chapter 1 first-order differential...

CHAPTER 1 First-Order Differential Equations

1.1 Dynamical Systems: Modeling

!!!! State Variables

1. Temperature, acidity, how much is in the glass, percent of glass full, color, maybe even taste if itcould be quantified

2. Height above the ground, distance fallen, velocity of the ball, acceleration, angular velocity,kinetic energy

3. Height above the ground, distance fallen, orientation of leaf, . . .

4. Coordinates (three of them) of the spaceship relative to the earth in some coordinate system,distance from the moon, fraction of the distance traveled

5. Current in the circuit, charge across the capacitor, . . .

6. Pain measured from 1 (very little) to 10 (severe), body temperature, blood pressure, preciselocation of pain

7. Coordinates of the center of the hurricane, radius of the hurricane, rotational speed, barometricpressure at the center, velocity at which the center is moving

8. Coordinates of the ball, speed, rotation

9. Orbital state of the electron, potential of the electron, electron spin

10. Number of students having the flu, daily change in the number coming down with the flu,number of students not able to attend classes

11. Temperature, barometric pressure, wind speed, rain or sun, etc.

12. Height, weight, blood pressure, IQ, . . . endless others

13. GNP, jobless rate, national debt, interest rates, endless economic and social indicators

1

2 CHAPTER 1 First-Order Differential Equations

!!!! Complexity

14. If we knew the answer to this problem, we would be very smart indeed. However, a general ruleof thumb might be that simple systems have fewer variables with straightforward interactions. Aball falling from a tree in a vacuum is a simple system. However, if we add air resistance androtation to the ball, then the system becomes more complicated. If the ball falls in a medium sothat it accumulates moisture and ice, then it becomes even more complicated. A falling leaf is avery complicated system because its motion is so sensitive to air disturbances. It would beliterally impossible to predict the path of a falling leaf (including how it turns) if you dropped itfrom a height of 100 feet. Sensitivity of interactions of the variables is one of the mainingredients that make systems complicated (i.e., like the weather).

The human body is an incredibly complex system, and medical science is trying tounderstand the relevant variables and the complex relations between them. This statement will nodoubt be just as valid a hundred years from now. Differential equations play an important role inunderstanding the human body because most variables in the body change over time.

The number of variables is also a measure of the complexity of the system. It is far easierto construct a model for finding the temperature at a single point than the temperature at manypoints. To determine the temperature at many points would require finding the temperature as afunction of time t and coordinates x, y, z. This model would require the introduction of partialdifferential equations.

!!!! Special Circumstances

15. Often, external influences determine the behavior of a system. One example: The number ofstudents coming down with the flu on a college campus might depend on factors outside themodel, such as actions of health authorities, gender, age, and eating habits. Other possibleexamples: The economic status of a country can be affected by government policies. Physicalsystems behave one way in the absence of external effects and another way in the presence ofexternal forces (forcing terms, in the language of differential equations). A person complainingabout stomach pains to a physician will hopefully have the relevant variables modified withmedication.

!!!! Different Observers

16. The choice of the variable state of a physical system depends on what the researcher is interestedin studying. The number of variable states can be anywhere from 1 (temperature of a cup ofcoffee) to many (physical measurements taken from a person’s body). In the case of a cup ofcoffee, one person may be interested in only the temperature, but a researcher for a coffee

SECTION 1.1 Dynamical Systems: Modeling 3

company might be interested in knowing how the acidity of coffee changes (and hence taste)over time and maybe as a function of temperature.

!!!! Models

17. Some possible goals: to determine lift, drag of the wing, or air turbulence it causes.

18. We might determine drug reactions, the feasibility of surgical operations (like heart transplants,etc.).

19. The equation gives the relationship between the time and distance the ball falls, without airresistance. A modeler might use the difference between predicted values and observed values toexamine air resistance.

20. The effect of heavy water runoff on the effectiveness of the dam, possible effects of earthquakes

!!!! A First Model

21. Because d t= υ where d = distance traveled, υ = average velocity, and t = time elapsed, we have

the model for the time elapsed as simply the equation t d=υ

. Now, if we measure the distance

traveled as 1 mile and the average velocity as 3 miles/hour, then our model predicts the time to

be t d= =υ

13

hr , or 20 minutes. If it actually takes 20 minutes to walk to the store, the model is

perfectly accurate. This model is so simple we generally don’t even think of it as a model.

!!!! A Second Model

22. (a) Galileo has given us the model for the distance s t( )a ball falls in a vacuum near the

surface of the earth as a function of time t as s t gt( ) =12

2 , where g is the acceleration of

the ball, determined to be g ≈ 32 2. ft sec2 . Using this model, we can estimate the time it

takes a ball to fall from 100 feet as the solution of the equation, 100 12

1612 2= =gt t. ,

which turns out to be t = 2 49. seconds. (We use 3 significant digits in the answer becauseg is also given to 3 significant digits.)

(b) On the surface of the earth the acceleration of the ball is a constant d sdt

g2

2= . Integrating

twice and using the conditions s 0 0( ) = , dsdt

0 0( )= , we find s t gt( ) =

12

2, which is the

Galileo model for the distance fallen. We can then find the time it takes to fall 100 feetusing the same strategy as in part (a).

(c) If the observed time it takes for a ball to fall 100 feet is 2.6 seconds, but the modelpredicts 2.49 seconds, the first thing that might come to mind is the fact that Galileo’s


model assumes the ball is falling in a vacuum, so some of the difference might be due toair friction.

!!!! The Malthus Rate Constant k

23. (a) Replacing

e0 03 103045. .≈

in Equation (3) gives

y t= ( )09 103045. . ,

which increases roughly 3% per year.

(b)

18601820

4

1800

6

8

2

1840t

y10

1880

3

5

7

1

9 Malthus

World population

(c) Clearly, Malthus’ rate estimate was far too high. The world population indeed rises, asdoes the exponential function, but at a far slower rate.

If y t ert( ) = 0 9. , you might try solving y e r200 09 60200( ) = =. . for r. Then

200 60 9

1897r = ≈ln.

.

so

r ≈ ≈1897200

00095. . ,

which is less than 1%.

!!!! Population Update

24. If we assume the world’s population in billions is currently following the unrestricted growthcurve

y e ekt t0

0 0176= . ,

then the population in the years 2010 t =( )10 , 2020 t =( )20 , and 2030 t =( )30 , will be, respec-

tively, the values

6 711186 8 42976 9 9917

0 017 10

0 017 20

0 017 30

eee

.

.

.

... .

( )

( )

( )

=

≈

≈

Because these values agree fairly closely to the estimates of the U.S. Census Bureau, this meansthat the U.S. Census Bureau thinks the world’s population is currently in a state of unrestrictedexponential growth.

SECTION 1.1 Dynamical Systems: Modeling 5

!!!! The Malthus Model

25. (a) Malthus thought the human population was increasing exponentially ekt , whereas thefood supply increases arithmetically according to a linear function a bt+ . This means the

number of people per food supply would be in the ratio ea bt

kt

+( ), which although not a

pure exponential function, is concave up. This means that the rate of increase in thenumber of persons per the amount of food is increasing.

(b) The model cannot last forever since its population approaches infinity; reality wouldproduce some limitation. The exponential model does not take under considerationstarvation, wars, diseases, and other influences that slow growth.

(c) A linear growth model for food supply will increase supply without bound and fails toaccount for technological innovations, such as mechanization, pesticides and geneticengineering. A nonlinear model that approaches some finite upper limit would be moreappropriate.

(d) An exponential model is sometimes reasonable with simple populations over shortperiods of time, e.g., when you get sick a bacteria might multiply exponentially untilyour body’s defenses come into action or you receive appropriate medication.

!!!! Discrete-Time Malthus

26. (a) Taking the 1798 population as y0 09= . (0.9 billion), we have the population in the years

1799, 1800, 1801, and 1802, respectively

y

y

y

y

1

22

33

44

103 09 0 927

103 0 9 0956

103 0 9 0983

103 0 9 1023

= ( ) =

= ( ) ( ) =

= ( ) ( ) =

= ( ) ( ) =

. . .

. . .

. . .

. . . .

(b) In 1990 we have t = 192 , hence

y192192103 0 9 262= ( ) ( ) ≈. . (262 billion).

(c) The discrete model will always give a value lower than the continuous model. Later,when we study compound interest, you will learn the exact relationship between discretecompounding (as in the discrete-time Malthus model) and continuous compounding (asdescribed by ′ =y ky).


!!!! Verhulst Model

27. dydt

y k cy= −( ) . The constant k affects the initial growth of the population whereas the constant c

controls the damping of the population for larger y. There is no reason to suspect the two valueswould be the same and so a model like this would seem to be promising if we only knew theirvalues. From the equation ′ = −( )y y k cy , we see that for small y the population closely obeys

′ =y ky , but reaches a steady state ′ =( )y 0 when y kc

= .

!!!! Suggested Journal Entry

28. Student Project

SECTION 1.2 Solutions and Direction Fields 7

1.2 Solutions and Direction Fields

!!!! Verification

1. If y t= 2 2tan , then ′ =y t4 22sec . Substituting ′y and y into ′ = +y y2 4 yields a trigonometric

identity

4 2 4 2 42 2sec tant t( ) ≡ ( ) + .

2. Substituting

y t ty t= +′ = +

33 2

2

into ′ = +yt

y t1 yields the identity

3 2 1 3 2+ ≡ + +tt

t t ta f .

3. Substituting

y t ty t t t=′ = +

2

2lnln

into ′ = +yt

y t2 yields the identity

2 2 2t t tt

t t tln ln+ ≡ +b g .

4. If y e ds e e dss tt t st= =− − −z z2

02 2

0

2 2 2 2b g , then, using the product rule and the fundamental theorem of

calculus, we have

′ = + = +− − −z zy e e te e ds te e dst t t st t st2 2 2 20

2 20

2 2 2 2 2 24 1 4 .

Substituting ′y and y into ′ −y ty4 yields

1 4 42 2 2 200

2 2 2 2+ −− −zzte e ds te e dst s t stt,

which is 1 as the differential equation requires.


!!!! IVPs

5. Here

y e e

y e e

t t

t t

= −

′ = − +

− −

− −

12

12

3

3

3 .

Substituting into the differential equation

′ + = −y y e t3

we get

− +FHG

IKJ + −FHG

IKJ

− − − −12

3 3 12

3 3e e e et t t t ,

which is equal to e t− as the differential equation requires. It is also a simple matter to see that

y 0 12

( ) = − , and so the initial condition is also satisfied.

6. Another direct substitution

!!!! Applying Initial Conditions

7. If y cet= 2 , then we have ′ =y ctet2 2 and if we substitute y and ′y into ′ =y ty2 , we get theidentity 2 22 2cte t cet t≡ d i . If y 0 2( ) = , then we have ce c02 2≡ = .

8. We have

y e t cey e t e t ce

t t

t t t

= +

′ = − +

coscos sin

and plugging y and ′y into ′ −y y yields

e t e t ce e t cet t t t tcos sin cos− + − +b g b g,which is −e tt sin . If y 0 1( ) = − , then − = +1 00 0e cecos yields c = −2 .


!!!! Using the Direction Field

9. ′ =y y2

2–2

–2

2y

t

Solutions are y ce t= 2 .

10. ′ = −y ty

–2

2y

2–2t

Solutions are y c t= − 2 .

11. ′ = −y t y

–2

2y

2–2t

Solutions are y t ce t= − + −1 .

!!!! Linear Solution

12. It appears from the direction field that there is a straight-line solution passing through (0, –1)with slope 1, i.e., the line y t= −1. Computing ′ =y 1, we see it satisfies the equation ′ = −y t ybecause 1 1≡ − −( )t t .

!!!! Match Game

13. (C) Because the slope is always the same

14. (D) Because the slope is always the value of y

15. (F) Because F is the only direction field that has vertical slopes when t = 0 and zero slopeswhen y = 0

16. (B) Because it is the only direction field that has all zero slopes when t = 0


17. (E) The slope is always positive and equal to the square of the distance from the origin.

18. (A) Because it is undefined when t = 0 and the directional field has slopes that areindependent of y and have the same sign as that of t

!!!! Isoclines

19. ′ =y t . The isoclines are vertical lines t c= , as

follows for c = 0; ±1, ±2 shown in the figure.

–2

2y

2–2t

20. ′ = −y y . Here the slope of the solution isnegative when y > 0 and positive for y < 0. The

isoclines for c = −1, 0, 1 are shown in the figure.

–2

2y

2–2t

slopes –1

slopes 1

slopes 0

21. ′ =y y2 . Here the slope of the solution is always

≥ 0. The isoclines where the slope is c > 0 arethe horizontal lines y c= ± ≥ 0. In other wordsthe isoclines where the slope is 4 are y = ±2 . The

isoclines for c = 0, 1, and 4 are shown in thefigure.

–2

2y

2–2t

slopes 4

slopes 4

slopes 1

slopes 0

slopes 1

Isoclines of ′ =y y2 for slopes 0, ±1

22. ′ = −y ty . Setting − =ty c, we see that the points

where the slope is c are along the curve y ct

= − ,

t ≠ 0 or hyperbolas in the ty plane.

For c = 1 the isocline is the hyperbola yt

= −1 .

For c = −1 the isocline is the hyperbola yt

=1.

2–2t

–2

2y

slopes 1

slopes 1

slopes 0


When t = 0 the slope is zero for any y; when y = 0 the slope is zero for any t, and y = 0 is in fact

a solution. See figure for the direction field for this equation with isoclines for c = 0, ±1.

23. ′ = −y t y2 . The isocline where ′ =y c is thestraight line y t c= −2 . The isoclines with slopes

c = −2 , –1, 0, 1, 2 are shown from left to right(see figure).

–2

2y

2–2t

24. ′ = −y y t2 . The isocline where ′ =y c is a parab-

ola that opens to the right. Three isoclines, withslopes c = 1, 0, –1, are shown from left to right(see figure).

–2

2y

2–2t

slopes –1

slopes 1

slopes 0

!!!! Second-Order Equations

25. (a) Direct substitution of y, ′y , and ′′y into the differential equation reduces it to an identity.

(b) Direct computation

(c) Direct computation

(d) Substituting

y t Ae Bey t Ae Be

t t

t t

( ) = +

′( ) = −

−

−

2

22

into the initial conditions gives

y A By A B

0 20 2 5( ) = + =′( ) = − = − .

Solving these equations, gives A = −1, B = 3, so y e et t= − +− −2 3 .


!!!! Logistic Population Model

26. We find the constant solutions by setting ′ =y 0

and solving for y. This gives ky y1 0−( ) = , hencethe constant solutions are y t( ) ≡ 0, 1. Notice

from the direction field or from the sign of thederivative that solutions starting at 0 or 1 remainat those values, and solutions starting between 0and 1 increase asymptotically to 1, solutionsstarting larger than 1 decrease to 1 asymptoti-cally. The following figure shows the directionfield of ′ = −( )y y y1 and some sample solutions.

30t0

2y

1 stable equilibrium

unstable equilibrium

Logistic model

!!!! Autonomous Equations

27. (a) Autonomous:####

#

9 213 11420

21 2

′ =′ =′ =′ = −

′ =

y yyy yy y

y y

The others are nonautonomous.

(b) Isoclines for autonomous equations consist of horizontal lines.

!!!! Comparison

28. (a) ′ =y y2

–2

2y

2–2t



(b) ′ = +( )y y 1 2

–2

2y

2–2t


(c) ′ = +y y2 1

–2

2y

2–2t

Equations (a) and (b) each have aconstant solution that is unstable for highervalues and stable for lower y values, but theseequilibria occur at different levels. Equation (c)has no equilibrium at all.

All three DEs are autonomous,so within each graph solutions from left andright are always horizontal translates.

(i) For y > 0 we have

y y y2 2 21 1< + < +( ) .

For the three equations ′ =y y2 , ′ = +y y2 1, and ′ = +( )y y 1 2, all with y 0 1( ) = ;the solution of ′ = +( )y y 1 2 will be the largest and the solution of ′ =y y2 will be

the smallest.

(ii) Because y tt

( ) =−1

1 is a solution of the initial-value problem ′ =y y2 , y 0 1( ) = ,

which blows up at t = 1. We then know that the solution of ′ = +y y2 1, y 0 1( ) =

will blow up (approach infinity) somewhere between 0 and 1. When we solvethis problem later using the method of separation of variables, we will find outwhere the solution blows up.


!!!! Basins of Attraction

29. ′ = −( )y y y1 . The constant solutions are found bysetting ′ =y 0, giving y t( ) ≡ 0, 1. Either by look-

ing at the direction field or by analyzing the signof the derivative, we conclude the constant solu-tion y t( ) ≡ 1 has a basin of attraction of 0, ∞( ),and y t( ) ≡ 0 has a basin attraction of the single

value {0}. When the solutions have negative in-itial conditions, the solutions approach –∞.

–1

2y

4t

30. ′ = −y y2 4. The constant solutions are the (real)roots of y2 4 0− = , or y = ±2 . For y > 2, wehave ′ >y 0. We, therefore, conclude solutions

with initial conditions greater than 2 increase; for− < <2 2y we have ′ <y 0, hence solutions with

initial conditions in this range decrease; and fory < 0, we have ′ >y 0, hence solutions with

initial conditions in this interval increase.

3t

–3

3y

We can therefore, conclude that the constant solution y = 2 has a basin of attraction of the singlevalue {2}, whereas the constant solution y = −2 has the basin of attraction of −∞( ), 2

31. ′ = −( ) −( )y y y y1 2 . Analyzing the sign of thederivative in each of the intervals −∞( ), 0 , 0 1, ( ) ,1 2, ( ) , 2, ∞( ) , we conclude that the constant

solutions y t( ) ≡ 0, 1, 2 have the following basinsof attraction: y t( ) ≡ 0 has the single point {0}basin of attraction; y t( ) ≡ 1 has the basin of at-traction 0 2, ( ); and y t( ) ≡ 2 has the single value

{2} basin of attraction.

–1

3y

2t


32. ′ = −( )y y1 2. Because the derivative ′y is always

zero or positive, we conclude the constant solu-tion y t( ) ≡ 1 has basin of attraction the interval−∞( , 1 .

2t0

2y

0

!!!! Computer or Calculator

The student can refer to Problems 35–39 as examples when working Problems 33, 34, and 40.

33. ′ =y y2

. Student Project 34. ′ = +y y t2 . Student Project

35. ′ =y ty . The direction field shows one constantsolution y t( ) ≡ 0, which is unstable (see figure).

For negative t solutions approach zero slope, andfor positive t solutions move away from zeroslope.

–2

2y

2–2tunstable equilibrium

36. ′ = +y y t2 . We see that eventually all solutions

approach plus infinity. In backwards time mostsolutions approach the top part of this parabola.There are no constant or periodic solutions tothis equation. You might also note that thenullcline y t2 0+ = is a parabola sitting on its

side for t < 0 . In backwards time most solutionsapproach the top part of this parabola.

–2

2y

2–2t

37. ′ =y tcos2 . The direction field indicates that the

equation has periodic solutions with the periodroughly 3. This estimate is fairly accurate be-

cause y t t c( ) = +12

2sin has period π.

–2

2y

2–2t


38. ′ = ( )y tysin . We have a constant solution y t( ) ≡ 0

and there is a symmetry between solutions aboveand below the t-axis. Note: This equation doesnot have a closed form solution.

–2

2y

4–4t

39. ′ = −y ysin . We can see from the direction fieldthat y = ± ±0 2, , , π π … are constant solutionswith 0 2 4, , , ± ±π π … being stable and± ±π π, , 3 … unstable. The solutions between

the equilibria have positive or negative slopesdepending on the y interval. From left to rightthese solutions are horizontal translates.

–5

5

y

5–5t



stable equilibrium

stable equilibrium

stable equilibrium

40. ′ = +y y t2 . Student Project


41. Student Project

SECTION 1.3 Separation of Variables: Quantitative Analysis 17

1.3 Separation of Variables: Quantitative Analysis

!!!! Separable or Not

1. ′ = +y y1 . Separable; dyy

dt1+

= ; constant solution y ≡ −1.

2. ′ = −y y y3 . Separable; dyy y

dt−

=3

; constant solutions y t( ) ≡ ±0 1, .

3. ′ = +( )y t ysin . Not separable; no constant solutions.

4. ′ = ( )y tyln . Not separable; no constant solutions.

5. ′ =y e et y . Separable; e dy e dty t− = ; no constant solutions.

6. ′ =+

+y yty

y1 . Not separable; no constant solutions.

7. ′ =+

y e ey

t y

1. Separable; e y dy e dty t− +( ) =1 ; no constant solutions.

8. ′ = + = +( )y t y t t ytln ln2 2 2 2 1b g . Separable; dyy

t dt2 1

2ln +

= ; constant solution y t e( ) ≡ −1 2 .

9. ′ = +y yt

ty

. Not separable; no constant solutions.

10. ′ =+y yt

1 2. Separable; dy

ydt t

1 2+= ; no constant solution.

!!!! Solving by Separation

11. ′ =y ty

2. Separating variables, we get y dy t dt= 2 . Integrating each side gives the implicit solution

12

13

2 3y t c= + .

Solving for y yields branches so we leave the solution in implicit form.

12. ty y′ = −1 2 . Separating variables, we get

dyy

dtt1 2−

= .

Integrating gives the implicit solution

sin ln− = +1 y t c.


Solving for y gives the explicit solution

y t c= +( )sin ln .

13. ′ =+−

y ty y

2

4 37

4. Separating variables we get the equation

y y dy t dt4 3 24 7− = +b g b g .


15

13

75 4 3y y t t c− = + + .

We cannot find an explicit solution for y.

14. ty y′ = 4 . Separating variables we get

dyy

dtt

= 4 .


ln lny t c= +4 .

Solving for y gives the explicit solution

y Ct= 4

where C is an arbitrary constant.

15. ′ =−y ty

1 2 , y 1 2( ) = − . Separating variables gives

y dy t dt= −( )1 2 .


12

2 2y t t c= − + .

Substituting in the initial condition y 1 2( ) = − gives c = 2. Hence, the implicit solution is given by

y t t2 22 2 4= − + .

Solving for y we get

y t t t( ) = − − + +2 2 42 .

Note that we take the negative square root so the initial condition is satisfied.


16. ′ =+

y ty

21 2

, y 2 0( ) = . Separating variables

1 2 2+( ) =y dy t dt .


y y t c+ = +2 2 .

Substituting in the initial condition y 2 0( ) = gives c = −4 . Solving for y the preceding quadratic

equation in y we get

y t=− + + −1 1 4 4

2

2a f .

17. ′ = −y y2 4 , y 0 0( ) = . Separating variables gives

dyy

dt2 4−

= .

Rewriting this expression as a partial fraction decomposition (see Appendix PF), we get

14 2

1 14 2y y

dy dt−( )

−+( )

LNM

OQP = .

Integrating we get

ln lny y t c− − + = +2 2 4

or

yy

e ec t−+

=22

4 .

Hence, the implicit solution is

yy

e e kec t t−+

= ± =22

4 4

where k is an arbitrary constant. Solving for y, we get the general solution

y tke

ke

t

t( ) =+−

2 11

4

4b g

.

Substituting in the initial condition y 0 0( ) = gives k = −1.

18. ′ = −++

y yt

11

2

2, y 0 1( ) = − . Separating variables, we get the equation

dyy

dtt1 12 2+

= −+

.


Integrating gives

tan tan− −= − +1 1y t c.

Substituting in the initial condition y 0 1( ) = − gives c = −( ) = −−tan 1 14π . Solving for y gives

y t= − −FH IK−tan tan 14π .

!!!! Equilibria and Direction Fields

19. (C) 20. (B) 21. (E) 22. (F) 23. (A) 24. (D)

!!!! Finding the Nonequilibrium Solutions

25. ′ = −y y1 2

We note first that y = ±1 are equilibrium solutions. To find the nonconstant solutions we divideby 1 2− y and rewrite the equation in differential form as

dyy

dt1 2−

= .

By a partial fraction decomposition (see Appendix PF), we have

dyy y

dyy

dyy

dt1 1 2 1 2 1−( ) +( )

=−( )

++( )

= .

Integrating, we find

12

1 12

1ln ln− + + = +y y t c

where c is any constant. Simplifying, we getln ln

ln1 1 2 2

1 1 2 21 1 2

− + + = +−( ) +( ) = +

−( ) +( ) =

y y t cy y t cy y Ce t

where C is any positive constant. Hence

1 1 2 2−( ) +( ) = ± =y y Ce ket t

where k is any nonzero real constant. Hence

y ke t2 21= + .

If we now solve for y, we find

y ke t= ± +1 2 ,


so there is an argument for leaving the solution in implicit form unless we are given initialconditions that tell which branch of the solution would be used.

26. ′ = −y y y2 2

We note first that y = 0, 2 are equilibrium solutions. To find the nonconstant solutions, we divideby 2 2y y− and rewrite the equation in differential form as

dyy y

dt2 −( )

= .

By a partial fraction decomposition (see Appendix PF),

dyy y

dyy

dyy

dt2 2 2 2−( )

= +−( )

= .


12

12

2ln lny y t c+ − = +

where c is any real constant. Simplifying, we getln ln

lny y t c

y y t cy y Ce t

+ − = +−( ) = +

−( ) =

2 2 22 2 22 2

where C is any positive constant. Hence

y y Ce ket t2 2 2−( ) = ± =

where k is any nonzero real constant. Hence

y y ke t2 22 0− − = .

If we solve for y, we get two solutions

y ke t= ± +1 1 2 ,

so there is an argument for leaving the solution in implicit form unless we are given initialconditions that tell which branch of the solution would be used.

27. ′ = −( ) +( )y y y y1 1

We note first that y = 0, ±1 are equilibrium solutions. To find the nonconstant solutions, wedivide by y y y−( ) +( )1 1 and rewrite the equation in differential form as

dyy y y

dt−( ) +( )

=1 1

.


By finding a partial fraction decomposition, (see Appendix PF)

dyy y y

dyy

dyy

dyy

dt−( ) +( )

= − +−( )

++( )

=1 1 2 1 2 1

.


− + − + + = +

− + − + + = +

ln ln

ln ln ln

y y y t c

y y y t c

12

1 12

1

2 1 1 2 2

or

ln

.

y yy

t c

y yy

Ce

y yy

Ce

ke

t

t

t

−( ) +( )= +

−( ) +( )=

−( ) +( )= ±

=

1 1 2 2

1 1

1 1

2

22

22

2

Multiplying each side of the above equation by y2 gives a quadratic equation in y, which can be

solved, getting

yke t

= ±+

11 2a f .

Initial conditions will tell which brand of this solution would be used.

28. ′ = −( )y y 1 2

We note that y = 1 is a constant solution. Seeking nonconstant solutions, we divide by y −( )1 2

getting dyy

dt−( )

=1 2 . This can be integrated to get

−−

= +

− =− +

= +− +

11

1 1

1 1

yt c

yt c

yt c

.

!!!! Integration by Parts

29. ′ =y y tcos ln2b g . Separating variables we get

dyy

tdtcos

ln2 = .



dyy

t dt c

y dy t dt c

y t t t cy t t t c

cosln

sec ln

tan lntan ln .

2

2

1

z zz z

= +

= +

= − +

= − +( )−

30. ′ = −y t t2 5 2b gcos . Separating variables we get

dy t t dt= −2 5 2b gcos .


y t t dt c

t t dt t dt c

t t t t t c

= − +

= − +

= − + − +

zz z

2

2

2

5 2

2 5 2

14

2 1 2 12

2 52

2

b g

b g

cos

cos cos

sin cos sin .

31. ′ = +y t ey t2 2 . Separating variables we get

dye

t e dtyt= 2 2 .


e dy t e dt c

e t t e e c

e t t e e c

y t

y t t

y t t

−

−

−

z z= +

− = − + +

= − − − +

2 2

2 2 2

2 2 2

12

14

12

14

b g

b g .

Solving for y, we get

y t t e e ct t= − − − +LNM

OQPln 1

214

2 2 2a f .

32. ′ = −y t ye t . Separating variables we get

dyy

te dtt= − .


dyy

te dt c

y te e cy te e c

t

t t

t t

z z= +

= − − +

= − − +

−

− −

− −

lnexp .b g


!!!! Help from Technology

33. ′ =y y , y 1 1( ) = , y −( ) = −1 1

The solution of ′ =y y , y 1 1( ) = is y et= −1. Thesolution of ′ =y y , y −( ) = −1 1 is y et= − +1. These

solutions are shown in the figure. 2t

–3

3y

–2

34. ′ =y tcos , y 1 1( ) = , y −( ) = −1 1

The solution of the initial-value problem

′ =y tcos , y 1 1( ) =

is y t t( ) = + − ( )sin sin1 1 . The solution of

′ =y tcos , y −( ) = −1 1

is y t= − + −( )sin sin1 1 . The solutions are shown

in the figure.

–2

2y

6–6t

35. dydt

ty t

=+2 21

, y 1 1( ) = , y −( ) = −1 1

Separating variables and integrating we find theimplicit solution

y dy tt

dt c221z z=

++

or

13

13 2y t c= + + .

–2

2y

2–2t

′ =+

y ty t2 21

Plugging in y 1 1( ) = , we find c = −13

2 . For y −( ) = −1 1 we find c = − −13

2 . These two curves

are shown in the figure.


36. ′ =y y tcos , y 1 1( ) = , y −( ) = −1 1

Separating variables we get

dyy

t dt= cos .

Integrating, we find the implicit solution

ln siny t c= + . –8

2y

6–6t

With y 1 1( ) = , we find c = − ( )sin 1 . With y −( ) = −1 1, we find c = ( )sin 1 . These two implicit solu-

tion curves are shown imposed on the direction field (see figure).

37. ′ =+( )y t y

y2 1 , y 1 1( ) = , y −( ) = −1 1

Separating variables and assuming y ≠ −1, we

find

yy

dy t dt+

=1

2

or

yy

dy t dt c+

= +z z12 .

–2

2y

2–2t

′ =+( )y t y

y2 1

Integrating, we find the implicit solution

y y t c− + = +ln 1 2 .

For y 1 1( ) = , we get 1 2 1− = +ln c or c = − ln2 . For y −( ) = −1 1 we can see even more easily thaty ≡ −1 is the solution. These two solutions are plotted on the direction field (see figure). Notethat the implicit solution involves branching. The initial condition y 1 1( ) = lies on the upper

branch, and the solution through that point does not cross the t-axis.

38. ′ = ( )y tysin , y 1 1( ) = , y −( ) = −1 1

This equation is not separable and has no closedform solution. However, we can draw its direc-tion field along with the requested solutions (seefigure).

6–6t

–2

2y


!!!! Making Equations Separable

39. Given

′ =+

= +y y tt

yt

1 ,

we let υ = yt

and get the separable equation υ υ υ+ = +t ddt

1 . Separating variables gives

dtt

d= υ .

Integrating gives

υ = +ln t c .

But,

y t t ct= +ln .

40. Letting υ = yt

, we write

′ =+

= + = +y y tyt

yt

ty

2 2 1υυ

.

But y t= υ so ′ = + ′y tυ υ . Hence, we have

υ υ υυ

+ ′ = +t 1

or

t ′ =υυ1

or

υ υd dtt

= .


12

2υ = +ln t c

or

υ = ± +2ln t c .

But υ = yt

, so

y t t c= ± +2ln .


The initial condition is y 1 2( ) = − requires the negative square root and gives c = 4. Hence,

y t t t( ) = − +2 4ln .

41. Given

′ =+

= + = +y y tty

yt

ty

2 2 2 14 4

3

3

3 3υ

υ.

with the new variable υ = yt

. Using ′ = + ′y tυ υ and separating variables gives

dtt

d d= =

++

υ υ υυυ

υ4

31

34 1

.

Integrating gives the solution

ln lnt c= + +14

14υa f

or

ln lnt yt

c= FH IK +LNM

OQP +

14

14

.

42. Given

′ =+ +

= + + = + +y y ty tt

yt

yt

2 2

2

2

221 1υ υ

with the new variable υ = yt

. Using ′ = + ′y tυ υ and separating variables, we get

d dtt

υυ2 1+

= .


ln tant c= +−1υ .

Solving for υ gives υ = +( )tan ln t c . Hence, we have the explicit solution

y t t c= +( )tan ln .

!!!! Autonomous Equations

43. (a) Problems 1, 2 and 17 are autonomous:###

1 1217 4

3

2

′ = +′ = −′ = −

y yy y yy y

All the others are nonautonomous.


(b) The isoclines of an autonomous equation are horizontal lines (i.e., if you follow along ahorizontal line y k= in the ty plane, the slopes of the line elements do not change).Another way to say this is that solutions for y t( ) through any y all have the same slope.

!!!! Orthogonal Trajectories

44. (a) Starting with f x y c, ( ) = , we differentiate implicitly getting the equation

∂∂

+∂∂

=fx

dx fy

dy 0

Solving for ′ =y dydx

, we have

dydx

fxfy

= −∂∂∂∂

.

These slopes are the slopes of the tangent lines.

(b) Taking the negative reciprocal of the slopes of the tangents, the orthogonal curves satisfy

dydx

fyfx

=∂∂∂∂

.

(c) Given f x y x y, ( ) = +2 2 , we have

∂∂

=fy

y2 and ∂∂

=fy

x2 ,

so our equation is dydx

yx

= . Hence, from part (b) the orthogonal trajectories satisfy the

differential equation

dydx

ff

yx

y

x= = ,

which is a separable equation having solution y kx= .


!!!! More Orthogonal Trajectories

45. y cx= 2. Here f x y yx

, ( ) = 2 so

f yxx = −2

3 , fxy =12 ,

and so the orthogonal trajectories satisfy

dydx

ff

xy

y

x= = −

2

or

2y dy x dx= − .

x

1

y

3–3

23

–1–2–3

–1–2 1 2

Orthogonal trajectories

Integrating, we have

y x K2 212

= − +

or

x y c2 22+ = .

Hence, we have a family of ellipses that are all orthogonal to members of the family y cx= 2.

Graphs of the orthogonal families are shown in the figure.

46. y cx

= 2 . Here f x y x y, ( ) = 2 so

f xyx = 2 , f xy = 2

and the orthogonal trajectories satisfy

dydx

ff

xy

y

x= =

2

or, in differential form, 2y dy x dx= . Integrating,

we have

y x C2 212

= + or 2 2 2y x K− = .

x

1

y

3–3

23

–1–2–3

–1–2 1 2


Hence, the preceding hyperbolas are orthogonal to the family of hyperbolas y cx

= 2 . Graphs of

the two orthogonal families of hyperbolas are shown.


47. xy c= . Here f x y xy, ( ) = so f yx = , f xy = . The

orthogonal trajectories satisfy

dydx

ff

xy

y

x= =

or, in differential form, y dy x dx= . Integrating,

we have the solution

y x C2 2− = .

Hence, the preceding family of hyperbolas areorthogonal to the hyperbolas xy c= . Graphs of

the orthogonal families are shown.

t

5y

5–5

–5

Orthogonal hyperbolas

!!!! Calculator or Computer

48. y c= . We know the orthogonal trajectories of

this family of horizontal lines is the family ofvertical lines x C= (see figure).

x

1

y

3–3

23

–1–2–3

–1–2 1 2


49. 4 2 2x y c+ = . Here

f x y x y, ( ) = +4 2 2

and f xx = 8 , f yy = 2 , so the orthogonal trajecto-

ries satisfy

dydx

ff

yx

yx

y

x= = =

28 4

or 4dyy

dxx

= , which has the implicit solution the

family y Cx4 = where C is any constant different

from zero. These orthogonal families are shownin the figure.

x

1

y

2–2

2

–1

–2

–1 1



50. x cy2 34= . Here

f x y xy

, ( ) =2

34

and f xyx = 2 3 , f x

yy = −34

2

4 . The differential equa-

tion of the orthogonal family is

dydx

ff

xy

y

x= =

−32

x

1

y

2–2

2

–1

–2

–1 1


or 2 3y dy x dx= − , which has the general solution 2 32 2y x C+ = , where C is any real constant.

These orthogonal families are shown in the figure.

51. x y cy2 2+ = . Here f x y x yy

, ( ) =+2 2

, so

f xyx =

2 , f y xyy =−2 2

2 .

The differential equations of the orthogonal family are

dydx

ff

y xxy

y

x

y xyxy

= = =−

−2 2

2

2

2 2

2.

We are unable to solve this equation analytically, so we use a different approach inspired bylooking at the graph of the original family.

Completing the square of the original equation, we can write x y cy2 2+ = as

x y c c22 2

2 4+ −FH IK = , and recognize that this describes a family of circles centered at 0

2, cFH IK

passing through the origin. We propose that the orthogonal family to the original family consists

of another set of circles, x C y C−FH IK + =

2 4

22

2 centered at C

20, FH IK and passing through the origin.


To verify this conjecture we rewrite thisequation for the second family of circles as

x y Cx2 2+ = , which gives C g x y x yx

= ( ) =+,

2 2

or g x yxx =−2 2

2 , g yxy =

2 . Hence the family

orthogonal to the proposed second familysatisfies the equation

dydx

gg

xyx y

y

x= =

−2

2 2 ,

x

y

2

–2

–2 2

Orthogonal circles

which indeed has slopes perpendicular to those orthogonal to the original family derived above.Hence the original family of circles (centered on the y-axis) and the second family of circles(centered on the x-axis) are indeed orthogonal. These families are shown in the figure.

!!!! The Sine Function

52. The general equation is

y y2 2 1+ ′( ) =

or

dydx

y= ± −1 2 .

Separating variables and integrating, we get

± = +−sin 1 y x c or y x c x c= ± + = ± +sin sinb gc h b g .This is the most general solution. Note that cos x is included because cos sinx x= −FH IK

π2

.

!!!! Disappearing Mothball

53. (a) We have dVdt

kA= − , where V is the volume, t is time, A is the surface area, and k is a

positive constant. Because V r=43

3π and A r= 4 2π , the differential equation becomes

4 42 2π πr drdt

k r= −

or

drdt

k= − .


Integrating, we find r t kt c( ) = − + . At t = 0, r = 12

; hence c = 12

. At t = 6, r = 14

; hence

k = 124

, and the solution is

r t t( ) = − +1

2412

,

where t is measured in months and r in inches. Because we can’t have a negative radiusor time, 0 12≤ ≤t .

(b) Solving − + =1

2412

0t gives t = 12 months or one year.

!!!! Four-Bug Problem

54. (a) According to the hint, the distance between the bugs is shrinking at the rate of 1 inch persecond, and the hint provides an adequate explanation why this is so. Because the bugsare L inches apart, they will collide in L seconds. Because their motion is constantlytowards each other and they start off in symmetric positions, they must collide at a pointequidistant from all the bugs (i.e., the center of the carpet).

(b) The bugs travel at 1 inch per second for L seconds, hence the bugs travel L inches each.

(c)

rd θ

0

Q

r P

A

r , θ r dr +

dr B

This sketch of text Figure 1.3.8(b) shows a typical bug atP r= ( ), θ and its subsequent position A r dr dx + +( ), θ θ

as it heads toward the next bug at Q r= +FH IK, θ π2

. Note

that dr is negative, and considers that dθ is a very smallangle that is exaggerated as the drawing shows.

Consider the small shaded triangle ABP. For small dθ :

• angle BAP is approximately a right angle,

• angle APB OQP= =angle π4

,

• side BP lies along QP

Hence triangle ABP is similar to triangle OQP, which is a right isoscelestriangle, so − ≈dr rdθ .

Solving this separable DE gives r ce= −θ , and the initial condition r 0 1( ) = gives

c = 1. Hence our bug is following the path r e= −θ , and the other bugs’ paths simply shift

θ by πe

for each successive bug.


!!!! Radiant Energy

55. Separating variables, we can write dTT M

kdt4 4−= − . We then write

1 1 12

124 4 2 2 2 2 2 2 2 2 2T M T M T M M T M M T M−

=+ −

=−

−+2b gb g b g b g .

Integrating

1 1 22 2 2 22

T M T MdT kM dt

−−

+RST

UVW = − ,

we find the implicit solution

12

1 21 2M

M TM T M

TM

kM t cln tan−+

− FH IK = − +−

or in the more convenient form

ln arctanM TM T

TM

kM t C+−

+ FH IK = +2 4 3 .


56. Student Project

SECTION 1.4 Euler’s Method: Numerical Analysis 35

1.4 Euler’s Method: Numerical Analysis

!!!! Easy by Calculator ′ =y ty

, y 0 1( ) =

1. (a) Using step size 0.1 we enter t0 and y0, then calculate row by row to fill in the following

table:

n t t hn n= +−1 y y hyn n n= + ′− −1 1 ′ =y tynn

n

0 0 1 01

0=

1 0.1 1 011

01. .=

2 0.2 1.01 02101

01980..

.=

3 0.3 1.0298 0310298

02913..

.=

y2 0 2 101. .( ) = , y3 03 10298. .( ) =

we get

Euler’s Method h =( )0.1

n tn yn ′yn

0 0 1 0

1 0.1 1 0.1

2 0.2 1.01 0.19802

3 0.3 1.0298 0.29132


(b) Using step size 0.05, we recalculate which requires twice as many steps. We get thefollowing results.


n tn yn ′yn

0 0 1 0

1 0.05 1 0.05

2 0.1 1.0025 0.0998

3 0.15 1.0075 0.1489

4 0.2 1.0149 0.1971

5 0.25 1.0248 0.2440

6 0.3 1.03698 0.2893

y4 0 2 10149. .( ) ≈ , y6 0 3 1037. .( ) ≈

(c) Solving the IVP ′ =y ty

, y 0 1( ) = by separation of variables, we get y dy t dt= .

Integration gives

12

12

2 2y t c= + .

The initial condition y 0 1( ) = gives c = 12

and the implicit solution y t2 2 1− = . Solving

for y gives the explicit solution

y t t( ) = +1 2 .

To four decimal place accuracy, the exact solutions are y 0 2 10198. .( ) = andy 03.( ) = 1.0440. Hence, the errors in Euler approximation are

h y yy y

h y yy y

= = ( ) − ( ) = − == ( ) − ( ) = − =

= = ( ) − ( ) = − == ( ) − ( ) = − =

01 02 0 2 10198 10100 0009803 0 3 00142,

0 05 02 0 2 10198 10149 0005003 0 3 10440 10370 0007

2

3

2

6

. : . . . . . ,. . .

. : . . . . . ,. . . . .

errorerror 1.0440 1.0298 errorerror

Euler approximations are both high, but the smaller stepsize gives smaller error.!!!! Calculator Again ′ =y ty , y 0 1( ) =

2. (a) For each value of h we calculate a table as in Problem 1, with ′ =y ty . The results are

summarized as follows.


Comparison of Step Sizes

h = 1 h = 0.5 h = 0.25 h = 0.125

t y t y t y t y

0 1 0 1 0 1 0 1

1 1 0.5 1 0.25 1 0.125 1

1 1.25 0.50 1.062 0.250 1.0156

0.75 1.195 0.375 1.0474

1 1.419 0.50 1.0965

0.625 1.1650

0.750 1.2560

0.875 1.3737

1 1.5240

(b) Solve the IVP ′ =y ty , y 0 1( ) = by separating variables to get dyy

tdt= . Integration yields

ln y t c= +2

2. Using the initial condition y 0 1( ) = gives c = 0. Solving for y gives the

explicit solution y t et( ) = 2 2/ , so y e1 164871 2( ) = ≈ . .

hhhh

= = − == = − == = − == = − =

1 16487 1 0648705 16487 125 03987025 16487 1419 022970125 16487 1524 01247

: . .. : . . .. : . . .. : . . .

errorerror errorerror

!!!! Computer Help Advisable

3. Using a spreadsheet and Euler’s method we obtain the following values:

Spreadsheet Instructions for Euler’s Method

A B C D

1 n tn y y hyn n n= + ′− −1 1 3 2t yn n−

2 0 0 1 = −3 22 2tB C3 = +A2 1 = +B2 1. = + ×C D2 1 2.

′ = −y t y3 2 , y 0 1( ) = ; 0 1,

Using step size h = 01. and Euler’s method we obtain the following results. (Table shows printout of only selected values.)



t y t y

0 1 0.6 0.6822

0.1 0.9 0.7 0.7220

0.2 0.813 0.8 0.7968

0.3 0.7437 0.9 0.9091

0.4 0.6963 1.0 1.0612

0.5 0.6747

Smaller steps give higher approximate values y tn na f. The DE is not separable so we have no

exact solution for comparison.

4. ′ = + −y t e y2 , y 0 0( ) = ; 0 2,

Using step size h = 0 01. , and Euler’s method we obtain the following results. (Table shows printout of only selected values.)


t y t y

0 0 1.2 1.2915

0.2 0.1855 1.4 1.6740

0.4 0.3568 1.6 2.1521

0.6 0.5355 1.8 2.7453

0.8 0.7395 2.0 3.4736

1.0 0.9858

Smaller steps give higher approximate values y tn na f. The DE is not separable so we have no

exact solution for comparison.


5. ′ = +y t y , y 1 1( ) = ; 1, 5

Using step size h = 0 01. and Euler’s method we obtain the following results. (Table shows printout of only selected values.)


t y t y

1 1 3.5 6.8792

1.5 1.8078 4 8.5696

2 2.8099 4.5 10.4203

2.5 3.9942 5 12.4283

3 5.3525

Smaller steps give higher y tn na f. The DE is not separable so we have no exact solution for

comparison.

6. ′ = −y t y2 2 , y 0 1( ) = ; 0, 5

Using step size h = 0 01. and Euler’s method we obtain following results. (Table shows print outof only selected values.)


t y t y

0 1 3 2.8143

0.5 0.6992 3.5 3.3464

1 0.7463 4 3.8682

1.5 1.1171 4.5 4.3843

2 1.6783 5 4.8967

2.5 2.2615

Smaller steps give higher approximate values y tn na f. The DE is not separable so we have noexact solution for comparison.


7. ′ = −y t y , y 0 2( ) =



t y t y

0 2 0.6 1.2211

0.1 1.8075 0.7 1.1630

0.2 1.6435 0.8 1.1204

0.3 1.5053 0.9 1.0916

0.4 1.3903 1 1.0755

0.5 1.2962

Smaller steps give higher y tn na f. The DE is not separable so we have no exact solution for

comparison.

8. ′ = −y ty

, y 0 1( ) =

Using step size h = 01. and Euler’s method we obtain the following results. (Table shows printout of only selected values.)


t y t y

0 1 0.6 0.8405

0.1 1.0000 0.7 0.7691

0.2 0.9900 0.8 0.6781

0.3 0.9698 0.9 0.5601

0.4 0.9389 1 0.3994

0.5 0.8963

The analytical solution of the initial-value problem is

y t t( ) = −1 2 ,

whose value at t = 1 is y 1 0( ) = . Note, however, that the solution to this IVP does not exist for

t > 1. Hence, the absolute error at t = 1 is 0.3994. You can experiment yourself to see how thiserror is diminished by decreasing the step size or by using a more accurate method like theRunge-Kutta method.


9. ′ =y yt

sin , y 2 1( ) =



t y t y

2 1 2.6 1.2366

2.1 1.0418 2.7 1.2727

2.2 1.0827 2.8 1.3079

2.3 1.1226 2.9 1.3421

2.4 1.1616 3 1.3755

2.5 1.1995

Smaller stepsize predicts lower value. The DE is not separable so we have no exact solution forcomparison.

10. ′ = −y ty , y0 1=



t y t y

0 1 0.6 0.8375

0.1 0.9955 0.7 0.7850

0.2 0.9812 0.8 0.7284

0.3 0.9574 0.9 0.6692

0.4 0.9249 1 0.6086

0.5 0.8845

Smaller step size predicts lower value. The analytical solution of the initial-value problem is

y t e t( ) = − 2 2

whose exact value at t = 1 is y 1 06065( ) = . . Hence, the absolute error at t = 1 is

error = − =06065 06086 0 0021. . . .


!!!! Stefan’s Law Again dTdt

T= −005 3004 4. a f , T 0 400( ) =

11. Solution diverges at the values of the parameters stated in the problem.

!!!! Nasty Surprise

12. ′ =y y2 , y 0 1( ) =

Using Euler’s method with h = 0 25. we obtain the following values.


t y ′y = y2

0 1 1

0.25 1.25 1.5625

0.50 1.6406 2.6917

0.75 2.3135 5.3525

1.00 3.6517

Euler’s method estimates the solution at t = 1 to be 3.6517, whereas from the analytical solution

y tt

( ) =−1

1 or the direction field, we can see that the solution blows up at 1. So Euler’s method

gives an approximation far too small.

!!!! Approximating e

13. ′ =y y , y 0 1( ) =

We have estimated the solution of this IVP at t = 1 using Euler’s method with different step sizesh. The true value of y et= for t = 1 is e ≈ 2 7182818. … .

Euler’s Method

h y 1( ) e y− ( )1

0.5 2.25 0.4683

0.1 2.5937 0.1245

0.05 2.6533 0.0650

0.025 2.6850 0.0332

0.01 2.7048 0.0135

0.005 2.7115 0.0068

0.0025 2.7149 0.0034

0.001 2.7169 0.0013


We now use the fourth-order Runge-Kutta method with the same values of h, getting thefollowing values.

Runge-Kutta Method

h y(1) e y− ( )1

0.5 2.717346191 0.00093

0.1 2.718279744 021 10 5. × −

0.05 2.718281693 013 10 6. × −

0.025 2.718281820 087 10 8. × −

0.01 2.718281828 0 22 10 11. × −

Note that even with a large step size of h = 05. the Runge-Kutta method gives y 1( ) correct to

within 0.001.

!!!! Double Trouble or Worse

14. y y= 1 3 , y 0 0( ) =

(a) The solution starting at the initial point y 0 0( ) = never gets off the ground (i.e., it returnsall zero values for yn ).

(b) Starting with y 0 0 01( ) = . , the solution increases. We have given a few values in the

following table on a spreadsheet.

Euler’s Method ′ =y y1 3 , y 0 0.01( ) = (h = 0.1)

t y t y

0 0.01 3.5 3.5187

0.5 0.2029 4 4.3005

1 0.5454 4.5 5.1336

1.5 0.9913 5 6.0151

2 1.5213 5.5 6.9424

2.5 2.1241 6 7.9134

3 2.7918


(c) Direction field of ′ =y y1 3 for− ≤ ≤3 3t , 0 10≤ ≤y .

–5

5y

5–5x

′ =y y3

!!!! Roundoff Problems

15. If a round off error of ε occurs, then the solution of the new IVP ′ =y y , y A0( ) = + ε is

y t A e Ae et t t( ) = +( ) = +ε ε .

The difference between this perturbed solution and Aet is ε et . This difference at various

intervals of time will bet et et e

= ⇒ =

= ⇒ = ≈

= ⇒ = =

110 22,02620 485165195

10

20

differencedifferencedifference

ε

ε ε

ε ε, , .

Hence, the accumulate roundoff error grows at an exponential rate.

!!!! Think Before You Compute

16. Because y = 2 and y = −2 are constant solutions, any initial conditions starting at these valuesshould remain there. On the other hand, a roundoff error in computations starting at y = −2 is

not as serious as the perturbed solution will move back towards the stable solution –2.

!!!! Euler’s Errors

17. (a) Differentiating ′ = ( )y f t y, gives

′′ = + ′ = +y f f y f f ft y t y .

Here we assume ft , f y and y f= are continuous, so ′′y is continuous as well, being

the sum of continuous functions.

(b) The expression

y t h y t y t h y t hn n h n+ = + ′ + ′′a f a f a f b g12

2*

is simply a statement of Taylor series for one term.


(c) Direct computation gives

e M hn+ ≤1

2

2.

(d) We can make the local discretization error en in Euler’s method less than a preassigned

value E by choosing h so it satisfies e Mh En ≤ ≤2

2, where M is the maximum of the

second derivative of ′′y on the interval t tn n, +1 . Hence, if h EM

≤2 , we have the

desired condition e En ≤ .

!!!! Three-Term Taylor Series

18. (a) Starting with ′ =y f t y,a f , and differentiating with respect to t, we get

′′ = + ′ = +y f t y f t y y f t y f t y f t yt y t y, , , , ,a f a f a f a f a f.Hence, we have the new rule

y y hf t y h f t y f t y f t yn n n n t n n y n n n n+ = + + +121

2, , , , .a f a f a f a f

(b) The local discretization error has order of the highest power of h in the approximation ofyn+1 , which in this case is 3.

(c) For the equation ′ = =y f t y ty

,a f we have f t yyt ,a f = 1 , f t y t

yy ,a f = − 2 and so the

preceding three-term Taylor series becomes

y y h ty

hy

tyn n

n

n n

n

n+ = + FHG

IKJ + −LNM

OQP1

22

312

1 .

Using this formula and a spreadsheet we get the following results.

Taylor’s Three-Term Series

Approximation of ′ =y ty

, y 0 1( ) =

t y t y

0 1 0.6 1.1667

0.1 1.005 0.7 1.2213

0.2 1.0199 0.8 1.2314

0.3 1.0442 0.9 1.3262

0.4 1.0443 1.0 1.4151

0.5 1.1185


The exact solution of the initial-value problem ′ =y ty

, y 0 1( ) = is y t t( ) = +1 2 , so we

have y 1 2 14142( ) = ≈ . …. Euler’s three-term method gave the value 1.4151, which has

an error of

2 14151 0 0009− ≈. . .

(d) For the differential equation ′ = =y f t y ty,a f we have f t y yt ,a f = , f t y ty ,a f = , so the

Euler three-term approximation becomes

y y ht y h y t yn n n n n n n+ = + + −12 21

2.

Using this formula and a spreadsheet, we arrive at the following results.

Taylor’s Three-Term SeriesApproximation of ′ =y ty , y 0 1( ) =

t y t y

0 1 0.6 1.1962

0.1 1.005 0.7 1.2761

0.2 1.0201 0.8 1.3749

0.3 1.0458 0.9 1.4962

0.4 1.1083 1.0 1.6444

0.5 1.1325

The solution of ′ =y ty , y 0 1( ) = is y t et( ) = 2 2 , so y e1 1649( ) = ≈ . …. Hence the error

at t = 1 using Euler’s three-term method is

e − ≈16444 0 0043. . .

!!!! Richardson’s Extrapolation

19. ′ =y y , y 0 1( ) =

We first estimate the solution y at t = 01. using Euler’s method twice: first using one step, thenagain using two steps.

One step, with h = 01.

y y hy y1 0 0 101 0 0 1 01 1 11 01. . . .( ) ≈ ( ) + ′( ) = + ( )( ) = = ( )


Two steps, with h = 0 05.

= + ( )( ) == + ( )( ) = = ( )

1 005 1 105105 0 05 105 11025 012

. .. . . . .y

We now have two approximations of y 01.( ) , and Richardson’s extrapolation gives a closer

estimate of y 01.( ).

y y y01 2 01 01 2 11025 112 1. . . . .( ) ≈ ( ) − ( ) = ( ) − = 1.105 .

We continue by estimating y at t = 02. using both one step and two steps, starting fromy 01 1105. .( ) ≈ .

One step, with h = 01.

= + ( )( ) = = ( )1105 01 1105 12155 021. . . . .y

Two steps, with h = 0 05.

= + ( )( ) == + ( )( ) = = ( )

1105 005 1105 116025116025 0 05 116025 12183 0 22

. . . .

. . . . .y

Using Richardson’s extrapolation, we have a final estimate of y 0 2.( ).

y y y0 2 2 02 0 2 2 12183 12155 122112 1. . . . . .( ) ≈ ( ) − ( ) = ( ) − =

The exact solution is y e0 2 122140 2. ..( ) = ≈ . Note that Richardson’s extrapolation is much closerto the exact solution than both y1 02.( ), y2 02.( ).

20. ′ =y ty , y 0 1( ) =

Using Euler’s method we first estimate y at t = 01. . Using step size h = 01. we get

y1 01 1000. .( ) = .

If we cut the step size in half to h = 005. , we get

y2 01 10025. .( ) = .

From the two estimates of y 01.( ) we use Richardson’s formula to estimate more precisely,

y 01 2 10025 1000 1005. . . .( ) = ( ) − = .

To continue we estimate y at t = 02. using y 01 1005. .( ) = and step size h = 01. , getting

y1 02 10150. .( ) = . We now cut the step size in half and estimate y at t = 02. in two steps usingh = 005. and getting y2 0 2 10176. .( ) = . From these two estimates of y 0 2.( ) , we use Richardson’s

formula to estimate more precisely

y 0 2 2 10176 10151 10201. . . .( ) = ( ) − = .The exact solution (see problem 2) is y t et( ) = 2 2/ , y e( . ..02) 102020 02= ≈ , so the error ofRichardson’s extrapolation is 1 10 4× − .


21. ′ =y y2 , y 0 1( ) =

Using Euler’s method we first estimate y at t = 01. using step size h = 01. , getting

y1 01 11. .( ) = .

We now cut the step size in half and estimate y at t = 01. in two steps using h = 005. and getting

y2 01 11051. .( ) = .

From the two estimates of y 01.( ) we use Richardson’s formula to estimate more precisely

y 01 2 11051 11 11102. . . .( ) = ( ) − = .

To continue we start at y 01 11102. .( ) = and estimate y at t = 02. using y 01 11102. .( ) = and step sizeh = 01. , getting y1 02 12335. .( ) = . Then we cut the step size in half and estimate y at t = 02. in twosteps using h = 005. , getting y2 0 2 12405. .( ) = . From these two estimates of y 0 2.( ) we use

Richardson’s formula to estimate more precisely y 0 2 2 12405 12335 12475. . . .( ) = ( ) − = . Solving the

IVP ′ =y y2 , y 0 1( ) = by separation of variables, we get dyy

dt2 = . Integration gives

yt c

= −+1 .

The initial condition y 0 1( ) = gives c = −1 and the exact solution is

yt

=−1

1.

y 0 2 11 0 2

125..

.( ) =−

= , so the error of Richardson’s extrapolation is 0.0025.

22. ′ = ( )y tysin , y 0 1( ) =

Using Euler’s method we first estimate y at t = 01. using step size h = 01. getting the valuey1 01 1.( ) = . We now cut the step size in half and estimate y at t = 01. in two steps using h = 005.and get y2 01 10025. .( ) = . From the two estimates of y 01.( ) we use Richardson’s formula to get

y 01 2 10025 1 10050. . .( ) = ( ) − = .

We now continue and estimate y at t = 02. using y 01 10050. .( ) = and step size h = 01. , gettingy1 02 10150. .( ) = . Next we now cut the step size in half and estimate y at t = 02. in two steps usingh = 005. , getting y2 0 2 10176. .( ) = . From these two estimates of y 0 2.( ) we use Richardson’s

formula to estimate

y 0 2 2 10176 10150 10202. . . . .( ) = ( ) − =

There is no exact solution for this problem.


!!!! Runge-Kutta Method

23. ′ = −y t y3 2 , y 0 1( ) = ; [0, 1]

Using the fourth-order Runge-Kutta method and h = 01. we arrive at the following table ofvalues.

Runge-Kutta Method, ′ = −y t y3 2 , y 0 1( ) =

t y t y

0 1 0.6 0.7359

0.1 0.9058 0.7 0.7870

0.2 0.8263 0.8 0.8734

0.3 0.7659 0.9 0.9972

0.4 0.7284 1.0 1.1606

0.5 0.7173

We compare this with #3 y 1 10612b g ≈ . for h = 01. . Exact solution is not possible at this stage.

24. ′ = −y t y , y 0 2( ) =


Runge-Kutta Method, ′ = −y t y , y 0 2( ) =

t y t y

0 2 0.6 1.2464

0.1 1.8145 0.7 1.1898

0.2 1.6562 0.8 1.148

0.3 1.5225 0.9 1.1197

0.4 1.4110 1.0 1.1036

0.5 1.3196

We compare this with #7. y 1 1046b g ≈ . for step h = 01. ; y 1 107545( ) ≈ . for step h = 0 05. . Exact

solution is not possible at this stage.

25. ′ = −y ty

, y 0 1( ) =



Runge-Kutta Method, ′ = −y ty

, y 0 1( ) =

t y t y

0 1 0.6 0.8000

0.1 0.9950 0.7 0.7141

0.2 0.9798 0.8 0.6000

0.3 0.9539 0.9 0.4358

0.4 0.9165 1.0 0.04880

0.5 0.8660

We compare this with #8. Euler’s method for step h = 01. ; y 1 03994( ) ≈ . . Exact y 1 0( ) = . Runge-

Kutta solution is much closer.

26. ′ = −y ty , y 0 1( ) =

Using the 4th-order Runge Kutta method and h = 0 01. to arrive at the following table. (Tableshows print out of only selected values.)

Runge-Kutta Method,′ = −y ty , y 0 1( ) =

t y t y

0 1 0.6 0.8353

0.1 0.9950 0.7 0.7827

0.2 0.9802 0.8 0.7261

0.3 0.9560 0.9 0.6670

0.4 0.9231 1 0.6065

0.5 0.8825

We compare this with #10. Euler’s method for step h = 01. ; y 1 06086( ) ≈ . . Exact y 1 06065( ) = . .

Runge-Kutta solution is exact within given accuracy.

!!!! Integral Equation

27. (a) Starting with

y t y f s y s dstt

( ) = + ( )z00

,a f

we differentiate respect to t, getting

′ = ( )y f t y t,a f .


We also have y t y0 0a f = . Conversely, starting with the initial-value problem

′ = ( )y f t y t,a f , y t y0 0a f =we integrate getting the solution

y t f s y s ds ctt

( ) = ( ) +z ,a f0

.

Using the initial condition y t y0 0a f = , we get the constant c y= 0 . Hence, the integral

equation is equivalent to IVP.

(b) The initial-value problem, ′ = ( )y f t , y y0 0( ) = , is transformed into the integral equation

y t y f s dst( ) = + ( )z0 0

.

To find the approximate value of the solution at t T= , we evaluate the precedingintegral at t T= using the Riemann sum with left endpoints, getting

y T y f s ds

y h f f h f T h

T( ) = + ( )

≈ + ( ) + ( ) + + −( )z0 0

0 0 … .

If we, however, write the expression asy T y h f f h f T h

y hf h hf T hy hf h hf T hy hf h hf T h y hf T h

y h T hy

n

n

n

( ) = + ( ) + ( ) + + −( )= + ( ) + + −( )= + ( ) + + −( )= + ( ) + + −( ) + + −( )

= + −( )=

−

−

0

1

2

3 1

1

0

23

…………

… …

.

we get the desired conclusion.

(c) The Riemann sum only holds for integrals of the form f t dtab( )z .

!!!! Computer Lab: Other Methods

28. Sample study of different numerical methods. Using step size h = 01. , we solve the IVP ofProblem 5 ′ = +y t y , y 1 1( ) = by several different methods. The table shows a print out for

selected values of y.


Fourth Order Runge-Kutta Method

t y t y

1 1 3.5 6.8910

1.5 1.8100 4 8.5840

2 2.8144 4.5 10.4373

2.5 4.0010 5 12.4480

3 5.3618

Instructors might also ask students to compare the results of Euler method with step h = 01. andh = 0 01. to other methods (for example Adams-Bashforth method or Dormand Prince method)andexplain which method is most accurate.

!!!! Suggested Journal Entry I

29. Student Project

!!!! Suggested Journal Entry II

30. Student Project

SECTION 1.5 Picard’s Theorem: Theoretical Analysis 53

1.5 Picard’s Theorem: Theoretical Analysis

!!!! Picard’s Conditions

1. (a) ′ = = −y f t y ty,a f 1

Hence f ty = − . The fact that f is

continuous for all t tells us a solutionexists passing through each point in thety plane. The further fact that thederivative f y is also continuous for all t

and y tells us that the solution is unique.Hence, there is a unique solution of thisequation passing through y 0 0( ) = . The

direction field is shown in the figure.

3–3t

–3

3y

(b) The condition holds in entire yt plane.

(c) Not applicable - the answer to part (a) is positive.

2. (a) ′ =−y yt

2

Here f t y yt

,a f = −2 , fty = −1. The functions f and f y are continuous for t ≠ 0 , so there

is a unique solution passing through any initial point y t y0 0a f = with t0 0≠ . When t0 0=

the derivative ′y is not only discontinuous, it isn’t defined. No solution of this DEpasses through points t y0 0,a f with t0 0= . In particular the DE with IC y 0 1( ) = does not

make sense.

(b) Not applicable - the answer to part (a) is negative.

(c) If we think of DEs as models for physi-cal phenomena, we might be tempted toreplace t0 in the IC by a small number

and examine the unique solution, whichwe know exists. It would also be usefulto draw the direction field of this equa-tion and see the big picture. The direc-tion field is shown in the figure.

3t

–2

6y


3. (a) ′ =y y4 3

Here

f t y y

f yy

,

.

a f ==

4 3

1 343

–4

4y

4–4t

Here f and f y are continuous for all t and y, so by Picard’s theorem we conclude that the

DE has a unique solution through any initial condition y t y0 0a f = . In particular, there willbe a unique solution passing through y 0 0( ) = , which we know to be y t( ) ≡ 0 . The

directions field of the equation is shown in the figure.

(b) The condition holds in entire yt plane.


4. (a) ′ =−+

y t yt y

Here both

f t y t yt y

f tt yy

,a f = −+

= −+( )2

2–4

4y

4–4t

are continuous for t and y except when y t= − . Hence, there is a unique solution passingthrough any initial condition y t y0 0a f = as long as y t0 0≠ − . When y t= − the derivative

′y is not only discontinuous but also not even defined, so there is really no need to

resort to Picard’s theorem to conclude there is no solution passing through such points.

(b) The condition holds for entire ty plane except the line y t= − .



5. (a) ′ =+

yt y

12 2

Here both

f t yt y

f t y yt y

y

,

,

a f

a f b g

=+

= −+

1

2

2 2

2 2 2–2

2y

2–2t

are continuous for all t and y except at the point y t= = 0 . Hence, there is a uniquesolution passing through any initial point y t y0 0a f = except y 0 0( ) = . In this case f does

not exist, so the IVP does not make sense. The direction field of the equation illustratesthese ideas (see figure).


(c) It may be useful to replace the initial condition y 0 0( ) = by y y0 0b g = with small but

nonzero y0 .

6. (a) ′ =y ytan

Here

f t y yf yy

, tansec

a f == 2

are both continuous except at the points

y = ± ±π π2

32

, ,… .

2–2t

–3 /2

y

π

3 /2π

Hence, there exists a unique solution passing through y t y0 0a f = except when

y = ± ±π π2

32

, ,… .

The IVP problem passing through π2

does not have a solution. It would be useful to look

at the direction field to get an idea of the behavior of solutions for nearby initial points.The direction field of the equation shows that where the DE does not work the slope hasbecome vertical (see figure).


(c) It may be useful to replace the initial condition y 02

( ) =π by y y0 0( ) = with y0 close

enough to π2

.


7. (a) ′ = −y yln 1

Here

f t y y

fyy

, lna f = −

=−

11

1

are both continuous for all t and y aslong as

y ≠ 1,

–4

4y

4–4t

where neither is defined. Hence, there is a unique solution passing through any initialpoint y t y0 0a f = with y0 1≠ . In particular, there is a unique solution passing throughy 0 2( ) = . The direction field of the equation illustrates these ideas (see figure).

(b) The condition holds for entire ty plane except the line y = 1.


8. (a) ′ =−

y yy t

Here

f t y yy t

f ty ty

,a f =−

= −−( )2 –4

4y

4–4t

are continuous for all t and y except when y t≠ where neither function exists. Hence,we can be assured there is a unique solution passing through y t y0 0a f = except whent y0 0= . When t y0 0= the derivative isn’t defined, so IVP problems with these IC doesnot make sense. Hence the IVP with y 1 1( ) = is not defined. See figure for the direction

field of the equation.

(b) Not applicable - the answer to (a) is negative.

(c) It may be useful to replace the initial condition y 1 1( ) = by y y1 0( ) = with y0 close enough

to 1.


!!!! Linear Equations

9. ′ + ( ) = ( )y p t y q t

For the first-order linear equation, we can write ′ = ( ) − ( )y q t p t y and so

f t y q t p t yf t y p ty

,, .a fa f

= ( ) − ( )

= − ( )

Hence, if we assume p t( ) and q t( ) are continuous, then Picard’s theorem holds at any pointy t y0 0a f = .

!!!! Eyeballing the Flows

For the following problems it appears from the figures given in the text that:

10. A unique solution will pass through each point A, B, C, and D and the solutions appear to existfor all t.

11. A unique solution passes through A and B defined for negative t; no unique solution passingthrough C where derivative is not uniquely defined; and a unique solution passes through D forpositive t.

12. Unique solutions exist passing through points B and C on intervals until the solution curvereaches the t-axis, where finite slope does not exist. Nonunique solutions at A; possibly uniquesolutions at D where t y= = 0 .

13. A unique solution will pass through each of the points A, B, C, and D. Solutions appear to existfor all t.


15. A unique solution will pass through each of the points B, C, and D. Solutions exist only for t tA>

or t tA< because all solutions appear to leave from or go toward A, where there is no unique

slope.

16. Unique solutions will pass through each of the points A, B, C, and D. Solutions appear to existfor all t.




!!!! Local Conclusions

19. (a) f t y y,a f = 2 , f yy = 2

are both continuous for all t, y so byPicard’s theorem there is a uniquesolution passing through any point t, y.Hence the region, where conditionholds is entire ty plane. The weaknessof Picard’s theorem is that it does notprovide the interval on which exis-tence and uniqueness holds.

(b)

–3 3t

–3

3y t = 1

Solution of ′ =y y2 , y 0 1( ) =

(c) The separated equation is y dy dt− =2 . Integrating gives the result − = +−y t c1 and

solving for y, we get −+1

t c. Substituting the initial conditions y 0 1( ) = , gives c = −1.

Hence, we have y tt

( ) =−1

1, t < 1, y > 0. The interval over which this solution is defined

cannot pass through t = 1, and the solution with IC y 0 1( ) = exists on the interval−∞, 1a f .

(d) Because Picard’s theorem holds for all t, y we conclude there exists a unique solution to′ =y y2 , y t y0 0a f = for any t y0 0,a f . We cannot determine the size of the interval of

existence. We can find it, however, for any IVP if we solve it getting

y tt t y

( ) = −− −

1

010

.

Hence, the interval over which this solution is defined cannot pass through t ty

= +00

1 ,

which implies an interval of

−∞ +FHG

IKJ, t

y00

1

for positive y0 and

ty0

0

1− ∞FHG

IKJ,

for negative y0.


!!!! Nonuniqueness

20. For t t< 0 , the solution is y t( ) ≡ 0 and hence satisfies ′ =y y . For t t> 0 , we have

y t t= −14 0

2a f . Hence

′ = −y t t12 0a f

satisfies the equation ′ =y y2 . At t t= 0 the left-hand derivative of y t( ) ≡ 0 is 0, and the right-

hand derivative of y t t t( ) = −14 0

2a f is 0, so they agree.

!!!! Nonuniqueness Again

21. Here

′ =y y1 3 , ∂∂

= −fy

y13

2 3.

Because f y= 1 3 is continuous for all t y,a f , Picard’s theorem says that there exists a solution

through any point y t y0 0a f = . However, f yy = −13

2 3 is not continuous when y = 0 , and so

Picard’s theorem does not guarantee a unique solution through any point where y = 0 .

In fact we can find an infinite number of solutions passing through the origin. We firstseparate variables, getting y dy dt− =1 3 , and integrating, we find

32

2 3y t c= + .

Picking the initial condition y 0 0( ) = , we find c = 0 . Hence, we have found one solution of the

initial-value problem

′ =y y1 3, y 0 0( ) =

as

y t t( ) = ±FH IK23

3 23 2 .


But clearly, y t( ) ≡ 0 is another solution. In fact,

we can paste these solutions together, getting aninfinite number of solutions to the initial-valueproblem as

y tt c

t c t cb g b g=<

±FHGIKJ − ≥

RS|T|0

23

3 23 2

where c is any non-negative real number. A fewof these solutions are plotted (see figure).

t

y

1

3

–3

2 3 4

Nonuniqueness of solutions through y 0 0( ) =

!!!! Hubbard’s Leaky Bucket

22. (a) f t h k hfh

kh

,a f = −∂∂

= −2

Because ∂∂fh

is not continuous at h = 0 , we cannot be sure of unique solutions passing

through points h t0 0a f = .

(b) Let us assume the bucket becomes emptyat t T= , and so we have h T( ) = 0 . We

can find an infinite number of solutions.

h t c kt t ck

h t t ck

( ) = −( ) <

( ) = >

140

2 for

for .t

h/k

1

1 2 3

2

Each one of these functions describes the bucket emptying. Hence, we don’t know whenthe bucket became empty.

(c) Setting 14

20c kt h−( ) = , we get

c kt h− = ±2 0 ,

then setting c = 0 , as it only represents a translation in the infinite number of ways thebucket can empty. We then solve for t, taking the positive root and get

tk

h=2

0 .


!!!! The Melted Snowball

23. (a) We are given ∂∂

= −Vt

kA , where A is the surface area of the snowball and k > 0 is the rate

at which the snowball decreases in volume. Given the relationships between the volume

of the snowball and its radius r, which is V r=43

3π , and between the surface area of the

snowball and its radius, given by A r= 4 2π , we can relate A and V by

A V V= FH IK =4 34

362 3

2 3 3 2 3ππ

π .

(b) Here

f t V kVfV

kV

,

.

a f = −∂∂

= − −

2 3

1 323

Because the uniqueness condition for Picard’s theorem does not hold when V = 0 , wecannot conclude that the IVP

∂∂

= −Vt

kV 2 3, V t0 0a f =

has a unique solution. Hence, we cannot tell when the snowball melted; the backwardssolution is not unique.

(c) Separating ∂∂

= −Vt

kV 2 3 where k > 0 ,

we have

V dV kdt− = −2 3 .


3 1 3V kt c= − + .

Using the initial condition V t0 0a f = wefind c kt= 0 , and hence

V t K t t( ) = −

−FH IK03

3

t

y

–2

20

10

–1 1 2–3

∂∂

= −Vt

kV 2 3. Solutions with y t0 0a f = .


where K k= 3 . But we also know V t( ) ≡ 0, is a solution of this initial-value problem, and

so we can piece together the nonzero solutions with the zero solution, getting the infinityfamily of solutions

V t K t t t t

t tb g = −

−FHGIKJ <

≥

RS|T|

03

0

0

30

.

(d) The function f t V V,a f = 2 3 does not satisfy the uniqueness condition of Picard’s

theorem when V = 0 .

!!!! The Accumulating Raindrop

24. (a) We are given dVdt

kA= , where A is the surface area of the raindrop and k > 0 is the rate

at which the raindrop increases in volume. We plug in the relationships

V r=43

3π , A r= 4 2π

between the volume V and area A of a raindrop and its radius r into dVdt

kA= getting

A V V= FHGIKJ =4 3

436

2 32 3 3 2 3π

ππ

hence

dVdt

kV= 2 3 .

(b) Separating variables in the above DE, we have

V dV kdt− =2 3 .


3 1 3V kt c= + .

Using the initial condition V t0 0a f = , we get the relation c kt= − 0 , and hence

V t K t t( ) =

−FH IK03

3


where K k= 3 . But clearly, V t( ) ≡ 0 is a solution of this initial-value problem, and so we

can piece together the nonzero solutions with the zero solution, getting the infinite familyof solutions

V tt t

K t t t tb g =<

−FHGIKJ ≥

RS|T|0

3

0

03

0

!!!! Other Drops

25. (a) dsdt

gs= 2 1 2

Separating variables, we can write

s ds gdt− =1 2 2 .


2 21 2s gt c= + .

Using the initial condition s 0 0( ) = , we get c = 0 , so we have a solution s t gt( ) =12

2 . But

clearly, s t( ) ≡ 0 is a solution, and we can “merge” these solutions together at any point

t0, getting the infinite number of solutions

s tt t

g t t t tb g b g=<

− ≥

RS|T|012

0

02

0

(b) The differential equation dsdt

gs= 2 1 2 does not provide a relationship between the

distance fallen s and time t, but a relationship between the velocity dsdt

of the object and

the distance fallen. Hence, the solution tells us the distance s the object has fallen fromthe time t0 when s was zero (i.e., when the object was dropped).

!!!! Picard Approximations

26. y t

y t s y ds s ds t t

y t s s s ds t t t

y t s s s s ds t t t t

t t

t

t

0

1 00 02

22

03 2

33 2

04 3 2

1

1 1 1 1 12

1 1 12

16

1

1 16

1 124

13

1

( ) ≡

( ) = + − = + −( ) = + −

( ) = + − + −FH IKRSTUVW = − + − +

( ) = + − − + − +LNM

OQP

RSTUVW = − + − +

z zzz

a f


27. y t t

y t s y ds s s ds t

y t s s ds t

y t s s ds t t

t t

t

t

0

1 00 0

2 0

3 02

1

1 1 1 1

1 1 1

1 1 1

( ) ≡ −

( ) = + − = + − −( ) = +

( ) = + − +( ) = − +

( ) = + − −( ) = − +

z zzz

a f

28. y t e

y t s y ds s e ds e t

y t s e s ds e t t

y t s e s s ds e t t t

t

t st t

st t

st t

0

1 00 02

22

03 2

33 2

04 3 2

1 1 12

1 12

16

12

1 16

12

124

16

12

( ) ≡

( ) = + − = + − = +

( ) = + − +FH IKLNM

OQP = − +

( ) = + − − +FH IKLNM

OQP = + − +

−

− −

− −

− −

z zzz

a f b g

29. y t t

y t s y ds s s ds ds t

y t s s ds t t

y t s s s ds t t t

t t t

t

t

0

1 00 00

2 02

32

03 2

1

1 1 1 1 1

1 1 1

1 1 13

1

( ) ≡ +

( ) = + − = + − +( ) = + − = −

( ) = + − −( )( ) = − +

( ) = + − − + = − + − +

z z zzz

a f

b g

!!!! Computer Lab

30. (a) We show how the computer algebra system Maple can be used to estimate the solution of#27 ′ = −y t y , y 0 1( ) = with starting function y t t0 1( ) = − . We leave for the reader the

other starting functions for #26, 28, and 29. In Maple open a new window and type theint() command. In this problem, because f t y t y,a f = − , y 0 1( ) = , we can find the

sequence of approximations

y t y f s y s ds s y s dsn nt

nt

+ ( ) = + ( ) = + − ( )z z1 0 0 01,a f a f

by typing

y ty t y ty t y ty t y ty t y ty t y ty t y t

0

1 0

2 1

3 2

4 3

5 4

6 5

1111111

: ;: int , ;: int , ;: int , ;: int , ;: int , ;: int , .

= −= + −= + −= + −= + −= + −= + −

a fa fa fa fa fa f


If you then hit the enter key you will see displayedy ty ty ty t t

y t t t

y t t t t

y t t t t t

0

1

2

32

43 2

54 3 2

65 4 3 2

11

11

13

1

112

13

11

601

1213

1

::::

:

:

:

= −= += − += − +

= − + − +

= − + − +

= − + − + − +

If you then type the Maple command

plot({2*exp(–t)+t–1, y6},

t=0..4, y=0..2),

you will get a plot of y6 and the solutiony t e ttb g = + −−2 1 as shown (see figure).

31

0.8

0

1.2

1.6

0.4

2t

y2

4

0.6

1

1.4

0.2

1.8

Picard Approximation

Picard’s sixth approximation to

′ = −y t y , y 0 1( ) =

(b) If you recall the Maclaurin series e t t tt− ≈ − + −( ) + −( ) +2 2 31 2 12

2 13

2!

… and carry out a

little algebra, you will convince yourself that these Picard’s approximations areconverging to the analytical solution y t e ttb g = + −−2 1. For most initial-value problems,

however, such a nice identification is not possible.

!!!! Calculator or Computer

31. ′ = =∂∂

= −

y f t y yfy

y

,a f 1 4

3 414

Note the direction field is only defined wheny ≥ 0 . Picard’s theorem guarantees existencethrough any point y t y0 0a f = , but not uniquenessfor points y t y0 0a f = when y0 0= . The direction

field shown illustrates these ideas.–3 3

t

2y

′ =y y1 4 ; DE does not exist for y < 0 .


32. ′ = = ( )∂∂

= ( )

y f t y tyfy

t ty

, sin

cos

a f

Picard’s theorem guarantees both existence anduniqueness for any point t y, 0a f. The direction

field shown also indicates these ideas.

5–5t

–2

2y

33. ′ = =∂∂

=

y f t y yfy

y

,a f 5 3

2 353

Picard’s theorem guarantees existence anduniqueness for all initial conditions y t y0 0a f = .

The direction field shown also illustrates thisfact.

–2

1 y

1 –1 t

34. ′ = = ( )∂∂

= ( ) =− −

y f t y tyfy

ty t t y

,

/ /

a f 1 3

2 3 1 3 2 313

13

The function f is continuous for all t y,a f , butf y is not continuous when y = 0 . Hence, we are

not guaranteed uniqueness through pointst y0 0,a f when y0 is zero. See figure for this

direction field.

3t

–3

3y

35. ′ = = −( )∂∂

= −( )−

y f t y y tfy

y t

,a f 1 3

2 313

The function f is continuous for all t y,a f , butf y is not continuous when y t= (it doesn’t

exist). Hence, the DE has a solution throughevery point t y0 0,a f but Picard’s theorem does

not guarantee uniqueness through points forwhich y t0 0= . See figure for this direction field.

4t

–4

4y

–4


36. ′ = = −

∂∂

= −

y f t y tt

y

fy t

,a f 6 3

3

2

The function f is continuous except when t = 0 ,hence there exists a solution through all pointst y0 0,a f except possibly when t0 0= . Also f y is

continuous except when t = 0, and so the DE hasa unique solution for all initial conditions exceptpossibly when t0 0= . The direction field of the

equation as shown indicates that no solutionspass through initial points of the form 0 0, ya f(see figure).

–1

1y

1–1t


37. Student Project

chapter 1 first-order differential...

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