A first course in differential equations 9th ed solutions

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 www.elsolucionario.net http://www.elsolucionario.net LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS www.elsolucionario.net 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 506 783 Gamma function II Matrices 785 www.elsolucionario.net 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 at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionreq uest@cenga ge com Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized [earning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd For your course and learning solutions: visit academic.cengage.com Purchase any of our products at your local college store or at our preferred online store www.ichapters.com NOTE’ UNDER NO C\RCUMST ANCES MAY TH\S MATERIAL OR AMY PORTION THEREOF BE SOLD, UCENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement' ) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the ‘Course’ ), you and your students who purchase the textbook may use the Supplement as described below Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you not accept these conditions, you must return the Supplement unused within 30 days of receipt All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors The Supplement is furnished by Cengage Learning on an "as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s 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 :"tcd in Canada 11 10 09 08 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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 www.elsolucionario.net 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)