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Solution Manuals OfADVANCED ENGINEERINGMATHEMATICSByERWIN KREYSZIG9TH EDITIONThis is Downloaded Fromwww.mechanical.tkVisitwww.mechanical.tkFor More Solution Manuals Hand BooksAnd Much Much More

imfm.qxd9/15/0512:06 PMPage iINSTRUCTOR’SMANUAL FORADVANCEDENGINEERINGMATHEMATICS

imfm.qxd9/15/0512:06 PMPage ii

imfm.qxd9/15/0512:06 PMPage iiiINSTRUCTOR’SMANUAL FORADVANCEDENGINEERINGMATHEMATICSNINTH EDITIONERWIN KREYSZIGProfessor of MathematicsOhio State UniversityColumbus, OhioJOHN WILEY & SONS, INC.

imfm.qxd9/15/0512:06 PMPage ivVice President and Publisher: Laurie RosatoneEditorial Assistant: Daniel GraceAssociate Production Director: Lucille BuonocoreSenior Production Editor: Ken SantorMedia Editor: Stefanie LiebmanCover Designer: Madelyn LesureCover Photo: John Sohm/Chromosohm/Photo ResearchersThis book was set in Times Roman by GGS Information Services and printed and bound byHamilton Printing. The cover was printed by Hamilton Printing.This book is printed on acid free paper.Copyright 2006 by John Wiley & Sons, Inc. All rights reserved.No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted underSections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of thePublisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center,222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4470. Requests to the Publisher forpermission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street,Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, E-Mail: PERMREQ@WILEY.COM.ISBN-13: 978-0-471-72647-0ISBN-10: 0471-72647-8Printed in the United States of America10 9 8 7 6 5 4 3 2 1

imfm.qxd9/15/0512:06 PMPage vPREFACEGeneral Character and Purpose of the Instructor’s ManualThis Manual contains:(I) Detailed solutions of the even-numbered problems.(II) General comments on the purpose of each section and its classroom use, withmathematical and didactic information on teaching practice and pedagogical aspects. Someof the comments refer to whole chapters (and are indicated accordingly).Changes in Problem SetsThe major changes in this edition of the text are listed and explained in the Preface of thebook. They include global improvements produced by updating and streamlining chaptersas well as many local improvements aimed at simplification of the whole text. Speedyorientation is helped by chapter summaries at the end of each chapter, as in the last edition,and by the subdivision of sections into subsections with unnumbered headings. Resultingeffects of these changes on the problem sets are as follows.The problems have been changed. The large total number of more than 4000 problemshas been retained, increasing their overall usefulness by the following: Placing more emphasis on modeling and conceptual thinking and less emphasis ontechnicalities, to parallel recent and ongoing developments in calculus. Balancing by extending problem sets that seemed too short and contracting othersthat were too long, adjusting the length to the relative importance of the materialin a section, so that important issues are reflected sufficiently well not only in thetext but also in the problems. Thus, the danger of overemphasizing minor techniquesand ideas is avoided as much as possible. Simplification by omitting a small number of very difficult problems that appearedin the previous edition, retaining the wide spectrum ranging from simple routineproblems to more sophisticated engineering applications, and taking into account the“algorithmic thinking” that is developing along with computers. Amalgamation of text, examples, and problems by including the large number ofmore than 600 worked-out examples in the text and by providing problems closelyrelated to those examples. Addition of TEAM PROJECTS, CAS PROJECTS, and WRITING PROJECTS,whose role is explained in the Preface of the book. Addition of CAS EXPERIMENTS, that is, the use of the computer in “experimentalmathematics” for experimentation, discovery, and research, which often producesunexpected results for open-ended problems, deeper insights, and relations amongpractical problems.These changes in the problem sets will help students in solving problems as well as ingaining a better understanding of practical aspects in the text. It will also enable instructorsto explain ideas and methods in terms of examples supplementing and illustratingtheoretical discussions—or even replacing some of them if so desired.

imfm.qxd9/15/0512:06 PMPage viviInstructor’s Manual“Show the details of your work.”This request repeatedly stated in the book applies to all the problem sets. Of course, it isintended to prevent the student from simply producing answers by a CAS instead of tryingto understand the underlying mathematics.Orientation on ComputersComments on computer use are included in the Preface of the book. Software systems arelisted in the book at the beginning of Chap. 19 on numeric analysis and at the beginningof Chap. 24 on probability theory.ERWIN KREYSZIG

im01.qxd9/21/0510:17 AMPage 1Part A. ORDINARY DIFFERENTIALEQUATIONS (ODEs)CHAPTER 1First-Order ODEsMajor ChangesThere is more material on modeling in the text as well as in the problem set.Some additions on population dynamics appear in Sec. 1.5.Electric circuits are shifted to Chap. 2, where second-order ODEs will be available.This avoids repetitions that are unnecessary and practically irrelevant.Team Projects, CAS Projects, and CAS Experiments are included in most problem sets.SECTION 1.1. Basic Concepts. Modeling, page 2Purpose. To give the students a first impression what an ODE is and what we mean bysolving it.Background Material. For the whole chapter we need integration formulas andtechniques, which the student should review.General CommentsThis section should be covered relatively rapidly to get quickly to the actual solutionmethods in the next sections.Equations (1)–(3) are just examples, not for solution, but the student will see thatsolutions of (1) and (2) can be found by calculus, and a solution y e x of (3) by inspection.Problem Set 1.1 will help the student with the tasks ofSolving y ƒ(x) by calculusFinding particular solutions from given general solutionsSetting up an ODE for a given function as solutionGaining a first experience in modeling, by doing one or two problemsGaining a first impression of the importance of ODEswithout wasting time on matters that can be done much faster, once systematic methodsare available.Comment on “General Solution” and “Singular Solution”Usage of the term “general solution” is not uniform in the literature. Some books use theterm to mean a solution that includes all solutions, that is, both the particular and thesingular ones. We do not adopt this definition for two reasons. First, it is frequently quitedifficult to prove that a formula includes all solutions; hence, this definition of a generalsolution is rather useless in practice. Second, linear differential equations (satisfying rathergeneral conditions on the coefficients) have no singular solutions (as mentioned in thetext), so that for these equations a general solution as defined does include all solutions.For the latter reason, some books use the term “general solution” for linear equations only;but this seems very unfortunate.1

im01.qxd9/21/0510:17 AMPage 22Instructor’s ManualSOLUTIONS TO PROBLEM SET 1.1, page 82.6.10.12.14.16.y e 3x/3 c 4. y (sinh 4x)/4 cSecond order. 8. First order.y ce0.5x, y(2) ce 2, c 2/e, y (2/e)e0.5x 0.736e0.5xy ce x x 1, y(0) c 1 3, c 2, y 2e x x 1y c sec x, y(0) c/cos 0 c 12 , y 12 sec xSubstitution of y cx c 2 into the ODE givesy 2 xy y c 2 xc (cx c 2) 0.Similarly,y 14x 2,y 12x,1 x 2 x( 1 x) 1 x 2 0.424thus18. In Prob. 17 the constants of integration were set to zero. Here, by two integrations,y g,v y gt c1,y 21gt 2 c1t c2,y(0) c2 y0,and, furthermore,v(0) c1 v0,hencey 12gt 2 v0 t y0,as claimed. Times of fall are 4.5 and 6.4 sec, from t 100/4. 9 and 200/4. 9 .20. y ky. Solution y y0 ekx, where y0 is the pressure at sea level x 0. Nowy(18000) y0 ek 18000 12y0 (given). From this,ek 18000 1 , y(36000) y ek 2 18000 y (ek 18000)2 y ( 1 )2 1 y .0200 24 022. For 1 year and annual, daily, and continuous compounding we obtain the valuesya(1) 1060.00,yd(1) 1000(1 0.06/365)365 1061.83,yc(1) 1000e0.06 1061.84,respectively. Similarly for 5 years,ya(5) 1000 1.065 1338.23,yd(5) 1000(1 0.06/365)365 5 1349.83,yc(5) 1000e0.06 5 1349.86.We see that the difference between daily compounding and continuous compoundingis very small.The ODE for continuous compounding is yc ryc.SECTION 1.2. Geometric Meaning of y ƒ(x, y). Direction Fields, page 9Purpose. To give the student a feel for the nature of ODEs and the general behavior offields of solutions. This amounts to a conceptual clarification before entering into formalmanipulations of solution methods, the latter being restricted to relatively small—albeitimportant—classes of ODEs. This approach is becoming increasingly important, especiallybecause of the graphical power of computer software. It is the analog of conceptualstudies of the derivative and integral in calculus as opposed to formal techniques ofdifferentiation and integration.Comment on IsoclinesThese could be omitted because students sometimes confuse them with solutions. In thecomputer approach to direction fields they no longer play a role.

im01.qxd9/21/0510:17 AMPage 3Instructor’s Manual3Comment on Order of SectionsThis section could equally well be presented later in Chap. 1, perhaps after one or twoformal methods of solution have been studied.SOLUTIONS TO PROBLEM SET 1.2, page 112. Semi-ellipse x 2/4 y 2/9 13/9, y 0. To graph it, choose the y-interval largeenough, at least 0 y 4.4. Logistic equation (Verhulst equation; Sec. 1.5). Constant solutions y 0 and y 12.For these, y 0. Increasing solutions for 0 y(0) 12, decreasing for y(0) 12.6. The solution (not of interest for doing the problem) is obtained by usingdy/dx 1/(dx/dy) and solvingx c 2/(tan 12 y 1);dx/dy 1/(1 sin y) by integration,thus y 2 arctan ((x 2 c)/(x c)).8. Linear ODE. The solution involves the error function.12. By integration, y c 1/x.16. The solution (not needed for doing the problem) of y 1/y can be obtained byseparating variables and using the initial condition; y 2/2 t c, y 2t . 118. The solution of this initial value problem involving the linear ODE y y t 2 isy 4e t t 2 2t 2.20. CAS Project. (a) Verify by substitution that the general solution is y 1 ce x.Limit y 1 (y(x) 1 for all x), increasing for y(0)1, decreasing fory(0) 1.(b) Verify by substitution that the general solution is x 4 y 4 c. More “squareshaped,” isoclines y kx. Without the minus on the right you get “hyperbola-like”curves y 4 x 4 const as solutions (verify!). The direction fields should turn out inperfect shape.(c) The computer may be better if the isoclines are complicated; but the computermay give you nonsense even in simpler cases, for instance when y(x) becomesimaginary. Much will depend on the choice of x- and y-intervals, a method of trialand error. Isoclines may be preferable if the explicit form of the ODE contains rootson the right.SECTION 1.3. Separable ODEs. Modeling, page 12Purpose. To familiarize the student with the first “big” method of solving ODEs, theseparation of variables, and an extension of it, the reduction to separable form by atransformation of the ODE, namely, by introducing a new unknown function.The section includes standard applications that lead to separable ODEs, namely,1. the ODE giving tan x as solution2. the ODE of the exponential function, having various applications, such as inradiocarbon dating3. a mixing problem for a single tank4. Newton’s law of cooling5. Torricelli’s law of outflow.

im01.qxd9/21/0510:17 AMPage 44Instructor’s ManualIn reducing to separability we consider6. the transformation u y/x, giving perhaps the most important reducible class ofODEs.Ince’s classical book [A11] contains many further reductions as well as a systematictheory of reduction for certain classes of ODEs.Comment on Problem 5From the implicit solution we can get two explicit solutionsy c (6x) 2representing semi-ellipses in the upper half-plane, andy c (6x) 2representing semi-ellipses in the lower half-plane. [Similarly, we can get two explicitsolutions x(y) representing semi-ellipses in the left and right half-planes, respectively.]On the x-axis, the tangents to the ellipses are vertical, so that y (x) does not exist. Similarlyfor x (y) on the y-axis.This also illustrates that it is natural to consider solutions of ODEs on open rather thanon closed intervals.Comment on SeparabilityAn analytic function ƒ(x, y) in a domain D of the xy-plane can be factored in D,ƒ(x, y) g(x)h(y), if and only if in D,ƒxyƒ ƒx ƒy[D. Scott, American Math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide,but this may save time in cases of more complicated ODEs, some of which may perhapsbe of practical interest. You may perhaps ask your students to derive such a criterion.Comments on ApplicationEach of those examples can be modified in various ways, for example, by changing theapplication or by taking another form of the tank, so that each example characterizes awhole class of applications.The many ODEs in the problem set, much more than one would ordinarily be willingand have the time to consider, should serve to convince the student of the practicalimportance of ODEs; so these are ODEs to choose from, depending on the students’interest and background.Comment on Footnote 3Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22.Philosophiae Naturalis Principia Mathematica was his most influential work.Leibniz invented calculus independently in 1675 and introduced notations that wereessential to the rapid development in this field. His first publication on differential calculusappeared in 1684.SOLUTIONS TO PROBLEM SET 1.3, page 182. dy/y 2 (x 2)dx. The variables are now separated. Integration on both sides gives 1 12x 2 2x c*.yHencey 2.x 2 4x c

im01.qxd9/21/0510:17 AMPage 5Instructor’s Manual54. Set y 9x v. Then y v 9x. By substitution into the given ODE you obtainy v 9 v2.dv dx.v 9By separation,2Integration gives1varctan x c*,33arctanv 3x c3and from this and substitution of y v 9x,v 3 tan (3x c),y 3 tan (3x c) 9x.6. Set u y/x. Then y xu, y u xu . Substitution into the ODE and subtractionof u on both sides givesy 4xy4 u xu u,yxuxu 4.uSeparation of variables and replacement of u with y/x yields2u du 8dx,xu2 8 ln x c,y 2 x 2 (8 ln x c).8. u y/x, y xu, y u xu . Substitute u into the ODE, drop xu on both sides,and divide by x 2 to getxy xu x 2u 12x 2u2 xu,u 12u2.Separate variables, integrate, and solve algebraically for u:du 12 dx,u2 1 12(x c*),uHencey xu u 2.c x2x.c x10. By separation, y dy 4x dx. By integration, y 2 4x 2 c. The initial conditiony(0) 3, applied to the last equation, gives 9 0 c. Hence y 2 4x 2 9.12. Set u y/x. Then y u xu . Divide the given ODE by x 2 and substitute u andu into the resulting equation. This gives2u(u xu ) 3u2 1.Subtract 2u2 on both sides and separate the variables. This givesdx2u du .2xu 12xuu u2 1,Integrate, take exponents, and then take the square root:ln (u2 1) ln x c*,u2 1 cx,u cx . 1Hencey xu x cx . 1From this and the initial condition, y(1) c 1 2, c 5. This gives the answer .y x 5x 1

im01.qxd9/21/0510:17 AMPage 66Instructor’s Manual14. Set u y/x. Then y xu, y u xu . Substitute this into the ODE, subtract u onboth sides, simplify algebraically, and integrate:xu 2x 2cos (x 2)uuu 2x cos (x 2),u2/2 sin (x 2) c.Hence y 2 2x 2(sin (x 2) c). By the initial condition, (sin 12 c), c 0,y xu x 2 sin ( x �5–6Problem Set 1.3. Problem 14. First five real branches of the solution16. u y/x, y xu, y u xu u 4x 4 cos2u. Simplify, separate variables, andintegrate:u 4x 3 cos2u,du/cos2u 4x 3 dx,tan u x 4 c.Hencey xu x arctan (x 4 c).From the initial condition, y(2) 2 arctan (16 c) 0, c 16. Answer:y x arctan (x 4 16).18. Order terms:dr(1 b cos ) br sin .dSeparate variables and integrate:drb sin r1 b cosd ,ln r ln (1 b cos ) c*.

im01.qxd9/21/0510:17 AMPage 7Instructor’s Manual7Take exponents and use the initial condition:r c(1 b cos ),r( ) c(1 b 0) ,2c .Hence the answer is r (1 b cos ).20. On the left, integrate g(w) over w from y0 to y. On the right, integrate ƒ(t) over t fromx0 to x. In Prob. 19, wey (t 1) dt.xw2dw 1022. Consider any straight line y ax through the origin. Its slope is y/x a. The slopeof a solution curve at a point of intersection (x, ax) is y g(y/x) g(a) const,independent of the point (x, y) on the straight line considered.24. Let kB and kD be the constants of proportionality for the birth rate and death rate,respectively. Then y kB y kD y, where y(t) is the population at time t. By separatingvariables, integrating, and taking exponents,dy/y (kB kD) dt,ln y (kB kD)t c*,y ce(kB k )tD.26. The model is y Ay ln y with A 0. Constant solutions are obtained fromy 0 when y 0 and 1. Between 0 and 1 the right side is positive (since ln y 0),so that the solutions grow. For y 1 we have ln y 0; hence the right side is negative,so that the solutions decrease with increasing t. It follows that y 1 is stable. Thegeneral solution is obtained by separation of variables, integration, and two subsequentexponentiations:dy/(y ln y) A dt, Atln y ce,ln (ln y) At c*,y exp (ce At).28. The temperature of the water is decreasing exponentially according to Newton’s lawof cooling. The decrease during the first 30 min, call it d1, is greater than that, d2,during the next 30 min. Thus d1 d2 190 110 80 as measured. Hence thetemperature at the beginning of parking, if it had been 30 min earlier, before the arrest,would have been greater than 190 80 270, which is impossible. Therefore Jackhas no alibi.30. The cross-sectional area A of the hole is multiplied by 4. In the particular solution,15.00 0.000332t is changed to 15.00 4 0.000332t because the second termcontains A/B. This changes the time t 15.00/0.000332 when the tank is empty, tot 15.00/(4 0.000332), that is, to t 12.6/4 3.1 hr, which is 1/4 of the originaltime.32. According to the physical information given, you haveS 0.15S .Now let * 0. This gives the ODE dS/d 0.15S. Separation of variables yieldsthe general solution S S0 e0.15 with the arbitrary constant denoted by S0. Theangle should be so large that S equals 1000 times S0. Hence e0.15 1000, (ln 1000)/0.15 46 7.3 2 , that is, eight times, which is surprisingly little.Equally remarkable is that here we see another application of the ODE y ky anda derivation of it by a general principle, namely, by working with small quantitiesand then taking limits to zero.

im01.qxd9/21/0510:17 AMPage 88Instructor’s Manual36. B now depends on h, namely, by the Pythagorean theorem,B(h) r 2 (R2 (R h)2 ) (2Rh h2).Hence you can use the ODEh 26.56(A/B) h in the text, with constant A as before and the new B. The latter makes the furthercalculations different f

ADVANCED ENGINEERING MATHEMATICS By ERWIN KREYSZIG 9TH EDITION This is Downloaded From www.mechanical.tk Visit www.mechanical.tk For More Solution Manuals Hand Books And Much Much More. INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS imfm.qxd 9/15/05 12:06 PM Page i. imfm.qxd 9/15/05 12:06 PM Page ii. INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS NINTH EDITION ERWIN .

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