations: A Systems Approach

282 ORDINARY DIFFERENTIAL EQUATIONS

7.

Page 1034: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 283

9. Decompose A as the direct sum [−2]⊕C⊕ [1], where C =

[0 −41 0

].

Since tr C = 0 and det C = 4, we have

eCt = (cos 2t) +12(sin 2t)C =

[cos 2t −2 sin 2t

12 sin 2t cos 2t

].

Therefore

eAt = [e−2t]⊕ eCt ⊕ [et] =

e−2t 0 0 0

0 cos 2t −2 sin 2t 00 1

2 sin 2t cos 2t 00 0 0 et

.

Return

Page 1035: Ordinary Differential Equations: A Systems Approach

284 ORDINARY DIFFERENTIAL EQUATIONS

11. The eigenvalues of A are the roots of f (s) = 0: s = 0 (a doubleroot) and s = 2. There are two independent eigenvectors belonging

to s = 0: ~b1 =

201

and ~b2 =

110

. Thus the algebraic mul-

tiplicity of this eigenvalue matches its geometric multiplicity, and

thus A is semisimple. We will take ~b1 =

1−12

as an eigenvec-

tor belonging to 2. Let P be the matrix with columns ~b1,~b2,~b3, andlet D = diag (0, 0, 2). Then AP = PD, so A = PDP−1. Taking theexponential,

eAt = PDP−1 =

12 (3− e2t) 1

2 (e2t − 1) e2t − 1

12 (e

2t − 1) 12 (3− e2t) 1− e2t

1− e2t e2t − 1 2e2t − 1

.

Return

Page 1036: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 285

13. The eigenvalues of A are the zeros of f (s): s = −1± i and s = −2(a double root). Since the eigenvalue −2 has algebraic multiplicity 2,we start by row-reducing A + 2I to get

1 0 0 − 34

0 1 0 − 34

0 0 1 −10 0 0 0

.

We find that~b1 =

343411

. is an eigenvector of A belonging to −2 and

that the geometric multiplicity of −2 is 1. Hence A is not semisimple.Let

B = (A + 2I)2 =

6 −22 16 −44 −12 8 −24 −12 8 −24 −12 8 −2

.

A vector~c 6=~0 is a belongs to the generalized eigenspace of A belong-ing to −2 if and only if B~c =~0. Row reduction applied to B yields

R =

1 0 −1 1

40 1 −1 1

40 0 0 00 0 0 0

.

Let~b2 =

1110

, and note that R~b2 = ~0. Thus B~b2 = ~0 too, and since

S1 = {~b1,~b2} is linearly independent, S1 is a basis for the generalizedeigenspace of A belonging to −2. Notice that

(A + 2I)~b2 =

−6−6−8−8

= −8~b1.

Thus A~b2 = 2~b2 − 8~b1.

Page 1037: Ordinary Differential Equations: A Systems Approach

286 ORDINARY DIFFERENTIAL EQUATIONS

Let C = A − (−1 + i)I =

3− i −8 −2 5

2 −3− i −6 62 −2 −9− i 82 −2 −8 7− i

. Apply

row reduction to C to get

R1 =

1 0 0 − 3

2 −12 i

0 1 0 −10 0 1 −10 0 0 0

.

To obtain an eigenvector of A we need a vector~b3 such that C~b3 =~0,

or equivalently, R1~b3 = ~0. We will take ~b3 =

32 +

12 i

111

. Let ~b4 be

the complex conjugate of~b3; it will be an eigenvector belonging to theconjugate eigenvalue −1− i. The matrix

P =

34 1 3

2 +12 i 3

2 −12 i

34 1 1 11 1 1 11 0 1 1

has as its columns the vectors ~b1,~b2,~b3,~b4. Thus the columns of APare −2~b1, −2~b2 − 8~b1, (−1 + i)~b3, (−1− i)~b4. Let

D =

−2 −8 0 00 −2 0 00 0 −1 + i 00 0 0 −1− i

,

and consider what happens when we form a product MD. where Mis any four-column matrix, with columns ~m1, . . . , ~m4. The four columnsof MD will be −2~m1, −2~m1 − 8~m2, (−1 + i)~m3, and (−1− i)~m4, re-spectively. Applying this rule to M = P we get PD = AP. ThereforeA = PDP−1 and we have accomplished the objective of finding an E-T-E matrix D that is conjugate to A. (As a direct sum of a 2× 2 matrixand a diagonal matrix, D is E-T-E.)

Let H =

[−2 −80 −2

], and K =

[ 32 +

12 i 0

0 32 −

12 i

], so that D = H⊕

K. We have H = −2I + N, where N =

[0 −80 0

]is nilpotent, with

Page 1038: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 287

N2 = 0. Thus eHt = e−2teNt = e−2t(I + Nt) =

[e−2t −8te−2t

0 e−2t

].

Since K = diag (−1+ i,−1− i), we have eKt = diag (e(−1+i)t, e(−1−i)t).Thus

eDt = eHt ⊕ eKt =

e−2t −8te−2t 0 0

0 e−2t 0 00 0 e(−1+i)t 00 0 0 e(−1−i)t

,

and

eAt = PeDtP−1 =

e−t

cos t + 3 sin t 3 cos t− 3e−t − 11 sin t

2 sin t 4 cos t− 3e−t − 6 sin t2 sin t 4 cos t− 4e−t − 6 sin t2 sin t 4 cos t− 4e−t − 6 sin t

4(e−t − cos t) + 8 sin t− 6te−t cos t− e−t + 6te−t

4(e−t − cos t)− 4 sin t− 6te−t cos t− e−t + 6te−t

5e−t − 4 cos t + 4 sin t− 8te−t cos t− e−t + 8te−t

4(e−t − cos t) + 4 sin t− 8te−t cos t + 8te−t

Return

Page 1039: Ordinary Differential Equations: A Systems Approach

288 ORDINARY DIFFERENTIAL EQUATIONS

15. Put ~v =

xyz

, and let A =

−2 1 21 −1 2−2 2 4

be the coefficient

matrix, so that our system is

~v′ = A~v +

1−11

.

In Exercise 11 we found that

eAt =

12 (3− e2t) 1

2 (e2t − 1) e2t − 1

12 (e

2t − 1) 12 (3− e2t) 1− e2t

1− e2t e2t − 1 2e2t − 1

.

Substitute ~v = eAt~w to get

eAt~w′ + AeAt~w = AeAt~w +

1−11

.

Thus

~w′ = e−At

1−11

=

12 (3− e−2t) 1

2 (e−2t − 1) e−2t − 1

12 (e−2t − 1) 1

2 (3− e−2t) 1− e−2t

1− e−2t e−2t − 1 2e−2t − 1

1−11

=

1−11

.

It follows that ~w =

t−tt

+~c, and hence

~v = eAt~w = t

1−11

+ eAt~c

Finally, set t = 0 for the initial condition to get ~v(0) =~0 = eAt~c. Since

eAt is nonsingular it follows that~c =~0. Thus ~v = t

1−11

.

Return

Page 1040: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 289

1. Let ~v =

[xy

]and A =

[2 11 2

]so that the IVP is equivalent

to ~v′ = Av;~v(0) =

[10

]. Since tr (A) = 4 and det(A) = 3, the

characteristic equation is s2 − 4s + 3 = 0 and the characteristic roots

are 3 and 1. Let ~b1 =

[hk

]be a characteristic vector belonging to

3. Then 2h + k = 3h, h + 2k = 3k. Each of these equations reduces to

h = k, so we can take~b1 =

[11

]Now suppose that~b2 =

[hk

]is

a characteristic vector belonging to 1. Then 2h + k = h, h + 2k = k,

so that h = −k. Hence we can set~b2 =

[1−1

]. The general solution

to the matrix equation is ~v = c1e3t[

11

]+ c2et

[1−1

]. Substituting

~v(0) =[

10

], we have c1 + c2 = 1, c1 − c2 = 0. Thus c1 = 1/2, c2 =

1/2. Therefore the solution of the IVP is ~v = 12 e3t

[11

]+ 1

2 et[

1−1

],

or x = 12 e3t + 1

2 et, y = 12 e3t − 1

2 et.Return

Page 1041: Ordinary Differential Equations: A Systems Approach

290 ORDINARY DIFFERENTIAL EQUATIONS

2.

(a) Let ~v =

[xy

]and A =

[1 −25 −1

]so that our system is equiv-

alent to ~v′ = A~v. Since tr (A) = 0 and det(A) = 9, the char-acteristic equation is s2 + 9 = 0 and the characteristic roots are±3i.

To find a characteristic vector~b =

[hk

](h and k will be complex

numbers) belonging to s = 3i, we need to solve[

1 −25 −1

] [hk

]=

(3i)[

hk

], That is h− 2k = 3ih, 5h− k = 3ik. These will all re-

duce to 5h = (3i + 1)k. Thus we will put~b =

[3i + 1

5

]. The

corresponding complex solution of ~v′ = A~v is

e(3i)t~b =

[(3i + 1)(cos(3t) + i sin(3t))

5 cos(3t) + 5i sin(3t)

]=

[cos(3t)− 3 sin(3t)5 cos(3t)

]+ i[

sin(3t) + 3 cos(3t)5 sin(3t)

]It follows that[

cos(3t)− 3 sin(3t) sin(3t) + 3 cos(3t)5 cos(3t) 5 sin(3t)

]is a fundamental matrix solution.

(b) Let ~v =

[xy

]and A =

[1 −11 1

]so that our system is equiva-

lent to ~v′ = A~v. Since tr (A) = 2 and det(A) = 2, the character-istic equation is s2 − 2s + 2 = 0 and the characteristic roots are1± i.

To find a characteristic vector~b =

[hk

](h and k will be complex

numbers) belonging to s = 1+ i, we need to solve[

1 −11 1

] [hk

]=

(1 + i)[

hk

], That is h− k = (1 + i)h, h + k = (1 + i)k. Each if

these equations reduces to h = ik. Thus we will put~b =

[i1

].

Page 1042: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 291

The corresponding complex solution of ~v′ = A~v is

e(1+i)t~b = et[

i(cos(t) + i sin(t))cos(t) + i sin(t)

]= et

([− sin(t)cos(t)

]+ i[

cos(t)sin(t)

])Thus a fundamental matrix solutions is[

−et sin(t) et cos(t)et cos(t) et sin(t)

]

(c) Let A =

[1 20 1

]so that our system is equivalent to ~v′ = A~v.

Notice that A = I + B, where B =

[0 20 0

], that I and B com-

mute, and that B2 = 0. As with any homogeneous linear systemof ODEs of the form v′ = A~v, the matrix exponential eAt is afundamental solution. We can use our observations about thisspecial matrix to compute eAt as follows:

eAt = e(I+B)t = et eBt

because I and B commute.

eBt = I + Bt

because B2 = 0. Thus

eAt = et(I + Bt) =[

et 2tet

0 et

]is a fundamental matrix solution.

(d) ~x′ = A~x, where

A =

1 2 30 2 30 0 3

.

We will use the fact that this system is actually uncoupled. Let

~x =

xyz

; then the system is the same as

x′ = x + 2y + 3zy′ = 2y + 3zz′ = 3z

Page 1043: Ordinary Differential Equations: A Systems Approach

292 ORDINARY DIFFERENTIAL EQUATIONS

Solving the third equation (which does not involve x or y), weget z = c3e3t, where c3 is a constant. Let’s substitute this ex-pression for z in the second equation to obtain y′ = 2y + 3c3e3t.This equation does not involve x or z; we can solve it and gety = c2e2t + 3c3e3t; c2 is another constant. Finally, substitute theseexpressions for y and z in the first equation. The resulting equa-tion, x′ = x + 2(c2e2t + 3c3e3t) + 3c3e3t = x + 2c2e2t + 9c3e3t.Solve it we obtain x = c1et + 2c2e2t + 9

2 c3e3t, where c1 is a thirdconstant. The general solution is x

yz

=

et 2e2t 92 e3t

0 e2t 3e3t

0 0 e3t

c1c2c3

and a fundamental matrix solution is et 2e2t 9

2 e3t

0 e2t 3e3t

0 0 e3t

Return

Page 1044: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 293

3. First we need find a particular solution. Suppose X (t)v is a solu-

tion, where ~v =

[v1v2

]. Substituting this expression for

[xy

]in the

system[

xy

]′= A

[xy

]+

[tan(2t)

0

], we have, after some cancel-

ing,X~v′ =[

tan(2t)0

]. That is

{(2 sin 2t− cos 2t)v′1 + (sin 2t + 2 cos 2t)v′2 = tan(2t)

cos(2t)v′1 − sin(2t)v′2 = 0

These equations can be solved to obtain v′1 =sin2(2t)2 cos(2t)

, v′2 = −14

cos(2t).

After integrating, we have v1 =14

ln(

1 + sin(2t)cos(2t)

)− 1

4sin(2t), v2 =

− 14 cos(2t). and we get a particular solution:

X (t)~v =

− 12 +

( 12 sin(2t)− 1

4 cos(2t))

ln(

1+sin(2t)cos(2t)

)14 cos(2t) ln

(1+sin(2t)

cos(2t)

) .

Therefore, the general solution is

x = −12+

(12

sin(2t)− 14

cos(2t))

ln(

1 + sin(2t)cos(2t)

)+ C1[2 sin(2t)− cos(2t)] + C2[sin(2t) + 2 cos(2t)]

y =14

cos(2t) ln(

1 + sin(2t)cos(2t)

)+ C1 cos(2t) + C2 sin(2t)

Return

Page 1045: Ordinary Differential Equations: A Systems Approach

294 ORDINARY DIFFERENTIAL EQUATIONS

4.

X (t) = V · E(t)

= [~v1,~v2, · · · , vn] ·

er1t 0 · · · 00 er2t · · · 0· · · · · · · · · · · ·0 0 0 ernt

=

[er1t~v1, er2t~v2, · · · , erntvn

]Then X (t) is the matrix for which column i is the vector erit~vi; wehave noted that such a matrix is a fundamental matrix solution. Itfollows that

eAt = X (t) · [X (0)]−1 = V · E(t) ·V−1.

Return

Page 1046: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 295

5.

Let r be an eigenvalue of a nilpotent matrix M, and let ~v be corre-sponding eigenvector. Since M is nilpotent, Mn is the zero matrixfor some n. Therefore Mn~v = ~0. But Mn~v = rn~v so it follows thatr = 0. Since all eigenvalues of M are equal to zero, and since the traceof any matrix is the sum of its eigenvalues (counted with algebraicmultiplicity), and the determinant of any matrix is the product of itseigenvalues, it follows that the determinant of M is also zero.

Conversely, if the trace and determinant of M are equal to zero, thenthe characteristic equation of M is s2 = 0. The Cayley-Hamilton theo-rem says that every matrix a satisfies its own characteristic equation;hence M2 is is equal to the zero matrix.

Page 1047: Ordinary Differential Equations: A Systems Approach

296 ORDINARY DIFFERENTIAL EQUATIONS

6. Since tr (A) = a + d, B =

(a− (a + d)/2 b

c d− (a + d)/2

)=(

(a− d)/2 bc (d− a)/2

). Thus

det(B) = −14(a− d)2 − bc

=−a2 + 2ad− d2

4− bc

= − a2 + 2add2

4+ ad− bc

= −14

tr (A)2 + det(A)

Return

Page 1048: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 297

1. (E). Because tr (A) = 0 and det(A) = 0, the characteristic equa-tion is s2 = 0 and which has a double root, 0. As this system is degen-erate, there is a stationary line, which narrows the choices down to(D) and (E). As the system corresponding to (D) has a negative eigen-value (indicated by the orbits that approach the stationary line), theonly system that matches is (E).

Here are more details about the system. The vector[

11

]is an eigen-

vector belonging to the eigenvalue 0; thus the line y = x, which isparallel to this vector, is the stationary line. The matrix A is nilpotent,so

eAt = I + At =[

1− t t−t 1 + t

].

put ~c =

[a0

]; then eAt~c = a

[1− t−t

], which traces a line of slope

1 passing through the point (a, 0), parallel to the stationary line. Itsdirection is northeast if a < 0 and southwest if a > 0.Return

Page 1049: Ordinary Differential Equations: A Systems Approach

298 ORDINARY DIFFERENTIAL EQUATIONS

3. (I). Notice first that det(A) = −2, which indicates that the eigen-values are of opposite sign. The phase portrait is therefore a sad-dle, which narrows the choices to (B), (F), (I), or (L). Noting thattr (A) = −1, the characteristic equation is s2 + s− 2 = 0, which has

roots −2, 1. The eigenvectors,[

12

]and

[11

], each have positive

slope. Thus the phase portrait is (I).The stable line is y = 2x, and theunstable line is y = x.Return

Page 1050: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 299

5. (The determinant of A is equal to −4, negative. Hence the eigen-values are opposite in sign and the phase portrait is a saddle. narrowsthe choices to (B), (F), (I), or (L). Noting that tr (A) = 3, the charac-teristic equation is s2 − 3s− 4 = 0, which has roots λ1 = −1, λ2 = 4.

The eigenvectors are[−32

], belonging to λ1, and

[11

], belonging

to λ2 Thus the stable line, −3y = 2x, is of negative slope, and theunstable line, y = x, has positive slope. The phase portrait is (F).Return

Page 1051: Ordinary Differential Equations: A Systems Approach

300 ORDINARY DIFFERENTIAL EQUATIONS

7. Here we have a saddle, because det(A) = −1 is negative, in-dicating eigenvalues of opposite sign. The saddles are choices (B),(F), (I), and (L). Noting thst tr (A) = 0, the characteristic equation iss2 − 1 = 0, which has roots λ1 = −1, and λ2 = 1. The eigenvectors

are[

11

], belonging to λ1; and

[1−1

], belonging to λ2. Thus the

stable line is y = x, the unstable line is y = −x, and hence the phaseportrait is (L).Return

Page 1052: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 301

9. (A). A = 2I. The phase portrait is an unstable proper node; (A) isthe only choice.Return

Page 1053: Ordinary Differential Equations: A Systems Approach

302 ORDINARY DIFFERENTIAL EQUATIONS

11. (D). The characteristic equation is s2 + 2s = 0, and hence theeigenvalues are 0,−2. This is a degenerate system with single 0 root,and hence there is a stationary line, with all orbits directed towardit. The phase portrait (D) is the only one that is consistent with thisinformation.

Here is the complete solution: The eigenvectors are[

11

], belonging

to the 0 root; and[

1−1

]belonging to −1. The stationary line is

y = x, and all non-stationary orbits approach it from the direction

indicated by the eigenvector[

1−1

].

Return

Page 1054: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 303

D EF

A BC

J

KL

G HI

Page 1055: Ordinary Differential Equations: A Systems Approach

304 ORDINARY DIFFERENTIAL EQUATIONS

13. The characteristic equation of the coefficient matrix A =

[1 11 1

]is s2 − 2s = 0, which has roots 0, 2. Hence the system is degenerate.

The eigenvectors are[

1−1

],[

11

]. The stationary line is y = −x.

The other orbits are half-lines, all with slope 1, directed away fromthe stable line.

-4 -2 2 4

-4

-2

2

4

Return

Page 1056: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 305

15. Obtain an integral of the system by integrating the exact ODE

(x + y) dx + x dy = 0

to get x2 + 2xy = c, which we can solve for y to get

y =12

( cx− x)

.

The graph of this equation is a hyperbola with asymptotes x = 0 andy = − 1

2 x. The asymptotes each contain two orbits and the stationarypoint at the origin. On the y-axis, we have y′ = y and x′ = 0; thisindicates that the orbits on this asymptote are directed away from theorigin. On the line y = − 1

2 x, we have x′ = −x and y′ = 12 x; thus the

orbits on this asymptote are directed toward the origin.Return

Page 1057: Ordinary Differential Equations: A Systems Approach

306 ORDINARY DIFFERENTIAL EQUATIONS

17. The coefficient matrix, A =

[2 13 0

], has eigenvalues of oppo-

site signs: −1, 3. Thus the system us a saddle. The eigenvectors are[1−3

], belonging to −1, and

[11

], belonging to 3. Thus the stable

line is y = −3x. The unstable line is y = x.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Return

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SOLUTIONS MANUAL 307

19. The eigenvalues of the coefficient matrix, A =

[−3 5− 5

2 2

], are

−1±5i2 These have negative real part; hence the phase portrait is a sta-

ble spiral node.

-3 -2 -1 1 2 3

-4

-2

2

4

Return

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308 ORDINARY DIFFERENTIAL EQUATIONS

21. The coefficient matrix A =

[1 −10 0

]has characteristic equation,

s2 − s = 0, and eigenvalues 0, 1. The eigenvectors are~b1 =

[11

], be-

longing to 0, and~b2 =

[10

], belonging to 1. The stationary line is

y = x, in the direction of~b1, and the non-stationary orbits are hori-zontal half-lines directed away from the stationary line.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Return

Page 1060: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 309

23. The trace of the coefficient matrix A =

[2 −517 −2

]is 0, and

the determinant is 81 ¿ 0; hence the phase portrait is a center. Thedirection is counterclockwise, as the velocity vector at a point (0, b)

on the positive y-axis is b[−5−2

], pointed southwest. We can find an

integral by integrating the exact equation

(17x− 2y) dx− (2x− 5y) dy = 0;

we obtain 5y2 − 4xy + 17x2

-4 -2 2 4

-7.5

-5

-2.5

2.5

5

7.5

Return

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310 ORDINARY DIFFERENTIAL EQUATIONS

25. The eigenvalues of the coefficient matrix, A =

[3 1−5 −3

], are

−2, 2. Hence the phase portrait is a saddle. The stable and unstable

lines are determined by the eigenvectors,[

1−5

], belonging to −1,

and[

1−1

], belonging to 1. Thus the stable line is y = −5x. The

unstable line is y = −x.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Return

Page 1062: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 311

27. The solution found in example 4.1.1 is x = c1et + c2e3t, y = c1et −c2e3t. After rotation the corresponding solution will be[

cos(π/4) sin(π/4)− sin(π/4) cos(π/4)

] [c1et + c2e3t

c1et − c2e3t

]=√

2[

c1et

−c2e3t

].

In other words, x =√

2c1et, y = −√

2c2e3t = kx3 where k = −c22c3

1.

Return

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312 ORDINARY DIFFERENTIAL EQUATIONS

29. If both eigenvalues of A are negative −r,−s, then the generalsolution is ~v(t) = c1e−rt~b1 + c2e−st~b2, As t→ ∞, ~v(t)→~0.

If the eigenvalues are −r± is, with negative real parts, put B = A−rI. Then tr B = 0 and det B = s2. Therefore the phase portrait of~w′ = B~w is a center. The general solution of ~v′ = A~v is ~v(t) = er teBt,and because r < 0, ~v(t)→~0 as t→ ∞.

In each of the above cases the phase portrait of ~v′ = a~v is a stablenode. The same reasoning shows that if the eigenvalues of A are pos-itive (or have positive real parts) then ~v(t) → ~0 as t → −∞ so thephase portrait is an unstable node.

If one of the eigenvalues is 0, the eigenvector belonging to 0 deter-mines the stationary line. Each point (other than the origin) on thestationary line is an orbit that does not approach the origin; hence thephase portrait is not a stable or unstable node.

If the eigenvalues are pure imaginary numbers, the phase portraitis a center. Each orbit is an ellipse, which does not converge to theorigin. If one of the eigenvalues is positive and the other is negative,the phase portrait is a saddle. The stable line will consist of orbitsdirected toward the origin as t → ∞; the unstable line consists oforbits that converge to the origin as t → −∞, and all other orbits goto ∞ as t → ±∞. Therefore the phase portrait is not a stable node inthese cases, either.Return

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SOLUTIONS MANUAL 313

31.

(a) If y = vx, then by the product rule, y′ = xv′ + vx′. The rightsides of the equations are obtained by factoring.

(c) To determine the stationary points, solve the equation −bv2 +(d − a)v + c = 0. Thus, we have two stationary points if (d −a)2 + 4bc > 0, one if (d− a)2 + 4bc = 0, and none if (d− a)2 +4bc < 0. In the case of two stationary points, when b > 0 the signof −bv2 + (d− a)v + c changes from − to + to − as v increases.This implies a phase diagram that looks like this:� �r r-

Thus the greater of the two stationary points is stable; the lesseris unstable. If b < 0 all arrows in the phase diagram reverse, andthe lesser of the two stationary points is stable.If there is only one stationary point then the quadratic expres-sion−bv2 +(d− a)v+ c does not change sign; the phase portraitis as follows if b > 0 (arrows reversed if b < 0):� �rFinally, if there are no stationary points, the phase diagram is adown arrow if b > 0, or an up arrow if b < 0.

(d) The characteristic equation is s2 − (a + d)s + (ad− bc) = 0. Theroots are 1

2 (a+ d±√(a + d)2 − 4(ad− bc)) = 1

2 (a+ d±√(a− d)2 + 4bc).

The corresponding eigenvectors are[

b12 (−a + d±

√(a− d)2 + 4bc)

].

The slopes of these vectors are

−a + d±√(a− d)2 + 4bc2b

,

the same as the the stationary points of the ODE (4.20).(e) Assume b > 0. If the equation (4.20) has two stationary points,

s1 < s2 then the two lines l1 : cx + dy = s1(ax + by) and l2 :cx + dy = s2(ax + by) are parallel to eigenvectors. Each lineconsists of the origin and two nonstationary orbits convergingeither toward or away from the origin. (If 0 is an eigenvalue, thecorresponding line is a stationary line.)The half-plane on one side of the line ax + by = 0 is divided intothree regions by l1 and l2. The region of the phase plane where

s1 <cx + dyax + by

< s2

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314 ORDINARY DIFFERENTIAL EQUATIONS

corresponds to the middle section of the first phase diagram inpart (c). The slope v of an orbit in this region increases as t in-creases, converging to s2 as t→ ∞ and to s1 as t→ −∞.

In the region wherecx + dyax + by

< s1 the slope v → −∞ when the

orbit crosses the line ax+ by = 0 with a vertical tangent; then thesign of v changes and the slope will decrease from ∞, convergingto s2 as t→ ∞.If the system (4.20) has just one stationary point s1, then the sys-tem (4.19) has a double eigenvalue. The slope of each orbit isdecreasing, but changes sign as it crosses the line ax + by = 0with a vertical tangent. Hence the slope converges to s1 both ast→ −∞ and as t→ +∞!

(f) Again we will assume that b > 0. If the equation (4.20) has nostationary points, then the system (4.18) has no real eigenvec-tors. The orbits will be either clockwise ellipses (if the eigenval-ues are purely imaginary) or clockwise spirals. In either case, theslope will be decreasing until the orbit assumes a vertical direc-tion, and then, as the orbit turns farther, the slope will decreaseagain until a vertical direction is assumed again. This is consis-tent with the picture we see from the phase diagram of (4.20),consisting of one down arrow. The slope changes from +∞ to−∞ twice in each revolution about the origin.

Return

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SOLUTIONS MANUAL 315

33. (D). The characteristic equation is s2 + 4 = 0. The eigenvalues are±2i.The phase portrait is a center, directed clockwise, because orbitsare directed to the right as they cross the positive y-axis. The integralis x2 + 4y2. This confirms the choice, because the orbits are ellipseswith horizontal major axes and vertical minor axes.Return

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316 ORDINARY DIFFERENTIAL EQUATIONS

35. (B). The characteristic equation is s2 + 2s + 17 = 0. The roots are−1± 4i. It’s a stable spiral node, directed clockwise, because it crossesthe positive y-axis traveling to the right.

Let B = A + I. Then tr B = 0 and det B = 16 = 42. Hence eBt =(cos 4t)I + 1

4 (sin 4t)B, and eAt = e−t[(cos 4t)I + 14 (sin 4t)B]. This shows

that our spiral has angular velocity of 4, and crosses the positive axis(or any raw emanating form the origin) in a geometric sequence withratio e−π/2 ≈ 5.8.

Note also that the system ~w′ = B~w has an integral, F(x, y) = x2 −2xy + 5y2. The level curves are ellipses which can be used to drawthe orbits of ~v′ = A~v.Return

Page 1068: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 317

37. (A). The characteristic equation is s2 + 4s + 8 = 0. The roots are−2± 2i. It’s a stable spiral node, directed counterclockwise, becauseit crosses the positive y-axis traveling to the left.Return

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318 ORDINARY DIFFERENTIAL EQUATIONS

1. The characteristic equation of A is s2 + s = 0, or s(s + 1) = 0. Theeigenvalues are 0 and −1. According theorem 8.1, A is stable but notasymptotically stable.

Page 1070: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 319

2. The characteristic equation of A is s2 − 2s− 3 = 0, or (s− 3)(s +1) = 0. The eigenvalues are 3 and −1. According theorem 8.1, A

is not stable. The eigenvector corresponding to 3 is[

11

]. Thus we

have an unbounded solution v = e3t[

11

].

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320 ORDINARY DIFFERENTIAL EQUATIONS

4. The characteristic equation of A is s2 − 2s + 2 = 0. The eigenval-ues are 1± i. According theorem 8.1, A is unstable.

The eigenvector corresponding to 1 + i is[

1i

]. Thus we have an

unbounded solution v = Re(

e(1+i)t[

1i

])= et

[cos(t)− sin(t)

]

Page 1072: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 321

5. The characteristic equation of A is s2 = 0. It has double eigen-value 0. According theorem 8.1, A is unstable. The eigenvector cor-

responding to 0 is[

1−1

]. Thus we have an unbounded solution

v =

[t + 1−t

].

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322 ORDINARY DIFFERENTIAL EQUATIONS

7.

The characteristic equation of A is (s2 + s)(s2 − 2s− 3) = 0. It haveeigenvalues 0,−1,−1, 3. According theorem 8.1, A is unstable. The

eigenvector corresponding to 3 is

0011

. Thus we have an unbounded

solution v = e3t

0011

.

Page 1074: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 323

8. The characteristic equation of A is s2 + 1 = 0. It have

eigenvalues ±i. According theorem 8.1, A is stable but not asymptot-ically stable.

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324 ORDINARY DIFFERENTIAL EQUATIONS

10. The characteristic equation of A is (s2 + 1)(s2 − 1) = 0. It

have eigenvalues ±i,±1. According theorem 8.1, A is unstable. The

eigenvector corresponding to 1 is

0011

. Thus we have an unbounded

solution v = et

0011

.

Page 1076: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 325

11. The characteristic equation of A is (s + 1)(s + 2)(s + 3) = 0. Ithave eigenvalues −1,−2,−3. According theorem 8.1, A is asymptot-ically stable.

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326 ORDINARY DIFFERENTIAL EQUATIONS

13.

The characteristic equation of A is (s2 + 1)(s2 + 1) = 0. It has doubleeigenvalues ±i.

It is easily checked that b =

0100

is not an eigenvector, and we put

c = (A − i I)b =

1−i00

, which is an eigenvector. Thus there is a

solutionv(t) = eit(tc + b)

This solution is unbounded. The same may be said for its real part,

Re(v) =

t cos(t)

cos(t) + t sin(t)00

.

Page 1078: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 327

14. The characteristic equation of A is (s2 + 1)(s2 + 4) = 0. It haveeigenvalues±i,±2i. According theorem 8.1, A is stable, but not asymp-totically stable.

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328 ORDINARY DIFFERENTIAL EQUATIONS

16. If s1 = a i where a 6= 0 is real, is an eigenvalue, then s2 = −a iis also an eigenvalue, According theorem 8.1, A is stable, but notasymptotically stable. Therefore the phase portrait of the system isa center, and thus the solution is periodic.

Page 1080: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 329

18. By theorem 8.1, A is stable, but not asymptotically stable. How-ever, some solutions may not be periodic. Let

A =

0 1 0 0−1 0 0 00 0 0 −20 0 1 0

The characteristic equation of A is (s2 + 1)(s2 + 2) = 0 and hence theeigenvalues are ±i,±

√2i. The vectors

b =

1i00

and c =

001

1/(√

2i)

are eigenvectors belonging to i and

√2i, respectively. The solution

v(t) = eitb + ei√

2tc is bounded but not periodic. To see this, note thatthe real part of this solution is

w =

cos(t)− sin(t)

cos(√

2t)sin(√

2t)/√

2

.

If this solution had a period T then T would have to be simultane-ously a period of cos(t) — thus T = 2Nπ, where N is an integer; anda period of cos(

√2t) — thus T = M

√2π, where M is also an integer.

Hence 2Nπ = M√

2π, and it follows that

2MN

=√

2,

which is impossible, since√

2 is an irrational number.

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330 ORDINARY DIFFERENTIAL EQUATIONS

20. If c could be expressed as the linear combination of the n − 1eigenvectors corresponding the left half-plane eigenvalues, then eAtcwill be bounded as t → ∞. However, the probability of randomlyselecting a vector with this property is equal to 0.

Page 1082: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 331

21. If c can be expressed as a linear combination of eigenvectors be-longing to pure imaginary eigenvalues of A, then eAtc is periodic andthus is bounded on R.

Assume c can not be expressed as a linear combination of eigenvec-tors belonging to pure imaginary eigenvalues of A, but that c can beexpressed as a linear combination of eigenvectors belonging to eigen-values of A, then there are terms belonging to the eigenvalues of theform a + bi, where a 6= 0. If a > 0, limt→∞ eAtc → ∞, If a < 0,limt→−∞ eAtc→ ∞. In either case eAtc can not be bounded.

If c cannot be expressed as a linear combination of eigenvectors thenA has repeated eigenvalues and that eAtc is unbounded.

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332 ORDINARY DIFFERENTIAL EQUATIONS

1. The repelled set consists of those points whose orbits lead backto( 7

2 , 0)

as t → −∞. Because the eigenvalues of A( 7

2 , 0)

are bothpositive, the repelled set contains a small disk centered at

( 72 , 0)

.

Inspection of figure 4.17 shows that the entire fourth quadrant is asubset of the repelled set, which extends into the first quadrant aswell. The upper boundary is repelled set (part of the separatrix) ofthe saddle point at (2,1).Return

Page 1084: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 333

3. This system is linear. The characteristic equation is s2 + 2s + 2 =0, and the eigenvalues are −1± i, which are located in the left half-plane. Therefore the phase portrait is a stable spiral node. The originis the only stationary point; it is asyptotically stable and its attracted

set is the entire plane.

-4 -3 -2 -1 1 2 3 4

-4

-2

2

4

.Return

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334 ORDINARY DIFFERENTIAL EQUATIONS

5. This system is linear; its characteristic equation is s2 + 35 = 0.The eigenvalues, ±

√35 i, are pure imaginary, so the phase portrait is

a center. The origin is the only stationary point, and it is neutrallystable.

-4 -3 -2 -1 1 2 3 4

-4

-2

2

4

.Return

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SOLUTIONS MANUAL 335

7. To obtain the stationary points, solve the equations

y(1− x2 − y2) = 0−x(1− x2 − y2) = 0.

We find that the origin is a isolated stationary point, and that everypoint on the unit circle x2 + y2 = 1 is stationary. We can performanalysis of the isolated stationary point by linear approximation, butnot of the non-isolated points on the circle.

The derivative matrix is

A(x, y) =

[∂

∂x (y(1− x2 − y2)) ∂∂y (y(1− x2 − y2))

∂∂x (−x(1− x2 − y2)) ∂

∂y (−x(1− x2 − y2))

]

=

[−2xy 1− x2 − 3y2

−1 + 3x2 + y2 2xy

].

A(0, 0) =

[0 1−1 0

]has eigenvalues ±i. Thus linear analysis is in-

conclusive for this stationary point, which could be a center or a spi-ral. Although the linear approximation is a center, the nonlinearitycould disrupt the closed orbits of the center, as we have seen in Fig-ure 4.16.

To analyze this stationary point, consider the ODE that is satisfied bythe orbits:

x(1− x2 − y2) dx + y(1− x2 − y2) dy = 0.

Divide through by the factor (1 − x2 − y2) to obtain the separableequation x dx + y dy = 0, and thus an integral x2 + y2 = C. It followsthat the stationary point at the origin is a center after all, and is thusneutrally stable, for the non-stationary orbits are circles. The orien-tation is determined by the direction of the orbits as they cross thepositive y-axis, which is to the right when x2 + y2 < 1 (inside the sta-tionary circle), and to the left when x2 + y2 > 1. Thus the orbits areclockwise inside the stationary circle, and counterclockwise outsideit. The stationary points on the unit circle are unstable, because theorbit of any point close to, but not on, the unit circle will be a circlethrough that point with center at the origin. Hence it will not stay ina circle of radius 1, centered at that point.

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336 ORDINARY DIFFERENTIAL EQUATIONS

-1.5 -1 -0.5 0.5 1 1.5

-1.5

-1

-0.5

0.5

1

1.5

.Return

Page 1088: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 337

9. Solve the equations

x(2− x− y) = 0y(x− y) = 0.

to obtain the stationary points (0, 0), (2, 0), (1, 1).

The derivative matrix is

A(x, y) =

[∂

∂x x(2− x− y) ∂∂y x(2− x− y)

∂∂x y(x− y) ∂

∂y y(x− y)

]

=

[2− 2x− y −x

y x− 2y

].

A(0, 0) =

[2 00 0

]has eigenvalues 2, 0. Because A(0, 0) is singular,

linear approximation is not conclusive. However, we note that thex- and y-axes are invariant sets. On the x-axis, the system reduces toa single logistic ODE, x = x(2 − x), found by setting y = 0 in thefirst equation. This shows that the origin is unstable, since the orbitson the x-axis lead away from it. Similarly, the system reduces to theODE y′ = −y2 on the y-axis; orbits are directed downward, towardthe origin on the positive y-axis, and away from it on the negativey-axis. Therefore the origin is unstable.

A(2, 0) =

[−2 −20 2

]has eigenvalues ±2, and therefore (2, 0) is

a saddle point with attracted set tangent to the eigenvector (1, 0) be-longing to−2 and with repelled set tangent to the eigenvector (1,−2)belonging to 2. It is also unstable.

The characteristic equation of A(1, 1) =[−1 −11 −1

]is s2 + 2s + 2 =

0. The roots, s = −1± i, are in the left half plane, and thus (1, 1) isa stable and a asymptotically stable equilibrium point. Orbits spiraltoward (1,1); to determine the direction of the spiral we note that forx = 1, we have x′ = 1− y. Hence orbits cross the vertical line x = 1from right to left when y > 1 and from left to right when y < 1. Thespiral is counterclockwise. The attracted set is the first quadrant. Tosee this, note that no orbit can cross either axis, since the positive axesare orbits. Therefore the any orbit that leads to the point (1,1) mustbe confined to the first quadrant. The portion of the x-nullcline in the

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338 ORDINARY DIFFERENTIAL EQUATIONS

first quadrant is the line y = −x + 2. Above this line, x′ > 0, andbelow it x′ < 0. The y-nullcline is the line y = x. Above it, y′ < 0,and below it y′ < 0. Thus no point in the first quadrant is attractedto either the origin or the point (2,0). Also, if y is large, then y′ < 0,while if x is large, x′ < 0 All orbits in the first quadrant thus spiraltoward (1,1).

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

.Return

Page 1090: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 339

11. Notice first that the x- and y-axes are invariant. On the x-axis, thefirst equation reduces to x′ = x(x2 − 10); hence the following phasediagram appears on the x-axisr r r

−√

10 0√

10

- �� -

The second equation reduces to y′ = −3y on the y-axis, and hencethere is an upward arrow on the negative y-axis, and a downwardarrow on the positive y-axis.

To find all of the stationary points, solve the equations

x(x2 + y2 − 10) = 0y(xy− 3) = 0.

In addition to (0, 0) and (±√

10, 0), we find the stationary points(3, 1), (1, 3) (−1,−3), and (−3,−1).

The derivative matrix is

A(x, y) =

[∂

∂x x(x2 + y2 − 10) ∂∂y x(x2 + y2 − 10)

∂∂x y(xy− 3) ∂

∂y y(xy− 3)

]

=

[3x2 + y2 − 10 2xy

y2 2xy− 3

].

As we might expect, the origin is asymptotically stable, since A(0, 0) =[−10 0

0 −3

]has eigenvalues −10,−3.

A(±√

10, 0) =[

20 00 −3

]has eigenvalues 20,−3. Therefore (±

√10, 0)

are saddle points. Their attracted sets are curves that is tangent to theeigenvector (0, 1) belonging to −3, and their repelled sets are inter-vals on the x-axis, as shown in the phase diagram that we consideredearlier.

The characteristic equation of A(1, 3) = A(−1,−3) =

[2 69 3

]is

s2− 5s− 48 = 0. Hence the stationary points±(1, 3) are saddle points.The eigenvalues are 1

2 (5±√

217). The attracted set of each stationarypoint is a curve tangent to the eigenvector (1−

√217, 12) belonging

to 12 (5−

√217) and with repelled set is a curve tangent to the eigen-

vector (1 +√

217, 12) belonging to 12 (5 +

√217).

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340 ORDINARY DIFFERENTIAL EQUATIONS

At the stationary points±(3, 1), the derivative matrix is A =

[18 61 3

].

The characteristic equation is s2 − 21s + 48 = 0, and hence the eigen-values, 1

2 (21±√

249), are positive. Therefore (3, 1), (−3,−1) are un-stable points.

The incoming separatrices of the saddle points at (1,3) and (√

10, 0)connect with the unstable node at (3,1). These separatrices form theright boundary of the attracted set of the origin. The left boundary isa symmetric curve in the second and third quadrants.

-4 -2 2 4

-4

-3

-2

-1

1

2

3

4

.Return

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SOLUTIONS MANUAL 341

13. Solve the equations

x(1− y) = 0y(x− 1) = 0.

to obtain the stationary points (0, 0), (1, 1).

The axes are invariant sets of this system. On the x-axis, the systemreduces to the ODE x′ = x, and on the y-axis, it reduces to y′ = −y.Thus the nonstationary orbits on the x-axis are directed away fromthe origin, and the nonstationary orbits on the y-axis are directed to-ward the origin. It follows that the origin is a saddle point, with they-axis as attracted set and the x-axis as repelled set.

The derivative matrix is

A(x, y) =

[∂

∂x x(1− y) ∂∂y x(1− y)

∂∂x y(x− 1) ∂

∂y y(x− 1)

]

=

[1− y −x

y x− 1

].

A(0, 0) =

[1 00 −1

]has eigenvalues 1,−1, confirming the saddle

point at the origin.

A(1, 1) =

[0 −11 0

]. The characteristic equation is s2 + 1 = 0, and

A(1, 1) thus has eigenvalues ±i. Because this is a nonlinear system,we cannot be sure that (1, 1) is a center because the nonlinearity maydisrupt the closed orbits that the linear approximation has.

The orbits of this system are curves that satisfy then ODE

(1− x)y dx + (1− y)x dy.

This ODE is separable. Divide through by xy to obtain (x−1− 1) dx +(y−1 − 1) dy− 0. We can then integrate to obtain solutions

F(x, y) = ln(x) + ln(y)− (x + y) = C.

The integral, F, has a critical point at (1, 1). The discriminant,

∂2F∂x2

∂2F∂y2 −

(∂2F

∂x∂y

)2

=

(−1x2

)(−1y2

),

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342 ORDINARY DIFFERENTIAL EQUATIONS

is positive at (x, y) = (1, 1). It follows that F has a local maximum atthe stationary point (1, 1), and hence the level curves of F near thatstationary point are closed. This verifies that (1,1) is indeed a center.The orbits are are oriented counterclockwise, to be consistent with thesaddle structure at the origin.

-1 -0.5 0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

.Return

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SOLUTIONS MANUAL 343

15.

The derivative matrix is

A(x, y) =

[∂

∂x [−y + kx(1− x2 − y2)] ∂∂y [−y + kx(1− x2 − y2)]

∂∂x [x + ky(1− x2 − y2)] ∂

∂y [x + ky(1− x2 − y2)]

]

=

[k− 3x2k− ky2 −1− 2kxy

1− 2kxy k− kx2 − 3y2k

].

The eigenvalues of A(0, 0) =

[k −11 k

]are k ± i. Because they de-

pend on k, the stability also depends on k. If k < 0, the origin is a sta-ble spiral node, and if k > 0 the origin is an unstable spiral node. Ineither case, the spirals are counterclockwise, because the orbits will bedirected to the left as they cross the positive y-axis. If k = 0 the eigen-values are pure imaginary, which means that linear approximation isinconclusive. However, when k = 0 the system is linear! Hence thephase portrait is a center when k = 0.

We have completed the local analysis at the stationary point of thissystem. We now turn to polar coordinates to see the big picture. Setr2 = x2 + y2; then

2rr′ = 2xx′ + 2yy′

= 2x[−y + kx(1− x2 − y2)] + 2y[x + ky(1− x2 − y2)]

= 2k(x2 + y2)(1− x2 − y2)

= 2kr2(1− r2)

Therefore r satisfies the ODE r′ = kr(1− r2). For k < 0, phase diagramfor this ODE is � �- -s s s , where the stationary pointsare at 0, and ±1. The orbits that start within the unit circle, so thatr < 1, will spiral toward the origin. If r 6= 0, 1, then as t → −∞ theorbits spiral toward the unit circle. Thus the orbit r = 1 is an unstableperiodic orbit.

If k > 0 the phase diagram - -� �s s s for r′ = kr(1− r2)indicates that the periodic orbit r = 1 is stable, because r → 1 ast→ ∞, provided that r 6= 0, 1.

If k < 0 the attracted set of the origin is the set of points inside theunit circle, and if k > 0 the repelled set of the origin is also the set ofpoints inside the unit circle.

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344 ORDINARY DIFFERENTIAL EQUATIONS

-2 2

-2

2k=-0.05 k=0

-2 2

-2

2k=0.05

.Return

Page 1096: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 345

17. Use the identities

rr′ = x x′ + y y′ and r2θ′ = x y′ − y x′

to convert the system to polar coordinates. We obtain an uncoupledsystem

r′ = r(1− r2)

θ′ = 1− cos(θ).

The stationary points are those with polar coordinates r = 0, 1 andθ = 0. Thus the points with cartesian coordinates (0, 0) and (1, 0) arethe only stationary points. To determine the stability, we will con-sider the phase diagrams of the polar ODEs. The phase diagram forr′ = r(1− r2) is �-s s , indicating that the circle r = 1 is an in-variant set, and that all nonstationary orbits converge to it. The phasediagram for θ′ = 1− cos(θ) is - - - -s s s s . . ., where thestationary points are 0, 2π, 4π, . . .. We thus see that every orbit con-verges toward the unit circle. The positive x-axis is an invariant set,and the orbits on that axis form the same phase diagram as the phasediagram just drawn for the variable r. Orbits that do not start on thataxis follow a counterclockwise path that does not cross the positivex-axis, but converges to the point (1, 0). Thus, (1,0) is asymptoticallystable. It is not stable, because orbits starting at points in the upperhalf-plane, no matter how close to (1, 0), must navigate counterclock-wise around the origin and approach (1,0) from the fourth quadrant.See the phase portrait.

-2 -1.5 -1 -0.5 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

.

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346 ORDINARY DIFFERENTIAL EQUATIONS

Return

Page 1098: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 347

1. The stationary points of the system are (0,0), (K, 0), (0, L), and,provided that BC 6= 1, a point (x∗, y∗), where

x∗ = K+BL1−BC and y∗ = L+CK

1−BC .

The point (x∗, y∗) is thus in the first quadrant if BC < 1, and in thethird quadrant (where it would represent negative populations, andhas no biological significance) if BC > 1.

The derivative matrix of the system is

D(x, y) =[

a(K− 2x− By) a B xd C y d(L + Cx− 2y)

]

Thus, D(0, 0) =

[a K 00 d L

], which has positive eigenvalues, in-

dicating that the origin is an unstable node. On the axes, we have

D(K, 0) =[−a K a B K

0 d(L + CK)

]and D(0, L) =

[a(K + BL) 0

d C L −d L

],

indicating that both stationary points are saddles. If we rememberthat K− x∗ + B y∗ = 0 and L + C x∗ − y∗ = 0, it is not hard to see that

D(x∗, y∗) =[−a x∗ a B x∗

d C y∗ −d y∗

]. This matrix has a negative trace, and

in the relevant case where BC < 1, det D(x∗, y∗) = a x∗ y∗(1− B C) >0. It follows that both eigenvalues are negative, and that the station-ary point in the first quadrant is a stable node. In the phase por-

traits shown below, (B, C) =

(2,

13

)and (B, C) =

(2,

23

), respec-

tively. The other parameters are the same for both: (a, d, K, L) =(0.05, 0.05, 2, 1).

2 4 6 8 10 12 14

2

4

6

8

2 4 6 8 10 12 14

2

4

6

8

Return

Page 1099: Ordinary Differential Equations: A Systems Approach

348 ORDINARY DIFFERENTIAL EQUATIONS

3. There are three stationary points: The origin, where both speciesare extinct, the point (C, 0), where the predator is extinct, and the preypopulation is equal to the carrying capacity, and (d, y∗), where y∗ =Ab (C − d). The axes are invariant sets, indicating that if one of thespecies is extinct, it will stay extinct. If x = 0 (the prey is extinct) thenthe ODE for the predator is y′ = −cdy, predicting that the predatorpopulation will decay exponentially. If y = 0 (no predators) the preypopulation follows the logistic growth equation, x′ = Ax(C− x). Bydrawing the single-species phase diagram on each axis, we obtain aframework on which to build the two-species phase portrait:

r r(C, 0)-

?

6

� �

We can see from this diagram that the origin is a saddle point, just asit is in the Lotka-Volterra system. This is confirmed by the derivativematrix,

D(x, y) =[

A(C− 2x)− b y −b xc y c(x− d)

].

Thus D(0, 0) is the diagonal matrix with entries A C and−c d indicat-ing that there is an saddle point at the origin.

For the stationary point on the x-axis,

D(C, 0) =[−AC −bC

0 c(C− d)

],

is upper triangular, so the eigenvalues are the diagonal entries, −ACand c(C− d).

(a) If C − d < 0 then both eigenvalues at (C, 0) are negative, andhence that stationary point is a stable node. Furthermore, y∗

is negative, so the stationary points on the x-axis are the onlyones with biological significance. It follows that y is decreasingthroughout the first quadrant, and that all nonstationary orbitsin the first quadrant converge to (C, 0). The small carrying ca-pacity for the prey means that there will never be enough prey inthis environment to support the predators. In the phase portrait

Page 1100: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 349

that follows, the parameters have been set as to be (A, C, b, c, d) =(0.5, 1500, 1, 0.125, 2000)

500 1000 1500 2000

100

200

300

400

500

(b) When C > d, the third stationary point, (d, y∗), is in the firstquadrant. When the predator population is equal to y∗, thereare just enough predators to limit the prey population to d, theminimum number needed to support any predators at all.The eigenvalues of derivative matrix D(C, 0) are of opposite sign,indicating that there is a saddle point at (C, 0). Furthermore, thecharacteristic polynomial of

D(d, y∗) =[−A d −b dc y∗ 0

].

is s2 + Ad s + bcdy∗, which has positive real eigenvalues when

A2d2 ≥ 4bcdy∗ = 4Acd(C − d); that is,Ad4c≥ C − d. Hence the

stationary point (d, y∗) is a stable node. In the phase portraitthat follows, the parameters have been set to be (A, C, b, c, d) =(0.5, 1500, 1, 0.125, 1000)

500 1000 1500 2000

100

200

300

400

500

Page 1101: Ordinary Differential Equations: A Systems Approach

350 ORDINARY DIFFERENTIAL EQUATIONS

(c) When C− d > Ad4c the stationary point at (d, y∗) is a stable spiral

node, because the real part each eigenvalue, − 12 Ad, is negative.

Orbits are directed counterclockwise. In the following phaseportrait, the parameters have been set to be (A, C, b, c, d) = (0.5, 1500, 2, 0.125, 500)

500 1000 1500 2000

100

200

300

400

500

Return

Page 1102: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 351

1. Suppose that (x1, 0) is asymptotically stable—we will reach a con-tradiction. Then there is a circle C, centered at (x1, 0), in the phaseplane, such that every orbit starting at a point inside C is attracted to(x1, 0). Let (x(t), y(t)) be a solution of (4.37) such that (x(0), y(0)) liesinside C. Because F is an integral, F(x(t), y(t)) is constant. The orbitof this solution is attracted to (x1, 0), so limt→∞(x(t), y(t)) = (x1, 0).By continuity, limt→∞ F(x(t), y(t)) = (x1, 0), and since F is constanton this orbit, F(x(0), y(0)) = F(x1, 0). But (x(0), y(0)) was arbitrarilychosen as a point within C; therefore F(x, y) = F(x1, 0) for all (x, y)inside C. In particular F(x, 0) = F(x1, 0) if |x − x1| < r, where rdenotes the radius of C. Since F(x, 0) = U(x) it follows that U is con-stant on the interval (x1 − r, x1 + r), and hence U′(x) = 0 there. Thiscontradicts our assumption that x1 was an isolated critical point.Return

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352 ORDINARY DIFFERENTIAL EQUATIONS

3. Since U′(x1) = −g(x1) = 0, and U′′(x1) = −g′(x1) > 0, it fol-lows that x1 is a critical point of U(x), and, by the second derivativetest, that U(x1) is a relative minimum. Therefore by proposition 4.4.2,(x1, 0) is a neutrally stable stationary point of the system (4.37).Return

Page 1104: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 353

5.U(x) = −

∫ x

0ax dx = − a

2x2

Potential Phase Portrait

Return

Page 1105: Ordinary Differential Equations: A Systems Approach

354 ORDINARY DIFFERENTIAL EQUATIONS

7.U(x) = −

∫ x

01 dx = −x

Potential Phase Portrait

Return

Page 1106: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 355

9.U(x) = −

∫ x

0(−x2) dx =

13

x3

Potential Phase Portrait

Return

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356 ORDINARY DIFFERENTIAL EQUATIONS

11.U(x) = −

∫ x

0(−4x3) dx = x4

Potential Phase Portrait

Return

Page 1108: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 357

13.U(x) = −

∫ x

0(3s|s|) ds = −x2|x|

There is one stationary point, located at (0,0). As U(x) has a localmaximum at X = 0, the stationary point is a saddle.

Potential Phase Portrait

Return

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358 ORDINARY DIFFERENTIAL EQUATIONS

15.

U(x) = −∫ x

0x2−1

(x2+1)2 dx = −Re∫ x

0

[1

(x+i)2

]dx = Re 1

x+i =x

x2+1

Because u(x) has a local minimum at x = −1 and a local maximum atx = 1, the phase portrait has a center at (−1, 0) and a saddle at (1, 0).

Potential Phase Portrait

Return

Page 1110: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 359

17.U(x) = −

∫ x

0(sin x− 1

2) dx = cos(x) +

12

x

There is a doubly infinite sequence of stationary points (xn, 0) wheresin(xn) =

12 . Thus, xn = . . .− 11

6 π,− 76 π, 1

6 π, 56 π, . . . . These stationary

points alternate between saddles and centers, with a saddle at ( 16 π, 0),

a center at ( 56 π, 0), and so on.

Potential Phase Portrait

Return

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360 ORDINARY DIFFERENTIAL EQUATIONS

19.

U(x) = −∫ x

0

[sin x−

(2π

)x]

dx = cos(x) +(

)x2

has critical points at x = 0,±π2 , with a local maximum at 0 and local

minima at ±π2 . Thus the phase portrait has a saddle at (0,0) and cen-

ters at (π2 , 0). The attracted set of the origin is the same as the repelled

set; all orbits except those attracted to the origin are closed.

Potential Phase Portrait

Return

Page 1112: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 361

1.

L(x, y) = ax2 + bxy + cy2

= a(x +b

2ay)2 + (c− b2

4a)y2

= a(x +b

2ay)2 +

14a

(4ac− b2)y2

Thus if b2 − 4ac < 0, and a > 0, then L(x, y) ≥ 0; equality holds ifand only if a(x + b

2a y)2 = 14a (4ac− b2)y2 = 0, and hence x = y = 0.

Therefore, L(x, y) is positive definite.

Conversely, if a ≤ 0, then L(1, 0) ≤ 0 so L is not positive definite. Ifa > 0 and b2 − 4ac ≥ 0, then L(−b, 2a) = (4ac− b2)a ≤ 0, and againL is not positive definite.Return

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362 ORDINARY DIFFERENTIAL EQUATIONS

3. The function L(x, y) = x2 + y2 is positive definite, and

L′(x, y) = 2x(−y + x3) + 2y(x + y3)

= 2(x4 + y4)

L′(x, y) is also positive definite, therefore L(x, y) is a Lyapunov func-tion for the given system and the origin is unstable.Return

Page 1114: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 363

5.

The function L(x, y) = x2 + y2 is positive definite, and

L′(x, y) = 2x(y− sin x) + 2y(−x− sin y)= −2x sin x− 2y sin y

When (x, y) is in a small disk centered at the origin, L′(x, y) ≈ −2x2−2y2; thus L′(x, y) is negative definite. It follows that L(x, y) is a Lya-punov function for the given system, and that the origin is stable andasymptotically stable.Return

Page 1115: Ordinary Differential Equations: A Systems Approach

364 ORDINARY DIFFERENTIAL EQUATIONS

7.

The function L(x, y) = x2 + y2 is positive definite, and

L′(x, y) = 2x(x3 + x + y) + 2y(−2x + y)= 2x4 + 2x2 − 2xy + 2y2

= 2x4 + x2 + y2 + (x− y)2

L′(x, y) is positive definite, therefore L(x, y) is a Lyapunov functionfor the given system and the origin is unstable.Return

Page 1116: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 365

9. Since the function L(x, y) = ax2 + cy2 is required to be positivedefinite, a and c must be positive. We will now calculate L′.

L′(x, y) = 2ax(−x + y) + 2cy(−x− y)= −2ax2 − 2cy2 + (2a− 2c)xy

If we take a = c = 1, then L′(x, y) is negative definite, and henceL(x, y) = x2 + y2 is a Lyapunov function that shows the origin to beasymptotically stable.Return

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366 ORDINARY DIFFERENTIAL EQUATIONS

11. Since the function L(x, y) = ax2 + cy2 is required to be positivedefinite, a and c must be positive. We will now calculate L′.

L′(x, y) = 2ax(2x− y) + 2cy(40x + 2y)= 4ax2 + (80c− 2a)xy + 4cy2

If we take a = 40, c = 1 then L′(x, y) is positive definite, and henceL(x, y) = 40x2 + y2 is a Lyapunov function that shows the origin tobe unstable.Return

Page 1118: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 367

13. Since the function L(x, y) = ax2 + bxy + cy2 is required to be pos-itive definite, a > 0 and b2 − 4ac < 0 . We will now calculate L′.

L′(x, y) = (2ax + by)(x− y) + (2cy + bx)(50x− y)= (2a + 50b)x2 + (−b− 2c)y2 + (b− 2a + 100c− b)xy

If we take a = 50, b = −2, c = 1 then L′(x, y) ≡ 0 is negative semidef-inite, and hence L(x, y) = 50x2− 2xy+ y2 is a Lyapunov function thatshows the origin to be neutrally stable but not asymptotically stable.Return

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368 ORDINARY DIFFERENTIAL EQUATIONS

15. Since the function L(x, y) = ax2 + bxy + cy2 is required to be pos-itive definite, a > 0 and b2 − 4ac < 0 . We will now calculate L′.

L′(x, y) = (2ax + by)[2x + 5(y2 − y)] + (2cy + bx)[5(x− x2)− 4y]= (4a + 5b)x2 + (−5b− 8c)y2 + (2b− 10a + 10c− 4b)xy

+5y2(2ax + by)− 5x2(2cy + bx)

By proposition 4.5.8, L′(x, y) will be positive or negative definite ifthe expression obtained by deleting the third and higher order termsis also positive or negative definite. We will therefore consider theexpression

L′2(x, y) = (4a + 5b)x2 + (−5b− 8c)y2 + (2b− 10a + 10c− 4b)xy.

If we put a = 5, b = −5 and c = 4, the coefficient of xy will vanish,giving the expression L′2(x, y) = −5x2 − 7y2, which is negative defi-nite. Therefore L′(x, y) is also negative definite, and hence L(x, y) =5x2 − 5xy + 4y2 is a Lyapunov function that shows the origin to beasymptotically stable.Return

Page 1120: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 369

17. Since the function L(x, y) = ax2 + cy2 is required to be positivedefinite, a and c must be positive. We will now calculate L′.

L′(x, y) = 2ax(3y− x3 − xy2) + 2cy(−4x− x2y− y3)

= (6a− 8c)xy− 2ax4 − 2ax2y2 − 2cx2y2 − 2cy4

If we take a = 4, c = 3 then L′(x, y) is negative definite, and henceL(x, y) = 4x2 + 3y2 is a Lyapunov function that shows the origin tobe asymptotically stable.Return

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370 ORDINARY DIFFERENTIAL EQUATIONS

19. If q(x, y) is negative semidefinite, then L′(x, y) is negative semidef-inite, so L(x, y) is still a Lyapunov function for the system. If q(x, y) ≡0, then L(x, y) is an integral of the system, and hence the origin is neu-traly stable. On the other hand, if q(x, y) = −y2 then L′(x, y) vanishesonly on the x-axis, and is negative for y 6= 0. Because the system 4.49reduces to x′ = 0, y′ = x in this case, orbit are vertical as they inter-sect the x-axis, and hence the x-axis contains no invariant sets otherthan the origin. It follows that the origin is a stable stationary pointin this case.

We conclude that if q(x, y) is only known to be negative semidefinite,then the origin is a stable stationary point of (4.49), but may or maynot be asymptotically stable.Return

Page 1122: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 371

21.

(a)

L′(x, y) = 2(x2 + y2)2(1− x2 − y2)

L′(x, y) is positive definite when (x, y) is in the unit disk. Thusthe origin is unstable.

(b) Since q(cos t, sin t) = 0, substituting x = cos t, y = sin t in thesystem (4.49) gives the identities

(cos t)′ ≡ − sin t + (cos(t))(0)(sin t)′ ≡ cos t + (sin(t))(0)

Thus x = cos t, y = sin t is a solution of the system ((4.49). Theorbit of this solution is the unit circle, traversed, counterclock-wise.

(c) Let r2 = x2 + y2, and θ = arg(x + iy) be polar coordinates. Then

2rr′ = 2xx′ + 2yy′

= 2x[−y + xq(x, y)] + 2y[x + yq(x, y)]= 2(x2 + y2)q(x, y)= 2(x2 + y2)2(1− x2 − y2)

= 2r4(1− r2)

Thus the radial coordinate satisfies the autonomous ODE r′ =r3(1− r2), which has stationary points 0,±1. The phase diagramof this ODE is

- s s s� - �

−1 0 1

It follows that if an orbit starts at a point other than the origin,the radial coordinate converges to 1.If we differentiate the relation tan(θ) = y

x , we obtain

sec2(θ)θ′ =x y′ − x′ y

x2 . (55)

Noting that q = r2(1− r2) in polar coordinates, we have

x y′ = r cos(θ)(−r cos(θ) + r sin(θ)r2(1− r2))

x′ y = (r sin(θ) + r cos(θ)r2(1− r2))r sin(θ)

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372 ORDINARY DIFFERENTIAL EQUATIONS

Combining these, we have

x y′ − x′ y = −r2.

Substitute this result in (55) to obtain

sec2(θ)θ′ = − r2

x2 = − sec2(θ),

from which it follows that θ′ = −1. We have thus establishedthat all orbits spiral clockwise toward the unit circle, which istherefore a limit cycle.

(d)

-2 -1.5 -1 -0.5 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

Return

Page 1124: Ordinary Differential Equations: A Systems Approach

SOLUTIONS MANUAL 373

23. The system has a limit cycle for any negative value of k. Hereare two phase portraits illustrating the limit cycle for k = −1 and fork = −10.

-2 -1 1 2

-4

-2

2

4k=-1

-2 -1 1 2

-20

-10

10

20

k=-10

Return

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374 ORDINARY DIFFERENTIAL EQUATIONS

25.

The matrix AT = −I + N, where N =

[0 0−1 0

]is nilpotent. There-

fore eNt = I + t N =

[1 0t 1

]and eAT t = e−teNt. Now we calculate

Y(t) = eAT t(eAT t)T = e−2t[

1 0t 1

] [1 t0 1

]= e−2t

[1 tt 1 + t2

]Integrating, we get

S =∫ ∞

0Y(t) dt = e−2t

[− 1

2 − 12 t− 1

4− 1

2 t− 14 −12t2 − 1

2 t− 34

]∣∣∣∣∞0

=

[ 12

14

14

34

].

We can directly verify that L is positive definite,

L(~x) = ~xTS~x =12

x21 +

34

x22 +

12

x1x2 =12(x1 +

12

x2)2 +

58

x22

is positive definite. Also,

ATS+SA =

[−1 01 −1

] [ 12

14

14

34

]+

[ 12

14

14

34

] [−1 10 −1

]=

[−1 00 −1

]is negative definite.Return