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36

Systems of Differential Equations:General Introduction and Basics

Thus far, we have been dealing with individual differentialequations. But there are manyapplications that lead to sets of differential equations sharing common solutions. In this chapterwe will start examining such sets — generally refered to as “systems”. In particular, we willdevelop and describe some of the most basic ideas and terminology, and will derive a selectcollection of sample systems corresponding to different applications. These systems will laterbe useful in illustrating the “systems analysis” we will later develop. We will also discover thatwe can convert most systems of interest (including a “system” involving only one differentialequation) into a relatively simple “standard” form. This fact will help shape virtually all of ourdiscussion of systems in the many chapters that follow.

36.1 General IntroductionBasic Terminology and Notions

A kth-order system of M differential equations with N unknownsis simply a collection ofMdifferential equations involvingN unknown functions withk being the highest order derivativeexplicitly appearing in the equations. For brevity, we may refer to a system ofM differentialequations withN unknowns as a “M×N system”.

!◮Example 36.1: Letting y1 , y2 and y3 be three unknown functions oft , the two differentialequations

d2y1

dt2+ 3

dy1

dt−

dy2

dt+ sin(t) [y2 − y1] = y3

andd2y2

dt2− t2 dy3

dt+ y1y2 = 0

make up a second-order system of two differential equationswith three unknown functions;that is, a second-order2×3 system.

In practice, the number of equations is often equal to the number of unknown functions.Also, in practice, these numbers are rarely stated since it’s usually assumed the reader can look

8/11/2012

Chapter & Page: 36–2 Systems of Differential Equations: Basics

at the given system and count the number of equations and unknowns themself.As suggested by the above example, we will find it convenient to start usingt as the variable

(so y′ = dy/dt , y′′ = d2y/dt2 , etc.). This will be the convention for the next several chapters. Thesymbols denoting the unknown functions in the differentialequations,however, will vary. Above,we usedy1 , y2 and y3 , but we will find it convenient to use other symbols, including x1 , x2 , . . .and xN to represent the unknown functions oft . In particular, when there are only two or threeunknown functions, these functions will often be denoted byx , y and, if needed,z ; and ifthe system arises from some application, then the symbols will often indicate the quantities ofinterest (as in usingv for a velocity).

Now suppose we have a system ofM differential equations with unknown functionsy1 ,y2 , . . . and yN , along with some interval of interestI . A (real)solutionto this system over theinterval I is any ordered set ofN specific real-valued functionsy1 , y2 , . . . and yN such thatall the equations in the system are satisfied for all values oft in the interval I when we let1

y1 = y1 , y2 = y2 , . . . , and yN = yN .

A general solutionto our system of differential equations (overI ) is any ordered set ofN formulas describing all possible such solutions. Typically, these formulas include arbitraryconstants.2

!◮Example 36.2: Consider the relatively simple system

x′ = x + 2y

y′ = 5x − 2y

over the entire real line (soI = (−∞,∞) ). If we let

x(t) = e3t + 2e−4t and y(t) = e3t − 5e−4t ,

and plug these formulas forx and y into the first differential equation in our system,

x′ = x + 2y ,

we getd

dt

[

e3t + 2e−4t]

=[

e3t + 2e−4t]

+ 2[

e3t − 5e−4t]

→ 3e3t − 2 · 4e−4t = [1 + 2]e3t + [2 − 2 · 5]e−4t

→ 3e3t − 8e−3t = 3e3t − 8e−3t ,

which is an equation valid for all values oft . So these two functions,x and y , satisfy thefirst differential equation in the system over(−∞,∞) .

Likewise, it is easily seen that these two functions also satisfy the second equation:

y′ = 5x − 2y

→d

dt

[

e3t − 5e−4t]

= 5[

e3t + 2e−4t]

− 2[

e3t − 5e−4t]

1 We could allow theyk(t)’s to be complex valued, but this will not gain us anything with the systems that we willhave interest in, and will complicate the “graphing” techniques that we’ll later develop and use.

2 And it is also typical that the precise interval of interest,I , is not explicitly stated, and may not even be preciselyknown.

General Introduction Chapter & Page: 36–3

→ 3e3t − 5(−4)e−4t = [5 − 2]e3t + [5 · 22(−5)]e−4t

→ 3e3t + 20e−4t = 3e3t + 20e−4t .

Thus, the pair

x(t) = e3t + 2e−4t and y(t) = e3t − 5e−4t

is a solution to our system (over(−∞,∞) ).More generally, you can easily verify that, for any choice ofconstantsc1 and c2 ,

x(t) = c1e3t + 2c2e

−4t and y(t) = c1e3t − 5c2e−4t

satisfies the given system (see exercise 36.4). Later, we will see that this pair of formulas isthe general solution for this system.

(Note that the two formulas in the general solution in the above example shared arbitraryconstants. This will be typical.)

If, in addition to our system of differential equations, we have the values of the solutions andsome of their derivatives specified at some single point, then we have aninitial-value problem,a solution of which is any solution to the system that also satisfies the given initial values.Unsurprisingly, we usually solve initial-value problems by first finding the general solution tothe system, and then applying the initial conditions to the general solution to determine the valuesof the ‘arbitrary’ constants.

!◮Example 36.3: Consider the initial-value problem consisting of the system from the previousexample,

x′ = x + 2y

y′ = 5x − 2y,

along with the initial conditions

x(0) = 0 and y(0) = 1 .

In the previous example, it was asserted that

x(t) = c1e3t + 2c2e

−4t and y(t) = c1e3t − 5c2e−4t (36.1)

is a solution to our system for any choice of constantsc1 and c2 . Using these formulas withthe initial conditions, we get

0 = x(0) = c1e3·0 + 2c2e

−4·0 = c1 + 2c2

and

1 = y(0) = c1e3·0 − 5c2e

−4·0 = c1 − 5c2 .

So, to findc1 and c2 , we solve the simple algebraic linear system

c1 + 2c2 = 0

c1 − 5c2 = 1.


Doing so however you wish, you should easily discover that

c1 =2

7and c2 = −

1

7.

which, after plugging these values back into the general formulas for x(t) and y(t) given inequation set (36.1), yields the solution to the given initial-value problem,

x(t) =2

7e3t −

2

7e−4t and y(t) =

2

7e3t +

5

7e−4t .

By the way, you will occasionally hear the term “coupling” describing the extent in whicheach equation of the system contains different unknown functions. A system iscompletelyuncoupledif each equation involves just one of the unknown functions,as in

x′ = 5x + sin(x)

y′′ = 4y,

and isweakly coupledor only partially coupledif at least one of the equations just involves onlyone unknown functions, as in

x′ = 5x + 2y

y′′ = 4y.

Such systems can be solved in the obvious manner: First solveeach equation involving a singleunknown function, and then plug those solutions into the other equations, and deal with them.

!◮Example 36.4: Consider the systemx′ = 5x + 2y

y′′ = 4y.

The second equation is the simple linear equation.

y′′ + 4y = 0

whose general solution you can readily find to be

y(t) = c1e2t + c2e−2t .

With this, the first equation in the system becomes

x′ = 5x + 2[

c1e2t + c2e−2t

]

,

a first-order linear equation that you should have little trouble solving (see chapter 5). Itsgeneral solution is

x(t) = −2

3c1e

2t −2

7c2e

−2t + c3e5t .

So, the general solution to our system is

x(t) = −2

3c1e2t −

2

7c2e−2t + c3e

5t and y(t) = c1e2t + c2e−2t .

For the most part, our interest will be in systems that are notweakly coupled.

A Few Illustrative Applications Chapter & Page: 36–5

36.2 A Few Illustrative Applications

Our first example is little more than an observation that the very first example considered in thistext naturally led to a system of two equations and two unknowns.

A Falling Object

Way back in section 1.2, we considered object of massm plummeting towards the ground underthe influence of gravity. As we did there, let us set

t = time (in seconds) since the object was dropped

y(t) = vertical distance (in meters) between the object and the ground at timet

v(t) = vertical velocity (in meters/second) of the object at timet

We can viewy and v as two unknown functions related by

dy

dt= v .

Now, in developing a “better model” describing the fall (seethe discussion starting on page 12),we took into account air resistance and obtained

dv

dt= −9.8 − κv

where κ is a positive constant describing how strongly air resistance acts on the falling object.This gives us a system of two differential equations with twounknown functions,

dy

dt= v

dv

dt= −9.8 − κv

.

Fortunately, this is a very weakly coupled system whose second equation is a simple first-orderequation involving only the functionv . We’ve already solved it (in example 4.7 on page 81),obtaining

v(t) = v0 + c1e−κt where v0 = −

9.8

κ.

Plugging this back into the first equation of the system yields

dy

dt= v0 + c1e

−κt ,

which we can now integrate:

y(t) =

∫

dy

dtdt =

∫

[

v0 + c1e−κt

]

dt = v0t −c1

κe−κt + c2 .

So, the general solution to this system is the pair

y(t) = v0t −c1

κe−κt + c2 and v(t) = v0 + c1e

−κt .


TTankA(500 gal.)

TankB(1,000 gal.)

γ gal./min.

R + γ gal./min.

R gal./min.

R gal./min.

Figure 36.1: A simple system of two tanks containing water/alcohol mixtures.

Mixing Problems with Multiple Tanks

Let us expand, slightly, our discussion of “mixing” from section 10.6 on page 226 by consideringthe situation illustrated in figure 36.1. Here we have two tanks,A andB. Each minuteR gallonsof a water/alcohol mix containingC0 gallons of alcohol per gallon of mix is added to tankA.At the same time,R gallons of the mix in tankB is drained out. In addition, the two tanks areconnected by two pipes, with one pumping liquid from tankA to tankB at a rate ofR + γ

gallons per minute, and with the other pumping liquid in the opposite direction, from tankB totankA, at a rate ofγ gallons per minute.

Following standard conventions, we will let

t = number of minutes since we started the pumping liquid between the two tanks

x = x(t) = gallons of pure alcohol in tankA at time t

and

y = y(t) = gallons of pure alcohol in tankB at time t .

Let us assume that tankA initially contains 500 gallons of pure water, while tankB initiallycontains 1,000 gallons of an alcohol-water mix with 90 percent of that mix being alcohol. Notethat the input and output flows for each tank cancel out, leaving the total amount of mix in eachtank constant. So, our initial conditions are

x(0) = 0 and y(0) =90

100× 1000 = 900 ,

and, at timet , the concentration of alcohol in the two tanks are given, respectively, byx

500and

y

1000.

In this system we have four “flows” affecting the rate the amount of alcohol varies in each tankover time, each corresponding to one of the pipes in figure 36.1. In each case the rate at whichalcohol is flowing is simply the total flow rate of the mix in thepipe times the concentration ofalcohol in that mix.

Thus,dx

dt= change in the amount of alcohol in tankA per minute

= rate alcohol is pumped into tankA from the outside

+ rate alcohol is pumped from tankB into tankA

− rate alcohol is pumped from tankA into tankB


= RC0 + γy

1000− (R + γ )

x

500

= RC0 +γ

1000y −

R + γ

500x ,

anddy

dt= change in the amount of alcohol in tankB per minute

= rate alcohol is pumped from tankA into tankB

− rate alcohol is pumped from tankB into tankA

− rate alcohol is drained from tankB

= (R + γ )x

500− γ

y

1000− R

y

1000

=R + γ

500x −

R + γ

1000y ,

giving us the system

x′ = −R + γ

500x +

γ

1000y + RC0

y′ =R + γ

500x −

R + γ

1000y

.

In particular, if we are adding a 50 percent mixture of alcohol and water at a rate of 3gallons/minute (soR = 3 , C0 = 1/2 and RC0 = 3/2 , and each minute we have 5 gallons of mixflowing from tankB to tankA (so γ = 5 and R + γ = 8 , then the above system is

x′ = −8

500x +

5

1000y +

3

2

y′ =8

500x −

8

1000y

.

Foxes in the Rabbit Ranch (A Predator-Prey Model)

In chapter 10, we considered how the number of rabbits in a large ranch varies with time. Let usnow assume foxes have entered the fields and, not being vegetarians, have begun dining on therabbits they can catch, and raising little pups of their own.Our interest is in determining both

R(t) = number of rabbits at timet

and

F(t) = number of foxes at timet .

For convenience, we’ll again take the basic unit of time to bemonths.Recall that the basic equation describing the rate of changein the number of rabbits is

d R

dt= βRR − δRR

where βR is the monthly birthrate per rabbit, andδR is the monthly deathrate per rabbit (thatis, δR is the fraction of the rabbit population that dies each month). Under ideal conditions —plenty of food and no predators —βR and δR are constants,βR,0 and δR,0 (according to the


information in chapter 10,βR,0 ≈ 5/4 and δR,0 ≈ 0 ). For simplicity, let’s takeδR,0 = 0 assumethere is plenty of food for the rabbits (soβR = βR,0 ). But with foxes around, the deathrate cannot be assumed constant; it will be a function of the number offoxes,

δR = δR(F) .

In particular, as the number of foxes increases, so does the deathrateδR . With a little thought,you will probably agree that the simplest formula describing a death rateδR that increases fromthe ideal rate of 0 as the number of foxes increases from 0 is

δR = δR(F) = δR,1F

where δR,1 is some positive constant. Thus,

d R

dt= βRR − δRR

= βR,0R −[

δR,1F]

R = βR,0R − δR,1F R .(36.2)

Likewise, the basic equation describing the rate of change in the number of foxes is

d F

dt= βF F − δF F

whereβF is the monthly birthrate per fox, andδF is the monthly deathrate per fox. In this case,however, both the birthrate and deathrate will depend on theamount of food available for thefoxes (i.e., the number of rabbits in the fields). In other words, βF and δF should be treated asfunctions of R ,

βF = βF(R) and δF = δF(R) .

If R = 0 , there are no rabbits, and, hence, no food for the foxes, which, in turn, means no newfoxes are born, and a large portion of the fox population diesfrom starvation each month. Thatis, we should have

βF(0) = 0 and δF(0) = δF,0 .

where δF,0 is the fraction of the fox population that would die from starvation each month ifthere is no food for them (hence,δF,0 is some positive number less than or equal to one). As thenumber of rabbits increases, there is more food and the birthrate will increase while the deathratewill decrease. The simplest formulas describing this are

βF(R) = βF,1R and δF(R) = δF,0 − δF,1R

whereβF,1 and δF,1 are positive constants. Thus,

d F

dt= βF F − δF F

=[

βF,1R]

F −[

δF,0 − δF,1R]

F

=[

βF,1 + δF,1]

RF − δF,0F .

Combining the last equation with (36.2) (and lettingγF,1 = βF,1 + δF,1 ) we now have thesystem

R′ = βR,0R − δR,1F R

F ′ = γF,1RF − δF,0F(36.3)

where βR,0 , δR,1 , γF,1 and δF,0 are positive constants that would have to be determined byexperiment.


X0 x1(t) x2(t)

m1 m2

L1 L2

Figure 36.2: A double mass/spring system with objects of massm1 and m2 located atpositionsx1(t) and x2(t) , respectively. The first spring, which has a naturallength of L1 and spring constantκ1 , connects the object of massm1 to thewall. The second spring, which has a natural length ofL2 and spring constantκ2 , connects the two objects together. In this snapshot, the first spring isstretched, and the second is compressed.

Double Mass-Spring System

Consider the spring system in figure 36.2 with the assumptionthat there are no frictional forces.Here the “natural length” of each spring —L1 and L2 , respectively — takes into account thehorizontal dimension of the object; that is, if the springs are neither compressed or stretched,then

x1 = L1 and x2 − x1 = L2 .

(If it helps, pretend that each mass is a ‘point mass’.)Now remember, if we have a horizontal spring with spring constant κ and natural length

L , then the force exerted by the spring on an object attached toits right end is

Fright = −κ × the ‘stretch’ in the spring= −κ × [current length of the spring− L] .

(The negative sign tells us that the force of the spring is in the negative direction if the spring isstretched beyond its natural length, and is positive if the spring is compressed to a length lessthan its normal length.)

Changing the sign then gives the corresponding force exerted by the spring at the left end,

Fleft = κ × ‘stretch’ in the spring= κ × [current length of the spring− L] .

Then applyingF = ma to the first object and noting how the “current length” of eachspring is computed fromx1(t) and x2(t) , we get

m1d2x1

dt2= force of spring 1 on object 1+ force of spring 2 on object 1

= F1,right + F2,left

= −κ1 [x1 − L1] + κ2 [(x2 − x1) − L]

= −[κ1 + κ2]x1 + κ2x2 + [κ1L1 − κ2L2] .

Since the second object is only attached to the second spring,

m2d2x2

dt2= force of spring 2 on object 2

= F2,right


= κ2 [L − (x2 − x1)]

= κ2x1 − κ2x2 + κ2L2 .

So the motion of the objects in this physical system is described by the solutions to the system

m1x1′′ = −(κ1 + κ2)x1 + κ2x2 + (κ1L1 − κ2L2)

m2x2′′ = κ2x1 − κ2x2 + κ2L2

. (36.4)

Equivalently,x1

′′ = a11x1 + a12x2 + b1

x2′′ = a21x1 + a22x2 + b2

wherea11 = −

κ1 + κ2

m1, a12 =

κ2

m1, b1 =

κ1L1 − κ2L2

m1,

a21 =κ2

m2, a22 = −

κ2

m2and b2 =

κ2L2

m2.

In particular, suppose the first spring has a natural length of L1 = 1 meter and springconstant ofκ1 = 1 kg./sec.2 , and is attached to an object of massm1 = 1 kg. , while thesecond spring is shorter and stiffer with natural lengthL2 = 0.2 meter and spring constantκ2 = 2.5 kg./sec.2 and is attached on the right to an object of massm2 = 0.1 kg. . Then (in unitsof sec.−2)

a11 = −1 + 2.5

1= −

7

2, a12 =

2.5

1=

5

2,

a21 =2.5

0.1= 25 and a22 = −

2.5

0.1= −25 ,

and (in units of meters·sec.−2)

b1 =1 · 1 − 2.5 · 0.2

1=

1

2and b2 =

2.5 · 0.2

0.1= 5 ,

and the above system governing the positions of the two objects as functions of time,x1(t) andx2(t) , is

x1′′ = −

7

2x1 +

5

2x2 +

1

2

x2′′ = 25x1 − 25x2 + 5

.

Converting to First-Order Systems Chapter & Page: 36–11

36.3 Converting High-Order Differential Equationsand Systems to Simple First-Order Systems

Converting Single Differential EquationsLet’s start with an example.

!◮Example 36.5: Consider the second-order differential equation

y′′ − 3y′ + 8 cos(y) = 0 ,

which we will rewrite asy′′ = 3y′ − 8 cos(y) .

Let us introduce two new “unknown” functionsy1 and y2 related toy by

y1 = y and y2 = y′ .

Theny1

′ = y′ = y2

and

y2′ = y′′ = 3y′ − 8 cos(y) = 3y1 − cos(y1) .

Cutting out the middle of each then gives us the2×2 system of first-order differential equations

y1′ = y2

y2′ = 3y1 − cos(y1)

.

In other words, we have converted the single second-order differential equation

y′′ − 3y′ + 8 cos(y) = 0 .

to the above2×2 system of first-order differential equations. If we can solve this system, thensetting y = y1 gives us the solution to the original single second-order differential equation.

In general, anyN th-order differential equation that can be written as

y(N) = F(

t, y, y′, y′′, . . . , y(N−1))

can be converted to a system ofN first-order differential equations withN unknowns byintroducing N new unknown functionsy1 , y2 , . . . and yN related toy via

y1 = y , y2 = y′ , y3 = y′′ , . . . and yN = y(N−1) .

Then we havey1

′ = y′ = y2 ,

y2′ =

(

y′)′

= y′′ = y3 ,

y3′ =

(

y′′)′

= y′′′ = y4 ,


...

yN−1′ =

(

y(N−2))′

= yN

and, finally

yN′ =

(

y(N−1))′

= y(N) .

Buty(N) = F

(

t, y, y′, y′′, . . . , y(N−1))

= F(t, y1, y2, y3, . . . , yN) .

So the last equation can be written as

yN′ = F(t, y1, y2, y3, . . . , yN)

and our original differential equation has been converted to the system of first-order differentialequations

y1′ = y2

y2′ = y3

y3′ = y4

...

yN−1′ = yN

yN′ = F(t, y1, y2, y3, . . . , yN)

.

For convenience, we will call this system thefirst-order system corresponding to the originaldifferential equation. If we can solve this system, then, sincey(t) = y1(t) , we automaticallyhave the solution to our original differential equation. And even if we cannot easily solve thesystem, we will find that some of the tools we’ll later developfor first-order systems will greatlyaid us in analyzing the possible solutions, especially whenthe original equation is not linear.

Do note that, using this procedure, anyN th order set of initial values

y(t0) = α1 , y′(t0) = α2 , y′′(t0) = α3 , . . . and y(n−1)(t0) = αN

is converted to values

y1(t0) = α1 , y2(t0) = α2 , y3(t0) = α3 , . . . and yN(t0) = αN .

Let us also note that, as a practical matter, it’s not necessary to use subscripted symbolsfor your new unknowns or to rename your original unknown function. It may be sufficient (andsimpler) to just introduce one or two new functions with any convenient names. For example,if y is, say, the height above ground of some falling object, thenit is natural to let a secondunknown function be the corresponding velocity,v = y′ (just as we did in the original fallingobject example in section 1.2).

On the other hand, the new functions may simply be any convenient letters of the alphabet.

!◮Example 36.6: Consider the third-order differential equation

y′′′ − 3y′′ + sin(y) y′ = 0 ,

which we will rewrite asy′′′ = 3y′′ − sin(y) y′ .

Converting to First-Order Systems Chapter & Page: 36–13

Introducing the functionsx and z related to each other and toy by

x = y′ and z = y′′

and observing thatx′ = y′′ = z , y′ = x

and

z′ = y′′′ = 3y′′ − sin(y) y′ = 3z − sin(y) x ,

we see that the first-order system of three equations corresponding to our original third-orderequation is

x′ = z

y′ = x

z′ = 3z − sin(y) x

.

Converting Higher-Order Systems

In a similar manner, we can convert just about any system of differential equations to a largersystem of first-order equations.

A single example should suffice.

!◮Example 36.7: Consider the system

x1′′ = −

7

2x1 +

5

2x2 +

1

2

x2′′ = 25x1 − 25x2 + 5

from our discussion of a double mass-spring system. Let

x3 = x1′ and x4 = x2

′ .

Thenx3

′ = x1′′ and x4

′ = x2′′ ,

allowing us to rewrite our second-order system of two equations as the first-order system

x1′ = x3

x3′ = −

7

2x1 +

5

2x2 +

1

2

x2′ = x4

x4′ = 25x1 − 25x2 + 5

.


T

θ

θ

mg

L

Fgrav,tan

Figure 36.3: The pendulum system with a weight of massm attached to a massless rod oflength L swinging about a pivot point under the influence of gravity.

36.4 The Pendulum

There is one more system that we will want for future use. Thatis the system describing themotion of the pendulum illustrated in figure 36.3 consistingof a small weight of massm attachedat one end of a rigid rod, with the other end of the rod attachedto a pivot so that the weight canswing around in a circle of radiusL in a vertical plane. The forces acting on this pendulum arethe downward force of gravity and, possibly, a frictional force from either friction at the pivotpoint or air resistance.

Let us describe the motion of this pendulum using the angleθ from the vertical downwardline through the pivot to the rod, measured (in radians) in the counterclockwise direction. Thismeans thatdθ/dt is positive when the pendulum is moving counterclockwise, and is negativewhen the pendulum is moving clockwise.

Since the motion is circular, and the ‘positive’ direction is counterclockwise, our interestis in the components of velocity, acceleration and force in the direction of vectorT illustratedfigure 36.3. This is the unit vector tangent to the circle of motion pointing in the counterclockwisedirection from the current location of the weight. From basic physics and geometry, we knowthese tangential components of the weight’s velocity and acceleration are

vtan = Ldθ

dtand atan = L

d2θ

dt2.

Using basic physics and trigonometry, we can see that the corresponding component of thegravitational force is

Fgrav,tan = −mgsin(θ) .

For the frictional force, we’ll use what we’ve used several times before,

Ffric,tan = −γ vtan = −γ Ldθ

dt

where γ is some nonnegative constant — either zero if this is an idealpendulum having nofriction, or a small to large positive value corresponding to a small to large frictional force actingon the pendulum.

Writing out the classic “ma = F ” equation, we have

mLd2θ

dt2= matan = Fgrav,tan + Ffric,tan = −mgsin(θ) − γ L

dθ

dt.

Additional Exercises Chapter & Page: 36–15

Cutting out the middle and dividing through bymL then gives us the slightly simpler equation

d2θ

dt2= −

g

Lsin(θ) − κ

dθ

dtwhere κ =

γ

m. (36.5)

Observe that this is a second-order differential equation that, because of the sin(θ) term, isnonlinear.

To convert this equation to a first-order system, we will letω be the angular velocity,dθ/dt .So

dθ

dt= ω

anddω

dt=

d2θ

dt2= −

g

Lsin(θ) − κ

dθ

dt= −

g

Lsin(θ) − κω ,

giving us the first-order system

dθ

dt= ω

dω

dt= −

g

Lsin(θ) − κω

. (36.6)

Additional Exercises

36.1. Consider the following system of differential equations:

x′ = 2y

y′ = 1 − 2x.

Now determine whether or not each of the following pairs of functions is a solution tothe associated system of differential equations.

a. x(t) = sin(2t) +1

2and y(t) = cos(2t)

b. x(t) = e2t − 1 and y(t) = e2t

c. x(t) = 3 cos(2t) +1

2and y(t) = −3 sin(2t)


x′ = 4x − 3y

y′ = 6x − 7y.


a. x(t) = 6e3t and y(t) = 2e3t

b. x(t) = 3e2t − e−5t and y(t) = 2e2t − 3e−5t


TTTankA(700 gal.)

TankB(1,000 gal.)

2 gal./min.

7 gal./min.

TankC(800 gal.)

5 gal./min.

10 gal./min.

5 gal./min.

5 gal./min.

Figure 36.4: The system of three tanks containing water/alcohol mixtures for exercise 36.7.In this scenario, each tank contains a mixture of water and alcohol, and eachminute five gallons of mix is added from the upper spigot, with40 % of thatadded mix being alcohol.

c. x(t) = 3e2t + e−5t and y(t) = 2e2t − 3e−5t


t x′ + 2x = 15y

ty′ = x.


a. x(t) = 3t3 and y(t) = −t3

b. x(t) = 3t3 and y(t) = t3

c. x(t) = −5t−5 and y(t) = t−5

36.4. Verify that, for any choice of constantsc1 and c2 , the corresponding pair of functions

x(t) = c1e9t − c2e−3t and y(t) = c1e

9t + 2c2e−3t

satisfies the systemx′ = 5x + 4y

y′ = 8x + y.

Then find the solution to this system that satisfies

x(0) = 0 and y(0) = 9 .

36.5. Verify that, for any choice of constantsc1 and c2 ,

x(t) = c1e−2t + 2c2e5t and y(t) = −3c1e

−2t + c2e5t

satisfies the systemx′ = 4x + 2y

y′ = 3x − y.

Then find the solution to this system that satisfies

x(0) = 0 and y(0) = −21 .


X0 Wx1(t) x2(t)

m1 m2

Spring #1 Spring #2 Spring #3

Figure 36.5: The mass/spring system for exercise 36.8 consisting of two objects with massesm1 and m2 located at positionsx1(t) and x2(t) , respectively, and attached toeach other and to walls atx = 0 and x = W by three springs as indicated.

36.6. Solve each of the following weakly coupled systems:

a.x′′ + x = 0

y′ = xb.

x′ = 2yx

ty′ = yc.

x′ + 2x = 10z

zy′ + 5zy = 15x

z′ − 3z = 0

36.7. Consider the tank system illustrated in figure 36.4. Letx , y and z be, respectively,the amount of alcohol in tanksA , B andC at time t (measured in minutes), and findthe first-order system of three differential equations describing how x , y and z variesover time.

36.8. Consider the mass/spring system illustrated in figure 36.5.Assume there are no frictionalforces, and letκ j and L j be, respectively, the spring constant and natural length forthe j th spring (for j = 1, 2, and3).

a. Derive the second-order system of two differential equations describing howx1 andx2 vary in time. (As in the derivation of system (36.4) on page 36–10, assume thewidths of the two objects are both zero.)

b. What, in particular, is the system just derived whenW = 3 meters ,

m1 = m2 =1

2(kilogram) ,

L1 = L3 = 1 (meter) , L2 =1

5(meter) ,

κ1 = κ3 = 1

(

kilogram

second2

)

and κ2 =5

2

(

kilogram

second2

)

?

36.9. Rewrite the following differential equations as systems offirst order equations:

a. y′′ + 4y′ + 2y = 0 b. y′′ − 8t2y′ − 32y = sin(t)

c. y′′ = 4 − y2 d. t2y′′ − 5t y′ + 8y = 0

e. t2y′′ − t y′ + 10y = 0 f. 4t2y′′ + y = 0

g. y′′ = 4t2 − sin(

y′)

y h. y′′′ + 2y′′ − 3y′ − 4y = 0

i. y′′′ + y′(

t2 + y2)

= 0 j. y(4) + y4 = 0


36.10. Rewrite each of the following second-order systems as first-order systems:

a.x′ − 7y′ = t x2

y′′ + 4y = 3xb.

x1′′ + 2x2x1

′ + 3x1x2′ = 0

x2′′ − 4x2

′ + 8x2 = (x1)2

c. The system of two second-order differential equations which is the answer to exercise36.8 b, above.

36.11 a. In section 36.3, we saw that we can convert any second-order differential equation ofthe form

ay′′ + by′ + cy = 0

to a first order system after introducing a new functionx related to y by x = y′ .While this is the “standard” approach, it is not the only approach. In particular,convert each of the following second-order Euler equationsto a first-order system byintroducing an a new functionx related toy by x = t y′ . (Also, compare the resultingsystems to those obtained for the same equations in exercise36.9, above.)

i. t2y′′ − 5t y′ + 8y = 0 ii. t2y′′ − t y′ + 10y = 0

iii. 4t2y′′ + y = 0

b. Show that, by introducing a new functionx related toy by x = t y′ , any second-orderEuler equation

αt2y′′ + βt2y′ + γ y = 0

can be converted to the first-order system

t x′ =[

1 −β

α

]

x −γ

αy

ty′ = x.

36.12 a. Convert each of the following third-order Euler equations to a first-order system byintroducing new functionsx and z satisfying x = t y′ and z = t x′ :

i. t3y′′′ + 2t2y′′ − 4t y′ + 4y = 0

ii. t3y′′′ + 4t2y′′ + 2t y′ − 3y = 0

b. Show that, by introducing new functionsx and z satisfying x = t y′ and z = t x′ ,any third-order Euler equation

αt3y′′′ + βt2y′′ + γ t y′ + ωy = 0

can be converted to the first-order system

t x′ = z

ty′ = x

tz′ =(

β − γ

α− 2

)

x −ω

αy +

(

3 −β

α

)

z

.


Some Answers to Some of the Exercises

WARNING! Most of the following answers were prepared hastily and late at night. Theyhave not been properly proofread! Errors are likely!

1a. yes, it is1b. no, it is not1c. yes, it is2a. no, it is not2b. yes, it is2c. no, it is not3a. no, it is not3b. yes, it is3c. yes, it is4. x(t) = 3e9t − 3e−3t and y(t) = 3e9t + 6e−3t

5. x(t) = 6e−2t − 6e5t and y(t) = −18c1e−2t − 3e5t

6a. x(t) = c1 cos(t) + c2 sin(t) and y(t) = c1 sin(t) − c2 cos(t) + c3

6b. x(t) = Aec1t2and y(t) = c1t

6c. x(t) = 2Ae3t + Be−2t , y(t) = 6 +[

15B

At + C

]

e−5t and z(t) = Ae3t

7.x′ = 2 −

1

100x +

1

500y

y′ =1

100x −

3

250y +

1

160z

z′ =1

100y −

1

80z

8a. m1x1′′ = −(κ1 + κ2)x1 + κ2x2 + (κ1L1 − κ2L2)

m2x2′′ = κ2x1 − (κ2 + κ3)x2 + κ2L2 + κ3(W − L3)

8b. x1′′ = −7x1 + 5x2 + 1

x2′′ = 5x1 − 7x2 + 5

9a. x′ = −2y − 4xy′ = x

9b. x′ = 32y + 8t2x + sin(t)y′ = x

9c. x′ = 4 − y2

y′ = x

9d. x′ = 5t−1x − 8t−2yy′ = x

9e. x′ = t−1x − 10t−2yy′ = x


9f. x′ = −1

4t2y

y′ = x

9g. x′ = 4t2 − sin(x) yy′ = x

9h.x′ = zy′ = xz′ = 4y + 3x − 2z

9i.x′ = zy′ = xz′ = −x

(

t2 + y2)

9j.y1

′ = y2 (with y1 = y)

y2′ = y3

y3′ = y4

y4′ = − (y1)

4

10a.x′ = 7z + t x2

y′ = zz′ = −4y + 3x

10b.x1

′ = x3

x2′ = x4

x3′ = −2x2x3 − 3x1x4

x4′ = 4x4 − 8x2 + (x1)

2

10c.x1

′ = x3

x2′ = x4

x3′ = −7x1 + 5x2 + 1

x4′ = 5x1 − 7x2 + 5

11a i. t x′ = 6x − 8yty′ = x

11a ii. t x′ = 2x − 10yty′ = x

11a iii. t x′ = x −1

4y

ty′ = x

12a i.t x′ = zty′ = xtz′ = 4x − 4y + z


12a ii.t x′ = zty′ = xtz′ = 3y − z

systems of differential equations: general …howellkb.uah.edu/public_html/detext/part6/systems...

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