Advanced Engineering Mathematics 10th Edition (1), Notas de estudo de Engenharia Mecânica

Baixe Advanced Engineering Mathematics 10th Edition (1) e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity! ETR Sp NE: Am ERWIN KREYSZIG Systems of Units. Some Important Conversion Factors The most important systems of units are shown in the table below. The mks system is also known as the International System of Units (abbreviated SI), and the abbreviations sec (instead of s), gm (instead of g), and nt (instead of N) are also used. System of units Length Mass Time Force cgs system centimeter (cm) gram (g) second (s) dyne mks system meter (m) kilogram (kg) second (s) newton (nt) Engineering system foot (ft) slug second (s) pound (lb) 1 inch (in.)  2.540000 cm 1 foot (ft)  12 in.  30.480000 cm 1 yard (yd)  3 ft  91.440000 cm 1 statute mile (mi)  5280 ft  1.609344 km 1 nautical mile  6080 ft  1.853184 km 1 acre  4840 yd2  4046.8564 m2 1 mi2  640 acres  2.5899881 km2 1 fluid ounce  1/128 U.S. gallon  231/128 in.3  29.573730 cm3 1 U.S. gallon  4 quarts (liq)  8 pints (liq)  128 fl oz  3785.4118 cm3 1 British Imperial and Canadian gallon  1.200949 U.S. gallons  4546.087 cm3 1 slug  14.59390 kg 1 pound (lb)  4.448444 nt 1 newton (nt)  105 dynes 1 British thermal unit (Btu)  1054.35 joules 1 joule  107 ergs 1 calorie (cal)  4.1840 joules 1 kilowatt-hour (kWh)  3414.4 Btu  3.6 • 106 joules 1 horsepower (hp)  2542.48 Btu/h  178.298 cal/sec  0.74570 kW 1 kilowatt (kW)  1000 watts  3414.43 Btu/h  238.662 cal/s °F  °C • 1.8  32 1°  60  3600  0.017453293 radian For further details see, for example, D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics. 9th ed., Hoboken, N. J: Wiley, 2011. See also AN American National Standard, ASTM/IEEE Standard Metric Practice, Institute of Electrical and Electronics Engineers, Inc. (IEEE), 445 Hoes Lane, Piscataway, N. J. 08854, website at www.ieee.org. fendpaper.qxd 11/4/10 12:05 PM Page 2 PLUS ncia www-wileyplus.com WileyPLUS is an innovative, research-based, online environment for effective teaching and learning. WileyPLUS... «.-motivates students with «supports instructors with confidence-boosting reliable resources that feedback and proof of reinforce course goals progress, 24/7. inside and ouiside of the classroom. IA [CS Textbook & Resources WILEY O PLUS Ce www.wileyplus.com ALL THE HELP, RESOURCES, AND PERSONAL SUPPORT YOU AND YOUR STUDENTS NEED! Wwww.wileyplus.com/resources DAY OF Tt ciass «o AND BEYOND! 2-Minute Tutorials and all of the resources you & your students need to get started. PLUS QuickStart Pre-loaded, ready-to-use assignments and presentations. Created by subject matter experts. PLUS Student Partner Program Student support from an experienced student user. Technical Support 24/7 FAQs, online chat, and phone support. www.wileyplus.com/support RR FACULTY Eos Collaborate with your colleagues, find a mentor, attend virtual and live events, and view resources. www.WhereFacultyConnect.com Your WileyPLUS Account Manager. Personal training and implementation support. ADVANCED ENGINEERING MATHEMATICS ffirs.qxd 11/8/10 3:50 PM Page iii PUBLISHER Laurie Rosatone PROJECT EDITOR Shannon Corliss MARKETING MANAGER Jonathan Cottrell CONTENT MANAGER Lucille Buonocore PRODUCTION EDITOR Barbara Russiello MEDIA EDITOR Melissa Edwards MEDIA PRODUCTION SPECIALIST Lisa Sabatini TEXT AND COVER DESIGN Madelyn Lesure PHOTO RESEARCHER Sheena Goldstein COVER PHOTO © Denis Jr. Tangney/iStockphoto Cover photo shows the Zakim Bunker Hill Memorial Bridge in Boston, MA. This book was set in Times Roman. The book was composed by PreMedia Global, and printed and bound by RR Donnelley & Sons Company, Jefferson City, MO. The cover was printed by RR Donnelley & Sons Company, Jefferson City, MO. This book is printed on acid free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. 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Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at: www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. ISBN 978-0-470-45836-5 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1  ffirs.qxd 11/4/10 10:50 AM Page vi P R E F A C E See also http://www.wiley.com/college/kreyszig Purpose and Structure of the Book This book provides a comprehensive, thorough, and up-to-date treatment of engineering mathematics. It is intended to introduce students of engineering, physics, mathematics, computer science, and related fields to those areas of applied mathematics that are most relevant for solving practical problems. A course in elementary calculus is the sole prerequisite. (However, a concise refresher of basic calculus for the student is included on the inside cover and in Appendix 3.) The subject matter is arranged into seven parts as follows: A. Ordinary Differential Equations (ODEs) in Chapters 1–6 B. Linear Algebra. Vector Calculus. See Chapters 7–10 C. Fourier Analysis. Partial Differential Equations (PDEs). See Chapters 11 and 12 D. Complex Analysis in Chapters 13–18 E. Numeric Analysis in Chapters 19–21 F. Optimization, Graphs in Chapters 22 and 23 G. Probability, Statistics in Chapters 24 and 25. These are followed by five appendices: 1. References, 2. Answers to Odd-Numbered Problems, 3. Auxiliary Materials (see also inside covers of book), 4. Additional Proofs, 5. Table of Functions. This is shown in a block diagram on the next page. The parts of the book are kept independent. In addition, individual chapters are kept as independent as possible. (If so needed, any prerequisites—to the level of individual sections of prior chapters—are clearly stated at the opening of each chapter.) We give the instructor maximum flexibility in selecting the material and tailoring it to his or her need. The book has helped to pave the way for the present development of engineering mathematics. This new edition will prepare the student for the current tasks and the future by a modern approach to the areas listed above. We provide the material and learning tools for the students to get a good foundation of engineering mathematics that will help them in their careers and in further studies. General Features of the Book Include: • Simplicity of examples to make the book teachable—why choose complicated examples when simple ones are as instructive or even better? • Independence of parts and blocks of chapters to provide flexibility in tailoring courses to specific needs. • Self-contained presentation, except for a few clearly marked places where a proof would exceed the level of the book and a reference is given instead. • Gradual increase in difficulty of material with no jumps or gaps to ensure an enjoyable teaching and learning experience. • Modern standard notation to help students with other courses, modern books, and journals in mathematics, engineering, statistics, physics, computer science, and others. Furthermore, we designed the book to be a single, self-contained, authoritative, and convenient source for studying and teaching applied mathematics, eliminating the need for time-consuming searches on the Internet or time-consuming trips to the library to get a particular reference book. vii fpref.qxd 11/8/10 3:16 PM Page vii viii Preface GUIDES AND MANUALS Maple Computer Guide Mathematica Computer Guide Student Solutions Manual and Study Guide Instructor’s Manual PART A Chaps. 1–6 Ordinary Differential Equations (ODEs) Chaps. 1–4 Basic Material Chap. 5 Chap. 6 Series Solutions Laplace Transforms PART B Chaps. 7–10 Linear Algebra. Vector Calculus Chap. 7 Chap. 9 Matrices, Vector Differential Linear Systems Calculus Chap. 8 Chap. 10 Eigenvalue Problems Vector Integral Calculus PARTS AND CHAPTERS OF THE BOOK PART C Chaps. 11–12 Fourier Analysis. Partial Differential Equations (PDEs) Chap. 11 Fourier Analysis Chap. 12 Partial Differential Equations PART D Chaps. 13–18 Complex Analysis, Potential Theory Chaps. 13–17 Basic Material Chap. 18 Potential Theory PART E Chaps. 19–21 Numeric Analysis Chap. 19 Chap. 20 Chap. 21 Numerics in Numeric Numerics for General Linear Algebra ODEs and PDEs PART F Chaps. 22–23 Optimization, Graphs Chap. 22 Chap. 23 Linear Programming Graphs, Optimization PART G Chaps. 24–25 Probability, Statistics Chap. 24 Data Analysis. Probability Theory Chap. 25 Mathematical Statistics fpref.qxd 11/8/10 3:16 PM Page viii to the field of engineering mathematics. It should encourage the students to be creative in their own interests and careers and perhaps also to make contributions to engineering mathematics. Further Changes and New Features • Parts of Chap. 1 on first-order ODEs are rewritten. More emphasis on modeling, also new block diagram explaining this concept in Sec. 1.1. Early introduction of Euler’s method in Sec. 1.2 to familiarize student with basic numerics. More examples of separable ODEs in Sec. 1.3. • For Chap. 2, on second-order ODEs, note the following changes: For ease of reading, the first part of Sec. 2.4, which deals with setting up the mass-spring system, has been rewritten; also some rewriting in Sec. 2.5 on the Euler–Cauchy equation. • Substantially shortened Chap. 5, Series Solutions of ODEs. Special Functions: combined Secs. 5.1 and 5.2 into one section called “Power Series Method,” shortened material in Sec. 5.4 Bessel’s Equation (of the first kind), removed Sec. 5.7 (Sturm–Liouville Problems) and Sec. 5.8 (Orthogonal Eigenfunction Expansions) and moved material into Chap. 11 (see “Major Changes” above). • New equivalent definition of basis (Sec. 7.4). • In Sec. 7.9, completely new part on composition of linear transformations with two new examples. Also, more detailed explanation of the role of axioms, in connection with the definition of vector space. • New table of orientation (opening of Chap. 8 “Linear Algebra: Matrix Eigenvalue Problems”) where eigenvalue problems occur in the book. More intuitive explanation of what an eigenvalue is at the begining of Sec. 8.1. • Better definition of cross product (in vector differential calculus) by properly identifying the degenerate case (in Sec. 9.3). • Chap. 11 on Fourier Analysis extensively rearranged: Secs. 11.2 and 11.3 combined into one section (Sec. 11.2), old Sec. 11.4 on complex Fourier Series removed and new Secs. 11.5 (Sturm–Liouville Problems) and 11.6 (Orthogonal Series) put in (see “Major Changes” above). New problems (new!) in problem set 11.9 on discrete Fourier transform. • New section 12.5 on modeling heat flow from a body in space by setting up the heat equation. Modeling PDEs is more difficult so we separated the modeling process from the solving process (in Sec. 12.6). • Introduction to Numerics rewritten for greater clarity and better presentation; new Example 1 on how to round a number. Sec. 19.3 on interpolation shortened by removing the less important central difference formula and giving a reference instead. • Large new footnote with historical details in Sec. 22.3, honoring George Dantzig, the inventor of the simplex method. • Traveling salesman problem now described better as a “difficult” problem, typical of combinatorial optimization (in Sec. 23.2). More careful explanation on how to compute the capacity of a cut set in Sec. 23.6 (Flows on Networks). • In Chap. 24, material on data representation and characterization restructured in terms of five examples and enlarged to include empirical rule on distribution of Preface xi fpref.qxd 11/8/10 3:16 PM Page xi xii Preface data, outliers, and the z-score (Sec. 24.1). Furthermore, new example on encription (Sec. 24.4). • Lists of software for numerics (Part E) and statistics (Part G) updated. • References in Appendix 1 updated to include new editions and some references to websites. Use of Computers The presentation in this book is adaptable to various degrees of use of software, Computer Algebra Systems (CAS’s), or programmable graphic calculators, ranging from no use, very little use, medium use, to intensive use of such technology. The choice of how much computer content the course should have is left up to the instructor, thereby exhibiting our philosophy of maximum flexibility and adaptability. And, no matter what the instructor decides, there will be no gaps or jumps in the text or problem set. Some problems are clearly designed as routine and drill exercises and should be solved by hand (paper and pencil, or typing on your computer). Other problems require more thinking and can also be solved without computers. Then there are problems where the computer can give the student a hand. And finally, the book has CAS projects, CAS problems and CAS experiments, which do require a computer, and show its power in solving problems that are difficult or impossible to access otherwise. Here our goal is to combine intelligent computer use with high-quality mathematics. The computer invites visualization, experimentation, and independent discovery work. In summary, the high degree of flexibility of computer use for the book is possible since there are plenty of problems to choose from and the CAS problems can be omitted if desired. Note that information on software (what is available and where to order it) is at the beginning of Part E on Numeric Analysis and Part G on Probability and Statistics. Since Maple and Mathematica are popular Computer Algebra Systems, there are two computer guides available that are specifically tailored to Advanced Engineering Mathematics: E. Kreyszig and E.J. Norminton, Maple Computer Guide, 10th Edition and Mathematica Computer Guide, 10th Edition. Their use is completely optional as the text in the book is written without the guides in mind. Suggestions for Courses: A Four-Semester Sequence The material, when taken in sequence, is suitable for four consecutive semester courses, meeting 3 to 4 hours a week: 1st Semester ODEs (Chaps. 1–5 or 1–6) 2nd Semester Linear Algebra. Vector Analysis (Chaps. 7–10) 3rd Semester Complex Analysis (Chaps. 13–18) 4th Semester Numeric Methods (Chaps. 19–21) Suggestions for Independent One-Semester Courses The book is also suitable for various independent one-semester courses meeting 3 hours a week. For instance, Introduction to ODEs (Chaps. 1–2, 21.1) Laplace Transforms (Chap. 6) Matrices and Linear Systems (Chaps. 7–8) fpref.qxd 11/8/10 3:16 PM Page xii Vector Algebra and Calculus (Chaps. 9–10) Fourier Series and PDEs (Chaps. 11–12, Secs. 21.4–21.7) Introduction to Complex Analysis (Chaps. 13–17) Numeric Analysis (Chaps. 19, 21) Numeric Linear Algebra (Chap. 20) Optimization (Chaps. 22–23) Graphs and Combinatorial Optimization (Chap. 23) Probability and Statistics (Chaps. 24–25) Acknowledgments We are indebted to former teachers, colleagues, and students who helped us directly or indirectly in preparing this book, in particular this new edition. We profited greatly from discussions with engineers, physicists, mathematicians, computer scientists, and others, and from their written comments. We would like to mention in particular Professors Y. A. Antipov, R. Belinski, S. L. Campbell, R. Carr, P. L. Chambré, Isabel F. Cruz, Z. Davis, D. Dicker, L. D. Drager, D. Ellis, W. Fox, A. Goriely, R. B. Guenther, J. B. Handley, N. Harbertson, A. Hassen, V. W. Howe, H. Kuhn, K. Millet, J. D. Moore, W. D. Munroe, A. Nadim, B. S. Ng, J. N. Ong, P. J. Pritchard, W. O. Ray, L. F. Shampine, H. L. Smith, Roberto Tamassia, A. L. Villone, H. J. Weiss, A. Wilansky, Neil M. Wigley, and L. Ying; Maria E. and Jorge A. Miranda, JD, all from the United States; Professors Wayne H. Enright, Francis. L. Lemire, James J. Little, David G. Lowe, Gerry McPhail, Theodore S. Norvell, and R. Vaillancourt; Jeff Seiler and David Stanley, all from Canada; and Professor Eugen Eichhorn, Gisela Heckler, Dr. Gunnar Schroeder, and Wiltrud Stiefenhofer from Europe. Furthermore, we would like to thank Professors John B. Donaldson, Bruce C. N. Greenwald, Jonathan L. Gross, Morris B. Holbrook, John R. Kender, and Bernd Schmitt; and Nicholaiv Villalobos, all from Columbia University, New York; as well as Dr. Pearl Chang, Chris Gee, Mike Hale, Joshua Jayasingh, MD, David Kahr, Mike Lee, R. Richard Royce, Elaine Schattner, MD, Raheel Siddiqui, Robert Sullivan, MD, Nancy Veit, and Ana M. Kreyszig, JD, all from New York City. We would also like to gratefully acknowledge the use of facilities at Carleton University, Ottawa, and Columbia University, New York. Furthermore we wish to thank John Wiley and Sons, in particular Publisher Laurie Rosatone, Editor Shannon Corliss, Production Editor Barbara Russiello, Media Editor Melissa Edwards, Text and Cover Designer Madelyn Lesure, and Photo Editor Sheena Goldstein for their great care and dedication in preparing this edition. In the same vein, we would also like to thank Beatrice Ruberto, copy editor and proofreader, WordCo, for the Index, and Joyce Franzen of PreMedia and those of PreMedia Global who typeset this edition. Suggestions of many readers worldwide were evaluated in preparing this edition. Further comments and suggestions for improving the book will be gratefully received. KREYSZIG Preface xiii fpref.qxd 11/8/10 8:51 PM Page xiii 5.3 Extended Power Series Method: Frobenius Method 180 5.4 Bessel’s Equation. Bessel Functions J(x) 187 5.5 Bessel Functions of the Y(x). General Solution 196 Chapter 5 Review Questions and Problems 200 Summary of Chapter 5 201 CHAPTER 6 Laplace Transforms 203 6.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting) 204 6.2 Transforms of Derivatives and Integrals. ODEs 211 6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting) 217 6.4 Short Impulses. Dirac’s Delta Function. Partial Fractions 225 6.5 Convolution. Integral Equations 232 6.6 Differentiation and Integration of Transforms. ODEs with Variable Coefficients 238 6.7 Systems of ODEs 242 6.8 Laplace Transform: General Formulas 248 6.9 Table of Laplace Transforms 249 Chapter 6 Review Questions and Problems 251 Summary of Chapter 6 253 P A R T B Linear Algebra. Vector Calculus 255 CHAPTER 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 256 7.1 Matrices, Vectors: Addition and Scalar Multiplication 257 7.2 Matrix Multiplication 263 7.3 Linear Systems of Equations. Gauss Elimination 272 7.4 Linear Independence. Rank of a Matrix. Vector Space 282 7.5 Solutions of Linear Systems: Existence, Uniqueness 288 7.6 For Reference: Second- and Third-Order Determinants 291 7.7 Determinants. Cramer’s Rule 293 7.8 Inverse of a Matrix. Gauss–Jordan Elimination 301 7.9 Vector Spaces, Inner Product Spaces. Linear Transformations. Optional 309 Chapter 7 Review Questions and Problems 318 Summary of Chapter 7 320 CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems 322 8.1 The Matrix Eigenvalue Problem. Determining Eigenvalues and Eigenvectors 323 8.2 Some Applications of Eigenvalue Problems 329 8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 334 8.4 Eigenbases. Diagonalization. Quadratic Forms 339 8.5 Complex Matrices and Forms. Optional 346 Chapter 8 Review Questions and Problems 352 Summary of Chapter 8 353 xvi Contents ftoc.qxd 11/4/10 11:48 AM Page xvi CHAPTER 9 Vector Differential Calculus. Grad, Div, Curl 354 9.1 Vectors in 2-Space and 3-Space 354 9.2 Inner Product (Dot Product) 361 9.3 Vector Product (Cross Product) 368 9.4 Vector and Scalar Functions and Their Fields. Vector Calculus: Derivatives 375 9.5 Curves. Arc Length. Curvature. Torsion 381 9.6 Calculus Review: Functions of Several Variables. Optional 392 9.7 Gradient of a Scalar Field. Directional Derivative 395 9.8 Divergence of a Vector Field 402 9.9 Curl of a Vector Field 406 Chapter 9 Review Questions and Problems 409 Summary of Chapter 9 410 CHAPTER 10 Vector Integral Calculus. Integral Theorems 413 10.1 Line Integrals 413 10.2 Path Independence of Line Integrals 419 10.3 Calculus Review: Double Integrals. Optional 426 10.4 Green’s Theorem in the Plane 433 10.5 Surfaces for Surface Integrals 439 10.6 Surface Integrals 443 10.7 Triple Integrals. Divergence Theorem of Gauss 452 10.8 Further Applications of the Divergence Theorem 458 10.9 Stokes’s Theorem 463 Chapter 10 Review Questions and Problems 469 Summary of Chapter 10 470 P A R T C Fourier Analysis. Partial Differential Equations (PDEs) 473 CHAPTER 11 Fourier Analysis 474 11.1 Fourier Series 474 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions 483 11.3 Forced Oscillations 492 11.4 Approximation by Trigonometric Polynomials 495 11.5 Sturm–Liouville Problems. Orthogonal Functions 498 11.6 Orthogonal Series. Generalized Fourier Series 504 11.7 Fourier Integral 510 11.8 Fourier Cosine and Sine Transforms 518 11.9 Fourier Transform. Discrete and Fast Fourier Transforms 522 11.10 Tables of Transforms 534 Chapter 11 Review Questions and Problems 537 Summary of Chapter 11 538 CHAPTER 12 Partial Differential Equations (PDEs) 540 12.1 Basic Concepts of PDEs 540 12.2 Modeling: Vibrating String, Wave Equation 543 12.3 Solution by Separating Variables. Use of Fourier Series 545 12.4 D’Alembert’s Solution of the Wave Equation. Characteristics 553 12.5 Modeling: Heat Flow from a Body in Space. Heat Equation 557 Contents xvii ftoc.qxd 11/4/10 11:48 AM Page xvii 12.6 Heat Equation: Solution by Fourier Series. Steady Two-Dimensional Heat Problems. Dirichlet Problem 558 12.7 Heat Equation: Modeling Very Long Bars. Solution by Fourier Integrals and Transforms 568 12.8 Modeling: Membrane, Two-Dimensional Wave Equation 575 12.9 Rectangular Membrane. Double Fourier Series 577 12.10 Laplacian in Polar Coordinates. Circular Membrane. Fourier–Bessel Series 585 12.11 Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential 593 12.12 Solution of PDEs by Laplace Transforms 600 Chapter 12 Review Questions and Problems 603 Summary of Chapter 12 604 P A R T D Complex Analysis 607 CHAPTER 13 Complex Numbers and Functions. Complex Differentiation 608 13.1 Complex Numbers and Their Geometric Representation 608 13.2 Polar Form of Complex Numbers. Powers and Roots 613 13.3 Derivative. Analytic Function 619 13.4 Cauchy–Riemann Equations. Laplace’s Equation 625 13.5 Exponential Function 630 13.6 Trigonometric and Hyperbolic Functions. Euler’s Formula 633 13.7 Logarithm. General Power. Principal Value 636 Chapter 13 Review Questions and Problems 641 Summary of Chapter 13 641 CHAPTER 14 Complex Integration 643 14.1 Line Integral in the Complex Plane 643 14.2 Cauchy’s Integral Theorem 652 14.3 Cauchy’s Integral Formula 660 14.4 Derivatives of Analytic Functions 664 Chapter 14 Review Questions and Problems 668 Summary of Chapter 14 669 CHAPTER 15 Power Series, Taylor Series 671 15.1 Sequences, Series, Convergence Tests 671 15.2 Power Series 680 15.3 Functions Given by Power Series 685 15.4 Taylor and Maclaurin Series 690 15.5 Uniform Convergence. Optional 698 Chapter 15 Review Questions and Problems 706 Summary of Chapter 15 706 CHAPTER 16 Laurent Series. Residue Integration 708 16.1 Laurent Series 708 16.2 Singularities and Zeros. Infinity 715 16.3 Residue Integration Method 719 16.4 Residue Integration of Real Integrals 725 Chapter 16 Review Questions and Problems 733 Summary of Chapter 16 734 xviii Contents ftoc.qxd 11/4/10 11:48 AM Page xviii 25.4 Testing Hypotheses. Decisions 1077 25.5 Quality Control 1087 25.6 Acceptance Sampling 1092 25.7 Goodness of Fit. 2-Test 1096 25.8 Nonparametric Tests 1100 25.9 Regression. Fitting Straight Lines. Correlation 1103 Chapter 25 Review Questions and Problems 1111 Summary of Chapter 25 1112 APPENDIX 1 References A1 APPENDIX 2 Answers to Odd-Numbered Problems A4 APPENDIX 3 Auxiliary Material A63 A3.1 Formulas for Special Functions A63 A3.2 Partial Derivatives A69 A3.3 Sequences and Series A72 A3.4 Grad, Div, Curl, 2 in Curvilinear Coordinates A74 APPENDIX 4 Additional Proofs A77 APPENDIX 5 Tables A97 INDEX I1 PHOTO CREDITS P1 Contents xxi ftoc.qxd 11/4/10 11:48 AM Page xxi ftoc.qxd 11/4/10 11:48 AM Page xxii C H A P T E R 1 First-Order ODEs C H A P T E R 2 Second-Order Linear ODEs C H A P T E R 3 Higher Order Linear ODEs C H A P T E R 4 Systems of ODEs. Phase Plane. Qualitative Methods C H A P T E R 5 Series Solutions of ODEs. Special Functions C H A P T E R 6 Laplace Transforms Many physical laws and relations can be expressed mathematically in the form of differential equations. Thus it is natural that this book opens with the study of differential equations and their solutions. Indeed, many engineering problems appear as differential equations. The main objectives of Part A are twofold: the study of ordinary differential equations and their most important methods for solving them and the study of modeling. Ordinary differential equations (ODEs) are differential equations that depend on a single variable. The more difficult study of partial differential equations (PDEs), that is, differential equations that depend on several variables, is covered in Part C. Modeling is a crucial general process in engineering, physics, computer science, biology, medicine, environmental science, chemistry, economics, and other fields that translates a physical situation or some other observations into a “mathematical model.” Numerous examples from engineering (e.g., mixing problem), physics (e.g., Newton’s law of cooling), biology (e.g., Gompertz model), chemistry (e.g., radiocarbon dating), environmental science (e.g., population control), etc. shall be given, whereby this process is explained in detail, that is, how to set up the problems correctly in terms of differential equations. For those interested in solving ODEs numerically on the computer, look at Secs. 21.1–21.3 of Chapter 21 of Part F, that is, numeric methods for ODEs. These sections are kept independent by design of the other sections on numerics. This allows for the study of numerics for ODEs directly after Chap. 1 or 2. 1 P A R T A Ordinary Differential Equations (ODEs) c01.qxd 7/30/10 8:14 PM Page 1 are ordinary differential equations (ODEs). Here, as in calculus, denotes , etc. The term ordinary distinguishes them from partial differential equations (PDEs), which involve partial derivatives of an unknown function of two or more variables. For instance, a PDE with unknown function u of two variables x and y is PDEs have important engineering applications, but they are more complicated than ODEs; they will be considered in Chap. 12. An ODE is said to be of order n if the nth derivative of the unknown function y is the highest derivative of y in the equation. The concept of order gives a useful classification into ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of second order, and (3) of third order. In this chapter we shall consider first-order ODEs. Such equations contain only the first derivative and may contain y and any given functions of x. Hence we can write them as (4) or often in the form This is called the explicit form, in contrast to the implicit form (4). For instance, the implicit ODE (where ) can be written explicitly as Concept of Solution A function is called a solution of a given ODE (4) on some open interval if is defined and differentiable throughout the interval and is such that the equation becomes an identity if y and are replaced with h and , respectively. The curve (the graph) of h is called a solution curve. Here, open interval means that the endpoints a and b are not regarded as points belonging to the interval. Also, includes infinite intervals (the real line) as special cases. E X A M P L E 1 Verification of Solution Verify that (c an arbitrary constant) is a solution of the ODE for all Indeed, differentiate to get Multiply this by x, obtaining thus, the given ODE. xyr  y,xyr  c>x;yr  c>x2.y  c>x x  0.xyr  yy  c>x a  x  ,   x     x  b,a  x  b a  x  b hryr h(x)a  x  b y  h(x) yr  4x3y2.x  0x3yr  4y2  0 yr  f (x, y). F(x, y, yr)  0 yr 0 2u 0x2  0 2u 0y2  0. ys  d2y>dx2, dy>dxyr 4 CHAP. 1 First-Order ODEs c01.qxd 7/30/10 8:14 PM Page 4 E X A M P L E 2 Solution by Calculus. Solution Curves The ODE can be solved directly by integration on both sides. Indeed, using calculus, we obtain where c is an arbitrary constant. This is a family of solutions. Each value of c, for instance, 2.75 or 0 or gives one of these curves. Figure 3 shows some of them, for 1, 0, 1, 2, 3, 4. c  3, 2,8, y  cos x dx  sin x  c, yr  dy>dx  cos x SEC. 1.1 Basic Concepts. Modeling 5 y x0 –4 2ππ–π 4 2 –2 Fig. 3. Solutions of the ODE yr  cos xy  sin x  c 0 0.5 1.0 1.5 2.5 2.0 0 2 4 6 8 10 12 14 t y Fig. 4B. Solutions of in Example 3 (exponential decay) yr  0.2y 0 10 20 30 40 0 2 4 6 8 10 12 14 t y Fig. 4A. Solutions of in Example 3 (exponential growth) yr  0.2y E X A M P L E 3 (A) Exponential Growth. (B) Exponential Decay From calculus we know that has the derivative Hence y is a solution of (Fig. 4A). This ODE is of the form With positive-constant k it can model exponential growth, for instance, of colonies of bacteria or populations of animals. It also applies to humans for small populations in a large country (e.g., the United States in early times) and is then known as Malthus’s law.1 We shall say more about this topic in Sec. 1.5. (B) Similarly, (with a minus on the right) has the solution (Fig. 4B) modeling exponential decay, as, for instance, of a radioactive substance (see Example 5).  y  ce0.2t,yr  0.2 yr  ky.yr  0.2y yr  dy dt  0.2e0.2t  0.2y. y  ce0.2t 1Named after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766–1834). c01.qxd 7/30/10 8:14 PM Page 5 We see that each ODE in these examples has a solution that contains an arbitrary constant c. Such a solution containing an arbitrary constant c is called a general solution of the ODE. (We shall see that c is sometimes not completely arbitrary but must be restricted to some interval to avoid complex expressions in the solution.) We shall develop methods that will give general solutions uniquely (perhaps except for notation). Hence we shall say the general solution of a given ODE (instead of a general solution). Geometrically, the general solution of an ODE is a family of infinitely many solution curves, one for each value of the constant c. If we choose a specific c (e.g., or 0 or ) we obtain what is called a particular solution of the ODE. A particular solution does not contain any arbitrary constants. In most cases, general solutions exist, and every solution not containing an arbitrary constant is obtained as a particular solution by assigning a suitable value to c. Exceptions to these rules occur but are of minor interest in applications; see Prob. 16 in Problem Set 1.1. Initial Value Problem In most cases the unique solution of a given problem, hence a particular solution, is obtained from a general solution by an initial condition with given values and , that is used to determine a value of the arbitrary constant c. Geometrically this condition means that the solution curve should pass through the point in the xy-plane. An ODE, together with an initial condition, is called an initial value problem. Thus, if the ODE is explicit, the initial value problem is of the form (5) E X A M P L E 4 Initial Value Problem Solve the initial value problem Solution. The general solution is ; see Example 3. From this solution and the initial condition we obtain Hence the initial value problem has the solution . This is a particular solution. More on Modeling The general importance of modeling to the engineer and physicist was emphasized at the beginning of this section. We shall now consider a basic physical problem that will show the details of the typical steps of modeling. Step 1: the transition from the physical situation (the physical system) to its mathematical formulation (its mathematical model); Step 2: the solution by a mathematical method; and Step 3: the physical interpretation of the result. This may be the easiest way to obtain a first idea of the nature and purpose of differential equations and their applications. Realize at the outset that your computer (your CAS) may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work.  y(x)  5.7e3xy(0)  ce0  c  5.7. y(x)  ce3x y(0)  5.7.yr  dy dx  3y, y(x0)  y0.yr  f (x, y), yr  f (x, y), (x0, y0) y0x0 y(x0)  y0, 2.01 c  6.45 6 CHAP. 1 First-Order ODEs c01.qxd 7/30/10 8:14 PM Page 6 1.2 Geometric Meaning of Direction Fields, Euler’s Method A first-order ODE (1) has a simple geometric interpretation. From calculus you know that the derivative of is the slope of . Hence a solution curve of (1) that passes through a point must have, at that point, the slope equal to the value of f at that point; that is, Using this fact, we can develop graphic or numeric methods for obtaining approximate solutions of ODEs (1). This will lead to a better conceptual understanding of an ODE (1). Moreover, such methods are of practical importance since many ODEs have complicated solution formulas or no solution formulas at all, whereby numeric methods are needed. Graphic Method of Direction Fields. Practical Example Illustrated in Fig. 7. We can show directions of solution curves of a given ODE (1) by drawing short straight-line segments (lineal elements) in the xy-plane. This gives a direction field (or slope field) into which you can then fit (approximate) solution curves. This may reveal typical properties of the whole family of solutions. Figure 7 shows a direction field for the ODE (2) obtained by a CAS (Computer Algebra System) and some approximate solution curves fitted in. yr  y  x yr(x0)  f (x0, y0). yr(x0) (x0, y0)y(x)y(x) yr(x) yr  f (x, y) yr  f (x, y). SEC. 1.2 Geometric Meaning of y  ƒ(x, y). Direction Fields, Euler’s Method 9 1 2 0.5 1–0.5–1–1.5–2 –1 –2 y x Fig. 7. Direction field of with three approximate solution curves passing through (0, 1), (0, 0), (0, ), respectively1 yr  y  x, c01.qxd 7/30/10 8:15 PM Page 9 If you have no CAS, first draw a few level curves const of , then parallel lineal elements along each such curve (which is also called an isocline, meaning a curve of equal inclination), and finally draw approximation curves fit to the lineal elements. We shall now illustrate how numeric methods work by applying the simplest numeric method, that is Euler’s method, to an initial value problem involving ODE (2). First we give a brief description of Euler’s method. Numeric Method by Euler Given an ODE (1) and an initial value Euler’s method yields approximate solution values at equidistant x-values namely, (Fig. 8) , etc. In general, where the step h equals, e.g., 0.1 or 0.2 (as in Table 1.1) or a smaller value for greater accuracy. yn  yn1  hf (xn1, yn1) y2  y1  hf (x1, y1) y1  y0  hf (x0, y0) x0, x1  x0  h, x2  x0  2h, Á , y(x0)  y0, f (x, y)f (x, y)  10 CHAP. 1 First-Order ODEs y xx0 x1 y 0 y 1 y(x 1 ) Solution curve Error of y 1 hf (x 0 , y 0 ) h Fig. 8. First Euler step, showing a solution curve, its tangent at ( ), step h and increment in the formula for y1hf (x0, y0) x0, y0 Table 1.1 shows the computation of steps with step for the ODE (2) and initial condition corresponding to the middle curve in the direction field. We shall solve the ODE exactly in Sec. 1.5. For the time being, verify that the initial value problem has the solution . The solution curve and the values in Table 1.1 are shown in Fig. 9. These values are rather inaccurate. The errors are shown in Table 1.1 as well as in Fig. 9. Decreasing h would improve the values, but would soon require an impractical amount of computation. Much better methods of a similar nature will be discussed in Sec. 21.1. y(xn)  yn y  ex  x  1 y(0)  0, h  0.2n  5 c01.qxd 7/30/10 8:15 PM Page 10 Table 1.1. Euler method for for with step h  0.2x  0, Á , 1.0 yr  y  x, y (0)  0 SEC. 1.2 Geometric Meaning of y  ƒ(x, y). Direction Fields, Euler’s Method 11 0.7 0.5 0.3 0.1 0 0.2 0.4 0.6 0.8 1 x y Fig. 9. Euler method: Approximate values in Table 1.1 and solution curve n Error 0 0.0 0.000 0.000 0.000 1 0.2 0.000 0.021 0.021 2 0.4 0.04 0.092 0.052 3 0.6 0.128 0.222 0.094 4 0.8 0.274 0.426 0.152 5 1.0 0.488 0.718 0.230 y(xn)ynxn 1–8 DIRECTION FIELDS, SOLUTION CURVES Graph a direction field (by a CAS or by hand). In the field graph several solution curves by hand, particularly those passing through the given points . 1. 2. 3. 4. 5. 6. 7. 8. 9–10 ACCURACY OF DIRECTION FIELDS Direction fields are very useful because they can give you an impression of all solutions without solving the ODE, which may be difficult or even impossible. To get a feel for the accuracy of the method, graph a field, sketch solution curves in it, and compare them with the exact solutions. 9. 10. (Sol. ) 11. Autonomous ODE. This means an ODE not showing x (the independent variable) explicitly. (The ODEs in Probs. 6 and 10 are autonomous.) What will the level curves const (also called isoclines curvesf (x, y)  1y  52 x  cyr  5y1>2 yr  cos px yr  2xy, (0, 12), (0, 1), (0, 2) yr  ey>x, (2, 2), (3, 3) yr  sin2 y, (0, 0.4), (0, 1) yr  x  1>y, (1, 12) yr  2y  y2, (0, 0), (0, 1), (0, 2), (0, 3) yr  1  y2, (0, 0), (2, 12) yyr  4x  0, (1, 1), (0, 2) yr  1  y2, (14 p, 1) (x, y) of equal inclination) of an autonomous ODE look like? Give reason. 12–15 MOTIONS Model the motion of a body B on a straight line with velocity as given, being the distance of B from a point at time t. Graph a direction field of the model (the ODE). In the field sketch the solution curve satisfying the given initial condition. 12. Product of velocity times distance constant, equal to 2, 13. 14. Square of the distance plus square of the velocity equal to 1, initial distance 15. Parachutist. Two forces act on a parachutist, the attraction by the earth mg (m mass of person plus equipment, the acceleration of gravity) and the air resistance, assumed to be proportional to the square of the velocity v(t). Using Newton’s second law of motion (mass acceleration resultant of the forces), set up a model (an ODE for v(t)). Graph a direction field (choosing m and the constant of proportionality equal to 1). Assume that the parachute opens when v Graph the corresponding solution in the field. What is the limiting velocity? Would the parachute still be sufficient if the air resistance were only proportional to v(t)?  10 m>sec.  g  9.8 m>sec2  1>12 Distance  Velocity Time, y(1)  1 y(0)  2. y  0 y(t) P R O B L E M S E T 1 . 2 c01.qxd 7/30/10 8:15 PM Page 11 Next we use the half-life to determine k. When , half of the original substance is still present. Thus, Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed), Answer: About 5300 years ago. Other methods show that radiocarbon dating values are usually too small. According to recent research, this is due to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing. E X A M P L E 5 Mixing Problem Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model involving a single tank. The tank in Fig. 11 contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate of 10 gal min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at 10 gal min. Find the amount of salt in the tank at any time t. Solution. Step 1. Setting up a model. Let denote the amount of salt in the tank at time t. Its time rate of change is Balance law. 5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This is of the total brine content in the tank, hence 0.01 of the salt content , that is, 0.01 . Thus the model is the ODE (4) Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on both sides gives Initially the tank contains 100 lb of salt. Hence is the initial condition that will give the unique solution. Substituting and in the last equation gives Hence Hence the amount of salt in the tank at time t is (5) This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically that should increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE? The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35) or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flow rates (in and out) may be different and known only very roughly.  y(t) y(t)  5000  4900e0.01t. c  4900.100  5000  ce0  c.t  0y  100 y(0)  100 y  5000  ce0.01t.ln ƒ y  5000 ƒ  0.01t  c*, dy y  5000  0.01 dt, yr  50  0.01y  0.01(y  5000). y(t)y(t)( 1%) 10>1000  0.01 yr  Salt inflow rate  Salt outflow rate y(t) > >  t  ln 0.525 0.0001 213  5312.ekt  e0.0001 213t  0.525, k  ln 0.5 H   0.693 5715  0.0001 213.ekH  0.5,y0e kH  0.5y0, t  HH  5715 14 CHAP. 1 First-Order ODEs 100 2000 3000 1000 5000 4000 1000 300200 400 500 Salt content y(t) t Tank y Fig. 11. Mixing problem in Example 5 c01.qxd 7/30/10 8:15 PM Page 14 E X A M P L E 6 Heating an Office Building (Newton’s Law of Cooling3) Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heating is shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M. was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. What was the temperature inside the building when the heat was turned on at 6 A.M.? Physical information. Experiments show that the time rate of change of the temperature T of a body B (which conducts heat well, for example, as a copper ball does) is proportional to the difference between T and the temperature of the surrounding medium (Newton’s law of cooling). Solution. Step 1. Setting up a model. Let be the temperature inside the building and TA the outside temperature (assumed to be constant in Newton’s law). Then by Newton’s law, (6) Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if a model seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information. To see how good a model is, the engineer will collect experimental data and compare them with calculations from the model. Step 2. General solution. We cannot solve (6) because we do not know TA, just that it varied between 50°F and 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. We solve (6) with the unknown function TA replaced with the average of the two known values, or 45°F. For physical reasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M. For constant (or any other constant value) the ODE (6) is separable. Separation, integration, and taking exponents gives the general solution Step 3. Particular solution. We choose 10 P.M. to be Then the given initial condition is and yields a particular solution, call it . By substitution, Step 4. Determination of k. We use where is 2 A.M. Solving algebraically for k and inserting k into gives (Fig. 12) Tp(t)  45  25e 0.056t.k  14 ln 0.8  0.056,e 4k  0.8,Tp(4)  45  25e 4k  65, Tp(t) t  4T(4)  65, Tp(t)  45  25e kt.c  70  45  25,T(0)  45  ce0  70, Tp T(0)  70t  0. (c  ec * ).T(t)  45  cektln ƒ T  45 ƒ  kt  c*, dT T  45  k dt, TA  45 dT dt  k(T  TA). T(t) SEC. 1.3 Separable ODEs. Modeling 15 62 64 68 70 60 y 2 4 6 80 t 66 61 65 Fig. 12. Particular solution (temperature) in Example 6 3Sir ISAAC NEWTON (1642–1727), great English physicist and mathematician, became a professor at Cambridge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopher GOTTFRIED WILHELM LEIBNIZ (1646–1716) invented (independently) the differential and integral calculus. Newton discovered many basic physical laws and created the method of investigating physical problems by means of calculus. His Philosophiae naturalis principia mathematica (Mathematical Principles of Natural Philosophy, 1687) contains the development of classical mechanics. His work is of greatest importance to both mathematics and physics. c01.qxd 7/30/10 8:15 PM Page 15 Step 5. Answer and interpretation. 6 A.M. is (namely, 8 hours after 10 P.M.), and Hence the temperature in the building dropped 9°F, a result that looks reasonable. E X A M P L E 7 Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law) This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from a cylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank at any time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when the hole is opened is 2.25 m. When will the tank be empty? Physical information. Under the influence of gravity the outflowing water has velocity (7) (Torricelli’s law4), where is the height of the water above the hole at time t, and is the acceleration of gravity at the surface of the earth. Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in water level to the outflow. The volume of the outflow during a short time is (A Area of hole). must equal the change of the volume of the water in the tank. Now (B Cross-sectional area of tank) where is the decrease of the height of the water. The minus sign appears because the volume of the water in the tank decreases. Equating and gives We now express v according to Torricelli’s law and then let (the length of the time interval considered) approach 0—this is a standard way of obtaining an ODE as a model. That is, we have and by letting we obtain the ODE , where This is our model, a first-order ODE. Step 2. General solution. Our ODE is separable. is constant. Separation and integration gives and Dividing by 2 and squaring gives . Inserting yields the general solution h(t)  (c  0.000 332t)2. 13.28A>B  13.28  0.52p>1002p 0.000 332h  (c  13.28At>B)2 21h  c*  26.56 A B t. dh 1h  26.56 A B dt A>B 26.56  0.60022  980. dh dt  26.56 A B 1h ¢t : 0 ¢h ¢t   A B v   A B 0.60012gh(t) ¢t B ¢h  Av ¢t. ¢V*¢V h(t)¢h ( 0) ¢V*  B ¢h ¢V*¢V ¢V  Av ¢t ¢t¢V h(t) g  980 cm>sec2  32.17 ft>sec2h(t) v(t)  0.60022gh(t)  Tp(8)  45  25e 0.056  8  613°F4. t  8 16 CHAP. 1 First-Order ODEs 4EVANGELISTA TORRICELLI (1608–1647), Italian physicist, pupil and successor of GALILEO GALILEI (1564–1642) at Florence. The “contraction factor” 0.600 was introduced by J. C. BORDA in 1766 because the stream has a smaller cross section than the area of the hole. c01.qxd 7/30/10 8:15 PM Page 16 19–36 MODELING, APPLICATIONS 19. Exponential growth. If the growth rate of the number of bacteria at any time t is proportional to the number present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks? After 4 weeks? 20. Another population model. (a) If the birth rate and death rate of the number of bacteria are proportional to the number of bacteria present, what is the population as a function of time. (b) What is the limiting situation for increasing time? Interpret it. 21. Radiocarbon dating. What should be the content (in percent of ) of a fossilized tree that is claimed to be 3000 years old? (See Example 4.) 22. Linear accelerators are used in physics for accelerating charged particles. Suppose that an alpha particle enters an accelerator and undergoes a constant acceleration that increases the speed of the particle from to sec. Find the acceleration a and the distance traveled during that period of sec. 23. Boyle–Mariotte’s law for ideal gases.5 Experiments show for a gas at low pressure p (and constant temperature) the rate of change of the volume equals . Solve the model. 24. Mixing problem. A tank contains 400 gal of brine in which 100 lb of salt are dissolved. Fresh water runs into the tank at a rate of The mixture, kept practically uniform by stirring, runs out at the same rate. How much salt will there be in the tank at the end of 1 hour? 25. Newton’s law of cooling. A thermometer, reading 5°C, is brought into a room whose temperature is 22°C. One minute later the thermometer reading is 12°C. How long does it take until the reading is practically 22°C, say, 21.9°C? 26. Gompertz growth in tumors. The Gompertz model is , where is the mass of tumor cells at time t. The model agrees well with clinical observations. The declining growth rate with increasing corresponds to the fact that cells in the interior of a tumor may die because of insufficient oxygen and nutrients. Use the ODE to discuss the growth and decline of solutions (tumors) and to find constant solutions. Then solve the ODE. 27. Dryer. If a wet sheet in a dryer loses its moisture at a rate proportional to its moisture content, and if it loses half of its moisture during the first 10 min of y  1 y(t)yr  Ay ln y (A  0) 2 gal>min. V>p V(p) 103 104 m>sec in 103103 m>sec y0 14 6 C SEC. 1.3 Separable ODEs. Modeling 19 drying, when will it be practically dry, say, when will it have lost 99% of its moisture? First guess, then calculate. 28. Estimation. Could you see, practically without calcu- lation, that the answer in Prob. 27 must lie between 60 and 70 min? Explain. 29. Alibi? Jack, arrested when leaving a bar, claims that he has been inside for at least half an hour (which would provide him with an alibi). The police check the water temperature of his car (parked near the entrance of the bar) at the instant of arrest and again 30 min later, obtaining the values 190°F and 110°F, respectively. Do these results give Jack an alibi? (Solve by inspection.) 30. Rocket. A rocket is shot straight up from the earth, with a net acceleration ( acceleration by the rocket engine minus gravitational pullback) of during the initial stage of flight until the engine cut out at sec. How high will it go, air resistance neglected? 31. Solution curves of Show that any (nonvertical) straight line through the origin of the xy-plane intersects all these curves of a given ODE at the same angle. 32. Friction. If a body slides on a surface, it experiences friction F (a force against the direction of motion). Experiments show that (Coulomb’s6 law of kinetic friction without lubrication), where N is the normal force (force that holds the two surfaces together; see Fig. 15) and the constant of proportionality is called the coefficient of kinetic friction. In Fig. 15 assume that the body weighs 45 nt (about 10 lb; see front cover for conversion). (corresponding to steel on steel), the slide is 10 m long, the initial velocity is zero, and air resistance is negligible. Find the velocity of the body at the end of the slide. a  30°,   0.20  ƒ F ƒ   ƒ N ƒ yr  g1y>x2. t  10 7t m>sec2  5ROBERT BOYLE (1627–1691), English physicist and chemist, one of the founders of the Royal Society. EDME MARIOTTE (about 1620–1684), French physicist and prior of a monastry near Dijon. They found the law experimentally in 1662 and 1676, respectively. 6CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer. v(t) W N Body α s(t) Fig. 15. Problem 32 c01.qxd 7/30/10 8:15 PM Page 19 33. Rope. To tie a boat in a harbor, how many times must a rope be wound around a bollard (a vertical rough cylindrical post fixed on the ground) so that a man holding one end of the rope can resist a force exerted by the boat 1000 times greater than the man can exert? First guess. Experiments show that the change of the force S in a small portion of the rope is proportional to S and to the small angle in Fig. 16. Take the proportionality constant 0.15. The result should surprise you! ¢ ¢S 20 CHAP. 1 First-Order ODEs this as the condition for the two families to be orthogonal (i.e., to intersect at right angles)? Do your graphs confirm this? (e) Sketch families of curves of your own choice and find their ODEs. Can every family of curves be given by an ODE? 35. CAS PROJECT. Graphing Solutions. A CAS can usually graph solutions, even if they are integrals that cannot be evaluated by the usual analytical methods of calculus. (a) Show this for the five initial value problems , , graphing all five curves on the same axes. (b) Graph approximate solution curves, using the first few terms of the Maclaurin series (obtained by term- wise integration of that of ) and compare with the exact curves. (c) Repeat the work in (a) for another ODE and initial conditions of your own choice, leading to an integral that cannot be evaluated as indicated. 36. TEAM PROJECT. Torricelli’s Law. Suppose that the tank in Example 7 is hemispherical, of radius R, initially full of water, and has an outlet of 5 cm2 cross- sectional area at the bottom. (Make a sketch.) Set up the model for outflow. Indicate what portion of your work in Example 7 you can use (so that it can become part of the general method independent of the shape of the tank). Find the time t to empty the tank (a) for any R, (b) for Plot t as function of R. Find the time when (a) for any R, (b) for R  1 m. h  R>2 R  1 m. yr y(0)  0, 1, 2yr  ex 2 S + ΔS Δ S Small portion of rope Fig. 16. Problem 33 34. TEAM PROJECT. Family of Curves. A family of curves can often be characterized as the general solution of (a) Show that for the circles with center at the origin we get (b) Graph some of the hyperbolas Find an ODE for them. (c) Find an ODE for the straight lines through the origin. (d) You will see that the product of the right sides of the ODEs in (a) and (c) equals Do you recognize1. xy  c. yr  x>y. yr  f (x, y). 1.4 Exact ODEs. Integrating Factors We recall from calculus that if a function has continuous partial derivatives, its differential (also called its total differential) is From this it follows that if then For example, if , then or yr  dy dx   1  2xy3 3x2y2 , du  (1  2xy3) dx  3x2y2 dy  0 u  x  x2y3  c du  0.u(x, y)  c  const, du  0u 0x dx  0u 0y dy. u(x, y) c01.qxd 7/30/10 8:15 PM Page 20 an ODE that we can solve by going backward. This idea leads to a powerful solution method as follows. A first-order ODE written as (use as in Sec. 1.3) (1) is called an exact differential equation if the differential form is exact, that is, this form is the differential (2) of some function . Then (1) can be written By integration we immediately obtain the general solution of (1) in the form (3) This is called an implicit solution, in contrast to a solution as defined in Sec. 1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solution can be converted to explicit form. (Do this for ) If this is not possible, your CAS may graph a figure of the contour lines (3) of the function and help you in understanding the solution. Comparing (1) and (2), we see that (1) is an exact differential equation if there is some function such that (4) (a) (b) From this we can derive a formula for checking whether (1) is exact or not, as follows. Let M and N be continuous and have continuous first partial derivatives in a region in the xy-plane whose boundary is a closed curve without self-intersections. Then by partial differentiation of (4) (see App. 3.2 for notation), By the assumption of continuity the two second partial derivaties are equal. Thus (5) 0M 0y  0N 0x . 0N 0x  0 2u 0x 0y . 0M 0y  0 2u 0y 0x , 0u 0y  N. 0u 0x  M, u(x, y) u(x, y) x2  y2  1. y  h(x) u(x, y)  c. du  0. u(x, y) du  0u 0x dx  0u 0y dy M(x, y) dx  N(x, y) dy M(x, y) dx  N(x, y) dy  0 dy  yrdxM(x, y)  N(x, y)yr  0, SEC. 1.4 Exact ODEs. Integrating Factors 21 c01.qxd 7/30/10 8:15 PM Page 21 This example gives the idea. All we did was to multiply a given nonexact equation, say, (12) by a function F that, in general, will be a function of both x and y. The result was an equation (13) that is exact, so we can solve it as just discussed. Such a function is then called an integrating factor of (12). E X A M P L E 4 Integrating Factor The integrating factor in (11) is Hence in this case the exact equation (13) is Solution These are straight lines through the origin. (Note that is also a solution of ) It is remarkable that we can readily find other integrating factors for the equation namely, and because (14) How to Find Integrating Factors In simpler cases we may find integrating factors by inspection or perhaps after some trials, keeping (14) in mind. In the general case, the idea is the following. For the exactness condition (5) is Hence for (13), the exactness condition is (15) By the product rule, with subscripts denoting partial derivatives, this gives In the general case, this would be complicated and useless. So we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one—the result may be useful (and may also help you later on). Hence we look for an integrating factor depending only on one variable: fortunately, in many practical cases, there are such factors, as we shall see. Thus, let Then and so that (15) becomes Dividing by FQ and reshuffling terms, we have (16) where R  1 Q a 0P 0y  0Q 0x b .1 F dF dx  R, FPy  FrQ  FQx. Fx  Fr  dF>dx,Fy  0,F  F(x). FyP  FPy  FxQ  FQx. 0 0y (FP)  0 0x (FQ). FP dx  FQ dy  0, 0M>0y  0N>0x.M dx  N dy  0  y dx  x dy x2  y2  d aarctan y x b .y dx  x dy xy  d aln x y b ,y dx  x dy y2  d ax y b , 1>(x2  y2),1>y2, 1>(xy), y dx  x dy  0, y dx  x dy  0.x  0y  cx y x  c.FP dx  FQ dy  y dx  x dy x2  d ay x b  0. F  1>x2. F(x, y) FP dx  FQ dy  0 P(x, y) dx  Q(x, y) dy  0, 24 CHAP. 1 First-Order ODEs c01.qxd 7/30/10 8:15 PM Page 24 This proves the following theorem. T H E O R E M 1 Integrating Factor F (x) If (12) is such that the right side R of (16) depends only on x, then (12) has an integrating factor which is obtained by integrating (16) and taking exponents on both sides. (17) Similarly, if then instead of (16) we get (18) where and we have the companion T H E O R E M 2 Integrating Factor F* (y) If (12) is such that the right side R* of (18) depends only on y, then (12) has an integrating factor , which is obtained from (18) in the form (19) E X A M P L E 5 Application of Theorems 1 and 2. Initial Value Problem Using Theorem 1 or 2, find an integrating factor and solve the initial value problem (20) Solution. Step 1. Nonexactness. The exactness check fails: but Step 2. Integrating factor. General solution. Theorem 1 fails because R [the right side of (16)] depends on both x and y. Try Theorem 2. The right side of (18) is Hence (19) gives the integrating factor From this result and (20) you get the exact equation (ex  y) dx  (x  ey) dy  0. F*(y)  ey. R*  1 P a 0Q 0x  0P 0y b  1 exy  yey (ey  exy  ey  yey)  1. R  1 Q a 0P 0y  0Q 0x b  1 xey  1 (exy  ey  yey  ey). 0Q 0x  0 0x (xey  1)  ey. 0P 0y  0 0y (exy  yey)  exy  ey  yey y(0)  1(exy  yey) dx  (xey  1) dy  0, F*(y)  expR*(y) dy. F*  F*(y) R*  1 P a 0Q 0x  0P 0y b1 F* dF* dy  R*, F*  F*(y), F(x)  expR(x) dx. F  F(x), SEC. 1.4 Exact ODEs. Integrating Factors 25 c01.qxd 7/30/10 8:15 PM Page 25 Test for exactness; you will get 1 on both sides of the exactness condition. By integration, using (4a), Differentiate this with respect to y and use (4b) to get Hence the general solution is Setp 3. Particular solution. The initial condition gives Hence the answer is Figure 18 shows several particular solutions obtained as level curves of obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast a solution into explicit form. Note the curve that (nearly) satisfies the initial condition. Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initial condition.  u(x, y)  c, ex  xy  ey  1  e  3.72. u(0, 1)  1  0  e  3.72.y(0)  1 u(x, y)  ex  xy  ey  c. k  ey  c*. dk dy  ey, 0u 0y  x  dk dy  N  x  ey, u  (ex  y) dx  ex  xy  k(y). 26 CHAP. 1 First-Order ODEs y x0–1–2–3 1 3 1 2 3 –1 –2 –3 2 Fig. 18. Particular solutions in Example 5 1–14 ODEs. INTEGRATING FACTORS Test for exactness. If exact, solve. If not, use an integrating factor as given or obtained by inspection or by the theorems in the text. Also, if an initial condition is given, find the corresponding particular solution. 1. 2. 3. 4. 5. 6. 7. 2x tan y dx  sec2 y dy  0 3(y  1) dx  2x dy, (y  1)x4 (x2  y2) dx  2xy dy  0 e3u(dr  3r du)  0 sin x cos y dx  cos x sin y dy  0 x3dx  y3dy  0 2xy dx  x2 dy  0 8. 9. 10. 11. 2 cosh x cos y 12. 13. 14. 15. Exactness. Under what conditions for the constants a, b, k, l is exact? Solve the exact ODE. (ax  by) dx  (kx  ly) dy  0 F  xayb (a  1)y dx  (b  1)x dy  0, y(1)  1, ey dx  ex(ey  1) dy  0, F  exy (2xy dx  dy)ex 2  0, y(0)  2 dx  sinh x sin y dy y dx  3y  tan (x  y)4 dy  0, cos (x  y) e2x(2 cos y dx  sin y dy)  0, y(0)  0 ex(cos y dx  sin y dy)  0 P R O B L E M S E T 1 . 4 c01.qxd 7/30/10 8:15 PM Page 26 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 29 This reduces solving (1) to the generally simpler task of evaluating integrals. For ODEs for which this is still difficult, you may have to use a numeric method for integrals from Sec. 19.5 or for the ODE itself from Sec. 21.1. We mention that h has nothing to do with in Sec. 1.1 and that the constant of integration in h does not matter; see Prob. 2. The structure of (4) is interesting. The only quantity depending on a given initial condition is c. Accordingly, writing (4) as a sum of two terms, (4*) we see the following: (5) E X A M P L E 1 First-Order ODE, General Solution, Initial Value Problem Solve the initial value problem Solution. Here and From this we see that in (4), and the general solution of our equation is From this and the initial condition, thus and the solution of our initial value problem is Here 3 cos x is the response to the initial data, and is the response to the input sin 2x. E X A M P L E 2 Electric Circuit Model the RL-circuit in Fig. 19 and solve the resulting ODE for the current A (amperes), where t is time. Assume that the circuit contains as an EMF (electromotive force) a battery of V (volts), which is constant, a resistor of (ohms), and an inductor of H (henrys), and that the current is initially zero. Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm’s law) and a voltage drop across the conductor, and the sum of these two voltage drops equals the EMF (Kirchhoff’s Voltage Law, KVL). Remark. In general, KVL states that “The voltage (the electromotive force EMF) impressed on a closed loop is equal to the sum of the voltage drops across all the other elements of the loop.” For Kirchoff’s Current Law (KCL) and historical information, see footnote 7 in Sec. 2.9. Solution. According to these laws the model of the RL-circuit is in standard form (6) Ir  R L I  E(t) L . LIr  RI  E(t), LIr  L dI>dt L  0.1R  11  E  48E(t) I(t)  2 cos2 xy  3 cos x  2 cos2 x. c  31  c # 1  2 # 12; y(x)  cos x a2sin x dx  cb  c cos x  2 cos2x. ehr  (sec x)(2 sin x cos x)  2 sin x,eh  cos x,eh  sec x, h  p dx   tan x dx  ln ƒ sec x ƒ . p  tan x, r  sin 2x  2 sin x cos x, y(0)  1.yr  y tan x  sin 2x, Total Output  Response to the Input r  Response to the Initial Data. y(x)  ehehr dx  ceh, h(x) c01.qxd 7/30/10 8:15 PM Page 29 30 CHAP. 1 First-Order ODEs We can solve this linear ODE by (4) with obtaining the general solution By integration, (7) In our case, and thus, In modeling, one often gets better insight into the nature of a solution (and smaller roundoff errors) by inserting given numeric data only near the end. Here, the general solution (7) shows that the current approaches the limit faster the larger is, in our case, and the approach is very fast, from below if or from above if If the solution is constant (48/11 A). See Fig. 19. The initial value gives and the particular solution (8)  I  E R (1  e(R>L)t), thus I  48 11 (1  e110t). c  E>RI(0)  E>R  c  0,I(0)  0 I(0)  48>11,I(0)  48>11.I(0)  48>11 R>L  11>0.1  110,R>LE>R  48>11 I  4811  ce 110t. E(t)  48>0.1  480  const;R>L  11>0.1  110 I  e(R>L)t aE L e1R>L2t R>L  cb  E R  ce(R>L)t. I  e(R>L)t a e(R>L)t E(t)L dt  c b. x  t, y  I, p  R>L, h  (R>L)t, Fig. 19. RL-circuit E X A M P L E 3 Hormone Level Assume that the level of a certain hormone in the blood of a patient varies with time. Suppose that the time rate of change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuous removal rate proportional to the level present. Set up a model for the hormone level in the blood and find its general solution. Find the particular solution satisfying a suitable initial condition. Solution. Step 1. Setting up a model. Let be the hormone level at time t. Then the removal rate is The input rate is where and A is the average input rate; here to make the input rate nonnegative. The constants A, B, K can be determined from measurements. Hence the model is the linear ODE The initial condition for a particular solution is with suitably chosen, for example, 6:00 A.M. Step 2. General solution. In (4) we have and Hence (4) gives the general solution (evaluate by integration by parts)eKt cos vt dt r  A  B cos vt.p  K  const, h  Kt, t  0ypart(0)  y0ypart yr(t)  In  Out  A  B cos vt  Ky(t), thus yr  Ky  A  B cos vt. A  Bv  2p>24  p>12A  B cos vt, Ky(t).y(t) L = 0.1 H Circuit Current I (t) I (t) E = 48 V R = 11  0.01 0.02 0.03 0.04 0.05 t 2 4 6 8 0 c01.qxd 7/30/10 8:15 PM Page 30 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 31 Fig. 20. Particular solution in Example 3 0 10 15 20 25 100 2000 5 t y The last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initial condition). The other part of is called the steady-state solution because it consists of constant and periodic terms. The entire solution is called the transient-state solution because it models the transition from rest to the steady state. These terms are used quite generally for physical and other systems whose behavior depends on time. Step 3. Particular solution. Setting in and choosing we have thus Inserting this result into we obtain the particular solution with the steady-state part as before. To plot we must specify values for the constants, say, and Figure 20 shows this solution. Notice that the transition period is relatively short (although K is small), and the curve soon looks sinusoidal; this is the response to the input 1  cos ( 112 pt). A  B cos ( 112 pt)  K  0.05. A  B  1ypart ypart(t)  A K  B K 2  (p>12)2 aK cos pt 12  p 12 sin pt 12 b  aA K  KB K 2  (p>12)2 b eK y(t), c   A K  KB K 2  (p>12)2 .y(0)  A K  B K 2  (p>12)2 u p K  c  0, y0  0,y(t)t  0 y(t)  A K  B K 2  (p>12)2 aK cos pt 12  p 12 sin pt 12 b  ceKt.  eKteKt c A K  B K 2  v2 aK cos vt  v sin vtb d  ceKt y(t)  eKteKt aA  B cos vtb dt  ceKt Reduction to Linear Form. Bernoulli Equation Numerous applications can be modeled by ODEs that are nonlinear but can be transformed to linear ODEs. One of the most useful ones of these is the Bernoulli equation7 (9) (a any real number).yr  p(x)y  g(x)ya 7JAKOB BERNOULLI (1654–1705), Swiss mathematician, professor at Basel, also known for his contribution to elasticity theory and mathematical probability. The method for solving Bernoulli’s equation was discovered by Leibniz in 1696. Jakob Bernoulli’s students included his nephew NIKLAUS BERNOULLI (1687–1759), who contributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI (1667–1748), who had profound influence on the development of calculus, became Jakob’s successor at Basel, and had among his students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIEL BERNOULLI (1700–1782) is known for his basic work in fluid flow and the kinetic theory of gases. c01.qxd 7/30/10 8:15 PM Page 31 34 CHAP. 1 First-Order ODEs y(x) x0 2–2 (a) y1 y2 y1 y2 y1 y2 (b) (c) 1–1 1.0 2.0 0.5 1.5 2.5 3.0 y x0 2.0 2.5 3.00.5 1.0 1.5 1.0 0.5 1.5 2.0 Fig. 22. Example 5. (A) Direction field. (B) “Phase line”. (C) Parabola f (y) A few further population models will be discussed in the problem set. For some more details of population dynamics, see C. W. Clark. Mathematical Bioeconomics: The Mathematics of Conservation 3rd ed. Hoboken, NJ, Wiley, 2010. Further applications of linear ODEs follow in the next section. 1. CAUTION! Show that and 2. Integration constant. Give a reason why in (4) you may choose the constant of integration in to be zero. 3–13 GENERAL SOLUTION. INITIAL VALUE PROBLEMS Find the general solution. If an initial condition is given, find also the corresponding particular solution and graph or sketch it. (Show the details of your work.) 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. yr  6(y  2.5) tanh 1.5x xyr  4y  8x4, y(1)  2 yr  (y  2) cot x yr cos x  (3y  1) sec x  0, y(14p)  4>3 yr  y sin x  ecos x, y(0)  2.5 yr  y tan x  e0.01x cos x, y(0)  0 xyr  2y  x3ex yr  2y  4 cos 2x, y(14p)  3 yr  ky  ekx yr  2y  4x yr  y  5.2 p dx eln(sec x)  cos x. eln x  1>x (not x) 14. CAS EXPERIMENT. (a) Solve the ODE Find an initial condition for which the arbitrary constant becomes zero. Graph the resulting particular solution, experimenting to obtain a good figure near (b) Generalizing (a) from to arbitrary n, solve the ODE Find an initial condition as in (a) and experiment with the graph. 15–20 GENERAL PROPERTIES OF LINEAR ODEs These properties are of practical and theoretical importance because they enable us to obtain new solutions from given ones. Thus in modeling, whenever possible, we prefer linear ODEs over nonlinear ones, which have no similar properties. Show that nonhomogeneous linear ODEs (1) and homo- geneous linear ODEs (2) have the following properties. Illustrate each property by a calculation for two or three equations of your choice. Give proofs. 15. The sum of two solutions and of the homogeneous equation (2) is a solution of (2), and so is a scalar multiple for any constant a. These properties are not true for (1)! ay1 y2y1y1  y2 yr  ny>x  xn2 cos (1>x). n  1 x  0. x1 cos (1>x). yr  y>x  P R O B L E M S E T 1 . 5 c01.qxd 7/30/10 8:15 PM Page 34 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 35 16. (that is, for all x, also written ) is a solution of (2) [not of (1) if !], called the trivial solution. 17. The sum of a solution of (1) and a solution of (2) is a solution of (1). 18. The difference of two solutions of (1) is a solution of (2). 19. If is a solution of (1), what can you say about 20. If and are solutions of and respectively (with the same p!), what can you say about the sum 21. Variation of parameter. Another method of obtaining (4) results from the following idea. Write (3) as where is the exponential function, which is a solution of the homogeneous linear ODE Replace the arbitrary constant c in (3) with a function u to be determined so that the resulting function is a solution of the nonhomogeneous linear ODE 22–28 NONLINEAR ODEs Using a method of this section or separating variables, find the general solution. If an initial condition is given, find also the particular solution and sketch or graph it. 22. 23. 24. 25. 26. 27. 28. 29. REPORT PROJECT. Transformation of ODEs. We have transformed ODEs to separable form, to exact form, and to linear form. The purpose of such transformations is an extension of solution methods to larger classes of ODEs. Describe the key idea of each of these transformations and give three typical exam- ples of your choice for each transformation. Show each step (not just the transformed ODE). 30. TEAM PROJECT. Riccati Equation. Clairaut Equation. Singular Solution. A Riccati equation is of the form (14) A Clairaut equation is of the form (15) (a) Apply the transformation to the Riccati equation (14), where Y is a solution of (14), and obtain for u the linear ODE Explain the effect of the transformation by writing it as y  Y  v, v  1>u. ur  (2Yg  p)u  g. y  Y  1>u y  xyr  g(yr). yr  p(x)y  g(x)y2  h(x). 2xyyr  (x  1)y2  x2ex (Set y2  z) yr  1>(6ey  2x) y(0)  12 pyr  (tan y)>(x  1), yr  3.2y  10y2 yr  y  x>y yr  xy  xy1, y(0)  3 yr  y  y2, y(0)  13 yr  py  r. y  uy* y* r  py*  0. y* cy*, y1  y2? y2r  py2  r2, y1r  py1  r1y2y1 cy1?y1 r(x)  0 y(x)  0y(x)  0y  0 (b) Show that is a solution of the ODE and solve this Riccati equation, showing the details. (c) Solve the Clairaut equation as follows. Differentiate it with respect to x, obtaining Then solve (A) and (B) separately and substitute the two solutions (a) and (b) of (A) and (B) into the given ODE. Thus obtain (a) a general solution (straight lines) and (b) a parabola for which those lines (a) are tangents (Fig. 6 in Prob. Set 1.1); so (b) is the envelope of (a). Such a solution (b) that cannot be obtained from a general solution is called a singular solution. (d) Show that the Clairaut equation (15) has as solutions a family of straight lines and a singular solution determined by where that forms the envelope of that family. 31–40 MODELING. FURTHER APPLICATIONS 31. Newton’s law of cooling. If the temperature of a cake is when it leaves the oven and is ten minutes later, when will it be practically equal to the room temperature of say, when will it be 32. Heating and cooling of a building. Heating and cooling of a building can be modeled by the ODE where is the temperature in the building at time t, the outside temperature, the temperature wanted in the building, and P the rate of increase of T due to machines and people in the building, and and are (negative) constants. Solve this ODE, assuming and varying sinusoidally over 24 hours, say, Discuss the effect of each term of the equation on the solution. 33. Drug injection. Find and solve the model for drug injection into the bloodstream if, beginning at a constant amount A g min is injected and the drug is simultaneously removed at a rate proportional to the amount of the drug present at time t. 34. Epidemics. A model for the spread of contagious diseases is obtained by assuming that the rate of spread is proportional to the number of contacts between infected and noninfected persons, who are assumed to move freely among each other. Set up the model. Find the equilibrium solutions and indicate their stability or instability. Solve the ODE. Find the limit of the proportion of infected persons as and explain what it means. 35. Lake Erie. Lake Erie has a water volume of about and a flow rate (in and out) of about 175 km2450 km3 t :  > t  0, Ta  A  C cos(2p>24)t. TaP  const, Tw  const, k2 k1 TwTa T  T(t) Tr  k1(T  Ta)  k2(T  Tv)  P, 61°F?60°F, 200°F300°F s  yr, gr(s)  x, y  cx  g(c) 2yr  x  0 ys  0ys(2yr  x)  0. yr2  xyr  y  0 y  x2y2  x4  x  1(2x3  1)yr  y  Y  x c01.qxd 7/30/10 10:01 PM Page 35 36 CHAP. 1 First-Order ODEs per year. If at some instant the lake has pollution concentration how long, approximately, will it take to decrease it to p 2, assuming that the inflow is much cleaner, say, it has pollution concentration p 4, and the mixture is uniform (an assumption that is only imperfectly true)? First guess. 36. Harvesting renewable resources. Fishing. Suppose that the population of a certain kind of fish is given by the logistic equation (11), and fish are caught at a rate Hy proportional to y. Solve this so-called Schaefer model. Find the equilibrium solutions and when The expression is called the equilibrium harvest or sustainable yield corre- sponding to H. Why? 37. Harvesting. In Prob. 36 find and graph the solution satisfying when (for simplicity) and What is the limit? What does it mean? What if there were no fishing? 38. Intermittent harvesting. In Prob. 36 assume that you fish for 3 years, then fishing is banned for the next 3 years. Thereafter you start again. And so on. This is called intermittent harvesting. Describe qualitatively how the population will develop if intermitting is continued periodically. Find and graph the solution for the first 9 years, assuming that and y(0)  2. A  B  1, H  0.2, H  0.2. A  B  1y(0)  2 Y  Hy2H  A. y2 ( 0)y1 y(t) > > p  0.04%, 39. Extinction vs. unlimited growth. If in a population the death rate is proportional to the population, and the birth rate is proportional to the chance encounters of meeting mates for reproduction, what will the model be? Without solving, find out what will eventually happen to a small initial population. To a large one. Then solve the model. 40. Air circulation. In a room containing of air, of fresh air flows in per minute, and the mixture (made practically uniform by circulating fans) is exhausted at a rate of 600 cubic feet per minute (cfm). What is the amount of fresh air at any time if After what time will 90% of the air be fresh?y(0)  0? y(t) 600 ft3 20,000 ft3 y(t) Fig. 23. Fish population in Problem 38 0.8 1 1.2 1.4 1.6 1.8 2 2 4 6 80 t y 1.6 Orthogonal Trajectories. Optional An important type of problem in physics or geometry is to find a family of curves that intersects a given family of curves at right angles. The new curves are called orthogonal trajectories of the given curves (and conversely). Examples are curves of equal temperature (isotherms) and curves of heat flow, curves of equal altitude (contour lines) on a map and curves of steepest descent on that map, curves of equal potential (equipotential curves, curves of equal voltage—the ellipses in Fig. 24) and curves of electric force (the parabolas in Fig. 24). Here the angle of intersection between two curves is defined to be the angle between the tangents of the curves at the intersection point. Orthogonal is another word for perpendicular. In many cases orthogonal trajectories can be found using ODEs. In general, if we consider to be a given family of curves in the xy-plane, then each value of c gives a particular curve. Since c is one parameter, such a family is called a one- parameter family of curves. In detail, let us explain this method by a family of ellipses (1) (c  0)12 x 2  y2  c G(x, y, c)  0 c01.qxd 7/30/10 8:15 PM Page 36 SEC. 1.7 Existence and Uniqueness of Solutions 39 Theorems that state such conditions are called existence theorems and uniqueness theorems, respectively. Of course, for our simple examples, we need no theorems because we can solve these examples by inspection; however, for complicated ODEs such theorems may be of considerable practical importance. Even when you are sure that your physical or other system behaves uniquely, occasionally your model may be oversimplified and may not give a faithful picture of reality. T H E O R E M 1 Existence Theorem Let the right side of the ODE in the initial value problem (1) be continuous at all points in some rectangle (Fig. 26) and bounded in R; that is, there is a number K such that (2) for all in R. Then the initial value problem (1) has at least one solution . This solution exists at least for all x in the subinterval of the interval here, is the smaller of the two numbers a and b K.>a ƒ x  x0 ƒ  a;ƒ x  x0 ƒ  a y(x) (x, y)ƒ f (x, y) ƒ K ƒ y  y0 ƒ  bR: ƒ x  x0 ƒ  a, (x, y) y(x0)  y0yr  f (x, y), f (x, y) has precisely one solution, namely, The initial value problem has infinitely many solutions, namely, where c is an arbitrary constant because for all c. From these examples we see that an initial value problem (1) may have no solution, precisely one solution, or more than one solution. This fact leads to the following two fundamental questions. Problem of Existence Under what conditions does an initial value problem of the form (1) have at least one solution (hence one or several solutions)? Problem of Uniqueness Under what conditions does that problem have at most one solution (hence excluding the case that is has more than one solution)? y(x0)  y0yr  f (x, y), y(0)  1 y  1  cx, y(0)  1xyr  y  1, y  x2  1. c01.qxd 7/30/10 8:15 PM Page 39 40 CHAP. 1 First-Order ODEs y x y 0 + b x 0 + ax 0 – a x 0 y 0 y 0 – b R Fig. 26. Rectangle R in the existence and uniqueness theorems (Example of Boundedness. The function is bounded (with ) in the square . The function is not bounded for . Explain!) T H E O R E M 2 Uniqueness Theorem Let f and its partial derivative be continuous for all in the rectangle R (Fig. 26) and bounded, say, (3) (a) (b) for all in R. Then the initial value problem (1) has at most one solution . Thus, by Theorem 1, the problem has precisely one solution. This solution exists at least for all x in that subinterval ƒ x  x0 ƒ  a. y(x) (x, y)ƒ fy(x, y) ƒ Mƒ f (x, y) ƒ K, (x, y)fy  0f>0y ƒ x  y ƒ  p>2 f (x, y)  tan (x  y)ƒ x ƒ  1, ƒ y ƒ  1 K  2f (x, y)  x2  y2 Understanding These Theorems These two theorems take care of almost all practical cases. Theorem 1 says that if is continuous in some region in the xy-plane containing the point , then the initial value problem (1) has at least one solution. Theorem 2 says that if, moreover, the partial derivative of f with respect to y exists and is continuous in that region, then (1) can have at most one solution; hence, by Theorem 1, it has precisely one solution. Read again what you have just read—these are entirely new ideas in our discussion. Proofs of these theorems are beyond the level of this book (see Ref. [A11] in App. 1); however, the following remarks and examples may help you to a good understanding of the theorems. Since , the condition (2) implies that that is, the slope of any solution curve in R is at least and at most K. Hence a solution curve that passes through the point must lie in the colored region in Fig. 27 bounded by the lines and whose slopes are and K, respectively. Depending on the form of R, two different cases may arise. In the first case, shown in Fig. 27a, we have and therefore in the existence theorem, which then asserts that the solution exists for all x between and . In the second case, shown in Fig. 27b, we have . Therefore, and all we can conclude from the theorems is that the solutiona  b>K  a, b>K  ax0  ax0  a a  a b>K  a Kl2l1 (x0, y0) Ky(x) ƒ yr ƒ K;yr  f (x, y) 0f>0y (x0, y0) f (x, y) c01.qxd 7/30/10 8:15 PM Page 40 and take the rectangle Then , and Indeed, the solution of the problem is (see Sec. 1.3, Example 1). This solution is discontinuous at , and there is no continuous solution valid in the entire interval from which we started. The conditions in the two theorems are sufficient conditions rather than necessary ones, and can be lessened. In particular, by the mean value theorem of differential calculus we have where and are assumed to be in R, and is a suitable value between and . From this and (3b) it follows that (4) ƒ f (x, y2)  f (x, y1) ƒ M ƒ y2  y1 ƒ . y2 y1y (x, y2)(x, y1) f (x, y2)  f (x, y1)  (y2  y1) 0f 0y ` yy ƒ x ƒ  5 p>2 y  tan x a  b K  0.3  a. ` 0f 0y `  2 ƒ y ƒ M  6, ƒ f (x, y) ƒ  ƒ 1  y2 ƒ K  10, a  5, b  3R; ƒ x ƒ  5, ƒ y ƒ  3. SEC. 1.7 Existence and Uniqueness of Solutions 41 y y x y 0 + b l 1 l 2 x 0 (a) y 0 y 0 – b R x y 0 + b l 1 l 2 x 0 (b) y 0 y 0 – b R a a = a = aα α α α Let us illustrate our discussion with a simple example. We shall see that our choice of a rectangle R with a large base (a long x-interval) will lead to the case in Fig. 27b. E X A M P L E 1 Choice of a Rectangle Consider the initial value problem y(0)  0yr  1  y2, exists for all x between and . For larger or smaller x’s the solution curve may leave the rectangle R, and since nothing is assumed about f outside R, nothing can be concluded about the solution for those larger or amaller x’s; that is, for such x’s the solution may or may not exist—we don’t know. x0  b>Kx0  b>K Fig. 27. The condition (2) of the existence theorem. (a) First case. (b) Second case c01.qxd 7/30/10 8:15 PM Page 41 44 CHAP. 1 First-Order ODEs 28. Mixing problem. The tank in Fig. 28 contains 80 lb of salt dissolved in 500 gal of water. The inflow per minute is 20 lb of salt dissolved in 20 gal of water. The outflow is 20 gal min of the uniform mixture. Find the time when the salt content in the tank reaches 95% of its limiting value (as ).t :  y(t) > Fig. 28. Tank in Problem 28 29. Half-life. If in a reactor, uranium loses 10% of its weight within one day, what is its half-life? How long would it take for 99% of the original amount to disappear? 30. Newton’s law of cooling. A metal bar whose temperature is is placed in boiling water. How long does it take to heat the bar to practically say, to , if the temperature of the bar after 1 min of heating is First guess, then calculate.51.5°C? 99.9°C 100°C, 20°C 237 97 U This chapter concerns ordinary differential equations (ODEs) of first order and their applications. These are equations of the form (1) or in explicit form involving the derivative of an unknown function y, given functions of x, and, perhaps, y itself. If the independent variable x is time, we denote it by t. In Sec. 1.1 we explained the basic concepts and the process of modeling, that is, of expressing a physical or other problem in some mathematical form and solving it. Then we discussed the method of direction fields (Sec. 1.2), solution methods and models (Secs. 1.3–1.6), and, finally, ideas on existence and uniqueness of solutions (Sec. 1.7). A first-order ODE usually has a general solution, that is, a solution involving an arbitrary constant, which we denote by c. In applications we usually have to find a unique solution by determining a value of c from an initial condition . Together with the ODE this is called an initial value problem (2) and its solution is a particular solution of the ODE. Geometrically, a general solution represents a family of curves, which can be graphed by using direction fields (Sec. 1.2). And each particular solution corresponds to one of these curves. A separable ODE is one that we can put into the form (3) (Sec. 1.3) by algebraic manipulations (possibly combined with transformations, such as ) and solve by integrating on both sides.y>x  u g(y) dy  f (x) dx (x0, y0 given numbers)y(x0)  y0yr  f (x, y), y(x0)  y0 yr  dy>dx yr  f (x, y)F(x, y, yr)  0 SUMMARY OF CHAPTER 1 First-Order ODEs c01.qxd 7/30/10 8:15 PM Page 44 An exact ODE is of the form (4) (Sec. 1.4) where is the differential of a function so that from we immediately get the implicit general solution This method extends to nonexact ODEs that can be made exact by multiplying them by some function called an integrating factor (Sec. 1.4). Linear ODEs (5) are very important. Their solutions are given by the integral formula (4), Sec. 1.5. Certain nonlinear ODEs can be transformed to linear form in terms of new variables. This holds for the Bernoulli equation (Sec. 1.5). Applications and modeling are discussed throughout the chapter, in particular in Secs. 1.1, 1.3, 1.5 (population dynamics, etc.), and 1.6 (trajectories). Picard’s existence and uniqueness theorems are explained in Sec. 1.7 (and Picard’s iteration in Problem Set 1.7). Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2 immediately after this chapter, as indicated in the chapter opening. yr  p(x)y  g(x)ya yr  p(x)y  r(x) F(x, y,), u(x, y)  c. du  0u(x, y), du  ux dx  uy dy M dx  N dy M(x, y) dx  N(x, y) dy  0 Summary of Chapter 1 45 c01.qxd 7/30/10 8:15 PM Page 45 46 C H A P T E R 2 Second-Order Linear ODEs Many important applications in mechanical and electrical engineering, as shown in Secs. 2.4, 2.8, and 2.9, are modeled by linear ordinary differential equations (linear ODEs) of the second order. Their theory is representative of all linear ODEs as is seen when compared to linear ODEs of third and higher order, respectively. However, the solution formulas for second-order linear ODEs are simpler than those of higher order, so it is a natural progression to study ODEs of second order first in this chapter and then of higher order in Chap. 3. Although ordinary differential equations (ODEs) can be grouped into linear and nonlinear ODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which many beautiful standard methods exist. Chapter 2 includes the derivation of general and particular solutions, the latter in connection with initial value problems. For those interested in solution methods for Legendre’s, Bessel’s, and the hypergeometric equations consult Chap. 5 and for Sturm–Liouville problems Chap. 11. COMMENT. Numerics for second-order ODEs can be studied immediately after this chapter. See Sec. 21.3, which is independent of other sections in Chaps. 19–21. Prerequisite: Chap. 1, in particular, Sec. 1.5. Sections that may be omitted in a shorter course: 2.3, 2.9, 2.10. References and Answers to Problems: App. 1 Part A, and App. 2. 2.1 Homogeneous Linear ODEs of Second Order We have already considered first-order linear ODEs (Sec. 1.5) and shall now define and discuss linear ODEs of second order. These equations have important engineering applications, especially in connection with mechanical and electrical vibrations (Secs. 2.4, 2.8, 2.9) as well as in wave motion, heat conduction, and other parts of physics, as we shall see in Chap. 12. A second-order ODE is called linear if it can be written (1) and nonlinear if it cannot be written in this form. The distinctive feature of this equation is that it is linear in y and its derivatives, whereas the functions p, q, and r on the right may be any given functions of x. If the equation begins with, say, then divide by to have the standard form (1) with as the first term. ysf (x)f (x)ys, ys  p(x)yr  q(x)y  r(x) c02.qxd 10/27/10 6:06 PM Page 46 Initial Value Problem. Basis. General Solution Recall from Chap. 1 that for a first-order ODE, an initial value problem consists of the ODE and one initial condition . The initial condition is used to determine the arbitrary constant c in the general solution of the ODE. This results in a unique solution, as we need it in most applications. That solution is called a particular solution of the ODE. These ideas extend to second-order ODEs as follows. For a second-order homogeneous linear ODE (2) an initial value problem consists of (2) and two initial conditions (4) These conditions prescribe given values and of the solution and its first derivative (the slope of its curve) at the same given in the open interval considered. The conditions (4) are used to determine the two arbitrary constants and in a general solution (5) of the ODE; here, and are suitable solutions of the ODE, with “suitable” to be explained after the next example. This results in a unique solution, passing through the point with as the tangent direction (the slope) at that point. That solution is called a particular solution of the ODE (2). E X A M P L E 4 Initial Value Problem Solve the initial value problem Solution. Step 1. General solution. The functions and are solutions of the ODE (by Example 1), and we take This will turn out to be a general solution as defined below. Step 2. Particular solution. We need the derivative . From this and the initial values we obtain, since and , This gives as the solution of our initial value problem the particular solution Figure 29 shows that at it has the value 3.0 and the slope , so that its tangent intersects the x-axis at . (The scales on the axes differ!) Observation. Our choice of and was general enough to satisfy both initial conditions. Now let us take instead two proportional solutions and so that . Then we can write in the form .y  c1 cos x  c2(k cos x)  C cos x where C  c1  c2k y  c1y1  c2y2y1/y2  1/k  const y2  k cos x,y1  cos x y2y1 x  3.0>0.5  6.0 0.5x  0 y  3.0 cos x  0.5 sin x. y(0)  c1  3.0 and yr(0)  c2  0.5. sin 0  0cos 0  1 yr  c1 sin x  c2 cos x y  c1 cos x  c2 sin x. sin xcos x ys  y  0, y(0)  3.0, yr(0)  0.5. K1(x0, K0) y2y1 y  c1y1  c2y2 c2c1 x  x0 K1K0 y(x0)  K0, yr(x0)  K1. y(x0)  y0 SEC. 2.1 Homogeneous Linear ODEs of Second Order 49 2 4 6 108 x –3 –2 –1 0 1 2 3 y Fig. 29. Particular solution and initial tangent in Example 4 c02.qxd 10/27/10 6:06 PM Page 49 Hence we are no longer able to satisfy two initial conditions with only one arbitrary constant C. Consequently, in defining the concept of a general solution, we must exclude proportionality. And we see at the same time why the concept of a general solution is of importance in connection with initial value problems. D E F I N I T I O N General Solution, Basis, Particular Solution A general solution of an ODE (2) on an open interval I is a solution (5) in which and are solutions of (2) on I that are not proportional, and and are arbitrary constants. These , are called a basis (or a fundamental system) of solutions of (2) on I. A particular solution of (2) on I is obtained if we assign specific values to and in (5). For the definition of an interval see Sec. 1.1. Furthermore, as usual, and are called proportional on I if for all x on I, (6) (a) or (b) where k and l are numbers, zero or not. (Note that (a) implies (b) if and only if ). Actually, we can reformulate our definition of a basis by using a concept of general importance. Namely, two functions and are called linearly independent on an interval I where they are defined if (7) everywhere on I implies . And and are called linearly dependent on I if (7) also holds for some constants , not both zero. Then, if , we can divide and see that and are proportional, or In contrast, in the case of linear independence these functions are not proportional because then we cannot divide in (7). This gives the following D E F I N I T I O N Basis (Reformulated) A basis of solutions of (2) on an open interval I is a pair of linearly independent solutions of (2) on I. If the coefficients p and q of (2) are continuous on some open interval I, then (2) has a general solution. It yields the unique solution of any initial value problem (2), (4). It includes all solutions of (2) on I; hence (2) has no singular solutions (solutions not obtainable from of a general solution; see also Problem Set 1.1). All this will be shown in Sec. 2.6. y2   k1 k2 y1.y1   k2 k1 y2 y2y1k1  0 or k2  0k2 k1y2y1 k1  0 and k2  0k1y1(x)  k2y2(x)  0 y2y1 k  0 y2  ly1y1  ky2 y2y1 c2 c1 y2y1 c2c1y2y1 50 CHAP. 2 Second-Order Linear ODEs c02.qxd 10/27/10 6:06 PM Page 50 E X A M P L E 5 Basis, General Solution, Particular Solution and in Example 4 form a basis of solutions of the ODE for all x because their quotient is (or ). Hence is a general solution. The solution of the initial value problem is a particular solution. E X A M P L E 6 Basis, General Solution, Particular Solution Verify by substitution that and are solutions of the ODE . Then solve the initial value problem . Solution. and show that and are solutions. They are not proportional, . Hence , form a basis for all x. We now write down the corresponding general solution and its derivative and equate their values at 0 to the given initial conditions, . By addition and subtraction, , so that the answer is . This is the particular solution satisfying the two initial conditions. Find a Basis if One Solution Is Known. Reduction of Order It happens quite often that one solution can be found by inspection or in some other way. Then a second linearly independent solution can be obtained by solving a first-order ODE. This is called the method of reduction of order.1 We first show how this method works in an example and then in general. E X A M P L E 7 Reduction of Order if a Solution Is Known. Basis Find a basis of solutions of the ODE . Solution. Inspection shows that is a solution because and , so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute into the ODE. This gives ux and –xu cancel and we are left with the following ODE, which we divide by x, order, and simplify, , This ODE is of first order in , namely, . Separation of variables and integration gives , .ln ƒ v ƒ  ln ƒ x  1 ƒ  2 ln ƒ x ƒ  ln ƒ x  1 ƒ x2 dv v   x  2 x2  x dx  a 1 x  1  2 x b dx (x2  x)vr  (x  2)v  0v  ur (x2  x)us  (x  2)ur  0.(x2  x)(usx  2ur)  x2ur  0 (x2  x)(usx  2ur)  x(urx  u)  ux  0. y  uy1  ux, yr  urx  u, ys  usx  2ur ys1  0yr1  1y1  x (x2  x)ys  xyr  y  0  y  2ex  4exc1  2, c2  4 y  c1e x  c2e x, yr  c1ex  c2ex, y(0)  c1  c2  6, yr(0)  c1  c2  2 exexex/ex  e2x  const exex(ex)s  ex  0(ex)s  ex  0 ys  y  0, y(0)  6, yr(0)  2 ys  y  0y2  exy1  ex y  3.0 cos x  0.5 sin x y  c1 cos x  c2 sin xtan x  constcot x  const ys  y  0sin xcos x SEC. 2.1 Homogeneous Linear ODEs of Second Order 51 1Credited to the great mathematician JOSEPH LOUIS LAGRANGE (1736–1813), who was born in Turin, of French extraction, got his first professorship when he was 19 (at the Military Academy of Turin), became director of the mathematical section of the Berlin Academy in 1766, and moved to Paris in 1787. His important major work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique analytique, Paris, 1788), differential equations, approximation theory, algebra, and number theory. c02.qxd 10/27/10 6:06 PM Page 51 is an exponential function . This gives us the idea to try as a solution of (1) the function (2) . Substituting (2) and its derivatives and into our equation (1), we obtain . Hence if is a solution of the important characteristic equation (or auxiliary equation) (3) then the exponential function (2) is a solution of the ODE (1). Now from algebra we recall that the roots of this quadratic equation (3) are (4) , (3) and (4) will be basic because our derivation shows that the functions (5) and are solutions of (1). Verify this by substituting (5) into (1). From algebra we further know that the quadratic equation (3) may have three kinds of roots, depending on the sign of the discriminant , namely,a2  4b y2  e l2xy1  e l1x l2  1 2 Aa  2a 2  4b B .l1  1 2 Aa  2a 2  4b B l2  al  b  0 l (l2  al  b)elx  0 ys  l2elxyr  lelx y  elx y  cekx 54 CHAP. 2 Second-Order Linear ODEs (Case I) Two real roots if , (Case II) A real double root if , (Case III) Complex conjugate roots if .a2  4b  0 a2  4b  0 a2  4b  0 Case I. Two Distinct Real-Roots and In this case, a basis of solutions of (1) on any interval is and because and are defined (and real) for all x and their quotient is not constant. The corresponding general solution is (6) .y  c1e l1x  c2e l2x y2y1 y2  e l2xy1  e l1x l2l1 c02.qxd 10/27/10 6:06 PM Page 54 E X A M P L E 1 General Solution in the Case of Distinct Real Roots We can now solve in Example 6 of Sec. 2.1 systematically. The characteristic equation is Its roots are and . Hence a basis of solutions is and and gives the same general solution as before, . E X A M P L E 2 Initial Value Problem in the Case of Distinct Real Roots Solve the initial value problem , , . Solution. Step 1. General solution. The characteristic equation is . Its roots are and so that we obtain the general solution . Step 2. Particular solution. Since , we obtain from the general solution and the initial conditions Hence and . This gives the answer . Figure 30 shows that the curve begins at with a negative slope but note that the axes have different scales!), in agreement with the initial conditions.  (5,y  4 y  ex  3e2xc2  3c1  1 yr(0)  c1  2c2  5. y(0)  c1  c2  4, yr(x)  c1ex  2c2e2x y  c1e x  c2e 2x l2  1 2 (1  19)  2l1  1 2 (1  19)  1 l2  l  2  0 yr(0)  5y(0)  4ys  yr  2y  0 y  c1e x  c2e x exexl2  1l1  1l 2  1  0. ys  y  0 SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 55 2 0 4 1 1.50.50 x 6 8 2 y Case II. Real Double Root If the discriminant is zero, we see directly from (4) that we get only one root, , hence only one solution, . To obtain a second independent solution (needed for a basis), we use the method of reduction of order discussed in the last section, setting . Substituting this and its derivatives and into (1), we first have .(usy1  2uryr1  uys1)  a(ury1  uyr1)  buy1  0 ys2yr2  ury1  uyr1 y2  uy1 y2 y1  e (a/2)x l  l1  l2  a/2 a2  4b l  a/2 Fig. 30. Solution in Example 2 c02.qxd 10/27/10 6:06 PM Page 55 Collecting terms in and u, as in the last section, we obtain . The expression in the last parentheses is zero, since is a solution of (1). The expression in the first parentheses is zero, too, since . We are thus left with . Hence . By two integrations, . To get a second independent solution , we can simply choose and take . Then . Since these solutions are not proportional, they form a basis. Hence in the case of a double root of (3) a basis of solutions of (1) on any interval is . The corresponding general solution is (7) WARNING! If is a simple root of (4), then with is not a solution of (1). E X A M P L E 3 General Solution in the Case of a Double Root The characteristic equation of the ODE is . It has the double root . Hence a basis is and . The corresponding general solution is . E X A M P L E 4 Initial Value Problem in the Case of a Double Root Solve the initial value problem , , . Solution. The characteristic equation is . It has the double root This gives the general solution . We need its derivative . From this and the initial conditions we obtain , ; hence . The particular solution of the initial value problem is . See Fig. 31. y  (3  2x)e0.5x c2  2yr(0)  c2  0.5c1  3.5y(0)  c1  3.0 yr  c2e0.5x  0.5(c1  c2x)e0.5x y  (c1  c2x)e 0.5x l  0.5.l2  l  0.25  (l  0.5) 2  0 yr(0)  3.5y(0)  3.0ys  yr  0.25y  0 y  (c1  c2x)e 3xxe3xe3xl  3 l2  6l  9  (l  3)2  0ys  6yr  9y  0 c2  0(c1  c2x)e lxl y  (c1  c2x)e ax/2. eax/2, xeax/2 y2  xy1u  x c1  1, c2  0y2  uy1 u  c1x  c2us  0usy1  0 2yr1  aeax/2  ay1 y1 usy1  ur(2yr1  ay1)  u(ys1  ayr1  by1)  0 us, ur, 56 CHAP. 2 Second-Order Linear ODEs 1412108642 x –1 0 1 2 3 y Fig. 31. Solution in Example 4 c02.qxd 10/27/10 6:06 PM Page 56 For later use we note that so that by addition and subtraction of this and (11), (12) . After these comments on the definition (10), let us now turn to Case III. In Case III the radicand in (4) is negative. Hence is positive and, using , we obtain in (4) with defined as in (8). Hence in (4), and, similarly, . Using (10) with and , we thus obtain We now add these two lines and multiply the result by . This gives as in (8). Then we subtract the second line from the first and multiply the result by . This gives as in (8). These results obtained by addition and multiplication by constants are again solutions, as follows from the superposition principle in Sec. 2.1. This concludes the derivation of these real solutions in Case III. y21/(2i) y1 1 2 el2x  e(a/2)xivx  e(a/2)x(cos vx  i sin vx). el1x  e(a/2)xivx  e(a/2)x(cos vx  i sin vx) t  vxr  12 ax l2  1 2 a  ivl1  1 2 a  iv v 1 22a 2  4b  122(4b  a 2)  2(b  14 a 2)  i2b  14 a 2  iv 11  i 4b  a2a2  4b cos t  12 (e it  eit), sin t  1 2i (eit  eit) eit  cos (t)  i sin (t)  cos t  i sin t, SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 59 1–15 GENERAL SOLUTION Find a general solution. Check your answer by substitution. ODEs of this kind have important applications to be discussed in Secs. 2.4, 2.7, and 2.9. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 9ys  30yr  25y  0 ys  9yr  20y  0 4ys  4yr  3y  0 100ys  240yr  (196p2  144)y  0 ys  1.8yr  2.08y  0 ys  yr  3.25y  0 ys  4.5yr  0 10ys  32yr  25.6y  0 ys  2pyr  p2y  0 ys  4yr  (p2  4)y  0 ys  6yr  8.96y  0 ys  36y  0 4ys  25y  0 P R O B L E M S E T 2 . 2 14. 15. 16–20 FIND AN ODE for the given basis. 16. , 17. , 18. , 19. , 20. , 21–30 INITIAL VALUES PROBLEMS Solve the IVP. Check that your answer satisfies the ODE as well as the initial conditions. Show the details of your work. 21. , 22. The ODE in Prob. , 23. , 24. , 25. , 26. , yr(0)  1ys  k2y  0 (k  0), y(0)  1 yr(0)  2ys  y  0, y(0)  2 yr(2)  e>24ys 4yr  3y  0, y(2)  e yr(0)  0ys  yr  6y  0, y(0)  10 yr(12)  24, y(12)  1 yr(0)  1.2ys  25y  0, y(0)  4.6 e3.1x sin 2.1xe3.1x cos 2.1x e(2i)xe(2i)xsin 2pxcos 2px xe25xe25xe4.3xe2.6x ys  ayr  by  0 ys  0.54yr  (0.0729  p)y  0 ys  2k2yr  k4y  0 c02.qxd 10/27/10 6:06 PM Page 59 27. The ODE in Prob. 5, , 28. , 29. The ODE in Prob. , 30. , 31–36 LINEAR INDEPENDENCE is of basic impor- tance, in this chapter, in connection with general solutions, as explained in the text. Are the following functions linearly independent on the given interval? Show the details of your work. 31. any interval 32. 33. 34. 35. 36. , 0, 37. Instability. Solve for the initial conditions , . Then change the initial conditions to , and explain why this small change of 0.001 at causes a large change later,t  0 yr(0)  0.999y(0)  1.001 yr(0)  1y(0)  1 ys  y  0 1 x 1ex cos 12 x sin 2x, cos x sin x, x  0 ln x, ln (x3), x  1 x2, x2 ln x, x  1 eax, eax, x  0 ekx, xekx, yr(0)  10.09ys  30yr  25y  0, y(0)  3.3 yr(0)  115, y(0)  0 yr(0)  0.3258ys  2yr  y  0, y(0)  0.2 4.5p  1  13.137 yr(0) y(0)  4.5 60 CHAP. 2 Second-Order Linear ODEs e.g., 22 at . This is instability: a small initial difference in setting a quantity (a current, for in- stance) becomes larger and larger with time t. This is undesirable. 38. TEAM PROJECT. General Properties of Solutions (a) Coefficient formulas. Show how a and b in (1) can be expressed in terms of and . Explain how these formulas can be used in constructing equations for given bases. (b) Root zero. Solve (i) by the present method, and (ii) by reduction to first order. Can you explain why the result must be the same in both cases? Can you do the same for a general ODE (c) Double root. Verify directly that with is a solution of (1) in the case of a double root. Verify and explain why is a solution of but is not. (d) Limits. Double roots should be limiting cases of distinct roots , as, say, . Experiment with this idea. (Remember l’Hôpital’s rule from calculus.) Can you arrive at ? Give it a try.xel1x l2 : l1l2l1 xe2xys  yr  6y  0 y  e2x a>2 l xelx ys  ayr  0? ys  4yr  0 l2l1 t  10 2.3 Differential Operators. Optional This short section can be omitted without interrupting the flow of ideas. It will not be used subsequently, except for the notations , etc. to stand for , etc. Operational calculus means the technique and application of operators. Here, an operator is a transformation that transforms a function into another function. Hence differential calculus involves an operator, the differential operator D, which transforms a (differentiable) function into its derivative. In operator notation we write and (1) . Similarly, for the higher derivatives we write , and so on. For example, etc. For a homogeneous linear ODE with constant coefficients we can now introduce the second-order differential operator , where I is the identity operator defined by . Then we can write that ODE as (2) .Ly  P(D)y  (D2  aD  bI )y  0 Iy  y L  P(D)  D2  aD  bI ys  ayr  by  0 D sin  cos, D2 sin  sin, D2y  D(Dy)  ys Dy  yr  dy dx D  ddx yr, ysDy, D 2 y c02.qxd 10/27/10 6:06 PM Page 60 P suggests “polynomial.” L is a linear operator. By definition this means that if Ly and exist (this is the case if y and w are twice differentiable), then exists for any constants c and k, and . Let us show that from (2) we reach agreement with the results in Sec. 2.2. Since and , we obtain (3) This confirms our result of Sec. 2.2 that is a solution of the ODE (2) if and only if is a solution of the characteristic equation . is a polynomial in the usual sense of algebra. If we replace by the operator D, we obtain the “operator polynomial” . The point of this operational calculus is that can be treated just like an algebraic quantity. In particular, we can factor it. E X A M P L E 1 Factorization, Solution of an ODE Factor and solve . Solution. because . Now has the solution . Similarly, the solution of is . This is a basis of on any interval. From the factorization we obtain the ODE, as expected, . Verify that this agrees with the result of our method in Sec. 2.2. This is not unexpected because we factored in the same way as the characteristic polynomial . It was essential that L in (2) had constant coefficients. Extension of operator methods to variable-coefficient ODEs is more difficult and will not be considered here. If operational methods were limited to the simple situations illustrated in this section, it would perhaps not be worth mentioning. Actually, the power of the operator approach appears in more complicated engineering problems, as we shall see in Chap. 6. P(l)  l2  3l  40P(D)  ys  5yr  8yr  40y  ys  3r  40y  0 (D  8I )(D  5I )y  (D  8I )(yr  5y)  D(yr  5y)  8(yr  5y) P(D)y  0y2  e 5x(D  5I )y  0y1  e 8x (D  8I)y  yr  8y  0I 2  ID2  3D  40I  (D  8I )(D  5I ) P(D)y  0P(D)  D2  3D  40I P(D) P(D) lP(l) P(l)  0 lelx  (l2  al  b)elx  P(l)elx  0. Lel(x)  P(D)el(x)  (D2  aD  bI)el(x) (D2el)(x)  l2elx(Del)(x)  lelx L(cy  kw)  cLy  kLw L(cy  kw)Lw SEC. 2.3 Differential Operators. Optional 61 1–5 APPLICATION OF DIFFERENTIAL OPERATORS Apply the given operator to the given functions. Show all steps in detail. 1. 2. 3. 4. 5. (D  2I )(D  3I ); e2x, xe2x, e3x (D  6I )2; 6x  sin 6x, xe6x (D  2I )2; e2x, xe2x, e2x D  3I; 3x2  3x, 3e3x, cos 4x  sin 4x D2  2D; cosh 2x, ex  e2x, cos x P R O B L E M S E T 2 . 3 6–12 GENERAL SOLUTION Factor as in the text and solve. 6. 7. 8. 9. 10. 11. 12. (D2  3.0D  2.5I )y  0 (D2  4.00D  3.84I )y  0 (D2  4.80D  5.76I )y  0 (D2  4.20D  4.41I )y  0 (D2  3I )y  0 (4D2  I )y  0 (D2  4.00D  3.36I )y  0 c02.qxd 10/27/10 6:06 PM Page 61 An alternative representation of (4), which shows the physical characteristics of amplitude and phase shift of (4), is (4*) with and phase angle , where . This follows from the addition formula (6) in App. 3.1. E X A M P L E 1 Harmonic Oscillation of an Undamped Mass–Spring System If a mass–spring system with an iron ball of weight nt (about 22 lb) can be regarded as undamped, and the spring is such that the ball stretches it 1.09 m (about 43 in.), how many cycles per minute will the system execute? What will its motion be if we pull the ball down from rest by 16 cm (about 6 in.) and let it start with zero initial velocity? Solution. Hooke’s law (1) with W as the force and 1.09 meter as the stretch gives ; thus . The mass is . This gives the frequency . From (4) and the initial conditions, . Hence the motion is (Fig. 35). If you have a chance of experimenting with a mass–spring system, don’t miss it. You will be surprised about the good agreement between theory and experiment, usually within a fraction of one percent if you measure carefully.  y(t)  0.16 cos 3t [meter] or 0.52 cos 3t [ft] y(0)  A  0.16 [meter] and yr(0)  v0B  0 v0>(2p)  2k>m>(2p)  3>(2p)  0.48 [Hz]  29 [cycles>min] m  W>g  98>9.8  10 [kg]98>1.09  90 [kg>sec2]  90 [nt>meter]k  W>1.09  W  1.09k W  98 tan d  B>AdC  2A2  B2 y(t)  C cos (v0t  d) 64 CHAP. 2 Second-Order Linear ODEs 102 4 6 8 t –0.1 –0.2 0 0.1 0.2 y Fig. 35. Harmonic oscillation in Example 1 ODE of the Damped System To our model we now add a damping force obtaining ; thus the ODE of the damped mass–spring system is (5) (Fig. 36) Physically this can be done by connecting the ball to a dashpot; see Fig. 36. We assume this damping force to be proportional to the velocity . This is generally a good approximation for small velocities. yr  dy>dt mys  cyr  ky  0. mys  ky  cyr F2  cyr, mys  ky Fig. 36. Damped system Dashpot Ball Springk m c c02.qxd 10/27/10 6:06 PM Page 64 SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 65 Case I. . Distinct real roots . (Overdamping) Case II. . A real double root. (Critical damping) Case III. . Complex conjugate roots. (Underdamping)c2  4mk c2  4mk l1, l2c 2  4mk They correspond to the three Cases I, II, III in Sec. 2.2. Discussion of the Three Cases Case I. Overdamping If the damping constant c is so large that , then are distinct real roots. In this case the corresponding general solution of (5) is (7) . We see that in this case, damping takes out energy so quickly that the body does not oscillate. For both exponents in (7) are negative because , and . Hence both terms in (7) approach zero as . Practically speaking, after a sufficiently long time the mass will be at rest at the static equilibrium position . Figure 37 shows (7) for some typical initial conditions.(y  0) t : b2  a2  k>m  a2 a  0, b  0t  0 y(t)  c1e (ab)t  c2e (ab)t l1 and l2c 2  4mk The constant c is called the damping constant. Let us show that c is positive. Indeed, the damping force acts against the motion; hence for a downward motion we have which for positive c makes F negative (an upward force), as it should be. Similarly, for an upward motion we have which, for makes positive (a downward force). The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solve it by the method in Sec. 2.2. The characteristic equation is (divide (5) by m) . By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2, (6) , where and . It is now interesting that depending on the amount of damping present—whether a lot of damping, a medium amount of damping or little damping—three types of motions occur, respectively: b  1 2m 2c2  4mka  c 2m l1  a  b, l2  a  b l2  c m l  k m  0 F2c  0yr  0 yr  0 F2  cyr c02.qxd 10/27/10 6:06 PM Page 65 66 CHAP. 2 Second-Order Linear ODEs t y 1 2 3 (a) y t 1 1 2 3 2 Positive Zero Negative Initial velocity 3 (b) Fig. 37. Typical motions (7) in the overdamped case (a) Positive initial displacement (b) Negative initial displacement Case II. Critical Damping Critical damping is the border case between nonoscillatory motions (Case I) and oscillations (Case III). It occurs if the characteristic equation has a double root, that is, if , so that . Then the corresponding general solution of (5) is (8) . This solution can pass through the equilibrium position at most once because is never zero and can have at most one positive zero. If both are positive (or both negative), it has no positive zero, so that y does not pass through 0 at all. Figure 38 shows typical forms of (8). Note that they look almost like those in the previous figure. c1 and c2c1  c2t eaty  0 y(t)  (c1  c2t)e at b  0, l1  l2  a c2  4mk y t 1 2 3 1 2 3 Positive Zero Negative Initial velocity Fig. 38. Critical damping [see (8)] c02.qxd 10/27/10 6:06 PM Page 66 SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 69 1–10 HARMONIC OSCILLATIONS (UNDAMPED MOTION) 1. Initial value problem. Find the harmonic motion (4) that starts from with initial velocity . Graph or sketch the solutions for , and various of your choice on common axes. At what t-values do all these curves intersect? Why? 2. Frequency. If a weight of 20 nt (about 4.5 lb) stretches a certain spring by 2 cm, what will the frequency of the corresponding harmonic oscillation be? The period? 3. Frequency. How does the frequency of the harmonic oscillation change if we (i) double the mass, (ii) take a spring of twice the modulus? First find qualitative answers by physics, then look at formulas. 4. Initial velocity. Could you make a harmonic oscillation move faster by giving the body a greater initial push? 5. Springs in parallel. What are the frequencies of vibration of a body of mass kg (i) on a spring of modulus , (ii) on a spring of modulus , (iii) on the two springs in parallel? See Fig. 41. k2  45 nt>m k1  20 nt>m m  5 v0 v0  p, y0  1 v0y0 P R O B L E M S E T 2 . 4 The cylindrical buoy of diameter 60 cm in Fig. 43 is floating in water with its axis vertical. When depressed downward in the water and released, it vibrates with period 2 sec. What is its weight? Fig. 41. Parallel springs (Problem 5) Fig. 42. Pendulum (Problem 7) 6. Spring in series. If a body hangs on a spring of modulus , which in turn hangs on a spring of modulus , what is the modulus k of this combination of springs? 7. Pendulum. Find the frequency of oscillation of a pendulum of length L (Fig. 42), neglecting air resistance and the weight of the rod, and assuming to be so small that practically equals .usin u u k2  12 s2k1  8 s1 10. TEAM PROJECT. Harmonic Motions of Similar Models. The unifying power of mathematical meth- ods results to a large extent from the fact that different physical (or other) systems may have the same or very similar models. Illustrate this for the following three systems (a) Pendulum clock. A clock has a 1-meter pendulum. The clock ticks once for each time the pendulum completes a full swing, returning to its original position. How many times a minute does the clock tick? (b) Flat spring (Fig. 45). The harmonic oscillations of a flat spring with a body attached at one end and horizontally clamped at the other are also governed by (3). Find its motions, assuming that the body weighs 8 nt (about 1.8 lb), the system has its static equilibrium 1 cm below the horizontal line, and we let it start from this position with initial velocity 10 cm/sec. 8. Archimedian principle. This principle states that the buoyancy force equals the weight of the water displaced by the body (partly or totally submerged). Fig. 44. Tube (Problem 9) 9. Vibration of water in a tube. If 1 liter of water (about 1.06 US quart) is vibrating up and down under the influence of gravitation in a U-shaped tube of diameter 2 cm (Fig. 44), what is the frequency? Neglect friction. First guess. Fig. 43. Buoy (Problem 8) L θ Body of mass m Water level ( y = 0) y y Fig. 45. Flat spring c02.qxd 10/27/10 6:06 PM Page 69 (c) Torsional vibrations (Fig. 46). Undamped torsional vibrations (rotations back and forth) of a wheel attached to an elastic thin rod or wire are governed by the equation , where is the angle measured from the state of equilibrium. Solve this equation for , initial angle and initial angular velocity .20° sec1 ( 0.349 rad # sec1) 30°( 0.5235 rad) K>I0  13.69 sec 2 uI0us  Ku  0 70 CHAP. 2 Second-Order Linear ODEs 11–20 DAMPED MOTION 11. Overdamping. Show that for (7) to satisfy initial condi- tions and we must have and . 12. Overdamping. Show that in the overdamped case, the body can pass through at most once (Fig. 37). 13. Initial value problem. Find the critical motion (8) that starts from with initial velocity . Graph solution curves for and several such that (i) the curve does not intersect the t-axis, (ii) it intersects it at respectively. 14. Shock absorber. What is the smallest value of the damping constant of a shock absorber in the suspen- sion of a wheel of a car (consisting of a spring and an absorber) that will provide (theoretically) an oscillation- free ride if the mass of the car is 2000 kg and the spring constant equals ? 15. Frequency. Find an approximation formula for in terms of by applying the binomial theorem in (9) and retaining only the first two terms. How good is the approximation in Example 2, III? 16. Maxima. Show that the maxima of an underdamped motion occur at equidistant t-values and find the distance. 17. Underdamping. Determine the values of t corre- sponding to the maxima and minima of the oscillation . Check your result by graphing . 18. Logarithmic decrement. Show that the ratio of two consecutive maximum amplitudes of a damped oscillation (10) is constant, and the natural logarithm of this ratio called the logarithmic decrement, y(t)y(t)  et sin t v0 v* 4500 kg>sec2 t  1, 2, . . . , 5, v0a  1, y0  1 v0y0 y  0 v0>b]>2 c2  [(1  a>b)y0 [(1  a>b)y0  v0>b]>2 c1 v(0)  v0y(0)  y0 equals . Find for the solutions of . 19. Damping constant. Consider an underdamped motion of a body of mass . If the time between two consecutive maxima is 3 sec and the maximum amplitude decreases to its initial value after 10 cycles, what is the damping constant of the system? 20. CAS PROJECT. Transition Between Cases I, II, III. Study this transition in terms of graphs of typical solutions. (Cf. Fig. 47.) (a) Avoiding unnecessary generality is part of good modeling. Show that the initial value problems (A) and (B), (A) (B) the same with different c and (instead of 0), will give practically as much information as a problem with other m, k, . (b) Consider (A). Choose suitable values of c, perhaps better ones than in Fig. 47, for the transition from Case III to II and I. Guess c for the curves in the figure. (c) Time to go to rest. Theoretically, this time is infinite (why?). Practically, the system is at rest when its motion has become very small, say, less than 0.1% of the initial displacement (this choice being up to us), that is in our case, (11) for all t greater than some . In engineering constructions, damping can often be varied without too much trouble. Experimenting with your graphs, find empirically a relation between and c. (d) Solve (A) analytically. Give a reason why the solution c of , with the solution of , will give you the best possible c satisfying (11). (e) Consider (B) empirically as in (a) and (b). What is the main difference between (B) and (A)? yr(t)  0 t2y(t2)  0.001 t1 t1ƒ y(t) ƒ  0.001 y(0), yr(0) yr(0)  2 ys  cyr  y  0, y(0)  1, yr(0)  0 1 2 m  0.5 kg ys  2yr  5y  0 ¢¢  2pa>v* Fig. 47. CAS Project 20 Fig. 46. Torsional vibrations θ 1 0.5 –0.5 –1 6 1084 y t2 c02.qxd 10/27/10 6:06 PM Page 70 2.5 Euler–Cauchy Equations Euler–Cauchy equations4 are ODEs of the form (1) with given constants a and b and unknown function . We substitute into (1). This gives and we now see that was a rather natural choice because we have obtained a com- mon factor . Dropping it, we have the auxiliary equation or (2) . (Note: , not a.) Hence is a solution of (1) if and only if m is a root of (2). The roots of (2) are (3) . Case I. Real different roots give two real solutions and . These are linearly independent since their quotient is not constant. Hence they constitute a basis of solutions of (1) for all x for which they are real. The corresponding general solution for all these x is (4) (c1, c2 arbitrary). E X A M P L E 1 General Solution in the Case of Different Real Roots The Euler–Cauchy equation has the auxiliary equation . The roots are 0.5 and . Hence a basis of solutions for all positive x is and and gives the general solution . (x  0)y  c11x  c2 x y2  1>xy1  x 0.51 m2  0.5m  0.5  0x2ys  1.5xyr  0.5y  0 y  c1x m1  c2x m2 y2(x)  x m2y1(x)  x m1 m1 and m2 m1  1 2 (1  a)  2 1 4 (1  a) 2  b, m2  12 (1  a)  2 1 4 (1  a) 2  b y  xm a  1m2  (a  1)m  b  0 m(m  1)  am  b  0xm y  xm x2m(m  1)xm2  axmxm1  bxm  0 y  xm, yr  mxm1, ys  m(m  1)xm2 y(x) x2ys  axyr  by  0 SEC. 2.5 Euler–Cauchy Equations 71 4LEONHARD EULER (1707–1783) was an enormously creative Swiss mathematician. He made fundamental contributions to almost all branches of mathematics and its application to physics. His important books on algebra and calculus contain numerous basic results of his own research. The great French mathematician AUGUSTIN LOUIS CAUCHY (1789–1857) is the father of modern analysis. He is the creator of complex analysis and had great influence on ODEs, PDEs, infinite series, elasticity theory, and optics. c02.qxd 10/27/10 6:06 PM Page 71