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A First Course in Differential Equations with Modeling Applications
A First Course in Differential Equations with Modeling Applications
Dennis G. Zill
11th Edition
Year:
2017
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english
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462
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9781305965720
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DIFFERENTIAL EQUATIONS with Modeling Applications A First Course in Dennis G. Zill 11E A First C o urse in D IFFE R E N TIA L E Q U A TIO N S w ith M o d eling A p p licatio ns ZILL 11E To register or access your online learning solution or purchase materials for your course, visit www.cengagebrain.com. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Dennis G. Zill Loyola Marymount University 11E A First Course in Differential Equations with Modeling Applications Australia ● Brazil ● Mexico ● Singapore ● United Kingdom ● United States Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 © 2018, 2013 Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced or distributed in any form or by any means, except as permitted by U.S. copyright law, without the prior written permission of the copyright owner. Library of Congress Control Number: 2016936077 Student Edition: ISBN: 9781305965720 Looseleaf Edition: ISBN: 9781337293129 Cengage Learning 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with employees residing in nearly 40 different countries and sales in more than 125 countries around the world. Find your local representative at www.cengage.com. Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Cengage Learning Solutions, visit www.cengage.com. Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com. A First Course in Differential Equations with Modeling Applications, Eleventh Edition Dennis G. Zill Product Director: Terry Boyle Product Manager: Kate Schrumpf Content Developer: Spencer Arritt Product Assistant: Stephen Previte Marketing Manager: Ana Albinson Content Project Manager: Jennifer Risden Art Director: Vernon Boes Manufacturing Planner: Becky Cross Production Service: MPS Limited Photo Researcher: Lumina Datamatics Text Researcher: Lumina Datamatics Copy Editor: Steven Summerlight Text Designer: Terri Wright Cover Designer: Terri Wright Cover Image: STS114 Crew, ISS Expedition 11 Crew, NASA Compositor: MPS Limited Portions of this text appear in Advanced Engineering Mathematics, Sixth Edition, Copyright 2018, Jones & Bartlett Learning, Burlington, MA 01803 and are used with the permission of the publisher. For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 18003549706. For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be emailed to permissionrequest@cengage.com. Printed in the United States of America Print Number: 01 Print Year: 2016 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 iii Preface vii 1 Introduction to Differential Equations 2 1.1 De�nitions and Terminology 3 1.2 InitialValue Problems 15 1.3 Differential Equations as Mathematical Models 22 CHAPTER IN RE VIE W 2 FirstOrder Differential Equations 36 2.1 Solution Curves Without a Solution 37 2.1.1 Direction Fields 37 2.1.2 Autonomous FirstOrder DEs 39 2.2 Separable Equations 47 2.3 Linear Equations 55 2.4 Exact Equations 64 2.5 Solutions by Substitutions 72 2.6 A Numerical Method 76 CHAPTER IN RE VIE W 3 Modeling with FirstOrder Differential Equations 84 3.1 Linear Models 85 3.2 Nonlinear Models 96 3.3 Modeling with Systems of FirstOrder DEs 107 CHAPTER IN RE VIE W 4 HigherOrder Differential Equations 118 4.1 Preliminary Theory—Linear Equations 119 4.1.1 InitialValue and BoundaryValue Problems 119 4.1.2 Homogeneous Equations 121 4.1.3 Nonhomogeneous Equations 127 4.2 Reduction of Order 132 4.3 Homogeneous Linear Equations with Constant Coef�cients 135 4.4 Undetermined Coef�cients—Superposition Approach 142 4.5 Undetermined Coef�cients—Annihilator Approach 152 4.6 Variation of Parameters 159 4.7 CauchyEuler Equations 166 Contents Ke vi n Ge or ge /S hu tte rs to ck .c om © J og gi e Bo tm a/ Sh ut te rs to ck .c om Fo to s5 93 /S hu tte rs to ck .c om Bi ll I ng al ls /N AS A Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 iv CONTENTS 4.8 Green’s Functions 173 4.8.1 InitialValue Problems 173 4.8.2 BoundaryValue Problems 179 4.9 Solving Systems of Linear DEs by Elimination 183 4.10 Nonlinear Differential Equations 188 CHAPTER IN RE VIE W 5 Modeling with HigherOrder Differential Equations 196 5.1 Linear Models: InitialValue Problems 197 5.1.1 Spring/Mass Systems: Free Undamped Motion 197 5.1.2 Spring/Mass Systems: Free Damped Motion 202 5.1.3 Spring/Mass Systems: Driven Motion 204 5.1.4 Series Circuit Analogue 207 5.2 Linear Models: BoundaryValue Problems 213 5.3 Nonlinear Models 222 CHAPTER IN RE VIE W 6 Series Solutions of Linear Equations 236 6.1 Review of Power Series 237 6.2 Solutions About Ordinary Points 243 6.3 Solutions About Singular Points 252 6.4 Special Functions 262 CHAPTER IN RE VIE W 7 The Laplace Transform 278 7.1 De�nition of the Laplace Transform 279 7.2 Inverse Transforms and Transforms of Derivatives 286 7.2.1 Inverse Transforms 286 7.2.2 Transforms of Derivatives 289 7.3 Operational Properties I 294 7.3.1 Translation on the SAxis 295 7.3.2 Translation on the tAxis 298 7.4 Operational Properties II 306 7.4.1 Derivatives of a Transform 306 7.4.2 Transforms of Integrals 307 7.4.3 Transform of a Periodic Function 313 7.5 The Dirac Delta Function 318 7.6 Systems of Linear Differential Equations 322 CHAPTER IN RE VIE W 8 Systems of Linear FirstOrder Differential Equations 332 8.1 Preliminary Theory—Linear Systems 333 8.2 Homogeneous Linear Systems 340 8.2.1 Distinct Real Eigenvalues 341 8.2.2 Repeated Eigenvalues 344 8.2.3 Complex Eigenvalues 348 Br ia n A Ja ck so n/ Sh ut te rs to ck .c om To dd D al to n/ Sh ut te rs to ck .c om Ra im un da s/ Sh ut te rs to ck .c om Pa ve l L P ho to a nd V id eo /S hu tte rs to ck .c om Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 CONTENTS v 8.3 Nonhomogeneous Linear Systems 355 8.3.1 Undetermined Coef�cients 355 8.3.2 Variation of Parameters 357 8.4 Matrix Exponential 362 CHAPTER IN RE VIE W 9 Numerical Solutions of Ordinary Differential Equations 368 9.1 Euler Methods and Error Analysis 369 9.2 RungeKutta Methods 374 9.3 Multistep Methods 378 9.4 HigherOrder Equations and Systems 381 9.5 SecondOrder BoundaryValue Problems 385 CHAPTER IN RE VIE W Appendices A IntegralDe�ned Functions APP3 B Matrices APP11 C Laplace Transforms APP29 Answers for Selected OddNumbered Problems ANS1 Index I1 Pa ul B . M oo re /S hu tte rs to ck .c om Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 vii TO THE STUDENT Authors of books live with the hope that someone actually reads them. Contrary to what you might believe, almost everything in a typical collegelevel mathematics text is written for you and not the instructor. True, the topics covered in the text are chosen to appeal to instructors because they make the decision on whether to use it in their classes, but everything written in it is aimed directly at you the student. So I want to encourage you—no, actually I want to tell you—to read this textbook! But do not read this text as you would a novel; you should not read it fast and you should not skip anything. Think of it as a workbook. By this I mean that mathematics should workbook. By this I mean that mathematics should work always be read with pencil and paper at the ready because, most likely, you will have to work your way through the examples and the discussion. Before attempting any problems in the section exercise sets, work through all the examples in that section. all the examples in that section. all The examples are constructed to illustrate what I consider the most important aspects of the section, and therefore, re�ect the procedures necessary to work most of the problems. When reading an example, copy it down on a piece of paper and do not look at the solution in the book. Try working it, then compare your results against the solution given, and, if necessary resolve any differences. I have tried to include most of the important steps in each example, but if something is not clear you should always try—and here is where the pencil and paper come in again—to �ll in the details or missing steps. This may not be easy, but it is part of the learning process. The accumulation of facts followed by the slow assimilation of understanding simply cannot be achieved without a struggle. Speci�cally for you, a Student Resource Manual (SRM) is available as an opSRM) is available as an opSRM  tional supplement. In addition to containing solutions of selected problems from the exercises sets, the SRM contains hints for solving problems, extra examples, and a review of those areas of algebra and calculus that I feel are particularly important to the successful study of differential equations. Bear in mind you do not have to purchase the SRM; you can review the appropriate mathematics from your old pre calculus or calculus texts. In conclusion, I wish you good luck and success. I hope you enjoy the text and the course you are about to embark on—as an undergraduate math major it was one of my favorites because I liked mathematics that connected with the physical world. If you have any comments, or if you �nd any errors as you read/work your way through the text, or if you come up with a good idea for improving either it or the SRM, please feel free to contact me through Cengage Learning: spencer.arritt@cengage.com. TO THE INSTRUCTOR In case you are examining this text for the �rst time, A First Course in Differen tial Equations with Modeling Applications, Eleventh Edition, is intended for a one semester or onequarter course in ordinary differential equations. The longer version of the text, Differential Equations with BoundaryValue Problems, Ninth Edition, can be used for either a one or twosemester course that covers ordinary and partial differential equations. This text contains six additional chapters. For a onesemester Preface Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 viii PREFACE course, it is assumed that the students have successfully completed at least two semes ters of calculus. Since you are reading this, undoubtedly you have already examined the table of contents for the topics that are covered. You will not �nd a “suggested syl labus” in this preface; I will not pretend to be so wise as to tell other teachers what to teach. I feel that there is plenty of material here to choose from and to form a course to your liking. The text strikes a reasonable balance between the analytical, qualita tive, and quantitative approaches to the study of differential equations. As far as my “underlying philosophy” goes, it is this: An undergraduate text should be written with the students’ understanding kept �rmly in mind, which means to me that the material should be presented in a straightforward, readable, and helpful manner, while keep ing the level of theory consistent with the notion of a “�rst course.” For those who are familiar with the previous editions, I would like to mention a few improvements made in this edition. Many exercise sets have been updated by the addition of new problems. Some of these problems involve new and, I think, interesting mathematical models. Additional examples, �gures, and remarks have been added to many sections. Throughout the text I have given a greater emphasis to the concepts of piecewiselinear differential equations and solutions that involve nonelementary integrals. Finally, Appendix A, IntegralDe�ned Functions, is new to the text. Student Resources ● Student Resource Manual (SRM), prepared by Warren S. Wright and Roberto Martinez (ISBN 9781305965737, accompanies A First Course in Differential Equations with Modeling Applications, Eleventh Edition, and ISBN 9781305965812 accompanies Differential Equations with Boundary Value Problems, Ninth Edition), provides important review material from algebra and calculus, the solution of every third problem in each exercise set (with the exception of the Discussion Problems and Computer Lab Assignments), relevant command syntax for the computer algebra systems Mathematica and Maple, and lists of important concepts, as well as helpful hints on how to start certain problems. ● MindTap for A First Course in Differential Equations with Modeling Applications, Eleventh Edition, is a digital representation of your course that provides you with the tools you need to better manage your limited time, stay organized, and be successful. You can complete assignments whenever and wherever you are ready to learn with course material specially customized for you by your instructor and streamlined in one proven, easytouse interface. With an array of study tools, you’ll get a true understanding of course concepts, achieve better grades, and set the groundwork for your future courses. Learn more at www.cengage.com/mindtap. Instructor Resources ● Instructor’s Solutions Manual (ISM(ISM( ), prepared by Warren S. Wright and Roberto Martinez, provides complete workedout solutions for all problems in the text. It is available through the Instructor Companion website at www .cengage.com. ● Cengage Learning Testing Powered by Cognero is a �exible, online system that allows you to author, edit, and manage test bank content, create multiple test versions in an instant, and deliver tests from your learning management system (LMS), your classroom, or wherever you want. Cognero is available online at www.cengage.com/login. ● Turn the light on with MindTap for A First Course in Differential Equations with Modeling Applications, Eleventh Edition. Through personalized paths of dynamic assignments and applications, MindTap is a digital learning solution and representation of your course. The Right Content: With MindTap’s carefully curated material, you get the precise content and groundbreaking tools you need for every course you teach. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 PREFACE ix Personalization: Customize every element of your course—from rearranging the Learning Path to inserting videos and activities. Improved Work�ow: Save time when planning lessons with all of the trusted, most current content you need in one place in MindTap. Tracking Students’ Progress in Real Time: Promote positive outcomes by tracking students in real time and tailoring your course as needed based on the analytics. Learn more at www.cengage.com/mindtap. ACKNOWLEDGEMENTS Compiling a mathematics textbook such as this and making sure that its thousands of symbols and hundreds of equations are accurate is an enormous task, but since I am called “the author” that is my job and responsibility. But many people besides myself have expended enormous amounts of time and energy in working towards its eventual publication. So I would like to take this opportunity to express my sincerest appreciation to everyone—most of them unknown to me—at Cengage Learning and at MPS North America who were involved in the publication of this edition. A spe cial word of thanks goes to Spencer Arritt, Kathryn Schrumpf, Jennifer Risden, Ver non Boes, and Jill Traut for their guidance in the labyrinth of the production process. Finally, over the years this text has been improved in a countless number of ways through the suggestions and criticisms of the reviewers. Thus it is �tting to conclude with an acknowledgement of my debt to the following generous people for sharing their expertise and experience. REVIEWERS OF PAST EDITIONS William Atherton, Cleveland State University Philip Bacon, University of Florida Bruce Bayly, University of Arizona William H. Beyer, University of Akron R. G. Bradshaw, Clarkson College Bernard Brooks, Rochester Institute of Technology Allen Brown, Wabash Valley College Dean R. Brown, Youngstown State University David Buchthal, University of Akron Nguyen P. Cac, University of Iowa T. Chow, California State University–Sacramento Dominic P. Clemence, North Carolina Agricultural and Technical State University Pasquale Condo, University of Massachusetts–Lowell –Lowell – Vincent Connolly, Worcester Polytechnic Institute Philip S. Crooke, Vanderbilt University Bruce E. Davis, St. Louis Community College at Florissant Valley Paul W. Davis, Worcester Polytechnic Institute Richard A. DiDio, La Salle University James Draper, University of Florida James M. Edmondson, Santa Barbara City College John H. Ellison, Grove City College Raymond Fabec, Louisiana State University Donna Farrior, University of Tulsa Robert E. Fennell, Clemson University W. E. Fitzgibbon, University of Houston Harvey J. Fletcher, Brigham Young University Paul J. Gormley, Villanova Layachi Hadji, University of Alabama Ruben Hayrapetyan, Kettering University Terry Herdman, Virginia Polytechnic Institute and State University Zdzislaw Jackiewicz, Arizona State University Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 x PREFACE S. K. Jain, Ohio University Anthony J. John, Southeastern Massachusetts University David C. Johnson, University of Kentucky–Lexington–Lexington– Harry L. Johnson, Virginia Polytechnic Institute and State University Kenneth R. Johnson, North Dakota State University Joseph Kazimir, East Los Angeles College J. Keener, University of Arizona Steve B. Khlief, Tennessee Technological University Helmut Knaust, The University of Texas at El Paso C. J. Knickerbocker, Sensis Corporation Carlon A. Krantz, Kean College of New Jersey Thomas G. Kudzma, University of Lowell Alexandra Kurepa, North Carolina A&T State University G. E. Latta, University of Virginia Cecelia Laurie, University of Alabama Mulatu Lemma, Savannah State University James R. McKinney, California Polytechnic State University James L. Meek, University of Arkansas Gary H. Meisters, University of Nebraska–Lincoln –Lincoln – Stephen J. Merrill, Marquette University Vivien Miller, Mississippi State University George Moss, Union University Gerald Mueller, Columbus State Community College Philip S. Mulry, Colgate University Martin Nakashima, California State Polytechnic University–Pomona C. J. Neugebauer, Purdue University Tyre A. Newton, Washington State University Brian M. O’Connor, Tennessee Technological University J. K. Oddson, University of California–Riverside –Riverside – Carol S. O’Dell, Ohio Northern University Bruce O’Neill, Milwaukee School of Engineering A. Peressini, University of Illinois, Urbana–Champaign J. Perryman, University of Texas at Arlington Joseph H. Phillips, Sacramento City College Jacek Polewczak, California State University–Northridge Nancy J. Poxon, California State University–Sacramento Robert Pruitt, San Jose State University K. Rager, Metropolitan State College F. B. Reis, Northeastern University Brian Rodrigues, California State Polytechnic University Tom Roe, South Dakota State University Kimmo I. Rosenthal, Union College Barbara Shabell, California Polytechnic State University Seenith Sivasundaram, Embry–Riddle Aeronautical University Don E. Soash, Hillsborough Community College F. W. Stallard, Georgia Institute of Technology Gregory Stein, The Cooper Union M. B. Tamburro, Georgia Institute of Technology Patrick Ward, Illinois Central College Jianping Zhu, University of Akron Jan Zijlstra, Middle Tennessee State University Jay Zimmerman, Towson University Dennis G. Zill Los Angeles, CA Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 A First Course in Differential Equations with Modeling Applications Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Kevin George/Shutterstock.com 2 1.1 Definitions and Terminology 1.2 InitialValue Problems 1.3 Differential Equations as Mathematical Models C H A P T E R 1 I N R E V I E W T he words differential and differential and differential equations suggest solving some kind of equation that contains derivatives y9, y0, Á . Analogous to a course in algebra, in which a good amount of time is spent solving equations such as x2 1 5x 1 4 5 0 for the unknown number x, in this course one of our tasks will be to solve differential equations such as y0 1 2y9 1 y 5 0 for an unknown function y 5 �(x). As the course unfolds, you will see there is more to the study of differential equations than just mastering methods that mathematicians over past centuries devised to solve them. But �rst things �rst. In order to read, study, and be conversant in a specialized subject you have to learn some of the terminology of that discipline. This is the thrust of the �rst two sections of this chapter. In the last section we brie�y examine the link between differential equations and the real world. 1 Introduction to Differential Equations Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 INTRODUCTION The derivative dydyd ydxdxd of a function y 5 �(x) is itself another function �9(x) found by an appropriate rule. The exponential function y 5 e0.1x 2 is differentiable on the interval (2`, `) and by the Chain Rule its �rst derivative is dydyd ydxdxd 5 0.2x.2x.2 e0.1x2. If we replace e0.1x2 on the righthand side of the last equation by the symbol y, the derivative becomes dydyd dxdxd 5 0.2x.2x.2 yxyx . (1) Now imagine that a friend of yours simply hands you equation (1)—you have no idea how it was constructed—and asks, What is the function represented by the symbol y? You are now face to face with one of the basic problems in this course: How do you solve an equation such as (1) for the function y = �(x)? A DEFINITION The equation that we made up in (1) is called a differential equation. Before proceeding any further, let us consider a more precise de�nition of this concept. 1.1 Definitions and Terminology DEFINITION 1.1.1 Differential Equation An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE). To talk about them, we shall classify differential equations according to type, order, and linearity. CLASSIFICATION BY TYPE If a differential equation contains only ordinary derivatives of one or more unknown functions with respect to a single independent variable, it is said to be an ordinary differential equation (ODE). An equation involving partial derivatives of one or more unknown functions of two or more inde pendent variables is called a partial differential equation (PDE). Our �rst example illustrates several of each type of differential equation. EXAMPLE 1 Types of Differential Equations (a) The equations an ODE can contain more than one unknown function p p dydyd dxdxd 1 5y 5 ex, d2y dxdxd 2 2 dydyd dxdxd 1 6y 5 0, and dxdxd dt 1 dydyd dt 5 2x 1 y (2) are examples of ordinary differential equations. (b) The following equations are partial differential equations:* −2u −x2 1 −2u −y2 5 0, −2u −x2 5 −2u −t2 2 2 −u −t , −u −y 5 2 −v −x . (3) *Except for this introductory section, only ordinary differential equations are considered in A First Course in Differential Equations with Modeling Applications, Eleventh Edition. In that text the word equation and the abbreviation DE refer only to ODEs. Partial differential equations or PDEs are considered in the expanded volume Differential Equations with BoundaryValue Problems, Ninth Edition. . DEFINITIONS AND TERMINOLOGY 3 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Notice in the third equation that there are two unknown functions and two indepen dent variables in the PDE. This means u and v must be functions of two or more independent variables. . NOTATION Throughout this text ordinary derivatives will be written by using either the Leibniz notation dyydx, d2yydx2, d3yydx3, . . . or the prime notation y9, y0, y, . . . . By using the latter notation, the �rst two differential equations in (2) can be written a little more compactly as y9 + 5y = ex and x and x y0 − y9 + 6y = 0. Actually, the prime notation is used to denote only the �rst three derivatives; the fourth derivative is written y(4) instead of y00. In general, the nth derivative of y is written dnyydxn or y(n). Although less convenient to write and to typeset, the Leibniz notation has an advantage over the prime notation in that it clearly displays both the dependent and independent variables. For example, in the equation d 2x2x2 ––– dt2 1 16x 5 0 unknown function or dependent variable independent variable it is immediately seen that the symbol x now represents a dependent variable, whereas the independent variable is t. You should also be aware that in physical sciences and engineering, Newton’s dot notation (derogatorily referred to by some as the “�yspeck” notation) is sometimes used to denote derivatives with respect to time t. Thus the differential equation d2sydt2 = −32 becomes s̈ = −32. Partial derivatives are often denoted by a subscript notation indicating the indepen dent variables. For example, with the subscript notation the second equation in (3) becomes uxx = utt − 2ut. CLASSIFICATION BY ORDER The order of a differential equation (either ODE or PDE) is the order of the highest derivative in the equation. For example, �rst ordersecond order 1 5( )3 2 4y 5 exdy–––dx dx d 2y2y2 –––– dx2 is a secondorder ordinary differential equation. In Example 1, the �rst and third equations in (2) are �rstorder ODEs, whereas in (3) the �rst two equations are secondorder PDEs. A �rstorder ordinary differential equation is sometimes written in the differential form M(x, y) dxdxd 1 N(N(N x, y) dydyd 5 0. EXAMPLE 2 Differential Form of a FirstOrder ODE If we assume that y is the dependent variable in a �rstorder ODE, then recall from calculus that the differential dy is de�ned to be dydyd 5 y9dxdxd . (a) By dividing by the differential dx an alternative form of the equation dx an alternative form of the equation dx (y 2 x) dxdxd 1 4x dydyd 5 0 is given by y 2 x 1 4x dydyd dxdxd 5 0 or equivalently 4x dydyd dxdxd 1 y 5 x. 4 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 (b) By multiplying the differential equation 6xy dydyd dxdxd 1 x2 1 y2 5 0 by dxdxd we see that the equation has the alternative differential form (x2 1 y2) dxdxd 1 6xy dydyd 5 0. . In symbols we can express an nthorder ordinary differential equation in one dependent variable by the general form F(x, y, y9, . . . , y(n)) 5 0, (4) where F is a realvalued function of F is a realvalued function of F n + 2 variables: x, y, y9, . . . , y(n). For both practical and theoretical reasons we shall also make the assumption hereafter that it is possible to solve an ordinary differential equation in the form (4) uniquely for the highest derivative y(n) in terms of the remaining n + 1 variables. The dif ferential equation dny dxdxd n 5 f (x, y, y9, . . . , y(n21)), (5) where f is a realvalued continuous function, is referred to as the f is a realvalued continuous function, is referred to as the f normal form of (4). Thus when it suits our purposes, we shall use the normal forms dy dx 5 f (x, y) and d2y dx2 5 f (x, y, y9) to represent general �rst and secondorder ordinary differential equations. EXAMPLE 3 Normal Form of an ODE (a) By solving for the derivative dydyd ydxdxd the normal form of the �rstorder differential equation 4x dydyd dxdxd 1 y 5 x is dydyd dxdxd 5 x 2 y 4x . (b) By solving for the derivative y0 the normal form of the secondorder differential equation y0 2 y9 1 6 5 0 is y0 5 y9 2 6y. . CLASSIFICATION BY LINEARITY An nthorder ordinary differential equation (4) is said to be linear if F is linear in F is linear in F y, y9, . . . , y(n). This means that an nthorder ODE is linear when (4) is an(x(x( )y(n) + an−1(x(x( )y(n−1) + Á + a1(x(x( )y9 + a0(x(x( )y − g(x(x( ) = 0 or an(x) dny dxdxd n 1 an21(x) dn21y dxdxd n21 1 Á 1 a1(x) dydyd dxdxd 1 a0(x)y 5 g(x). (6) Two important special cases of (6) are linear �rstorder (n 5 1) and linear second order (n = 2) DEs: a1(x) dy dx 1 a0(x)y 5 g(x) and a2(x) d2y dx2 1 a1(x) dy dx 1 a0(x)y 5 g(x). (7) In the additive combination on the lefthand side of equation (6) we see that the char acteristic two properties of a linear ODE are as follows: ● The dependent variable y and all its derivatives y9, y0, . . . , y(n) are of the �rst degree, that is, the power of each term involving y is 1. ● The coef�cients a0, a1, . . . , an of y, y9, . . . , y(n) depend at most on the independent variable x. . DEFINITIONS AND TERMINOLOGY 5 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 A nonlinear ordinary differential equation is simply one that is not linear. Nonlinear functions of the dependent variable or its derivatives, such as sin y or ey9, cannot appear in a linear equation. EXAMPLE 4 Linear and Nonlinear ODEs (a) The equations (y 2 x) dxdxd 1 4x dydyd 5 0, y0 2 2y 1 y 5 0, x3 d3y dxdxd 3 1 x dydyd dxdxd 2 5y 5 ex are, in turn, linear �rst, second, and thirdorder ordinary differential equations. We linear �rst, second, and thirdorder ordinary differential equations. We linear have just demonstrated in part (a) of Example 2 that the �rst equation is linear in the variable y by writing it in the alternative form 4xy9 + y = x. (b) The equations nonlinear term: coef�cient depends on y nonlinear term: nonlinear function of y nonlinear term: power not 1 (1 2 y)y9 1 2y 5 ex, 1 sin y 5 0, and d 2y2y2 –––– dx2 1 y 2 5 0 d 4y4y4 –––– dx 4 are examples of nonlinear �rst, second, and fourthorder ordinary differential equanonlinear �rst, second, and fourthorder ordinary differential equanonlinear  tions, respectively. . SOLUTIONS As was stated on page 2, one of the goals in this course is to solve, or �nd solutions of, differential equations. In the next de�nition we consider the con cept of a solution of an ordinary differential equation. DEFINITION 1.1.2 Solution of an ODE Any function f, de�ned on an interval I and possessing at least I and possessing at least I n derivatives that are continuous on I, which when substituted into an nthorder ordinary differential equation reduces the equation to an identity, is said to be a solution of the equation on the interval. In other words, a solution of an nthorder ordinary differential equation (4) is a function f that possesses at least n derivatives and for which F(x, �(x), �9(x), . . . , �(n)(x)) 5 0 for all x ix ix n I. We say that f satis�es the differential equation on I. For our purposes we shall also assume that a solution f is a realvalued function. In our introductory discussion we saw that y 5 e0.1x 2 is a solution of dyydx = 0.2xy on the interval (−`, `). Occasionally, it will be convenient to denote a solution by the alternative symbol y(x). INTERVAL OF DEFINITION You cannot think solution of an ordinary differential equation without simultaneously thinking interval. The interval I in De�nition 1.1.2 I in De�nition 1.1.2 I is variously called the interval of de�nition, the interval of existence, the interval of validity, or the domain of the solution and can be an open interval (a, b), a closed interval [a, b], an in�nite interval (a, `), and so on. 6 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 1 x y 1 (a) function y 5 1/x, x ? 0 (b) solution y 5 1/x, (0, ∞) 1 x y 1 FIGURE 1.1.1 In Example 6 the function y = 1yx is not the same as the solution x is not the same as the solution x y = 1yx EXAMPLE 5 Verification of a Solution Verify that the indicated function is a solution of the given differential equation on the interval (−`, `). (a) dydyd dxdxd 5 xyxyx 1/2; y 5 116 x 4 (b) y0 2 2y9 1 y 5 0; y 5 xex SOLUTION One way of verifying that the given function is a solution is to see, after substituting, whether each side of the equation is the same for every x in the interval.x in the interval.x (a) From lefefe tftf {hand side: dydyd dxdxd 5 1 16 (4 (4 ? x3) 5 1 4 x3, right{hand side: xy1/2 5 x ? 1 116 x42 1/2 5 x ? 114 x22 5 14 x3, we see that each side of the equation is the same for every real number x. Note that y1/2 5 14 x 2 is, by de�nition, the nonnegative square root of 116 x 4. (b) From the derivatives y9 = xex + ex and x and x y0 = xex + 2ex we have, for every real x we have, for every real x number x, lefefe tftf {hand side: y0 2 2y9 1 y 5 (xex 1 2ex) 2 2(xex 1 ex) 1 xex 5 0, right{hand side: 0. . Note, too, that each differential equation in Example 5 possesses the constant solution y 5 0, −` < x < `. A solution of a differential equation that is identically zero on an interval I is said to be a I is said to be a I trivial solution. SOLUTION CURVE The graph of a solution f of an ODE is called a solution curve. Since f is a differentiable function, it is continuous on its interval I of de�niI of de�niI  tion. Thus there may be a difference between the graph of the function f and the graph of the solution f. Put another way, the domain of the function f need not be the same as the interval I of de�nition (or domain) of the solution I of de�nition (or domain) of the solution I f. Example 6 illustrates the difference. EXAMPLE 6 Function versus Solution (a) The domain of y = 1yx, considered simply as a function, is the set of all real numbers x except 0. When we graph y = 1yx, we plot points in the xyplane corresponding to a judicious sampling of numbers taken from its domain. The rational function y = 1yx is discontinuous at 0, and its graph, in a neighborhood of the origin, is given in Figure 1.1.1(a). The function y = 1yx is not differen tiable at x = 0, since the yaxis (whose equation is x = 0) is a vertical asymptote of the graph. (b) Now y = 1yx is also a solution of the linear �rstorder differential equation x is also a solution of the linear �rstorder differential equation x xy9 + y = 0. (Verify.) But when we say that y = 1yx is a x is a x solution of this DE, we mean that it is a function de�ned on an interval I on which it is differentiable and I on which it is differentiable and I satis�es the equation. In other words, y = 1yx is a solution of the DE on x is a solution of the DE on x any interval that does not contain 0, such as (−3, −1), _ y _ y 1 2, 10+, (−`, 0), or (0, `). Because the solution curves de�ned by y = 1yx for x for x −3 < x < −1 and 12 , x , 10 are simply segments, or pieces, of the solution curves de�ned by y = 1yx for x for x −` < x < 0 and . DEFINITIONS AND TERMINOLOGY 7 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 0 < x < `, respectively, it makes sense to take the interval I to be as large as posI to be as large as posI  sible. Thus we take I to be either (I to be either (I −`, 0) or (0, `). The solution curve on (0, `) is shown in Figure 1.1.1(b). . EXPLICIT AND IMPLICIT SOLUTIONS You should be familiar with the terms explicit functions and implicit functions from your study of calculus. A solution in which the dependent variable is expressed solely in terms of the independent variable and constants is said to be an explicit solution. For our purposes, let us think of an explicit solution as an explicit formula y = f(x) that we can manipulate, evaluate, and differentiate using the standard rules. We have just seen in the last two examples that y 5 116 x 4, y = xex, and y = 1yx are, in turn, explicit solutions of dyydx = xy1/2, y0 − 2y9 + y = 0, and xy9 + y = 0. Moreover, the trivial solu tion y = 0 is an explicit solution of all three equations. When we get down to the business of actually solving some ordinary differential equations, you will see that methods of solution do not always lead directly to an explicit solution y = f(x). This is particularly true when we attempt to solve nonlinear �rstorder differential equations. Often we have to be content with a relation or expression G(x, y) = 0 that de�nes a solution f implicitly. DEFINITION 1.1.3 Implicit Solution of an ODE A relation G(x, y) = 0 is said to be an implicit solution of an ordinary differen tial equation (4) on an interval I, provided that there exists at least one function f that satis�es the relation as well as the differential equation on I. It is beyond the scope of this course to investigate the conditions under which a relation G(x, y) = 0 de�nes a differentiable function f. So we shall assume that if the formal implementation of a method of solution leads to a relation G(x, y) = 0, then there exists at least one function f that satis�es both the relation (that is, G(x, f(x)) = 0) and the differential equation on an interval I. If the implicit solution G(x, y) = 0 is fairly simple, we may be able to solve for y in terms of x and obtain x and obtain x one or more explicit solutions. See (iv) in the Remarks. EXAMPLE 7 Verification of an Implicit Solution The relation x2 + y2 = 25 is an implicit solution of the differential equation dydyd dxdxd 5 2 x y (8) on the open interval (−5, 5). By implicit differentiation we obtain d dxdxd x2 1 d dxdxd y2 5 d dxdxd 2 25 or 2x2x2 1 2y dydyd dxdxd 5 0. (9) Solving the last equation in (9) for the symbol dyydx gives (8). Moreover, solving dx gives (8). Moreover, solving dx x2 + y2 = 25 for y in terms of x yields x yields x y 5 6Ï25 2 x2Ï . The two functions y 5 �1(x) 5 Ï25 2 x2 and y 5 �2(x) 5 2Ï25 2 x2 satisfy the relation (that is, x2 + �1 2 = 25 and x2 + �2 2 = 25) and are explicit solutions de�ned on the interval (−5, 5). The solution curves given in Figures 1.1.2(b) and 1.1.2(c) are segments of the graph of the implicit solution in Figure 1.1.2(a). 8 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 FIGURE 1.1.3 Some solutions of DE in part (a) of Example 8 y x c . 0 c , 0 c 5 0 Because the distinction between an explicit solution and an implicit solution should be intuitively clear, we will not belabor the issue by always saying, “Here is an explicit (implicit) solution.” FAMILIES OF SOLUTIONS The study of differential equations is similar to that of integral calculus. When evaluating an antiderivative or inde�nite integral in calculus, we use a single constant c of integration. Analogously, we shall see in Chapter 2 that when solving a �rstorder differential equation F(x, y, y9) 5 0 we usually obtain a solution containing a single constant or parameter c. A solu tion of F(x, y, y9) 5 0 containing a constant c is a set of solutions Gsx, y, cd 5 0 called a oneparameter family of solutions. When solving an nthorder differen tial equation F(x, y, y9, Á , y(n)) 5 0 we seek an nparameter family of solutions G(x, y, c1, c2, Á , cn) 5 0. This means that a single differential equation can possess an in�nite number of solutions corresponding to an unlimited number of choices for the parameter(s). A solution of a differential equation that is free of parameters is called a particular solution. The parameters in a family of solutions such as G(x, y, c1, c2, Á , cn) 5 0 are arbitrary up to a point. For example, proceeding as in (9) a relation x2 1 y2 5 c formally satis�es (8) for any constant c. However, it is understood that the relation should always make sense in the real number system; thus, if c 5 225 we cannot say that x2 1 y2 5 225 is an implicit solution of the differential equation. EXAMPLE 8 Particular Solutions (a) For all real values of c, the oneparameter family y 5 cxcxc 2 x cx cx os x is an explicit x is an explicit x solution of the linear �rstorder equation xyxyx 9 2 y 5 x2 sin x on the interval (−`, `). (Verify.) Figure 1.1.3 shows the graphs of some particular solutions in this family for various choices of c. The solution y = −x cos x, the blue graph in the �gure, is a particular solution corresponding to c = 0. (b) The twoparameter family y = c1ex + c2xex is an explicit solution of the linear x is an explicit solution of the linear x secondorder equation y0 − 2y9 + y = 0 in part (b) of Example 5. (Verify.) In Figure 1.1.4 we have shown seven of the “dou ble in�nity” of solutions in the family. The solution curves in red, green, and blue are the graphs of the particular solutions y = 5xex (x (x cl = 0, c2 = 5), y = 3ex (x (x cl = 3, c2 = 0), and y = 5ex − 2xex (x (x c1 = 5, c2 = 2), respectively. . FIGURE 1.1.2 An implicit solution and two explicit solutions of (8) in Example 7 . y x 5 5 (b) explicit solution y1 5 25 2ÏÏ255 2255 2Ï25Ï5 2Ï5 2255 2Ï5 2 x2, 52, 52 , x , 5 y x 5 5 25 (c) explicit solution y2 5 2ÏÏ25Ï25Ï 2 x2, 25 , x , 5 y x 5 5 x2 1 y2 5 25 implicit solution(a)(a)(a) FIGURE 1.1.4 Some solutions of DE in part (b) of Example 8 y x . DEFINITIONS AND TERMINOLOGY 9 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 FIGURE 1.1.5 Some solutions of DE in Example 10 (a) two explicit solutions (b) piecewisede�ned solution c 5 21 c 5 21, c 5 1 c 5 1, x y x , 0 x $ 0 x y Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation—that is, a solution that cannot be obtained by specializing any of the parameters in the family of solutions. Such an extra solution is called a singular solution. For example, we have seen that y 5 116 x 4 and y = 0 are solutions of the differential equation dyydx = xy1/2 on (−`, `). In Section 2.2 we shall demonstrate, by actually solving it, that the differential equation dyydx = xy1/2 possesses the oneparameter family of solutions y 5 _ we shall demonstrate, by actually solving it, that the differential equation _ we shall demonstrate, by actually solving it, that the differential equation 1 4 x 2 1 c+ we shall demonstrate, by actually solving it, that the differential equation + we shall demonstrate, by actually solving it, that the differential equation 2 we shall demonstrate, by actually solving it, that the differential equation 2 we shall demonstrate, by actually solving it, that the differential equation , c $ 0. When c = 0, the resulting particular solution is y 5 116 x 4. But notice that the trivial solution y = 0 is a singular solution since it is not a member of the family y 5 _14 x2 1 c+ 2; there is no way of assigning a value to the constant c to obtain y = 0. In all the preceding examples we used x and x and x y to denote the independent and dependent variables, respectively. But you should become accustomed to seeing and working with other symbols to denote these variables. For example, we could denote the independent variable by t and the dependent variable by t and the dependent variable by t x. EXAMPLE 9 Using Different Symbols The functions x = c1 cos 4t and t and t x = c2 sin 4t, where c1 and c2 are arbitrary constants or parameters, are both solutions of the linear differential equation x0 1 16x 5 0. For x = c1 cos 4t the �rst two derivatives with respect to t the �rst two derivatives with respect to t t are t are t x9 = −4c1 sin 4t and x0 = −16c1 cos 4t. Substituting x0 and x then givesx then givesx x0 1 16x 5 216c1 cos 4t 1 16(c1 cos 4t) 5 0. In like manner, for x = c2 sin 4t we have t we have t x0 = −16c2 sin 4t, and so x0 1 16x 5 216c2 sin 4t 1 16(c2 sin 4t) 5 0. Finally, it is straightforward to verify that the linear combination of solutions, or the twoparameter family x = c1 cos 4t + c2 sin 4t, is also a solution of the differential equation. . The next example shows that a solution of a differential equation can be a piecewisede�ned function. EXAMPLE 10 PiecewiseDefined Solution The oneparameter family of quartic monomial functions y = cx4 is an explicit solu tion of the linear �rstorder equation xy9 − 4y = 0 on the interval (−`, `). (Verify.) The blue and red solution curves shown in Figure 1.1.5(a) are the graphs of y = x4 and y = −x4 and correspond to the choices c = 1 and c = −1, respectively. The piecewisede�ned differentiable function y 5 52x 4, x , 0 x4, x . 0 is also a solution of the differential equation but cannot be obtained from the family y = cx4 by a single choice of c. As seen in Figure 1.1.5(b) the solution is constructed from the family by choosing c = −1 for x < 0 and c = 1 for x $ 0. . SYSTEMS OF DIFFERENTIAL EQUATIONS Up to this point we have been dis cussing single differential equations containing one unknown function. But often 10 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 in theory, as well as in many applications, we must deal with systems of differ ential equations. A system of ordinary differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable. For example, if x and y denote dependent variables and t denotes the independent variable, then a system of two �rstorder differential t denotes the independent variable, then a system of two �rstorder differential t equations is given by dx dt 5 f (t, x, y) dy dt 5 g(t, x, y). (10) A solution of a system such as (10) is a pair of differentiable functions x = f1(t), y = f2(t), de�ned on a common interval I, that satisfy each equation of the system on this interval. REMARKS (i) It might not be apparent whether a �rstorder ODE written in differential form M(x, y) dxdxd 1 N(N(N x, y) dydyd 5 0 is linear or nonlinear because there is nothing in this form that tells us which symbol denotes the dependent variable. See Problems 9 and 10 in Exercises 1.1. (ii) We will see in the chapters that follow that a solution of a differential equa tion may involve an integralde�ned function. One way of de�ning a function F of a single variable F of a single variable F x by means of a de�nite integral is:x by means of a de�nite integral is:x F(x) 5 #x a # a # g(t) dt. (11) If the integrand g in (11) is continuous on an interval [a, b] and a # x # b, then the derivative form of the Fundamental Theorem of Calculus states that F is F is F differentiable on (a, b) and F9(x) 5 d dxdxd # x a # a # g(t) dt 5 g(x) (12) The integral in (11) is often nonelementary, that is, an integral of a function g that does not have an elementaryfunction antiderivative. Elementary functions include the familiar functions studied in a typical precalculus course: constant, polynomial, rational, exponential, logarithmic, trigonometric, and inverse trigonometric functions, as well as rational powers of these functions; �nite combinations of these func tions using addition, subtraction, multiplication, division; and function com positions. For example, even though e2t 2t2t ,Ï1 1 t3Ï , and cos t2 are elementary functions, the integrals ee2t2t2t dt, eÏ1 1 t3Ï dt, and e cos t2 dt are nonelementary. t are nonelementary. t See Problems 25–28 in Exercises 1.1. Also see Appendix A. (iii) Although the concept of a solution of a differential equation has been emphasized in this section, you should be aware that a DE does not necessarily have to possess a solution. See Problem 43 in Exercises 1.1. The question of whether a solution exists will be touched on in the next section. (continued)continued)continued . DEFINITIONS AND TERMINOLOGY 11 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 EXERCISES 1.1 In Problems 1–8 state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with (6). 1. (1 − x)y0 − 4xy9+ 5y = cos x 2. x d 3y dxdxd 3 2 Sdydxdxd D4 1 y 5 0 3. t5y(4) − t3y0 + 6y = 0 4. d 2u dr 2 1 du dr 1 u 5 cos(r 1 u) 5. d 2y2y2 dxdxd 2 5Î1 1 1dydyddxdxd 22Î 6. d 2R dt 2 5 2 k R2 7. (sin u)y − (cos u)y9 = 2 8. x $ 2 11 2 x ?2 3 2x? 1 x 5 0 In Problems 9 and 10 determine whether the given �rstorder difIn Problems 9 and 10 determine whether the given �rstorder difIn Problems 9 and 10 determine whether the given �rstorder dif ferential equation is linear in the indicated dependent variable by matching it with the �rst differential equation given in (7). 9. (y2 − 1) dx + x dy = 0; in y; in x 10. u dv + (v + uv − ueu) du = 0; in v; in u In Problems 11–14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of de�nition for each solution.I of de�nition for each solution.I 11. 2y9 + y = 0; y = e−x/2x/2x Answers to selected oddnumbered problems begin on page ANS1. (iv) A few last words about implicit solutions of differential equations are in order. In Example 7 we were able to solve the relation x2 + y2 = 25 for y in terms of x to get two explicit solutions, x to get two explicit solutions, x �1(x) 5 Ï25 2 x2Ï and �2(x) 5 2Ï25 2 x2Ï , of the differential equation (8). But don’t read too much into this one example. Unless it is easy or important or you are instructed to, there is usually no need to try to solve an implicit solution G(x(x( , y) = 0 for y explicitly in terms of x. Also do not misinterpret the second sentence following De�nition 1.1.3. An implicit solution G(x(x( , y) = 0 can de�ne a perfectly good differentiable function f that is a solution of a DE, yet we might not be able to solve G(x(x( , y) = 0 using analyti cal methods such as algebra. The solution curve of f may be a segment or piece of the graph of G(x, y) = 0. See Problems 49 and 50 in Exercises 1.1. Also, read the discussion following Example 4 in Section 2.2. (v) It might not seem like a big deal to assume that F(x, y, y9, . . . , y(n)) = 0 can be solved for y(n), but one should be a little bit careful here. There are exceptions, and there certainly are some problems connected with this assumption. See Problems 56 and 57 in Exercises 1.1. (vi) If every solution of an nthorder ODE F(x(x( , y, y9, . . . , y(n)) = 0 on an inter val I can be obtained from an I can be obtained from an I nparameter family G(x(x( , y, c1, c2, . . . , cn) = 0 by appropriate choices of the parameters ci, i = 1, 2, . . . , n, we then say that the family is the general solution of the DE. In solving linear ODEs, we shall im pose relatively simple restrictions on the coef�cients of the equation; with these restrictions one can be assured that not only does a solution exist on an interval but also that a family of solutions yields all possible solutions. Nonlinear ODEs, with the exception of some �rstorder equations, are usually dif�cult or impos sible to solve in terms of elementary functions. Furthermore, if we happen to obtain a family of solutions for a nonlinear equation, it is not obvious whether this family contains all solutions. On a practical level, then, the designation “general solution” is applied only to linear ODEs. Don’t be concerned about this concept at this point, but store the words “general solution” in the back of your mind—we will come back to this notion in Section 2.3 and again in Chapter 4. 12 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 12. dy dt 1 20y20y20 5 24; y 5 6 5 2 6 5 e220t 13. y0 − 6y9 + 13y = 0; y = e3x cos 2x cos 2x x cos 2x cos 2 14. y0 + y = tan x; y = −(cos x) ln(sec x + tan x) In Problems 15–18 verify that the indicated function y = f(x) is an explicit solution of the given �rstorder differential equation. Pro ceed as in Example 6, by considering f simply as a function and give its domain. Then by considering f as a solution of the differential equation, give at least one interval I of de�nition.I of de�nition.I 15. (y 2 x)y9 5 y 2 x 1 8; y 5 x 1 4ÏxÏxÏ 1 2Ï 16. y9 = 25 + y2; y = 5 tan 5x 17. y9 = 2xy2xy2 2; y = 1y(4 − x2) 18. 2y9 = y3 cos x; y = (1 − sin x)−1/2 In Problems 19 and 20 verify that the indicated expression is an im plicit solution of the given �rstorder differential equation. Find at least one explicit solution y = f(x) in each case. Use a graphing util ity to obtain the graph of an explicit solution. Give an interval I of I of I de�nition of each solution f. 19. dX dt 5 (X 2 1)(1 2 2X2X2 );X);X ln12X2X2 2 1X 2 1 2 5 t 20. 2xy dx + (x2 − y) dy = 0; −2x2x2 2y + y2 = 1 In Problems 21–24 verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of de�nition for each solution.I of de�nition for each solution.I 21. dP dt 5 P(1 2 P); P 5 c1et 1 1 c1et 22. dy dx 1 4xy4xy4 5 8x3; y 5 2x2x2 2 2 1 1 c1e22x2x2 2 23. d 2y dx2 2 4 dy dx 1 4y 5 0; y 5 c1e2x 1 c2xe2xe2 2x 24. x3 d3y dx3 1 2x2x2 2 d 2y dx2 2 x dy dx 1 y 5 12x12x12 2; y 5 c1x21 1 c2x2x2 1 c3x ln x ln x x 1 4x4x4 2 In Problems 25–28 use (12) to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of de�nition of each solution.I of de�nition of each solution.I 25. x dy dx 2 3xy 5 1; y 5 e3x#x 1 # 1 # e 23t t dt 26. 2x2x2 dy dx 2 y 5 2x2x2 cos x cos x x; y 5 ÏxÏxÏÏ #x 4 # 4 # cos tÏtÏ dt 27. x2 dy dx 1 xy 5 10 sin x; y 5 5 x 1 10 x # x 1 # 1 # sin tt dt 28. dy dx 1 2xy2xy2 5 1; y 5 e2x 2 1 e2x 2#x 0 # 0 # et2t2t dt 29. Verify that the piecewisede�ned functionVerify that the piecewisede�ned functionV y 5 52x 2, x , 0 x2, x $ 0 is a solution of the differential equation xy9 − 2y = 0 on (−`, `). 30. In Example 7 we saw that y = f1(x) = Ï25 2 x2Ï and y 5 �2(x) 5 2Ï25 2 x2 In Example 7 we sa Ï are solutions of dyydx = −xyy on the interval (−5, 5). Explain why the piecewisede�ned function y 5 5Ï25 2 x 2Ï 25 , x , 0 2Ï25 2 x2Ï , 0 # x , 5 is not a solution of the differential equation on the interval (not a solution of the differential equation on the interval (not −5, 5). In Problems 31–34 �nd values of m so that the function y = emx is a mx is a mx solution of the given differential equation. 31. y9 + 2y = 0 32. 5y9 = 2y 33. y0 − 5y9 + 6y = 0 34. 2y0 + 7y9 − 4y = 0 In Problems 35 and 36 �nd values of m so that the function y = xm is a solution of the given differential equation. 35. xy0 + 2y9 = 0 36. x2y2y2 0 − 7xy9 + 15y = 0 In Problems 37–40 use the concept that y = c, −` < x < `, is a constant function if and only if y9 = 0 to determine whether the given differential equation possesses constant solutions. 37. 3xy9 + 5y = 10 38. y9 = y2 + 2y − 3 39. (y − 1)y9 = 1 40. y0 + 4y9 + 6y = 10 In Problems 41 and 42 verify that the indicated pair of functions is a solution of the given system of differential equations on the interval (−`, `). 41. dx dt 5 x 1 3y 42. d 2x2x2 dt 2 5 4y 1 et dy dt 5 5x 1 3y; d 2y2y2 dt 2 5 4x 2 et; x 5 e22t 1 3e6t, x 5 cos 2t 1 sin 2t 1 15 e t, y 5 2e22t 1 5e6t y 5 2cos 2t 2 sin 2t 2 15 e t Discussion Problems 43. Make up a differential equation that does not possess any real solutions. 44. Make up a differential equation that you feel con�dent possesses only the trivial solution y = 0. Explain your reasoning. 45. What function do you know from calculus is such that its �rst derivative is itself? Its �rst derivative is a constant multiple k of k of k itself? Write each answer in the form of a �rstorder differential equation with a solution. 46. What function (or functions) do you know from calculus is such that its second derivative is itself? Its second derivative is the negative of itself? Write each answer in the form of a second order differential equation with a solution. . DEFINITIONS AND TERMINOLOGY 13 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 47. The function y 5 sin x is an explicit solution of the �rstorder x is an explicit solution of the �rstorder x differential equation dydyd dxdxd 5 Ï1 2 y2Ï . Find an interval I of I of I de�nition. [Hint: I is I is I not the interval (not the interval (not −`, `).] 48. Discuss why it makes intuitive sense to presume that the linear differential equation y0 + 2y9 + 4y = 5 sin t has a solution of t has a solution of t the form y = A sin t + B cos t, where A and B are constants. Then �nd speci�c constants A and B so that y = A sin t + B cos t is a particular solution of the DE. In Problems 49 and 50 the given �gure represents the graph of an implicit solution G(x, y) = 0 of a differential equation dyydx = f (x, y). In each case the relation G(x, y) = 0 implicitly de �nes several solutions of the DE. Carefully reproduce each �gure on a piece of paper. Use different colored pencils to mark off seg ments, or pieces, on each graph that correspond to graphs of so lutions. Keep in mind that a solution f must be a function and differentiable. Use the solution curve to estimate an interval I of I of I de�nition of each solution f. 49. 50. 51. The graphs of members of the oneparameter family x3 + y3 = 3cxy are called folia of Descartes. Verify that this family is an implicit solution of the �rstorder differential equation dydyd dxdxd 5 y(y3 2 2x2x2 3) x(2y2y2 3 2 x3) . 52. The graph in Figure 1.1.7 is the member of the family of folia in Problem 51 corresponding to c = 1. Discuss: How can the DE in Problem 51 help in �nding points on the graph of x3 + y3 = 3xy where the tangent line is vertical? How does knowing where a tangent line is vertical help in determining an interval I of I of I de�nition of a solution f of the DE? Carry out your ideas and compare with your estimates of the intervals in Problem 50. 53. In Example 7 the largest interval I over which the explicit I over which the explicit I solutions y = f1(x) and y = f2(x) are de�ned is the open interval (−5, 5). Why can’t the interval I of de�nition be the I of de�nition be the I closed interval [−5, 5]? 54. In Problem 21 a oneparameter family of solutions of the DE P9 = P(1 − P) is given. Does any solution curve pass through the point (0, 3)? Through the point (0, 1)? 55. Discuss, and illustrate with examples, how to solve differential equations of the forms dyydx = f (x) and d2yydx2 = f (x). 56. The differential equation x(y9)2 − 4y9 − 12x12x12 3 = 0 has the form given in (4). Determine whether the equation can be put into the normal form dyydx = f (x, y). FIGURE 1.1.7 Graph for Problem 50 1 x 1 y 57. The normal form (5) of an nthorder differential equation is equivalent to (4) whenever both forms have exactly the same solutions. Make up a �rstorder differential equation for which F(x, y, y9) = 0 is not equivalent to the normal form dyydx = f (x, y). 58. Find a linear secondorder differential equation F(x(x( , y, y9, y0) = 0 for which y = c1x + c2x2x2 2 is a twoparameter family of solutions. Make sure that your equation is free of the arbitrary parameters c1 and c2. Qualitative information about a solution y = f(x) of a differential equation can often be obtained from the equation itself. Before working Problems 59–62, recall the geometric signi�cance of the derivatives dyydx and dx and dx d2d2d y2y2 ydx2. 59. Consider the differential equation dydyd ydxdxd 5 e2x2. (a) Explain why a solution of the DE must be an increasing function on any interval of the xaxis. (b) What are lim x S 2` dydyd ydxdxd and lim x S ` dydyd ydxdxd ? What does this suggest about a solution curve as x S 6`? (c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up. (d) Sketch the graph of a solution y = f(x) of the differential equation whose shape is suggested by parts (a)– (c). 60. Consider the differential equation dyydx = 5 − y. (a) Either by inspection or by the method suggested in Problems 37–40, �nd a constant solution of the DE. (b) Using only the differential equation, �nd intervals on the yaxis on which a nonconstant solution y = f(x) is increasing. Find intervals on the yaxis on which y = f(x) is decreasing. 61. Consider the differential equation dyydx = y(a − by), where a and b are positive constants. (a) Either by inspection or by the method suggested in Problems 37–40, �nd two constant solutions of the DE. (b) Using only the differential equation, �nd intervals on the yaxis on which a nonconstant solution y = f(x) is increas ing. Find intervals on which y = f(x) is decreasing. (c) Using only the differential equation, explain why y = ay2b is the ycoordinate of a point of in�ection of the graph of a nonconstant solution y = f(x). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the xyplane into three regions. In each region, sketch the graph of a nonconstant solution y = f(x) whose shape is suggested by the results in parts (b) and (c). 62. Consider the differential equation y9 = y2 + 4. (a) Explain why there exist no constant solutions of the DE. (b) Describe the graph of a solution y = f(x). For example, can a solution curve have any relative extrema? FIGURE 1.1.6 Graph for Problem 49 y x 1 1 14 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 FIGURE 1.2.1 Solution curve of �rstorder IVP xI solutions of the DE (x0, y0) y (c) Explain why y 5 0 is the ycoordinate of a point of in�ection of a solution curve. (d) Sketch the graph of a solution y = f(x) of the differential equation whose shape is suggested by parts (a)–(c). Computer Lab Assignments In Problems 63 and 64 use a CAS to compute all derivatives and to carry out the simplifications needed to verify that the indicated function is a particular solution of the given differential equation. 63. y(4) − 20y + 158y0 − 580y9 + 841y = 0; y = xe5x cos 2x cos 2x x cos 2x cos 2 64. x3y90 1 2x2y0 1 20xy9 2 78y 5 0; y 5 20 cos(5 ln x) x 2 3 sin(5 ln x) x INTRODUCTION We are often interested in problems in which we seek a solution y(x) of a differential equation so that y(x) also satis�es certain prescribed side conditions, that is, conditions that are imposed on the unknown function y(x) and its derivatives at a number x0. On some interval I containing I containing I x0 the problem of solving an nthorder differential equation subject to n side conditions speci�ed at x0: Solve: d ny dxdxd n 5 f (x, y, y9, Á , y(n21)) (1) Subject to: ysx0d 5 y0, y9sx0d 5 y1, Á , ysn21dsx0d 5 yn21, where y0, y1, Á , yn21 are arbitrary constants, is called an nthorder initialvalue problem (IVP). The values of y(x) and its �rst n21 derivatives at x0, y(x0) 5 y0, y9(x0) 5 y1, Á , y(n21)(x0) 5 yn21 are called initial conditions (IC). Solving an nthorder initialvalue problem such as (1) frequently entails �rst �nding an nparameter family of solutions of the differential equation and then using the initial conditions at x0 to determine the n constants in this family. The resulting particular solution is de�ned on some interval I containing the number I containing the number I x0. GEOMETRIC INTERPRETATION The cases n 5 1 and n 5 2 in (1), Solve: dy dx 5 f sx, yd Subject to: ysx0d 5 y0 (2) and Solve: d 2y dx2 5 f (x, y, y9) Subject to: y(x0) 5 y0, y9(x0) 5 y1 (3) are examples of �rst and secondorder initialvalue problems, respectively. These two problems are easy to interpret in geometric terms. For (2) we are seeking a solution y(x) of the differential equation y9 = f (x, y) on an interval I containing I containing I x0 so that its graph passes through the speci�ed point (x0, y0). A solution curve is shown in blue in Figure 1.2.1. For (3) we want to �nd a solution y(x) of the differential equation y0 = f (x, y, y9) on an interval I containing I containing I x0 so that its graph not only passes through (x0, y0) but the slope of the curve at this point is the number y1. A solution curve is shown in blue in Figure 1.2.2. The words initial conditions derive from physical systems where the independent variable is time t and where t and where t y(t0) = y0 1.2 InitialValue Problems FIGURE 1.2.2 Solution curve of secondorder IVP xI solutions of the DE (x0, y0) y m 5 y1 . INITIALVALUE PROBLEMS 15 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 FIGURE 1.2.4 Graphs of function and solution of IVP in Example 2 (0, 21) x y 121 x y 121 (a) function de�ned for all x ex ex xcept x = x = x 61 (b) solution de�ned on interval containing x = x = x 0 and y9(t0) = y1 represent the position and velocity, respectively, of an object at some beginning, or initial, time t0. EXAMPLE 1 Two FirstOrder IVPs (a) In Problem 45 in Exercises 1.1 you were asked to deduce that y = cex is a onex is a onex parameter family of solutions of the simple �rstorder equation y9 = y. All the so lutions in this family are de�ned on the interval (−`, `). If we impose an initial condition, say, y(0) = 3, then substituting x = 0, y = 3 in the family determines the constant 3 = ce0 = c. Thus y = 3ex is a solution of the IVPx is a solution of the IVPx y9 5 y, y(0) 5 3. (b) Now if we demand that a solution curve pass through the point (1, −2) rather than (0, 3), then y(1) = −2 will yield −2 = ce or c = −2e−1. In this case y = −2ex−1 is a solution of the IVP y9 5 y, y(1) 5 22. The two solution curves are shown in dark blue and dark red in Figure 1.2.3. . The next example illustrates another �rstorder initialvalue problem. In this ex ample notice how the interval I of de�nition of the solution I of de�nition of the solution I y(x) depends on the initial condition y(x0) = y0. EXAMPLE 2 Interval I of Definition of a SolutionI of Definition of a SolutionI In Problem 6 of Exercises 2.2 you will be asked to show that a oneparameter family of solutions of the �rstorder differential equation y9 + 2xy2 = 0 is y = 1y(x2 + c). If we impose the initial condition y(0) = −1, then substituting x = 0 and y = −1 into the family of solutions gives −1 = 1yc or c = −1. Thus y = 1y(x2 − 1). We now emphasize the following three distinctions: ● Considered as a function, the domain of y = 1y(x2 − 1) is the set of real numbers x for which x for which x y(x) is de�ned; this is the set of all real numbers except x = −1 and x = 1. See Figure 1.2.4(a). ● Considered as a solution of the differential equation y9 + 2xy2 = 0, the interval I of de�nition of I of de�nition of I y = 1y(x2 − 1) could be taken to be any interval over which y(x) is de�ned and differentiable. As can be seen in Figure 1.2.4(a), the largest intervals on which y = 1y(x2 − 1) is a solution are (−`,−1), (−1, 1), and (1, `). ● Considered as a solution of the initialvalue problem y9 + 2xy2 = 0, y(0) = −1, the interval I of de�nition of I of de�nition of I y = 1y(x2 − 1) could be taken to be any interval over which y(x) is de�ned, differentiable, and contains the and contains the and initial point x = 0; the largest interval for which this is true is (−1, 1). See the red curve in Figure 1.2.4(b). . See Problems 3–6 in Exercises 1.2 for a continuation of Example 2. EXAMPLE 3 SecondOrder IVP In Example 9 of Section 1.1 we saw that x = c1 cos 4t + c2 sin 4t is a twot is a twot parameter family of solutions of x0 + 16x = 0. Find a solution of the initialvalue problem x0 1 16x 5 0, x 1�2 2 5 22, x91�2 2 5 1. (4) FIGURE 1.2.3 Solution curves of two IVPs in Example 1 y x (0, 3) (1,22) 16 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 y y 5 x 4 / 16 (0, 0) 1 x y 5 0 FIGURE 1.2.5 Two solution curves of the same IVP in Example 4 SOLUTION We �rst apply x(py2) =−2 to the given family of solutions: c1 cos 2p + c2 sin 2p =−2. Since cos 2p = 1 and sin 2p = 0, we �nd that c1 =−2. We next apply x9(py2) = 1 to the oneparameter family x(t)t)t =−2 cos 4t + c2 sin 4t. Differentiating and then setting t = py2 and x9 = 1 gives 8 sin 2p + 4c2 cos 2p = 1, from which we see that c2 5 1 4. Hence x 5 22 cos 4t 1 1 4 sin 4t is a solution of (4). . EXISTENCE AND UNIQUENESS Two fundamental questions arise in considering an initialvalue problem: Does a solution of the problem exist? If a solution exists, is it unique? For the �rstorder initialvalue problem (2) we ask: Existence 5Does the differential equation dyydx = f (x, y) possess solutions? Do any of the solution curves pass through the point (x0, y0)? Uniqueness 5When can we be certain that there is precisely one solution curve passing through the point (x0, y0)? Note that in Examples 1 and 3 the phrase “a solution” is used rather than “the solution” of the problem. The inde�nite article “a” is used deliberately to suggest the possibility that other solutions may exist. At this point it has not been demonstrated that there is a single solution of each problem. The next example illustrates an initial value problem with two solutions. EXAMPLE 4 An IVP Can Have Several Solutions Each of the functions y = 0 and y 5 116 x4 satis�es the differential equation dyydx = xy1/2 and the initial condition y(0) = 0, so the initialvalue problem dydyd dxdxd 5 xyxyx 1/2, y(0) 5 0 has at least two solutions. As illustrated in Figure 1.2.5, the graphs of both functions, shown in red and blue pass through the same point (0, 0). . Within the safe con�nes of a formal course in differential equations one can be fairly con�dent that most differential equations will have solutions and that solumost differential equations will have solutions and that solumost  tions of initialvalue problems will probably be unique. Real life, however, is not so idyllic. Therefore it is desirable to know in advance of trying to solve an initialvalue problem whether a solution exists and, when it does, whether it is the only solution of the problem. Since we are going to consider �rstorder differential equations in the next two chapters, we state here without proof a straightforward theorem that gives conditions that are suf�cient to guarantee the existence and uniqueness of a solution of a �rstorder initialvalue problem of the form given in (2). We shall wait until Chapter 4 to address the question of existence and uniqueness of a secondorder initialvalue problem. THEOREM 1.2.1 Existence of a Unique Solution Let R be a rectangular region in the xyplane de�ned by a # x # b, c # y # d that contains the point (xthat contains the point (xthat contains the point ( 0, y0) in its interior. If f (x(x( , y) and −f−f− y−y are continuous on R, then there exists some interval I0: (x: (x: ( 0 − h, x0 + h), h . 0, contained in [a, b], and a unique function y(x(x( ), de�ned on I0, that is a solution of the initial value problem (2). . INITIALVALUE PROBLEMS 17 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 The foregoing result is one of the most popular existence and uniqueness theo rems for �rstorder differential equations because the criteria of continuity of f (x, y) and −f−f− yfyf −y are relatively easy to check. The geometry of Theorem 1.2.1 is illustrated in Figure 1.2.6. EXAMPLE 5 Example 4 Revisited We saw in Example 4 that the differential equation dyydx = xy1/2 possesses at least two solutions whose graphs pass through (0, 0). Inspection of the functions f (x, y) 5 xyxyx 1/2 and −f−f− −y 5 x 2y1/2 shows that they are continuous in the upper halfplane de�ned by y > 0. Hence Theorem 1.2.1 enables us to conclude that through any point (x0, y0), y0 > 0 in the upper halfplane there is some interval centered at x0 on which the given differential equation has a unique solution. Thus, for example, even without solving it, we know that there exists some interval centered at 2 on which the initialvalue problem dyydx = xy1/2, y(2) = 1 has a unique solution. . In Example 1, Theorem 1.2.1 guarantees that there are no other solutions of the initialvalue problems y9 = y, y(0) = 3 and y9 = y, y(1) = −2 other than y = 3ex and y = −2ex−1, respectively. This follows from the fact that f (x, y) = y and −f−f− yfyf −y = 1 are continuous throughout the entire xyplane. It can be further shown that the interval I on which each solution is de�ned is (I on which each solution is de�ned is (I −`, `). INTERVAL OF EXISTENCE/UNIQUENESS Suppose y(x) represents a solution of the initialvalue problem (2). The following three sets on the real xaxis may not be the same: the domain of the function y(x), the interval I over which the soluI over which the soluI  tion y(x) is de�ned or exists, and the interval I0I0I of existence and uniqueness. Exand uniqueness. Exand  ample 6 of Section 1.1 illustrated the difference between the domain of a function and the interval I of de�nition. Now suppose (I of de�nition. Now suppose (I x0, y0) is a point in the interior of the rectangular region R in Theorem 1.2.1. It turns out that the continuity of the function f (x, y) on R by itself is suf�cient to guarantee the existence of at least one solution of dyydx = f (x, y), y(x0) = y0, de�ned on some interval I. The interval I of de�nition for this initialvalue problem is usually taken to be the largest interval containing x0 over which the solution y(x) is de�ned and differentiable. The interval I depends on both I depends on both I f (x, y) and the initial condition y(x0) = y0. See Problems 31–34 in Exercises 1.2. The extra condition of continuity of the �rst partial derivative −f−f− yfyf −y on R enables us to say that not only does a solution exist on some interval I0I0I con taining x0, but it is the only solution satisfying y(x0) = y0. However, Theorem 1.2.1 does not give any indication of the sizes of intervals I and I and I I0I0I ; the interval I of de�nition need not be as wide as the region R, and the interval I0de�nition need not be as wide as the region R, and the interval I0de�nition need not be as wide as the region R, and the interval I of existence and uniqueness may not be as large as I. The number h > 0 that de�nes the interval I0I0I : (x0 − h, x0 + h) could be very small, so it is best to think that the solution y(x) is unique in a local sense—that is, a solution de�ned near the point (x0, y0). See Problem 51 in Exercises 1.2. x0 R a bIa bI0a b0 c d a ba ba ba ba ba b (x0, y0) y FIGURE 1.2.6 Rectangular region R REMARKS (i) The conditions in Theorem 1.2.1 are suf�cient but not necessary. This means that when f (x(x( , y) and −f−f− yfyf −y are continuous on a rectangular region R, it must always follow that a solution of (2) exists and is unique whenever (xalways follow that a solution of (2) exists and is unique whenever (xalways follow that a solution of (2) exists and is unique whenever ( 0, y0) is a point interior to R. However, if the conditions stated in the hypothesis of 18 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Theorem 1.2.1 do not hold, then anything could happen: Problem (2) may still have a solution and this solution may be unique, or (2) may have several solu tions, or it may have no solution at all. A rereading of Example 5 reveals that the hypotheses of Theorem 1.2.1 do not hold on the line y = 0 for the differen tial equation dyydx = xy1/2, so it is not surprising, as we saw in Example 4 of this section, that there are two solutions de�ned on a common interval (2h, h) satisfying y(0) = 0. On the other hand, the hypotheses of Theorem 1.2.1 do not hold on the line y = 1 for the differential equation dyydx = uy − 1u. Nevertheless it can be proved that the solution of the initialvalue problem dyydx = uy − 1u, y(0) = 1, is unique. Can you guess this solution? (ii) You are encouraged to read, think about, work, and then keep in mind Problem 50 in Exercises 1.2. (iii) Initial conditions are prescribed at a single point x0. But we are also interested in solving differential equations that are subject to conditions speci�ed on y(x) or its derivative at two different points x0 and x1. Condi tions such as y(1) 5 0, y(5) 5 0 or y(�y2) 5 0, y9(�) 5 1 are called boundary conditions (BC). A differential equation together with boundary conditions is called a boundaryvalue problem (BVP). For example, y0 1 y 5 0, y9(0) 5 0, y9(�) 5 0 is a boundaryvalue problem. See Problems 39–44 in Exercises 1.2. When we start to solve differential equations in Chapter 2 we will solve only �rstorder equations and �rstorder initialvalue problems. The math ematical description of many problems in science and engineering involve secondorder IVPs or twopoint BVPs. We will examine some of these problems in Chapters 4 and 5. EXERCISES 1.2 In Problems 1 and 2, y 5 1y(1 1 c1e2x) is a oneparameter family of solutions of the �rstorder DE y9 = y − y2. Find a solution of the �rstorder IVP consisting of this differential equation and the given initial condition. 1. y(0) 5 213 2. y(−1) = 2 In Problems 3–6, y = 1y(x2 + c) is a oneparameter family of so lutions of the �rstorder DE y9 + 2xy2xy2 2 = 0. Find a solution of the �rstorder IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution I over which the solution I is de�ned. 3. y(2) 5 13 4. y(22) 5 1 2 5. y(0) = 1 6. y_12+ 5 24 In Problems 7–10, x = c1 cos t + c2 sin t is a twoparameter family t is a twoparameter family t of solutions of the secondorder DE x0 + x = 0. Find a solution of the secondorder IVP consisting of this differential equation and the given initial conditions. 7. x(0) = −1, x9(0) = 8 8. x(py2) = 0, x9(py2) = 1 9. x(�y6) 5 12, x9(�y6) 5 0 10. x(�y4) 5 Ï2Ï , x9(�y4) 5 2Ï2Ï In Problems 11–14, y = c1ex + c2e−x is a twoparameter family of x is a twoparameter family of x solutions of the secondorder DE y0 − y = 0. Find a solution of the secondorder IVP consisting of this differential equation and the given initial conditions. 11. y(0) 5 1, y9(0) 5 2 12. y(1) = 0, y9(1) = e 13. y(−1) = 5, y9(−1) = −5 14. y(0) = 0, y9(0) = 0 Answers to selected oddnumbered problems begin on page ANS1. . INITIALVALUE PROBLEMS 19 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 In Problems 15 and 16 determine by inspection at least two solutions of the given �rstorder IVP. 15. y9 = 3y2/3, y(0) = 0 16. xy9 = 2y, y(0) = 0 In Problems 17–24 determine a region of the xyplane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. 17. dydyd dxdxd 5 y2/3 18. dydyd dxdxd 5 ÏxÏxÏ yxyxÏ 19. x dydyd dxdxd 5 y 20. dydyd dxdxd 2 y 5 x 21. (4 − y2)y9 = x2 22. (1 + y3)y9 = x2 23. (x2 + y2)y9 = y2 24. (y − x)y9 = y + x In Problems 25–28 determine whether Theorem 1.2.1 guarantees that the differential equation y9 5 ÏyÏyÏ 2 2 9 In Problems 25–28 determine whether Theorem 1.2.1 guarantees that Ï possesses a unique solution through the given point. 25. (1, 4) 26. (5, 3) 27. (2, −3) 28. (−1, 1) 29. (a) By inspection �nd a oneparameter family of solutions of the differential equation xy9 = y. Verify that each member of the family is a solution of the initialvalue problem xy9 = y, y(0) = 0. (b) Explain part (a) by determining a region R in the xyplane for which the differential equation xy9 = y would have a unique solution through a point (x0, y0) in R. (c) Verify that the piecewisede�ned function y 5 50, x , 0x, x $ 0 satis�es the condition y(0) = 0. Determine whether this function is also a solution of the initialvalue problem in part (a). 30. (a) Verify that y = tan (x + c) is a oneparameter family of solutions of the differential equation y9 = 1 + y2. (b) Since f (x, y) = 1 + y2 and −f−f− yfyf −y = 2y are continuous ev erywhere, the region R in Theorem 1.2.1 can be taken to be the entire xyplane. Use the family of solutions in part (a) to �nd an explicit solution of the �rstorder initialvalue problem y9 = 1 + y2, y(0) = 0. Even though x0 = 0 is in the interval (−2, 2), explain why the solution is not de�ned on this interval. (c) Determine the largest interval I of de�nition for the solution I of de�nition for the solution I of the initialvalue problem in part (b). 31. (a) Verify that y = −1y(x + c) is a oneparameter family of solutions of the differential equation y9 = y2. (b) Since f (x, y) = y2 and −f−f− yfyf −y = 2y are continuous every where, the region R in Theorem 1.2.1 can be taken to be the entire xyplane. Find a solution from the family in part (a) that satis�es y(0) = 1. Then �nd a solution from the family in part (a) that satis�es y(0) = −1. Determine the largest interval I of de�nition for the solution of each I of de�nition for the solution of each I initialvalue problem. (c) Determine the largest interval I of de�nition for the solution I of de�nition for the solution I of the �rstorder initialvalue problem y9 = y2, y(0) = 0. [Hint: The solution is not a member of the family of solutions in part (a).] 32. (a) Show that a solution from the family in part (a) of Problem 31 that satis�es y9 = y2, y(1) = 1, is y = 1y(2 − x). (b) Then show that a solution from the family in part (a) of Problem 31 that satis�es y9 = y2, y(3) = −1, is y = 1y(2 − x). (c) Are the solutions in parts (a) and (b) the same? 33. (a) Verify that 3x2 − y2 = c is a oneparameter family of solutions of the differential equation y dyydx = 3x. (b) By hand, sketch the graph of the implicit solution 3x2 − y2 = 3. Find all explicit solutions y = f(x) of the DE in part (a) de�ned by this relation. Give the interval I of de�nition of each explicit solution. (c) The point (−2, 3) is on the graph of 3x2 − y2 = 3, but which of the explicit solutions in part (b) satis�es y(−2) = 3? 34. (a) Use the family of solutions in part (a) of Problem 33 to �nd an implicit solution of the initialvalue problem ydyydx = 3x, y(2) = −4. Then, by hand, sketch the graph of the explicit solution of this problem and give its interval I of de�nition.I of de�nition.I (b) Are there any explicit solutions of y dyydx = 3x that pass x that pass x through the origin? In Problems 35–38 the graph of a member of a family of solutions of a secondorder differential equation d2d2d y2y2 ydx2 = f (x, y, y9) is given. Match the solution curve with at least one pair of the following initial conditions. (a) y(1) = 1, y9(1) = −2 (b) y(−1) = 0, y9(−1) = −4 (c) y(1) = 1, y9(1) = 2 (d) y(0) = −1, y9(0) = 2 (e) y(0) = −1, y9(0) = 0 (f) y(0) = −4, y9(0) = −2 35. FIGURE 1.2.7 Graph for Problem 35 y x 5 25 5 20 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 FIGURE 1.2.12 Two solutions of the IVP in Problem 51 (a) (2, 1) y x (b) (2, 1) y x FIGURE 1.2.11 Graphs for Problem 47 ((0, )12 1 1 x y 36. 37. 38. In Problems 39–44, y 5 c1 cos 2x2x2 1 c2 sin 2x2x2 is a twoparameter x is a twoparameter x family of solutions of the secondorder DE y0 1 4y 5 0. If possible, �nd a solution of the differential equation that satis�es the given side conditions. The conditions speci�ed at two different points are called boundary conditions. 39. y(0) 5 0, y(�y4) 5 3 40. y(0) 5 0, y(�) 5 0 41. y9(0) 5 0, y9(�y6) 5 0 42. y(0) 5 1, y9(�) 5 5 43. y(0) 5 0, y(�) 5 2 44. y9(�y2) 5 1, y9(�) 5 0 Discussion Problems In Problems 45 and 46 use Problem 55 in Exercises 1.1 and (2) and (3) of this section. 45. Find a function whose graph at each point (x, y) has the slope given by 8e2x2x2 + 6x and has the x and has the x yintercept (0, 9). 46. Find a function whose second derivative is y0 = 12x12x12 − 2 at each point (x, y) on its graph and y = −x + 5 is tangent to the graph at the point corresponding to x = 1. 47. Consider the initialvalue problem y9 = x − 2y, y(0) 5 12. Determine which of the two curves shown in Figure 1.2.11 is the only plausible solution curve. Explain your reasoning. 48. Show that x 5 #y 0 # 0 # 1Ït3 1 1Ï dt is an implicit solution of the initialvalue problem 2 d 2y2y2 dxdxd 2 2 3y2 5 0, ys0d 5 0, y9s0d 5 1. Assume that y $ 0. [Hint: The integral is nonelementary. See (ii) in the Remarks at the end of Section 1.1.] 49. Determine a plausible value of x0 for which the graph of the solution of the initialvalue problem y9 + 2y2y2 = 3x − 6, y(x(x( 0) = 0 is tangent to the xaxis at (xaxis at (xaxis at ( 0, 0). Explain your reasoning. 50. Suppose that the �rstorder differential equation dyydx = f (x, y) possesses a oneparameter family of solutions and that f (x, y) satis�es the hypotheses of Theorem 1.2.1 in some rectangular region R of the xyplane. Explain why two different solution curves cannot intersect or be tangent to each other at a point (x0, y0) in R. 51. The functions y(x) 5 116 x 4, 2` , x , ` and y(x) 5 50,1 16 x 4, x , 0 x $ 0 have the same domain but are clearly different. See Figures 1.2.12(a) and 1.2.12(b), respectively. Show that both functions are solutions of the initialvalue problem dyydx = xy1/2, y(2) = 1 on the interval (−`, `). Resolve the apparent contradiction between this fact and the last sentence in Example 5. FIGURE 1.2.9 Graph for Problem 37 y x 5 25 5 FIGURE 1.2.8 Graph for Problem 36 y x 5 25 5 FIGURE 1.2.10 Graph for Problem 38 x5 y 5 25 . INITIALVALUE PROBLEMS 21 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Express assumptions in terms of DEs Display predictions of model (e.g., graphically) Solve the DEs If necessary, alter assumptions or increase resolution of model Assumptions and hypotheses Mathematical formulation Obtain solutions Check model predictions with known facts FIGURE 1.3.1 Steps in the modeling process with differential equations INTRODUCTION In this section we introduce the notion of a differential equation as a mathematical model and discuss some speci�c models in biology, chemistry, and physics. Once we have studied some methods for solving differential equations in Chapters 2 and 4, we return to and solve some of these models in Chapters 3 and 5. MATHEMATICAL MODELS It is often desirable to describe the behavior of some reallife system or a phenomenon—whether physical, sociological, or even economic—in mathematical terms. The mathematical description of a system or a phenomenon is called a mathematical model and is constructed with certain goals in mind. For example, we may wish to understand the mechanisms of a certain eco system by studying the growth of animal populations in that system, or we may wish to date fossils by means of analyzing the decay of a radioactive substance, either in the fossil or in the stratum in which it was discovered. Construction of a mathematical model of a system starts with (i) identi�cation of the variables that are responsible for changing the sys tem. We may choose not to incorporate all these variables into the model at �rst. In this step we are specifying the level of resolution of the model. Next (ii) we make a set of reasonable assumptions, or hypotheses, about the system we are trying to describe. These assumptions will also include any empirical laws that may be applicable to the system. For some purposes it may be perfectly within reason to be content with low resolution models. For example, you may already be aware that in beginning physics courses, the retarding force of air friction is sometimes ignored in modeling the motion of a body falling near the surface of the Earth, but if you are a scientist whose job it is to accurately predict the �ight path of a longrange projectile, you have to take into account air resistance and other factors such as the curvature of the Earth. Since the assumptions made about a system frequently involve a rate of change of one or more of the variables, the mathematical depiction of all these assumptions may be one or more equations involving derivatives. In other words, the mathemati cal model may be a differential equation or a system of differential equations. Once we have formulated a mathematical model that is either a differential equa tion or a system of differential equations, we are faced with the not insigni�cant problem of trying to solve it. If we can solve it, then we deem the model to be reasonIf we can solve it, then we deem the model to be reasonIf  able if its solution is consistent with either experimental data or known facts about the behavior of the system. But if the predictions produced by the solution are poor, we can either increase the level of resolution of the model or make alternative as sumptions about the mechanisms for change in the system. The steps of the model ing process are then repeated, as shown in the diagram in Figure 1.3.1. 1.3 Differential Equations as Mathematical Models 22 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 *If two quantities u and v are proportional, we write u ~ v. This means that one quantity is a constant multiple of the other: u = kv. Of course, by increasing the resolution, we add to the complexity of the mathemati cal model and increase the likelihood that we cannot obtain an explicit solution of the differential equation. A mathematical model of a physical system will often involve the variable time t. A solution of the model then gives the state of the system; in other words, the values of the dependent variable (or variables) for appropriate values of t describe the system t describe the system t in the past, present, and future. POPULATION DYNAMICS One of the earliest attempts to model human population growth by means of mathematics was by the English clergyman and economist Thomas Malthus (1766–1834) in 1798. Basically, the idea behind the Malthusian model is the assumption that the rate at which the population of a country grows at a certain time is proportional* to the total population of the country at that time. In other words, the more people there are at time t, the more there are going to be in the future. In mathematical terms, if P(t) denotes the total population at time t, then this assumption can be expressed as dPdPd dt ~ P or dP dt 5 kP, (1) where k is a constant of proportionality. This simple model, which fails to take into k is a constant of proportionality. This simple model, which fails to take into k account many factors that can in�uence human populations to either grow or decline (immigration and emigration, for example), nevertheless turned out to be fairly accu rate in predicting the population of the United States during the years 1790–1860. Populations that grow at a rate described by (1) are rare; nevertheless, (1) is still used to model growth of small populations over short intervals of time (bacteria growing in a petri dish, for example). RADIOACTIVE DECAY The nucleus of an atom consists of combinations of protons and neutrons. Many of these combinations of protons and neutrons are unstable—that is, the atoms decay or transmute into atoms of another substance. Such nuclei are said to be radioactive. For example, over time the highly radioactive radium, Ra226, transmutes into the radioactive gas radon, Rn222. To model the phenomenon of radioactive decay, it is assumed that the rate dAydt at which the nuclei of a subdt at which the nuclei of a subdt  stance decay is proportional to the amount (more precisely, the number of nuclei) A(t) of the substance remaining at time t: dAdAd dt ~ A or dA dt 5 kA. (2) Of course, equations (1) and (2) are exactly the same; the difference is only in the in terpretation of the symbols and the constants of proportionality. For growth, as we expect in (1), k > 0, and for decay, as in (2), k < 0. The model (1) for growth can also be seen as the equation dSydt = rS, which describes the growth of capital S when an annual rate of interest S when an annual rate of interest S r is compounded r is compounded r continuously. The model (2) for decay also occurs in biological applications such as determining the halflife of a drug—the time that it takes for 50% of a drug to be eliminated from a body by excretion or metabolism. In chemistry the decay model (2) appears in the mathematical description of a �rstorder chemical reaction. The point is this: A single differential equation can serve as a mathematical model for many different phenomena. . DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 23 Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 Mathematical models are often accompanied by certain side conditions. For ex ample, in (1) and (2) we would expect to know, in turn, the initial population P0 and the initial amount of radioactive substance A0 on hand. If the initial point in time is taken to be t = 0, then we know that P(0) = P0 and A(0) = A0. In other words, a mathematical model can consist of either an initialvalue problem or, as we shall see later on in Section 5.2, a boundaryvalue problem. NEWTON’S LAW OF COOLING/WARMING According to Newton’s empirical law of cooling/warming, the rate at which the temperature of a body changes is pro portional to the difference between the temperature of the body and the temperature of the surrounding medium, the socalled ambient temperature. If T(T(T t) represents the temperature of a body at time t, TmTmT the temperature of the surrounding medium, and dTydTydT dt the rate at which the temperature of the body changes, then Newton’s law of dt the rate at which the temperature of the body changes, then Newton’s law of dt cooling/warming translates into the mathematical statement dT dt ~ T 2 TmTmT or dT dt 5 k(T 2 Tm), (3) where k is a constant of proportionality. In either case, cooling or warming, if k is a constant of proportionality. In either case, cooling or warming, if k TmTmT is a constant, it stands to reason that k < 0. SPREAD OF A DISEASE A contagious disease—for example, a �u virus—is spread throughout a community by people coming into contact with other people. Let x(t) denote the number of people who have contracted the disease and y(t) denote the number of people who have not yet been exposed. It seems reasonable to assume that the rate dxydt at which the disease spreads is proportional to the number of encoundt at which the disease spreads is proportional to the number of encoundt  ters, or interactions, between these two groups of people. If we assume that the num ber of interactions is jointly proportional to x(t) and y(t)—that is, proportional to the product xy—then dxdxd dt 5 kxy, (4) where k is the usual constant of proportionality. Suppose a small community has a k is the usual constant of proportionality. Suppose a small community has a k �xed population of n people. If one infected person is introduced into this commu nity, then it could be argued that x(t) and y(t) are related by x + y = n + 1. Using this last equation to eliminate y in (4) gives us the model dx dt 5 kx(n 1 1 2 x). (5) An obvious initial condition accompanying equation (5) is x(0) = 1. CHEMICAL REACTIONS The disintegration of a radioactive substance, governed by the differential equation (1), is said to be a �rstorder reaction. In chemistry a few reactions follow this same empirical law: If the molecules of substance A decompose into smaller molecules, it is a natural assumption that the rate at which this decomposition takes place is proportional to the amount of the �rst substance that has not undergone conversion; that is, if X(t) is the amount of substance A remaining at any time, then dXydt = kX, where k is a negative constant since k is a negative constant since k X is X is X decreasing. An example of a �rstorder chemical reaction is the conversion of tbutyl chloride, (CH3)3CCl, into tbutyl alcohol, (CH3)3COH: (CH3)3CCl 1 NaOH S (CH3)3COH 1 NaCl. Only the concentration of the tbutyl chloride controls the rate of reaction. But in the reaction CH3Cl 1 NaOH S CH3OH 1 NaCl one molecule of sodium hydroxide, NaOH, is consumed for every molecule of methyl chloride, CH3Cl, thus forming one molecule of methyl alcohol, CH3OH, and 24 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02200203 input rate of brine 3 gal/min output rate of brine 3 gal/min constant 300 gal FIGURE 1.3.2 Mixing tank *Don’t confuse these symbols with Rin and Rout, which are input and output rates of salt. one molecule of sodium chloride, NaCl. In this case the rate at which the reaction proceeds is proportional to the product of the remaining concentrations of CH3Cl and NaOH. To describe this second reaction in general, let us suppose one molecule of a substance A combines with one molecule of a substance B to form one molecule of a substance C. If X denotes the amount of chemical X denotes the amount of chemical X C formed at time C formed at time C t and if t and if t a and b are, in turn, the amounts of the two chemicals A and B at t = 0 (the initial amounts), then the instantaneous amounts of A and B not converted to chemical C are C are C a − X and b − X, respectively. Hence the rate of formation of C is given byC is given byC dX dt 5 k(� 2 X)(� 2 X), (6) where k is a constant of proportionality. A reaction whose model is equation (6) is k is a constant of proportionality. A reaction whose model is equation (6) is k said to be a secondorder reaction. MIXTURES The mixing of two salt solutions of differing concentrations gives rise to a �rstorder differential equation for the amount of salt contained in the mix ture. Let us suppose that a large mixing tank initially holds 300 gallons of brine (that is, water in which a certain number of pounds of salt has been dissolved). Another brine solution is pumped into the large tank at a rate of 3 gallons per minute; the concentration of the salt in this in�ow is 2 pounds per gallon. When the solution in the tank is well stirred, it is pumped out at the same rate as the entering solution. See Figure 1.3.2. If A(t) denotes the amount of salt (measured in pounds) in the tank at time t, then the rate at which A(t) changes is a net rate: dA dt 5 1input rateof salt 2 2 1 output rate of salt 2 5 Rin 2 Rout. (7) The input rate Rin at which salt enters the tank is the product of the in�ow concentra tion of salt and the in�ow rate of �uid. Note that Rin is measured in pounds per minute: concentration of salt in in�ow input rate of brine input rate of salt Rin 5 (2 lb/gal) ? (3 gal/min) 5 (6 lb/min). Now, since the solution is being pumped out of the tank at the same rate that it is pumped in, the number of gallons of brine in the tank at time t is a constant t is a constant t 300 gallons. Hence the concentration of the salt in the tank as well as in the
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