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Solution a first course in differential equations with modeling applications 9th ziill Solution a first course in differential equations with modeling applications 9th ziill Solution a first course in differential equations with modeling applications 9th ziill Solution a first course in differential equations with modeling applications 9th ziill Solution a first course in differential equations with modeling applications 9th ziill Solution a first course in differential equations with modeling applications 9th ziill
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 Introduction to Differential Equations First-Order Differential Equations 27 Modeling with First-Order Differential Equations 86 Higher-Order Differential Equations 137 Modeling with Higher-Order Differential Equations 231 Series Solutions of Linear Equations 274 The Laplace Transform 352 Systems of Linear First-Order Differential Equations 419 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 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 i 2009 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, Information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below For product information and technology assistance, contact us 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conflict of law rules Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool Introduction to Differential Equations Second order; linear Third order; nonlinear because of (dy/dx)4 Fourth order; linear Second order; nonlinear bccausc of cos(r + u) Second order; nonlinear because of (dy/dx)2 or + (dy/dx)2 Second order: nonlinear bccausc of R~ Third order: linear Second order; nonlinear because of x2 Writing the differential equation in the form x(dy/dx) -f y2 = 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 = 12 From y = | — |e-20* we obtain dy/dt = 24e-20t, so that % + 20y = 24e~m + 20 - |e_20t) = 24 clt \'o / 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 = 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 + > 0, is [—2, oo) From y’ = + 2(x + 2)_1/2 we 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 + 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 —sinx ^ or the domain is {z |x ^ tt/2 + 2?i7r} From y' = —1(1 —sin x) = 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 X - dt dX X - l dt 2X - - 2X + d X _ (2X - 1)(X - 1) dt IX — = -C2X - 1)(X - 1) = (X - 1)(1 - 2X A : Exercises 1.1 Definitions and Terminology x Exponentiating both sides of the implicit solution we obtain 2X-1 - = el X - l X - = X el - ef -4 (e* - 1) = (e‘ - 2)X X = -2 -2 ef' — e* - ' -4 Solving e* — = we get t = In 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 oc) is solid 20 Implicitly differentiating the solution, we obtain y —x dy — 2xy dx + y dy — 2xy dx + (a;2 —y)dy = Using the quadratic formula to solve y2—2x2y —1 = for y, we get y = (2x2 ± V4;c4 + 4)/2 = a’2 ± v V 1+ Thus, two explicit solutions are y\ = x2 + \A'4 + and y-2 = x2 — V.x4 + 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] + cief + cie( CiC = P( - P) + ci ef Ci + CI&- 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 2xe / e dt — 2c\xe =1 Jo Substituting into the differential equation, we have _ r x J 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 = Jo Jo 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 — d:r dx 24 From y — Cix-1 + c^x + c%x ]n x + 4a;2 we obtain ^ = —c\x + C2 + c$ + C3 In x + 8rc, dx d2y = 2cix,_3 + C3;r_1 -f dx2 and = —6cix -4 - c3a r 2, so that dx'3 + _r “J/ J' dx + V ~ ^_6ci + 4ci + Cl + Cl^x + ^_ °3 + 2cs ~~02 ~ C3 + C2^X 2a'2 dx2 ~ X t (—C3 + cz)x In a; + (16 - + 4)x2 = 12x2 ( —x2, x < , f —2x, 25 From y = < ' we obtain y' = < ^ tx x> { 2x, x< ^ „ so that x> - 2/y = 26 The function y(x) is not continuous at x = sincc lim y(x) = and lim y(x) = —5 Thus y’(x) x —>0“ x —>0+ does not exist at x = 27 From y = emx we obtain y' = mernx Then yf + 2y — implies rnemx + 2emx = (m + 2)emx = Since emx > 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 = Thus y = e2:c/5 > is a solution 29 From y = emx we obtain y' = memx and y" = rn2emx Then y" —5y' + Qy = implies m 2emx - 5rnemx + 6emx = (rn - 2)(m - 3)emx = Since ema! > for all x, rn = and m = Thus y = e2x and y = e3:r are solutions 30 From y = emx we obtain y1= rnemx an for ;r > m = and m = —1 Thus y = 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 — 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 = Since xm > for x > m = and m = Thus y — x^ and y = xa are solutions In Problems 33-86 we substitute y = c into the differential equations and use y' — and y" — 33 Solving 5c = 10 we see that y ~ is a constant solution 34 Solving c2 + 2c —3 = (c + 3)(c — 1) = we see that, y = —3 and y = are constant solutions 35 Since l/(c — 1) = 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) + (- 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 cos 22 + d.t and d2:r , „ ^ ^ , = —4 cos 2t — sm 22 + re and dt2 Id we obtain ^ = sin 22 — cos 2t — -e* dt ^2V , / -r-^- = cos 2t + sin 22 e d22 Then cPx and 1 4y + et = 4(—cos 21 — sin 21 — pef) + el — —4 cos 21 — sin 22 + -el = -7-^ o dt Exercises 1.1 Definitions and Terminology 4x — ef = 4(cos 21 + sin 21 -I- ^e*) — e* = cos 2£ + sin 2t — \ef — 39 (t/ ) + = has no real solutions becausc {y')2 + is positive for all functions y = 4>(x) 40 The only solution of (?/)2 + y2 = is y = 0, since if y ^ 0, y2 > and (i/ ) + y2 > y2 > 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 = 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 = 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 = — X=7T but \/l - y2\x=7r = V - sin2 7r = vT = Thus, y = sin re will only be a solution of y' - y l —y2 when cos x > 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 = sin t + TT7« Thus 3A — 2B = and 2A + 3B = 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 Exercises 1.1 Definitions and Terminology 47 Differentiating (V1+ y^)/xy = 3c we obtain xy(3x2 + 3y2y') - (a?3 + y*)(xi/ + y) = x?y2 3x3y + 3xy^y' —x'^y' — x% —xy^y’ — yA — (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) = This gives x = and 2y3 = a:3 Substituting y?J — a;3/2 into ;r3 + y3 = 3xy we get x + h = 3x { w x) -x3 = — r 2 2V3a a:3 = 22/ V z 2(.x - 22/3) = Thus, there are vertical tangent lines at x = 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 = and P = in the equation P = c-ie1/ (1 + eye*) This gives = c j/(l + ci) or c\= —| Thus, the solution curve (—3/2)e* = —3e* - (3/2)eL - 3e{ passes through the point, (0,3) Similarly, letting = and P = in the equation for the oneparameter family of solutions gives = c t/(l + ci) or ci = + 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 — y 4-3a?^^ , so the differential equation cannot be put in the form dy/dx = f(x,y) Exercises 9.2 R/unge-Kutta Methc Figure 1: Plot of y(t) versus time for N=5000 Therefore, the numerical solution docs not seem to capture the physics involved after : since there are no oscillations Note that the constant solution y = is a solution initial-value problem However, the solution is not physical (e) First separate variables and integrate I ■— — = j Lodt J \/l-y2 ' to obtain arcsiny = ojt -I- Cq Upon using the initial condition, we find y(t) = sin(u;t + arcsin y^) The analytic solution does capture the oscillations of the spring (f) Differentiate both sides of equation (2) with respect to time to obtain djl dy\ dt2 ^ Vdt) v/i _ y2 ! and then use the fact that dy/dt = wy'l —y2 From equation (2), we have y(Q) = yo and from equation again, we have y'{0 ) = a;^/l —yg (g) First create the following function file (name it spring2.m) function out=spring2 (t ,v); omega=4; out(l)=y( ); 491 Exercises 9.2 Runge-Kutta Methods out(2 )=-o;2 * 2/( 1); out=out?; then in the Matlab window, type the following commands: » M = 3; k = 48: a; = Jk/M: F = 10 : » y() = F/yJ(M 2£J2 + F2) » 2/1 = w y - 2/0 >> [t, y] = ode45('spring? , [0 ,pz/2], [yo, 2/i] : » plot (t y(:, 1)) where y\ - dy/dt at t = The resulting plot is shown in figure The graph is consist:: with the analytical solution y(t) = sin(o>t + arcsinyo) from part (e) Time Figure : Plot of y(t) versus time using ODE45 The second-order differential equation has constant coefficients The analytic solutio: easily be obtained, , x(t) VF2 + M2oj2 ( =— — , , yi , (,»cos^ ) + tr sm(a,t)J " 492 F ■ Exercises 9.3 Multistep Methods r V •; '"; • •: : /• Multistep Methods the tables in this section £:A BM‘>stands for' Adams-Bashforth-Moulton Writing the differential equation in the form y1 —y = x —1 we see that an integrating factor is e- J dx _ g0 -ie-'y] = (x - 1)