Differential equations

Student Solutions Manual

for Blanchard, Deuaney, and Hall’s

Student Solutions Manual

for

Blanchard, Devaney, and Hall's

Differential Equations

Th ird Edition

Paul Blanchard

Boston University

; V BROOKS/COLE

CENGAGE Learning'

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BROOKS/COLE

CENGAGE Learning'

Student Solutions Manual for

Blanchard, Devaney, and Hall's

Differential Equations, Third Edition

Paul Blanchard

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CONTENTS

1 First-Order Differential Equations I

1.1 Modeling via Differential Equations.........................................................................................2

1.2 Analytic Technique: Separation of Variables......................................................................... 5

1.3 Qualitative Technique: Slope Fields...................................................................................... 15

1.4 Numerical Technique: Euler’s M ethod.............................................................................18

1.5 Existence and Uniqueness of Solutions................................................................................ 21

1.6 Equilibria and the Phase Line................................................................................................. 26

1.7 Bifurcations..................................................................................................................................31

1.8 Linear Equations.........................................................................................................................37

1.9 Integrating Factors for Linear Equations...............................................................................45

Review Exercises ............................................................................................................................. 54

2 First-Order Systems 63

2.1 Modeling via System s............................................................................................................... 64

2.2 The Geometry of Systems.........................................................................................................66

2.3 Analytic Methods for Special System s..................................................................................72

2.4 Euler’s Method for System s.....................................................................................................77

2.5 The Lorenz Equations............................................................................................................... 80

Review Exercises ..............................................................................................................................81

3 Linear Systems 85

3.1 Properties of Linear Systems and The Linearity Principle............................................... 86

3.2 Straight-Line Solutions.............................................................................................................92

3.3 Phase Planes for Linear Systems with Real Eigenvalues.............................................. 103

3.4 Complex Eigenvalues...........................................................................................................114

3.5 Special Cases: Repeated and Zero Eigenvalues................................................................. 121

3.6 Second-Order Linear Equations.........................................................................................130

3.7 The Trace-Determinant P lan e.............................................................................................140

3.8 Linear Systems in Three Dimensions...................................................................................144

Review Exercises ........................................................................................................................... 152

iv CONTENTS

4 Forcing and Resonance 159

4.1 Forced Harmonic Oscillators................................................................................................. 160

4.2 Sinusoidal Forcing................................................................................................................... 173

4.3 Undamped Forcing and R esonance.................................................................................... 180

4.4 Amplitude and Phase of the Steady State............................................................................186

4.5 The Tacoma Narrows Bridge..................................................................................................188

Review Exercises ...........................................................................................................................189

5 Nonlinear Systems 195

5.1 Equilibrium Point A nalysis....................................................................................................196

5.2 Qualitative Analysis................................................................................................................ 209

5.3 Hamiltonian Systems...............................................................................................................215

5.4 Dissipative System s.................................................................................................................219

5.5 Nonlinear Systems in Three Dim ensions............................................................................226

5.6 Periodic Forcing of Nonlinear Systems and C haos............................................................228

Review Exercises ...........................................................................................................................230

6 Laplace Transforms 239

6.1 Laplace Transform s................................................................................................................ 240

6.2 Discontinuous Functions........................................................................................................ 248

6.3 Second-Order Equations.........................................................................................................254

6.4 Delta Functions and Impulse Forcing..................................................................................264

6.5 Convolutions............................................................................................................................ 266

6.6 The Qualitative Theory of Laplace Transforms................................................................. 270

Review Exercises ...........................................................................................................................271

7 Numerical Methods 279

7.1 Numerical Error in Euler’s M ethod..................................................................................... 280

7.2 Improving Euler’s M ethod.....................................................................................................290

7.3 The Runge-Kutta M ethod.......................................................................................................294

7.4 The Effects of Finite Arithmetic...........................................................................................296

Review Exercises .......................................................................................................................... 296

8 Discrete Dynamical Systems 299

8.1 The Discrete Logistic Equation.............................................................................................300

8.2 Fixed Points and Periodic Points...........................................................................................302

8.3 Bifurcations.............................................................................................................................. 305

8.4 Chaos..........................................................................................................................................307

Review Exercises ...........................................................................................................................308

Appendices 311

A Changing Variables.................................................................................................................... 312

B The Ultimate G uess...................................................................................................................322

CHAPTER I

First-Order Differential Equations

2 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

EXERCISES FOR SECTION 1.1

1. We know the solution of the differential equation is P(t) = Poekl. It is a little easier to use the year

1800 as / = 0, so 1900 corresponds to / = 1 (X) and 2000 corresponds to t = 200. To compute k, we

5.3 = P( 0) = P0eko = P0

and

75 = P (I00) = P0ek]0° = 5.3ei00k

from Table 1.1. Therefore, 75 = 5.3e100*, which yields

",(75/5.3)

100

Hence, the prediction for the year 2000 is

/ * ( 2 0 0 ) = 5 .3 e <0-026498)<200) r . 1 0 6 |

Since the data is given in millions, this “prediction” is more than one billion people. It is too high by

a factor of almost 4.

3. (a) The equilibrium solutions correspond to the values of P for which dP/dt = 0 for all t. For this

equation, dP/dt = 0 for all t if P = 0 or P = 230.

(b) The population is increasing if dP/dt > 0. That is, P(1 — P / 230) > 0. Hence, 0 < P < 230.

(c) The population is decreasing if dP/dt < 0. That is, P(1 — P / 230) < 0. Hence, P > 230 or

P < 0. Since this is a population model, P < 0 might be considered “nonphysical.”

5. In order to answer the question, we first need to analyze the sign of the polynomial y 3 — y 2 — \2y.

Factoring, we obtain

y 3 - y 2 - 12 v = y ( y 2 - y - 1 2 ) = y(y - 4)0- + 3).

(a) The equilibrium solutions correspond to the values o f y for which d y /d t = 0 for all t. For this

equation, dy/dt = 0 for all / if y = —3, y = 0, or y = 4.

(b) The solution y(/) is increasing if dy/dt > 0. That is, — 3 < y < 0 or y > 4.

(c) The solution >(/) is decreasing if dy/dt < 0. That is, y < — 3 or 0 < y < 4.

7. The rate of learning is dL/dt. Thus, we want to know the values of L between 0 and 1 for which

dL/dt is a maximum. As k > 0 and d L / d t = k(\ — L), d L / d t attains it maximum value at L = 0.

9. (a) We have 0) = L j( 0) = 0. So Jillian’s rate of learning at / = 0 is d L j/d t evaluated at

t = 0. At t = 0, we have

d L j

= 2(1 - L j ) = 2.

dt

Beth’s rate of learning at t = 0 is

— 3 0 — L b )2 = 3.

d t

Hence Beth’s rate is larger.

1.1 Modeling via Differential Equations 3

(b) In this case, Z>js(0) = L j (0) = 1 /2. So Jillian’s rate of learning at / = 0 is

dL j

- ~ = 2 ( \ - L j ) = \

at

because L j = 1 /2 at t = 0. Beth’s rate of learning at / = 0 is

dL B o 3

- f - = 3(1 - L B)2 = -

dt 4

because L b = 1/2 at / = 0 . Hence Jillian’s rate is larger.

(c) In this case, L b (0) = Lj (0) = 1 /3. So Jillian’s rate of learning at / = 0 is

d L j 4

— = 2 ( 1 - / . , ) = 5 .

Beth’s rate of learning at / = 0 is

d Ln , 4

— = 3 (1 - L b)2 =

They arc both learning at the same rate when t = 0.

11. The general solution of the differential equation dr/dt = — Xr is r(t) = roe~kl where r(0) = ro is

the initial amount.

(a) We have r(t) = roe~kl and r (5230) = ro/2. Thus

r° r — = A.-5230 r0e

1

2

In = - X • 5230

2

- In 2 = -X • 5230

because In 1/2 = — In 2. Thus,

A = — «0.000132533.

5230

(b) We have r ( t ) = roe~Xt and r (8) = ro/2. By a computation similar to the one in part (a), we

have

In 2

X = ---- % 0.0866434.

8

(c) If r(t) is the number o f atoms of C-14, then the units for dr/dt is number of atoms per year.

Since dr/dt = — kr, A. is “per year.” Similarly, for 1-131, A. is “per day.” The unit of m easure￾ment o f r does not matter.

(d) We get the same answer because the original quantity, ro, cancels from each side of the equa￾tion. We are only concerned with the proportion remaining (one-half of the original amount).

4 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

13. The most important difference between Carbon-14 and Carbon-12 is that C-14 is radioactive. The

models we constructed above are based on the assumption that the number of C-14 atoms that decay

in a particular time period is proportional to the number of C-14 atoms present where the constant of

proportionality A. is determined from the half-life. Hence we can determine the amount of C-14 in a

sample by counting the number of atoms that decay in a given time period. Divide this number by

the time elapsed to obtain an approximation for dr/dt. Then, using d r /d t = —Xr, divide by A. to get

r (making sure the units agree).

15. Let P(t) be the population at time t, k be the growth-rate parameter, and N be the carrying capacity.

The modified models are

(a) dP/dt = k( 1 - P /N )P - 100

(b) dP/dt = *(1 - P / N ) P - P/7,

(c) d P / d t = k (\ — P / N ) P — a>fP, where a is a positive parameter.

17. Several different models are possible. Let R(t) denote the rhinoceros population at time /. The basic

assumption is that there is a minimum threshold that the population must exceed if it is to survive. In

terms of the differential equation, this assumption means that dR/dt must be negative if R is close

to zero. Three models that satisfy this assumption are:

• If k is a growth-rate parameter and M is a parameter measuring when the population is ‘‘too

small”, then

• If k is a growth-rate parameter and b is a parameter that determines the level the population will

start to decrease ( R < b /k), then

• If k is a growth-rate parameter and b is a parameter that determines the extinction threshold,

then

In each case, if R is below a certain threshold, dR/dt is negative. Thus, the rhinos will eventually

die out. The choice of which model to use depends on other assumptions. There are other equations

that are also consistent with the basic assumption.

19. (a) The term governing the effect o f the interaction of x and y on the rate of change of x is +ftxy.

Since this term is positive, the presence of y's helps the x population grow. Hence, x is the

predator. Similarly, the term —Sxy in the dy/dt equation implies that when x > 0, y's grow

more slowly, so y is the prey. If y = 0, then dx/dt < 0, so the predators will die out; thus, they

must have insufficient alternative food sources. The prey has no limits on its growth other than

the predator since, if x = 0, then dy/dt > 0 and the population increases exponentially.

(b) Since —flxy is negative and +<$*>» is positive, x suffers due to its interaction with y and y ben￾efits from its interaction with x. Hence, x is the prey and y is the predator. The predator has

other sources of food than the prey since dy/dt > 0 even if x = 0. Also, the prey has a limit

on its growth due to the —a x 2/ N term.

21. (a) The independent variable is /, and x and y are dependent variables. Since each jcy-term is

positive, the presence of either species increases the rate of change of the other. Hence, these

1.2 Analytic Technique: Separation of Variables 5

species cooperate. The parameter a is the growth-rate parameter for x, and y is the growth-rate

parameter for y. The parameter N represents the carrying capacity for x, but y has no carrying

capacity. The parameter fi measures the benefit to x of the interaction of the two species, and 5

measures the benefit to y of the interaction.

(b) The independent variable is /, and x and y are the dependent variables. Since both xy-term s are

negative, these species compete. The parameter y is the growth-rate coefficient for x, and a is

the growth-rate parameter for y. Neither population has a carrying capacity. The parameter 5

measures the harm to x caused by the interaction of the two species, and p measures the harm

to y caused by the interaction.

EXERCISES FOR SECTION 1.2

1. (a) Let's check Bob’s solution first. Sinee d y / d t = 1 and

y(t) + I _ / + 1 _

/ + l ~ t + 1 _ ’

Bob’s answer is correct.

Now let’s check Glen’s solution. Since dy/dt = 2 and

y(t) + 1 = 2t + 2

/ + 1 t + 1

Glen’s solution is also correct.

Finally let’s check Paul’s solution. We have dy/dt = 21 on one hand and

y(t) + 1 = t2 - 1

/ + I t + I ’

on the other. Therefore, Paul is wrong.

(b) At first glance, they should have seen the equilibrium solution y(t) = —1 for all t because

dy/dt = 0 for any constant function and y = — 1 implies that

* ± 1 = 0

/ + 1

independent of t.

Strictly speaking the differential equation is not defined for / = —1, and hence the solutions are not

defined for t = —1.

3. In order to find one such / ( / , y), we compute the derivative of y(t). We obtain

dy de‘

= 3 l V .

dt dt

Now we replace e' in the last expression by y and get the differential equation

CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

5. The constant function y(/) = 0 is an equilibrium solution.

For y / Owe separate the variables and integrate

/?-/■ *

t 2

In |>>| = — + c

|>| = C|i>'2/2

where c\ = ec is an arbitrary positive constant.

If y > 0, then |y| = y and we can just drop the absolute value signs in this calculation. If y < 0,

then |>>| = — y, so —y = c\e‘ ,2. Hence, y = — c\e‘ ^2. Therefore,

y = k e'2' 2

where k = ± C | . Moreover, if A: = 0 , we get the equilibrium solution. Thus, y = ke ,2^2 yields

all solutions to the differential equation if we let k be any real number. (Strickly speaking we need

a theorem from Section 1.5 to justify the assertion that this formula provides all solutions.)

7. We separate variables and integrate to obtain

f & - b

We get

1

- In \2y + l| = ( + c

\2y + l| = c ,e 2',

where cj = t,lf . As in Exercise 5, we can drop the absolute value signs by replacing ± c i with a new

constant k | . Hence, we have

2y+ 1 = k \ e 2'

y = \ { ^ e 2' - l ) ,

and letting k = k \/2, y(t) = ke2i — 1/2. Note that, for A: = 0, we get the equilibrium solution.

9. We separate variables and integrate to obtain

J ey d y = J dt

ey = t + c,

where c is any constant. We obtain y(t) = ln(/ -f c).

11. First note that the differential equation is not defined if y = 0.

In order to separate the variables, we write the equation as

1.2 Analytic Technique: Separation of Variables 7

where k = e2t (hence any positive constant). We have

y(t) = ±y/ln (k ( t 2 + 1)),

where k is any positive constant and the sign is determined by the initial condition.

13. First note that the differential equation is not defined for y = —\/2. We separate variables and

integrate to obtain

where c\ = k + 1 /4. If k can be any possible constant, then C| can be as well.

15. First of all, the equilibrium solutions are y = 0 and y = 1. Now suppose y / 0 and y / 1. We

separate variables to obtain

dy _ t

dt y (/2 + l )

to obtain

2 |

— = - H i 2 + l) + c,

where c is any constant. So we get

y 2 = In (k(t2 + 1)) ,

y 2 + y = l + k ,

where k is any constant. So

y(t)

- I ± V4l + 4k + 1 - 1 ± V 4 7 T 7

2 2

where c is any constant and the ± sign is determined by the initial condition.

We can rewrite the answer in the more simple form

y ( t ) = - - ± y / t + C\

where c is any constant. To integrate, we use partial fractions. Write

1 _ A + B

y ( \ - y ) y 1 - y '

8 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

We must have A = 1 and — A + B = 0. Hence, A = B = 1 and

y(i - y) y + i - y '

Consequently,

S ^ h y ) dy = XnW l n | l - , | = l„ - 2 -

1 1 - y

After integration, we have

In —- — = / + c

I > -y\

where c\ = ec is any positive constant. To remove the absolute value signs, we replace the positive

constant C| with a constant k that can be any real number and get

where k = ±C |. If A: = 0, we get the first equilibrium solution. The formula y (t) = kel/(1 -I- ke1)

yields all the solutions to the differential equation except for the equilibrium solution y(t) = I.

17. The equation can be written in the form

and we note that v(f) = — 1 for all t is an equilibrium solution. Separating variables and integrating,

we obtain

where cj = ec. We can dispose of the absolute value signs by allowing the constant c\ to be any real

number. In other words.

dv ->

— = ( > ;+ I) « 2 - 2 ),

at

r

In |u -1- 1| = — - 2t + c,

where c is any constant. Thus,

|» + 11 = ci<--2' +,3/3,

v(t) = - I + k e - 2,+,,/3,

where k = ± C |. Note that, if k = 0, we get the equilibrium solution.

19. The function y(t) = 0 for all / is an equilibrium solution.

Suppose y / 0 and separate variables. We get

1.2 Analytic Technique: Separation of Variables 9

where c is any real constant. We cannot solve this equation for y, so we leave the expression for y

in this implicit form. Note that the equilibrium solution y = 0 cannot be obtained from this implicit

equation.

21. The constant function w(t) = 0 is an equilibrium solution. Suppose w / 0 and separate variables.

We get

/ ? - / *

In \u>\ = In |/| + c

= ln c i|f|,

where c is any constant and c 1 = e‘ . Therefore,

| III | = C ||» |.

We can eliminate the absolute value signs by allowing the constant to assume positive or negative

values. We have

y + In | v| = e ' +c.

w = kt,

where k = ± C |. Moreover, if k = 0 we get the equilibrium solution.

23. From Exercise 7, we already know that the general solution is

y(t) = ke2' -

so we need only find the constant k for which >>(0) = 3. We solve

3 = k e ° - \

for A: and obtain k = 7/2. The solution of the initial-valuc problem is

y(t) = \ e 2' -

25. Separating variables and integrating, we obtain

- t + c .

y

So we get

1

y = — e￾Now we need to find the constant c so that y(0) = 1 /2. To do this we solve

1 _ 1

2 “ 0 — c

and get c = —2. The solution of the initial-value problem is

* ')= r h -

27. We do not need to do any computations to solve this initial-value problem. We know that the constant

function _y(/) = 0 for all t is an equilibrium solution, and it satisfies the initial condition.

29. We write the equation in the form

dx _ t2

dt jt(/3 + 1)

and separate variables to obtain

I xdx = f j T - \ d‘

X 2 1

~2 = 3 In ^ + * * + C'

where c is a constant. Hence,

x 2 = ? In li-1 + 11 + 2c.

The initial condition a:(0) = —2 implies

4 = ( —2)2 = | In |1| + 2c.

Thus, c = 2. Solving for * (/), we choose the negative square root because x(0) is negative, and we

drop the absolute value sign because t + 1 > 0 for t near 0. The result is

Jr(/) = - v/ ^ |n((3 + | ) + 4.

31. We separate variables to obtain

/ * - / -

10 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

i2

arctan y = — + c.