Complete Solution Manual - A First Course in Differential Equations with Modeling Applications 9th, Differential Equations with Boundary-Value Problems 7th

Complete Solutions Manual

A First Course in Differential Equations with Modeling Applications Ninth Edition

Dennis G. Zill

Loyola Marymount University

Differential Equations with Boundary-Vary Problems Seventh Edition

Dennis G. Zill

Loyola Marymount University

Michael R. Cullen

Late of Loyola Marymount University

By

Warren S. Wright

Loyola Marymount University

Carol D. Wright

*

; BROOKS/COLE C E N G A G E Learning-

Australia • Brazil - Japan - Korea • Mexico • Singapore • Spain • United Kingdom • United States

Table of Contents 1 Introduction to Differential Equations

2 First-Order Differential Equations

27

3 Modeling with First-Order Differential Equations

86

4 Higher-Order Differential Equations

137

5 Modeling with Higher-Order Differential Equations

231

6 Series Solutions of Linear Equations

274

7 The Laplace Transform

352

8 Systems of Linear First-Order Differential Equations

419

9 Numerical Solutions of Ordinary Differential Equations

478

10 Plane Autonomous Systems 11 Fourier Series

538

12 Boundary-Value Problems in Rectangular Coordinates

586

13 Boundary-Value Problems in Other Coordinate Systems

675

14 Integral Transforms

717

15 Numerical Solutions of Partial Differential Equations

761

Appendix I A ppendix

1

Gamma function

II Matrices

506

783 785

3.ROOKS/COLE C 'N G A G E L e a rn in g ”

ISBN-13. 978-0-495-38609-4 ISBN-10: 0-495-38609-X

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1 Introduction to Differential Equations

1 . Second order; linear 2 . Third order; nonlinear because of (dy/dx)4

3. Fourth order; linear 4. Second order; nonlinear bccausc of cos(r + u) 5. Second order; nonlinear because of (dy/dx)2 or

1 + (dy/dx)2

6 . Second order: nonlinear bccausc of R~

7. Third order: linear 8 . Second order; nonlinear because of x2

9. Writing the differential equation in the form x(dy/dx) -f y2 = 1. we sec that it is nonlinear in y because of y2. However, writing it in the form (y2 —1)(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu wc see that it is linear in v. However, writing it in the form (v + uv —ueu)(du/dv) + u — 0, we see that it, is nonlinear in ■ Ji­

ll.

From y = e-*/2 we obtain y' = —\e~x'2. Then 2y' + y = —e~X//2 + e-x/2 = 0.

12 . From y = | — |e-20* we obtain dy/dt = 24e-20t, so that

% + 20y = 24e~m + 20 - |e_20t) = 24. clt \'o 5 / 13. R'om y = eix cos 2x we obtain y1= 3e^x cos 2x — 2e3* sin 2a? and y” = 5e3,xcos 2x — 12e3,xsin 2x, so that y" — (k/ + l?>y = 0. 14. From y = —cos:r ln(sec;r + tanrc) we obtain y’ — —1 + sin.Tln(secx + tana:) and y" = tan x + cos x ln(sec x + tan a?). Then y" -f y = tan x. 15. The domain of the function, found by solving x + 2 > 0, is [—2, oo). From y’ = 1 + 2(x + 2)_1/2 we

1

Exercises 1.1 Definitions and Terminology have

= (y - ®)[i + (20 + 2)_1/2]

{y - x)y'

= y — x + 2(y - x)(x + 2)-1/2 = y - x + 2[x + 4(z + 2)1/2 - a;](a: + 2)_1/2 = y — x + 8(ac + 2)1;/i(rr + 2)~1/2 = y —x + 8. An interval of definition for the solution of the differential equation is (—2, oo) because y defined at x = —2. 16. Since tan:r is not defined for x = 7r/2 + mr, n an integer, the domain of y = 5t£v. {a; |5x ^ tt/2 + 7i-7r} or {;r |x ^ tt/IO + mr/5}. From y' — 25sec2 §x we have y' = 25(1 + tan2 5x) = 25 + 25 tan2 5a: = 25 + y2. An interval of definition for the solution of the differential equation is (—7r/ 10, 7T/10 . A:, interval is (7r/10, 37t/10). and so on. 17. The domain of the function is {x \4 —x2 ^ 0} or {x\x ^ —2 or x ^ 2}. Prom y' — 2.::

-=-

we have

An interval of definition for the solution of the differential equation is (—2, 2). Other (—oc,—2) and (2, oo). 18. The function is y — l/y /l — s in s . whose domain is obtained from 1 —sinx ^ 0 or the domain is {z |x ^ tt/2 + 2?i7r}. From y' = —1(1 —sin x)

. = 1 T

2(—cos.x) we have

2y' = (1 —sin;r)_ ‘?/’2 cos# = [(1 — sin:r)~1//2]3 cos:r - f/3 cosx.

An interval of definition for the solution of the differential equation is (tt/2. 5tt/2 is (57r/2, 97r/2) and so on. 19. Writing ln(2X — 1) —ln(X — 1) = t and differentiating implicitly we obtain dX 2 X - 1 dt 2

dX X - l dt 1

2X - 2 - 2X + 1 d X _ (2X - 1)(X - 1) dt IX — = -C2X - 1)(X - 1) = (X - 1)(1 - 2X .

2

A :..

.

Exercises 1.1 Definitions and Terminology x

Exponentiating both sides of the implicit solution we obtain 2X-1 ----- = el X - l 2 X - 1 = X el - ef -4

(e* - 1) = (e‘ - 2)X X =

-2

-2

ef' — 1 e* - 2 '

-4

Solving e* — 2 = 0 we get t = In 2. Thus, the solution is defined on (—oc.ln2) or on (In 2, oo). The graph of the solution defined on (—oo,ln2) is dashed, and the graph of the solution defined on (In 2. oc) is solid. 20. Implicitly differentiating the solution, we obtain

y

—x 2 dy — 2xy dx + y dy — 0 2xy dx + (a;2 —y)dy = 0.

Using the quadratic formula to solve y2—2x2y —1 = 0 for y, we get y = (2x2 ± V4;c4 + 4)/2 = a’2 ± v V 1+ 1 . Thus, two explicit solutions are y\ = x2 + \A'4 + 1 and y-2 = x2 — V.x4 + 1. Both solutions are defined on (—oo. oc). The graph of yj (x) is solid and the graph of y-2 is daalied. 21 . Differentiating P = c\?}} ( l + cie^ we obtain

dP _ ( l + cie*) cie* - cie* • cie* _ (1 + cie*)"

eft

Cie«

[(l + cie‘) - cie4]

1 + cief

1 + cie(

CiC = P( 1 - P). 1 + ci ef

Ci

1 + CI&-

2 PX ,2 ,2 22 . Differentiating y = e~x / e: dt + c\e~x we obtain Jo 2* f X *22 -r.2 y/ = e-*2er 2 2xe / e dt — 2c\xe =1 Jo Substituting into the differential equation, we have ___

r x

J

2

2xe x

rx +2 e dt — 2cixe —X Jo

y' + 2xy = l — 2xe x I e* dt — 2cixe x +2xe x [ e* dt 4- 2cio;e x = 1. Jo Jo

3

Exercises 1.1 Definitions and Terminology

23. From y — ci e2x+c.2 xe2x we obtain ^

- (2c\+C2 )e2x-r2c2xe2x and —| = (4cj + 4c;2)e2x + 4c2xe2j'.

so that n .T. dx2

—4 ^ + Ay = (4ci + 4co - 8ci - 4c2 + 4ci)e2x + (4c2 — Sc2 + 4e2)xe2x — 0. d:r dx

24. From y — Cix-1 + c^x + c%x ]n x + 4a;2 we obtain ^ = —c\x 2 + C2 + c$ + C3 In x + 8rc, dx d2y = 2cix,_3 + C3;r_1 -f 8. dx2

and

= —6cix -4 - c3a r 2, so that dx'3 + _r “J/ J' dx + V ~ ^_6ci + 4ci + Cl + Cl^x 3 + ^_ °3 + 2cs ~~02 ~ C3 + C2^X 2a'2 dx2 ~ X t (—C3 + cz)x In a; + (16 - 8 + 4)x2 = 12x2. ( —x2, x < 0 , f —2x, 25. From y = < ' we obtain y' = < ^ tx . x> 0 { 2x,

x< 0 ^ „ so that x> 0

- 2/y = 0.

26. The function y(x) is not continuous at x = 0 sincc lim y(x) = 5 and lim y(x) = —5. Thus. y’(x) x —>0“

x —>0+

does not exist at x = 0. 27. From y = emx we obtain y' = mernx. Then yf + 2y — 0 implies rnemx + 2emx = (m + 2)emx = 0. Since emx > 0 for all x} m = —2. Thus y = e~2x is a solution. 28. From y = emx we obtain y1= mernx. Then by' — 2y implies brriemx = 2e"lx or

m =

5

Thus y = e2:c/5 > 0 is a solution. 29. From y = emx we obtain y' = memx and y" = rn2emx. Then y" —5y' + Qy = 0 implies m 2emx - 5rnemx + 6emx = (rn - 2)(m - 3)emx = 0. Since ema! > 0 for all x, rn = 2 and m = 3. Thus y = e2x and y = e3:r are solutions.

30.

From y = emx we obtain y1= rnemx an 0 for ;r > 0. m = 0 and m = —1. Thus y = 1 and y — x~l are solutions. 32. From y = xm we obtain y' = mxm~1 and y" = m(m — l)xm~2. Then x2y" —7xy' + 15y — 0 implies x2rn{rn — l)xrn~2 — lxm xm~A+ 15:em =

[m(m — 1) — 7m + 15]xm

= (ro2 - 8m + 15)a,m = (m - 3) (to - 5)xm = 0. Since xm > 0 for x > 0. m = 3 and m = 5. Thus y — x^ and y = xa are solutions. In Problems 33-86 we substitute y = c into the differential equations and use y' — 0 and y" — 0 33. Solving 5c = 10 we see that y ~ 2 is a constant solution. 34. Solving c2 + 2c —3 = (c + 3)(c — 1) = 0 we see that, y = —3 and y = 1 are constant solutions. 35. Since l/(c — 1) = 0 has no solutions, the differential equation has no constant solutions. 36. Solving 6c = 10 we see that y = 5/3 is a constant solution. 37. From x — e~2t + 3ec< and y — —e~2t + 5ew we obtain ^ = —2e~2t + 18e6* and dt

dt

= 2e~2t + 30e6*.

Then

and

x-+ 3y = (e~2t + 3e6t) + 3 (- e '2* + oe6t) = -2e"2* + 18e6t = ^ \ Jub 5:r + 3y = 5(e~2* + 3eet) + 3(-e~2* + 5e6') = 2e~2t + 30e6* = ^ . at

38. From x = cos 21 + sin 21 +

and

and y — —cos 21 —sin 21 —

— = —2 sin 2t -f 2 cos 22 + d.t 5

and

d2:r , „ . . ^ 1 ^ , = —4 cos 2t — 4 sm 22 + re and dt2 Id

we obtain

^ = 2 sin 22 — 2 cos 2t — -e* dt 5 ^2V , 1 / -r-^- = 4 cos 2t + 4 sin 22-- e . d22 5

Then

cPx

and

1 1 4y + et = 4(—cos 21 — sin 21 — pef) + el — —4 cos 21 — 4 sin 22 + -el = -7-^ 0 o dt

5

Exercises 1.1 Definitions and Terminology

4x — ef = 4(cos 21 + sin 21 -I- ^e*) — e* = 4 cos 2£ + 4 sin 2t — \ef — 39. (t/ ) 2 + 1 = 0 has no real solutions becausc {y')2 + 1 is positive for all functions y = 4>(x). 40. The only solution of (?/)2 + y2 = 0 is y = 0, since if y ^ 0, y2 > 0 and (i/ ) 2 + y2 > y2 > 0. 41. The first derivative of f(x ) = ex is eT. The first derivative of f{x) = ekx is kekx. The differential equations are y' — y and y' = k.y, respectively. 42. Any function of the form y = cex or y = ce~x is its own sccond derivative. The corresponding differential equation is y" — y = 0. Functions of the form y = c sin x or y — c cos x have sccond derivatives that are the negatives of themselves. The differential equation is y" -+ -y = 0. 43. We first note that yjl —y2 = \/l — sin2 x = Vcos2 x = |cos.-r|. This prompts us to consider values of x for which cos x < 0, such as x = tt. In this case %

dx

i {sklx)

= c o s x l ^ , . = COS7T =

— 1.

X=7T

but \/l - y2\x=7r = V 1 - sin2 7r = vT = 1. Thus, y = sin re will only be a solution of y' - y l —y2 when cos x > 0. An interval of definition is then (—tt/ 2, tt/ 2). Other intervals are (3tt/ 2, 5tt/ 2), (77t/ 2, 9tt/ 2). and so on. 44. Since the first and second derivatives of sint and cos t involve sint and cos t, it is plausible that a linear combination of these functions, Asint+ B cos t. could be a solution of the differential equation. Using y' — A cos t —B sin t and y" = —A sin t —B cos t and substituting into the differential equation we get y" + 2y' + 4y = —A sin t — B cos t + 2A cos t — 2B sin t + 4A sin t + 4B cos t = (3A — 2B) sin t + (2A + 3B) cos t = 5 sin t. -- --+ TT7« Thus 3A — 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we ^ITirl find AA = j# and 13

B = —

. A particular solution is y =

sint — ^ cost.

45. One solution is given by the upper portion of the graph with domain approximately (0,2.6). The other solution is given by the lower portion of the graph, also with domain approximately (0. 2.6). 46. One solution, with domain approximately (—oo, 1.6) is the portion of the graph in the second quadrant together with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0,1.6) is the upper part of the graph in the first quadrant. The third solution, with domain (0, oo), is the part of the graph in the fourth quadrant.

6

Exercises 1.1 Definitions and Terminology 47. Differentiating (V1+ y^)/xy = 3c we obtain xy(3x2 + 3y2y') - (a?3 + y*)(xi/ + y) = 0 x?y2 3x3y + 3xy^y' —x'^y' — x% —xy^y’ — yA — 0 (3:ry3 - xA - xyz}i/ = -3x3y + xi y + y4 , = y4 - 2x3y _ y(y[i - 2x3) ^

2.ry3 —x4

rt:(2y3 —a:3)

48. A tangent line will be vertical where y' is undefined, or in this case, where :r(2y3 — x3) = 0. This gives x = 0 and 2y3 = a:3. Substituting y?J — a;3/2 into ;r3 + y3 = 3xy we get

x 3 + h 3 = 3x { w x) -x3 = — r 2 2 2V3a a:3 = 22/ V z 2(.x - 22/3) = 0. Thus, there are vertical tangent lines at x = 0 and x = 22/3, or at (0, 0) and (22/ 3, 21'/3). Since 22/3 ~ 1.59. the estimates of the domains in Problem 46 were close.

49. The derivatives of the functions are ^(.x) — —xf a/25 —x2 and ^ { x ) = x/\/25 —x2, neither of which is defined at x = ±5. 50. To determine if a solution curve passes through (0,3) we let 2 = 0 and P = 3 in the equation P = c-ie1/ (1 + eye*). This gives 3 = c j/(l + ci) or c\= —| . Thus, the solution curve (—3/2)e* = —3e* 1 - (3/2)eL 2 - 3e{ passes through the point, (0,3). Similarly, letting 2 = 0 and P = 1 in the equation for the oneparameter family of solutions gives 1 = c t/(l + ci) or ci = 1 + c-|. Since this equation has no solution, no solution curve passes through (0. 1). 51. For the first-order differential equation integrate f(x). For the second-order differential equation integrate twice. In the latter case we get y = f ( f f(x)dx)dx + cja: + C2 52. Solving for y’ using the quadratic formula we obtain the two differential equations y>= — ^2 + 2\J1 + 3ar®^

and

y1= — ^2 — 2 y 1 4-3a?^^ ,

so the differential equation cannot be put in the form dy/dx = f(x,y).

7

Exercises 1.1 Definitions and Terminology 53. The differential equation yy'—xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a solution of the first differential equation but not a solution of the second. 54. Differentiating we get y' = c\+ 3c2%2 and y" = 602x. Then C2 - y"/(>x and

~ 1/ —xy"f 2, so

v=iy'-^-)x+{t)x3=xy'-rv and the differential equation is x2y" —3xy' + Sy = 0. 2

55. (a) Since e~x is positive for all values of x. dy/dx > 0 for all x, and a sohition. y(x), of the differential equation must be increasing on any interval. (b) lim ^ = lim e~x‘ = 0 and lim ^ = lim e~x = 0. Since dy/dx approaches 0 as x v ' x^-cc dx x-+-x dx approaches —oc and oc, the solution curve has horizontal asymptotes to the left and to the right. (c) To test concavity we consider the second derivative d2y

d (dy\ d { *\ _ \ dr.)-dx\ e'

2

Since the sccond derivative is positive for x < 0 and negative for x > 0, the solution curve is concave up on (—00. 0) and concave down 011 (0. 00). x

56. (a) The derivative of a constant solution y — c is 0, so solving 5 — c = 0 we see that, c — 5 and so y = 5 is a constant sohition. (b) A solution is increasing where dyjdx = 5 — y > 0 or y < 5. A solution is decreasing where dy/dx = 5 —y < 0 or y > 5. 57. (a) The derivative of a constant solution is 0, so solving y(a — by) = 0 we see that y = 0 and y = a/b are constant solutions. (b) A solution is increasing where dy/dx = y(a —by) = by(a/b —y) > 0 or 0 < y < a/b. A solution is decreasing where dy/dx = by(a/b — y) < 0 or y < 0 or y > a/b. (c) Using implicit differentiation we compute = y(-by') + y'{a - by) = y'(a - 2by). Solving d2y/dx2 = 0 we obtain y = a/2b. Since dl y/dx2 > 0 for 0 < y < a/2b and d2y/dx2 < 0 for a/26 < y < a/b, the graph of y =