# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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3. Compare your final solution with the one provided to see whether yours is more or less efficient than the guide, since there is frequently more than one correct way to solve a problem.

4. Ask yourself why that particular problem was assigned.

The use of a symbolic computational software package, in many cases, would greatly simplify finding the solution to a given problem, but the details given in this solutions manual are important for the student's understanding of the underlying mathematical principles and applications. In other cases, these software packages are essential for completing the given problem, as the calculations would be overwhelming using analytical techniques. In these cases, some steps or hints are given and then reference made to the use of an appropriate software package.

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In order to simplify the text, the following abbreviations have been used:

D.E. differential equation(s)

0.D.E. ordinary differential equation(s)

P.D.E. partial differential equation(s)

1.C. initial condition(s)

I.V.P. initial value problem(s)

B.C. boundary condition(s)

B.V.P. boundary value problem(s)

I wish to express my appreciation to Mrs. Susan A. Hickey for her excellent typing and proofreading of all stages of the manuscript. Dr. Josef Torok has also provided invaluable assistance with the preparation of the figures as well as

assistance with many of the solutions involving the use of

symbolic computational software.

Charles W. Haines Professor of Mathematics and Mechanical Engineering Rochester Institute of Technology Rochester, New York June 2000

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CONTENTS

CHAPTER 1 1

CHAPTER 2 6

CHAPTER 3 35

CHAPTER 4 64

CHAPTER 5 ............................................... 73

CHAPTER 6 .............................................. 103

CHAPTER 7 121

CHAPTER 8 ............................................. 16 0

CHAPTER 9 .............................................. 174

CHAPTER 10 .............................................. 2 03

CHAPTER 11

235

CHAPTER 1

1

Section 1.1, Page 8

2. For y > 3/2 we see that y' > 0

and thus y(t) is increasing there. For y < 3/2 we have y' < 0 and thus y(t) is decreasing there. Hence y(t) diverges from 3/2 as t^ro.

4. Observing the direction field, we see that for y>-1/2 we have y'<0, so the solution is decreasing here. Likewise, for y<-1/2 we have y'>0 and thus y(t) is increasing here.

Since the slopes get closer to zero as y gets closer to -1/2, we conclude that y^-1/2 as t^ro.

If all solutions approach 3, then 3 is the equilibrium

dy dy

solution and we want ---------- < 0 for y > 3 and > 0 for

dt dt

dy

y < 3. Thus -------- = 3-y.

dt

11. For y = 0 and y = 4 we have y' = 0 and thus y = 0 and

y = 4 are equilibrium solutions. For y > 4, y' < 0 so if y(0) > 4 the solution approaches y = 4 from above. If 0 < y(0) < 4, then y' > 0 and the solutions "grow" to y = 4 as t. For y(0) < 0 we see that y' < 0 and the solutions diverge from 0.

13. Since y' = y2, y = 0 is the equilibrium solution and y' > 0 for all y. Thus if y(0) > 0, solutions will diverge from 0 and if y(0) < 0, solutions will aproach y = 0 as tro.

15a. Let q(t) be the number of grams of the substance in the

dq 300q

water at any time. Then = 300(.01)

dt 1,000,000

300(10 2- 10 6q).

/ 4

15b. The equilibrium solution occurs when q = 0, or c = 10 gm,

independent of the amount present at t = 0 (all solutions

approach the equilibrium solution).

Section 1.2

dV

16. The D.E. expressing the evaporation is ---------------- = - aS, a > 0.

dt

4 3 2

Now V = pr and S = 4pr , so S = 4P 3

\2/3

V4PZ

V

2/3

dv

2/3

Thus --------- = - kV , for k > 0.

dt

21.

22.

24.

Section 1.2 Page 14

2

3

1b. dy/dt = -2y+5 can be rewritten as = -2dt. Thus

y-5/2

ln I y-5/2 I = -2t + c1, or y-5/2 = ce-2t. y(0) = y0 yields

c = y0 - 5/2, so y = 5/2 + (y0-5/2)e-2t.

If y0 > 5/2, the solution starts above the equilibrium solution and decreases exponentially and approaches 5/2 as t. Conversely, if y < 5/2, the solution starts below 5/2 and grows exponentially and approaches 5/2 from

below as t<^>.

3

dy/dt | |

4a. Rewrite Eq.(ii) as ---------- = a and thus ln | y | = at+c1(- or

y

y = ceat.

dy dy1

4b. If y = y1(t) + k, then ------ = ----. Substituting both

dt dt

dy1

these into Eq.(i) we get = a(y1+k) - b. Since

dt

dy1

= ay1, this leaves ak - b = 0 and thus k = b/a.

dt

Hence y = y1(t) + b/a is the solution to Eq(i).

6b. From Eq.(11) we have p = 900 + cet/2. If p(0) = p0, then

c = p0 - 900 and thus p = 900 + (p0 - 900)et/2. If p0 < 900, this decreases, so if we set p = 0 and solve for T (the time of extinction) we obtain eT/2 = 900/(900-p0), or T = 2ln[900/(900-p0)].

8a. Use Eq.(26).

8b. Use Eq.(29).

dQ dQ/dt I I

10a. = -rQ yields ---------- = -r, or ln | Q | = -rt + c1. Thus

dt Q 1

Q = ce-rt and Q(0) = 100 yields c = 100. Hence Q = 100e-rt.

Setting t = 1, we have 82.04 = 100e-r, which yields

r = .1980/wk or r = .02828/day.

dQ/dt -1

13a. Rewrite the D.E. as -------- = , thus upon integrating and

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