Published on: **Mar 3, 2016**

Published in:
Technology Education

Source: www.slideshare.net

- 1. FUNDAMENTALS OF DIFFERENTIAL EQUATIONS SEVENTH EDITION AND FUNDAMENTALS OF DIFFERENTIAL EQUATIONS AND BOUNDARY VALUE PROBLEMS FIFTH EDITION R. Kent Nagle University of South Florida Edward B. Saff Vanderbilt University A. David Snider University of South Florida INSTRUCTOR’S SOLUTIONS MANUAL 388445_Nagle_ttl.qxd 1/9/08 11:53 AM Page 1
- 2. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or trans- mitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, with- out the prior written permission of the publisher. ISBN-13: 978-0-321-38844-5 ISBN-10: 0-321-38844-5 This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permit- ted.The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes.All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. 388445_Nagle_ttl.qxd 1/9/08 11:53 AM Page 2
- 3. Contents Notes to the Instructor 1 Software Supplements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Computer Labs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Group Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Technical Writing Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Student Presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Homework Assignments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Syllabus Suggestions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Numerical, Graphical, and Qualitative Methods . . . . . . . . . . . . . . . . . . 4 Engineering/Physics Applications . . . . . . . . . . . . . . . . . . . . . . . . . 5 Biology/Ecology Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Supplemental Group Projects 9 Detailed Solutions & Answers to Even-Numbered Problems 17 CHAPTER 1 Introduction 17 Exercises 1.1 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 17 Exercises 1.2 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 18 Exercises 1.3 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 22 Exercises 1.4 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 25 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 CHAPTER 2 First Order Diﬀerential Equations 35 Exercises 2.2 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 35 Exercises 2.3 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 41 Exercises 2.4 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 48 Exercises 2.5 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 56 Exercises 2.6 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 61 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 iii
- 4. CHAPTER 3 Mathematical Models and Numerical Methods Involving First Order Equations 73 Exercises 3.2 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 73 Exercises 3.3 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 81 Exercises 3.4 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 87 Exercises 3.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Exercises 3.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Exercises 3.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 CHAPTER 4 Linear Second Order Equations 101 Exercises 4.1 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 101 Exercises 4.2 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 103 Exercises 4.3 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 111 Exercises 4.4 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 119 Exercises 4.5 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 125 Exercises 4.6 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 137 Exercises 4.7 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 144 Exercises 4.8 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 157 Exercises 4.9 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 160 Exercises 4.10 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 167 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 CHAPTER 5 Introduction to Systems and Phase Plane Analysis 177 Exercises 5.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Exercises 5.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Exercises 5.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 Exercises 5.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Exercises 5.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Exercises 5.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Exercises 5.8 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 CHAPTER 6 Theory of Higher-Order Linear Diﬀerential Equations 193 Exercises 6.1 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Exercises 6.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Exercises 6.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Exercises 6.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 iv
- 5. Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 CHAPTER 7 Laplace Transforms 197 Exercises 7.2 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 197 Exercises 7.3 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 201 Exercises 7.4 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 206 Exercises 7.5 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 215 Exercises 7.6 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 224 Exercises 7.7 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 239 Exercises 7.8 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 247 Exercises 7.9 Detailed Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 252 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 CHAPTER 8 Series Solutions of Diﬀerential Equations 267 Exercises 8.1 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Exercises 8.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Exercises 8.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Exercises 8.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 Exercises 8.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Exercises 8.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 Exercises 8.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Exercises 8.8 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 CHAPTER 9 Matrix Methods for Linear Systems 277 Exercises 9.1 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 Exercises 9.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 Exercises 9.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 Exercises 9.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Exercises 9.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 Exercises 9.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Exercises 9.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 Exercises 9.8 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 CHAPTER 10 Partial Diﬀerential Equations 291 Exercises 10.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Exercises 10.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Exercises 10.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 v
- 6. Exercises 10.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Exercises 10.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 Exercises 10.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 CHAPTER 11 Eigenvalue Problems and Sturm-Liouville Equations 297 Exercises 11.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 Exercises 11.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Exercises 11.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Exercises 11.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 Exercises 11.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Exercises 11.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 Exercises 11.8 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 CHAPTER 12 Stability of Autonomous Systems 305 Exercises 12.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Exercises 12.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Exercises 12.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 Exercises 12.5 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 Exercises 12.6 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Exercises 12.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 CHAPTER 13 Existence and Uniqueness Theory 317 Exercises 13.1 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Exercises 13.2 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Exercises 13.3 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Exercises 13.4 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Review Problems Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 vi
- 7. Notes to the Instructor One goal in our writing has been to create ﬂexible texts that aﬀord the instructor a variety of topics and make available to the student an abundance of practice problems and projects. We recommend that the instructor read the discussion given in the preface in order to gain an overview of the prerequisites, topics of emphasis, and general philosophy of the text. Software Supplements Interactive Diﬀerential Equations CD-ROM: By Beverly West (Cornell University), Steven Strogatz (Cornell University), Jean Marie McDill (California Polytechnic State Uni- versity – San Luis Obispo), John Cantwell (St. Louis University), and Hubert Hohn (Mas- sachusetts College of Arts) is a popular software directly tied to the text that focuses on helping students visualize concepts. Applications are drawn from engineering, physics, chemistry, and biology. Runs on Windows or Macintosh and is included free with every book. Instructor’s MAPLE/MATHLAB/MATHEMATICA manual: By Thomas W. Po- laski (Winthrop University), Bruno Welfert (Arizona State University), and Maurino Bautista (Rochester Institute of Technology). A collection of worksheets and projects to aid instruc- tors in integrating computer algebra systems into their courses. Available via Addison-Wesley Instructor’s Resource Center. MATLAB Manual ISBN 13: 978-0-321-53015-8; ISBN 10: 0-321-53015-2 MAPLE Manual ISBN 13: 978-0-321-38842-1; ISBN 10: 0-321-38842-9 MATHEMATICA Manual ISBN 13: 978-0-321-52178-1; ISBN 10: 0-321-52178-1 Computer Labs A computer lab in connection with a diﬀerential equations course can add a whole new di- mension to the teaching and learning of diﬀerential equations. As more and more colleges and universities set up computer labs with software such as MAPLE, MATLAB, DERIVE, MATHEMATICA, PHASEPLANE, and MACMATH, there will be more opportunities to in- clude a lab as part of the diﬀerential equations course. In our teaching and in our texts, we have tried to provide a variety of exercises, problems, and projects that encourage the student to use the computer to explore. Even one or two hours at a computer generating phase plane diagrams can provide the students with a feeling of how they will use technology together 1
- 8. with the theory to investigate real world problems. Furthermore, our experience is that they thoroughly enjoy these activities. Of course, the software, provided free with the texts, is especially convenient for such labs. Group Projects Although the projects that appear at the end of the chapters in the text can be worked out by the conscientious student working alone, making them group projects adds a social element that encourages discussion and interactions that simulate a professional work place atmosphere. Group sizes of 3 or 4 seem to be optimal. Moreover, requiring that each individual student separately write up the group’s solution as a formal technical report for grading by the instructor also contributes to the professional ﬂavor. Typically, our students each work on 3 or 4 projects per semester. If class time permits, oral presentations by the groups can be scheduled and help to improve the communication skills of the students. The role of the instructor is, of course, to help the students solve these elaborate problems on their own and to recommend additional reference material when appropriate. Some additional Group Projects are presented in this guide (see page 9). Technical Writing Exercises The technical writing exercises at the end of most chapters invite students to make documented responses to questions dealing with the concepts in the chapter. This not only gives students an opportunity to improve their writing skills, but it helps them organize their thoughts and better understand the new concepts. Moreover, many questions deal with critical thinking skills that will be useful in their careers as engineers, scientists, or mathematicians. Since most students have little experience with technical writing, it may be necessary to return ungraded the ﬁrst few technical writing assignments with comments and have the students redo the the exercise. This has worked well in our classes and is much appreciated by the students. Handing out a “model” technical writing response is also helpful for the students. Student Presentations It is not uncommon for an instructor to have students go to the board and present a solution 2
- 9. to a problem. Diﬀerential equations is so rich in theory and applications that it is an excellent course to allow (require) a student to give a presentation on a special application (e.g., almost any topic from Chapter 3 and 5), on a new technique not covered in class (e.g., material from Section 2.6, Projects A, B, or C in Chapter 4), or on additional theory (e.g., material from Chapter 6 which generalizes the results in Chapter 4). In addition to improving students’ communication skills, these “special” topics are long remembered by the students. Here, too, working in groups of 3 or 4 and sharing the presentation responsibilities can add substantially to the interest and quality of the presentation. Students should also be encouraged to enliven their communication by building physical models, preparing part of their lectures on video cassette, etc. Homework Assignments We would like to share with you an obvious, non-original, but eﬀective method to encourage students to do homework problems. An essential feature is that it requires little extra work on the part of the instructor or grader. We assign homework problems (about 10 of them) after each lecture. At the end of the week (Fridays), students are asked to turn in their homework (typically, 3 sets) for that week. We then choose at random one problem from each assignment (typically, a total of 3) that will be graded. (The point is that the student does not know in advance which problems will be chosen.) Full credit is given for any of the chosen problems for which there is evidence that the student has made an honest attempt at solving. The homework problem sets are returned to the students at the next meeting (Mondays) with grades like 0/3, 1/3, 2/3, or 3/3 indicating the proportion of problems for which the student received credit. The homework grades are tallied at the end of the semester and count as one test grade. Certainly, there are variations on this theme. The point is that students are motivated to do their homework with little additional cost (= time) to the instructor. Syllabus Suggestions To serve as a guide in constructing a syllabus for a one-semester or two-semester course, the prefaces to the texts list sample outlines that emphasize methods, applications, theory, partial diﬀerential equations, phase plane analysis, computation, or combinations of these. As a further guide in making a choice of subject matter, we provide below a listing of text material dealing with some common areas of emphasis. 3
- 10. Numerical, Graphical, and Qualitative Methods The sections and projects dealing with numerical, graphical, and qualitative techniques of solving diﬀerential equations include: Section 1.3: Direction Fields Section 1.4: The Approximation Method of Euler Project A for Chapter 1: Taylor Series Project B for Chapter 1: Picard’s Method Project D for Chapter 1: The Phase Line Section 3.6: Improved Euler’s Method, which includes step-by-step outlines of the im- proved Euler’s method subroutine and improved Euler’s method with tolerance. These outlines are easy for the student to translate into a computer program (cf. pages 135 and 136). Section 3.7: Higher-Order Numerical Methods : Taylor and Runge-Kutta, which includes outlines for the Fourth Order Runge-Kutta subroutine and algorithm with tolerance (see pages 144 and 145). Project H for Chapter 3: Stability of Numerical Methods Project I for Chapter 3: Period Doubling an Chaos Section 4.8: Qualitative Considerations for Variable Coeﬃcient and Nonlinear Equa- tions, which discusses the energy integral lemma, as well as the Airy, Bessel, Duﬃng, and van der Pol equations. Section 5.3: Solving Systems and Higher-Order Equations Numerically, which describes the vectorized forms of Euler’s method and the Fourth Order Runge-Kutta method, and discusses an application to population dynamics. Section 5.4: Introduction to the Phase Plane, which introduces the study of trajectories of autonomous systems, critical points, and stability. 4
- 11. Section 5.8: Dynamical Systems, Poincar`e Maps, and Chaos, which discusses the use of numerical methods to approximate the Poincar`e map and how to interpret the results. Project A for Chapter 5: Designing a Landing System for Interplanetary Travel Project B for Chapter 5: Things That Bob Project D for Chapter 5: Strange Behavior of Competing Species – Part I Project D for Chapter 9: Strange Behavior of Competing Species – Part II Project D for Chapter 10: Numerical Method for ∆u = f on a Rectangle Project D for Chapter 11: Shooting Method Project E for Chapter 11: Finite-Diﬀerence Method for Boundary Value Problems Project C for Chapter 12: Computing Phase Plane Diagrams Project D for Chapter 12: Ecosystem of Planet GLIA-2 Appendix A: Newton’s Method Appendix B: Simpson’s Rule Appendix D: Method of Least Squares Appendix E: Runge-Kutta Procedure for Equations The instructor who wishes to emphasize numerical methods should also note that the text contains an extensive chapter of series solutions of diﬀerential equations (Chapter 8). Engineering/Physics Applications Since Laplace transforms is a subject vital to engineering, we have included a detailed chapter on this topic – see Chapter 7. Stability is also an important subject for engineers, so we have included an introduction to the subject in Chapter 5.4 along with an entire chapter addressing this topic – see Chapter 12. Further material dealing with engineering/physic applications include: Project C for Chapter 1: Magnetic “Dipole” 5
- 12. Project B for Chapter 2: Torricelli’s Law of Fluid Flow Section 3.1: Mathematical Modeling Section 3.2: Compartmental Analysis, which contains a discussion of mixing problems and of population models. Section 3.3: Heating and Cooling Buildings, which discusses temperature variations in the presence of air conditioning or furnace heating. Section 3.4: Newtonian Mechanics Section 3.5: Electrical Circuits Project C for Chapter 3: Curve of Pursuit Project D for Chapter 3: Aircraft Guidance in a Crosswind Project E for Chapter 3: Feedback and the Op Amp Project F for Chapter 3: Band-Bang Controls Section 4.1: Introduction: Mass-Spring Oscillator Section 4.8: Qualitative Considerations for Variable-Coeﬃcient and Nonlinear Equations Section 4.9: A Closer Look at Free Mechanical Vibrations Section 4.10: A Closer Look at Forced Mechanical Vibrations Project B for Chapter 4: Apollo Reentry Project C for Chapter 4: Simple Pendulum Chapter 5: Introduction to Systems and Phase Plane Analysis, which includes sections on coupled mass-spring systems, electrical circuits, and phase plane analysis. Project A for Chapter 5: Designing a Landing System for Interplanetary Travel Project B for Chapter 5: Things that Bob Project C for Chapter 5: Hamiltonian Systems 6
- 13. Project D for Chapter 5: Transverse Vibrations of a Beam Chapter 7: Laplace Transforms, which in addition to basic material includes discussions of transfer functions, the Dirac delta function, and frequency response modeling. Projects for Chapter 8, dealing with Schr¨odinger’s equation, bucking of a tower, and again springs. Project B for Chapter 9: Matrix Laplace Transform Method Project C for Chapter 9: Undamped Second-Order Systems Chapter 10: Partial Diﬀerential Equations, which includes sections on Fourier series, the heat equation, wave equation, and Laplace’s equation. Project A for Chapter 10: Steady-State Temperature Distribution in a Circular Cylinder Project B for Chapter 10: A Laplace Transform Solution of the Wave Equation Project A for Chapter 11: Hermite Polynomials and the Harmonic Oscillator Section 12.4: Energy Methods, which addresses both conservative and nonconservative autonomous mechanical systems. Project A for Chapter 12: Solitons and Korteweg-de Vries Equation Project B for Chapter 12: Burger’s Equation Students of engineering and physics would also ﬁnd Chapter 8 on series solutions particularly useful, especially Section 8.8 on special functions. Biology/Ecology Applications Project D for Chapter 1: The Phase Plane, which discusses the logistic population model and bifurcation diagrams for population control. Project A for Chapter 2: Diﬀerential Equations in Clinical Medicine Section 3.1: Mathematical Modeling 7
- 14. Section 3.2: Compartmental Analysis, which contains a discussion of mixing problems and population models. Project A for Chapter 3: Dynamics for HIV Infection Project B for Chapter 3: Aquaculture, which deals with a model of raising and harvesting catﬁsh. Section 5.1: Interconnected Fluid Tanks, which introduces systems of equations. Section 5.3: Solving Systems and Higher-Order Equations Numerically, which contains an application to population dynamics. Section 5.5: Applications to Biomathematics: Epidemic and Tumor Growth Models Project D for Chapter 5: Strange Behavior of Competing Species – Part I Project E for Chapter 5: Cleaning Up the Great Lakes Project D for Chapter 9: Strange Behavior of Competing Species – Part II Problem 19 in Exercises 10.5 , which involves chemical diﬀusion through a thin layer. Project D for Chapter 12: Ecosystem on Planet GLIA-2 The basic content of the remainder of this instructor’s manual consists of supplemental group projects, answers to the even-numbered problems, and detailed solutions to the even-numbered problems in Chapters 1, 2, 4, and 7 as well as Sections 3.2, 3.3, and 3.4. The answers are, for the most part, not available any place else since the text only provides answers to odd- numbered problems, and the Student’s Solutions Manual contains only a handful of worked solutions to even-numbered problems. We would appreciate any comments you may have concerning the answers in this manual. These comments can be sent to the authors’ email addresses below. We also would encourage sharing with us (= the authors and users of the texts) any of your favorite group projects. E. B. Saﬀ A. D. Snider Edward.B.Saﬀ@Vanderbilt.edu snider@eng.usf.edu 8
- 15. Group Projects for Chapter 3 Delay Diﬀerential Equations In our discussion of mixing problems in Section 3.2, we encountered the initial value problem x (t) = 6 − 3 500 x (t − t0) , (0.1) x(t) = 0 for x ∈ [−t0, 0] , where t0 is a positive constant. The equation in (0.1) is an example of a delay diﬀer- ential equation. These equations diﬀer from the usual diﬀerential equations by the presence of the shift (t − t0) in the argument of the unknown function x(t). In general, these equations are more diﬃcult to work with than are regular diﬀerential equations, but quite a bit is known about them.1 (a) Show that the simple linear delay diﬀerential equation x = ax(t − b), (0.2) where a, b are constants, has a solution of the form x(t) = Cest for any constant C, provided s satisﬁes the transcendental equation s = ae−bs . (b) A solution to (0.2) for t > 0 can also be found using the method of steps. Assume that x(t) = f(t) for −b ≤ t ≤ 0. For 0 ≤ t ≤ b, equation (0.2) becomes x (t) = ax(t − b) = af(t − b), and so x(t) = t 0 af(ν − b)dν + x(0). Now that we know x(t) on [0, b], we can repeat this procedure to obtain x(t) = t b ax(ν − b)dν + x(b) for b ≤ x ≤ 2b. This process can be continued indeﬁnitely. 1 See, for example, Diﬀerential–Diﬀerence Equations, by R. Bellman and K. L. Cooke, Academic Press, New York, 1963, or Ordinary and Delay Diﬀerential Equations, by R. D. Driver, Springer–Verlag, New York, 1977 9
- 16. Use the method of steps to show that the solution to the initial value problem x (t) = −x(t − 1), x(t) = 1 on [−1, 0], is given by x(t) = n k=0 (−1)k [t − (k − 1)]k k! , for n − 1 ≤ t ≤ n , where n is a nonnegative integer. (This problem can also be solved using the Laplace transform method of Chapter 7.) (c) Use the method of steps to compute the solution to the initial value problem given in (0.1) on the interval 0 ≤ t ≤ 15 for t0 = 3. Extrapolation When precise information about the form of the error in an approximation is known, a technique called extrapolation can be used to improve the rate of convergence. Suppose the approximation method converges with rate O (hp ) as h → 0 (cf. Section 3.6). From theoretical considerations, assume we know, more precisely, that y(x; h) = φ(x) + hp ap(x) + O hp+1 , (0.3) where y(x; h) is the approximation to φ(x) using step size h and ap(x) is some function that is independent of h (typically, we do not know a formula for ap(x), only that it exists). Our goal is to obtain approximations that converge at the faster rate O (hp+1 ). We start by replacing h by h/2 in (0.3) to get y x; h 2 = φ(x) + hp 2p ap(x) + O hp+1 . If we multiply both sides by 2p and subtract equation (0.3), we ﬁnd 2p y x; h 2 − y(x; h) = (2p − 1) φ(x) + O hp+1 . Solving for φ(x) yields φ(x) = 2p y (x; h/2) − y(x; h) 2p − 1 + O hp+1 . Hence, y∗ x; h 2 := 2p y (x; h/2) − y(x; h) 2p − 1 has a rate of convergence of O (hp+1 ). 10
- 17. (a) Assuming y∗ x; h 2 = φ(x) + hp+1 ap+1(x) + O hp+2 , show that y∗∗ x; h 4 := 2p+1 y∗ (x; h/4) − y∗ (x; h/2) 2p+1 − 1 has a rate of convergence of O (hp+2 ). (b) Assuming y∗∗ x; h 4 = φ(x) + hp+2 ap+2(x) + O hp+3 , show that y∗∗∗ x; h 8 := 2p+2 y∗∗ (x; h/8) − y∗∗ (x; h/4) 2p+2 − 1 has a rate of convergence of O (hp+3 ). (c) The results of using Euler’s method (with h = 1, 1/2, 1/4, 1/8) to approximate the solution to the initial value problem y = y, y(0) = 1 at x = 1 are given in Table 1.2, page 27. For Euler’s method, the extrapolation procedure applies with p = 1. Use the results in Table 1.2 to ﬁnd an approximation to e = y(1) by computing y∗∗∗ (1; 1/8). [Hint: Compute y∗ (1; 1/2), y∗ (1; 1/4), and y∗ (1; 1/8); then compute y∗∗ (1; 1/4) and y∗∗ (1; 1/8).] (d) Table 1.2 also contains Euler’s approximation for y(1) when h = 1/16. Use this additional information to compute the next step in the extrapolation procedure; that is, compute y∗∗∗∗ (1; 1/16). Group Projects for Chapter 5 Eﬀects of Hunting on Predator–Prey Systems As discussed in Section 5.3 (page 277), cyclic variations in the population of predators and their prey have been studied using the Volterra-Lotka predator–prey model dx dt = Ax − Bxy , (0.4) dy dt = −Cy + Dxy , (0.5) 11
- 18. where A, B, C, and D are positive constants, x(t) is the population of prey at time t, and y(t) is the population of predators. It can be shown that such a system has a periodic solution (see Project D). That is, there exists some constant T such that x(t) = x(t+T) and y(t) = y(t + T) for all t. The periodic or cyclic variation in the population has been observed in various systems such as sharks–food ﬁsh, lynx–rabbits, and ladybird beetles–cottony cushion scale. Because of this periodic behavior, it is useful to consider the average population x and y deﬁned by x := 1 T t 0 x(t)dt , y := 1 T t 0 y(t)dt . (a) Show that x = C/D and y = A/B. [Hint: Use equation (0.4) and the fact that x(0) = x(T) to show that T 0 [A − By(t)] dt = T 0 x (t) x(t) d = dt 0. ] (b) To determine the eﬀect of indiscriminate hunting on the population, assume hunting reduces the rate of change in a population by a constant times the population. Then the predator–prey system satisﬁes the new set of equations dx dt = Ax − Bxy − εx = (A − ε)x − Bxy , (0.6) dy dt = −Cy + Dxy − δy = −(C + δ)y + Dxy , (0.7) where ε and δ are positive constants with ε < A. What eﬀect does this have on the average population of prey? On the average population of predators? (c) Assume the hunting was done selectively, as in shooting only rabbits (or shooting only lynx). Then we have ε > 0 and δ = 0 (or ε = 0 and δ > 0) in (0.6)–(0.7). What eﬀect does this have on the average populations of predator and prey? (d) In a rural county, foxes prey mainly on rabbits but occasionally include a chicken in their diet. The farmers decide to put a stop to the chicken killing by hunting the foxes. What do you predict will happen? What will happen to the farmers’ gardens? 12
- 19. Limit Cycles In the study of triode vacuum tubes, one encounters the van der Pol equation2 y − µ 1 − y2 y + y = 0 , where the constant µ is regarded as a parameter. In Section 4.8 (page 224), we used the mass-spring oscillator analogy to argue that the nonzero solutions to the van der Pol equation with µ = 1 should approach a periodic limit cycle. The same argument applies for any positive value of µ. (a) Recast the van der Pol equation as a system in normal form and use software to plot some typical trajectories for µ = 0.1, 1, and 10. Re-scale the plots if necessary until you can discern the limit cycle trajectory; ﬁnd trajectories that spiral in, and ones that spiral out, to the limit cycle. (b) Now let µ = −0.1, −1, and −10. Try to predict the nature of the solutions using the mass-spring analogy. Then use the software to check your predictions. Are there limit cycles? Do the neighboring trajectories spiral into, or spiral out from, the limit cycles? (c) Repeat parts (a) and (b) for the Rayleigh equation y − µ 1 − (y ) 2 y + y = 0 . Group Project for Chapter 13 David Stapleton, University of Central Oklahoma Satellite Altitude Stability In this problem, we determine the orientation at which a satellite in a circular orbit of radius r can maintain a relatively constant facing with respect to a spherical primary (e.g., a planet) of mass M. The torque of gravity on the asymmetric satellite maintains the orientation. 2 Historical Footnote: Experimental research by E. V. Appleton and B. van der Pol in 1921 on the oscillation of an electrical circuit containing a triode generator (vacuum tube) led to the nonlinear equation now called van der Pol’s equation. Methods of solution were developed by van der Pol in 1926–1927. Mary L. Cartwright continued research into nonlinear oscillation theory and together with J. E. Little- wood obtained existence results for forced oscillations in nonlinear systems in 1945. 13
- 20. Suppose (x, y, z) and (x, y, z) refer to coordinates in two systems that have a common origin at the satellite’s center of mass. Fix the xyz-axes in the satellite as principal axes; then let the z-axis point toward the primary and let the x-axis point in the direction of the satellite’s velocity. The xyz-axes may be rotated to coincide with the xyz-axes by a rotation φ about the x-axis (roll), followed by a rotation θ about the resulting y-axis (pitch), and a rotation ψ about the ﬁnal z-axis (yaw). Euler’s equations from physics (with high terms omitted3 to obtain approximate solutions valid near (φ, θ, ψ) = (0, 0, 0)) show that the equations for the rotational motion due to gravity acting on the satellite are Ixφ = −4ω2 0 (Iz − Iy) φ − ω0 (Iy − Iz − Ix) ψ Iyθ = −3ω2 0 (Ix − Iz) θ Izψ = −4ω2 0 (Iy − Ix) ψ + ω0 (Iy − Iz − Ix) φ , where ω0 = (GM)/r3 is the angular frequency of the orbit and the positive constants Ix, Iy, Iz are the moments of inertia of the satellite about the x, y, and z-axes. (a) Find constants c1, . . . , c5 such tha these equations can be written as two systems d dt φ ψ φ θ = 0 0 1 0 0 0 0 1 c1 0 0 c2 0 c3 c4 0 φ ψ φ ψ and d dt θ θ = 0 1 c5 0 θ θ . (b) Show that the origin is asymptotically stable for the ﬁrst system in (a) if (c2c4 + c3 + c1)2 − 4c1c3 > 0 , c1c3 > 0 , c2c4 + c3 + c1 > 0 and hence deduce that Iy > Ix > Iz yields an asymptotically stable origin. Are there other conditions on the moments of inertia by which the origin is stable? 3 The derivation of these equations is found in Attitude Stabilization and Control of Earth Satellites, by O. H. Gerlach, Space Science Reviews, #4 (1965), 541–566 14
- 21. (c) Show that, for the asymptotically stable conﬁguration in (b), the second system in (a) becomes a harmonic oscillator problem, and ﬁnd the frequency of oscillation in terms of Ix, Iy, Iz, and ω0 . Phobos maintains Iy > Ix > Iz in its orientation with respect to Mars, and has angular frequency of orbit ω0 = 0.82 rad/hr. If (Ix − Iz) /Iy = 0.23, show that the period of the libration for Phobos (the period with which the side of Phobos facing Mars shakes back and forth) is about 9 hours. 15
- 22. CHAPTER 1: Introduction EXERCISES 1.1: Background 2. This equation is an ODE because it contains no partial derivatives. Since the highest order derivative is d2 y/dx2 , the equation is a second order equation. This same term also shows us that the independent variable is x and the dependent variable is y. This equation is linear. 4. This equation is a PDE of the second order because it contains second partial derivatives. x and y are independent variables, and u is the dependent variable. 6. This equation is an ODE of the ﬁrst order with the independent variable t and the dependent variable x. It is nonlinear. 8. ODE of the second order with the independent variable x and the dependent variable y, nonlinear. 10. ODE of the fourth order with the independent variable x and the dependent variable y, linear. 12. ODE of the second order with the independent variable x and the dependent variable y, nonlinear. 14. The velocity at time t is the rate of change of the position function x(t), i.e., x . Thus, dx dt = kx4 , where k is the proportionality constant. 16. The equation is dA dt = kA2 , where k is the proportionality constant. 17
- 23. Chapter 1 EXERCISES 1.2: Solutions and Initial Value Problems 2. (a) Writing the given equation in the form y2 = 3 − x, we see that it deﬁnes two functions of x on x ≤ 3, y = ± √ 3 − x. Diﬀerentiation yields dy dx = d dx ± √ 3 − x = ± d dx (3 − x)1/2 = ± 1 2 (3 − x)−1/2 (−1) = − 1 ±2 √ 3 − x = − 1 2y . (b) Solving for y yields y3 (x − x sin x) = 1 ⇒ y3 = 1 x(1 − sin x) ⇒ y = 1 3 x(1 − sin x) = [x(1 − sin x)]−1/3 . The domain of this function is x = 0 and sin x = 1 ⇒ x = π 2 + 2kπ, k = 0, ±1, ±2, . . . . For 0 < x < π/2, one has dy dx = d dx [x(1 − sin x)]−1/3 = − 1 3 [x(1 − sin x)]−1/3−1 d dx [x(1 − sin x)] = − 1 3 [x(1 − sin x)]−1 [x(1 − sin x)]−1/3 [(1 − sin x) + x(− cos x)] = (x cos x + sin x − 1)y 3x(1 − sin x) . We also remark that the given relation is an implicit solution on any interval not containing points x = 0, π/2 + 2kπ, k = 0, ±1, ±2, . . . . 4. Diﬀerentiating the function x = 2 cos t − 3 sin t twice, we obtain x = −2 sin t − 3 cos t, x = −2 cos t + 3 sin t. Thus, x + x = (−2 cos t + 3 sin t) + (2 cos t − 3 sin t) = 0 for any t on (−∞, ∞). 6. Substituting x = cos 2t and x = −2 sin 2t into the given equation yields (−2 sin 2t) + t cos 2t = sin 2t ⇔ t cos 2t = 3 sin 2t . Clearly, this is not an identity and, therefore, the function x = cos 2t is not a solution. 18
- 24. Exercises 1.2 8. Using the chain rule, we have y = 3 sin 2x + e−x , y = 3(cos 2x)(2x) + e−x (−x) = 6 cos 2x − e−x , y = 6(− sin 2x)(2x) − e−x (−x) = −12 sin 2x + e−x . Therefore, y + 4y = −12 sin 2x + e−x + 4 3 sin 2x + e−x = 5e−x , which is the right-hand side of the given equation. So, y = 3 sin 2x + e−x is a solution. 10. Taking derivatives of both sides of the given relation with respect to x yields d dx (y − ln y) = d dx x2 + 1 ⇒ dy dx − 1 y dy dx = 2x ⇒ dy dx 1 − 1 y = 2x ⇒ dy dx y − 1 y = 2x ⇒ dy dx = 2xy y − 1 . Thus, the relation y−ln y = x2 +1 is an implicit solution to the equation y = 2xy/(y−1). 12. To ﬁnd dy/dx, we use implicit diﬀerentiation. d dx x2 − sin(x + y) = d dx (1) = 0 ⇒ 2x − cos(x + y) d dx (x + y) = 0 ⇒ 2x − cos(x + y) 1 + dy dx = 0 ⇒ dy dx = 2x cos(x + y) − 1 = 2x sec(x + y) − 1, and so the given diﬀerential equation is satisﬁed. 14. Assuming that C1 and C2 are constants, we diﬀerentiate the function φ(x) twice to get φ (x) = C1 cos x − C2 sin x, φ (x) = −C1 sin x − C2 cos x. Therefore, φ + φ = (−C1 sin x − C2 cos x) + (C1 sin x + C2 cos x) = 0. Thus, φ(x) is a solution with any choice of constants C1 and C2. 16. Diﬀerentiating both sides, we obtain d dx x2 + Cy2 = d dx (1) = 0 ⇒ 2x + 2Cy dy dx = 0 ⇒ dy dx = − x Cy . 19
- 25. Chapter 1 Since, from the given relation, Cy2 = 1 − x2 , we have − x Cy = xy −Cy2 = xy x2 − 1 . So, dy dx = xy x2 − 1 . Writing Cy2 = 1 − x2 in the form x2 + y2 1/ √ C 2 = 1, we see that the curves deﬁned by the given relation are ellipses with semi-axes 1 and 1/ √ C and so the integral curves are half-ellipses located in the upper/lower half plane. 18. The function φ(x) is deﬁned and diﬀerentiable for all values of x except those satisfying c2 − x2 = 0 ⇒ x = ±c. In particular, this function is diﬀerentiable on (−c, c). Clearly, φ(x) satisﬁes the initial condition: φ(0) = 1 c2 − 02 = 1 c2 . Next, for any x in (−c, c), dφ dx = d dx c2 − x2 −1 = (−1) c2 − x2 −2 c2 − x2 = 2x c2 − x2 −1 2 = 2xφ(x)2 . Therefore, φ(x) is a solution to the equation y = 2xy2 on (−c, c). Several integral curves are shown in Fig. 1–A on page 29. 20. (a) Substituting φ(x) = emx into the given equation yields (emx ) + 6 (emx ) + 5 (emx ) = 0 ⇒ emx m2 + 6m + 5 = 0. Since emx = 0 for any x, φ(x) satisﬁes the given equation if and only if m2 + 6m + 5 = 0 ⇔ m = −1, −5. 20
- 26. Exercises 1.2 (b) We have (emx ) + 3 (emx ) + 2 (emx ) = 0 ⇒ emx m3 + 3m2 + 2m = 0 ⇒ m(m2 + 3m + 2) = 0 ⇔ m = 0, −1, −2. 22. We ﬁnd φ (x) = c1ex − 2c2e−2x , φ (x) = c1ex + 4c2e−2x . Substitution yields φ + φ − 2φ = c1ex + 4c2e−2x + c1ex − 2c2e−2x − 2 c1ex + c2e−2x = (c1 + c1 − 2c1) ex + (4c2 − 2c2 − 2c2) e−2x = 0. Thus, with any choice of constants c1 and c2, φ(x) is a solution to the given equation. (a) Constants c1 and c2 must satisfy the system 2 = φ(0) = c1 + c2 1 = φ (0) = c1 − 2c2 . Subtracting the second equation from the ﬁrst one yields 3c2 = 1 ⇒ c2 = 1/3 ⇒ c1 = 2 − c2 = 5/3. (b) Similarly to the part (a), we obtain the system 1 = φ(1) = c1e + c2e−2 0 = φ (1) = c1e − 2c2e−2 which has the solution c1 = (2/3)e−1 , c2 = (1/3)e2 . 24. In this problem, the independent variable is t, the dependent variable is y. Writing the equation in the form dy dt = ty + sin2 t , we conclude that f(t, y) = ty + sin2 t, ∂f(t, y)/∂y = t. Both functions, f and ∂f/∂y, are continuous on the whole ty-plane. So, Theorem 1 applies for any initial condition, in particular, for y(π) = 5. 21
- 27. Chapter 1 26. With the independent variable t and the dependent variable x, we have f(t, x) = sin t − cos x, ∂f(t, x) ∂x = sin x , which are continuous on tx-plane. So, Theorem 1 applies for any initial condition. 28. Here, f(x, y) = 3x − 3 √ y − 1 and ∂f(x, y) ∂y = ∂ ∂y 3x − (y − 1)1/3 ) = − 1 3 3 (y − 1)2 . The function f is continuous at any point (x, y) while ∂f/∂y is deﬁned and continuous at any point (x, y) with y = 1 i.e., on the xy-plane excluding the horizontal line y = 1. Since the initial point (2, 1) belongs to this line, there is no rectangle containing the initial point, on which ∂f/∂y is continuous. Thus, Theorem 1 does not apply. 30. Here, the initial point (x0, y0) is (0, −1) and G(x, y) = x + y + exy . The ﬁrst partial derivatives, Gx(x, y) = (x + y + exy )x = 1 + yexy and Gy(x, y) = (x + y + exy )y = 1 + xexy , are continuous on the xy-plane. Next, G(0, −1) = −1 + e0 = 0, Gy(0, −1) = 1 + (0)e0 = 1 = 0. Therefore, all the hypotheses of Implicit Function Theorem are satisﬁed, and so the relation x + y + exy = 0 deﬁnes a diﬀerentiable function y = φ(x) on some interval (−δ, δ) about x0 = 0. EXERCISES 1.3: Direction Fields 2. (a) Starting from the initial point (0, −2) and following the direction markers we get the curve shown in Fig. 1–B on page 30. Thus, the solution curve to the initial value problem dy/dx = 2x + y, y(0) = −2, is the line with slope dy dx (0) = (2x + y)|x=0 = y(0) = −2 and y-intercept y(0) = −2. Using the slope-intercept form of an equation of a line, we get y = −2x − 2. 22
- 28. Exercises 1.3 (b) This time, we start from the point (−1, 3) and obtain the curve shown in Fig. 1–C on page 30. (c) From Fig. 1–C, we conclude that lim x→∞ y(x) = ∞, lim x→−∞ y(x) = ∞. 4. The direction ﬁeld and the solution curve satisfying the given initial conditions are sketched in Fig. 1–D on page 30. From this ﬁgure we ﬁnd that the terminal velocity is limt→∞ v(t) = 2. 6. (a) The slope of the solution curve to the diﬀerential equation y = x + sin y at a point (x, y) is given by y . Therefore the slope at (1, π/2) is equal to dy dx x=1 = (x + sin y)|x=1 = 1 + sin π 2 = 2. (b) The solution curve is increasing if the slope of the curve is greater than zero. From the part (a), we know that the slope is x + sin y. The function sin y has values ranging from −1 to 1; therefore if x is greater than 1 then the slope will always have a value greater than zero. This tells us that the solution curve is increasing. (c) The second derivative of every solution can be determined by diﬀerentiating both sides of the original equation, y = x + sin y. Thus d dx dy dx = d dx (x + sin y) ⇒ d2 y dx2 = 1 + (cos y) dy dx (chain rule) = 1 + (cos y)(x + sin y) = 1 + x cos y + sin y cos y = 1 + x cos y + 1 2 sin 2y . (d) Relative minima occur when the ﬁrst derivative, y , is equal to zero and the second derivative, y , is positive (Second Derivative Test). The value of the ﬁrst derivative at the point (0, 0) is given by dy dx = 0 + sin 0 = 0. This tells us that the solution has a critical point at the point (0, 0). Using the second derivative found in part (c) we have d2 y dx2 = 1 + 0 · cos 0 + 1 2 sin 0 = 1. 23
- 29. Chapter 1 This tells us that the point (0, 0) is a point of relative minimum. 8. (a) For this particle, we have x(2) = 1, and so the velocity v(2) = dx dt t=2 = t3 − x3 t=2 = 23 − x(2)3 = 7. (b) Diﬀerentiating the given equation yields d2 x dt2 = d dt dx dt = d dt t3 − x3 = 3t2 − 3x2 dx dt = 3t2 − 3x2 t3 − x3 = 3t2 − 3t3 x2 + 3x5 . (c) The function u3 is an increasing function. Therefore, as long as x(t) < t, x(t)3 < t3 and dx dt = t3 − x(t)3 > 0 meaning that x(t) increases. At the initial point t0 = 2.5 we have x(t0) = 2 < t0. Therefore, x(t) cannot take values smaller than 2.5, and the answer is “no”. 10. Direction ﬁelds and some solution curves to diﬀerential equations given in (a)–(e) are shown in Fig. 1–E through Fig. 1–I on pages 31–32. (a) y = sin x. (b) y = sin y. (c) y = sin x sin y. (d) y = x2 + 2y2 . (e) y = x2 − 2y2 . 12. The isoclines satisfy the equation f(x, y) = y = c, i.e., they are horizontal lines shown in Fig. 1–J, page 32, along with solution curves. The curve, satisfying the initial condition, is shown in bold. 14. Here, f(x, y) = x/y, and so the isoclines are deﬁned by x y = c ⇒ y = 1 c x. These are lines passing through the origin and having slope 1/c. See Fig. 1–K on page 33. 24
- 30. Exercises 1.4 16. The relation x+2y = c yields y = (c−x)/2. Therefore, the isoclines are lines with slope −1/2 and y-intercept c/2. See Fig. 1–L on page 33. 18. The direction ﬁeld for this equation is shown in Fig. 1–M on page 33. From this picture we conclude that any solution y(x) approaches zero, as x → +∞. EXERCISES 1.4: The Approximation Method of Euler 2. In this problem, x0 = 0, y0 = 4, h = 0.1, and f(x, y) = −x/y. Thus, the recursive formulas given in equations (2) and (3) of the text become xn+1 = xn + h = xn + 0.1 , yn+1 = yn + hf(xn, yn) = yn + 0.1 − xn yn , n = 0, 1, 2, . . . . To ﬁnd an approximation for the solution at the point x1 = x0 + 0.1 = 0.1, we let n = 0 in the last recursive formula to ﬁnd y1 = y0 + 0.1 − x0 y0 = 4 + 0.1(0) = 4. To approximate the value of the solution at the point x2 = x1 + 0.1 = 0.2, we let n = 1 in the last recursive formula to obtain y2 = y1 + 0.1 − x1 y1 = 4 + 0.1 − 0.1 4 = 4 − 1 400 = 3.9975 ≈ 3.998 . Continuing in this way we ﬁnd x3 = x2 + 0.1 = 0.3 , y3 = y2 + 0.1 − x2 y2 = 3.9975 + 0.1 − 0.2 3.9975 ≈ 3.992 , x4 = 0.4 , y4 ≈ 3.985 , x5 = 0.5 , y5 ≈ 3.975 , where all of the answers have been rounded oﬀ to three decimal places. 4. Here x0 = 0, y0 = 1, and f(x, y) = x + y. So, xn+1 = xn + h = xn + 0.1 , yn+1 = yn + hf(xn, yn) = yn + 0.1 (xn + yn) , n = 0, 1, 2, . . . . Letting n = 0, 1, 2, 3, and 4, we recursively ﬁnd x1 = x0 + h = 0.1 , y1 = y0 + 0.1 (x0 + y0) = 1 + 0.1(0 + 1) = 1.1 , 25
- 31. Chapter 1 x2 = x1 + h = 0.2 , y2 = y1 + 0.1 (x1 + y1) = 1.1 + 0.1(0.1 + 1.1) = 1.22 , x3 = x2 + h = 0.3 , y3 = y2 + 0.1 (x2 + y2) = 1.22 + 0.1(0.2 + 1.22) = 1.362 , x4 = x3 + h = 0.4 , y4 = y3 + 0.1 (x3 + y3) = 1.362 + 0.1(0.3 + 1.362) = 1.528 , x5 = x4 + h = 0.5 , y5 = y4 + 0.1 (x4 + y4) = 1.5282 + 0.1(0.4 + 1.5282) = 1.721 , where all of the answers have been rounded oﬀ to three decimal places. 6. In this problem, x0 = 1, y0 = 0, and f(x, y) = x − y2 . So, we let n = 0, 1, 2, 3, and 4, in the recursive formulas and ﬁnd x1 = x0 + h = 1.1 , y1 = y0 + 0.1 x0 − y2 0 = 0 + 0.1(1 − 02 ) = 0.1 , x2 = x1 + h = 1.2 , y2 = y1 + 0.1 x1 − y2 1 = 0.1 + 0.1(1.1 − 0.12 ) = 0.209 , x3 = x2 + h = 1.3 , y3 = y2 + 0.1 x2 − y2 2 = 0.209 + 0.1(1.2 − 0.2092 ) = 0.325 , x4 = x3 + h = 1.4 , y4 = y3 + 0.1 x3 − y2 3 = 0.325 + 0.1(1.3 − 0.3252 ) = 0.444 , x5 = x4 + h = 1.5 , y5 = y4 + 0.1 x4 − y2 4 = 0.444 + 0.1(1.4 − 0.4442 ) = 0.564 , where all of the answers have been rounded oﬀ to three decimal places. 8. The initial values are x0 = y0 = 0, f(x, y) = 1 − sin y. If number of steps is N, then the step h = (π − x0)/N = π/N. For N = 1, h = π, x1 = x0 + h = π, y1 = y0 + h(1 − sin y0) = π ≈ 3.1416 . For N = 2, h = π/2, x1 = x0 + π/2 = π/2, y1 = y0 + h(1 − sin y0) = π/2 ≈ 1.571 , x2 = x1 + π/2 = π, y2 = y1 + h(1 − sin y1) = π/2 ≈ 1.571 . We continue with N = 4 and 8, and ﬁll in Table 1 on page 28, where the approximations to φ(π) are rounded to three decimal places. 10. We have x0 = y(0) = 0, h = 0.1. With this step size, we need (1 − 0)/0.1 = 10 steps to approximate the solution on [0, 1]. The results of computation are given in Table 1 on page 28. 26
- 32. Exercises 1.4 Next we check that y = e−x + x − 1 is the actual solution to the given initial value problem. y = e−x + x − 1 = −e−x + 1 = x − e−x + x − 1 = x − y, y(0) = e−x + x − 1 x=0 = e0 + 0 − 1 = 0. Thus, it is the solution. The solution curve y = e−x + x − 1 and the polygonal line approximation using data from Table 1 are shown in Fig. 1–N, page 34. 12. Here, x0 = 0, y0 = 1, f(x, y) = y. With h = 1/n, the recursive formula (3) of the text yields y(1) = yn = yn−1 + yn−1 n = yn−1 1 + 1 n = yn−2 1 + 1 n 1 + 1 n = yn−2 1 + 1 n 2 = . . . = y0 1 + 1 n n = 1 + 1 n n . 14. Computation results are given in Table 1 on page 29. 16. For this problem notice that the independent variable is t and the dependent variable is T. Hence, in the recursive formulas for Euler’s method, t will take the place of x and T will take the place of y. Also we see that h = 0.1 and f(t, T) = K (M4 − T4 ), where K = 40−4 and M = 70. Therefore, the recursive formulas given in equations (2) and (3) of the text become tn+1 = tn + 0.1 , Tn+1 = Tn + hf (tn, Tn) = Tn + 0.1 40−4 704 − T4 n , n = 0, 1, 2, . . . . From the initial condition T(0) = 100 we see that t0 = 0 and T0 = 100. Therefore, for n = 0, we have t1 = t0 + 0.1 = 0 + 0.1 = 0.1 , T1 = T0 + 0.1(40−4 )(704 − T4 0 ) = 100 + 0.1(40−4 )(704 − 1004 ) ≈ 97.0316, where we have rounded oﬀ to four decimal places. For n = 1, t2 = t1 + 0.1 = 0.1 + 0.1 = 0.2 , 27
- 33. Chapter 1 T2 = T1 + 0.1(40−4 )(704 − T4 1 ) = 97.0316 + 0.1(40−4 )(704 − 97.03164 ) ≈ 94.5068. By continuing this way, we ﬁll in Table 1 on page 29. From this table we see that T(1) = T(t10) ≈ T10 = 82.694 , T(2) = T(t20) ≈ T20 = 76.446 , where we have rounded to three decimal places. TABLES NNN hhh φφφ(πππ) 1 π 3.142 2 π/2 1.571 4 π/4 1.207 8 π/8 1.148 Table 1–A: Euler’s approximations to y = 1 − sin y, y(0) = 0, with N steps. nnn xxxnnn yyynnn nnn xxxnnn yyynnn 0 0 0 6 0.6 0.131 1 0.1 0 7 0.7 0.178 2 0.2 0.01 8 0.8 0.230 3 0.3 0.029 9 0.9 0.287 4 0.4 0.056 10 1.0 0.349 5 0.5 0.091 Table 1–B: Euler’s approximations to y = x − y, y(0) = 0, on [0, 1] with h = 0.1. 28
- 34. Figures hhh yyy(2) 0.5 24.8438 0.1 ≈ 6.4 · 10176 0.05 ≈ 1.9 · 10114571 0.01 > 101030 Table 1–C: Euler’s method approximations of y(2) for y = 2xy2 , y(0) = 1. nnn tttnnn TTTnnn nnn tttnnn TTTnnn 1 0.1 97.0316 11 1.1 81.8049 2 0.2 94.5068 12 1.2 80.9934 3 0.3 92.3286 13 1.3 80.2504 4 0.4 90.4279 14 1.4 79.5681 5 0.5 88.7538 15 1.5 78.9403 6 0.6 87.2678 16 1.6 78.3613 7 0.7 85.9402 17 1.7 77.8263 8 0.8 84.7472 18 1.8 77.3311 9 0.9 83.6702 19 1.9 76.8721 10 1.0 82.6936 20 2.0 76.4459 Table 1–D: Euler’s approximations to the solution of T = K (M4 − T4 ), T(0) = 100, with K = 40−4 , M = 70, and h = 0.1. FIGURES K2 K1 0 1 2 1 2 3 4 5 Figure 1–A: Integral curves in Problem 18. 29
- 35. Chapter 1 K1 0 1 K4 K2 2 Figure 1–B: The solution curve in Problem 2(a). K3 K2 K1 0 1 2 3 4 Figure 1–C: The solution curve in Problem 2(b). v(t) 0 1 2 3 4 Figure 1–D: The direction ﬁeld and solution curves in Problem 4. 30
- 36. Figures x K4 K2 0 2 4 y(x) K2 2 4 Figure 1–E: The direction ﬁeld and solution curves in Problem 10(a). x K2.5 0 2.5 y(x) K2 2 Figure 1–F: The direction ﬁeld and solution curves in Problem 10(b). x K4 K2 0 2 4 y(x) K4 K2 2 4 Figure 1–G: The direction ﬁeld and solution curves in Problem 10(c). 31
- 37. Chapter 1 x K2 0 2 y(x) K2 2 Figure 1–H: The direction ﬁeld and solution curves in Problem 10(d). x K2 0 2 y(x) K2 2 Figure 1–I: The direction ﬁeld and solution curves in Problem 10(e). K2 0 2 K2 2 Figure 1–J: The isoclines and solution curves in Problem 12. 32
- 38. Figures K3 K2 K1 0 1 2 3 K3 K2 K1 1 2 3 Figure 1–K: The isoclines and solution curves in Problem 14. K3 K2 K1 0 1 2 3 K3 K2 K1 1 2 3 Figure 1–L: The isoclines and solution curves in Problem 16. x 0 2.5 5.0 y(x) K2 2 Figure 1–M: The direction ﬁeld in Problem 18. 33
- 39. Chapter 1 0 0.5 1.0 0 0.2 0.4 Figure 1–N: Euler’s method approximations to y = e−x +x−1 on [0, 1] with h = 0.1. 34
- 40. CHAPTER 2: First Order Diﬀerential Equations EXERCISES 2.2: Separable Equations 2. This equation is not separable because sin(x + y) cannot be expressed as a product g(x)p(y). 4. This equation is separable because ds dt = t ln s2t + 8t2 = t(2t) ln |s| + 8t2 = 2t2 (ln |s| + 4). 6. Writing the equation in the form dy dx = 2x xy2 + 3y2 = 2x (x + 3)y2 = 2x x + 3 · 1 y2 , we see that the equation is separable. 8. Multiplying both sides of the equation by y3 dx and integrating yields y3 dy = dx x ⇒ y3 dy = dx x ⇒ 1 4 y4 = x ln |x| + C1 ⇒ y4 = 4 ln |x| + C ⇒ y = ± 4 4 ln |x| + C , where C := 4C1 is an arbitrary constant. 10. To separate variables, we divide the equation by x and multiply by dt. Integrating yields dx x = 3t2 dt ⇒ ln |x| = t3 + C1 ⇒ |x| = et3+C1 = eC1 et3 ⇒ |x| = C2et3 ⇒ x = ±C2et3 = Cet3 , where C1 is an arbitrary constant and, therefore, C2 := eC1 is an arbitrary positive constant, C = ±C2 is any nonzero constant. Separating variables, we lost a solution x ≡ 0, which can be included in the above formula by taking C = 0. Thus, x = Cet3 , C – arbitrary constant, is a general solution. 35
- 41. Chapter 2 12. We have 3vdv 1 − 4v2 = dx x ⇒ 3vdv 1 − 4v2 = dx x ⇒ − 3 8 du u = dx x u = 1 − 4v2 , du = −8vdv ⇒ − 3 8 ln 1 − 4v2 = ln |x| + C1 ⇒ 1 − 4v2 = ± exp − 8 3 ln |x| + C1 = Cx−8/3 , where C = ±eC1 is any nonzero constant. Separating variables, we lost constant solutions satisfying 1 − 4v2 = 0 ⇒ v = ± 1 2 , which can be included in the above formula by letting C = 0. Thus, v = ± √ 1 − Cx−8/3 2 , C arbitrary, is a general solution to the given equation. 14. Separating variables, we get dy 1 + y2 = 3x2 dx ⇒ dy 1 + y2 = 3x2 dx ⇒ arctan y = x3 + C ⇒ y = tan x3 + C , where C is any constant. Since 1 + y2 = 0, we did not lose any solution. 16. We rewrite the equation in the form x(1 + y2 )dx + ex2 ydy = 0, separate variables, and integrate. e−x2 xdx = − ydy 1 + y2 ⇒ e−x2 xdx = − ydy 1 + y2 ⇒ e−u du = − dv v u = x2 , v = 1 + y2 ⇒ −e−u = − ln |v| + C ⇒ ln 1 + y2 − e−x2 = C is an implicit solution to the given equation. Solving for y yields y = ± C1 exp [exp (−x2)] − 1, where C1 = eC is any positive constant, 36
- 42. Exercises 2.2 18. Separating variables yields dy 1 + y2 = tan xdx ⇒ dy 1 + y2 = tan xdx ⇒ arctan y = − ln | cos x| + C. Since y(0) = √ 3, we have arctan √ 3 = − ln cos 0 + C = C ⇒ C = π 3 . Therefore, arctan y = − ln | cos x| + π 3 ⇒ y = tan − ln | cos x| + π 3 is the solution to the given initial value problem. 20. Separating variables and integrating, we get (2y + 1)dy = 3x2 + 4x + 2 dx ⇒ y2 + y = x3 + 2x2 + 2x + C. Since y(0) = −1, substitution yields (−1)2 + (−1) = (0)3 + 2(0)2 + 2(0) + C ⇒ C = 0, and the solution is given, implicitly, by y2 + y = x3 + 2x2 + 2x or, explicitly, by y = − 1 2 − 1 4 + x3 + 2x2 + 2x. (Solving for y, we used the initial condition.) 22. Writing 2ydy = −x2 dx and integrating, we ﬁnd y2 = − x3 3 + C. With y(0) = 2, (2)2 = − (0)3 3 + C ⇒ C = 4, and so y2 = − x3 3 + 4 ⇒ y = − x3 3 + 4. We note that, taking the square root, we chose the positive sign because y(0) > 0. 37
- 43. Chapter 2 24. For a general solution, we separate variables and integrate. e2y dy = 8x3 dx ⇒ e2y 2 = 2x4 + C1 ⇒ e2y = 4x4 + C. We substitute now the initial condition, y(1) = 0, and obtain 1 = 4 + C ⇒ C = −3. Hence, the answer is given by e2y = 4x4 − 3 ⇒ y = 1 2 ln 4x4 − 3 . 26. We separate variables and obtain dy √ y = − dx 1 + x ⇒ 2 √ y = − ln |1 + x| + C = − ln(1 + x) + C, because at initial point, x = 0, 1 + x > 0. Using the fact that y(0) = 1, we ﬁnd C. 2 = 0 + C ⇒ C = 2, and so y = [2 − ln(1 + x)]2 /4 is the answer. 28. We have dy dt = 2y(1 − t) ⇒ dy y = 2(1 − t)dt ⇒ ln |y| = −(t − 1)2 + C ⇒ y = ±eC e−(t−1)2 = C1e−(t−1)2 , where C1 = 0 is any constant. Separating variables, we lost the solution y ≡ 0. So, a general solution to the given equation is y = C2e−(t−1)2 , C2 is any. Substituting t = 0 and y = 3, we ﬁnd 3 = C2e−1 ⇒ C2 = 3e ⇒ y = 3e1−(t−1)2 = 3e2t−t2 . The graph of this function is given in Fig. 2–A on page 71. Since y(t) > 0 for any t, from the given equation we have y (t) > 0 for t < 1 and y (t) < 0 for t > 1. Thus t = 1 is the point of absolute maximum with ymax = y(1) = 3e. 38
- 44. Exercises 2.2 30. (a) Dividing by (y + 1)2/3 , multiplying by dx, and integrating, we obtain dy (y + 1)2/3 = (x − 3)dx ⇒ 3(y + 1)1/3 = x2 2 − 3x + C ⇒ y = −1 + x2 6 − x + C1 3 . (b) Substituting y ≡ −1 into the original equation yields d(−1) dx = (x − 3)(−1 + 1)2/3 = 0, and so the equation is satisﬁed. (c) For y ≡ −1 for the solution in part (a), we must have x2 6 − x + C1 3 ≡ 0 ⇔ x2 6 − x + C1 ≡ 0, which is impossible since a quadratic polynomial has at most two zeros. 32. (a) The direction ﬁeld of the given diﬀerential equation is shown in Fig. 2–B, page 72. Using this picture we predict that limx→∞ φ(x) = 1. (b) In notation of Section 1.4, we have x0 = 0, y0 = 1.5, f(x, y) = y2 − 3y + 2, and h = 0.1. With this step size, we need (1 − 0)/0.1 = 10 steps to approximate φ(1). The results of computation are given in Table 2 on page 71. From this table we conclude that φ(1) ≈ 1.26660 . (c) Separating variables and integrating, we obtain dy y2 − 3y + 2 = dx ⇒ dy y2 − 3y + 2 = dx ⇒ ln y − 2 y − 1 = x + C , where we have used a partial fractions decomposition 1 y2 − 3y + 2 = 1 y − 2 − 1 y − 1 to evaluate the integral. The initial condition, y(0) = 1.5 , implies that C = 0, and so ln y − 2 y − 1 = x ⇒ y − 2 y − 1 = ex ⇒ y − 2 y − 1 = −ex . (We have chosen the negative sign because of the initial condition.) Solving for y yields y = φ(x) = ex + 2 ex + 1 The graph of this solution is shown in Fig. 2–B on page 72. 39
- 45. Chapter 2 (d) We ﬁnd φ(1) = e + 2 e + 1 ≈ 1.26894 . Thus, the approximate value φ(1) ≈ 1.26660 found in part (b) diﬀers from the actual value by less than 0.003 . (e) We ﬁnd the limit of φ(x) at inﬁnity writing lim x→∞ ex + 2 ex + 1 = lim x→∞ 1 + 1 ex + 1 = 1 , which conﬁrms our guess in part (a). 34. (a) Separating variables and integrating, we get dT T − M = −kdt ⇒ dT T − M = − kdt ⇒ ln |T − M| = −kt + C1 ⇒ |T − M| = eC1 e−kt ⇒ T − M = ±eC1 e−kt = Ce−kt , where C is any nonzero constant. We can include the lost solution T ≡ M into this formula by letting C = 0. Thus, a general solution to the equation is T = M + Ce−kt . (b) Given that M = 70◦ , T(0) = 100◦ , T(6) = 80◦ , we form a system to determine C and k. 100 = 70 + C 80 = 70 + Ce−6k ⇒ C = 30 k = −(1/6) ln[(80 − 70)/30] = (1/6) ln 3. Therefore, T = 70 + 30e−(t ln 3)/6 = 70 + (30)3−t/6 , and after 20 min the reading is T(20) = 70 + (30)3−20/6 ≈ 70.77◦ . 36. A general solution to the cooling equation found in Problem 34, that is, T = M +Ce−kt . Since T(0) = 100◦ , T(5) = 80◦ , and T(10) = 65◦ , we determine M, C, and k from the system M + C = 100 M + Ce−5k = 80 M + Ce−10k = 65 ⇒ C(1 − e−5k ) = 20 Ce−5k (1 − e−5k ) = 15 ⇒ e−5k = 3/4. 40
- 46. Exercises 2.3 To ﬁnd M, we can now use the ﬁrst two equations in the above system. M + C = 100 M + (3/4)C = 80 ⇒ M = 20. 38. With m = 10, g = 9.81, and k = 5, the equation becomes 100 dv dt = 100(9.81) − 5v ⇒ 20 dv dt = 196.2 − v. Separating variables and integrating yields dv v − 196.2 = − 1 20 dt ⇒ ln |v − 196.2| = − t 20 + C1 ⇒ v = 196.2 + Ce−t/20 , where C is an arbitrary nonzero constant. With C = 0, this formula also gives the (lost) constant solution v = 196.2. From the initial condition, v(0) = 10, we ﬁnd C. 196.2 + C = 10 ⇒ C = −186.2 ⇒ v(t) = 196.2 − 186.2e−t/20 . The terminal velocity of the object can be found by letting t → ∞. v∞ = lim t→∞ 196.2 − 186.2e−t/20 = 196.2 (m/sec). EXERCISES 2.3: Linear Equations 2. Neither. 4. Linear. 6. Linear. 8. Writing the equation in standard form, dy dx − y x = 2x + 1, we see that P(x) = − 1 x ⇒ µ(x) = exp − 1 x dx = exp (− ln x) = 1 x . Multiplying the given equation by µ(x), we get d dx y x = 2 + 1 x ⇒ y = x 2 + 1 x dx = x (2x + ln |x| + C) . 41
- 47. Chapter 2 10. From the standard form of the given equation, dy dx + 2 x y = x−4 , we ﬁnd that µ(x) = exp (2/x)dx = exp (2 ln x) = x2 ⇒ d dx x2 y = x−2 ⇒ y = x−2 x−2 dx = x−2 −x−1 + C = Cx − 1 x3 . 12. Here, P(x) = 4, Q(x) = x2 e−4x . So, µ(x) = e4x and d dx e4x y = x2 ⇒ y = e−4x x2 dx = e−4x x3 3 + C . 14. We divide the equation by x to get to get its standard form. dy dx + 3 x y = x2 − 2x + 4. Thus, P(x) = 3/x, Q(x) = x2 − 2x + 4, µ(x) = exp 3 x dx = x3 ⇒ x3 y = x3 x2 − 2x + 4 dx = x6 6 − 2x5 5 + x4 + C ⇒ y = x3 6 − 2x2 5 + x + Cx−3 . 16. We divide by x2 + 1 both sides of the given equation to get its standard form, dy dx + 4x x2 + 1 y = x2 + 2x − 1 x2 + 1 . Thus, P(x) = (4x)/(x2 + 1), Q(x) = (x2 + 2x − 1)/(x2 + 1), µ(x) = exp 4x x2 + 1 dx = exp 2 ln(x2 + 1) = (x2 + 1)2 ⇒ (x2 + 1)2 y = (x2 + 1)(x2 + 2x − 1)dx = x5 5 + x4 2 + x2 − x + C ⇒ y = x5 5 + x4 2 + x2 − x + C x2 + 1 −2 . 42
- 48. Exercises 2.3 18. Since µ(x) = exp 4dx = e4x , we have d dx e4x y = e4x e−x = e3x ⇒ y = e−4x e3x dx = e−x 3 + Ce−4x . Substituting the initial condition, y = 4/3 at x = 0, yields 4 3 = 1 3 + C ⇒ C = 1, and so y = e−x /3 + e−4x is the solution to the given initial value problem. 20. We have µ(x) = exp 3dx x = exp (3 ln x) = x3 ⇒ x3 y = x3 (3x − 2) dx = 3x5 5 − x4 2 + C ⇒ y = 3x2 5 − x 2 + Cx−3 . With y(1) = 1, 1 = y(1) = 3 5 − 1 2 + C ⇒ C = 9 10 ⇒ y = 3x2 5 − x 2 + 9 10x3 . 22. From the standard form of this equation, dy dx + y cot x = x, we ﬁnd µ(x) = exp cot x dx = exp (ln sin x) = sin x. (Alternatively, one can notice that the left-hand side of the original equation is the derivative of the product y sin x.) So, using integration by parts, we obtain y sin x = x sin x dx = −x cos x + sin x + C ⇒ y = −x cot x + 1 + C csc x. We ﬁnd C using the initial condition y(π/2) = 2: 2 = − π 2 cot π 2 + 1 + C csc π 2 = 1 + C ⇒ C = 1, and the solution is given by y = −x cot x + 1 + csc x. 43
- 49. Chapter 2 24. (a) The equation (12) on of the text becomes dy dt + 20y = 50e−10t ⇒ µ(t) = e20t ⇒ y = e−20t 50e10t dt = 5e−10t + Ce−20t . Since y(0) = 40, we have 40 = 5 + C ⇒ C = 35 ⇒ y = 5e−10t + 35e−20t . The term 5e−10t will eventually dominate. (b) This time, the equation (12) has the form dy dt + 10y = 50e−10t ⇒ µ(t) = e10t ⇒ y = e−10t 50dt = e−10t (50t + C). Substituting the initial condition yields 40 = y(0) = C ⇒ y = e−10t (50t + 40). 26. Here P(x) = sin x cos x 1 + sin2 x ⇒ µ(x) = exp sin x cos x dx 1 + sin2 x = exp 1 2 ln 1 + sin2 x = 1 + sin2 x. Thus, y 1 + sin2 x = x 0 1 + sin2 tdt ⇒ y = (1 + sin2 x)−1/2 x 0 (1 + sin2 t)1/2 dt and y(1) = (1 + sin2 1)−1/2 1 0 (1 + sin2 t)1/2 dt. We now use the Simpson’s Rule to ﬁnd that y(1) ≈ 0.860. 28. (a) Substituting y = e−x into the equation (16) yields d(e−x ) dx + e−x = −e−x + e−x = 0. 44
- 50. Exercises 2.3 So, y = e−x is a solution to (16). The function y = x−1 is a solution to (17) because d(x−1 ) dx + (x−1 )2 = (−1)x−2 + x−2 = 0. (b) For any constant C, d(Ce−x ) dx + Ce−x = −Ce−x + Ce−x = 0. Thus y = Ce−x is a solution to (16). Substituting y = Cx−1 into (17), we obtain d(Cx−1 ) dx + (Cx−1 )2 = (−C)x−2 + C2 x−2 = C(C − 1)x−2 , and so we must have C(C − 1) = 0 in order that y = Cx−1 is a solution to (17). Thus, either C = 0 or C = 1. (c) For the function y = Cˆy, one has d(Cˆy) dx + P(x) (Cˆy) = C dˆy dx + C (P(x)ˆy) = C dˆy dx + P(x)ˆy = 0 if ˆy is a solution to y + P(x)y = 0. 30. (a) Multiplying both sides of (18) by y2 , we get y2 dy dx + 2y3 = x. If v = y3 , then v = 3y2 y . Thus, y2 y = v /3, and we have 1 3 dv dx + 2v = x, which is equivalent to (19). (b) The equation (19) is linear with P(x) = 6 and Q(x) = 3x. So, µ(x) = exp 6dx = e6x ⇒ v(x) = e−6x (3xe6x )dx = e−6x 2 xe6x − e6x dx = e−6x 2 xe6x − e6x 6 + C1 = x 2 − 1 12 + Ce−6x , where C = C1/2 is an arbitrary constant. The back substitution yields y = 3 x 2 − 1 12 + Ce−6x. 45
- 51. Chapter 2 32. In the given equation, P(x) = 2, which implies that µ(x) = e2x . Following guidelines, ﬁrst we solve the equation on [0, 3]. On this interval, Q(x) ≡ 2. Therefore, y1(x) = e−2x (2)e2x dx = 1 + C1e−2x . Since y1(0) = 0, we get 1 + C1e0 = 0 ⇒ C1 = −1 ⇒ y1(x) = 1 − e−2x . For x > 3, Q(x) = −2 and so y2(x) = e−2x (−2)e2x dx = −1 + C2e−2x . We now choose C2 so that y2(3) = y1(3) = 1 − e−6 ⇒ −1 + C2e−6 = 1 − e−6 ⇒ C2 = 2e6 − 1. Therefore, y2(x) = −1 + (2e6 − 1)e−2x , and the continuous solution to the given initial value problem on [0, ∞) is y(x) = 1 − e−2x , 0 ≤ x ≤ 3, −1 + (2e6 − 1)e−2x , x > 3. The graph of this function is shown in Fig. 2–C, page 72. 34. (a) Since P(x) is continuous on (a, b), its antiderivatives given by P(x)dx are con- tinuously diﬀerentiable, and therefore continuous, functions on (a, b). Since the function ex is continuous on (−∞, ∞), composite functions µ(x) = e R P(x)dx are continuous on (a, b). The range of the exponential function is (0, ∞). This implies that µ(x) is positive with any choice of the integration constant. Using the chain rule, we conclude that dµ(x) dx = e R P(x)dx d dx P(x)dx = µ(x)P(x) for any x on (a, b). (b) Diﬀerentiating (8), we apply the product rule and obtain dy dx = −µ−2 µ µQ dx + C + µ−1 µQ = −µ−1 P µQ dx + C + Q , and so dy dx + Py = −µ−1 P µQ dx + C + Q + P µ−1 µQ dx + C = Q . 46
- 52. Exercises 2.3 (c) Suggested choice of the antiderivative and the constant C yields y(x0) = µ(x)−1 x x0 µQdx + y0µ(x0) x=x0 = µ(x0)−1 y0µ(x0) = y0 . (d) We assume that y(x) is a solution to the initial value problem (15). Since µ(x) is a continuous positive function on (a, b), the equation (5) is equivalent to (4). Since, from the part (a), the left-hand side of (5) is the derivative of the product µ(x)y(x), this function must be an antiderivative of the right-hand side, which is µ(x)Q(x). Thus, we come up with (8), where the integral means one of the antiderivatives, for example, the one suggested in the part (c) (which has zero value at x0). Substituting x = x0 into (8), we conclude that y0 = y(x0) = µ(x0)−1 µQdx + C x=x0 = Cµ(x0)−1 , and so C = y0µ(x0) is uniquely deﬁned. 36. (a) If µ(x) = exp Pdx and yh(x) = µ(x)−1 , then dyh dx = (−1)µ(x)−2 dµ(x) dx = −µ(x)−2 µ(x)P(x) = −µ(x)−1 P(x) and so dyh dx + P(x)yh = −µ(x)−1 P(x) + P(x)µ(x)−1 = 0, i.e., yh is a solution to the equation y + Py = 0. Now, the formula (8) yields y = µ(x)−1 µ(x)Q(x)dx + C = yh(x)v(x) + Cyh(x) = yp(x) + Cyh(x), where v(x) = µ(x)Q(x)dx. (b) Separating variables in (22) and integrating, we obtain dy y = − 3dx x ⇒ dy y = − 3dx x ⇒ ln |y| = −3 ln x + C. Since we need just one solution yh, we take C = 0 ln |y| = −3 ln x ⇒ y = ±x−3 , and we choose, say, yh = x−3 . 47
- 53. Chapter 2 (c) Substituting yp = v(x)yh(x) = v(x)x−3 into (21), we get dv dx yh + v dyh dx + 3 x vyh = dv dx yh + v dyh dx + 3 x yh = dv dx yh = x2 . Therefore, dv/dx = x2 /yh = x5 . (d) Integrating yields v(x) = x5 dx = x6 6 . (We have chosen zero integration constant.) (e) The function y = Cyh + vyh = Cx−3 + x3 6 is a general solution to (21) because dy dx + 3 x y = d dx Cx−3 + x3 6 + 3 x Cx−3 + x3 6 = −3Cx−4 + x2 2 + 3Cx−4 + x2 2 = x2 . 38. Dividing both sides of (6) by µ and multiplying by dx yields dµ µ = Pdx ⇒ dµ µ = Pdx ⇒ ln |µ| = Pdx ⇒ µ = ± exp Pdx . Choosing the positive sign, we obtain (7). EXERCISES 2.4: Exact Equations 2. This equation is not separable because the coeﬃcient x10/3 − 2y cannot be written as a product f(x)g(y). Writing the equation in the form x dy dx − 2y = −x10/3 , we see that the equation is linear. Since M(x, y) = x10/3 − 2y, N(x, y) = x, ∂M ∂y = −2 = ∂N ∂x = 1, and so the equation is not exact. 48
- 54. Exercises 2.4 4. First we note that M(x, y) = −2y − y2 depends only on y and N(x, y) = 3 + 2x − x2 depends only on x. So, the equation is separable. It is not linear with x as independent variable because M(x, y) is not a linear function of y. Similarly, it is not linear with y as independent variable because N(x, y) is not a linear function of x. Computing ∂M ∂y = 1 2 −2y − y2 −1/2 (−2 − 2y) = − 1 + y −2y − y2 , ∂N ∂x = 2 − 2x, we see that the equation (5) in Theorem 2 is not satisﬁed. Therefore, the equation is not exact. 6. It is separable, linear with x as independent variable, and not exact because ∂M ∂y = x = ∂N ∂x = 0. 8. Here, M(x, y) = 2x+y cos(xy), N(x, y) = x cos(xy)−2y. Since M(x, y)/N(x, y) cannot be expressed as a product f(x)g(y), the equation is not separable. We also conclude that it is not linear because M(x, y)/N(x, y) is not a linear function of y and N(x, y)/M(x, y) is not a linear function of x. Taking partial derivatives ∂M ∂y = cos(xy) − xy sin(xy) = ∂N ∂x , we see that the equation is exact. 10. In this problem, M(x, y) = 2x + y, N(x, y) = x − 2y. Thus, My = Nx = 1, and the equation is exact. We ﬁnd F(x, y) = (2x + y)dx = x2 + xy + g(y), ∂F ∂y = x + g (y) = N(x, y) = x − 2y ⇒ g (y) = −2y ⇒ g(y) = (−2y)dy = −y2 ⇒ F(x, y) = x2 + xy − y2 , and so x2 + xy − y2 = C is a general solution. 12. We compute ∂M ∂y = ex cos y = ∂N ∂x . 49
- 55. Chapter 2 Thus, the equation is exact. F(x, y) = ex sin y − 3x2 dx = ex sin y − x3 + g(y), ∂F ∂y = ex cos y + g (y) = ex cos y − 1 3 y−2/3 ⇒ g (y) = − 1 3 y−2/3 ⇒ g(y) = 1 3 y−2/3 dy = y1/3 . So, ex sin y − x3 + 3 √ y = C is a general solution. 14. Since M(t, y) = et (y − t), N(t, y) = 1 + et , we ﬁnd that ∂M ∂y = et = ∂N ∂t . Then F(t, y) = (1 + et )dy = (1 + et )y + h(t), ∂F ∂t = et y + h (t) = et (y − t) ⇒ h (t) = −tet ⇒ h(t) = − tet dt = −(t − 1)et , and a general solution is given by (1 + et )y − (t − 1)et = C ⇒ y = (t − 1)et + C 1 + et . 16. Computing ∂M ∂y = ∂ ∂y yexy − y−1 = exy + xyexy + y−2 , ∂N ∂x = ∂ ∂x xexy + xy−2 = exy + xyexy + y−2 , we see that the equation is exact. Therefore, F(x, y) = yexy − y−1 dx = exy − xy−1 + g(y). So, ∂F ∂y = xexy + xy−2 + g (y) = N(x, y) ⇒ g (y) = 0. Thus, g(y) = 0, and the answer is exy − xy−1 = C. 50
- 56. Exercises 2.4 18. Since ∂M ∂y = ∂N ∂x = 2y2 + sin(x + y), the equation is exact. We ﬁnd F(x, y) = 2x + y2 − cos(x + y) dx = x2 + xy2 − sin(x + y) + g(y), ∂F ∂y = 2xy − cos(x + y) + g (y) = 2xy − cos(x + y) − ey ⇒ g (y) = −ey ⇒ g(y) = −ey . Therefore, F(x, y) = x2 + xy2 − sin(x + y) − ey = C gives a general solution. 20. We ﬁnd ∂M ∂y = ∂ ∂y [y cos(xy)] = cos(xy) − xy sin(xy), ∂N ∂x = ∂ ∂x [x cos(xy)] = cos(xy) − xy sin(xy). Therefore, the equation is exact and F(x, y) = x cos(xy) − y−1/3 dy = sin(xy) − 3 2 y2/3 + h(x) ∂F ∂x = y cos(xy) + h (x) = 2 √ 1 − x2 + y cos(xy) ⇒ h (x) = 2 √ 1 − x2 ⇒ h(x) = 2 arcsin x, and a general solution is given by sin(xy) − 3 2 y2/3 + 2 arcsin x = C. 22. In Problem 16, we found that a general solution to this equation is exy − xy−1 = C. Substituting the initial condition, y(1) = 1, yields e − 1 = C. So, the answer is exy − xy−1 = e − 1. 51
- 57. Chapter 2 24. First, we check the given equation for exactness. dM dx = et = ∂N ∂t . So, it is exact. We ﬁnd F(t, x) = et − 1 dx = x et − 1 + g(t), ∂F ∂t = xet + g (t) = xet + 1 ⇒ g(t) = dt = t ⇒ x et − 1 + t = C is a general solution. With x(1) = 1, we get (1) (e − 1) + 1 = C ⇒ C = e, and the solution is given by x = e − t et − 1 . 26. Taking partial derivatives My and Nx, we ﬁnd that the equation is exact. So, F(x, y) = (tan y − 2) dx = x(tan y − 2) + g(y), ∂F ∂y = x sec2 y + g (y) = x sec2 y + y−1 ⇒ g (y) = y−1 ⇒ g(y) = ln |y|, and x(tan y − 2) + ln |y| = C is a general solution. Substituting y(0) = 1 yields C = 0. Therefore, the answer is x(tan y − 2) + ln y = 0. (We removed the absolute value sign in the logarithmic function because y(0) > 0.) 28. (a) Computing ∂M ∂y = cos(xy) − xy sin(xy), which must be equal to ∂N/∂x, we ﬁnd that N(x, y) = [cos(xy) − xy sin(xy)] dx = [x cos(xy)]x dx = x cos(xy) + g(y). 52
- 58. Exercises 2.4 (b) Since ∂M ∂y = (1 + xy)exy − 4x3 = ∂N ∂x , we conclude that N(x, y) = (1 + xy)exy − 4x3 dx = xexy − x4 + g(y). 30. (a) Diﬀerentiating, we ﬁnd that ∂M ∂y = 5x2 + 12x3 y + 8xy , ∂N ∂x = 6x2 + 12x3 y + 6xy . Since My = Nx, the equation is not exact. (b) Multiplying given equation by xn ym and taking partial derivatives of new coeﬃ- cients yields d dy 5xn+2 ym+1 + 6xn+3 ym+2 + 4xn+1 ym+2 = 5(m + 1)xn+2 ym + 6(m + 2)xn+3 ym+1 + 4(m + 2)xn+1 ym+1 d dx 2xn+3 ym + 3xn+4 ym+1 + 3xn+2 ym+1 = 2(n + 3)xn+2 ym + 3(n + 4)xn+3 ym+1 + 3(n + 2)xn+1 ym+1 . In order that these polynomials are equal, we must have equal coeﬃcients at similar monomials. Thus, n and m must satisfy the system 5(m + 1) = 2(n + 3) 6(m + 2) = 3(n + 4) 4(m + 2) = 3(n + 2). Solving, we obtain n = 2 and m = 1. Therefore, multiplying the given equation by x2 y yields an exact equation. (c) We ﬁnd F(x, y) = 5x4 y2 + 6x5 y3 + 4x3 y3 dx = x5 y2 + x6 y3 + x4 y3 + g(y). Therefore, ∂F ∂y = 2x5 y + 3x6 y2 + 3x4 y2 + g (y) 53
- 59. Chapter 2 = 2x5 y + 3x6 y2 + 3x4 y2 ⇒ g(y) = 0, and a general solution to the given equation is x5 y2 + x6 y3 + x4 y3 = C. 32. (a) The slope of the orthogonal curves, say m⊥, must be −1/m, where m is the slope of the original curves. Therefore, we have m⊥ = Fy(x, y) Fx(x, y) ⇒ dy dx = Fy(x, y) Fx(x, y) ⇒ Fy(x, y) dx − Fx(x, y) dy = 0. (b) Let F(x, y) = x2 + y2 . Then we have Fx(x, y) = 2x and Fy(x, y) = 2y. Plugging these expressions into the ﬁnal result of part (a) gives 2y dx − 2x dy = 0 ⇒ y dx − x dy = 0. To ﬁnd the orthogonal trajectories, we must solve this diﬀerential equation. To this end, note that this equation is separable and thus 1 x dx = 1 y dy ⇒ ln |x| = ln |y| + C ⇒ eln |x|−C = eln |y| ⇒ y = kx, where k = ±e−C . Therefore, the orthogonal trajectories are lines through the origin. (c) Let F(x, y) = xy. Then we have Fx(x, y) = y and Fy(x, y) = x. Plugging these expressions into the ﬁnal result of part (a) gives x dx − y dy = 0. To ﬁnd the orthogonal trajectories, we must solve this diﬀerential equation. To this end, note that this equation is separable and thus x dx = y dy ⇒ x2 2 = y2 2 + C ⇒ x2 − y2 = k , where k := 2C. Therefore, the orthogonal trajectories are hyperbolas. 34. To use the method described in Problem 32, we rewrite the equation x2 + y2 = kx in the form x + x−1 y2 = k. Thus, F(x, y) = x + x−1 y2 , ∂F ∂x = 1 − x−2 y2 , ∂F ∂y = 2x−1 y. 54
- 60. Exercises 2.4 Substituting these derivatives in the equation given in Problem 32(b), we get the re- quired. Multiplying the equation by xn ym , we obtain 2xn−1 ym+1 dx + xn−2 ym+2 − xn ym dy = 0. Therefore, ∂M ∂y = 2(m + 1)xn−1 ym , ∂N ∂x = (n − 2)xn−3 ym+2 − nxn−1 ym . Thus, to have an exact equation, n and m must satisfy n − 2 = 0 2(m + 1) = −n . Solving, we obtain n = 2, m = −2. With this choice, the equation becomes 2xy−1 dx + 1 − x2 y−2 dy = 0, and so G(x, y) = M(x, y)dx = 2xy−1 dx = x2 y−1 + g(y), ∂G ∂y = −x2 y−2 + g (y) = N(x, y) = 1 − x2 y−2 . Therefore, g(y) = y, and the family of orthogonal trajectories is given by x2 y−1 +y = C. Writing this equation in the form x2 + y2 − Cy = 0, we see that, given C, the trajectory is the circle centered at (0, C/2) and of radius C/2. Several given curves and their orthogonal trajectories are shown in Fig. 2–D, page 72. 36. The ﬁrst equation in (4) follows from (9) and the Fundamental Theorem of Calculus. ∂F ∂x = ∂ ∂x x x0 M(t, y)dt + g(y) = M(t, y)|t=x = M(x, y). For the second equation in (4), ∂F ∂y = ∂ ∂y x x0 M(t, y)dt + g(y) 55
- 61. Chapter 2 = ∂ ∂y x x0 M(t, y)dt + g (y) = ∂ ∂y x x0 M(t, y)dt + N(x, y) − ∂ ∂y x x0 M(t, y)dt = N(x, y). EXERCISES 2.5: Special Integrating Factors 2. This equation is neither separable, nor linear. Since ∂M ∂y = x−1 = ∂N ∂x = y, it is not exact either. But My − Nx N = x−1 − y xy − 1 = 1 − xy x(xy − 1) = − 1 x is a function of just x. So, there exists an integrating factor µ(x), which makes the equation exact. 4. This equation is also not separable and not linear. Computing ∂M ∂y = 1 = ∂N ∂x , we see that it is exact. 6. It is not separable, but linear with x as independent variable. Since ∂M ∂y = 4 = ∂N ∂x = 1, this equation is not exact, but it has an integrating factor µ(x), because My − Nx N = 3 x depends on x only. 8. We ﬁnd that ∂M ∂y = 2x, ∂N ∂x = −6x ⇒ Nx − My M = −8x 2xy = − 4 y depends just on y. So, an integrating factor is µ(y) = exp − 4 y dy = exp (−4 ln y) = y−4 . 56
- 62. Exercises 2.5 So, multiplying the given equation by y−4 , we get an exact equation 2xy−3 dx + y−2 − 3x2 y−4 dy = 0. Thus, F(x, y) = 2xy−3 dx = x2 y−3 + g(y), ∂F ∂y = −3x2 y−4 + g (y) = y−2 − 3x2 y−4 ⇒ g (y) = y−2 ⇒ g(y) = −y−1 . This yields a solution F(x, y) = x2 y−3 − y−1 = C, which together with the lost solution y ≡ 0, gives a general solution to the given equation. 10. Since ∂M ∂y = 1, ∂N ∂x = −1, and My − Nx N = 2 −x , the equation has an integrating factor µ(x) = exp − 2 x dx = exp (−2 ln x) = x−2 . Therefore, the equation x−2 x4 − x + y dx − xdy = x2 − x−1 + x−2 y dx − x−1 dy = 0 is exact. Therefore, F(x, y) = −x−1 dy = −x−1 y + h(x), ∂F ∂x = x−2 y + h (x) = x2 − x−1 + x−2 y ⇒ h (x) = x2 − x−1 ⇒ h(x) = x3 3 − ln |x| ⇒ − y x + x3 3 − ln |x| = C ⇒ y = x4 3 − x ln |x| − Cx . Together with the lost solution, x ≡ 0, this gives a general solution to the problem. 12. Here, M(x, y) = 2xy3 + 1, N(x, y) = 3x2 y2 − y−1 . Since ∂M ∂y = 6xy2 = ∂N ∂x , 57
- 63. Chapter 2 the equation is exact. So, we ﬁnd that F(x, y) = 2xy3 + 1 dx = x2 y3 + x + g(y), ∂F ∂y = 3x2 y2 + g (y) = 3x2 y2 − y−1 ⇒ g (y) = −y−1 ⇒ g(y) = − ln |y|, and the given equation has a general solution x2 y3 + x − ln |y| = C. 14. Multiplying the given equation by xn ym yields 12xn ym + 5xn+1 ym+1 dx + 6xn+1 ym−1 + 3xn+2 ym dy = 0. Therefore, ∂M ∂y = 12mxn ym−1 + 5(m + 1)xn+1 ym , ∂N ∂x = 6(n + 1)xn ym−1 + 3(n + 2)xn+1 ym . Matching the coeﬃcients, we get a system 12m = 6(n + 1) 5(m + 1) = 3(n + 2) to determine n and m. This system has the solution n = 3, m = 2. Thus, the given equation multiplied by x3 y2 , that is, 12x3 y2 + 5x4 y3 dx + 6x4 y + 3x5 y2 dy = 0, is exact. We compute F(x, y) = 12x3 y2 + 5x4 y3 dx = 3x4 y2 + x5 y3 + g(y), ∂F ∂y = 6x4 y + 3x5 y2 + g (y) = 6x4 y + 3x5 y2 ⇒ g (y) = 0 ⇒ g(y) = 0, and so 3x4 y2 + x5 y3 = C is a general solution to the given equation. 58
- 64. Exercises 2.5 16. (a) An equation Mdx + Ndy = 0 has an integrating factor µ(x + y) if and only if the equation µ(x + y)M(x, y)dx + µ(x + y)N(x, y)dy = 0 is exact. According to Theorem 2, Section 2.4, this means that ∂ ∂y [µ(x + y)M(x, y)] = ∂ ∂x [µ(x + y)N(x, y)] . Applying the product and chain rules yields µ (x + y)M(x, y) + µ(x + y) ∂M(x, y) ∂y = µ (x + y)N(x, y) + µ(x + y) ∂N(x, y) ∂x . Collecting similar terms yields µ (x + y) [M(x, y) − N(x, y)] = µ(x + y) ∂N(x, y) ∂x − ∂M(x, y) ∂y ⇔ ∂N/∂x − ∂M/∂y M − N = µ (x + y) µ(x + y) . (2.1) The right-hand side of (2.1) depends on x + y only so the left-hand side does. To ﬁnd an integrating factor, we let s = x + y and denote G(s) = ∂N/∂x − ∂M/∂y M − N . Then (2.1) implies that µ (s) µ(s) = G(s) ⇒ ln |µ(s)| = G(s) ds ⇒ |µ(s)| = exp G(s) ds ⇒ µ(s) = ± exp G(s) ds . (2.2) In this formula, we can choose either sign and any integration constant. (b) We compute ∂N/∂x − ∂M/∂y M − N = (1 + y) − (1 + x) (3 + y + xy) − (3 + x + xy) = 1 . Applying formula (2.2), we obtain µ(s) = exp (1)ds = es ⇒ µ(x + y) = ex+y , Multiplying the given equation by µ(x + y), we get an exact equation ex+y (3 + y + xy)dx + ex+y (3 + x + xy)dy = 0 59
- 65. Chapter 2 and follow the procedure of solving exact equations, Section 2.4. F(x, y) = ex+y (3 + y + xy) dx = ey (3 + y) ex dx + y xex dx = ey [(3 + y)ex + y(x − 1)ex ] + h(y) = ex+y (3 + xy) + h(y) . Taking the partial derivative of F with respect to y, we ﬁnd h(y). ∂F ∂y = ex+y (3 + xy + x) + h (y) = ex+y N(x, y) = ex+y (3 + x + xy) ⇒ h (y) = 0 ⇒ h(y) = 0 . Thus, a general solution is ex+y (3 + xy) = C . 18. The given condition, xM(x, y) + yN(x, y) ≡ 0, is equivalent to yN(x, y) ≡ −xM(x, y). In particular, substituting x = 0, we obtain yN(0, y) ≡ −(0)M(0, y) ≡ 0. This implies that x ≡ 0 is a solution to the given equation. To obtain other solutions, we multiply the equation by x−1 y. This gives x−1 yM(x, y)dx + x−1 yN(x, y)dy = x−1 yM(x, y)dx − x−1 xM(x, y)dy = xM(x, y) x−2 ydx − x−1 dy = −xM(x, y)d x−1 y = 0 . Therefore, x−1 y = C or y = Cx. Thus, a general solution is y = Cx and x ≡ 0. 20. For the equation e R P(x)dx [P(x)y − Q(x)] dx + e R P(x)dx dy = 0, we compute ∂M ∂y = ∂ ∂y e R P(x)dx [P(x)y − Q(x)] = e R P(x)dx P(x), ∂N ∂x = ∂ ∂x e R P(x)dx = e R P(x)dx d dx P(x)dx = e R P(x)dx P(x). Therefore, ∂M/∂y = ∂N/∂x, and the equation is exact. 60
- 66. Exercises 2.6 EXERCISES 2.6: Substitutions and Transformations 2. We can write the equation in the form dx dt = x2 − t2 2tx = 1 2 x t − t x , which shows that it is homogeneous. At the same time, it is a Bernoulli equation because it can be written as dx dt − 1 2t x = − t 2 x−1 , 4. This is a Bernoulli equation. 6. Dividing this equation by θdθ, we obtain dy dθ − 1 θ y = 1 √ θ y1/2 . Therefore, it is a Bernoulli equation. It can also be written in the form dy dθ = y θ + y θ , and so it is homogeneous too. 8. We can rewrite the equation in the form dy dx = sin(x + y) cos(x + y) = tan(x + y). Thus, it is of the form dy/dx = G(ax + by) with G(t) = tan t. 10. Writing the equation in the form dy dx = xy + y2 x2 = y x + y x 2 and making the substitution v = y/x, we obtain v + x dv dx = v + v2 ⇒ dv v2 = dx x ⇒ dv v2 = dx x ⇒ − 1 v = ln |x| + C ⇒ − x y = ln |x| + C ⇒ y = − x ln |x| + C . In addition, separating variables, we lost a solution v ≡ 0, corresponding to y ≡ 0. 61
- 67. Chapter 2 12. From dy dx = − x2 + y2 2xy = − 1 2 x y + y x , making the substitution v = y/x, we obtain v + x dv dx = − 1 2 1 v + v = − 1 + v2 2v ⇒ x dv dx = − 1 + v2 2v − v = − 1 + 3v2 2v ⇒ 2v dv 1 + 3v2 = − dx x ⇒ 2v dv 1 + 3v2 = − dx x ⇒ 1 3 ln 1 + 3v2 = − ln |x| + C2 ⇒ 1 + 3v2 = C1|x|−3 , where C1 = e3C2 is any positive constant. Making the back substitution, we ﬁnally get 1 + 3 y x 2 = C1 |x|3 ⇒ 3 y x 2 = C1 |x|3 − 1 = C1 − |x|3 |x|3 ⇒ 3|x|y2 = C1 − |x|3 ⇒ 3|x|y2 + |x|3 = C1 ⇒ 3xy2 + x3 = C , where C = ±C1 is any nonzero constant. 14. Substituting v = y/θ yields v + θ dv dθ = sec v + v ⇒ θ dv dθ = sec v ⇒ cos v dv = dθ θ ⇒ cos v dv = dθ θ ⇒ sin v = ln |θ| + C ⇒ y = θ arcsin (ln |θ| + C) . 16. We rewrite the equation in the form dy dx = y x ln y x + 1 and substitute v = y/x to get v + x dv dx = v (ln v + 1) ⇒ x dv dx = v ln v ⇒ dv v ln v = dx x ⇒ ln | ln v| = ln |x| + C1 ⇒ ln v = ±eC1 x = Cx ⇒ v = eCx , where C = 0 is any constant. Note that, separating variables, we lost a solution, v ≡ 1, which can be included in the above formula by letting C = 0. Thus we have v = eCx . where C is any constant. Substituting back y = xv yields a general solution y = xeCx to the given equation. 62
- 68. Exercises 2.6 18. With z = x + y + 2 and z = 1 + y , we have dz dx = z2 + 1 ⇒ dz z2 + 1 = dx ⇒ dz z2 + 1 = dx ⇒ arctan z = x + C ⇒ x + y + 2 = z = tan(x + C) ⇒ y = tan(x + C) − x − 2. 20. Substitution z = x − y yields 1 − dz dx = sin z ⇒ dz dx = 1 − sin z ⇒ dz 1 − sin z = dx ⇒ dz 1 − sin z = dx = x + C. The left-hand side integral can be found as follows. dz 1 − sin z = (1 + sin z)dz 1 − sin2 z = (1 + sin z)dz cos2 z = sec2 z + tan z sec z dz = tan z + sec z. Thus, a general solution is given implicitly by tan(x − y) + sec(x − y) = x + C. 22. Dividing the equation by y3 yields y−3 dy dx − y−2 = e2x . We now make a substitution v = y−2 so that v = −2y−3 y , and get dv dx + 2v = −2e2x . This is a linear equation. So, µ(x) = exp 2dx = e2x , v(x) = e−2x −2e2x e2x dx = −(1/2)e−2x e4x + C = − e2x + Ce−2x 2 . Therefore, 1 y2 = − e2x + Ce−2x 2 ⇒ y = ± − 2 e2x + Ce−2x . Dividing the equation by y3 , we lost a constant solution y ≡ 0. 63
- 69. Chapter 2 24. We divide this Bernoulli equation by y1/2 and make a substitution v = y1/2 . y−1/2 dy dx + 1 x − 2 y1/2 = 5(x − 2) ⇒ 2 dv dx + 1 x − 2 v = 5(x − 2) ⇒ dv dx + 1 2(x − 2) v = 5(x − 2) 2 . An integrating factor for this linear equation is µ(x) = exp dx 2(x − 2) = |x − 2|. Therefore, v(x) = 1 |x − 2| 5(x − 2) |x − 2| 2 dx = 1 |x − 2| |x − 2|5/2 + C = (x − 2)2 + C|x − 2|−1/2 . Since y = v2 , we ﬁnally get y = (x − 2)2 + C|x − 2|−1/2 2 . In addition, y ≡ 0 is a (lost) solution. 26. Multiplying the equation by y2 , we get y2 dy dx + y3 = ex . With v = y3 , v = 3y2 y , the equation becomes 1 3 dv dx + v = ex ⇒ dv dx + 3v = 3ex ⇒ d dx e3x v = 3e4x ⇒ v = e−3x 3e4x dx = 3ex 4 + Ce−3x . Therefore, y = 3 √ v = 3 3ex 4 + Ce−3x . 28. First, we note that y ≡ 0 is a solution, which will be lost when we divide the equation by y3 and make a substitution v = y−2 to get a linear equation y−3 dy dx + y−2 + x = 0 ⇒ dv dx − 2v = 2x. 64
- 70. Exercises 2.6 An integrating factor for this equation is µ(x) = exp (−2)dx = e−2x . Thus, v = e2x 2xe−2x dx = e2x −xe−2x + e−2x dx = e2x −xe−2x − e−2x 2 + C = −x − 1 2 + Ce2x . So, y = ±v−1/2 = ± 1 −x − 1 2 + Ce2x . 30. We make a substitution x = u + h, y = v + k, where h and k satisfy the system (14) in the text, i.e., h + k − 1 = 0 k − h − 5 = 0. Solving yields h = −2, k = 3. Thus, x = u − 2 and y = v + 3. Since dx = du, dy = dv, this substitution leads to the equation (u + v)du + (v − u)dv = 0 ⇒ dv du = u + v u − v = 1 + (v/u) 1 − (v/u) . This is a homogeneous equation, and a substitution z = v/u (v = z + uz ) yields z + u dz du = 1 + z 1 − z ⇒ u dz du = 1 + z 1 − z − z = 1 + z2 1 − z ⇒ (1 − z)dz 1 + z2 = du u ⇒ arctan z − 1 2 ln 1 + z2 = ln |u| + C1 ⇒ 2 arctan v u − ln u2 1 + z2 = 2C1 ⇒ 2 arctan v u − ln u2 + v2 = C. The back substitution yields 2 arctan y − 3 x + 2 − ln (x + 2)2 + (y − 3)2 = C. 65
- 71. Chapter 2 32. To obtain a homogeneous equation, we make a substitution x = u + h, y = v + k with h and k satisfying 2h + k + 4 = 0 h − 2k − 2 = 0 ⇒ h = − 6 5 , k = − 8 5 . This substitution yields (2u + v)du + (u − 2v)dv = 0 ⇒ dv du = v + 2u 2v − u = (v/u) + 2 2(v/u) − 1 . We now let z = v/u (so, v = z + uz ) and conclude that z + u dz du = z + 2 2z − 1 ⇒ u dz du = z + 2 2z − 1 − z = −2z2 + 2z + 2 2z − 1 ⇒ (2z − 1)dz z2 − z − 1 = −2 du u ⇒ ln z2 − z − 1 = −2 ln |u| + C1 ⇒ ln u2 z2 − u2 z − u2 = C1 ⇒ ln v2 − uv − u2 = C1 ⇒ ln y + 8 5 2 − y + 8 5 x + 6 5 − x + 6 5 2 = C1 ⇒ (5y + 8)2 − (5y + 8) (5x + 6) − (5x + 6)2 = C, where C = ±25eC1 = 0 is any constant. Separating variables, we lost two constant solutions z(u), which are the zeros of the polynomial z2 − z − 1. They can be included in the above formula by taking C = 0. Therefore, a general solution is given by (5y + 8)2 − (5y + 8) (5x + 6) − (5x + 6)2 = C, where C is an arbitrary constant. 34. In Problem 2, we found that the given equation can be written as a Bernoulli equation, dx dt − 1 2t x = − t 2 x−1 . Thus, 2x dx dt − 1 t x2 = −t ⇒ v = x2 dv dt − 1 t v = −t. The latter is a linear equation, which has an integrating factor µ(t) = exp − dt t = 1 t . 66
- 72. Exercises 2.6 Thus, v = t (−1)dt = t(−t + C) = −t2 + Ct ⇒ x2 + t2 − Ct = 0, where C is an arbitrary constant. We also note that a constant solution, t ≡ 0, was lost in writing the given equation as a Bernoulli equation. 36. Dividing the equation by y2 yields y−2 dy dx + 1 x y−1 = x3 ⇒ v = y−1 , v = −y−2 y ⇒ − dv dx + 1 x v = x3 ⇒ dv dx − 1 x v = −x3 ⇒ µ(x) = exp − dx x = 1 x ⇒ v = −x x2 dx = −x x3 3 + C1 = − x4 + Cx 3 , where C = 3C1 is an arbitrary constant. Thus, y = v−1 = − 3 x4 + Cx . Together with the constant (lost) solution y ≡ 0, this gives a general solution to the original equation. 38. Since this equation is a Bernoulli equation (see Problem 6), we make a substitution v = y1/2 so that 2v = y−1/2 y and obtain a linear equation 2 dv dθ − 1 θ v = θ−1/2 ⇒ dv dθ − 1 2θ v = 1 2 θ−1/2 . An integrating factor for this equation is µ(θ) = exp − dθ 2θ = θ−1/2 . So, v = θ1/2 1 2 θ−1/2 θ−1/2 dθ = θ1/2 2 (ln |θ| + C). Therefore, y = v2 = θ 4 (ln |θ| + C)2 . Dividing the given equation by θ dθ, we lost a constant solution θ ≡ 0. 67
- 73. Chapter 2 40. Using the conclusion made in Problem 8, we make a substitution v = x + y, v = 1 + y , and obtain a separable equation dv dx = tan v + 1 ⇒ dv tan v + 1 = dx. The integral of the left-hand side can be found, for instance, as follows. dv tan v + 1 = cos v dv sin v + cos v = 1 2 cos v − sin v sin v + cos v + 1 dv = 1 2 d(sin v + cos v) sin v + cos v + dv = 1 2 (ln | sin v + cos v| + v) . Therefore, 1 2 (ln | sin v + cos v| + v) = x + C1 ⇒ ln | sin(x + y) + cos(x + y)| + x + y = 2x + C2 ⇒ ln | sin(x + y) + cos(x + y)| = x − y + C2 ⇒ sin(x + y) + cos(x + y) = ±eC2 ex−y = Cex−y , where C = 0 is an arbitrary constant. Note that in procedure of separating variables we lost solutions corresponding to tan v + 1 = 0 ⇒ x + y = v = − π 4 + kπ, k = 0, ±1, ±2, . . . , which can be included in the above formula by letting C = 0. 42. Suggested substitution, y = vx2 (so that y = 2xv + x2 v ), yields 2xv + x2 dv dx = 2vx + cos v ⇒ x2 dv dx = cos v. Solving this separable equation, we obtain dv cos v = dx x2 ⇒ ln | sec v + tan v| = −x−1 + C1 ⇒ sec v + tan v = ±eC1 e−1/x = Ce−1/x ⇒ sec y x2 + tan y x2 = Ce−1/x , where C = ±eC1 is an arbitrary nonzero constant. With C = 0, this formula also includes lost solutions y = π 2 + (2k + 1)π x2 , k = 0, ±1, ±2, . . . . 68
- 74. Exercises 2.6 So, together with the other set of lost solutions, y = π 2 + 2kπ x2 , k = 0, ±1, ±2, . . . , we get a general solution to the given equation. 44. From dy dx = − a1x + b1y + c1 a2x + b2y + c2 , using that a2 = ka1 and b2 = kb1, we obtain dy dx = − a1x + b1y + c1 ka1x + kb1y + c2 = − a1x + b1y + c1 k(a1x + b1y) + c2 = G (a1x + b1y) , where G(t) = − t + c1 kt + c2 . 46. (a) Substituting y = u + 1/v into the Riccati equation (18) and using the fact that u(x) is a solution, we obtain d dx u + v−1 = P(x) u + v−1 2 + Q(x) u + v−1 + R(x) ⇔ du dx − v−2 dv dx = P(x)u2 + 2P(x)u(x)v−1 + P(x)v−2 +Q(x)u + Q(x)v−1 + R(x) ⇔ du dx − v−2 dv dx = P(x)u2 + Q(x)u + R(x) + [2P(x)u(x) + Q(x)] v−1 + P(x)v−2 ⇔ −v−2 dv dx = [2P(x)u(x) + Q(x)] v−1 + P(x)v−2 ⇔ dv dx + [2P(x)u(x) + Q(x)] v = −P(x), which is indeed a linear equation with respect to v. (b) Writing dy dx = x3 y2 + −2x4 + 1 x y + x5 and using notations in (18), we see that P(x) = x3 , Q(x) = (−2x4 + 1/x), and R(x) = x5 . So, using part (a), we are looking for other solutions to the given equation of the form y = x + 1/v, where v(x) satisﬁes dv dx + 2 x3 x + −2x4 + 1 x v = dv dx + − 1 x v = −x3 . 69
- 75. Chapter 2 Since an integrating factor for this linear equation is µ(x) = exp dx x = x, we obtain v = x−1 −x4 dx = −x5 + C 5x ⇒ 1 v = 5x C − x5 , and so a general solution is given by y = x + 5x C − x5 . REVIEW PROBLEMS 2. y = −8x2 − 4x − 1 + Ce4x 4. x3 6 − 4x2 5 + 3x 4 − Cx−3 6. y−2 = 2 ln |1 − x2 | + C and y ≡ 0 8. y = (Cx2 − 2x3 ) −1 and y ≡ 0 10. x + y + 2y1/2 + arctan(x + y) = C 12. 2ye2x + y3 ex = C 14. x = t2 (t − 1) 2 + t(t − 1) + 3(t − 1) ln |t − 1| + C(t − 1) 16. y = cos x ln | cos x| + C cos x 18. y = 1 − 2x + √ 2 tan √ 2 x + C 20. y = Cθ−3 − 12θ2 5 1/3 22. (3y − 2x + 9)(y + x − 2)4 = C 24. 2 √ xy + sin x − cos y = C 26. y = Ce−x2/2 28. (y + 3)2 + 2(y + 3)(x + 2) − (x + 2)2 = C 70
- 76. Figures 30. y = Ce4x − x − 1 4 32. y2 = x2 ln(x2 ) + 16x2 34. y = x2 sin x + 2x2 π2 36. sin(2x + y) − x3 3 + ey = sin 2 + 2 3 38. y = 2 − 1 4 arctan x 2 2 40. y = 8 1 − 3e−4x − 4x TABLES nnn xxxnnn yyynnn nnn xxxnnn yyynnn 1 0.1 1.475 6 0.6 1.353368921 2 0.2 1.4500625 7 0.7 1.330518988 3 0.3 1.425311875 8 0.8 1.308391369 4 0.4 1.400869707 9 0.9 1.287062756 5 0.5 1.376852388 10 1.0 1.266596983 Table 2–A: Euler’s approximations to y = x − y, y(0) = 0, on [0, 1] with h = 0.1. FIGURES K1 0 1 2 3 y(t) 2 4 6 8 Figure 2–A: The graph of the solution in Problem 28. 71
- 77. Chapter 2 K3 K2 1 2 3 K3 K2 K1 1 2 3 Figure 2–B: The direction ﬁeld and solution curve in Problem 32. 2 4 6 K1.0 K0.5 0 0.5 1.0 Figure 2–C: The graph of the solution in Problem 32. K2 0 2 4 K2 2 4 Figure 2–D: Curves and their orthogonal trajectories in Problem 34. 72
- 78. CHAPTER 3: Mathematical Models and Numerical Methods Involving First Order Equations EXERCISES 3.2: Compartmental Analysis 2. Let x(t) denote the mass of salt in the tank at time t with t = 0 denoting the moment when the process started. Thus we have x(0) = 0.5 kg. We use the mathematical model described by equation (1) of the text to ﬁnd x(t). Since the solution is entering the tank with rate 6 L/min and contains 0.05 kg/L of salt, input rate = 6 (L/min) · 0.05 (kg/L) = 0.3 (kg/min). We can determine the concentration of salt in the tank by dividing x(t) by the volume of the solution, which remains constant, 50 L, because the ﬂow rate in is the same as the ﬂow rate out. Therefore, the concentration of salt at time t is x(t)/50 kg/L and output rate = x(t) 50 (kg/L) · 6 (L/min) = 3x(t) 25 (kg/min). Then the equation (1) yields dx dt = 0.3 − 3x 25 ⇒ dx dt + 3x 25 = 0.3 , x(0) = 0.5 . This equation is linear, has integrating factor µ(t) = exp (3/25)dt = e3t/25 , and so d e3t/25 x dt = 0.3e3t/25 ⇒ e3t/25 x = 0.3 25 3 e3t/25 + C = 2.5e3t/25 + C ⇒ x = 2.5 + Ce−3t/25 . Using the initial condition, we ﬁnd C. 0.5 = x(0) = 2.5 + C ⇒ C = −2 , and so the mass of salt in the tank after t minutes is x(t) = 2.5 − 2e−3t/25 . 73