ERWIN KREYS7I G l 'tt "C+ i tt t., i}t.;rl=i| a ,-a. t- ,i i :: B. .- i" :,' it 'a - ..a tr. REsTRlCTED! Not for Sale in the united statesTH EDlTloN t Ť-: 1t{ :]". *"1 a * i." {'\ System of units Length Mass Time Force cgs system centimeter (cm) gram (gm) second (sec) dyne mks system meter (m) kilogram (kg) second (sec) newton (nt) Engineering system foot (ft) slug second (sec) pound (lb) Systems oíUnits. Some lmportant 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 S1), and the abbreviations s (instead of sec), g (instead of gm), and N (instead of nt) are also used. 1 inch (in.) : 2.540000 cm 1 yard (yd) : 3 ft : 91,440000 cm 1 nautical mile : 6080 ft : 1.853184 km 1 acre : 4840 yď2 : 4046.8564 m2 1 slug : 14.59390 kg 1 pound (1b; : 4.448444 nt 1 British thermal unit (Btu) : 1054.35 joules 1 calorie (cal) : 4.1840joules 1 newton (nt) : 105 dynes 1 joule : 107 ergs 1 foot (ft) : 12 in. 1 statute mile (mi) : 30.480000 cm : 5280 ft : 1 ,609344 km 1 mi2 : 640 acres :2.5899881 km2 1 fluid ounce : IlI28 U.S. gallon: 23lll28 in.3 : 29.573730 cm3 1 U.S. gallon : 4 quarts (liq) : 8 pints (liq) : I28 fl oz : 3785.4118 cm3 1 British Imperial and Canadian gallon : I.2OO949 U.S. gallons : 4546.08J cms 1 kilowatt-hour (kwh) : 3414.4 Btu : 3.6 , 106 joules 1 horsepower (hp) : 2542.48 Btu/h : 178.298 callsec : 0.74570 kW 1 kilowatt (kW) : 1000 watts :3414.43 Btu/h :238.662 callsec oF:oC,1.8+32 1o : 60' :3600" : 0.017453293 radian For further details see, for example, D. Halliday, R. Resnick, and J. Walker, Fundamentals of Phlsics. 7th ed.. New York: Wiley. 2005. See also AN American National Standard, ASTMiIEEE Standard Metric Practice. Institute of Electrical and Electronics Engineers, Inc., 445 Hoes Lane, Piscataway, N. J. 08854. (ru)' (u+ (ru)' (:) du Diííerentiation : c1,1' (c constant) t,)':u,+U, : bl'l) + tl'u ll _uU-Uu rl2 dy (Chain rule) dy dx Integration Irr'dx- uI) - f u'ra* r xn*7 |x"dx: _ lc (n+-l)J n-fl II*:hlxl +c r1 Jr"*dx--eo*+c r J sinx dx - -cos x l c r Jcosxdx:sinx-|c r J tan xdx: -ln |cosx| + c r J cot xdx: ln |sin x| + c /.". xdx :ln |secx -t tan x| + c r J csc xdx : ln |cscx - cotx| + c rdxIx l ", .,:-arctan-*cJ x'+a' a a rdxx | ,..---_:arcsin--rc,J Yao-X' a rdxxl--rrcsinh--rc J v?-+-F u o rdxx l ,.---- : arccosh- + c J \/x" - a" a /rin' xdx: lx - jsinLx * c f .or' x dx : ž*+ lsin2x -l c Itun'xdx:tanx-x-|c ["ot'xdx:-cotx- xlc r Jlnxdx-xlnx-x*c r J ,* sínbx dx eo* : o\ ur(asin bx - bcos bx) * c r J n"'cosbx dx eo* : a2 + b2(acos bx -| b sinbx) * c (rn)' : nť-r (e*)' : e* (o*)' : a* In a (sin x)' : cos.t (cos x)' : _ sin x (tan x)' : sec2, (cot x)' : - csc2 .t (Sinh x)' : cosh x (cosh x)' : sinh x ,I(lnx)' ; (1ogo x)' Iogo e (arccot x)' (arcsin x)' : -L\-'- ----- --l x/ 1 - x2 (arccosx)':-# ,1(arctan x)' : -:---- I+x2 1 Ilx2 o/rH EDlTIoN Entineerin8 JoHN WlLEY & soNS, lNc. Mathematics ERWIN KREY'ZIG professor of Mathematics Ohio State University Columbus, Ohio Advanced @\urlLEY vice president and publisher: Laurie Rosatone Editorial Assistant: Daniel Grace Associate Production Director: Lucille Buonocore senior production Editor: ken santor Media Editor: Stefanie Liebman Cover Designer: Madelyn Lesure Cover Photo: O John Sohm/Chromosohm/Photo Researchers This book was set in Times Roman by GGS Information Services Copyright O 2006 John Wiley & Sons, Inc. Al1 rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4410. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (20I) 148-6011, fax (201) 748-6008, E-Mail: PERMREQ@WILEY.COM. Kreyszig, Erwin. Advanced engineering mathematics / Erwin Kreyszig.-9th ed. p. cm. Accompanied by instructor's manual. Includes bibliographical references and index. 1. Mathematical physics. 2.Engineering mathematics. 1. Title. ISBN- 1 3 : 9] 8-0-47 I--/ 2891 -9 ISBN- 10: 0-41 1-7 2897 -7 Printed in Singapore 1098765432 Z PREFACE See also http: //www.wiley.com,/colle ge/kreyszig/ Goal of the Book. Arrangement of Material This new edition continues the tradition of providing instructors and students with a comprehensive and up-to-date resource for teaching and learning engineering mathematics, that is, applied mathematics for engineers and physicists, mathematicians and computer scientists, as well as members of other disciplines. A course in elementary calculus is the sole prerequisite. The subject matter is arranged into seven parts A-G: A Ordinary Differential Equations (ODEs) (Chaps. 1-6) B Linear Algebra. Vector Calculus (Chaps. 7-9) C Fourier Analysis. Partia[ Differential Equations (PDEs) (Chaps. 1I-I2) D Complex Analysis (Chaps. 13-18) E Numeric Analysis (Chaps. I9-2I) F Optimization, Graphs (Chaps. 22-23) G Probability, Statistics (Chaps. 24-25). This is followed by five appendices: App. 1 References (ordered by parts) App. 2 Answers to Odd-Numbered Problems App. 3 Auxiliary Material (see also inside covers) App. 4 Additional Proofs App. 5 Tables of Functions. This book has helped to pave the way for the present development of engineering mathematics. By a modern approach to those areas A-G, this new edition will prepare the student for the tasks of the present and of the future. The latter can be predicted to some extent by a judicious look at the present trend. Among other features, this trend shows the appearance of more complex production processes, more extreme physical conditions (in space travel, high-speed communication, etc.), and new tasks in robotics and communication systems (e.g., fiber optics and scan statistics on random graphs) and elsewhere. This requires the refinement of existing methods and the creation of new ones. It follows that students need solid knowledge of basic principles, methods, and results, and a clear view of what engineering mathematics is all about, and that it requires proficiency in all three phases of problem solving: . Modeling, that is, translating a physical or other problem into a mathematical form, into a mathematlcal model; this can be an algebraic equation, a differential equation, a graph, or some other mathematical expression. . Solving the model by selecting and applying a suitable mathematical method, often requiring numeric work on a computer. . lnterpreting the mathematical result in physical or other terms to see what it practically means and implies. It would make no sense to overload students with all kinds of little things that might be of occasional use. Instead they should recognize that mathematics rests on relatively few basic concepts and involves powerful unifying principles. This should give them a firm grasp on the interrelations among theory, computing, and (physical or other) experimentation. vI Preface PART A Chaps. 1-6 Ordinary Differential Equations (ODEs) Chaps. 1-4 Basic Material Ý chap.5 series solutions Chap.6 Ý Laplace Transforms PART B Chaps. 7-1O Linear Algebra. Vector Calculus Chap.7 Matrices, Linear Systems Chap. 9 vector Differential Catculus Ý chap.8 Eigenvalue Problems Chap. 10 Ý Vector !ntegral Calculus PART c Chaps. 11-12 Fourier Analysis. Partial Differential Equations (PDEs) Chap. 11 Fourier Analysis Chap. 12 Partial Differentia[ Equations PART D Chaps. 13-18 Complex Analysis, Potential Theory Chaps. 13-17 Basic Material Ch.pl8 Potential Theory PART F Chaps. 22-23 Optimization, Graphs Chap.22 l Chap.23 Linear Programming I Graphs, Optimization GUIDES AND MANUALS Maple Computer Guide Mathematica Computer Guide student solutions Manual Instructor's Manual l I I PART E Chaps. 19-21 Numeric Analysis Chap.'l9 Numerics in General Chap.20 Numeric Linear Algebra Chap.21 Numerics for ODEs and PDEs PART G Chaps. 24-25 Probabi Iity, Statistics Chap.24 Data Analysis. Probabitity Theory Chap. 25 Mathematical statistics Preface vll General Features of the Book lnclude: . Simplicity of examples, to make the book teachable-why choose complicated examples when simple ones are as instructive or even better? . Independence of chapters, to proviďeflexibifu in tailoring courses to special 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. . Modern standard notation, to help students with other courses, modern books, and mathematical and engineering journals. Many sections were rewritten in a more detailed fashion, to make it a simpler book. This also resulted in a better balance between theory and applicati.ons. Use of Computers The presentation is adaptable to various levels of technology and use of a computer or graphing calculator., very little or no use, medium use, or intensive use of a graphing calculator or of an unspecified CÁS (Computer Algebra System, Maple, Mathematica, or Matlab being popular examples). In either case texts and problem sets form an entity without gaps or jumps. And many problems can be solved by hand or with a computer or both ways. (For software, see the beginnings of Part E on Numeric Analysis and Part G on Probability and Statistics.) More specifically, this new edition on the one hand gives more prominence to tasks the computer c&nnol do, notably, modeling and interpreting results. On the other hand, it includes CAS projects, CAS problems, and CÁS expeňments, which do require a computer and show its power in solving problems that are difficult or impossible to access otherwise. Here our goal is the combination of intelligent computer use with high-quality mathematics. This has resulted in a change from a formula-centered teaching and learning of engineering mathematics to a more quantitative, project-oriented, and visual approach. CAS experiments also exhibit the computer as an instrument for observations and experimentations that may become the beginnings of new research, for "proving" or disproving conjectures, or for formalizing empirical relationships that are often quite useful to the engineer as working guidelines. These changes will also help the student in discovering the experimental aspect of modern applied mathematics. Some routine and dňll work is retained as a necessity for keeping firm contact with the subject matter. In some of it the computer can (but must not) give the student a hand, but there are plenty of problems that are more suitable for pencil-and-paper work. Major Changes 1. New Problem Sets. Modern engineering mathematics is mostly teamwork.It usually combines analytic work in the process of modeling and the use of computer algebra and numerics in the process of solution, followed by critical evaluation of results. Our problems-some straightforward, some more challenging, some "thinking problems" not accessible by a CAS, some open-ended-reflect this modern situation with its increased emphasis on qualitative methods and applications, and the problem sets take care of this novel situation by including team projects, CAS projects, and writing projects. The latter will also help the student in writing general reports, as they are required in engineering work quite frequently. 2. Computer Experiments, using the computer as an instrument of 'oexperimental mathematics" for exploration and research (see also above). These are mostly open-ended vlII Preface experiments, demonstrating the use of computers in experimentallY finding results, which máy be provable afterward or may be valuable heuristic qualitative guidelines to the engineer, in particular in complicated problems. 3. More on modeling and selecting methods, tasks that usually cannot be automated. 4. Student Solutions Manual and Study Guide enlarged, upon exPlicit requests of the users. This Manual contains worked-out solutions to carefully selected odd-numbered problems (to which App. 1 gives only the final answers) as well as general comments and hints on studying the text and working further problems, including exPlanations on the significance and character of concepts and methods in the various sections of the book. Further Chantes, New Features . Electric circuits moved entirely to Chap. 2, to avoid duplication and rePetition . Second-order ODEs and Higher Order ODEs placed into two separate chapters (2 and 3) . In Chap . 2, app\ications presented before variation of parameters . series solutions somewhat shortened, without changing the order of sections . Material on Laplace transforms brought into a better logical order: Partial fractions used earlier in a more practical approach, unit step and Dirac's delta Put into seParate subsequent sections, differentiation and integration of transforms (not of functions!) moved to a later section in favor of practically more important toPics . second- and third-order determinants made into a separate section for reference throughout the book . Complex matrices made optional . Three sections on curves and their application in mechanics combined in a single section . First two sections on Fourier series combined to provide a better, more direct start . Discrete and Fast Fourier Transforms included . conformal mapping presented in a separate chapter and enlarged . Numeric analysis updated . Backward Euler method included . stiffness of oDEs and systems discussed . List of software (in Part E) updated; another list for statistics software added (in Pat G) . References updated, now including about 75 books published or reprinted after 1990 Suggestions for Courses: A Four-Semester Sequence The material, when taken in sequence, is suitable for four consecutive semester courses, meeting 3-4 hours a week: 1st Semester. ODEs (Chaps. 1-5 or 6) 2nd Semester. Linear Algebra. Vector Analysis (Chaps. 7_10) 3rd Semester. Complex Analysis (Chaps. 13-18) 4th Semester. Numeric Methods (Chaps. I9-2I) Preface ix Suggestions for lndependent 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, Sec. 2I.I) Laplace Transforms (Chap. 6) Matrices and Linear Systems (Chaps. 7-8) Vector Algebra and Calculus (Chaps. 9-10) Fourier Series and PDEs (Chaps. 11-12, Secs. 2I.4-21.1) Introduction to Complex Analysis (Chaps, 13-11) Numeric Analysis (Chaps. 19,2I) Numeric Linear Algebra (Chap. 20) Optimization (Chap s. 22-23) Graphs and Combinatorial Optimization (Chap. 23) Probability and Statistics (Chaps. 24-25) Acknowledgments I am indebted to many of my former teachers, colleagues, and students who helped me directly or indirectly in preparing this book, in particular, the present edition. I profited greatly from discussions with engineers, physicists, mathematicians, and computer scientists, and from their written comments, I want to mention particularly Y. Antipov, D. N. Buechler, S. L. Campbell, R. Carr, P. L. Chambre, V. F. Connolly,Z. Davis, J. Delany, J. W. Dettman, 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, G. Lamb, M. T. Lusk, H. B. Mann, I. Marx, K. Millet, J. D. Moore, W. D. Munroe, A. Nadim, B. S. Ng, J. N. Ong, Jr., D. Panagiotis, A. Plotkin, P. J. Pritchard, W. O. Ray, J. T. Scheick, L. F. Shampine, H. A. Smith, J. Todd, H. Unz, A. L. Villone, H. J. Weiss, A. Wilansky, C. H. Wilcox, H. Ya Fan, and A. D. Zieblr, all from the United States, Professors E. J. Norminton and R. Vaillancourt from Canada, and Professors H. Florian and H. Unger from Europe. I can offer here only an inadequate acknowledgment of my gratitude and appreciation. Special cordial thanks go to Privatdozent Dr. M. Kracht and to Mr. Herbert Kreyszig, MBA, the coauthor of the Student Solutions Manual, who both checked the manuscript in all details and made numerous suggestions for improvements and helped me proofread the galley and page proofs. Furthermore, I wish to thank John Wiley and Sons (see the list on p, iv) as well as GGS Information Services, in particular Mr. K. Bradley and Mr. J. Nystrom, for their effective cooperation and great care in preparing this new edition. Suggestions of many readers worldwide were evaluated in preparing this edition. Further comments and suggestions for improving the book will be gratefully received. ERWIN KREYSZIG C,O N, ,T, ,,E, N,T,S PART A Ordinary Differential Equations (ODEs) 1 cHApTER t First-Order ODEs 2 1.1 Basic Concepts. Modeling 2 1.2 Geometric Meaning of y' : í(x, y). Direction Fields 9 1.3 Separable ODEs. Modeling 12 1.4 Exact ODEs. Integrating Factors 19 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 26 1.6 Orthogonal Trajectories. Optional 35 1.7 Existence and Uniqueness of Solutions 37 Chapter 1 Review Questions and Problems 42 Summary of Chapter 1 43 CHApTER 2 Second-Order Linear ODEs 45 2.1 Homogeneous Linear ODEs of Second Order 45 2.2 Homogeneous Linear ODEs with Constant Coefficients 53 2.3 Differential Operators. Optional 59 2.4 Modeling: Free Oscillations. (Mass-Spring System) 61 2.5 Euler-Cauchy Equations 69 2.6 Existence and Uniqueness of Solutions. Wronskian 73 2J Nonhomogeneous ODEs 78 2.8 Modeling: Forced Oscillations. Resonance 84 2.9 Modeling: Electric Circuits 91 2.10 Solution by Variation of Parameters 98 Chapter 2 Review Questions and Problems 1O2 Summary of Chapter 2 103 CHAITER l Higher Order Linear ODEs l05 3.1 Homogeneous Linear ODEs l05 3.2 Homogeneous Linear ODEs with Constant Coefficients l1l 3.3 Nonhomogeneous Linear ODEs 116 Chapter 3 Review Questions and Problems 122 Summary of Chapter 3 123 cHApTER 4 Systems of oDEs. Phase Plane. Qualitative Methods 4.0 Basics of Matrices and Vectors 124 4.1 Systems of ODEs as Models 130 4.2 Basic Theory of Systems of ODEs 136 4.3 Constant-Coefficient Systems. Phase Plane Method l39 4.4 Criteria for Critical Points. Stability 147 4.5 Qualitative Methods for Nonlinear Systems l5l 4.6 Nonhomogeneous Linear Systems of ODEs l59 Chapter 4 Review Questions and Problems 163 Summary of Chapter 4 164 cHAITER s Series Solutions of ODEs. Special Functions 166 5.1 Power Series Method 167 5.2 Theory of the Power Series Method 17O 124 xl xll Contents 5.3 Legendre's Equation. Legendre Polynomials P-(x) 177 5.4 Frobenius Method l82 5.5 Bessel's Equation. Bessel Functions -I,(r) l89 5.6 Bessel Functions of the Second Kind Y"(x) 198 5.7 Sturm-Liouville Problems. Orthogonal Functions 203 5,8 Orthogonal Eigenfunction Expansions 2l0 Chapter 5 Review Questions and Problems 217 Summary of Chapter 5 2l8 CHAITER 6 Laplace Transforms 22O 6.1 Laplace Transform. Inverse Transform. Linearity. s-Shifting 221 6.2 Transforms of Derivatives and Integrals. ODEs 227 6.3 Unit Step Function. r-Shifting 233 6.4 Short Impulses. Dirac's Delta Function. Partial Fractions 241 6.5 Convolution. Integral Equations 248 6.6 Differentiation and Integration of Transforms. 254 6J Systems of ODEs 258 6.8 Laplace Transform: General Formulas 264 6.9 Table of Laplace Transforms 265 Chapter 6 Review Questions and Problems 267 Summary of Chapter 6 269 PART B Linear Algebra. Vector Calculus 271 CHAeTER 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 272 7.1 Matrices, Vectors: Addition and Scalar Multiplication 272 7.2 Matrix Multiplication 278 7.3 Linear Systems of Equations. Gauss Elimination 287 7.4 Linear Independence. Rank of a Matrix. Vector Space 296 7.5 Solutions of Linear Systems: Existence, Uniqueness 302 7.6 For Reference: Second- and Third-Order Determinants 306 7.7 Determinants. Cramer's Rule 308 7.8 Inverse of a Matrix. Gauss-Jordan Elimination 3l5 7.9 Vector Spaces, Inner Product Spaces. Linear Transformations. Optional 323 Chapter 7 Review Questions and Problems 330 Summary of Chapter 7 33l cHAnTER 8 Linear Algebra: Matrix Eigenvalue Problems 8.1 Eigenvalues, Eigenvectors 334 8.2 Some Applications of Eigenvalue Problems 340 8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 345 8.4 Eigenbases. Diagonalization. Quadratic Forms 349 8.5 Complex Matrices and Forms. Optional 356 Chapter 8 Review Questions and Problems 362 Summary of Chapter 8 363 333 Contents CHApTER 9 Vector Differential Calculus. Grad, Div, Curl 364 9.1 Vectors in 2-Space and 3-Space 364 9.2 Inner Product (Dot Product) 371 9.3 Vector Product (Cross Product) 377 9.4 Vector and Scalar Functions and Fields, Derivatives 384 9.5 Curves. Arc Length. Curvature. Torsion 389 9.6 Calculus Review: Functions of Several Variables. Optional 4OO 9.7 Gradient of a Scalar Field. Directional Derivative 403 9.8 Divergence of a Vector Field 41O 9.9 Curl of a Vector Field 414 Chapter 9 Review Questions and Problems 416 Summary of Chapter 9 417 CHAnTER lo Vector lntegral Calculus. lntegral Theorems 42O l0.1 Line Integrals 42O l0.2 Path Independence of Line Integrals 426 l0.3 Calculus Review: Double Integrals. Optional 433 l0.4 Green's Theorem in the Plane 439 l0.5 Surfaces for Surface Integrals 445 10.6 Surface Integrals 449 10.7 Triple Integrals. Divergence Theorem of Gauss 458 l0.8 Further Applications of the Divergence Theorem 463 10.9 Stokes's Theorem 468 Chapter 10 Review Questions and Problems 473 Summary of Chapter 10 474 PART c Fourier Analysis. Partial Differential Equations (PDEs) 477 CHAnTER ll Fourier Series, lntegrals, and Transforms 478 l1.1 Fourier Series 478 11.2 Functions of Any Period p : 2L 487 l1.3 Even and Odd Functions. Half-Range Expansions 49O 11.4 Complex Fourier Series. Optional 496 11.5 Forced Oscillations 499 11.6 Approximation by Trigonometric Polynomials 502 11.7 Fourier Integral 506 11.8 Fourier Cosine and Sine Transforms 5l3 11.9 Fourier Transform. Discrete and Fast Fourier Transforms 5l8 11.10 Tables of Transforms 529 Chapter 11 Review Questions and Problems 532 Summary of Chapter 11 533 cHAITER 12 Partial Differential Equations (PDEs) 535 12.1 Basic Concepts 535 12.2 Modeling: Vibrating String, Wave Equation 538 12.3 Solution by Separating Variables. Use of Fourier Series 54O 12.4 D'Alembert's Solution of the Wave Equation. Characteristics 548 12.5 Heat Equation: Solution by Fourier Series 552 xllI xlv Contents 12.6 Heat Equation: Solution by Fourier Integrals and Transforms 562 12.7 Modeling: Membrane, Two-Dimensional Wave Equation 569 12.8 Rectangular Membrane. Double Fourier Series 571 12.9 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series 579 12.10 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential 587 12.11 Solution of PDEs by Laplace Transforms 594 Chapter 12 Review Questions and Problems 597 Summary of Chapter 12 598 Complex Analysis 601 CHAPTER 13 Complex Numbers and Functions 602 13.1 Complex Numbers. Complex Plane 602 13.2 Polar Form of Complex Numbers. Powers and Roots 607 13.3 Derivative. Analytic Function 612 13.4 Cauchy-Riemann Equations. Laplace's Equation 618 l3.5 Exponential Function 623 13.6 Trigonometric and Hyperbolic Functions 676 13.7 Logarithm. General Power 630 Chapter 13 Review Questions and Problems 634 Summary of Chapter 13 635 CHAPTER 14 Complex lntegration 637 14.1 Line Integral in the Complex Plane 637 14.2 Cauchy's Integral Theorem 646 14.3 Cauchy's Integral Formula 654 14.4 Derivatives of Analytic Functions 658 Chapter 14 Review Questions and Problems 662 Summary of Chapter 14 663 CHAPTER 15 Power Series, Taylor Series 664 15.1 Sequences, Series, Convergence Tests 664 15.2 Power Series 673 15.3 Functions Given by Power Series 678 15.4 Taylor and Maclaurin Series 683 l5.5 Uniform Convergence. Optional 691 Chapter 15 Review Questions and Problems 698 Summary of Chapter 15 699 CHAPTER 16 Laurent Series. Residue lntegration 7O1 16.1 Laurent Series 7O1 16.2 Singularities and Zeros.Infinity 7O7 16.3 Residue Integration Method 712 16.4 Residue Integration of Real Integrals 718 Chapter 16 Review Questions and Problems 726 Summary of Chapter 16 727 CHAPTER 17 Conformal Mapping 728 17.1 Geometry of Analytic Functions: Conformal Mapping 729 17.2 Linear Fractional Transformations 734 17.3 Special Linear Fractional Transformations 737 PART D Contents 17.4 Conformal Mapping by Other Functions 742 17.5 Riemann Surfaces. Optional 746 Chapter 17 Review Questions and Problems 747 Summary of Chapter 17 748 CHArTER 18 Complex Analysis and Potential Theory 749 l8.1 Electrostatic Fields 75O 18.2 Use of Conformal Mapping. Modeling 754 l8.3 Heat Problems 757 l8.4 Fluid Flow 761 l8.5 Poisson's Integral Formula for Potentials 768 18.6 General Properties of Harmonic Functions 771 Chapter 18 Review Questions and Problems 775 Summary of Chapter 18 776 PART E Numeric Analysis 777 778Software CHAPTER 19 Numerics in General 780 19.1 Introduction 78O 19.2 Solution of Equations by Iteration 787 l9.3 Interpolation 797 19.4 Spline Interpolation 8l0 l9.5 Numeric Integration and Differentiation 817 Chapter 19 Review Questions and Problems 83O Summary of Chapter 19 83l CHAnTER 20 Numeric Linear Algebra 833 20.1 Linear Systems: Gauss Elimination 833 20.2 Linear Systems: LU-Factorization, Matrix Inversion 84O 20.3 Linear Systems: Solution by lteration 845 20/ Linear Systems: Ill-Conditioning, Norms 85l 20.5 Least Squares Method 859 20.6 Matrix Eigenvalue Problems: Introduction 863 20.7 Inclusion of Matrix Eigenvalues 866 20.8 Power Method for Eigenvalues 872 20.9 Tridiagonalization and QR-Factorization 875 Chapter 20 Review Questions and Problems 883 Summary of Chapter 20 884 CHAPTER 2l Numerics for oDEs and PDEs 886 21.1 Methods for First-Order ODEs 886 21.2 Multistep Methods 898 213 Methods for Systems and Higher Order ODEs 9O2 21.4 Methods for Elliptic PDEs 9O9 21.5 Neumann and Mixed Problems. Irregular Boundary 917 21.6 Methods for Parabolic PDEs 922 21.7 Method for Hyperbolic PDEs 928 Chapter 21 Review Questions and Problems 93O Summary of Chapter 21 932 xvi PART F PART G Contents Optimization, Graphs 935 cHAnTER 22 Unconstrained Optimization. Linear Programming 22.1 Basic Concepts. Unconstrained Optimizatíon 936 222 Linear Programming 939 22.3 Simplex Method 944 22.4 Simplex Method: Difficulties 947 Chapter 22 Review Questions and Problems 952 Summary of Chapter 22 953 CHArTER 23 Graphs. Combinatorial optimization 954 23.1 Graphs and Digraphs 954 23.2 Shortest Path Problems. Complexity 959 23.3 Bellman's Principle. Dijkstra's Algorithm 963 23.4 Shortest Spanning Trees. Greedy Algorithm 966 23.5 Shortest Spanning Trees. Prim's Algorithm 97O 23.6 Flows in Networks 973 23J Maximum Flow: Ford-Fulkerson Algorithm 979 23.8 Bipartite Graphs. Assignment Problems 982 Chapter 23 Review Questions and Problems 987 Summary of Chapter 23 989 Probability, Statistics 991 cHAeTER 24 Data Analysis. Probability Theory 993 24.1 Data Representation. Average. Spread 993 24.2 Experiments, Outcomes, Events 997 24.3 Probability 1000 24.4 Permutations and Combinations 1006 24.5 Random Variables. Probability Distributions 'l0l0 24.6 Mean and Variance of a Distribution 1016 24.7 Binomial, Poisson, and Hypergeometric Distributions 1020 24.8 Normal Distribution 1026 249 Distributions of Several Random Variables 1032 Chapter 24 Review Questions and Problems 104l Summary of Chapter 24 1042 CHApTER 25 Mathematica[ Statistics 1044 25.1 Introduction. Random Sampling 1044 25.2 Point Estimation of Parameters 1046 253 Confidence Intervals 1049 25.4 Testing Hypotheses. Decisions 1058 25.5 Quality Control l0ó8 25.6 Acceptance Sampling 1073 25J Goodness of Fit. 72-Test 1076 25.8 Nonparametric Tests 1080 259 Regression. Fitting Straight Lines. Correlation l083 Chapter 25 Review Questions and Problems 1092 Summary of Chapter 25 1093 936 Contents xYll AppENDlx t References A1 AppENDlx 2 Answers to Odd-Numbered Problems A4 AppENDlx r Auxiliary Material A60 A3.1 Formulas for Special Functions A60 A3.2 Partial Derivatives A66 A3.3 Sequences and Series A69 A3.4 Grad, Div, Curl, V2 in Curvilinear Coordinates A71 AppENDlx 4 Additional Proofs A74 AppENDlx s Tables A94 PHoTo CREDITS Pl |NDEX l1 PART Ordinary Differentia1 Equations (ODEs) First-Order ODEs Second-Order Linear ODEs Higher Order Linear ODEs Systems of ODEs. Phase Plane. Qualitative Methods Series Solutions of ODEs. Special Functions Laplace Transforms Differential equations are of basic importance in engineering mathematics because many physical laws and relations appear mathematically in the form of a differential equation. In Part A we shall consider various physical and geometric problems that lead to differential equations, with emphasis on modeling, that is, the transition from the physical situation to a "mathematical model." In this chapter the model will be a differential equation, and as we proceed we shall explain the most important standard methods for solving such equations. Part A concerns ordinary differential equations (ODEs), whose unknown functions depend on a single vartab(e. Partial differential equations (PDEs), involving unknown functions of several variables, follow in Part C. ODEs are very well suited for computers. Numeric methods for ODEs can be studied directly after Chaps. l or 2. See Secs. 2I.I-2I.3, which are independent of the other sections on numerics. cHAPTER cHAPTER cHAPTER cHAPTER cHAPTER cHAPTER 1 2 3 4 5 6 l ,Cl,,H',,*,.P.,T''Et, ,,_ l First-Order ODEs In this chapter we begin our program of studying ordinary differential equations (ODEs) by deriving them from physical or other problems (modeling), solving them by standard methods, and interpreting solutions and their graphs in terms of a given problem. Questions of existence and uniqueness of solutions will also be discussed (in Sec. 1.7). We begin with the simplest ODEs, called ODEs of the first order because they involve only the first derivative of the unknown function, no higher derivatives. Our usual notation for the unknown function will be y(x), or y(t) if the independent variable is time t. If you wish, use your computer algebra system (CAS) for checking solutions, but make sure that you gain a conceptual understanding of the basic terms, such as ODE, direction field, and initial value problem. COMMENT. Numerics for first-order ODEs can be studied immediately after this chapter. See Secs. 2I.I-2í.2, which are independent of other sections on numerics. P rere quis ite : Integral calculus. Sections that may be omitted in a shorter course: 1.6, I.7. References and Answers to Problems; App. 1 Part A, and App.2 l.] Basic Concepts. Modelint If we want to solve an engineering problem (usually of a physical nature), we first have to formulate the problem as a mathematical expression in terms of variables, functions, equations, and so forth. Such an expression is known as a mathematical model of the given problem. The process of setting up a model, solving it mathematically, and interpreting the result in physical or other terms is called mathematical modeling or, briefly, modeling. We shall illustrate this process by various examples and problems because modeling requires experience. (Your computer may help you in solving but hardly in setting up models.) Since many physical concepts, such as velocity and acceleration, are derivatives, a model is very often an equation containing derivatives of an unknown function. Such a model is called a differential equation. Of course, we then want to find a solution (a function that satisfies the equation), explore its properties, graph it, find values of it, and interpret it in physical terms so that we can understand the behavior of the physical system in our given problem. However, before we can turn to methods of solution we must first define basic concepts needed throughout this chapter. SEC. 1.1 Basic Concepts. Modeling Falling stone !" :8 = const. (Sec.1.1) Velocity U ,l, parachuttst mu'=mg-buz (Sec. 1.2) Outflowing water h, = -hli (Sec. 1.3) water level á l, Displacement y Vibrating mass on a spring my"+lzy-O (Secs. 2,4, 2,8) Beats of a vibrating system y' + @oy = cos (Dí,@O= @ (Sec. 2.8) current 1 in an RLC circuit LI" + RI'* Lt = n'() (Sec. 2.9) Deformation of a beam EIyi' = f(x) (Sec. 3.3) -Pendu l u m L0"+gsin0=0 (Sec.4.5) Lotka-Volterra predator-prey model y'.=ay.-by.y^ "I"z ťr= fu,y r- ly, (Sec. 4.5) Fig. t. Some applications of differential equations CHAP. l First-Order ODEs An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function, which we usually call y(x) (or sometimes y(/) if the independent variable is time r). The equation may also contain y itself, known functions of x (or t), and constants. For example, (1) y' : cos.tr, (2) y" +9y:O, (3) *'y"'y' + 2e*y" : (x2 + 2)y2 are ordinary differential equations (ODEs). The term ordinary dístinguishes them from partial dffirential 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 .r and y is E2u ó2u a*'- 6,,rz :0' PDEs are more complicated than ODEs; 'rr.v rvirr be considered in Chap. t2. An ODE is said to be of order n tf 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 intoODEsof firstorder,secondorder,andsoon.Thus,(1)isof firstorder, (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 y' and may contain y and any given functions of x. Hence we can write them as (4) or often in the form F(x, y, y') : 0 y' : f(x, y). This is called the explicit form, in contrast with the implicit form (4). For instance, the implicit ODE ,-'y' - 4y' : 0 (wherc x * 0) can be written explicitly as y, - 4r"y,. Concept of Solution A function y : h(x) iscalledasolutionofagivenODE(4)onsomeopenintervalalxlbifh(x)isdefined and differentiable throughout the interval and is such that the equation becomes an identity if yand y'arereplaced withhandh',respectively.Thecurve(thegraph)of hiscalled a solution curye. Here, open interval a 1x 1b means that the endpoints a andb are not regarded as pointsbelongingtotheinterval. Also, alx < Ďincludesinfinite intervals -m ( x 1b, a 1 x { @, -co < x 1 m (the real line) as special cases. - SEC. 1.1 Basic Concepts. Modeling ExAMPLE I ExAMPLE 2 verification of solution y : h(x) : clx (c an arbitrary constant, r * 0) is y' : h'(x) : -clx2, and multiply by -r to get ry' - - 5 a solution of ry' To verify this, differentiate, -clx: -y. Thus, ry' : -y, the given ODE. t solution curves The ODE y' : dyldx : cos í can be solved directly by integration on both sides. Indeed, using calculus, we obtain y : I cos x dx : sin.r * c, where c is an arbitrary constant. This is afamily of solutions. Each value of c, for instance, 2.75 or 0 or -8, gives one of these curves. Figure 2 shows some of them, for c : -3, -2, -1,0, I,2,3,4. Fig.2. Solutions y : sinx * c of the oDEyl : cosx Exponential Growth, Exponential Decay From calculus we know that y : ,"a 1, any constant) has the derivative (chain rule!) ,':*:3ce3t:3y. This shows that y is a solution oíy' :3y. Hence this ODE can model exponential growth, for instance, of animal populations or colonies of bacteria. 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 lnw.l We shall say more about this topic in Sec. 1.5. Similarly, y' - -0.2y (with a minus on the right!) has the solution y : ce-o'Zt Hence this ODE models exponential decay, for instance, of a radioactive substance (see Example 5). Figure 3 shows solutions for some positive c. Can you find what the solutions look like for negative c? t y 2.5 2 1.5 1 0.5 0-o 2 4 6 8 10 12 14 t Fig.3. Solutions of yl : -0.2y in Example 3 lNamed after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766-1834). l ExAMPLE 3 CHAP. l First-Order ODEs 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., c : 6.45 or 0 or -2.01) 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. lnitial 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 y(.ro) : }o, with given values x9 and y6, 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 (xg, y6) in the xy-plane. An ODE together with an initial condition is called an initial value problem. Thus, if the ODE is explicit,y' : í(x,y), the initial value problem is of the form (5) y' : f (x, y), EXAMPLE 4 lnitialValue Problem Solve the initial value problem ,},,(.rro) : .}o, : 3"y, y(0) : 5.7. SOlUtiOn. The general solution is y(r) - ce3Í; see Example 3. From this solution and the initial condition we obtain,v(0) : ,uo : c : 5.J. Hence the initial value problem has the solution y(x) : 5.Je3*. This is a particular solution. 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 typical steps of modeling in detail: 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 CÁS) may perhaps give you a hand in Step Z,bllt Steps 1 and 3 are basically your work. And Step 2 ,dY "dx l SEC. 1.1 Basic Concepts. Modeling E XÁ.MPI_ E] 5 requires a solid knowledge and good understanding of solution methods available to youyou have to choose the method for your work by hand or by the computer. Keep this in mind, and always check computer results for errors (which may result, for instance, from false inputs). Radioactivity. Exponential Decay Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time. Physical Infotmation Experiments show that at each instant a radioactive substance decomposes at a íate proportional to the the amount present. Step 1. Setting up a mathematical model (a dffirential equation) of the physical process. Denote by y(r) the amount of substance still present at any time r. By the physical law, the time rate of change y'(t) : dyldt is proportional to y(r). Denote the constant of proportionality by k. Then The value of k is known fiom experiments fbr various radioactive substances (e.g., t : _ 7.4, l0-11sec-1, approximately, for radium uuRu"u). k is negative because y(l) decreases with time. The given initial amount is 0.5 g. Denote the corresponding time by / : 0. Then the initial condition is _y(0) : 0.5. This is the instant at which the process begins; this motivates the term initial condition (whích, however, is also used more generally when the independent variable is not time or when you choose a / other than t : 0). Hence the model of the process is the initial value problem dy dt : ky, y(0) : 0.5. Step 2. Mathematical solution. As in Example 3 we conclude that the ODE (6) models exponential decay and has the general solution (with arbitrary constant c but definite given k) Y(í) : cekt _ We now use the initial condition to determine c. Since }(0) : c from (8), this gives y(0) : c : 0.5. Hence the particular solution governing this process is (9) ,y(/) : 0.5ekt (Fig. a). Always check your result-it may involve human or computer errors! Verify by differentiation (chain rule!) that your solution (9) satisfies (7) as well as y(0) : 0.5: dy dt : O.5kekt : k. 0.5ekt : tcy, y(0):0.5eo:0.5. Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time /. It starts from the correct given initial amount and decreases with time because k (the constant of proportionality, depending on the kind of substance) is negative. The limit of y as t ---> cn is zero. l y 0.5 o.4 0.3 o.2 0.1 0 2,5 Fig. 4. Radioactivity (Exponential decay, y : 0.5 ekt, with k : -].5 as an example) dy dt : ky.(6) (1) (8) 1.5U.5 CHAP. l First-Order ODEs EXAMPLE 6 A Geometric Application Geometric problems may also lead to initial value problems. For instance, find the curve through the point (1, 1) in thexy-plane having at each of its points the slope -ylx. Solution. The slope y' should equal -ylx. This gives the ODE ,l : -ylx.Its general solution isy : clx (see Example 1). This is a family of hyperbolas with the coordinate axes as asymptotes. Now, for the curve to pass through (1, 1), we must have y : l when x : 1. Hence the initial condition is y(l):1.Fromthisconditionandy:clxwegety(1):cll:1;thatis,c:l.Thisgivestheparticular solution y : Ilx (drawn somewhat thicker in Fig, 5). \ _3 l _4 l Fig. 5. Solutions of yl : -y/x ( las) / hyperbo l y l 4 l ? l 2 1 _1 \-r -Z Fig. 6. Particular solutions and singular solution in Problem ]6 I _2-3-4- cALcULUs Solve the ODE by integration. L. y' : - sin n,x 2. 3. y' : *n*2l2 4. , -a-y :e Y' : corh 4" E vERlFlcATloN oF soluTloN State the order of the ODE. Verify that the given function is a solution. (a, b, c are arbitrary constants.) 5.y':Ily2, y:tan(x+c) 6. y" l r2y : g, y : acos Ťx * b stnrrx 7. y" ,t- 2y' * 10y : 0, ! : 4e-I sin 3x 8. y' l2y:4(x * t)2, !:5e-2* * 2x2 1_ 2x _l I 9.y"':cos.tr, y- -sinxi ax2+bx-lc @ INITIAL vALuE pRoBLEMs Verify that y is a solution of the ODE. Determine from y the particular solution satisfying the given initial condition. Sketch or graph this solution. 10. y/ : 0.5}, ! : ce0,5*, y(2) : 2 11. y' : I * 4y', y : *tan (2x l c), y(0) : 0 12. y' : y - r, !: ce'* x l l, y(0):3 13, y' * Zxy : 0, y : ce-*', y(I) : Ile 14.y':ytanx, y:c secí, y(D:Žrr 15. (Existence) (A) Does the ODE y'2 - - 1 have a (real) solution? (B) Does the ODE |y'| + lyl : 0 have a general solution? 16. (Singular solution) An ODE may sometimes have an additional solution that cannot be obtained from the general solution and is then called a singular solution, The ODE y'' - xy' + y : 0 is of the kind. Show by differentiation and substitution that it has the general solution y : cx - c2 andthe singular solution y : x2l4. Explain Fig. 6. @ MoDELlNG,AppLlcATloNs The following problems will give you a first impression of modeling. Many more problems on modeling follow throughout this chapter. 17. (Falling body) If we drop a stone, we can assume air resistance ("drag") to be negligible. Experiments show that under that assumption the acceleration y" : d2yldtz of this motion is constant (equal to the so-called acceleration of gravity g : 9.80 m/sec2 : 32 ftlsec2), State this as an ODE for y(r), the distance fallen as a function of time r. Solve the ODE to get the familiar law of free fall, y : 8 12. _ - SEC. 1.2 Geometric Meaning of y' : í(r, y).Direction Fields 18. (Falling body) If in Prob. l] the stone starls at t : 0 from initial position y6 with initial velocity u : ,)n, show that the solution is y : 8 12 + uot i y6. How long does a fall of 100 m take if the body falls from rest? A fall of 200 m? (Guess first.) 19. (Airplane takeoff) If an airplane has a run of 3 km, starts with a speed 6 m/sec, moves with constant acceleration, and makes the run in 1 min, with what speed does it take off? 20. (Subsonic flight) The efficiency of the engines of subsonic airplanes depends on air pressure and usually is maximum near about 36 000 ft. Find the air pressure y(x) at this height without calculation. Physical information The rate of change y'1"; is proportional to the pressure, and at 18 000 ft the pressure has decreased to half its value y6 at sea level. 21. (Half-life) The half-life of a radioactive substance is the time in which half of the given amount disappears. Hence it measures the rapidity of the decay. What 9 is the half-life of radium uuRu"u (in years) in Example 5? (Interest rates) Show by algebra that the investment y(r) from a deposit yo after / years at an interest rate r is y.,!) : y6[1 -F r]t (Interest compounded annually) ya(/) : }o[1 + (rl365))365t (Interest compounded daily). Recall from calculus that LI + (Iln)ln -> e as n -> ml hence |I + (rln)lnt ---> ert; thus y"(t) : yoeň (Interest compounded continuously). What ODE does the last function satisfy? Let the initial investment be $1000 and r : 6Vo. Compute the value of the investment after i year and after 5 years using each of the three formulas. Is there much difference? y' : f(x,y) Direction Fields A first-order ODE (1) has a simple geometric interpretation. From calculus you know that the derivative y'(x) of y(x) is the slope of y(x). Hence a solution curve of (1) that passes through a point (xo, yo) must have at that point the slope y' (x equal to the value of / at that point; that is, y'(x : í(xo, yo). Read this paragraph again before you go on, and think about it. It follows that you can indicate directions of solution curves of (1) by drawing short straight-line segments (lineal elements) in the xy-plane (as in Fig. 7a) and then fitting (approximate) solution curves through the direction field (or slope freld) thus obtained. This method is important for two reasons. 1. You need not solve (1). This is essential because many ODEs have complicated solution formulas or none at all. 2. The method shows, in graphical form, the whole family of solutions and their typical properties. The accuracy is somewhat limited, but in most cases this does not matter. Let us illustrate this method for the ODE I y:Xy.(2) 1o CHAP. l First-Order ODEs Direction Fields by a CAS (Computer Algebra System). A CAS plots lineal elements at the points of a square grid, as in Fig. J a for (2), into which you can fit solution curves. Decrease the mesh size of the grid in regions where f (x, y) varies rapidly. Direction Fields by Using Isoclines (the Older Method). Graph the curves f (x, y) : k : const, called isoclines (meaning curves of equal inclination). For (2) these are the hyperbolas .f(x, y) : xy : k : const (and the coordinate axes) in Fig. 7b. By (1), these are the curves along which the derivative y' is constant. These are not yet solution curves-don't get confused. Along each isocline draw many parallel line elements of the coíTesponding slope k. This gives the direction field, into which you can now graph approximate solution curves. We mention that for the ODE (2) tnFig.7 we would not need the method, because we shall see in the next section that ODEs such as (2) can easily be solved exactly. For the time being, let us verify by substitution that (2) has the general solution y(x) : ,nr2l2 (c arbitrary). Indeed, by differentiation (chainrule!) we get y' : x(cet2lz\ : xy. Of course, knowing the solution, we now have the advantage of obtaining a feel for the accuracy of the method by comparing with the exact solution. The particular solution in Fig.7 through (x, y): (1, 2) must satisfy y(1) : 2. Thus, 2 : cetl2, c : 2l\/i : L2I3, and the particular solution is y("r) : I.2I3e"l2. A famous oDE for which we do need direction fields is y' : 0.I(I - x2) (It is related to the van der Pol equation of electronics, which we shall discuss in Sec. 4.5.) The direction field in Fig. 8 shows lineal elements generated by the computer. We have also added the isoclines for k: -5, -3,i,1 as well as three typical solution curves, one that is (almost) a circle and two spirals approaching it from inside and outside. (a) By a CAS (b) By isoclines Fig. 7. Direction field of y' : xy y (3) l \ \\ lll 1 \\ \ \ \\\\ \\ \ \ \\\\ \\ \ \ \\\\ \\\\ -1//,z.zz / /,/./,/_ / / / // / / / // / / / // l/ / // ll/ // SEC. 1.2 Geometric Meaning of y' -- f (*, y).Direction Fields --\\.-:\\ -\\\ Fig.8. Direction field of y' : o.1t'' - n - ; On Numerics Direction fields give"all" solutions, but with limited accuracy. If we need accurate numeric values of a solution (or of several solutions) for which we have no formula, we can use a numeric method. If you want to get an idea of how these methods work, go to Sec. 2I.I and study the first two pages on the Euler-Cauchy method, which is typical of more accurate methods later in that section, notably of the classical Runge-Kutta method. It would make little sense to intemrpt the present flow of ideas by including such methods here; indeed, it would be a duplication of the material in Sec. 2I.I. For an excursion to that section you need no extra prerequisites; Sec. 1.1just discussed is sufficient. y : X =1 4 k -,/ r\\ E DlREcTloN FIELDs, soLUTIoN cURvEs Graph a direction field (by a CAS or by hand). In the field graph approximate solution curves through the given point or points (x, y) by hand. I. y' : * - !, (0, 0), (0, 1) 2. 4yy' * -9x, (2,2) 3. y' : I l y', (Žr, D 4. y' : y - 2y2, (0,0), (0,0.25), (0,0.5), (0, 1) 5. y' : x2 - Lly, (I, -2) 6. y' : 1 -l sin y, (- 1, 0), (I, -4) 7. y' : y3 + x3, (0, 1) 8. y' : Zxy -| 1, (- I,2), (0, 0), (I, -2) 9. y' : y tanh x - 2, (-1, -2), O, O), (I,2) 10. y' - gal', (1, 1), (2,2), (3,3) EL15l AccuRAcy Direction fields are very useful because you can see solutions (as many as you want) without solving the ODE, which may be difficult or impossible in terms of a formula. 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. |1. y' : sínlrrx 12. y' : Ilx2 13. y' : -2y (Sol. y : ,r-'*) 14. y' : 3ylx (So1. y : cx3) 15. y' : -In x E6JE-] MoT!oNs A body moves on a straight line, with velocity as given, and y(r) is its distance from a fixed point 0 and r time. Find a model of the motion (an ODE). Graph a direction field. 1l R=+ k=I, h=-5 h=-3 _4 \ at:i CHAP. l First-Order ODEs In it sketch a solution curve corresponding to the given initial condition. 16. Velocity equal to the reciprocal of the distance, y(1) : 1 17. Product of velocity and distance equal to -t,y(3) : -3 1,8. Velocity plus distance equal to the square of time, y(0) : 6 19. (Skydiver) Two forces act on a parachutist, the attraction by the earth mg (n : mass of person plus equipment, 8 : 9.8 m/sec2 the acceleration of gravity) and the air resistance, assumed to be proportional to the square of the velocity u(r). Using Newton's second law of motion (mass X acceleration : resultant of the forces), set up a model (an ODE for u(r)). Graph a direction field (choosing m and the constant of proporlionality equal to 1). Assume that the parachute opens when u : 10 m/sec. Graph the corresponding solution in the field. What is the limiting velocity? 20. CAS PROJECT. Direction Fields. Discuss direction fields as follows. (a) Graph a direction field for the ODE y' : I - y and in it the solution satisfying y(0) : 5 showing exponential approach. Can you see the limit of any solution directly from the ODE? For what initial condition will the solution be increasing? Constant? Decreasing? (b) What do the solution curves of y' - -x3ly3 look like, as concluded from a direction field. How do they seem to differ from circles? what are the isoclines? What happens to those curves when you drop the minus on the right? Do they look similar to familiar curves? First, guess. (c) Compare, as best as you can, the old and the computer methods, their advantages and disadvantages. Write a short report. (1) (2) ' dx: [r-l dx l c. /rrrl dy : Irn dx l c.(3) 1.3 Separable ODEs. Modelint Many practically useful ODEs can be reduced to the form s1)y' : í(x) by purely algebraic manipulations. Then we can integrate on both sides with respectto x, obtaining f s{l) l On the left we can switch to y as the variable of integration. By calculus, }' dx - dy, so that If f and g aíe continuous functions, the integrals in (3) exist, and by evaluating them we :j;xffi ff 1:T*il::';;.Tíi]Ji::T::i"*;;,:"J;,-,,:":t:.:,ť[i jfi ,:i,""t,:: are now separated: .r appears only on the right and y only on the left. ExAMPLE, :,,"::":'" :o,,The ODE y' : I + y2 is separable because it can be written I,|y2 :dx. Byintegration, arctany:x+ c or y:tan(x-|c). dy SEC. 1.3 Separable ODEs. Modeling l3 l ExAMPLE 2 ExAMPLE 3 kt -o.ooo1213,e :e :0.525. ln 0.525 r -----,-- _- - -- JJ l L. -0.000121 3 It is very important to introduce the constant of integration immediately when the integration is performed. Ifwewrotearctanj:x,theny:tanx,andthenintroducedc,wewouldhaveobtainedy:tanxlc,which is not a solution (when c * 0). Verify this. Modeling The impoťtance of modeling was emphasized in Sec. 1.1, and separable equations yield various useful models. Let us discuss this in terms of some typical examples. Radiocarbon Dating2 In Septembet I99l the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found in the ice of the Oetztal Alps (hence the name "Oetzi") in Southern Tyrolia near the Austrian-Italian border, caused a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon 6Cla to carbon uC12 in this mummy is 52.57o of that of a living organism? Physical Information, In the atmosphere and in living organisms, the ratio of radioactive carbon 6Cla (made radioactive by cosmic rays) to ordinary carbon uC12 is constant. When an organism dies, its absorption of 6Cla by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive carbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of 6Cla, which is 5715 years (CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page II-52,line 9). SOlUtiOn. Modeling. Radioactive decay is governed by the ODE y' : ty (see Sec. 1.1, Example 5).By separation and integration (where r is time and y6 is the initial ratio of ucla to uC121 : kdt, lnlyl :kt+c, kt !:loe Next we use the half-life H: 5715 to determine k. When t - H, half of the original substance is still present. Thus, yo"kH : 0.5yo, ekH : 0.5, ln 0.5 0.693 H 5715 Finally, we use the ratio 52.5Va for determining the time / when Oetzi died (actually, was killed), dy y 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. l 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. 9 contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate of 10 gallmin, and each gallon contains 5 lb of dissoved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at 10 gallmin. Find the amount of salt in the tank at any time r. SOlUtiOn. Sfup 1. Setting up a model. Let y(r) denote the amount of salt in the tank at time r. IB time rate of change is y' : Salt inflow rate - Salt outflow rate "Balanc law". 5lb times 10 gal gives an inflow of 50lb of salt. Now, the outflow is 10 gal of brine. This is 10/1000 : 0.0l (: 17o) of the total brine content in the tank, hence 0.01 of the salt content y(r), that is, 0.0ly(l). Thus the model is the ODE y' : 50 - 0.01y : -0.01(y - 5000). 2Method by WILLAR.D FRANK LIBBY (1908-1980), American chemist, who was awarded for this work the 1960 Nobel Prize in chemistry. (4) 14 CHAP. l First-Order ODEs Step 2. Solution of the model. The ODE (4) is separable. Separation, integration. and taking exponents on both sides gives y - 5000 : -0.01 dt. ln .y - 5000| : -0.01/ + c" y-5000:ce-o,olt Initially the tank contains 100 lb of salt. Hence y(0) : l00 is the initial condition that will give the unique solution.Substitutingy:l00andr:Ointhelastequationgives100-5000:rno-c.Hencec:-4900. Hence the amount of salt in the tank at time r is (5) _v(r) : 5000 - 490Oe-o'oÍt. This function shows an exponential approach to the limit 5000 lb; see Fig. 9. Can you explain physically that y(r) 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. 27) 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. l d1" y 5000 4000 3000 2000 1000 ..----.> 100 ExAMP,LE 4 100 2oo 300 400 Salt content y(l) 500 Tank Fig. 9. Mixing problem in Example 3 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.v. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 e.v. was found to be 65'F. The outside temperature was 50"F at l0 p.lt. and had dropped to 40'F by 6 e.v. What was the temperature inside the building when the heat was turned on at 6 R.na.? Physical infonnaíion. Experiments show that the time rate of change of the temperature Z of a body B (which conducts heat well, as, for example, a copper ball does) is proportional to the difference between 7 and the temperature of the surrounding medium (Newton's law of cooling). Solution. Sfup l. Setting up a model. Let T(r) be the temperature inside the building anď T6 the outside temperature (assumed to be constant in Newton's law). Then by Newton's law, - Ti. 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 stil1 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. 3Si. ISAAC NEWTON (1642-112'7), 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-I'1|6) 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. dT á:Kr(6) - SEC. 1.3 Separable ODEs. Modeling 15 Step 2. General solution. We cannot solve (6) because we do not know Ia, just that it varied between 50oF and 40'F, so we follow the Golden Rule: If vou cannot solve your problem, try to solve a simpler one.'We solve (6) with the unknown function Za replaced with the average of the two known values, or 45oF. For physical reasons we may expect that this will give us a reasonable approximate value of Z in the building at 6 a.na. For constant Ta : 45 (or any oíher constlnt yalue) the ODE (6) is separable. Separation, integration, and taking exponents gives the general solution T-45 : kdt, h|r-45|:ftt 1r* r@: 45 -| cekt , c, (c:e ). Step 3. Particular solution. We choose 10 p.v. to be r : 0. Then the given initial condition is Z(0) : 70 and yields a particular solution, call lt To. By substitution, (0) : 45 -| ceo : J0, c:10 - 45 :25. TeG) : 45 -l 25ekt. Step 4. Determination of k. We use rG) : 65, where t : 4 is 2 e.v. Solving algebraically for k and inserting k inío To(t) gives (Fig. 10) Te(4) : 45 -l 25eak : 65, e4k : 0.B, k: ltn0.8 : -0.056, TeG): 45 + 25e-O'O56t. Step 5. Answer and interpretation.6 a.v. is r : 8 (namely, 8 hours after l0 e.v.), and Ze(B): 45 + 25e-o'o56'8:6l["F]. Hence the temperature in the building dropped 9oF, a result that looks reasonable. l y 7o 68 66 65 64 60 o 2 4 6 8t Fig. 10. Particular solution (temperature) in Example 4 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. 1 1). 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) u(r) : 0,6o0Ýklt(t\ (Torricelli's tawa), where h(.t) is the height of the water above the hole at time t, and q : 980 cmlsec2 : 32.11 ftlsec2 is the acceleration of gravity at the surface of the earth. Solution. Sfup 1. Setting up the model. To get an equation, we relate the decrease in water level /z(r) to the outflow. The volume AV of the outflow during a short time Ar is LV: Au Ll (Á : Area of hole). apvlxGBLISTA TORRICELLI (l608-1647), Italian physicist, pupil and Successor of GALILEo GALILEI (1564-1642) at Florence. The "contraction factor" 0.600 was introduced by J. C. BORDA in 1166 because the stream has a smaller cross section than the area of the hole. dT t6 CHAP. l First-Order ODEs AV must equal the change AVx of the volume of the water in the tank. Now LV* : -B ^h (B : Cross-sectional area of tank) where ^h (> 0) is the decrease of the height h(t) of the water. The minus sign appears because the volume of the water in the tank decreases. Equating AV and AV* gives -B ^h: Au Lt. We now express u according to Tonicelli's law and then let A/ (the length of the time interval considered) approach O-this is a standard way of obtaining an ODE as a model. That is, we have lhAA ^, : - E u : - B 0.600\/2sh(t). and by letting At --+ 0 we obtain the ODE dt -"'"" B where 26.56 :0.600 \,e , 980. This is our model, a first-order ODE. Step 2. General solution. Our ODE is separable. AlB is constant. Separation and integration gives and 2\/i: c* Dividing by 2 anďsquaring gives h: (c - l3.2BAtlB)2.Inserting I3.28A|B : yields the general solution h(t):(c-0.00O332t)2. Step 3. Particular solution. The initial height (the initial condition) is á(0) : 225 cm. Substitution of r : 0 andh:225 givesfrom the general solution c2 :225, c : 15.00 and thus the particular solution (Fig. 11) hp(t): (15.00 * 0.00O332t)2. Step 4. Tank empty. he(t) : 0 if / : 15.00/0.000332: 45 181 [sec] : 12.6 [hours]. Here you see distinctly the importance of the choice of units-we have been working with the Cgs system, in which time is measured in seconds! We used g : 98O cm,/sec2. Step 5. Checking. Check the result. I25m ]_ h(t __l Fig. ll. Example 5. Outflow from a cylindricaltank ("leaking tank"). Torricelli's law Extended Method: Reduction to Separable Form Certain nonseparable ODEs can be made separable by transformations that introduce for y a new unknown function. We discuss this technique for a class of ODEs of practical dhA th:-26.56 udt A 26,56 - t. B 13.28, 0.52 rlIOO2 n : 0.000332 I water level at time l h 25o 20o 150 100 50 0 2. Outf lowing water Tank 0 10000 30000 Water level h(t) in tank _- - SEC. 1.3 Separable ODEs. Modeling (10) importance, namely, for equations (8) ),, : íía) \í/ 17 Here, / is any (differentiable) function of ylx, such as sin(y/r), (yl*)n, and so on. (Such an ODE is sometimes called a homogeneous ODE, a term we shall not use but reserve for a more important purpose in Sec. 1.5.) The form of such an ODE suggests that we set ylx : z; thus, (9) y : ux and by product differentiation y' : u'x * u. Substitution into y' : f(ylx) then gives u'x + u: f(u) or u'x: f(u) - u.We see that this can be separated: du dx :f(u)-u x EXAMPLE 6 Reduction to Separable Form Solve ^l22zxyy - x SOlUtiOn. To get the usual explicit form, divide the given equation by 2ry, 22 2xy 2x 2y' Now substitutey andy'from (9) and then simplify by subtractinguonboth sides, l , u I I u 7 -u2-IuxTu:r- 2u, uX:-r- 2r: 2, You see that in the last equation you can now separate the variables, 2udu dx ltl I_"r: -;. Byintegration, ln(1 + u2): -ln|x| +cx:h|]l+.*. |"I Take exponents on both sides to get 1 + u2 : clx or I + (ylx)2 : clx. Multiply the last equation by x2 to obtain (Fig. 12) ..2 , ..2 _ _.. Thus (_ _ :\' * .., : "x -Ťy:cx. \ ') *r-: 4 This general solution represents a family of circles passing through the origin with centers on the x-axis. l Fig. 12. General solution (family of circles) in Example 6 18 CHAP. ] First-Order ODEs 1,. (Constant of integration) An arbitrary constant of integration must be introduced immediately when the integration is performed. Why is this important? Give an example of your own. @ GENERAL soluTloN Find a general solution. Show the steps of derivation. Check your answer by substitution. 2.y'+(x*2)y2:g 3. y' : 2 sec}y 4. y' : (y + 9x)2 (y -l 9x : u) 5.yy'*36x:0 6. y' : (4x2 + yz)l@y) 7. y' sin lrx : y cos nrx 8.ry':iy'*y 9. y'rn' : y2 * I Eo-tr l lNlTlAL vALuE pRoBLEMs Find the particular solution. Show the steps of derivation, beginning with the general solution. (L, R, Ď are constants.) 10. yy' ]- 4x: 0, y(0) : 3 1,1,. drldt - -Ztr, r(0) : 16 12. 2xyy' : 3y2 * x', y(I) : 2 13. L dlldt + RI : 0, 1(0) : 16 14. y' : ylx 15. e2*y' : + (2x3ly) cos(xz), y6/rrD) : {, 2(x + 2)y", y(0) : Il\/5 : 0,45 l 4x5 coszly/x;, y(2) : 0 : y, }(3) : ln 81 of individuals present, what is the population as a function of time? Figure out the limiting situation for increasing time and interpret it. 25. (Radiocarbon dating) If a fossilized tree is claimed to be 4000 years old, what should be its 6Cla content expressed as a percent of the ratio of uCla to uC12 in a living organism? 26. (Gompertz growth in tumors) The Gomperlz model is y' : -Ay |ny (A > 0), where y(r) is the mass of tumor cells at time r. The model agrees well with clinical observations. The declining growth rate with increasing y > 1 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 wet laundry loses half of its moisture during the first 5 minutes of drying in a dryer and if the rate of loss of moisture is proportional to the moisture content, when will the laundry be practically dry, say, when will it have lost 957o of its moisture? First guess. 28. (Alibi?) Jack, arrested when leaving abar, 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 entfance of the bar) at the instant of arrest and again 30 minutes later, obtaining the values 190'F and 110'F, respectively. Do these results give Jack an alibi? (Solve by inspection.) 29. (Law of cooling) A thermometer, reading 10oC, is brought into a room whose temperature is 23'C. Two minutes later the thermometer reading is 18"C. How long will it take until the reading is practically 23"C, say, 22.8"C? First guess. 30. (Torricelli's law) How does the answef in Example 5 (the time when the tank is empty) change if the diameter of the hole is doubled? First guess. 31. (Torricelli's law) Show that (7) looks reasonable inasmuch u, ÝZgh(Ď is the speed a body gains if it falls a distance h (and air resistance is neglected). 32. (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 one thousand times greater than the man can exert? First guess. Experiments show that the change AS of the force S in a small portion of the rope is proportional to ^ and to the small angle A@ in Fig. 13. Take the proportionality constant 0.15. 16. 17. 18. 19. Xy:y y'x In x dr/d0 : 0 Bto maketheinputnonnegative. (The constants A, B, and K can be determined by measurements.) Hence the model is y'(t): In - Out: A + Bcos($rr) - Ky(t) or y'+Ky:A*Bcos($nfl. The initial condition for a particular solution lporl is lport(O) : y6 with r : 0 suitably chosen, e.g., 6:00 a.v. Step2.Generalsolution.In(4)wehavep:K:const,h:Kt,andr:A*Bcost$ttl.Hence(4)gives the general solution 29 A K _Kt Kt e B n,l44K *c:0, 144Ko l ro B/rt ' ^or4 * l,l++/(CoS |2 , _Kt -Ť ce AB(,--,l++nK I44K" + 7T" - i^_, _ *-L ('.O".o, -.\l2zrsin '"'| - ro-n' 12 l y(t): ,-xt[rxt(o - u "o, ff) r, n ,"-K' ,r\l-l l2rr,'" , /_]t+ Ťt 12 Ťrí 12 The last term decreases to 0 as / increases, practically after a short time and regardless of c (that is, of the initial condition). The other part of y(t) 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 / : 0 in y(t) and choosing yo : 0, we have ],(0) A --l I K A --l I K thus Inserting this result into y(t), we obtain the particular solution !eort(t) with the steady-state paft as before. To plot !pa6 we must specify values for the constants, say, Á : B : 1 and K : 0.05. Figure 17 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 Á -| B cos ($n1 : 1 * cos (bril. l y 25 20 15 10 5 0 0 100 2oo t Fig. 17. Particular solution in Example 3 Bl A4K, + ", (l++rcos 12rrsin#) - (+ - #_ "---),-"' 30 CHAP. l First-Order ODEs Reduction to Linear Form. Bernoul[i Equation I.{umerous 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 equation5 y' + p(x)y : g(x)y" (a any real number). If a : 0 or a : 1, Equation (6) is linear. Otherwise it is nonlinear. Then we set u(,x) : [y(")]'-". We differentiate this and substitute y' from (6), obtaining u' : (I - a)y-oy' : (1 - a)y-"(gy" - py). Simplification gives u,:(I-a)(s-pyl-"), where.}1-o : u on the right, so that we get the linear ODE u' + 1I - a)pu: 11 - a)g. For further ODEs reducible to linear from, see Ince's classic tA1l] listed in App. 1. See also Team Project 44 tn Problem Set 1.5. EXAMPLE 4 Logistic Equation Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation6): y':Ay-By' Solution. Write (B) in the form (6), that is, y' - Ay: -ur' to see that cL:2, so that tt: y|-o: ),-'. Differentiate this u and substitutey' from (B), ,/ : -y-ry, -- -y-r(Ay - ay1 - B - Ay-r. The last term is -Ay-' : -Au. Hence we have obtained the linear ODE 5JAron 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 the Leibniz in l696. Jakob Bernoulli's students included his nephew NIKLAUS BERNOULLI (1681-1159), who contributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI(1667-I'748), who had profound influence on the development of calculus, became Jakob's successor at Basel, and had among his students GABRIEL CRAMER (see Sec. ].1) anď 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. GpIERRB-FRANQOIS VERHULST, Belgian statistician, who introduced Eq. (B) as a model for human population growth in 183B. (6) (7) (8) SEC. 1.5 Linear ODEs. BernouIli Equation. Population Dynamics The general solution is tby (a)] u' + Au: B. ,,:rn-At+BlA. Since u : lly,this gives the general solution of (8), \ Directly from (8) we see that.y = 0 (_l(r) : 0 for all r) is also a solution. (Fig. 18). l 0 1 2 3 4 Timet Fig. 18. Logistic population model. Curves (9) in Example 4 with A/B : 4 Population Dynamics The logistic equation (8) plays an important role in population dynamics, a field that models the evolution of populations of plants, animals, or humans over time r. If B : 0, then (8) i, y' - dylctt : Ay.In this case its solution (9) is y : (Ilc)eAt andgives exponential growth, as for a small population in a large country (the United States in early times!). This is called Malthus's law. (See also Example 3 in Sec. 1.1.) The term -By'in (8) is a "braking term" that prevents the population from growing without bound. Indeed, if we write y' : Áyt1 - (.BlA)yl, we see that if y 1AlB,then y' > 0, so that an initially small population keeps growing as long as y { AlB. But if y žAlB, then y' ( 0 and the population is decreasing as long as y > AlB. The limit is the same in both cases, namely, Á/B. See Fig. 1B. We see that in the logistic equation (8) the independent variable / does not occur explicitly. An ODE y' : í(t, y) in which r does not occur explicitly is of the form y' : í(y) and is called an autonomous ODE. Thus the logistic equation (8) is autonomous. Equation (10) has constant solutions, called equilibrium solutions or equilibrium points. These are determined by the zeros of /(y), because í(y):0 gives y' : 0 by (10); hence y : consr. These zeros are known as critical points of (10). An equilibrium solution is called stable if solutions close to it for some / remain close to it for all further /. It is called unstable if solutions initially close to it do not remain close to it as r increases. For instance, } : 0 in Fig. 18 is an unstable equilibrium solution, and y : 4 is a stable one. 31 (9) l1 \:-: re-A\BlA 6 A^ B 2 (10) Population y ^r- J l _++++++ .ž.?-).)-).?.-.?.) ./ .r'.r' .r' ./ ./ ././ ,ž ///,/////v. ///////////l///íí// / /í/í/,//í ExAMPLE 5 CHAP. l First-Order ODEs Stable and Unstable Equilibrium Solutions. "Phase Line Plot" The oDE y' : (y - 1Xy - 2) has the stable equilibrium solution yr : 1 and the unstable jz : 2, as the direction field in Fig. 19 suggests. The values y1 and yzare the zeros of the parabola /(y) : (y - I)(y - 2) in the figure. Now, since the ODE is autonomous, we can "condense" the direction field to a "phase line plot" giving }1 and !z, andthe direction (upward or downward; of thnarrows in the field, and thus giving information about the stability or instability of the equilibrium solutions. \ l llíííl/l/í //////////////////// 1 y2 l Vr t ///////// !z(A) (B) (C) Fig. 19. Example 5. (A) Direction field. (B) "Phase line". (C) Parabola /(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, New York, Wiley, 1976. Further important applications of linear ODEs follow in the next section. ííl / / / / í1. /////////////////// / / / / / / //. .r'./././ / ///./ ,ž-ž.).r'.?/..).ž,ž ./././ ./,r'.r' ,r'.r' ./.r' ,ž).----ž.-.ž.ž.?-? 1. (CAUTION!) Show that e ,-ln(sec ,) : cos ,. : Ilx (not -x) and 2. (Integration constant) Give a reason why in (4) you ;1:..n""re the constant of integration in Jp dx tobe E GENERAL soLuTloN. lNlTlAL 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. y' + 3.5y : 2.3 4.y':4y*x 5. y' + L25y : 5, y(0) : 6.6 6. ,'y' + 3xy : I/x, }(1) : _1 7. y' + ky : "'o* 8. y' + 2y : 4 cos 2x, y(iň : 2 9. y' : 6(y - 2-5) tanhI.5x 10. y' + 4x2y : 14x2 - x)g-r2tz 11. y' + 2y sin2x: 2e'o"2*, y(0) : 0 12. y' tan x : 2y - 8, y(Žrů : 0 13. y' + 4y cotZx:6 cos2x, y(irr):2 14. y' + y tan x : ,-o,olr cos ,t, y(0) : 0 15. y' + ylx2:zxell', y(1): 13.86 16. y' cos2 x -l 3y : 1, y(iň : t 17. x3y' + 3x2y: 5 sinh 10x y(x) SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics NONLlNEAR 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. 18.y'+y:j2, y(0):-1 19. yt : 5.Jy _ 6.5y, 20. (x2 + 1)y' : -tany, y(0) : irr 21. y' + (* + l)y : ,*'y', y(0) : 0.5 22. y' sin 2y * x cos 2y : 2x 23. Zyy' -| y2 sin í : sin.t, y(Q : \E 24. y' l *'y : (g-í3 sinh -r) l(3y') FURTHER APPL!cATloNs 25. (Investment programs) Bill opens a retirement savings account with an initial amount y6 and then adds $k to the account at the beginning of every year until retirement at age 65. Assume that the interest is compounded continuously at the same rate R over the years. Set up a model for the balance in the account and find the general solution as well as the particular solution, letting t : 0 be the instant when the account is opened. How much money will Bill have in the account at age 65 if he starts at 25 and invests $1000 initially as well as annually, and the interest rate R is 6Va? How much should he invest initially and annually (same amounts) to obtain the same final balance as before if he starts at age 45? First, guess. 26. (Mixing problem) A tank (as in Fig. 9 in Sec. 1.3) contains 1000 gal of water in which 200lb of salt is dissolved. 50 gal of brine, each gallon containing (1 * cos r) lb of dissolved salt, runs into the tank per minute. The mixture, kept uniform by stirring, runs out at the same rate. Find the amount of salt in the tank at any time t (Fig.20). y 1000 500 20o 050100t Fi6. 20. Amount of salt y(t) in the tank in Problem 26 27. (Lake Erie) Lake Erie has a water volume of about 450 km3 and a flow rate (in and out) of about 175 kmS per year. If at some instant the lake has pollution concentration p : 0.04Vo, how long, approximately, will it take to decrease it to pl2, assuming that the inflow is much cleaner, say, it has pollution concentration pl4, and the assumption that is only very 8uesS. 28. (Heating and qooling of a cooling of a builct}ng can be 33 mixture is uniform (an imperfectly true)? First, building) Heating and modeled by the ODE T' : kt(T - T) + kz(T - Tu) + P, where T : T(t) is the temperature in the building at tíme t, To the outside temperature, T- the temperature wanted in the building, and P the rate of increase of 7 due to machines and people in the building, and k1 and k2 are (negative) constants. Solve this ODE, assuming P : const, T- : const, and To varying sinusoidally over24 hours, say,To- A - Ccos (2nl24)t. Discuss the effect of each term of the equation on the solution. 29. (Drug injection) Find and solve the model for drug injection into the bloodstream if, beginning at t : 0, a constant amount A glmin is injected and the drug is simultaneously removed at a rate proportional to the amount of the drug present at time t. 30. (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 / -+ oo and explain what it means. 31. (Extinction vs. unlimited growth) If in a population y(r) 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, 32. (Harvesting renewable resources. Fishing) Suppose that the population y(r) of a certain kind of fish is given by the logistic equation (8), and fish are caught at a rate Hy proportional to y, Solve this so-called Schaefer model. Find the equilibrium solutions y1 and }z (> 0) when H < A. The expression : Hyz is called the equilibrium harvest or sustainable yield corresponding to H. Why? 33. (Harvesting) In Prob. 32 find and graph the solution satisfying y(0) : 2 when (for simplicity) Á : B : I and H : 0.2. What is the limit? What does it mean? What if there were no fishing? 34. (Intermittent harvesting) In Prob. 32 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 34 CHAP. l First-Order ODEs continued periodically. Find and graph the solution for the first 9 years, assuming thatA: B : I, H:0.2, and y(0) : 2. 43. CAS EXPERIMENT. (a) Solve the ODE y' - ylx : -x-l cos (1/x). Find an initial condition for which the arbitrpry constant is zero. Graph the resulting particular s\ution, experimenting to obtain a good figure near x : 0. (b) Generalizing (a) from n : I to arbitrary n, solve the ODE y' - nylx: -ť-2 cos (1/x). Find an initial condition as in (a), and experiment with the graph. 44. TEAM PROJECT. Riccati Equation, Clairaut Equation. A Riccati equation is of the form (11) y' + p(x)y : 8@)y2 + hlx}. A Clairaut equation is of the form (I2) y:xy'+8(y,), (a) Apply the transformation y : Y + llu to the Riccati equation (11), where is a solution of (11), and obtain for z the linear ODE u' + (2Yg - Ďu - -8. Explain the effect of the transformation by writing it asy:Ylu,u:llu. (b) Show thaty : Y : x is a solution of y' - (2x3 + 1)y - -x'y' - x4 - x * I and solve this Riccati equation, showing the details. (c) Solve y' + (3 - 2x2 sin x)y : -y2 sin x * 2x + 3x2 - x4 sin x, using (and verifying) that y : x2 is a solution. (d) By working "backward" from the u-equation find further Riccati equations that have relatively simple solutions. (e) SolvetheClairautequation y: xy' + 7lyl .Hint. Differentiate this ODE with respect to.t. (f) Solve the Clairaut equation y'' - xy' + y : 0 in Prob. 16 of Problem Set 1.1. (g) Show that the Clairaut equation (I2) has as solutions a family of straight lines y : cx * g(c) and a singular solution determined by g'(r) - -x) where s : !' , that forms the envelope of that family. 45. (Variation of parameter) Another method of obtaining (4) results from the following idea. Write (3) as cyx, where yx is the exponential function, which is a solution of the homogeneous linear ODE y*' + py* :0. Replace the arbitrary constant c in (3) with a function u tobe determined so that the resulting function y : uy* is a solution of the nonhomogeneous linear ODE y' l py : r. 46. TEAM PROJECT. Transformations 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 examples of your choice for each transformation, showing each step (not just the transformed ODE). y 2 1.8 1.6 1.4 1.2 1 0.8 o2468t rig. ž1.Fish population in Problem 34 35. (Harvesting) If a population of mice (in multiples of 1000) follows the logistic law withA : 1 and B :0.25, and if owls catch at a time rate of L07o of the population pfesent, what is the model, its equilibrium harvest for that catch, and its solution? 36. (Harvesting) Do you save work in Prob. 34 if you first transform the ODE to a linear ODE? Do this transformation, Solve the resulting ODE. Does the resulting y(t) agree with that in Prob. 34? Fr_4,i GEI{ERAL pRopER.TlĚs 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 homogeneous linear ODEs (2) have the following properties. Illustrate each property by a calculation for two or three equations of your choice, Give proofs. 37. The sum )1 -l y2 of two solutions y1 and y, of the homogeneous equation (2) is a solution of (2), and so is a scalar multiple a,y1 for any constant a. These properties are not true for (1)! 38. y : 0 (that is, y(x) : 0 for all x, also written y(x) = 0) is a solution of (2) [not of (1) if r(x) + 0!], called the trivial solution. 39. The sum of a solution of (1) and a solution of (2) is a solution of (1). 40. The difference of two solutions of (1) is a solution ot (2). 41,. If y1 is a solution of (1), what can you say about cy1? 42. It y1 and y2 are solutions of yi * pyt : í1áild yL + pyz: f2, fespectively (with the same p!), what can you say about the sum yt * yz? - SEC. 1.6 Orthogonal Trajectories. Optional 35 1.6 Orthogonat Trajectories. Optional An important type of problem in physics or geometry is to find a family of curves that intersect 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 c:urves, curves of equal voltage-the concentric circles in Fig. 22), anď curves of electric force (the straight radial segments inFig.22). Fig. 22. Equipotential lines and curves of electric force (dashed) between two concentric (black) circles (cylinders in space) 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 by using ODEs, as follows. Let (1) G(x,y, c):0 be a given family of curves in the xy-plane, where each curve is specified by some value of c. This is called a one-parameter family of curves, and c is called the parameter of the family. For instance, a one-parameter family of quadratíc parabolas is given by (Fig.23) ! : cxz or, written as in (1), G(x,y, c): y - cxz :0. Step 1. Find an ODE for which the given family is a general solution. Of course, this ODE must no longer contain the parameter c.In our example we solve algebraically for c and then differentiate and simplify; thus, l, y x'- ZXy :0, hence ,2Y X x4 y , - Ll x' ,1 (2) CHAP. l First-Order ODEs The last of these equations is the ODE of the given family of curves. It is of the form y' : í(x, y). Step 2. Write down the ODE of the orthogonal trajectories, that is, the ODE whose general solution gives the orthogonal trajectories of the given curves. This ODE is with the same / as in (2). Why? Well, a given curve passing through a point (xg, y6) has slope .f(xo, yo) at that point, by (2). The trajectory through (xo, yo) has slope -Ilf(x6, yg) by (3). The product of these slopes is - 1, as we see. From calculus it is known that this is the condition for orthogonality (perpendicularity) of two straight lines (the tangents at (xo, yo)), hence of the curve and its orthogonal trajectory at (x9, yg). Step 3. Solve (3). For our parabolas ! : cx2 we have y' : 2ylx. Hence their orthogonal trajectories are obtained from' : -xlry or 2' -l x 0. By integration,' + ir' : c*. These are just the origin, and c * < 0 gives no real solution at allFig. 23. Parabolas and orthogonal trajectories (ellipses) in the text (3) @ oRTHoGoNALTRAJEcToRtEs Sketch or graph some of the given curves. Guess what their orthogonal trajectories may look like. Find these trajectories. (Show the details of your work.) Ly:4x*c 2.y:clx 3.y:cx 4.y':2x2+c 5.xzy:, 6.y:ce-3* FJn oTHER FoRMs oF THE oDEs (2) AND (3) 13. (J as independent variable) Show that (3) may be written dxld : -í(x, }). Use this form to find the orthogonal trajectories ofy : 2x l ce-". 7, y : "r2l2 9.4x2*y':, II. x - cea|4 8.x2-y2:c 10.x:, 12. x2 + () - c)2 : c2 36 -|y:- f(x,) x a '.. -z - -_4 SEC. 1.7 Existence and Uniqueness of Solutions 14. (Family 8(r,J) = c) Show that if a family is given as 8@, y) : c, then the orthogonal trajectories can be obtained from the following ODE, and use the latter to solve Prob. 6 written in the form g(x, !) : c. d _ asla dx dgl\x 15. (Cauchy-Riemann equations) Show that for a family u(x, y) : c : const the orthogonal trajectories u(x,y) : c* : const canbe obtained from the following Cauchy-Riemann equations (which aíebasic in complex analysis in Chap. 13) and use them to find the orthogonal trajectories of e* siny : const. (Here, subscripts denote partial derivatives.) U*: Ua. Ua: -U* Eo-ro-l AppllcAT!oNs 16. (Fluid flow) Suppose that the streamlines of the flow (paths of the particles of the fluid) in Fig. 24 are Ý(x, y) : xy : const. Find their orthogonal trajectories (called equipotential lines, for reasons given in Sec. 18.4). -Fig.24. Flow in a channel in Problem ]6 17. (Electric íield) Let the electric equipotential lines (curves of constant potential) between two concentric cylinders (Fig.22) be given by u(x,y): x2 l y' : ,. Use the method in the text to find their orthogonal trajectories (the curves of electric force). 37 18. (Electric field) The lines of electric force of two opposite charges of the same strength at (-1, 0) and (1, 0) are the circles through (-l, 0) and (1, 0). Show that these circles are given by x2 + (y - ,)' : l * c2. Show that the equipotential lines (orthogonal trajectories of those circles) are the circles given by (x -| gx12 * ' : c*2 - 1 (dashed in Fig. 25). Fig. 25. Electric field in Problem'l8 19. (Temperature field) Let the isotherms (curves of constant temperature) in a body in the upper half-plane y > 0 be given by 4x2 -| 9y' : c. Find the orthogonal trajectories (the curves along which heat will flow in regions filled with heat-conducting material and free of heat sources or heat sinks). 20. TEAM PROJECT. Conic Sections. (A) State the main steps of the present method of obtaining orthogonal trajectories. @) Find conditions under which the orthogonal trajectories of families of ellipses x2la2 + y2lb2 : c are again conic sections. Illustrate your result graphically by sketches or by using your CAS. What happens if a ---> 0? If b ---> 0? (C) Investigate families of hyperbolas x2la2 - y2lb2 : c in asimilar fashion. (D) Can you find more complicated curves for which you get ODEs that you can solve? Give it a try. 1.7 Existence and Uniqueness of solutions The initial value problem ly'l +lyl :0, y(0) : 1 has no solution because } : 0 (that is, y(x) : 0 for all x) is the only solution of the ODE. The initial value problem Y' :2x, y(0) : 1 38 CHAP. l First-Order ODEs has precisely one solution, has infinitely many solutions, namely, } : 1 + y(0) : 1 for all c. From these examples we see that an initial namely, y : x2 * 1. The initial value probiem xl-' : y - 1, y(0) : l (1) y' : f(x, y), 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 condítions does an initial value problem of the form (I) 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)? Theorems that state such conditions are called existence theorems, respectively. Of course, for our simple examples we need no theorems examples by inspection; however, for complicated ODEs considerable practical importance. Even when you are sure system behaves uniquely, occasionally your model may be give a faithful picture of the reality. cx, where c is an arbitrary constant because value problem y(xo) : yo theorems and uniqueness because we can solve these such theorems may be of that your physical or other oversimplified and may not THEoREM l Existence Theorem Let the right side f(x, y) of the ODE in the initial value problem (1) y' : f(.x, y), y(xo) : yo be continuous at all points (x, y) in some rectangle R: |x-xol 1a, ly-yol a and therefore (I : a in the existence theorem, which then asserts that the solution exists for all x between Xg - a and, ío* a. In the second case, shown in Fig. 27b, wehave blK < a. Therefore, q. : blK < a, and all we can conclude from the theorems is that the solution exists for all x between xg - blK and, xo * b/K. For larger or smaller r,s the solution curve maY leave the rectangle R, and since nothing is aszumed about / outside R, nothing can be concluded about the solution for those Iarger or smaller r's; that is, for such x's the solution may or may not exist-we don't know. 39 . T'H E O R"E.M' 2 R Y l I 40 CHAP. ] First-Order ODEs y yo+ b yo v^-b (4) y y^+b yo y^-b (a) (b) Fig.27. The condition (2) of the existence theorem. (a) First case. (b) Second case Let us illustrate our discussion with a simple example. We shall see that our choice of a rectangle R with aIarge base (a long x-interval) will lead to the case in Fíg.27b. E x A M P L E l :::;::;:ffi::ťi.ou.y'-I+y2, y(0):0 and take the rectangle n; l"l < 5, ly| < 3. Then a : 5, b : 3, and i/(,r,y)l : |r * y'l= r : rc, lar l lŤl:2lyl=M:6.ld)l b K Indeed, the solution of the problem is y : tanx (see Sec. 1.3, Example 1). This solution is discontinuous at +rl2,and there is no continuous solution valid in the entire interval l"l < S from which we started. l 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 f (x, y) - f(x, r) : 0z - yr) + l d! lu:u where (x, y) and (x, y2) are assumed to be in R, and ! is a suitable value between y1 and y2. From this and (3b) it follows that lf @, yr) - í(x,yr)l = Mly, - yrl. E It can be shown that (3b) may be replaced by the weaker condition (4), which is known as a Lipschitz condition.7 However, continuity of f(x, y) is not enough to guarantee the uniqueness of the solution. This may be illustrated by the following example. SEC. 1.7 Existence and Uniqueness of Solutions ExAMPLE 2 Nonuniqueness The initial value problem has the two solutions 41 y' : \/Ň, and y* : y(0) : 0 y=0 [ ,'t+ |-*'t+ if x>0 if -r{0 although í(x, y): \,4' is continuous for all y. The Lipschitz condition (4) is violated in any region that includes the line y : 0, because for y1 : 0 and positive y2 we have |í@,lz) , "/--/tx.y1)| _Yyz _ l tr6 r ol ly, - yrl jz \E and this can be made as large as we please by choosing y2 sufficiently small, whereas (4) requires that quotient on the left side of (5) should not exceed a fixed constant M. (5) the l 1. (Vertical strip) If the assumptions of Theorems I and 2 are satisfied not merely in a rectangle but in a vertical infinite strip |x - xol { a, in what interval will the solution of (1) exist? 2. (Existence?) Does the initial value problem (x - I)y' :2!, y(1) : 1have a solution? Does your result contradict our present theorems? 3. (Common points) Can two solution curves of the same ODE have a common point in a rectangle in which the assumptions of the present theorems are satisfied? 4. (Change of initial condition) What happens in Prob. 2 if you replace y(1) : 1 with y(I) : k? 5. (Linear ODE) If p and r in y' + p(x)y : r(x) are continuous for all x in an interval l" - ,ol š a, show that f (x, y) in this ODE satisfies the conditions of our present theorems, so that a conesponding initial value problem has a unique solution, Do you actually need these theorems for this ODE? 6. (Three possible cases) Find all initial conditions such that(x2 - 4x)y' : (2x - 4)y has no solution, precisely one solution, and more than one solution. 7. (Length of r-interval) In most cases the solution of an initial value problem (1) exists in an x-interval larger than that guaranteed by the present theorems. Show this fact for y' : 2y2,y(1) : 1 by finding the best possible a (choosing b optimally) and comparing the result with the actual solution. 8. PROJECT. Lipschitz Condition. (A) State the definition of a Lipschitz condition. Explain its relation to the existence of a partial derivative. Explain its significance in our present context. Illustrate your statements by examples of your own, (B) Showthatfor alinearODEy' + p(x)y: r(í)with continuous p andrin|x - xol Éa aLipschitz condition holds. This is remarkable because it means that for a linear ODE the continuity of f (x, y) guarantees not only the existence but also the uniqueness of the solution of an initial value problem. (Of course, this also follows directly from (4) in Sec. 1.5.) (C) Discuss the uniqueness of solution for a few simple ODEs that you can solve by one of the methods considered, and find whether a Lipschitz condition is satisfied. 9. (Maximum c) What is the largest possible a in Example 1 in the text? 10. CAS PROJECT. Picard Iteration. (A) Show that by integrating the ODE in (1) and observing the initial condition you obtain ť "ru,yGD dt.(6) y(x) : yo -| zRUooLp LIPSCHITZ (1832-1903), German mathematician. Lipschitz and similar conditions are important in modern theories, for instance, in partial differential equations. 42 CHAP. l First-Order ODEs This form (6) of (1) suggests Picard's method8, which is defined by (B) Apply the iteration to y' : x l y, y(0) : 0. Also solve the problem exactly. (C) Apply the iteration to y' : 2y2, y(0) : 1. Also solve the problem exactly. (D) Find all solutions of y' :2\5.y(1) : 0. Which of them does Picard's iteration approximate? (E) Experiment with the conjecture that Picard's iteration converges to the solution of the problem for any initial choice of y in the integrand in (7) (leaving y6 outside the integral as it is). Begin with a simple ODE and see what happens. When you are reasonably sure, take a slightly more complicated ODE and give it a try. -I l (1) jn(x) : yo + J f G, yn_11G)) dt. n : ro iteration ,l 1...Ll L; It gives approximations y1, !z, !3,. . . of the unknown solution y of (1). Indeed, you obtain y, by substituting } : )o on the right and integrating-this is the first step-, then y, by substituting y : }r o11 the right and integrating-this is the second step-, and so on. Write a program of the iteration that gives a printout of the first approximations !o, !t,..., .h,l as well as their graphs on common axes, Try your program on two initial value problems of your own choice. 1. Explain the terms ordinary dffirential eclucttion (ODE), partial dffirential equation (PDE), order, general solution, and particular solution Give examples. Why are these concepts of importance? 2. what is an initial condition? How is this condition used in an initial value problem? 3. What is a homogeneous linear ODE? A nonhomogeneous linear ODE? Why are these equations simpler than nonlinear ODEs? 4. What do you know about direction fields and their practical importance? 5. Give examples of mechanical problems that lead to ODEs. 6. Why do electric circuits lead to ODEs? 7. Make a list of the solution methods considered. Explain each method with a few short sentences and illustrate it by a typical example. 8. Can certain ODEs be solved by more than one method? Give three examples. 9. What are integrating factors? Explain the idea. Give examples. 10. Does every first-order ODE have a solution? A general solution? What do you know about uniqueness oť solutions? E1-141 DIREcTIoN FIELDs Graph a direction field (by a CAS or by hand) and sketch some of the solution curves. Solve the ODE exactly and compare. |1. y' : I * 4y2 12. y| : 3y - 2x 13. y' : 4y * y2 14. y| : I6xly rl_rd GENERAL soluTloN Find the general solution. Indicate which method in this chapter you are using. Show the details of your work. 15. y| : x2(t + y,) 16.y':x(j_x2+11 17. yy' l xy2 : , 18. - zr sin n,x cosh 3y dx * 3 cos zrx sinh 3y dy : 0 19. y' + ysinx : sinx 20. y' - y : Ily 21. 3 sin 2y dx -l 2x cos 2y dy : 0 22. xy' : x tan (ylx) + y 23. (y cos -ry - 2x) dx -f (x cos .ry + 2y) dy : 0 24. xy' : (y - 2x)2 + y (Set y - 2x : z.) 25. sin (y - x)dx * fcos (y - x) - sin (y - x)] dy : 0 26. xy' : 1ylx;3 + y W4 lNlTlAL vALuE pRoBLEMs Solve the following initial value problems. Indicate the method used. Show the details of your work. 27. yy' * ,r : 0, y(3) : 4 28. y' - 3y : -I2y2, y(O) : 2 29. y' : I * y2, y(žň: 0 30. y' -l rry : 2b cos Ťx, y(0) : 0 31,. (2xy2 - sin x) dx + (2 + 2*'y) dy:0, y(0): 1 32. |2y + y2lx -l e*(I + Ilx)l dx -l (x + 2y) dy : 0, y(1) : 1 BEMILB PICARD (1856-194l), French mathematician, also known for his important contributions to complex analysis (see Sec. 16.2for his famous theorem). Picard used his method to prove Theorems l anď2 as well as the convergence of the sequence (7) to the solution of (1). In precomputer times the iteration was of littlep ractical value because of the intesrations. E T!ONS AND PROBLEMS Summary of Chapter 1 F}fi] AppLlcATloNs,MoDELlNG 33. (Heat flow) If the isotherms in a region are x2 - y2 : c, what are the curves of heat flow (assuming orthogonality)? 34. (Law of cooling) A thermometer showing 10"C is brought into a room whose temperature is 25'C. After 5 minutes it shows 20"C. When will the thermometer practically reach the room temperature, say, 24.9"C? 35. (Half-life) If I)%a of aradioactive substance disintegrates in 4 days, what is its half-life? 36. (Half-life) What is the half-life of a substance if after 5 days, 0.020 g is present and after 10 days, 0.015 g? 37. (Half-life) When w1|| 997o of the substance in Prob. 35 have disintegrated? 38. (Air circulation) In a room containing 20 000 ft3 of air, 600 ft3 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 y(t) at any time if y(0) : 0? After what time wi|I907o of the air be fresh? 39. (Electric field) If the equipotential lines in a region of the xy-plan e are 4x2 * y' : c, what are the curves of the electrical force? sketch both families of curves. 43 40. (Chemistry) In a bimolecular reaction A + B ---> M, a moles per liter of a substance Á and b moles per liter of a substance B aíe combined, under constant temperature the rate of reaction is y' : k(a - y)(b - y) (Law of mass action); that is, y' is proportional to the product of the concentrations of the substances that are reacting, whore y(r) is the number of moles per liter which have reacted after time r. Solve this ODE, assuming that a * b. 41. (Population) Find the population y(0 if the birth rate is proportional to y(r) and the death rate is proportional to the square of y(r). 42. (Carves) Find all curves in the first quadrant of the xyplane such that for every tangent, the segment between the coordinate axes is bisected by the point of tangency. (Make a sketch.) 43. (Optics) Lambert's law of absorptiong states that the absorption of light in a thin transparent layer is proportional to the thickness of the layer and to the amount of light incident on that layer. Formulate this law as an ODE and solve it. This chapter concerns ordinary differential equations (ODEs) of first order and their applications. These are equations of the form (1) F(x, y, )') : O or in explicit form y' : í(x, y) involving the derivative y' : dylclx of an unknown function y, given functions of x, and, perhaps, y itself. If the independent variable x is time, we denote itby 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). glogaNN HEINRICH LAMBERT (1128-1171), German physicist and mathematician. First-Order ODEs 44 CHAP. ] First-Order ODEs (2) 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 y(xo) : yo. Together with the ODE this is called an initial yalue problem y' : f(x, y), y(xo,) : yo (xo, yo given numbers) 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 coffesponds to one of these curves. A separable ODE is one that we can put into the form sO) dy : f(x) dx (Sec. 1.3) by algebraic manipulations (possibly combined with transformations, such as ylx : u) and solve by integrating on both sides. An exact ODE is of the form M(x, y) dx -| N(-r, y) dy : 0 (Sec. 1.4) where M dx + N dy is the differential du:u*dx-luody of a functionu(x, }), so that from du:0 we immediately get the implicit general solution u(x, y) : c. This method extends to nonexact ODEs that can be made exact by multiplying them by some function F(x, y), called an integrating factor (Sec. 1.4). Linear ODEs (5) y' + p(x)y: r(x) 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 y' + p(x)y: g(x)y" (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. 2I.I and 2I.2 immediately after this chapter, as indicated in the chapter opening. (3) (4) ,) (tl 7\PTER L Second-Order Linear ODEs Ordinary differential equations (ODEs) may be divided into two large classes, linear ODEs and nonlinear ODEs. Whereas nonlinear ODEs of second (and higher) order generally are difficult to solve, linear ODEs are much simpler because various properties of their solutions can be characterizeďin a general way, and there are standard methods for solving many of these equations. Linear ODEs of the second order are the most important ones because of their applications in mechanical and electrical engineering (Secs. 2.4,2.8,2.9). Andtheir theory is typical of that of all linear ODEs, but the formulas are simpler than for higher order equations. Also the transition to higher order (in Chap. 3) will be almost immediate. This chapter includes the derivation of general and particular solutions, the latter in connection with initial value problems. (Boundary value problems follow in Chap. 5, which also contains solution methods for Legendre's, Bessel's, and the hypergeometric equations.) COMMENT. Numerics for second-order ODEs can be studied immediately after this chapter. See Sec. 2L3, which is independent of other sections in Chaps. I9-2L Prerequisite: Chap. 1, in particular, Sec. 1.5. Sections that may be omitted in a shorter course: 2.3,2.9,2.I0. References and Answers to Problems; App. 1 Part A, and App.2. 2.1 Homoteneous 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 y" + p(x)y' + q(x)y: r(x) and nonlinear if it cannot be written in this form. The distinctive feature of this equation is that itis linear iny and its derivatiyes, whereas the functions p, q, and r on the right may be any given functions of x. If the equation begins with, say, f(x)y", then divide by f(x) to have the standard form (1) with y" as the first term, which is practical. (1) 45 46 CHAP.2 Second-Order Linear ODEs If r(x) = 0 (that is, r(x) : 0 for all x considered; read"r(x) is identically zero"), then (1) reduces to (2) y" + p(x)y' * q(x)y:O and is called homogeneous. If r(x) * 0, then (1) is called nonhomogeneous. This is similar to Sec. 1.5. For instance, a nonhomogeneous linear ODE is y" + 25y : e-* cos x, and a homogeneous linear ODE is xy" + y' + xy : O, instandardform y" + ! r' *y : 0. An example of a nonlinear ODE is y"y*y'2:0. The functions p and q in (1) and (2) are called the coefficients of the ODEs. Solutions are defined similarly as for first-order ODEs in Chap. 1. A function y : h(x) is called a solution of a (linear or nonlinear) second-order ODE on some open interval I if h ísdefined and twice differentiable throughout that interval and is such that the ODE becomes an identity if we replace the unknowny by h, the derivativey' by h', and the second derivative y" by h". Examples are given below. Homogeneous Linear ODEs: Superposition Principle Sections 2.I-2.6 will be devoted to homogeneous linear ODEs (2) and the remaining sections of the chapter to nonhomogeneous linear ODEs. Linear ODEs have a rich solution structure. For the homogeneous equation the backbone of this structure is the superposition principle or linearity principle, which says that we can obtain further solutions from given ones by adding them or by multiplying them with any constants. Of course, this is a great advantage of homogeneous linear ODEs. Let us first discuss an example. ExAMpLE t Homogeneous Linear ODEs: Superposition of Solutions Ihe functions y : cos x and y : sin x are solutions of the homogeneous linear ODE y"+):0 for all x. We verify this by differentiation and substitution. We obtain (cos x)" : -cos x; hence 1," +.y: (cosx)" + cos.T: -cosl -| cosx: 0. Similarly fory : sinx (veriíy!). We can go an important step further. We multiply cosx by any constant, for instance, 4.'7, and sinx by, say, -2, and take the sum of the results, claiming that it is a solution. Indeed, differentiation and substitution gives (4.7cos x-2 sinr)" + (4,1cos. - 2sinl): -4.1cosr-1- 2sinx-]- 4.7cosr- 2sinx:0, SEC. 2.1 Homogeneous Linear ODEs of Second Order 47 In this example we have obtained from y, (: cos x) and lz (: sin x) a function of the form (3) ! : ctll * czyz (cy c2 arbitrary constants). This is called a linear combination of y1 and yz.In terms of this concept we can now formulate the result suggested by our example, often called the superposition principle or linearity principle. THEoREM l Fundamental Theorem for the Homoteneous Linear ODE (2) For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution oí(2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions. P R O O F Let y1 and y2 be solutions of (2) on 1. Then by substituting ! : ctlt derivatives into (2), and using the familiar rule (cůtl ,ryr)' : cry't get l czyz and its -Ť ,ryl, etc., we y" + py' + qy : (cryr l ,ryr)" -| p(c t l ,ryr)' -| q(cůt -l czyz) : ců'l,+ ,ry'J, + p(cl! + ,ry!) -| q(c t-l czyz) : cíy| + py', + qyr) + c2OU + pyl l qyz) : 0, since in the last line, (, , ,) : 0 because y, and y2 are solutions, by assumption. This shows that y is a solution of (2) on I. l CAUTION! Don't forget that this highly important theorem holds for homogeneous Ir:?;?":,'"!n'J :::ťrtr,"rir1!:':""'nonhomogeneous 'inear or nonlinear oDEs, as EXAMPLE 2 A Nonhomoteneous Linear ODE Verifybysubstitutionthatthefunctionsy: l tcosxandy: l *sinxaresolutionsofthenonhomogeneous linear ODE y" * y : 1, but their sum is not a solution. Neither is, for instance,2(I i cosx) or 5(1 i sin.r). EXAMPLE 3 A NonlinearODE Verify by substitution that the functions y : x2 and y : 1 are solutions of the nonlinear ODE y"y - xy' :0, but their sum is not a solution. Neither ls -x2, so you cannot even multiply by -1! lnitial Value Prob[em. 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 y(xo) : yo. 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 solutiolz of the ODE. These ideas extend to second-order equations as follows. l tr 48 CHAP.2 Second-Order Linear ODEs For a second-order homogeneous linear ODE (2) an initial value problem consists of (2) and two initial conditions y(xo) : Ko, y'(x : Kt. These conditions prescribe given values Ko and K, of the solution and its first derivative (the slope of its curve) at the same given x : xo in the open interval considered. The conditions (4) are used to determine the two arbitrary constants c, and c2 ína general solution (4) ExAMPLE 4 (5) !:ctjtlczyz of the ODE; here, y1 and y2 aíesuitable solutions of the ODE, with "suitable" to be explained after the next example. This results in a unique solution, passing through the point (xo, Ko) with K, as the tangent direction (the slope) at that point. That solution is called a particular solution of the ODE (2). !nitial value problem Solve the initial value problem y"-l-y:0, y(0) : 3.0, y'(o) : -o.s. Solution. Sfup 1. General solution. The functions cos rr and sin x are solutions of the ODE (by Example 1), and we take y: CI cosx i c2sinx. This will turn out to be a general solution as defined below. Step2. Particularsolution. Weneedthederivativey' : -c1 sin xl c2cosx.Fromthisandtheinitialvalues we obtain, since cos 0 : 1 and sin 0 : 0, y(0) : c1 : 3.0 and y'(0) : cz: _0.5. This gives as the solution of our initial value problem the particular solution y : 3.0 cos,r - 0,5 sin x. Figure 28 shows that at x : 0 it has the value 3.0 and the slope -0.5, so that its tangent intersects the x-axis at x : 3.0i0.5 : 6.0. (The scales on the axes differ!) l y 3 2 1 0 _1 -Z 6\P Io l ,c Fig. 28. Particular solution and initial tantent in Example 4 Observation. Our choice of y1 anď y, was general enough to satisfy both initial conditions. Now let us take instead two proportional solutions lt : cos -r and -l SEC. 2.1 Homogeneous Linear ODEs of Second Order lz : k cos.tr, so that ylly2 : Ilk : const. Then we can write ! : ctlt * czyz in the form j : ctcos.r * cr(kcosx) : Ccosx where C : ct* c2k. 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. DEFINlTloN General Solution, Basis, Particular Solution A general solution of an ODE (2) on an open interval 1is a solution (5) in which y1 and y2 aíe solutions of (2) on 1 that are not proportional, and c1 and c2 2.íe arbitrary constants. These lylz are called a basis (or a fundamental system) of solutions of (2) on I A particular solution of (2) on 1is obtained if we assign specific values to c1 and c2 in (5). For the definition of an interval see Sec. 1.1. Also, clanďc2 must sometimes be restricted to some interval in order to avoid complex expressions in the solution. Furthermore, as usual, y1 and lz Te called proportional on 1 if for aII x on I, (a) lt: k}z (b) !z: Ljt where k and / are numbers, zero or not. (Note that (a) implies (b) if and only If k + 0). Actually, we can reformulate our definition of a basis by using a concept of general importance. Namely, two functions y1 and y2 ate called linearly independent on an interval 1where they are defined if (7) klt@) -l k2y2@) :0 everywhere on / implies kr : 0 and k2 : 0. 49 (6) DEFlNlTloN And y1 artď y2 are called linearly dependent ky kz not both zero. Then if kl + 0 or k2 * 0, proportional, k2 .'l't:-, lzÁ1 If the coefficients p and q general solution. It yields on 1 if (7) also holds for some constants we can divide and see that y1 anď y2 are kr )2- , )7. K2 In contrast, in the case of linear independence these functions are not proportional because then we cannot divide in (7). This gives the following Basis (Reformulated) A basis of solutions of (2) on an open interval 1is a pair of linearly independent solutions of (2) on 1. ot (2) are continuous on some open interval I, then (2) has a the unique solution of any initial value problem (2), (4). It 5o CHAP. 2 Second-Order Linear ODEs includes all solutions of (2) on 1; hence (2) has no singular solutions (solutions not obtainable from of a general solution; see also Problem Set 1.1). Al1 this will be shown in Sec. 2.6. Basis, General Solution, Particular Solution cosxand sinxin Example 4 form abasis of solutions of the ODE y" + y:0 for allxbecause theirquotient iscotx*const(ortanx*const).Hence!:ctcos_r+c2sin-risageneralsolution.Thesolution ExAMPLE 5 ExAMPLE 6 ExAMPLE 7 y : 3,0 cos.x - 0.5 sin"r of the initial value problem is a particular solution. Basis, General Solution, Particular Solution Verify by substitution that ll : er and y2 : e-* are solutions of the ODE y" value problem Y"-y:0, yt0t : 6. y'{0) : -2. Solution. (r')" - e* : 0 anď (e-')" - e-r : 0 shows thaíeÍanď e-' are solutions. They are not proportional , e'le-* : "2* + const. Hence e', -* 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, !: cler * C2e-Í, I I -.T !:C -Cze y(0) : c1 l c2: 6, y'(0): ct- cz: By addition and subtraction, c1 : 2, cz: 4, so that the answer is y : 2rl í 4e-r. 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.l We first show this method for an example and then in general. Reduction of order if a solution ls known. Basis Find a basis of solutions of the ODE (*2 -,)y" - xy' + y:0, Solution. Inspection shows that y1 : x is a solution because y! : t and y'!: 0, so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute ll ll ^ l y:uX+lu l 0. Then solve the initial l ! : U!1: UX, y':u'x+tt, into the ODE. This gives (x2 - x)(u"x + 2u') - x(u'x -l w) * ux:0. ltx anď -xu cancel and we are left with the following ODE, which we divide by x, order, and simplify, (x2 - x)(u"x + 2u') - x2w' : O, (x2-x)w" *(x-2)u|:O. 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 ínI] 66, and moved to Paris in 1181 . His important ma.jor work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique anaLytiqlte, Paris, 1788), differential equations, approximation theory, algebra, and number theory. SEc. 2.1 Homogeneous Linear ODEs of Second Order ThisODEisof firstorderin u: l,|,',namely, (x2 - x)u'+ (.r - 2)u: O.Separationof variablesandintegration gives du x-2 / t 2\ ;:- xr-x'*:l,í-l -;)'lx - ll lnu| :ln|x- |1 -2ln|x| :ln-?We need no constant of integration because we want to obtain a particular solution; similarly in the next integration. Taking exponents and integrating again, we obtain x-| 1 | f t ,: _, :;-?. u:Judx:lnl*l +r, hence jz:Ltx:xh|x| +1. Since y1 : r and lz: x ln |x| + 1 are linearly independent (their quotient is not constant), we have obtained a basis of solutions, valid for all positive x. l In this example we applied reduction of order to a homogeneous linear ODE [see (2)] y"+p(x)y'*q(x)y:O. Note that we now take the ODE in standard form, with y", not f (x)y"-this is essential in applying our subsequent formulas. We assume a solution }r of (2) on an open interval 1to be known and want to find a basis. For this we need a second linearly independent solution lz of (2) on 1. To Eet !z, we substitute j : 1z: u!-]., y' : yL: Lt'y7 * ,y'r, y" : y'J,: u"y, l 2u'y', + ,yi into (2). This gives (8) u"y, l 2r'y'",, * uy'], f p(r'y, + "y'r) * quy1 : 0. Collecting terms ifl l,t", u' , anď u, we have u"y, * u'(2yl + py) + u(yi + py'r-| qy) : O. Now comes the main point, Since yl is a solution of (2), the expression in the last parentheses is zero. Hence z is gone, and we are left with an ODE in u' and u" .We divide this remaining ODE by y, and set bl' : (J, Ll" : (J', 1,1" + u' 2Y| -| PY' - o, thus {J' + ("l * ,) U : O. lt \y, 'l This is the desired first-order ODE, the reduced ODE. Separation of variables and integration gives +: (+ -,) and By taking exponents we finally obtain rn |u| - -2k lyrl - Iu o*. (9) 1 U: ,-ejt-Ip dr 5t 52 CHAP. 2 Second-Order Linear ODEs Here IJ : 6l , so that u: JU dx.Hence the desired second solution is jz:jtu:,rIUdx, The quotient y2ly. : Ll : IU dx cannot be constant (since U > 0), so that y1 and y2 form a basis of solutions. r GENERAL soLuTloN. lNlTlAL vALuE PRoBLEM (More in the next problem set.) Verify by substitution that the given functions form a basis. Solve the given initial value problem. (Show the details of your work.) 1. y" - 16y : g, e4*, e-4', y(0) : 3, y'10; : 3 2. y" + 25y : 0, cos 5x, sin 5x, y(0) : 0.8, ,.^. y (U) : _ó,5 3. y" -l 2y' * 2y : 0, e-' cos x, y(0): l,_y'(0): -l 4. y" - 6y' * 9y : O, e3', xe3,, y'(0) : 4.6 -rlll5. í-v + xv y'it) : -6 6. ,'y" - Jry' + y'(l) : 1.0 4y : O, x2, x-2, y(1) : 11, 15y : O, x3, x5, y(7) : 0.4, @ LINEAR INDEIENDENcE AND DEIENDENcE Are the following functions linearly independent on the given interval? 7.x,xlnx(O 0) |2.x-2,x+2(-2 0) 14, 0, sinh rrx (x > 0) REDUCTION OF ORDER is important because it gives a simpler ODE. A second-order ODE F(x,y,y' ,y"): 0, linear or not, can be reduced to first order if y does not occur explicitly (Prob. 15) or if x does not occur explicitly (Prob. 16) or if the ODE is homogeneous linear and we know a solution (see the text). 15. (Reduction) Show that F(x, y' , y") : 0 can be reduced to first order in z : y' (from which y follows by integration). Give two examples of your own. 16. (Reduction) Show that F(y, y' , y"): 0 can be reduced to a first-order ODE with y as the independent variable andy" : (clztdy)z, where z: y|; derive this by the chain rule. Give two examples. @ Reduce to first order and solve (showing each step in detail). I7. y" : ky' 18.y":I*y', 19. yy" - 4y'' 20. xy" + 2y' * xy : 0, lt : x-1 cosx 21. y" + y'3 siny : 0 22. (t - *')y" - Zxy' + 2y : g, !t: x 23. (Motion) A small body moves on a straight line. Its velocity equals twice the reciprocal of its acceleration, If at t : 0 the body has distance 1 m from the origin and velocity 2 mlsec, what are its distance and velocity after 3 sec? 24. (Hanging cable) It can be shown that the curve y(x) of an inextensible flexible homogeneous cable hanging between two fixed points is obtained by solving y" : 1r1/ 1 a ,n, where the constant k depends on the weight. This curve is called a catenary (from Latin catena : the chain). Find and graph y(x), assuming k : I and those fixed points are (- 1, 0) and (1, 0) in a vertical xy-plane. 25. (Curves) Find and sketch or graph the curves passing through the origin with slope 1 for which the second derivative is proportional to the first. 26. WRITING PROJECT. General Properties of Solutions of Linear ODEs. Write a short essay (with proofs and simple examples of your own) that includes the following. (a) The superposition principle. (b) y - 0 is a solution of the homogeneous equation (2) (called the trivial solution). (c) The sum ! : lt * y2 of a solution y1 of (1) and y2 of (2) is a solution of (1). (d) Explore possibilities of making further general statements on solutions of (1) and (2) (sums, differences, multiples). 27. CAS PROJECT. Linear Independence. Write a program for testing linear independence and dependence. Try it out on some of the problems in this problem set and on examples of your own. e-* Sin x, y(0) : -7.4, SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 53 with constant coefficients We shall now consider second-order homogeneous linear ODEs whose coefficients a and b are constant, (1) 2.2 Homoteneous Linear ODEs y"+oy'+by:0. These equations have important applications, especially in connection with mechanical and electrical vibrations, as we shall see in Secs. 2.4,2.8, and 2.9. How to solve (1)? We remember from Sec. 1.5 that the solution of thefirst-order linear oDE with a constant coefficient k y'+lq:0 is an exponential function ! : ce-k*. This gives us the idea to try as a solution of (1) the function (2) y : e^r Substituting (2) and its derivatives y' : Le^* and into our equation (1), we obtain Y" : 72rxr (^2+ ah*b)e^":g. Hence if .[ is a solution of the important characteristic equation (or auxiliary equation) (3) ^2+ah-|b:0 then the exponential function (2) is a solution of the ODE (1). Now from elementary algebra we recall that the roots of this quadratic equation (3) are (4) ,\t : Lr_ o + \/o' - +u), hz: žr-o - 1"z - +u1. (3) and (4) will be basic because our derivation shows that the functions lt : enr* and !2 : e^Z* 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 a2 - 4b,namely, (Case I) Two real roots iío' - 4b > 0, (Case II) A real double root if a2 - 4b : 0, (Case III) Complex conjugate roots iío' - 4b < 0. (5) 54 CHAP. 2 Second-Order Linear ODEs Case l. Two Distinct Rea[ Roots i, and ^2In this case, a basis of solutions of (1) on any interval is !7 : eÁr* and !2: enzr because y1 and y2 aíedefined (and real) for all x and their quotient is not constant. The coffesponding general solution is (6) j: CIe^'* + Cre^'*. ExAMPLE' ;::':::'::::;:"; T,:::;:i:"::i"5systematically Thecharacteristicequationis ^2 - I: 0. Its roots are ir : 1 and r\2 : -1. Hence a basis of solutions is e'and e-* and gives the same general solution as before, !: cteÍ * c2e-*. EXAMPLE 2 lnitial Value Problem in the Case of Distinct Real Roots Solve the initial value problem y" + y' - 2y : O, y(O) : 4, y'(0) : _5. Solution. Sfup 1. Its roots are General solution. The characteristic equation is ^2+^-2:O. nr:á(-t+x51 :l and ^z: i,-t - x6l : -z so that we obtain the general solution |: c|er + c2e-2'. Step 2. Particular solution. Sincey'(x) : cl Í - 2c2e-2',we obtain from the general solution and the initial conditions y(0) : c1 l c2: 4, y'(O) : c1- 2c2: -5. Hence ct: 7 and c2: 3. This gives the answer y : er + 3e-2*. Figure 29 shows that the curve begins at j: 4 with a negative slope (-5, but note that the axes have different scales!), in agreement with the initial conditions. y 8 6 4 2 0 1 1,5 2 x Solution in Example 2 '0 0.5 Fig. 29. l l SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients Case ll. Rea[ Double Root ^ - -a/2 If the discriminant a2 - 4b is zero, we see directly from (4) that we get only one root, i : i, : lz: -al2, hence only one solution, lt - e-@lDr, To obtain a second independent solution y, (needed for a basis), we use the method of reduction of order discussed in the last section, setting lz : ujy Substituting this and its derivative s yL : u' yt l uy'1 and y'/ tnto (1), we first have (u"y, + 2u' y_ + uy']) l a(u'y. + "y'r) * buy, : 0. Collecting terms In LI" , Lr' , anď u, as in the last section, we obtain u"y, * u'(2y'I -l ayr) + uO'|, + oy'-,, -l by1) : 0. The expression in the last parentheses is zero, since y, is a solution of (1). The expression in the first parentheses is zero, too, since 2y'r: -or-alcl2 : -a!l. We are thus left with u"y1: 0. Hence Lr" : 0. By two integrations, u : clx * c2. To geta second independent solution lz: uly,wa can simply choose cl,: I, c2:0 and take u : x. Then lz : xlr 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 n-arl2, *r-arl2. The comesponding general solution is ):(cr +c2x)e-o*l2 Warning. If ), is asimple rootof (4), then (cr i crx)e^' with c, * 0 is not a solution of (1). General solution in the case of a Double Root Thecharacteristic equationof theODEy" + 6y' + 9y: Ois ^2 + 6^ + 9: (^ + 3)2:0. Ithasthedouble root ), : -3. Hence abasis is e-3' anďxe-3*. The corresponding general solution is,y : (c1 -| c2x)e-3'. I lnitial value problem in the case of a Double Root Solve the initial value problem y" + y' + 0.25y : Q, .v(0) : 3.0, y'(o) : -:.s. Solution. Thecharacteristicequationisi2 + ^ + 0.25: (^ + 0.5)2:0.IthasthedoublerootA: -0,5. This gives the general solution , -0.5r We need its derivative ) : (cr Ť C2X)e y' : rzr-o'5' - 0.5{., + cr*\e-o,s* 55 (7) ExAMPLE 3 ExAMPLE 4 56 CHAP. 2 Second-Order Linear ODEs From this and the initial conditions we obtain )(0) : c1 - 3,0, )'(0) : c2 - 0.5c1: The particular solution of the initial value problem is y : (3 - 3.5; hence - 2x)g-o,s', See Fig. 30. ^ - -.|L2 - L, l y 3 2 1 0 _1 Fig. 30. Solution in Example 4 Case lll. Complex Roots -}a + irrl and -}a iol This case occurs if the discriminant a2 - 4b of the characteristic equation (3) is negative. In this case, the roots of (3) and thus the solutions of the ODE (1) come at first out complex. However, we show that from them we can obtain a basis of real solutions -^*Ljt : *"'- cos trl,r. (r,l > 0) where 0)2 : b - io'. It can be verified by substitution that these are solutions in the present case. We shall derive them systematically after the two examples by using the complex exponential function. They form a basis on any interval since their quotient cot alx is not constant. Hence a real general solution in Case III is j : e-o'l2 1Á cos olx * B sin ax) (A, B arbitrary). (8) (9) E XA M P L E 5 Complex Roots. lnitial Value Problem Solve the initial value problem y" + 0.4v' + 9.04y : 0, y(0) : 0, }'(0) : 3. Solution. Sfup 1. General solution. The characteristic equation is,tr2 + 0.4^ + g.O4: 0. It has the roots -0.2 -ť 3i. Hence a: 3, and a general solution (9) is y - "-o,"(A cos 3x i B sin 3x). Step 2. Particular solution. The first initial condition gives y(0) : A : 0. The remaining expression is y : 3"-o,2r sin 3x. We need the derivative (chain rule!) y' : B(-0.2r-o,2t sin 3r -]- 3r-o,2r cos 3x). From this and the second initial condition we obtain y'(0) : 38 :3. Hence B : I. Our solution is y : ,-o,2r sin 3x. Figure 31 shows y and the curves of e-o,zI anď -e-o2'(dashed), between which the curve of y oscillates. Such "damped vibrations" (with -t : r being time) have important mechanical and electrical applications, as we shall soon see (in Sec. 2.4). t l SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients Fig. 3t. Solution in Example 5 EXAMPLE 6 ComplexRoots A general solution of the ODE is (ol constant, not zero)y" + tl2l,: O y : A cos oJr * B sin rr.rx. With ro : 1 this confirms Example 4 in Sec. 2.1. It is very interesting that in applications to mechanical systems or electrical circuits, these three cases coíTespond to three different forms of motion or flows of current, respectively. We shall discuss this basic relation between theory and practice in detail in Sec. 2.4 (and again in Sec. 2.8). Derivation in Case lll. Complex Exponential Function If verification of the solutions in (8) satisfies you, skip the systematic derivation of these real solutions from the complex solutions by means of the complex exponential function e' of a complex variable z : r + i/. We write r l it,notx * i because x and y occur in the ODE. The definition of e' in terms of the real functions e', cos /, and sin r is (10) uz - ur+it : e'eit: e'(cos t + i sinr). l 57 Summary of Cases 1-1II Case Roots of (2) Basis of (1) General Solution of (1) I Distinct real ly lz g^l*, g^zr !: CI7X7' * Cre^zr il Real double root ^: -ia ,-arl2, ,r-arl2 /:(cr lcrx)e-o*/2 ilI Complex conjugate lr: -ta l ia, lr: -la - io ,-arl2 cos r,,l.r u-arl2 sin r..rx ! - e-o'l21Á cos ax l B sin cox) ,1 58 CHAP. 2 Second-Order Linear ODEs This is motivated as follows. For real z: r,hence t : 0, cos 0 : 1, sin 0 : 0, we get the real exponential function e'.It can be shown that e"*"' - e"e"',just as in real. (Proof in Sec. 13.5.) Finally, if we use the Maclaurin series of e" with z : it as well as i2 : -I, i3 : -i, i4 : I, etc., and reorder the terms as shown (this is permissible, as can be proved), we obtain the series pl| : l + it + 2! fíl -|| 2l, 4! :cos/-llsin/. Gt)4 4! .-,(t_ , ,\, (ir)5 -+.. 5! f -+3! 5! , (ir)3 -Ť-- 3! -+.. (Look up these real series in your calculus book if necessary.) We see that we have obtained the formula (11) eit:cos/+isin/, called the Euler formula. Multiplication by e'gives (10). For later use we note that e-it : cos (-r) * i sin (-t) -- cos / - l sin /, so that by addition and subtraction of this and (11), (I2) cos/: (nn' + e-i'), l , After these comments on the definition (10), let us now turn to Case III. In Case III the radicand c - 4b tn (4) is negative. Hence 4b - a2 is positive and, using \/-: i, we obtain in (4) +\/r' - +u: +\,t-w - "'): \[:@ -rn"1 : i\/b - iď : ia with ro defined as in (8). Hence in (4), it:}a+ia and, similarly, lz: ta - ia. Using (10) with , : -lax and t : @x, we thus obtain elr* : n-k1.12)r+žor - n-(a/2)r(cos olx l i sin rr.l"r) exr* : n-(al2)r-iar - ,-(a/2)tr(cos @x - i sin rr.rx). We now add these two lines and multiply the result bv L.This gives y1 as in (8). Then we subtract the second line from the first and multiply the result by ll(zi). This gives y, 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. SEC. 2.3 Differential Operators. Optional 59 @ GENERAL soluTloN Find a general solution. Check your answer by substitution. l.y"-6y'-Jy:O 2. I}y" - Jy' -l I.2y : Q 3. 4y" - zoy' * z5y : g 4. y" * 4rry' * 4n2y : g 5. 100y" + 20yl - 99y: g 6.y"*2y't5y:g 7.y"-y'+Z.5y:g 8. y" + 2.6y' + I.69y : 0 9. y" - 2y' - 5.25y : g 10.}"-2y:0 11. y" * 9rr2y : g 12. y" -l 2.4yl* 4.0y:6 13. y" - I44y : g 14. y" * y' - 0.96y : g @ FIND oDE Find an ODE y" l oy' l by :0 for the given basis. W4 lNlTtAL vALuE pRoBLEMs Solve the initial value problem. Check that your answer satisfies the ODE as well as the initial conditions. (Show the details of your work.) 21. y" - 2y' - 3y : 0, y(0) : 2, !'(0) : t4 22. y" + 2y' + y: 0, y(0) :4,!' (0) : -6 23. y" + 4y' + 5y: 0, y(0) :2,!'(0) : -5 24. IOy" - 50y' + 65y: 0, y(0) : 1.5, y'(0) : 1.5 25. y" + ,y' : 0, y(0) :3,y'(0): -, 26. I\y" + 18y' + 5.6y: O, y(0) : 4,!'(0) : _3.8 27. I\y" -f 5y' + 0.625y:0,y(0) :2,!'(0): _4.5 28. y" - 9y: 0, }(0) - -2, y'(0) : -I2 29.ZOy" + 4y' * } : 0, y(0) :3.2, y'(0) : 0 30. y" + zky' + (k2 + ,')y: 0, y(0) : 1, y'(0) - -k 31. y" - 25y: 0, y(0) : 0, )'(0) : 40 32. y" - 2y' * 24y : 0, y(0) : 0, }'(0) : 20 33. (Instability) Solve y" - y: 0 for the initial conditions y(0) : 1, y'(0) - - 1. Then change the initial conditions to y(0) : 1.001, y'(0) : -0,999 and explain why this small change of 0.001 at x : 0 causes alarge change later, e.g., 22 at x : 10. 34. TEAM PROJECT. General Properties of Solutions (A) Coefficient formulas. Show how a and b in (1) can be expressed in terms of ,[1 and ,tr2. Explain how these formulas can be used in constructing equations for given bases. (B) Root zero. Solve y" + 4y' : 0 (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 y" l ayl : 97 (C) Double root. Verify directly that xen* with )" : -al2 is a solution of (1) in the case of a double root. Verify and explain why y : g-2r is a solution of Y" - y' - 6y : O btí xe-2" is not. (D) Limits. Double roots should be limiting cases of distinct roots i1, r\2 &s, say, ,\2 - ňr. Experiment with this idea. (Remember l'Hópital's rule from calculus.) Can you arrive at xe^t*? Give it a try. 35. (Verification) Show by substitution that y1 in (8) is a solution of (1). 15. e2'. e' 17. e* 5, ,"* 19. e4*, e*4* 1,6. e0.5*, u-3.5r 18. I, e-3* 20. eGl+i)í. e-(L+i)r 2.3 Differential Operators. Optional This short section can be omitted without interrupting the flow of ideas; it will not be used in the sequel (except for the notations Dy, D'y, etc., 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 Dy:y(1) ,:dy dx 60 CHAP. 2 Second-Order Linear ODEs Similarly, for the higher derivatives we write D'y : D(Dy) : !", and so on. For example, D sin : cos, D2 sin: -sin, etc. For a homogeneous linear ODE y" + ay' + by :0 with constant coefficients we can now introduce the second-order differential operator L:P(D):D2-1 aD+bI, where 1is the identity operator defined by Iy : y. Then we can write that ODE as Ly : P(D)y : (D2 -f aD * bl)y :0. P suggests "polynomia1." L is a linear operator. By definition this means that if Ly and Lw exist (this is the case if y and w are twice differentiable), then L(cy + kw) exists for any constants c and k, and (2) show that from (2) : le^* and (Dze^)(x) L(cy + kw) : cLy -l kLw. we reach agreement with the results in Sec. 2.2. Slnce -- ).'en*, we obtain Let us (ne\@) (3) Le^(x) : p(D)e^(x) : (D2 + aD * bDe^(x) : (^2 -l aX -l b)e^" : p(i)e^" : 0. This confirms our result of Sec. 2.2 that e^* is a solution of the ODE (2) iíand only if )" is a solution of the characteristic equation P(,\) : 9. P(^) is a polynomial in the usual sense of algebra. If we replace ,\ by the operator D, we obtain the "operator polynomial" P(D).The point of this operational calculws is that P(D) can be treated just like an algebraic quantity. In particular, we can factor it. EXAMPLE l Factorization, Solution of an ODE Factor P(D) : o2 - 3o - 4OI and solve p(D)y : 0. Solution. D2 -3D- 40I: (D - 81)(D + 51)because 12: I.Now(D - 81)y:y' - 8y: Ohasthe solution y! : ea* . Similarly, the solution of (D + 5Dy : 0 is y2 : "-5* . This is a basis of P(D)y : 0 on any interval, From the factorization we obtain the ODE,, as expected, (D - 81XD + 51)y : (D - 81)(y/+ 5y): D(y'+ 5y) - 8(y'+ 5y) : y" +5y' - 8y' - 40y : y" - 3y' - 40y : g. Verify that this agrees with the result of our method in Sec. 2.2. This is not unexpected because we factored P(D) in the same way as the characteristic polynomial P(n) : 12 - 3^ - 40. l [t was essential that L in (2) has constant coefficients. Extension of operator methods to variable-coefficient ODEs is more difficult and wi1l 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. SEC. 2.4 Modeling: Free Oscillations (Mass-Sprin6 System) 6l E AppLlcATloN oF DIFFERENTIAL OPERATORS Apply the given operator to the given functions. (Show all steps in detail.) 1. (D - I)'; *, xe', sin x 2. 8D2 + 2D - I: cosh }x, sinh }x, ,ll2 3. D - 0.4I; 2x3 - I, eo,4*, ,uo,4r 4. (D + 51)(D - I); e-5* sinx, e5*, x2 5. (D - 4I)(D + 31); x3 - x2, sin 4x, e-3* @ GENERAL soluTloN Factor as in the text and solve. (Show the details.) 6. (D2 - 5.5D + 6,66I)y : O 7. (D ,l 2I)2y : g 8. (D' - 0.49I)y : g 9. (D2 + 6D + 131)y : 6 10. (10D2 + 2D 1- L7I)y : O II. (D2 + 4,ID + 3.11)y : Q 12. GD2 ]- 4rrD -l r2l)y : g 13. (D2 * ú.64u'I)y : 0 14. (Double root) If D2 -l aD -f bI has distinct roots t-L and ^, show that a particular solution is y : (e@ - "o*)l(p - ,\). Obtain from this a solution xe^'by letting p -+ ^ and applying l'Hópital's rule. 15. (Linear operator) Illustrate the linearity of L in (2) by taking c : 4, k : -6, y : e2', andw : cos2x. Prove that L is linear. 16. (Definition of linearity) Show that the definition of linearity in the text is equivalent to the following. If L|yl and l,[w] exist, then Lly + w] exists and L|cyl and L|kw] exist for all constants c and k, and Lly + w] : Lly] + L|wl as well as L|cy] : cL|ll and Llkw): kLlw]. 2.4 Modelint: Free Oscillations (Mass-Sprint System) Linear ODEs with constant coefficients have important applications in mechanics, as we show now (and in Sec. 2.8), and in electric circuits (to be shown in Sec. 2.9).In this section we consider a basic mechanical system, a mass on an elastic spring ("mass-spring system," Fig.32), which moves up and down. Its model will be a homogeneous linear ODE. Setting Up the Model We take an ordinary spring that resists compression as well extension and suspend it vertically from a fixed support, as shown in Fig. 32. At the lower end of the spring we System in equilibrium System in motion (a) (b) (c) Fig. 32. Mechanical mass-spring system 62 CHAP.2 Second-Order Linear ODEs attach a body of mass m, We assume m to be so large that we can neglect the mass of the spring. If we pull the body down a certain distance and then release it, it starts moving. We assume that it moves strictly vertically. How can we obtain the motion of the body, say, the displacement y(r) as function of time t? Now this motion is determined by Newton's second law Mass X Acceleration : m " : Force where y" : d2yldt2 and "Force" is the resultant of all the forces acting on the body. (For systems of units and conversion factors, see the inside of the front cover.) We choos e the downward direction as the positive direction, thus regarding downward forces as positive and upward forces as negative. Consider Fig.32. The spring is first unstretched. We now attach the body. This stretches the spring by an amount s6 shown in the figure. It causes an upward force Fo in the spring. Experiments show that F6 is proportional to the stretch ,6, s&}, Fo : -kso (Hooke's lawz). k (> 0) is called the spring constant (or spring modulus). The minus sign indicates that Fg points upward, in our negative direction. Stiff springs have large k. (Explain!) The extension s6 is such that F6 in the spring balances the weight W : mg of the body (where g : 980 cm/sec2 : 32.17 ftlsecz is the gravitational constant). Hence Fo * W : -kso -l mg : 0. These forces will not affect the motion. Spring and body are agatn at rest. This is called the static equilibrium of the system (Fig. 32b). We measure the displacement y(r) of the body from this 'equilibrium point' as the origin y : 0, downward positive and upward negative. From the position y : 0 we pull the body downward. This further stretches the spring by some amount y > 0 (the distance we pull it down). By Hooke's law this causes an (additional) upward force F1 in the spring, Ft: -kj. F, is a restoring force. It has the tendency to restore the system, that is, to pull the body backtoy:0. Undamped System: ODE and Solution Every system has damping-otherwise it would keep moving forever. But practically, the effect of damping may often be negligible, for example, for the motion of an iron ball on a spring during a few minutes. Then F1 is the only force in (1) causing the motion. Hence (1) gives the model my" : -lry o, my"+lcy:0. (1) (2) (3) 2RogBRt HooKE (1635-1703), English physicist, gravitation. a íbrerunner of Newton with respect to the law of SEC. 2.4 Modeling: Free Oscillations (Mass-Spring System) By the method in Sec. 2.2 (see Example 6) we obtain as a general solution y(t): Á cos agt * B sinclgt, The corresponding motion is called a harmonic oscillation. Since the trigonometric functions in (4) have the period 2Ťlog, the body executes agl2nr cycles per second. This is the frequency of the oscillation, which is also called the natural frequency of the system. It is measured in cycles per second. Another name for cycles/sec is hertz (Hz).3 The sum in (a) can be combined into a phase-shifted cosine with amplitude C : Y@ 1 3z and phase angle 6 : arctan (BlA), 63 (4) (4*) y(t): Ccos(cl6r- 6). To verify this, apply the addition formula for the cosine [(6) in App. 3.1] to (4*) and then compare with (4). Equation (4) is simpler in connection with initial value problems, whereas (4*) is physically more informative because it exhibits the amplitude and phase of the oscillation. Figure 33 shows typical forms of (4) and (4*), all corresponding to some positive initial displacement y(0) [which determines Á : y(0) in (4)] and different initial velocities y'(0) lwhich determine B : y' (0)/rr.ro]. |_.z @ Positive ,A|4 Lero @ Negative Fig. 33. Harmonic oscillations Undamped Motion. Harmonic Oscillation If an iron ball of weight W : 98 nt (about 22lb) stretches a spring 1.09 m (about 43 in.), how many cycles per minute will this mass-spring system execute? What will its motion be if we pull down the weight an additional 16 cm (about 6 in.) and let it start with zero initial velocity? SOlUtiOn. Hooke's law (2) with I4l as the force and 0.09meter as the stretch gives Iť : 1.09k; thus k : W/t.09 : 9BlI.0g : 9Olkg/sec'] : gO [nt/meter]. The mass is m : Wls : 98/9.8 : 10 tkg]. This gives the frequency agl(2T): l,/tt*t(Zr):3l(2rr): 0.48 tHz] : 29 [cycles/min]. 3HBtxRIcH HERTZ (1857-1894), German physicist, who discovered electromagnetic WaVeS, as the basis of wireless communication developed by GUGLIELMO MARCONI (l874-1937), Italian physicist (Nobel prize in 1909). lnitial velocity ExAM PLE 1 64 CHAP. 2 Second-Order Linear ODEs From (4) and the initial conditions, y(0) : A : 0.16 [meter] and y ooB :0, Hence the motion is y(t) :0.16 cos 3t [meter] or 0.52 cos 3t [ft] ig. 3a). 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. l y o.2 0.1 0 _0.1 _o.2 Damped System: ODE and Solutions We now add a damping force Fz: -C! to our model mryl| (5) : -kj, so that we have my" : -lq - ,y my"+cy'+b:0. Physically this can be done by connecting the body this new force to be proportional to the velocity y' a good approximation, at least for small velocities. c is called the damping constant. We show that c is positive. If at some instant, y' is positive, the body is moving downward (which is the positive direction). Hence the damping force Fz : -c!' , always acting against the direction of motion, must be an upward force, which means that it must be negative, Fz - -r!' ( 0, so that -c ( 0 and c ) 0. For an upward motion, y' < 0 and we have a downward Fz : -cy > 0; hence -c >ň R Spring <<< ínz I M Body aryll:. .WW&* 35. Damped Fig. 34. Harmonic oscillation in Example 1 Fit. SEC. 2.4 Modeling: Free Osci[lations (Mass-Spring System) By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2, 65 (6) it:-a+P, lz:-a-B, where dIt is now most interesting that depending on the amount of damping (much, medium, or little) there will be three types of motion corresponding to the three Cases I, II, II in Sec.2.2: Distinct real roots hu hz. (Overdamping) A real double root. (Critical damping) Complex conjugate roots. (Underdamping) C and 2m Case I. c2 > 4mk. Case II. c2 : 4mk. Case III. c2 < 4mk, (7) Discussion of the Three cases Case l. Overdampint If the damping constant c is so large that c2 } 4mk, then ),1 and ),2 are distinct real roots. In this case the coffesponding general solution of (5) is y(r) : cre-@-F)t * cre-(o+ F)t We see that in this case, damping takes out energy so quickly that the body does not oscillate. For r > 0 both exponents in (7) are negative because a } 0, B > 0, and F' : a2 - klm < a2. Hence both terms in (7) approach zero as t ---> @. Practically speaking, after a sufficiently long time the mass will be at rest at the static equilibrium position (y : 0). Figure 36 shows (7) for some typical initial conditions. Case ll. 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 c2 : 4mk, @ Positive ] @z",o I lni,iut velocity @ Negative J Fig.36. Typical motions (7) in the overdamped case (a) Positive initial displacement (b) Negative initial displacement (b) 66 CHAP. 2 Second-Order Linear ODEs so that F : O, Ar : Xz : - a. Then the coresponding general solution of (5) is y(t): (cr* c2t)e-'t This solution can pass through the equilibrium position y : 0 at most once because e-ot is never zero and c' l c2t can have at most one positive zero. If both c1 and c2 ďí9 positive (or both negative), it has no positive zeío, so that y does not pass through 0 at all. Figure 37 shows typical forms of (8). Note that they look almost like those in the previous figure. Case lll. Underdamping This is the most interesting case. It occurs if the damping constant c is so small that c2 < 4mk.Then B in (6) is no longer real but pure imaginary, say, (8) (9) F:ia* where (We write ar* to reserve a.r for driving and electromotive forces in Secs. 2.8 anď2.9.)The roots of the characteristic equation are now complex conjugate, ir: -alia*, hz:-a-ia* with c : cl(2m), as given in (6). Hence the corresponding general solution is (10) y(t) : e-'t(A cos rrr*/ -l B sin o*t) : Ce-ot cos (ro*r - 6) where C2 : A2 + 82 and tan 6 : BlA, as in (4*). This represents damped oscillations. Their curve lies between the dashed curves ! : Ce-ot andy - -Ce-ot in Fig. 38, touching them when a*t - 6 is an integer multiple of rrbecause these are the points at which cos (rr.l*/ - 6) equals 1 or -1. The frequency is a l(Zrr)Hz (hertz, cycles/sec). From (9) we see that the smaller c (> 0) is, the larger is c,.l* and the more rapid the oscillations become. If c approaches 0, then rrr* approaches do : \Elor, giving the harmonic oscillation (4), whose frequency agl(2Ťr) is the natural frequency of the system. Fig. 37. Critical damping [see (8)] Fig. 38. Damped oscillation in Case lll [see (10)] Ce-o' .o/ -.---- r /a\/\AL) / \- ,/'"Ý /\ \ ---.- / '. \ --- 67 The Three Cases of Damped Motion How does the motion in Example 1 change if we change the damping constant c to one of the following three values, with y(0) : 0.16 and y'10) : O as before? (I)c:100kg/sec, (II)c:60kg/sec, (III)c: l0kg/sec, SolutiOn. It is interesting to see how the behavior of the system changes due to the effect of the damping, which takes energy from the system, so that the oscillations decrease in amplitude (Case III) or even disappear (Cases II and I). (I) With m: 10 and k: 90, as in Example 1, the model is the initial value problem lOy" + 100y' + 90y: 0, y(0) : 0.16 [meter], }'(0) : 0. The characteristic equation is l0),2 + l00^ + 90 : 10(^ + 9X^ + 1) : 0. It has the roots -9 and -1. This gives the general solution -9t , _t j:C ŤC2 we also need y' : -9c -9t - c2e-t. The initial conditions give c1 l cz: 0.16, *9c1 - c2:0. The solution is c1 : *0.02, c2 : 0.18, Hence in the overdamped case the solution is !: -0.02e-9t + 0.1Be-ú. It approaches 0 as / --+ co. The approach is rapid; after a few seconds the solution is practically O, that is, the iron ball is at rest. (II) The model is as before, with c : 60 instead of 100. The characteristic equation now has the form 10),2 + 60^ + 90 : 10(^ + '2 :0. It has the double root -3. Hence the corresponding general solution is }:(cr+c2t)e-3t. we also need y' : (cz - 3rt - 3c2t)e-3t. The initial conditions give y(0) - c1 0.16, ),'(0) - c2 3rr: 0, c2:0.48. Hence in the critical case the solution is y:(0,16 +0.48t)e-3t. It is always positive and decreases to 0 in a monotone fashion. (III) The model now is 10y" + 10y' + 90y : 0. Since c : 10 is smaller than the critical c, we shall get oscillations.ThecharacteristicequationislOiz+10^+90:rO[1,1 +L)'ar-á] :0. Ithasthecomplex roots [see (4) in Sec. 2.2 with a : I and b : 9] ^: _0.5 ,x6.s\ g: _0.5 -| 2.g6i. This gives the general solution , : n-o,st1{ cos2.96t * B sin 2.96t). Thus y(0) : A : 0.16. We also need the derivative y' : "-o'5t(-0.5Á cos2.96t - 0.5B sin2.96t - 2.96Asin2.96t + 2.968 cos2.96t). Hence y'(0) : -0.5A + 2.968 : 0, B : 0.5A12.96 : 0.027. This gives the solution y: e-o,5t(0.16 cos 2.96t + 0.027 sin 2.96t): O.t62e-o,5t cos(2.96t - 0.17). We see that these damped oscillations have a smaller frequency than the harmonic oscillations in Example 1 by abo IVo (since 2.96 is smaller than 3.00 by about l7o). Their amplitude goes to zero. See Fig. 39. l y 0.15 0.1 0.05 0 -0.05 _0,1 Fig. 39. The three solutions in Example 2 SEC. 2.4 Modelin6: Free Oscillations (Mass-Spring System) 68 CHAP.2 Second-Order Linear ODEs This section concerned free motions of mass-spring systems. Their models are homogeneous lineat ODEs. Nonhomo7eneous linear ODEs will arise as models of forced motions, that is, motions under the influence of a "driving force". We shall study them in Sec. 2.8, after we have learned how to solve those ODEs. MoTloN WITHouT DAMP|NG (HARMoNlc osclLLATloNs) 1. (Initial value problem) Find the harmonic motion (4) that starts from y6 with initial velocity u6. Graph or sketch the solutions for @o : Ťí,.}o : 1, and various u6 of your choice on common axes. At what r-values do all these curves intersect? Why? 2. (Spring combinations) Find the frequency of vibration of a ball of mass m : 3 kg on a spring of modulus (i) frr : 2J nílm, (11) k2: 75 ntlm, (iii) on these springs in parallel (see Fig. 40), (iv) in series, that is, the ball hangs on one spring, which in turn hangs on the other spring. 3. (Pendulum) Find the frequency of oscillation of a pendulum of length L (Fig. 4I), neglecting air resistance and the weight of the rod, and assuming 0 to be so small that sin 9 practically equals 0. 4. (Frequency) What is the frequency of a harmonic oscillation if the static equilibrium position of the ball is 10 cm lower than the lower end of the spring before the ball is attached? 5. (Initial velocity) Could you make a harmonic oscillation move faster by giving the body a Ereater initial push? 6. (Archimedian principle) This principle states that the buoyancy force equals the weight of the water displaced by the body (partly or totally submerged). The cylindrical buoy of diameter 60 cm in Fig. 42 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? Body of maSS /7l (Frequency) How does the frequency of a harmonic motion change if we take (i) a spring of three times the modulus, (ii) a heavier ball? TEAM PROJECT. Harmonic Motions in Different Physical Systems. Different physical or other systems may have the same or similar models, thus showing the unifying power of mathematical methods. Illustrate this for the systems in Figs. 4345, (a) Flat spring (Fig. 43). The spring is horizontally clamped at one end, and a body of weight 25 nt (about 5.6 lb) is attached at the other end. Find the motion of the system, assuming that its static equilibrium is 2 cm below the horizontal line, we let the system start from this position with initial velocity 15 cm,/sec, and damping is negligible. (b) Torsional vibrations (Fig. 44), Undamped torsional vibrations (rotations back and forth) of a wheel attached to an elastic thin rod are modeled by the ODE Io0" + K0 : 0, where 0 is the angle measured from the state of equilibrium, 16 is the polar moment of inertia of the wheel about its center, and kis the torsional stiffness of the rod. Solve this ODE fot KlIg: I] .64 sec-2, initial angle 45o, and initial angular velocity 15" sec-l. (c) Water in a tube (Fig. a5). What is the frequency of vibration of 5liters of water (about 1.3 gal) in a U-shaped tube of diameter 4 cm, neglecting friction? 7. 8. Fig. a0. Parallel springs (Problem 2) Fig.4l. Pendulum (Problem 3) Fig.42. Buoy (Problem 6) Fig. 43. Flat spring (Project 8a) (y=0) :ž Fig.44. Torsional vibrations (Project 8b) Fig. a5. Tube (Project 8c) DAMPED MOTION 9. (Frequency) Find an approximation formula for ro* in terms of rr.16 by applying the binomial theorem in (9) and retaining only the first two terms. How good is the approximation in Example 2,III? -\eL-\ SEC.2.5 Euler-CauchyEquations 10. (Extrema) Find the location of the maxima and minima of y : e-2t cos r obtained approximately from a graph of y and compare it with the exact values obtained by calculation. 11. (Maxima) Show that the maxima of an underdamped motion occur at equidistant /-values and find the distance. 12. (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, equals L, : Zrala*. Find A for the solutions of y" + 2y' + 5y : 0. 13. (Shock absorber) What is the smallest value of the damping constant of a shock absorber in the suspension 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 4500 kg/sec2? 14. (Damping constant) Consider an underdamped motion of a body of mass m : 2 kg. If the time between two consecutive maxima is 2 sec and the maximum amplitude decreases to } of its initial value after 15 cycles, what is the damping constant of the system? 15. (Initial value problem) Find the critical motion (8) that starts from y6 with initial velocity u6. Graph solution curves for a : 1, }o : 1 and several u6 such that (i) the curve does not intersect the /-axis, (ii) it intersects it at t : I, 2,. . ., 5, respectively. 16. (Initial value problem) Find the overdamped motion (1) that starts from y6 with initial velocity uo. 17. (Overdamping) Show that in the overdamped case, the body can pass through y : 0 at most once. 18. CAS PROJECT. Transition Between Cases I,II, ilI. Study this transition in terms of graphs of typical solutions. (Cf. Fig. a6.) 69 (a) Avoiding unnecessary generaw is part of good modeling. Decide that the initial value problems (A) and (B), (A) y" + ,y' i y : 0, y(0) : 1, y'(0) : 0 (B) the same with different c and }'(0) : -2 (instead of 0), will give practically as much information as a problem with other m, k, y(O), y'(0). (b) Consider (A). Choose suitable values of c, perhaps better ones than in Fig. 46 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 0.1Vo of the initial displacement (this choice being up to us), that is in our case, (11) |y(l)| < 0.001 for all t greateí than some /r. 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 y(tr) : -0.001, with t, the solution of y' (t) :0, will give you the best possible c satisfying (1 1). (e) Consider (B) empirically as in (a) and (b). What is the main difference between (B) and (A)? 0.5 _0,5 _1 Fig.46. CAS Project 18 y 1 2.5 Euler-Cauchy Equations (1) Buler-Cauchy equationsa are ODEs of the form *'y"+axy'+by:0 aLBONHARD EULER (I101-I783)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. 7o CHAP.2 Second-Order Linear ODEs with given constants a and b and unknown y(x). We substitute (2) y:x* General solution in the case of Different Real Roots The E,uler-Cauchy equation ,'y" + 1.5xy'-0.5_v:0 has the auxiliary equation ,n2 + 0.5- - 0.5:0. The roots are 0.5 and - l. Hence a basis of solutions for all positive x is 1,1 : x general solution C2 r' : crV"{ _]r (6) and its derivatives y' : mx'n-l and y" : m(m - 1)7*-z into (1), This gives We now factor x- (3) xzm(m - 1)a*-z l axmx--l + bx* :0, see that (2) was a rather natural choice because we have obtained a common . Dropping it, we have the auxiliary equation m(m - I) * am * b : 0 or m2+(a-I)m-lb:0. (Note: a - I,not'a.) Hence y : xln is a solution of (1) if and only tf mis a root of (3). The roots of (3) are (4) l7ll:*rr - ol*\f'nrt u>'_ u, ffi2:ltl - a)-\F- ">'-t, Case I. trf the roots m1 and n,Iz are real and different, then solutions are }r(x) : x-' and !z(x) : x-' They 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 coffesPonding general solution for all these x is (5) !:CLX*'*C2x*' (cy cz arbitrary). EXA,MPtE t (Note: 0.5, not 1.5|) 'O'5 and y2: Ilx and gives the (x>0). I Case II. Equation (4) shows that the auxiliary equation (3) has a double root t,ll: Lo - a) if and only if (I - o)' - 4b: 0. The Euler-Cauchy equation (1) then has the form ,'y" + axy' + árt - a)zy:0. A solution i, y, : *(I-cl)l2. To obtain a second linearly independent solution, we aPPlY the method of reduction of order from Sec. 2.1 as follows. Starting from lz : Uly Wa obtain for u the expression (9), Sec. 2.1, namely, r u:]Udx where U# "^o(-í,*) SEC. 2.5 Euler-Cauchy Equations Here it is crucial that p is taken from the ODE written in standard form, in our case, (1 - o)' (6*) 4x2 This shows that p : a/x (not ax). Hence its integral is a ln x : In (x"), the exponential functionin Uis llxo,anddivisionbyyt'- xl-o gives (J: !lx,andu: lnxbyintegration. Thus, in this "critical case," a basis of solutions for positive x is y1 : x* and lz : x* Inx, where * : ŽQ - a). Linear independence follows from the fact that the quotient of these solutions is not constant. Hence, for all x for which y1 and y2 are defined and real, a general solution is ):(cr*c2lnx)x- m:*(I-a), EXAMPLE 2 General Solution in the Case of a Double Root TheEuler-Cauchyequattonxzy"-5"y'+9),:0hastheauxiliaryequationrr2_ 6*+9:0. Ithasthe double rooí m: 3, so that a general solution for all positive x is y:(cr *c2lnx)xs. Case III. The case of complex roots is of minor practical importance, and it suffices to present an example that explains the derivation of real solutions from complex ones. E x A M P L E 3 T:,o;::ffilT:;:in the Case of Complex Roots ,'y"+0.6xy'+l6.04y:Q has the auxiliary equation -2 - 0.4* + 16.04 : 0. The roots are complex conjugate, mt : 0.2 1- 4i anď i,"*?.'";Í;;,ffi j*fJ;iff #il"Tlí:;iT*iH#lT,xTl'J':ffi#coef f icientshascomplex m. 0-2. /1i 0.2, In r,4i 0.2 (4 In .r) i ,r -,r :jr (e ) :x e , m9 0.2-4i 0.2, In r.-4i 0.2 -(4 ln I) i X :x \e ) :X e Next apply Euler's formula (l1) in Sec,2.2 with r : 4lnx to these two formulas. This gives ť"l : "o,21"os (4 ln,r) i i sin (4 ln x)], x-' : "o,2;.o* (4 ln -r) - i sin (4 ln r)]. Add these two formulas, so that the sine drops out, and divide the result by 2. Then subtract the second formula from the first, so that the cosine drops out, and divide the result by 2i. This yields ,o'2 .o, (4 ln x) xo,2 sin (41n x) respectively. By the superposition principle in Sec. 2.2 these are solutions of the Euler-Cauchy equation (1). Since their quotient cot (4 ln x) is not constant, they are linearly independent. Hence they form a basis of solutions, and the corresponding real general solution for all positive x is (8) y : ,o,'íA cos (4 ln r) + B sin (4 ln x)]. Figure 47 shows typical solution curves in the three cases discussed, in particular the basis functions in Examples l and 3. l 71 a x (7) l 72 CHAP. 2 Second-Order Linear ODEs y 3.0 2.o 1.0 lnr 5lnr Case II: Double root Fig. 47. Euler-Cauchy equations _0.5 _1.0 -1.5 u(10):.r+fr:o. 0 2 sin (4 lnl) Case III: Complex roots y 1.5 1,0 0.5 EXAM PLE 4 Boundary Value Problem. Electric Potential Field Between Two Concentric Spheres Find the electrostatic potential u : u(r) between two concentric spheres of radii rt : 5 cm and 12 : 70 cm kept at potentials ut : 110 V and uz: 0, respectively. Physical Information u(r) is a solution of the Euler-Cauchy equation ru" + 2u' : 0, where u' : dulclr. Solution. The auxiliary equation is m2 + m: O.It has the roots 0 and -1. This gives the general solution u(r) : c1-| c2lr. From the "boundary conditions" (the potentials on the spheres) we obtain u(5):r-,+?:ll0,-J By subtract1,on, c2l10: 110, c2: 1100. From the second equation, c1 : -c2l10 : -Il0. Answer: u(r) : - 1 10 + 1 100ir V. Figure 48 shows that the potential is not a straight line, as it would be for a potential between two parallel plates. For example, on the sphere of radius 7.5 cm it is not 1I0l2:55 V, but considerably less. (What is it?) l U 100 B0 60 4o 2o nl l 5t) Fig.48. 7 B 9 10 r Potential v(r) in Example 4 _0.5 -1.0 _1.5 @ GENERAL soluTloN Find a real general solution, showing the details of your work. 1. r'y" - 6, _ O 2. 4x2y" + 4xy' - y : 0 3. ,'y" - J*y' * 16y : 9 4. r'y" + 3xy' * y : 0 5. r'y" - ry' + 2y : g 6.2x2y"+4xy'+5y:6 7. (IOxZD2 - 20xD + 22,4I)y : Q 8. (4x2D2 * 1)y : 6 9. (I00x2D2 1- 9I1y : g I0. (I0x2D2 + 6xD + 0.51)y : Q @ lNlTlAL vALuE IRoBLEM Solve and graph the solution, showing the details of your work. 1,1,. x2y" - 4ry' + 6y: 0,y(1) : 1,}'(1) : 0 |2. xzy" + 3xy' + y: 0, y(I) : 4, y'(1) _ _2 13. (x2D2 -l ZxD + I00,25I)y : 0, y(I) : 2, y'(1) : - 11 14. (x2D2 - ZxD + 2.25I)y: 0, }(I):2,2, ),'(1) :2,5 15. (xD2 + 4D)y: 0,y(1) : 12,y'(1) : _6 -l Case I: Real roots 1 1.4 2 x r0,2 cos (4 ln r) y 1.5 1.0 0.5 x0,5 SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 16. TEAM PROJECT. Double Root (A) Derive a second linearly independent solution of (1) by reduction of order; but instead of using (9), Sec. 2.I, perform all steps directly for the present ODE (1). (B) Obtain x* Inxby considering the solutions x- and x**' of a suitable Euler-Cauchy equation and letting s-+0. 73 (C) Verify by substitution that x* ln x) m : (I - a) 12, is a solution in the critical case. (D) Transform the Euler-Cauchy equation (1) into an ODE with constant coefficients by setting x: et (x > 0). (E) Obtain a second linearly independent solution of the Euler-Cauchy equation in the "critical case" from that of a constant-coefficient ODE. wronskian (1) (2) (3) THEoREM 1 2.6 Existence and Uniqueness of Solutions. In this section we shall discuss the general theory of homogeneous linear ODEs y"+p(x)y'*q(x)y:0 with continuous, but otherwise arbitrary variable coefftcients p and q. This will concern the existence and form of a general solution of (1) as well as the uniqueness of the solution of initial value problems consisting of such an ODE and two initial conditions y(xo) : Ko, y'(xo) : Kt with given xg, Kg, and K1. The two main results will be Theorem 1, stating that such an initial value problem always has a solution which is unique, and Theorem 4, stating that a general solution !:cl,ltlczyz (cy cz arbitrary) have no " singularincludes all solutions. Hence linear ODEs with continuous coefficients solutions" (solutions not obtainable from a general solution). Clearly, no such theory was needed for constant-coefficient or Euler-Cauchy equations because everything resulted explicitly from our calculations. Central to our present discussion is the following theorem. Existence and Uniqueness Theorem for lnitial Value Problems Iíp(x) and q(x) are continuous functions on some open interval I (see Sec. 1 .I) and xg is in I, then the initial value problem consisting oí(I) and (2) has a unique solution y(x) on the interval L The proof of existence uses the same prerequisites as the existence proof in Sec. 1.7 and will not be presented here; it can be found in Ref. tA11] listed in App. 1. Uniqueness proofs are usually simpler than existence proofs. But for Theorem 1, even the uniqueness proof is long, and we give it as an additional proof in App. 4. 74 CHAP.2 Second-Order Linear ODEs Linear lndependence of Solutions Remember from Sec. 2.1 that a general solution on an open interval 1is made up from a basis jt jz on 1, that is, from a pair of linearly independent solutions on 1. Here we call jt, lz linearly independent on 1 if the equation (4) klt@) * kry2(x) : 0 on 1 implies kr : 0, 0. We call !t, lz linearly dependent on 1 if this equation also holds for constants k1, k2 not both 0. In this case, and only in this case, y1 and !2 ere proportional on I, that is (see Sec. 2.1), (a) lt : k}z or (b) lz : llt for all x on 1.(5) For our discussion the following criterion of linear independence and dependence of solutions will be helpful. Linear Dependence and Independence of Solutions Let the ODE (I) have continuows cofficients p(x) and q(x) on an open interval I. Then two solutions !1and y2 oí(I) on I are linearly dependent on I if and only if their 'oWronskian" is0 at some xginI. Furthermore, líW:0 at anx: xoinI, thenW = 0 onI; hence if there is an xl in I at which W is not 0, then !t, jz are linearly independent on I. (6) P R O O F (a) Let yl andlzbe linearly dependent on l Then (5a) or (5b) holds on I If (5a) holds, then W(yt, jz) : yůL- yzy't : lqzyl - yztcyl : 0. Similarly if (5b) holds. (b) Conversely, we let W(yr, !z) : 0 for some x : xo and show that this implies linear dependence of y1 anďy2 on 1. We consider the linear system of equations in the unknowns kr kz kůt@ *k2y2(xg):0 kryi@ +kry!@o):0. To eliminate k2, multiply the first equation by yl and the second by -y, and add the resulting equations. This gives kl{x yl(xo) - kl\@ yz(ro) : klW(y{xg), yz(xo)) : 0. Similarly, to eliminate kr, multiply the first equation by -yi and the second by y, and add the resulting equations. This gives k2W(yl@g), yr(xo)) : 0. (]) THEoREM 2 W(yl,, jz) : yll - yry', SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian If W were not 0 at xg, we could divide by W and conclude that kl, : kz: 0. Since W is 0, division is not possible, and the system has a solution for which k1 and k2 are not both 0. Using these numbers ky k2, we introduce the function y(x):k t@)+k2y2@). Since (1) is homogeneous linear, Fundamental Theorem 1 in Sec. 2.I (the superposition principle) implies that this function is a solution of (1) on 1. From (7) we see that it satisfies the initial conditions y(xo) : 0, y'(xo) : 0. Now another solution of (1) satisfying the same initial conditions is y* = 0. Since the coefficients p and q of (I) are continuous, Theorem 1 applies and gives uniqueness, that is, y = y*, written out k t*kzjz=0 on 1. Now since k1 and k2 are not both zero, this means linear dependence of y1, y, on I. (c) We prove the last statement of the theorem. If W(xo) : 0 at an Jo in 1, we have linear dependence of yr, }z oíl I by part (b), hence W = 0 by part (a) of this proof. Hence in the case of linear dependence it cannot happen that W(xr) * 0 at aíxr in 1. If it does happen, it thus implies linear independence as claimed. Remark. Determinants. Students familiar with second-order determinants may have noticed that 7s l W(y-,, yz) : jzl ltl t l : )'l )'z - )'z)'r . !zlIi This determinant is called the Wronski determinants or, briefly, the Wronskian, of two solutions y1 and lz of (1), as has already been mentioned in (6). Note that its four entries occupy the same positions as in the linear system (7). lllustration of Theorem 2 The functions h : cos (dJ and y2: sin crx are solutions of y " + ,2y : 0. Their Wronskian is EXAMPLE ll E X A M,P L,E ,,2 I cos rr.,x W(cos rr,rr, sin a,rx) : |-, sin ox sin c,-,x l | : YrYL - Yzy'l: ,.o,2 ax l rosin2 ax: ,,_l. 0.,cos@'l '-'' ílxel l : t, + l)e2* - *o2* : 12'+ O. (x + l)eJ| Theorem 2 shows that these solutions are linearly independent if and only if a * 0. Of course, we can see this directly from the quotient yzlyt : tan ax. For r,.l : 0 we have y2 = 0, which implies linear dependence (why?). l lllustration of Theorem 2 for a Double Root Ageneral solutionofy"-2y'+y:Oonanyintervalisy:(c1 -|c2x)e".(Verify!).Thecorresponding Wronskian is not 0, which shows linear independence of e* and xe* on any interval. Namely, W(x, xer) : r"-_ l 5lntroduced by WRONSKI (JOSEF MARIA HÓNE, 1'7'76_1853), polish mathematician. _-1 76 CHAP.2 Second-Order Linear ODEs A General Solution of (l) lncludes All Solutions This will be our second main result, as announced at the beginning. Let us start with existence. THEoREM 3 Existence of a General solution Iíp(x) and q(x) are continuous on an open interval I, then (I) has a general solution on I. p Ro o F By Theorem 1, the ODE (1) has a solution yr(x) on 1 satisfying the initial conditions yi(xo) : 0yr(xo) : 1, and a solution yz@) on 1 satisfying the initial conditions yz(xo) : 0, yL@ : I. The Wronskian of these two solutions has at x : x6 the value W(yr(O), yz(O)) : y{x yr(xo) - y2@ y\@o) : 1. Hence, by Theorem 2, these solutions are linearly independent on 1. They form a basis of solutions of (1) on 1, and ! : ctlt l cry, with arbitrzíy ct, cris a general solution of (1) on 1, whose existence we wanted to prove. We finally show that a general solution is as general as it can possibly be. THEoREM 4 A General solution lncludes All solutions If the ODE (I) has continuous cofficients p(x) and q(x) on some open interval I, then every solution y : Y(x) oí(I) on I is of the form (8) Y(x) : Clt@) + C2y2@) where !t, jz is any basis of solutions o/ (1) on I and Ct, Cz are suitable constants. Hence (l) does not have singular solutions (that is, solutions not obtainable from a general solution). pRooF Lety : Y(x) be any solution of (1) on 1. Now, by Theorem 3 the ODE (1) has a general solution y(x) : cqt@) -l c2y2@)(9) (10) on 1. We have to find suitable values of c1, c2 such that y(x) : Y(x) on 1. We choose any xg in 1and show first that we can find values of c1, c2 such that we reach agreement at xg, that is, y(xg) : (xo) and y'(xo) : Y'(xo). Written out in terms of (9), this becomes cgt(x *c2y2(xg): (xo) cl|@ + crylr@g) : Y'(x . (a) (b) SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian We determine the unknowns c1 and c2. To eliminate c2,,vla multiply (10a) by y!@) and (10b) by -yr(xo) and add the resulting equations. This gives an equation for c1. Then we multiply (10a) by -yi("o) and (10b) by yr("o) and add the resulting equations. This gives an equation for cr. These new equations are as follows, where we take the values of }r, y'r, yr, y;, Y, Y| at xg, ctOll - yzyb : ctW(!t, !z) : Yy! - yrY' czjll - yzyb : czW(!l, jz) : yrY' - Yy],. Since jy jz is a basis, the Wronskian W in these equations is not 0, and we can solve for clarLd c2. We call the (unique) solution c1 : Cy cz: Cz. By substituting it into (9) we obtain from (9) the particular solution y*(x):Cůt(x)+C2y2@). Now since C1, C2is a solution of (10), we see from (10) that y*(xo) : Y(xo), y*'(xo) : Y'(xo). From the uniqueness stated in Theorem 1 this implies that y* and Y must be equal everywhere on I, and the proof is complete. l Looking back at he content of this section, we see that homogeneous linear ODEs with continuous variable coefficients have a conceptually and structurally rather transparent existence and uniqueness theory of solutions. Important in itself, this theory will also provide the foundation of an investigation of nonhomogeneous linear ODEs, whose theory and engineering applications we shall study in the remaining four sections of this chapter. E BAsEs oF soluTloNs. CORRESPOND|NG ODEs. WRONSKIANS Find an ODE (1) for which the given functions are solutions. Show linear independence (a) by considering quotients, (b) bv Theorem 2. 1. "o,'*, ,-o,5x 2. cos rrx, Sin rrx 3. ek*, xek* 4. x3, x-2 5. *o,", xo,25 In x 6. "",n*, e-2,5r 7. cos (2In x), sin (2 ln x) 8. ,-r*, xe-2* 9. *r-r, x-o.5 1,0. x-3, x-3 ln x 11. cosh 2.5x, sinhZ.Sx 12. e-2* cos (r-tr, e-2* Sin alx 13. e-' cos 0.8x, e-' sin 0.8x 14. x- 1 cos (ln x), x- 1 sin (ln x) 15. ,-2,5r cos 0.3x, "-2,5:t sin 0.3x 16. e-k' cos Ťrx, e-k* sin rrx 17. u-3.8rt, ag-S.8nr 18. TEAM PROJECT. Consequences of the Present Theory. This concerns some noteworthy general properties of solutions. Assume that the coefficients p and q of the ODE (1) are continuous on some open interval 1, to which the subsequent statements refer. (A) Solve y" - y : 0 (a) by exponential functions, (b) by hyperbolic functions. How are the constants in the corresponding general solutions related? (B) Prove that the solutions of a basis cannot be 0 at the same point. (C) Prove that the solutions of a basis cannot have a maximum or minimum at the same point. (D) Express (yzly)' by a formula involving the Wronskian W. Why is it likely that such a formula should exist? Use it to find I4z in Prob. 10. (E) Sketch yt(x) : x3 If x > 0 and 0 if x { 0, yz(x) : 0 if x > 0 and x3 If x < 0. Show linear Wronskian? What Euler-Cauchy equation do !t, lz satisfy? Is there a contradiction to Theorem 2? 77 78 CHAP.2 Second-Order Linear ODEs (F) Prove Abel's formulaG W(yr(x), yz(x)) : c exp |- ť,rr' "1 where c : W(7-{xo),,yz(.ro)). Apply it to Prob. 12. Hint: Write (1) for y1 and íbry2. Eliminate q a\gebraically from these two ODEs, obtaining a first-order linear ODE. Solve it. Method of undetermined coefficients In this section we proceed from homogeneous to nonhomogeneous linear ODEs y" + p(,x)y' + q(x)y : r(x) (2) DEFlNlTloN THEoREM 1 2.7 Nonhomoteneous ODEs (1) where r(x) * 0. We shall see that a "general solution" of (1) is the sum of a general solution of the conesponding homogeneous ODE y"+p(x)y'+q(x)y -t-, terms "general solution of ( 1)" andand a "particular solution" of (1). These two new "particular solution of (1)" are defined as follows. General Solution, Particular Solution A general solution of the nonhomogeneous ODE (1) on an open interval 1 is a solution of the form (3) _v("r) : yn@) + yo("); here, yn : ctlt l czyz is a general solution of the homogeneous ODE (2) on I and }p is any solution of (1) on 1 containing no arbitrary constants. A particular solution of (1) on 1 is a solution obtained from (3) by assigning specific values to the arbitrary constants c1 and c2 in y6. Our task is now twofold, first to justify these definitions and then to develop a method for finding a solution yu of (1). Accordingly, we first show that a general solution as just defined satisfies (1) and that the solutions of (1) and (2) are related in a very simple way. Relations of Solutions of (1) to Those of (2} (a) The sum of a solution y oí(I) on some open interval I and a solution of (2) on I is a solution oí (I) on I. In particular, (3) 's a solution of (I) on I. (b) The dffirence of two solutions of (I) on I is a solution of (2) on I. GNIELS HENRIK ABEL (18O2-1 829), Norwegian mathematician. SEC.2.7 Nonhomoseneous ODEs P R O O F (a) Let L[y] denote the 1eft side of (1). Then for any solutions y of (1) and of (2) on I, L[y + l : Llyl + Ll ] -- r l 0 : r. (b) For any solutions y and y* of (1) on 1we have Lly - y*l : LIyl Now for homogeneous ODEs (2) we know that general solutions We show that the same is true for nonhomogeneous ODEs (1). -Líy*]-r-r:0. l include all solutions. THEoREM z A General Solution of a Nonhomoteneous ODE lncludes All Solutions If the cofficients p(x), q(x), and the function r(x) in (I) are continuous on some open interval I, then every solution oí(I) on I is obtained by assigning suitable values to the arbitrary constants cl and c2 in a general solution (3) oí (I) on I. P R O O F Let y* be any solution of (1) on 1 and xg an! x in I. Let (3) be any general solution of (1) on 1. This solution exists. Indeed, lh : ctlt l c2y2 exists by Theorem 3 in Sec. 2.6 because of the continuity assumption, and )p exists according to a construction to be shown in Sec. 2.I0. Now, by Theorem 1(b) just proved, the difference Y : !* - }p is a solution of (2) on 1. At x6 we have Y(x : y*(xo) - !o(xo), Y'(x : y*'(ro) - yL@ . Theorem 1 in Sec. 2.6 impltes that for these conditions, as for any other initial conditions in 1, there exists a unique particular solution of (2) obtained by assigning suitable values to cr, c2inyn. From this and y* : Y * lp the statement follows. l Method of undetermined coefficients Our discussion suggests the following . To solve the nonhomo7eneous ODE (I) or an initial value problem for (I), we have to solve the homogeneous ODE (2) and find any solution lp oí (1), so that we obtain a general solution (3) of (1). How can we find a solution lo of (1)? One method is the so-called method of undetermined coefficients. It is much simpler than another, more general method (to be discussed in Sec. 2.I0). Since it applies to models of vibrational systems and electric circuits to be shown in the next two sections, it is frequently used in engineering. More precisely, the method of undetermined coefficients is suitable for linear ODEs with constant coefftcients a and b y" + ay' + by: r(x) when r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of such functions. These functions have derivatives similar to r(x) itself. This gives the idea. We choose a form for yo similar to r(x), but with unknown coefficients to be determined by substituting that yo and its derivatives into the ODE. Table 2.I on p. 80 shows the choice of }p for practically important forms of r(x). Corresponding rules are as follows. (4) 79 80 CHAP.2 Second-Order Linear ODEs choice Rules for the Method of undetermined coefficients (a) Basic Rule. Iír(x) in (4) is one of the fwnctions in the first column in Table 2.I, choose yo in the same line and determine its undetermined cofficients by substituting lp and its derivatives into (4). (b) Modification Rule. If a term in your choice íor yo happens to be a solution of the homogeneous ODE corresponding to (4), multiply your choice oí yp by r (.or by ,' if this solution corresponds to a double root of the characteristic eqwation of the homogeneous ODE). (c) Sum Rule. If r(x) is a sum of functions in the first column of Table 2.I, choose íor yp the sum of the functions in the corresponding lines of the second column. The Basic Rule applies when r(x) is a single term. The Modification Rule helps indicated case, and to recognize such a case, we have to solve the homogeneous first. The Sum Rule follows by noting that the sum of two solutions of (1) with r and r : 12 (and the same left side!) is a solution of (1) with T : TI l rz. erify!) The method is self-correcting. A false choice for yo or one with too few terms will lead to a contradiction. A choice with too many terms will give a correct result, with superfluous coefficients coming out zero. Let us illustrate Rules (a)-(c) by the typical Examples Table 2.1 Method of Undetermined Coefficients Application of the Basic Rule (a} Solve the initial value problem y" + y: 0.001x2 .v(0) ),'(0) : Solution. Sfup 1. General solution of the homogeneous ODE. The ODE y" + y: 0 has the general solution yn : A cos,t + -B sin r. Step 2. Solution y, of the nonhomogeneous ODE We flrst try ),, : Kxz. Then ),'; : 2K.By substitution, 2K + Kxz : 0.00l;2. For this to hold for all x, the coefficient of each power of r (x2 and ,ro) must be the same on both sides; thus K : 0.001 and2K: 0, 2 contradiction. The second line in Table 2.I suggests the choice !p: K2x2 -l Kě * Kg. Then yi + yo: 2Kz + K2x2 f Kl-r i 0.00 |;2. Equating the coefficients of *2, *, *o on both sides, we have K2: 0,00l, Kl : 0, 2K2 + Ko : 0. Hence Ko: -2Kz: -0.002. This gives ),, : 0.00l;r2 - 0.002, and ,),: )h *.,,o - Á cos_r * B sinr i 0.0O1;r2 - 0.002. in the oDE :TI ExAMPLE l (5) Term in r(x) Choice for yo(x) keY* kx" (n : 0, fr cos a;x k sin ax keo' cos ax keo' sin ax CeY* Knxn + Kn_lť-1 + . . . l Klx t Ko l lKcos ulx * Msínax ) }."(K cos rax -l M sin ax) ] SEC.2J Nonhomo8eneous ODEs ExAMPLE 2 8l Step 3. Solution of the initial value problen. Setting x : 0 and using the first initial condition gives y(0) : A - 0.002 : 0, hence A : 0.002. By differentiation and from the second initial condition, ) : }n + y; : -Á sin x -| B cos x -t- 0,002x This gives the answer (Fig. a9) and .y'(O):B:1.5. y : 0.002 cosx * 1.5 sinx i 0.001"12 - 0.002. Figure 49 shows y as well as the quadratic parabola }p about which y is oscillating, practically like a sine curve since the cosine term is smaller by a factor of about 1/1000. l Application of the Modification Rule (b) Solve the initial value problem (6) y" + 3y' + 2.25y: -l0 "-t,Sr, y(0): 1, )'(0):0. Solution. Sfup 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous ODE is i2 + 3^ + 2.25: (^ + 1.5)2 : O. Hence the homogeneous ODE has the general solution yh: (c7 + c2x)e-|,5*, Step 2. Solutionypoíthe nonhomogeneous ODE The function r*Í,5r on the right would normally require the choice g"-l'5r. But we see from y1 that this function is a solution of the homogeneous ODE, which corresponds to a double root of the characteristic equation. Hence, according to the Modification Rule we have to multiply our choice function by *'.That is, we choose ^ 2 -I.5.r -\'p : L,\, e Then yL -- c(zx - 1.5X2)e-|,5*, :,i _ clz 3x 3x i 2.25x2)g-Lsa. We substitute these expressions into the given ODE and omit the factor u-1,5r. This yields C(2 - 6x -f 2.25x2) + 3C(2x - 1.5x2) + 2.25Cx2: -10. Comparing the coefficients of x2, r, xo gives 0 : 0,0 : 0, 2C : _10, hence C : -5. This gives the solution lp : -5x2e-1,5'. H"n." the given ODE has the general solution ! : ln + yp (cI -| c2x)e-1.5r - 5x2e-1.5I . Step 3. Solution of the initial value problem. Setting x : 0 in y and using the first initial condition, we obtain y(0) : c1 : 1. Differentiation of y gives y' :(cz- 1.5c1 - 1.5c2x)e-!,5r * IOxe-Lí'+J.5x2e-1,5'. From this and the second initial condition we have y'(0) - c2 1.5c1 : 0. Hence cz: 1.5.1 : 1.5. This gives the answer (Fig. 50) y: (1 + 1.5r) e 1,5I - 5*2"-|,5r: (1 + 1.5r - Sxzlg-Ls*. Thecurvebeginswithahorizontaltangent,crossesthex-axis atx:0.6211 (where l * 1.5x - 5x2:0)and approaches the axis from below as x increases. l y 2 1 0 -i Fig.49. Solution in Example 1 82 CHAP.2 Second-Order Linear ODEs y 1.0 0.5 0 _0.5 _1,0 Fig. 50. Solution in Example 2 Application of the Sum Rule (c) Solve the initial value problem (7) y" + 2y' + 5y : eo,5t + 40 cos 10x - 190 sin 10x, ,y(0) - 0.16, .l"(0) - 40.08. Solution. Step l. General solution of the homogeneous ODE. The characteristic equation ^2 + 2^ + 5 : (^ + 1 + 2i)(^ + l - 2i):0 shows that a real general solution of the homogeneous ODE is lh: e-Í (Á cos 2x * B sin 2x), Step 2. Solution of the nonhomogeneous ODE We write _1rp : }p1 * :*pz,where yo1 corresponds to the exponential term and )pz to the sum of the other two terms. We set jp| : gro,5x. Then yLr: 0.5ceo'5' and y'i, - 0.25ceo,5*. Substitution into the given oDE and omission of the exponential Í'actor gives (0.25 + 2'0.5 + 5)C : l, hence C: 116.25: 0.16, andyr1 :0.16eo'5'. We now setyo2 : Kcos lOx -l M sin lOx, as in Table 2,1,and obtain yLr: - 10K sin 10x + I\M cos 10"r, ),'Jr: -100Kcos 10x - |00M sin lOx. Substitution into the given ODE gives for the cosine terms and for the sine terms -100K + 2,10M + 5K: 40, or, by simplification, -IjoM - 2.I)K + 5M: -190 -95K + 20M :40, -2oK - 95M: -l90. The solution is K : 0, M : 2. Hence !p2: 2 sin 10r. Together, |: ln* ypr * !p2: e-'(Acos2x i Bsin 2x) + 0.16eo,5'* 2 sin 10x. Step 3. Solution of the initial value problem. From y and the first initial condition, y(0) : Á + 0.16 : 0.16, hence Á : 0. Differentiation gives y' : e-*(-A cos2r- B sin 2x - 2Asin 2,r * 2'B cos 2i) + 0-08eo,5r * 20 cos lOx. Fromthisandthesecondinitialconditionwehavey'(0):-A+28+0.08+20:40.08,henceB:10. This gives the solution (Fig. 51) ! : l\e-Í sin 2x -| 0.I6eo,5* Ť 2 sin lOx. The ílrstterm goes to O relatively fast. When x : 4,it is practically 0, as the dashed curves + I}e-' + 0.16eo'5' show. From then on, the last term,2 sin 10,r, gives an oscillation about 0,16eo'5', the monotone increasing dashed curve. - l ExAMPLE 3 SEC.2.7 Nonhomogeneous ODEs 83 y i0 8 6 4 2 678r Fig. 51. Solution in Example 3 Stability. The following is important. If (and only if) all the roots of the characteristic equation of the homogeneous ODE y" + oy' * by :0 in (4) are negative, or have a negative real part, then a general solution ln of this ODE goes to 0 as x @, so that the 6'transient solution" ! : ln * y, of (4) approaches the 66steady-state solutig1|') !n.In this case the nonhomogeneous ODE and the physical or other system modeled by the ODE are called stable; otherwise they are called unstable. For instance, the ODE in Example 1 is unstable. Basic applications follow in the next two sections. @ GENERAL soluTloNs oF NONHOMOGENEOUS ODES Find a (real) general solution. Which rule are you using? (Show each step of your calculation.) l. y" * 3y' * 2y : 30e2* 2. y" -l 4y' + 3.75y: 109 cos 5x 3. y" - I6y : I9.2ea" * 60e" 4.y",|9y: cosxf jcos3x 5. y" l y' - 6y : 6r' _ 3x2 * I2x 6. y" -l 4y' -l 4y : e-2* sinzx 7. y" * 6y' + 73y : 8Oe* cos 4x 8. y" + 10y' + 25y: 100 sinh 5x 9. y" - 0.1 6y : 32 cosh 0.4x I0. y" -l 4y' + 6,25y : 3.I25(x + I)2 11. y" + I.44y : 24 cos L2x 12. y" l 9y : 18x * 36 sin 3x 13. y" ,f 4y' * 5y : 25x2 + 13 sin2x 14. y" * 2y' * y : 2x sinx E5J0l lNlTlAL vALuE pRoBLEMs FoR NONHOMOGENEOUS ODES Solve the initial value problem. State which rules you are using. Show each step of your calculation in detail. 15, y" l 4y : 16 cos2x, y(0) : 0, y'(0) : 0 16. y" - 3y' + 2.25y : 2J(x2 - x), y(0) : 20, y'(0) : 30 17. y" + 0.2y' + 0.26y : I.22eo,5', y(0) : 3.5, y'(0) : 0,35 18. y" - 2y' : I2e2, - 8u-'*, y(O) : -2, y'(0) : 12 19. y" - y' - I2y : 144x3 + I2.5, y(0) : 5, y'(0) : _0.5 20. y" + 2y' + 10y : IJ sinx - 3J sin3x, y(0) : 6.6, y'(0) : _2.2 21. WRITING PROJECT.Initial Value Problem. Write out all the details of Example 3 in your own words. Discuss Fig. 51 in more detail. Why is it that some of the "half-waves" do not reach the dashed curves, whereas others preceding them (and, of course, all later ones) excede the dashed curves? 22. TF,A]NII PROJECT. Extensions of the Method of Undetermined Coefficients. (a) Extend the method to products of the function in Table 2.1. (b) Extend the method to Euler-Cauchy equations. Comment on the practical significance of such extensions. 23. CAS PROJECT. Structure of Solutions of Initial Value Problems. Using the present method, find, graph, and discuss the solutions y of initial value problems of your own choice. Explore effects on solutions caused by CHAP. 2 Second-Order Linear ODEs changes of initial conditions. Graph !p, !, j - lp separately, to see the separate effects. Find a problem in which (a) the part of y resulting from y7, decreases to zero, (b) increases, (c) is not present in the answer y. Study a problem with y(0) : 0, y'(0) : 0. Consider a problem in which you need the Modification Rule (a) for a simple root, (b) for a double root. Make sure that your problems cover all three Cases I, II, III (see Sec. 2.2). motions of a mass-spring system (vibration of a mass 32 and 52) and modeled it by the homogeneous linear (1) my" + cy' + ky -- 0. Here y(r) as a function of time / is the displacement of the body of mass ru from rest. These were free motions, that is, motions in the absence of external forces (outside forces) caused solely by internal forces, forces within the system. These are the force of inertia my" , the damping force cy' 1tf c } 0), and the spring force lcy acting as a restoring force. We now extend our model by including an external force, call it r(t), on the right. Then we have (2*) my" + cy' + tcy : r(t). Mechanically this means that at each instant / the resultant of the internal forces is in equilibrium with r(r). The resulting motion is called a forced motion with forcing function r(r), which is also known as input or driving force, and the solution y(r) to be obtained is called the output or the response of the system to the driving force. Of special interest are periodic external forces, and we shall consider a driving force of the form r(t) : Fo cos a.r/ Then we have the nonhomogeneous ODE (Fo > 0, ro ) 0). my" + cy' + lry : Fo cos d/. Its solution will famlllartze us with further interesting facts fundamental in engineering mathematics, in particular with resonance. (2) 2.8 Modelin8: Forced Oscillations. Resonance In Sec. 2.4 we considered vertical m on an elastic spring, as in Figs. oDE Fig. 52. Mass on a spring -l 84 SEC. 2.8 Modeling: Forced Oscillations. Resonance 85 Solving the Nonhomogeneous ODE (2) From Sec. 2.] we know that a general solution of (2) is the sum of a general solution y7, of the homogeneous ODE (1) plus any soluti6n )n of (2). To find )p, we use the method of undetermined coefficients (Sec. 2.7), starting from (3) }p(/) : a cos at * b sin rr;/. By difterentiating this function (chain rule!) we obtain yL: -aa sin cor -f ab cos at, ),';: -o2r.L cos r,l/ - a}b sin cor. SubstitutinE )p, !L, and yi into (2) and collecting the cosine and the sine terms, we get [rr - m@2)u ,ť acbf cos r,,l/ + [- roca * (k - ma2)b] sin 0)t : F6 cos rrrl. The cosine terms on both sides must be equal, and the coefficient of the sine term on the left must be zero since there is no sine term on the right. This gives the two equations (4) 'r -_:::'": *::,,), -: fbr determining the unknown coefficients a and b. This is a linear system. We can solve it by elimination. To eliminate b, multiply the first equation by k - ma} and the second by - ac and add the results, obtaining (k - mu )2ct * a2c2a : Fg(k - maz). Similarly, to eliminate a, multiply the first equation by olc and the second by k - mo and add to get a2c2b + (k - ma2lzb : Fgcoc. If the factor (k - ma2')2 + tl2c2 is not zero, we can divide by this factor and solve for a and b, r k-tnu. (Í r"(k-,,ta2 12+u<,2 If we set \,[iň : a)o (> 0) as in Sec, (5) o:Fn _ tlt(t,lo2-*t,l2\ u tn2(ulo2 - u2|2 + a2r,2 l-_E'@C ' U 'u (k - m@2)2 + a2c2 2.4, then k : m@g2 and we obtain l,:I} @C ' U' u,n2(0)n2 - a2\2 + ul2c2 We thus obtain the general solution of the nonhomogeneous ODE (2) in the form y(/):yn(t)+y.p(t).(6) 86 CHAP.2 Second-Order Linear ODEs Here ynis a general solution of the homogeneous ODE (1) and )o is given by (3) with coefficients (5). We shall now discuss the behavior of the mechanical system, distinguishing between the two cases c : O (no damping) and c } 0 (damping). These cases will correspond to two basically different types of output. Case 1. Undamped Forced Oscillations. Resonance If the damping of the physical system is so small that its effect can be neglected over the time interval considered, we can set c : 0. Then (5) reduces to a : Fgl|m(ao' - ,\] and b : 0. Hence (3) becomes (use a42 -- klm; Here we must assume that co2 + @o2; physically, the frequency al(2rr) [cycles/sec] of the driving force is different from the natural frequency agl(2nr) of the system, whiCh is the frequency of the free undamped motion [see (4) in Sec. 2.4]. From (7) and from (4*) in Sec.2.4 we have the general solution of the "undamped system" (1) !o(t) : *:\-.T cos t,lí:' m(oo' - t,l') (9) oo:?o where Fo cos (,/. Fo (8) !ft): Ccos(ooí- 6)+ , , - cosoí.v ' m(@g'-uf) kll - (al@ 2] 1 U- - ó ' | - (alag)' We see that this output is a superposition of two harmonic oscillations of the frequencies .just mentioned. Resonance. We discuss (7). We see that the maximum amplitude of le is (put cos c,l/ : 1) ag depends on ct,l and rr.ro. If a ---> a.rg, then p and ag tend to infinity. This excitation of large oscillations by matching input and natural frequencies Qo : ao) is called resonance. p is called the resonance factor (Fig. 53), and from (9) we see that plk: aglFg is the ratio of the amplitudes of the particular solution lo and of the input F6 cos rrll. We shall see later in this section that resonance is of basic importance in the study of vibrating systems. In the case of resonance the nonhomogeneous ODE (2) becomes ,,qFO Y' * ao'y : cos @6í. m Then (7) is no longer valid, and from the Modification Rule in Sec.2.] we conclude that a particular solution of (10) is of the form !oQ) : /(a cos agt l b sin agt). (10) SEc. 2.8 : Forced OsciIlations. Resonance Fig. 53. Resonance factor p(a.r) By substituting this into (10) we finď a : 0 and b : Fgl(Z*ro). Hence (Fig. 5a) 87 (11) (I2) We see that because of the factor / the amplitude of the vibration becomes larger and larger. Practically speaking, systems with very little damping may undergo large vibrations that can destroy the system. We shall return to this practical aspect of resonance later in this section. Fig. 54. Particular solution in the case of resonance Beats. Another interesting and highly important type of oscillation is obtained if r,l is close to arg. Take, for example, the particular solution [see (8)] y-p Fo y(t) : -,--r: \ (cos rr;l - cos (t)oí) ' m(ag' - a') (a * ruog). Using (I2) tn App. 3.1, we may write this as ),(í): ^#:6*"(r+,) ,i" ('+,) Since r_o is close to og, the differeflc rrlg - ro is small. Hence the period of the last sine function is large, and we obtain an oscillation of the type shown in Fig. 55, the dashed curve resulting from the first sine factor. This is what musicians are listening to when they tune their instruments. Fo )o(í): ^ rSln d0/. ' Znl@g 88 CHAP.2 Second-Order Linear ODEs Fig. 55. Forced undamped oscillation when the difference or tn" input and natural frequencies is small ("beats") THEoREM l Case 2. Damped Forced Oscillations (13) If the damping of the mass-spring system is not negtigibly small, we have c ) 0 and a damping te.m .y' in (1) and (2). Then the general solution )n of the homogeneous ODE (1) approaches zero as / goes to infinity, as we know from Sec,2.4. PracticallY, it is zero after a sufficiently long time. Hence the "transient solution" (6) of (2), given by ! : jn * !p, approaches the "steady-state solution" }p. This proves the following. Steady-State Solution After a sfficiently long time the output of a damped vibrating system under a pwrely sinusoidal driving force Lsee (2)] will practically be a harmonic oscillation whose frequency is that of the inputAmplitude of the Steady_State Solution. Practical Resonance Whereas in the undamped case the amplitude of }p approaches infinitY aS r',l aPProaches cr.lg, this wi1l not happenin the damped case. In this case the amplitude will alwaYs be finite. But it may have a maximum for some ro depending on the damPing constant c. This maY be called practical resonance. It is of great importance because if c is not too large, then some input may excite oscillations large enough to damage or even destroY the sYstem. Such cases happened, in particular in earlier times when less was known about resonance. Machines, cars, ships, airplanes, bridges, and high-rising buildings are vibrating mechanical systems, and it is sometimes rather difficult to find constructions that are comPletelY free of undesired resonance effects, caused, for instance, by an engine or bY strong winds. To study the amplitude of _yp as a function of ro, we write (3) in the form !oG) : C* cos (at - D. c* is called the amplitude of .}o and 4 the phase angle or phase lag because it measures the lag of the output behind the input. According to (5), these quantities are C*(ai) : \F+ bŽ : (14) tan rlQ,l) mk)o2 - @2) F,o :L:0 SEC. 2.8 Modeling: Forced Oscillations. Resonance 89 Let us see whether C*(c,l) has a maximum and, if so, find its location and then its size. We denote the radicand in the second root in C* by R. Equating the derivative of C* to zero, we obtain dC* l l \- d, : Fo (- Ž o-''' )|2n2t uo2 - ,,2:{-2uls + 2oc2f. The expression in the brackets [. . .] is zero if (15) (15*) c2:2m2(@o2-@2) (@o2 : klm). By reshuffling terms we have 2m2a2 :2m2oo2 - c2 :2mk - c2. The right side of this equation becomes negativeif c2 ž2mk, so that then (l5) has no real solution and C* decreases monotone as r-o increases, as the lowest curve in Fig. 56 on p. 90 shows. If c is smaller, c2 < zmk, then (15) has areal solution @: @^u*, where 22c2@max-{Xg - ^ .r 2m' From (15*) we see that this solution increases as c decreases and approaches 0)g as c approaches zero. See also Fig. 56. The size of C*(al-.*) is obtained from (14), with a2 : @'^u* given by (15*). For this a2 we obtain in the second radicand in (14) from (15*) 4 tn2{roz - .'^ur)' : ;' ,+ma2^u*c2: (ro' -and ,*L)* The sum of the right sides of these two formulas is (rn + 4m2o42c2 - Zc4)t(+m2) : c2l4m2ao2 - c21l14m21_ Substitution into (14) gives (16) C*(c,r-"r) : We see that C*(a*.*) is always finite when c ) 0. Furthermore, since the expression c24m2ao2 - c4 : czl4mk - c2) in the denominator of (16) decreases monotone to zero as c2 (< 2mk) goes to zero, the maximum amplitude (16) increases monotone to infinity, in agreement with our result in Case 1. Figure 56 shows the amplification C*lFo (ratio of the amplitudes of output and input) as a function of c,.r for ffi : I, k : l, hence ao: 1, and various values of the damping constant c. 2mF6 ,\m%'- ,' 90 CHAP. 2 Second-Order Linear ODEs Figure 57 shows the phase angle (the lag of the output behind the input), which is less than rrl2 when al 1 @g, and greater than rrl2 for a žag. 2 (l) C:l: % 4 Fig. 56. Amplificat ion C*/Foas a function of afor ffi : 1, k :1, and various values of the damping constant c Fig. 57. Phase lag 4 as a function of a f or ffi:1, k : ], thus c,.lo : 1, and various values of the damping constant c 0 E sTEADv-sTATE soluTloNs Find the steady-state oscillation of the mass-spring system modeled by the given ODE. Show the details of your calculations. I. y" -f 6y' * 8y : 130 cos 3t 2. 4y" * 8y' + I3y: 8 sin 1.5t 3. y" + y' + 4.25y : 221 cos 4.5r 4. y" * 4y' i 5y : cos/ - sin/ 5. (D' + 2D + I)y : -sin2t 6. (D' + 4D + 3I)y : cos t + +cos3r 7. (D' + 6D + 181)y : cos 3t - 3 sin 3r 8. (D' + ZD + 101)y : -25 sin 4r E TRANSIENT soluTloNs Find the transient motion of the mass-spring system modeled by the given ODE. (Show the details of your work.) 9. y" -l 2y' + 0.75y : 13 sin r I0. y" -l 4y' l 4y : cos 4t 11. 4y" + I2y' + 9y : J5 sin3t 12. (D2 + 5D + 4I)y : sin}t 13. (D2 + 3D + 3.25I)y : 13 - 39 cos}t 1,4. (D2 + 2D + 51)y : 1 * sin r @ INITIAL vALuE pRoBLEMs Find the motion of the mass-spring system modeled by the ODE and initial conditions. Sketch or graph the solution curve. In addition, sketch or graph the curve of ! - lp to see when the system practically reaches the steady state. L5. y" -l 2y' + 26y: 13 cos 3í, y(0) : 1, y'(0) : 0,4 16. y" + 64y: cos /, }(0) : 0, y'(0) : 1 17. y" ,l 6y' + 8y : 4 sin1t, y(0) : 0.7, y'(0) : - 11.8 18. (D2 + 2D + I)y : 75(sin t - }sín1t * } sin3r), y(0) : 0, y'(0) : 1 19. GD2 + I2D + I3I)y : 12 cos / - 6 sin /, y(0) : 1, y'(0) - -1 20. y" + 25y:99 cos4.9t, y(0):2, y,(0):0 21. (Beats) Derive the formula after (12) from (I2). Can there be beats if the system has damping? 22. (Beats) How does the graph of the solution in Prob. 20 change ifyou change (a) y(0), (b) the frequency ofthe driving force? 23. WRITING PROJECT. Free and Forced Vibrations. Write a condensed report of 2-3 pages on the most important facts about free and forced vibrations. 24. CAS EXPERIMENT. Undamped Vibrations. (a) Solve the initial value problem y" + y: cos (r/, a2 + I, y(0) : 0, y'(0) : 0. Show that the solution can be written [i ,, * ,l,] * - ,)í] 2 y(t): , ,|-uItsin l - 11 L2 (11) SEC. 2.9 Modeling: Electric Circuits (b) Experiment with (17) by changing rr; to see the change of the curves from those for small a,l (> 0) to beats, to resonance and to large values of ro (see Fig. 58). a= 0.2 cO = 0.9 rrl=6 Fig. 58. Typical solution curves in CAS Experiment 24 9t 25. TEAM PROJECT. Practical Resonance. (a) Give a detailed derivation of the crucial formula (16). (b) By considertng dC*ldc show that C*(.,;*.*) increases as c (< ÝZ*t decreases. (c) Illustrate practical resonance with an ODE of your own in which you vary c) and sketch or graph corresponding curves as in Fig. 56. (d) Take your ODE with c fixed and an input of two terms, one with frequency close to the practical resonance frequency and the other not. Discuss and sketch or graph the output. (e) Give other applications (not in the book) in which resonance is important. 26. (Gun barrel) Solve ll , |1 - lŤ2 ifo=tarr y -1_ y:1 [ 0 iít>tr, y(0):y'(0):0, This models an undamped system on which a force F acts during some interval of time (see Fig, 59), for instance, the force on a gun barrel when a shell is fired, the barrel being braked by heavy springs (and then damped by a dashpot, which we disregard for simplicity). Hint. At rrboth y and y' must be continuous. Fig.59. Problem 26 2.9 Modelint: Electric Circuits Designing good models is a task the computer cannot do. Hence setting up models has become an important task in modern applied mathematics. The best way to gain experience is to consider models from various fields. Accordingly, modeling electric circuits to be discussed will be profitable for all students, not just for electrical engineers and computer scientists. We have just seen that linear ODEs have important applications in mechanics (see also Sec. 2.4). Similarly, they are models of electric circuits, as they occur as portions of large networks in computers and elsewhere. The circuits we shall consider here are basic building blocks of such networks. They contain three kinds of components, namely, resistors, inductors, and capacitors. Figure 60 on p. 92 shows such an RLC-circuit, as they are called. In it a resistor of resistance R C) (ohms), an inductor of inductance L H (henrys), and a capacitor of capacitance C F (farads) are wired in series as shown, and connected to an electromotive force E(t) Y (volts) (a generator, for instance), sinusoidal as in Fig. 60, or of some other kind. R, L, C, anď E are given and we want to find the current I(t) A (amperes) in the circuit. CHAP.2 Second-Order Linear ODEs "il]"E(t) = Eosinat Fig. 60. RLC-circuit An oDE for the current 1(r) in the RLC-circuit in Fig. 60 is obtained from the following law (which is the analog of Newton's second law, as we shall see later). Kirchhoffrs Voltage Law (KVL).? The voltage (the electromotive force) impressed on a closed loop is equal to the sum of the voltage drops across the other elements of the loop. In Fig. 60 the circuit is a closed loop, and the impressed voltage E(r) equals the sum of the voltage drops across the three elements R, L, C of the loop. Voltage Drops. Experiments show that a cunent 1flowing through a resistor, inductor or capacitor causes a voltage drop (voltage difference, measured in volts) at the two ends; these drops are (Ohm's law) dI :LeVoltagedrop RI LI' O C Voltage drop Voltage drop for a resistor of resistance R ohms (O), for an inductor of inductance L henrys (H), for a capacitor of capacitance C farads (F). current byHere Q coulombs is the do I(t): . . dí charge on the capacitor, equivalently, Q(t) related to the : fry> at, This is summarized in Fig. 61. According to KVL we thus have in Fig. 60 for an RLC-circllit with electromotive force E(t) : Eg sin at (Eg constant) as a model the o'integro-differential equation" (1') LI' + R1 + Ž I, dt : E(t) : Eosin cr.l/. ?Gustev R9BERT KIRCHHOFF (1824-1887), German physicist. Later we shall also need Kirchhoff's current taw (KCL): At any point of a circuit, íhe sum of the inflowing cwrrents is equal to the swn of the outflowing currents. The units of measurement of electrical quantities are named after ANDRB I\/IezuP AMPĚRE (1775-1836), French physicist, CHARLES AUGUSTIN DE COULOMB (1736-1806), French physicist and engineer, MICHAEL FARADAv (|'t91_1861), English physicist, JOSEPH HENRY (1'797-1878), American physicist, GE9RG SIMQN oHM (1789-1854), German physicist, and ALESSANDRO VOLTA (1'745-1827), Italian physicist. 92 SEC. 2.9 Modeling: Electric Circuits To get rid of the integral, we differentiate (1') with respect to t, obtaining 93 (1) LI" + RI'+Lr:C E'lts : Ena cos at. This shows that the current in an RLc-circuit is obtained as the solution of this nonhomogeneous second-order ODE (1) with constant coefficients. From (l'), using I: Q', hence I' : Q", we also have directly ( 1") Lg" + Rg' Eg sin at, But in most practical problems the current (r) is more important than the charge Q(t), and for this reason we shall concentrate on (1) rather than on (1"). Solving the ODE (1) for the Current. Discussion of Solution A general solution of (1) is the sum ,í: ln * lo, where In ts a general solution of the homogeneous ODE coíTesponding to (1) and 1, is a particular solution of (1). We first determine Io by the method of undetermined coefficients, proceeding as in the previous section. we substitute (2) Ip:acosol*bstnat tl lp : @(-a sín at -l b cos iuot) I'i : ,'(-a cos at - b sin ror) into (1). Then we collect the cosine terms and equate them to Ega cos rrl/ on the right, and we equate the sine terms to zero because there is no sine term on the right, 1 + -O:C* La2(-a) * Rab -l alC : Eoa ra26b1 -l Ro\-a) + blC : 0 1 ,C (Cosine terms) (Sine terms). To solve this system for a and b, we first introduce a combination of L and C, called the reactance (3) ohm's resistor Inductor Capacitor -{/VW- R Ohm's resistance /-0'trf0^\- L Inductance =C Capacitance ohms (í)) henrys (H) farads (F) RI L#t QlC Fig. 6l. Elements in an RLC-circuit S:aLUnit Voltage Drop 94 CHAP. 2 Second-Order Linear ODEs Dividing the previous two equations by ,, ordering them, and substituting S gives -Sa*Rb:Eg -Ra-SĎ:0. We now eliminate b by multiplying the first equation by S and the second bY R, and adding. Then we elimin ate a by multiplying the first equation by R and the second bY -S, and adding. This gives -(S2 + R')o: EoS, (R2+ S2lb:EoR. In any practical case the resistance R is different from zero, so that we can solve for a and b, (4) -EoSn:- R2 +s2' - EoR h: U R2+s2 Equation (2) with coefficients a and b given by (4) is the desired particular solution Io of the nonhomogeneous oDE (1) governing the current 1in an RLC-círcuit with sinusoidal electromotive force. Using (4), we can write 1o in terms of "physically visible" quantities, namely, amplitude 1o and phase lag 0 of the current behind the electromotive force, that is, (5) IeG) :16 sin @t - 0) where [see (14) in App. A3.1] Io: o' +b': -L, t.un|: -+ : + \/F1 b R The quantitv Vn' + S'is called the impedance. Our formula shows that the impedance equals the ratio Eollo. This is somewhat analogous to ElI: R (Ohm's law). A general solution of the homogeneous equation corresponding to (1) is Ih: Cte^" + 'ru^" where .tr1 and h2 are the roots of the characteristic equation 1 LC We can write these roots in the form it : -a * B and Lz : -0t - B, where R a: -_=. F:2L, Now in an actual circuit, R is never zeío (hence R > 0). From this it follows that In approaches zero, theoretically as t ---ž @, but practically after a relatively short time. (ThiS is as for the motion in the previous section.) Hence the transient current I : In -l /o tends ^2 + i n * r-' , lliP- rc- 2L SEC.2.9 Modeling: ExAMPLE l 95 to the steady-state current Io, and after some time the output will practically be a harmonic oscillation, which is given by (5) and whose frequency is that of the input (of the electromotive force). R[C-Circuit Findthecurrentl(r)inanRtC-circuitwithR:11()(ohms),L:0.1 H(henry),C:l0-2F(farad),which is connected to a source of vo|tage E(r) 1O0 sin4OOr (hence $lHz : $3 cycles/sec, because 400 : $l , 2Tr1. Assume that current and charge are zeío when r : 0. Solution. Sfup l. General solution of the homogeneous ODE. Substituting R, L, C, and the derivative Ě'(r) into (1), we obtain 1.1I" + 11I' + 1001 : l00.400 cos 400r. Hence the homogeneous ODE is 0.l/" + l11/ + l001 :0. lts characteristic equation is 0.1,t2 + l1^ + 100:0. The roots are ),1 : -l0 and.tr2 : -l00. The corresponding genera| solution oťthe homogeneous ODE is InQl : r,lr-lo' + r2r-loot. Step 2. Particular solution Io of (7). We calculate the reactance ^S : 40 , 1l4 : 39.75 and the steady-state current Ie@ : cr cos 400r * Ď sin 400r with coefficients obtained fiom (4) 100.39.75 100. l l u- --2.3368, b: _ =0.6-167. 1l'+ 39.]5' Il'_ 39.]5' Hence in our present case, a general solution of the nonhomogeneous ODE (l) is (6) I(t) : ,,ť-1o' + ,,2r-Io0' - 2.33ó8 cos 400l + 0.6467 sin 400l. Step 3. Particular solution satisfying the initial conditions. How to use QQ') = 0? We finally determine c1 and c2 from the initial conditions 1(0) : 0 and Q(0) : 0. From the first condition and (6) we have (,7) 1(0) : C1 l C2 2.3368 : 0, hence cz: 2,3368 - ct Furthermore, using 11'; wittr r: 0 anrl noting that the integral equals QG) (.see the formula befbre (1')), we obtain LI'Q)+R.0-: 0:0, hence /'(0) : 0. Differentiating (6) and setting / : 0, we thus obtain r'(o): -l0c1 - 100c2+0+ 0.6461,400:0, hence -10c1 :100(2.3368 _.r) _ 258.68. The solution of this and (7) is c1 : -0.27'76, cz : 2.6144. Hence the answer is I(t) : -0.27J6e-Iot + 2.6l44e-1ool - 2.3368 cos 4OOr + 0.6461sin 4OOr. Figure 62 on p. 96 shows 1(0 as well as 1o(r), which practically coincide, except for a very short time near t - o because the exponential terms go to zero very rapidly. Thus atter a very short time the current will practically execute harmonic oscillations of the input frequency $lHz, : $3cycles/sec. Its maximum amplitude and phase lag can be seen tiom (5), which here takes the íbrm Ip(t) : 2.4246 sin (400r - 1.3008). l 96 CHAP.2 Second-Order Linear ODEs Example 1 Analog y of Electrical and Mechanical Quantities Entirely dffirent physical or other systems may have the same mathematical model, For instance, we hur," .""n this from the various aPPlications of the oDE Y' -- lrY in Chap. 1. Another impressive demonstration of this unifying power of mathematics ts giu"n by the oDE (ti for an electric RLC_circuit and the oDE (2) in the last section for a mass-spring system. Both equations LI" + RI' Egul cos ult and my" + cy' + lry : Fo cos @t are of the same form. Table 2.2 shows the analogy between the various quantities involved, The inductance L corresponds to the mass m and, indeed, an inductor opposes a change in current, having an "inertia effect" similar to that of a mass. The resistance R conesPonds to the damping constant c, anda resistor causes loss of energy, just as a damping dashpot does. And so on. This ana1o gy is strictly quantitative in the sense that to a given mechanical sYstem we can construct an electric circuit whose current will give the exact values of the disPlacement in the mechanical system when suitable scale factors are introduced, The practical iiportance ofthis analogy is almost obvious. The analogY maY be used for constructing an i,electrical model" of a given mechanical model, resulting in substantial savings of time and money because electric circuits are easy to assemble, and electric qountiti", can be measured much more quickly and accurately than mechanical ones, Table 2.2 Analogy of Electrical and Mechanical Quantities Electrical System Mechanical System Inductance .L Mass lz Resistance R Damping constant c Reciproca| IlC of capacitance Spring modulus k Derivative E.ídCoS at of } priving force Focos aí electromotive force ) Cunent (r) Displacement y(r) 0 _1 1 -|-I: C Fig.62. Transient and steady-state currents in ,, SEC. 2.9 Modeling: Electric Circuits 97 1. (RL-circuit) Model the Rl-circuit in Fig. 63. Find the general solution when R, L, E are any constants. Graph or sketch solutions when l : 0.1 H, R : 5 O, E : 12Y. 2. (RL-circuit) Solve Prob. 1 when E : Eo sin oíand R, L, Eg, rr.r are arbitrary. Sketch a typical solution. 3. (RC-circuit) Model the RC-circuit in Fig. 66. Find the current due to a constant E. 4. (RC-circuit) Find the current in the RC-circuit in Fig. 66 with E : E6 sin at anď arbitrary R, C, Eo, and rr;. L Fig. 63. RL-circuit o.o2 0.04 0.06 0.08 0.1 Fig. 64. Currents in Problem ] 1.5 1 0.5 _0.5 _1 Fig. 65. Typical current l - e-o,lt -F sin (f - in) in Problem 2 C Fig.66. RC-circuit Fig.67. Current 1 in Problem 3 5. (LC-circuit) This is an RlC-circuit with negligibly small R (analog of an undamped mass-spring system). Find the current when l : 0.2H, C : 0.05 F, and E : sin / V, assuming zero initial current and charge. 6. (ZC-circuit) Find the current when L : 0.5 H, '*:.r. 10-4 F, E : t2 Y and,initial current and charge @ R[c-clRculTs (F!G. 60, p. 92) 7. (Tuning) In tuning a stereo system to a radio station, we adjust the tuning control (turn a knob) that changes C (or perhaps L) in an RLC-circuit so that the amplitude of the steady-state current (5) becomes maximum. For what C will this happen? 8. (Transient current) Prove the claim in the text that if R + 0 (hence R > 0), then the transient current approaches Io as t --> co. 9. (Cases of damping) What are the conditions for an RlC-circuit to be (I) overdamped, (II) critically damped, (III) underdamped? What is the critical resistance Rcrit (the analog of the critical damping constant ZX6if | 10-12 | Find the steady-state current in the RLC-ciralit in Fig. 60 on p.92 for the given data. (Show the details of your work.) 10. R : 8 O, L : 0.5H, C : 0.1 F, E : 100 sin2rV 11. R : 1 í), L: 0.25 H, C :5. 10-5 F, E : 110 V 12. R _ 2 d), L : I H, C : 0.05 F, n : # sin3rV |13-15l Find the transient current (a general solution) in the RLC-circuit in Fig. 60 for the given data. (Show the details of your work.) 13. R : 6 í),L : 0.2H, C : 0.025F, E : 110 sin 10rV 14. R:0.2 O, L:0,1 H, C :2F, E: ]54 sinO.5rV 15. R : 1/10 a, L - Il2H,C : 100/13 F, E : e-at(L 932 cos lt + 0,246 sin }r) V l 1ó-18 | Solve the initial value problem for the RlC-circuit in Fig. 60 with the given data, assuming zero initial current alld charge" Graph or sketch the solution. (Show, ttre t]etllils ltl'vtlltr u,ork") Current /(t) Current 1( ) Current( ) .,,il 5 4 3 2 1 98 CHAP.2 Second-Order Linear ODEs 16. R : 4 í),L: 0.L H, C : 0.025 F, E: 10 sin 10rV 17. R : 6 O, L - tH, C : 0.04 F, E : 600(cos r + 4 sin r) V 18. R :3.6 a, L:0,2H, C : 0.0625 F, E: 164 cos 10rV 19. WRITING PROJECT. Analogy of RlC,Circuits and Damped Mass-,Spring Systems. (a) Write an essay of 2-3 pages based on Table 2,2. Describe the analogy in more detail and indicate its practical significance. (b) What RlC-circuit with L : \H is the analog of the mass*spring system with mass 5 kg, damping constant 10 kg/sec, spring constant 60 kg/sec2, and driving force220 cos 10r? (c) Illustrate the analogy with another example of your own choice. 20. TEAM PROJECT. Complex Method for Particular Solutions. (a) Find aparticular solution of the complex oDE Lí"+ Ri'+ G: \/=) by substituting 7o : Y,iolt (K unknown) and its derivatives into (8), and then take the real part Io of Ío, showing that loagrees with (2), (4). Hint- Use the Euler tbrmula uit>y - x3l2 cos J. Hint.Tofindy1, y2 set !: ux-Ilz. (r'D' -l xD -| (r' - i>tly - x3l2 sin x. Hint: As in Prob. 16, TEAM PROJECT. Comparison of Methods. The undetermined-coefficient method should be used whenever possible because it is simpler. Compare it with the present method as follows. (a) Solve y" + 2y' 15y : 17 sin 5x by both methods, showing all details, and compare. (b) Solve y" + 9y : f1 + f2, ť1 : sec 3x, 12 : sin 3x by applying each method to a suitable function on the right. (c) Invent an undetermined-coefficient method for nonhomogeneous Euler-Cauchy equations by experimenting, 10l 1o2 CHAP.2 Second-Order Linear ODEs 1. What general properties make linear ODEs particularly attractive? 2. What is a general solution of a linear ODE? A basis of solutions? 3. How would you obtain a general solution of a nonhomogeneous linear ODE if you knew a general solution of the corresponding homogeneous ODE? 4. What does an initial value problem for a second-order ODE look like? 5. What is a particular solution and why is it more common than a general solution as the answer to practical problems? 6. Why are second-order ODEs more important in modeling than ODEs of higher order? 7. Describe the applications of ODEs in mechanical vibrating systems. What are the electrical analogs of those systems? 8. If a construction, such as a bridge, shows undesirable resonance, what could you do? GENEMI soLUTloN Find a general solution. Indicate the method you are using and show the details of your calculation. 9. y" - 2y' - 8): 52 cos6x 10. y" -l 6y' -| 9y : ,-3r - 2712 11. y" * 8}' + 25y : 26 sin 3x 12. yy" * 2y'' 13. (x2D2 * 2xD - I2I)y : Ilxs 14. (x2D2 -f 6xD * 6I)y : 7z 15. (D2 - 2D + 1)y - x-3e* 16. (D2 - 4D + 5I)y : e2* csc x t7. (D2 - 2D + 2I)y : e* csc x 18. (4x2D2 - 24xD + 49I)y : 36x5 ELrs] tNFtAL vALuE pRoBLEMs Solve the following initial value problems. Sketch or graph the solution. (Show the details of your work.) 19. y" * 5y' - l4y:6, y(0) : 6, }'(0) : _6 20. y" + 6y' * 18y:0, y(0):5, y'(0) : _2I 2l. x2y" - ry' - 24y: g, _v(1): 15, y'(1):0 22. x2y" * I5xy' l 49y:0, y(I):2, }'(1) : _11 23. y" -l 5y' f 6y: 108x2, }(0): 18, y'(:0): -26 24. y" + y' + 2.5y : 13 cos J, y(0) : 8.0, y'(0) : 4,5 25. (x2D2 * xD - 4I)y : x3, y(1) : _4l5, y'(l) :9315 V641 AppLIcATloNs 26. Find the steady-state solution of the system in Fig. 70 when ffi : 4, c : 4, k : I7 and the driving force is 202 cos 3t. 27. Find the motion of the system in Fig. 70 with mass 0.25 kg, no damping, spring constant 1 kg/sec2, and driving fbrce 15 cos 0.5r - 7 sin 1.5t nt, assuming zero initial displacement and velocity. For what frequency of the driving force would you get resonance? 28. In Prob. 26 find the solution corresponding to initial displacement 10 and initial velocity 0. 29. Show that the system in Fig, 70 with ffi : 4, c : 0, k : 36, and driving force 6l cos 3.1r exhibits beats. Hint: choose zero initial conditions. 30. In Fig. 70 let m : 2, c : 6, k : 2J, and r(t) : 10 cos at.For what ra will you obtain the steadystate vibration of maximum possible amplitude? Determine this amplitude. Then use this ro and the undetermined-coefficient method to see whether you obtain the same amplitude. 31. Find an electrical analog of the mass-spring system in Fig, 70 with mass 0.5 kg, spring constant 40 kg/sec2, damping constant 9 kg/sec, and driving force l02 cos 6rnt. Solve the analog, assuming zero initial curent and charge. 32. Find the current in the RlC-circuit in Fig. 7I when L:0.1 H, R :20 í),C - 2.10-4 F, and E(t) : 110 sin 4I5t Y (66 cycles/sec). 33. Find the current in the RLC-circuit when L : 0.4 H, R : 40 í),C : 10-4 F, and E(t) : 220 sin 314r V (50 cycles/sec), 34. Find a particular solution in Prob. 33 by the complex method. (See Team Project 20 in Sec. 2.9.) Spring Mass Dash pot Fig. 70. Mass-spring system "il},E(t) Fig. 7l. RLC-circuit Summary of Chapter 2 l] l03 (2) Second-order linear ODEs are particularly important in applications, for instance, in mechanics (Secs, 2.4,2.8) and electrical engineering (Sec. 2.9). A second-order ODE is called linear if it can be written (1) l y"'I+ p@)y' + q(x)y : r(x) (Sec, 2,1), l,(If the first term is. sa},. í(*)y",divide by í(x) to get the '(standard form" (1) with y" us the first term.) b,quation (1) is called homogeneous if r(x) is zero for aII x considered, usually in some open interval; this is written r(x) :0. Then )," + p(.x)),' + q(x)y : O, Equation (1) is called nonhomogeneous if r(x) * 0 (meaning r(x) is not zero for some x considered). For the homogeneous ODE (2) we have the important superposition principle (Sec. 2.I) that a linear combination y : lqt + ly2 of two solutiofls }1, y2 is again a solution. Two linearly independent solutions jy lz of (2) on an open interval 1form a basis (or fundamental system) of solutions on I, anď j : ctlt l czy, with arbitrary constants c., c2 is a general solution of (2) on 1. From it we obtain a particular solution if we specify numeric values (numbers) for c, and c2, usually by prescribing two initial conditions (3) y(xo) : Ko, y'(x : Kt (xg, Kg, K1 given numbers; Sec. 2.1). (2) and (3) together form an initial value problem. Similarly for (1) and (3). For a nonhomogeneous ODE (1) a general solution is of the form !:ln*jp (Sec. 2.7). Here yn is a general solution of (2) andyo is a particular solution of (1). Such a yo can be determined by a general method (variation of parameters, Sec. 2.10) or in many practicalcases by the method of undetermined coefficients. The latter applies when (1) has constant coefficients p and q, anď r(x) is a power of x, sine, cosine, etc. (Sec. 2.7).Then we write (1) as (5) y" + ay' + by : r(x) (Sec. 2.7). The corresponding homogeneous ODE y' + oy' + by : 0 has solutions y : et*, where ,\ is a root of ^2+il"*b:O. (4) (6) Hence there are three cases (Sec. 2.2): I il il Distinct real .i.1, .tr2 Double -}a Complex -ta + iu !: CIe^r* l Cre^zr }:(cr +c2x)e-o*l2 ! - e-arlz(Á cos a*x * B sin o*x) Important applications of (5) in mechanical and electrical engineering in connection with vibrations and resonance aíe discussed in Secs. 2.4, 2.7, and 2.8. Another large class of ODEs solvable "algebraically" consists of the Euler-Cauchy equations ,'y" + axy' + by: O(1) (Sec. 2.5). These have solutions of the form j : x*, where m is a solution of the auxiliary equation (8) m2+(a-I)m-lb:O. Existence and uniqueness of solutions of (1) and (2) is discussed in Secs. 2.6 and 2.7 , and reduction of order in Sec. 2.1. 104 CHAP.2 Second-Order Linear ODEs Case Type of Roots General Solution ,,|{, A,,,P"T",E,,.R,.. 3 Higher Order Linear ODEs \ In this chapter we extend the concepts and methods of Chap. 2 for linear ODEs from order n : 2 to arbitrary order n. This will be straightforward and needs no new ideas. However, the formulas become more involved, the variety of roots of the characteristic equation (in Sec. 3.2) becomes much larger with increasing n, and the Wronskian plays a more prominent role. Prerequisire., Secs. 2.I, 2.2, 2.6, 2.7, 2.I0. References and Answers to Problems: App. 1 Part A, and App.2. Recall from Sec. 1.1 that an ODE is of nth order if the nthderivativey@) : dnyldxn of the unknown function y(x) is the highest occurring derivative. Thus the ODE is of the form (,''': #) (1) (2) 3.1 Homoteneous Linear ODEs F(x,y,!',,",y(')):0 where lower order derivatives and y itself may or may not occur. Such an ODE is called linear if it can be written y'n' + pn_t@)y(n-l) a * p{x)y' + po@)y : r(x). (For n : 2 this is (1) in Sec. 2.1 with pt : p andpo : q). The coefficients po,,,,, pn_I and the function r on the right are any given functions of x, and y is unknown. y(') has coefficient 1. This is practical. We call this the standard form. (If you have pn(x)y'n', divide by p.(x) to get this form.) An nth-order ODE that cannot be written in the form (1) is called nonlinear. If r(x) is identically zero, r(x) : 0 (zero for all x considered, usually in some open interval 1), then (1) becomes y(n) + pn_{x)y(n-7) 1 -| p{x)y' -l pg(x)y : 0 and is called homogeneous. If r(x) is not identically zero, then the ODE is called nonhomogeneous. This is as in Sec. 2.1. A solution of an rrth-order (linear or nonlinear) ODE on some open interval 1 is a function y : h(x) that is defined anď n times differentiable on I anď is such that the oDE becomes an identity if we replace the unknown function y and its derivatives by h and its coffesponding derivatives. l05 106 CHAP. 3 Higher Order Linear ODEs Homogeneous Linear ODE: Superposition Princip[e, General Solution Sections 3.1-3.2 will be devoted to homogeneous linear ODEs and Sec. 3.3 to nonhomogeneous linear ODEs. The basic superposition or linearity principle in Sec. 2.1 extends to nth order homogeneous linear ODEs as follows. ( Fundamental Theorem for the Homoteneous Linear ODE (2) For a homogeneous linear ODE (2), sums and constant multiples of solutions on some open interval I are again solutions on I. (This does not hold for a nonhomogeneous or nonlinear ODE!) The proof is a simple generalization of that in Sec. 2.I and we leave it to the student. Our further discussion parallels and extends that for second-order ODEs in Sec. 2.1. So we define next a general solution of (2), which will require an extension of linear independence from 2 to n functions. General Solution, Basis, Particular Solution A general solution of (2) on an open interval 1 is a solution of (2) on 1of the form (3) y(x): cr}r(x) + ",* cnyn(x) (ct",,cnarbitrary) where }t, . , ln is a basis (or fundamental system) of solutions of (2) on 1,, that is, these solutions are linearly independent on 1, as defined below. A particular solution of (2) on / is obtained if we assign specific values to the n constantS c1, , , , , crl. in (3). Linear lndependence and Dependence n functions }r(x), , , , ,!n(x) are called linearly independent on some interval I where they are defined if the equation (4) k t@)+",-lknyn(x):O on1 implies that all kr, . . . , kn are zero. These functions are called linearly dependent on 1 if this equation also holds on 1for some kr, , , , , kn not all zero. (As in Secs. 1.1 and 2.I,the arbitrary constants cb, , , , cnmust sometimes be restricted to some interval.) If and only if lr , , , , ln are linearly dependent on 1, we can express (at least) one of these functions on 1 as a "linear combination" of the other n - t functions, that is, as a sum of those functions, each multiplied by a constant (zero or not). This motivates the term "linearly dependent." For instance, if (4) holds with k, * 0, we can divide by kl and express }r as the linear combination THEoREM t DEFlNlTloN DEFlNlTloN SEC.3.1 Homogeneous Linear ODEs 1o7 1 }t: - , (kzlz + ... * kn!). Á1 Note that when fl:2,these concepts reduce to those defined in Sec. 2.1. Linear Dependence Show that the functions }r : ,', yr:5.r, ys : 2x are linearly dependent on any interval. Solution. }z : 0h -l 2.5y. This proves linear dependence on any interval. l Linear lndependence Show that)1 : í,!2: x2,.!s : í3 are linearly independent on any interval, for instance, on -1 < x < 2. Solution. Equation(4)iskl_r + k2x2 + kgx3:0.Taking(a).r: -1,(b) x: I, (c)l:2,weget (a)-kr *kz- ks:0, (b)k1 + k2-1 kg:0, (c)2k1+ 4k2+ 8 3:9. kz: O from (a) + (b). Then ks : O from (c) -2(b). Then Ř1 : 0 from (b). This proves linear independence, A better method for testing linear independence of solutions of ODEs will soon be explained. l EX A M Pt E 3 General Solution. Basis Solve the fourth-order ODE yi' - 5y" + 4y :0 (where yi" : dayldxa), Solution. As in Sec. 2.Z we try and substitute y : e^r. Omitting the common factor e^', we obtain the characteristic equation ^4-5^2+4:0. This is a quadratic equation in p : ),2, namely, p2 - 5p-l 4: (p - 1)(p - 4) : 0. The roots are p: 1 and 4, Hence h: -2, -1, 1, 2. This gives four solutions. A general solution on any interval is y - (,|e'2* l c2e-Í l cger + cne2' provided those four solutions are linearly independent. This is true but will be shown later, l lnitial Value Problem. Existence and Uniqueness An initial value problem for the ODE (2) consists of (2) and n initial conditions y(Jo) : Ko, y'(x : K!, y'n-"(*o) : Kn_I with given x6 in the open interval 1 considered, and given Ko, , , , , Kn_' In extension of the existence and uniqueness theorem in Sec. 2.6 we now have the following. Existence and Uniqueness Theorem for lnitial Value Problems If the cofficients pg(x), . . , pn_l@) oí(2) are continuous on some open interval I and xg is in I, then the initial value problem (2), (5) has a unique solution y(x) on I. Existence is proved in Ref. tA11] in App. 1. Uniqueness can be proved by a slight generalization of the uniqueness proof at the beginning of App. 4. (5) THEoREM 2 lO8 CHAP. 3 Higher Order Linear ODEs ExAMpLE 4 lnitial Value Problem for a Third-Order Euler-Cauchy Equation Solve the following initial value problem on any open interval 1on the positive x-axis containing x : 1. 3"'ll - 3x2y" + 6ry' - 6.y: O, y(I):2, y'(1): l, y"(1): _4, xy Solution. Sfup 1. General solution. As in Sec. 2.5 we try y : x*.By differentiation and substitution, m(m - 1)(m - 2)** - 3m(m - t)x* + 6mx* - 6x* :0. \ Dropping x* andordering gives llz3 - 6m2 í l|m - 6:0. If we can g^ues^s the root m: I, we can divide by m - 1 and find the other roots 2 and 3, thus obtaining the solutions ,, ,2, *3, which are linearly independent on 1(see Example 2). [In general one shall need a root-finding method, such as Newton's (Sec. 19.2), also available in a CAS (Computer Algebra System).] Hence a general solution is ):Clx+c2x2*a3*3 valid on any interval I, eyen when it includes x : 0 where the coefficients of the ODE divided by x3 (to have the standard lorm) are not Continuous. Step 2. Particular solution. The derivatives are y' : ct -| 2c2x * 3cgx2 and y" : 2c2 -| 6cax. From this and y and the initial conditions we get by setting x : 1 (a) "v(1) : c1 i cz l c3: 2 (b) y'(1) : ct * 2c2-| 3ca: l (c) _y"lt; : 2c2 * 6ca: -4. This is solved by Cramer's rule (Sec. 7.6), or by elimination, which is simple, as follows. (b) - (a) gives (d) c2 + 2c": -1. Then (c) - 2(d) gives ca - -1. Then (c) gives c2 : 1. Finally ct:2 from (a). Answer: ! :2x + x2 - x3. l Linear lndependence of Solutions. Wronskian Linear independence of solutions is crucial for obtaining general solutions. Although it can often be seen by inspection, it would be good to have a criterion for it. Now Theorem 2 in Sec. 2.6 extends from order n: 2 to any n. This extended criterion uses the Wronskian W of ,? solutions }r, , , ,ln defined as the lzth order determinant W(yr",,!n): y?-r, yy-r, .r, (rz - 1) Jn Note thatW depends on_r since !t,. . , , jndoes. The criterion states that these solutions form a basis if and only if W is not zero1, more precisely: THEoREM 3 Linear Dependence and lndependence of Solutions Let the ODE (2) have continuous cofficients pg(x), , , , , pn_l(x) on an open interval I. Then n solution }t, , " , ln oí(Z) on I are linearly dependent on I if and only if their Wronskian is zero for some x : xo in I. FurtherTnore, if W is zero for x : xg; then W is identically zero on I. Hence if there is an xl in I at which W is not Zero, then y1, . . , ln are linearly independent on I, so that they form a basis of solutions of (2) on I. }r lz ln lll lt lz ln (6) SEC. 3.1 Homogeneous Linear ODEs PROOF (a)Letlb",, are constants k1, lo9 lnbe linearly dependent solutions of (2) on 1. Then, by definition, there , , , , kn not all zero, such that for all x in I, kůtt",+knln:0. By , - 1 differentiations of (7) we obtain for all x in I \ ktYí+ kry?-') * + k.Y'. - 0 + k-y*-l) : 0, (7), (8) is a homogeneous linear system of algebraic equations with a nontrivial solution kr, . . . , kn. Hence its coefficient determinant must be zero for every x on I, by Cramer's theorem (Sec. 7 .7). Bttt that determinant is the Wronskian W, as we see from (6). Hence W is zero for every x on I. (b) Conversely, if W is zero at an xo in 1, then the system (7), (8) with x : x6 has a solution kr*, . . . , kn*, not all zero, by the same theorem. With these constants we define the solutiony*: k yt +,,, l kn*ynof (2) on1.By(7),(8)thissolutionsatisfiesthe initial conditions y*(xo) - 0, . , l )l*(n-"("o) : 0. But another solution satisfying the same conditions is y = 0. Hence y* = y by Theorem2, which applies since the coefficients of (2) arecontinuous. Together,y*: k,,,*yt + ", l kn*ln= 0onl.Thismeanslinear dependence of y1,,,,, yn on I. (c) If W is zero at an x6 in 1, we have linear dependence by (b) and then W = 0 by (a). Hence if I4z is not zero at an xl in 1, the solutions !t, , , , , y,, must be linearly independent on1. l EXAMPLE 5 Basis,Wronskian We can now prove that in Example 3 we do have a basis. In evaluating I pull out the exponential functions columnwise. In the result, subtract Column 1 from Columns 2, 3, 4 (without changing Column 1). Then expand by Row 1. In the resulting third-order determinant, subtract Column 1 from Column 2 and expand the result by Row 2: (1) (8) -2r -I Ieee ^ -2x -_r "r -Ze -e " e n -2r -.r Ja +e e" e _8e-2* _ e-r e, e 2e2' 4e2* 8e2' 11 -2 -1 4I -8 -1 1l I2 14 l8 13 -3 -3 79 :72. l ,lI A General Solution of (2) Includes A Let us first show that general solutions always exist. Indeed, as follows. l1 Solutions Theorem 3 in Sec. 2.6 extends TH Eo REM 4 Existence of a General solution If the cofficients p6(x), , , , , pn_l@) oí(2) are continuous on some open interval I, then (2) has a general solution on I. W: 11o CHAP.3 Higher Order Linear ODEs PROOF We choose any fixedx6 in 1.By Theorem 2the ODE (2) has n solutions )t, ", jn, where }3. satisfies initial conditions (5) with Kj_t: 1 and all other K's equal to zero. Their WronsÉanat x6 equals 1. For instance, when n : 3, then yr(xo) : 1, y!@|) : l, yÍ(xo) : 1, and the other initial values are zeío.Thus, as claimed, |l,,txol yzTol }s(xo)l l' 0 lV(yl(x6). y2(x6). r,3(x6)l : |l{t*o) yj(rol ljrr6)l : IO I |yit*ol y'JGo) y]tx6ll lo 0 :l 1| - 1. Hence for any n those solutions }t, , , , ln are linearly independent on 1, by Theorem 3. Theyformabasisonl,and): cr)r + ", l cry,'isageneralsolution of (2) on1. l We can now prove the basic property that from a general solution of (2) every solution of (2) can be obtained by choosing suitable values of the arbitrary constants. Hence an nth order linear ODE has no singular solutions, that is, solutions that cannot be obtained from a general solution. THEoREM 5 General solution lncludes All solutions If the ODE (2) has continuous cofficients pg(x), , , pn_I@) on some open interval I, then evetry solution y : Y(x) oí (2) on I is of the form (9) Y(x):Cl{x) + ",-f Cnyn@) where }1, , !,n is a basis oJ' solutions oí(2) on I and Ct, , , , Cn are suitable constants. PROOF LetY beagivensolutionandy: c1lt + ", l cny',.ageneralsolution of (2) onl.We choose any fixed x6 in 1 and show that we can find constants cb , , ,, cn for which y and its first n - I derivatives agree with and its coíTesponding derivatives at xg. That is, we should have at x : xo (10) cr}r*",+ cnjn -Y cl!+ ",+ c-y|" :Y' : cry?-'' + . . . * rry*-l) - y{n-1) But this is a linear system of equations in the unknowfls c1, , , , , cn. Its coefficient determinantis the Wronskian Wof }t, , , !-ndtx6. Since.}t, , ,, yr. form abasis, they are linearly independent, so thatW is not zeroby Theorem 3. Hence (10) has a unique solution c1 : C1, " , , ctt,: Cr, (by Cramer's theorem in Sec. 7.7). With these values we obtain the particular solution y*(x) : Cryr(x) +,,, * Cnyn(x) on 1. Equation (10) shows that _y* and its íirstn - 1 derivatives agree at xo with Y and its corresponding derivatives. That is, )* and Y satisfy at x6 the same initial conditions. SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients The uniqueness theorem (Theorem 2) now implies that y* = on 1. This proves theorem. This completes our theory of the homogeneous linear ODE (2). Note that for n : 2 tt is identical with that in Sec. 2.6. This had to be expected. the l E TyplcAL ExAMpLEs oF BAsEs To get a feel for higher order ODEs, show that the given functions are solutions and form a basis on any interval. Use Wronskians. (In Prob. 2, x } 0.) l. l. x, x2, x3. yiu : 0 2. I, x2, x4, *'y"' - 3xy" + 3y' : 0 3. e', xe*, x2e', y"' - 3y" + 3y' - y : 0 4. e2* cos.]r, e2* sin x, e-2* cos jr, e-2* Sin x, yi'-6y"+25y:O. 5. l, x, cos 3x, sin 3x, }iu + 9y" : 0 6. TEAM PROJECT. General Properties of Solutions of Linear ODEs. These properties are important in obtaining new solutions from given ones. Therefore extend Team Project 34 in Sec. 2.2 tonth-order ODEs. Explore statements on sums and multiples of solutions of (1) and (2) systematically and with proofs. Recognize clearly that no new ideas are needed in this extension from n : 2 to general n. @ LINEAR INDEIENDENcE AND DEPENDENCE Are the given íunctions linearly independent or dependent on the positive x-axis? (Give a reason.) 7. I,e*,e-* 8.xf I,xl2,x 9. ln x, ln x2, (ln x)2 |0. e' , e-" , sinh 2x 11. x2, x|.r|, x 12. x, Ilx, 0 13. sin 2x, sin J, cos -tr 14. cos2 x, sinz x, cos 2,r 15. tanJ, cot.r, l 16. (,r - I)', (x * 1)2, -r 17. sin x, sin tx 18. cosh -r, sinh x, cosh2 x 19. cos2,r, sinz x,2rr 20. TEAM PROJECT. Linear Independence and Dependence. (a) Investigate the given question about a set ,S of ftrnctions on an interval 1. Give an example. Prove yollr answer. (1) If ,S contains the zero function, can ,S be linearly independent? (2) If .S is linearly independent on a subinterval J of I, is it linearly independent on /? (3) If S is linearly dependent on a subinterval J oí I, is it linearly dependent on 1? (4) If S is linearly independent on 1, is it linearly independent on a subinterval -/? (5) lf S is linearly dependent on 1, is it linearly independent on a subinterval ./? (6) If ,S is linearly dependent on 1, and if 7 contains S, is Z linearly dependent on 1'l (b) In what cases can you use the Wronskian for testing linear independence? By what other means can you perform such a test? (1) (2) ln + an_rl@-D + 3.2 Homogeneous Linear ODEs with Constant coefficients In this section we consider nth-order homogeneous linear ODEs with constant coefficients, which we write in the form y(n) + an_ty@-l) l where y@) : dny/dxn, etc. We shall see that Sec.2.2. Substituting y : e^* 1as in Sec. 2.2), *ory'lagy:0 this extends the case n : 2 discussed in we obtain the characteristic equation " l a1h t a6:0 112 CHAP. 3 Higher Order Linear ODEs of (1). If ,\ is a root of (2), then y : el* is a solution of (1). To find these roots, you may need a numeric method, such as Newton's in Sec. 19.2, also available on the usual CASs. For genera| n there are more cases than for n : 2. We shall discuss all of them and illustrate them with typical examples. Distinct Real If all the n roots i1, (3) constitute a basis for all x. The corresponding irr:Cl Ť Roots , , ln of (2) are real and different, then the n solutions Átr !t: ) ln: general solution of (1) is , ÁnI .+cne(4) y Indeed, the solutions in (3) are ExAMPLE 1 linearly independent, as we shall see after the example. Distinct Real Roots Solve the ODE y"' - 2y" - y' + 2y : 0. Solution. The characteristic equation is i3 - 2^2 - ^ + 2: O. It has the roots -Í,I,2; if You find one of them by inspection, you can obtain the other two roots by^ solving a quadratic equation (exPlain!). The corresponding general solution (4) is y : cI't * c2e* + cge2', l Linear Independence of (3). Students familiar with nth-order determinants may verify that by pulling out all exponential functions from the columns and denoting their product by E,thus E : exp [(nr + . . . + hn)xf,the Wronskian of the solutions in (3) becomes WXtr Xzr Xre^r* hze^'* l12g^'* X22g^,* lnr X,'g^nr Xnzg^n* Xff-lrl,,r 1 ^n Xr' ^t-' : 0 if and only if the determinant on or Cauchy determinantl. It can be (5) h!-lrx'r 1 ^1 Xr' ^T-' X!-Irxzr 1 .. ^2 lr'-E ^t-' The exponential function E is never zero. Hence W the right is zero. This is a so-called Vandermonde shown that it equals 'ALEXANDRE THEOPHILE VANDERMONDE (1135-1]96), French mathematician, who worked on solution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5, SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients lt3 (6) (-lsntz, , ], .,T]l{ E o,R E M, l THEoREM 2 where 7is the product of all factors Xi - howith7 < k(= n);for instance, when n:3 we get -V : -(ir - izXr\r - isXnz - ir). This shows that the Wronskian is not zero if and only if all the n roots of (2) are different and thus gives the following. Basis Solutions !7 : eX'* ln : nn"* of (1) (with any real or complex ),i s) form a basis of solutions o/ (1) on any open interval if and only if all n roots oÍ (2) are dffirent, Actually, Theorem 1 from (5) and (6): is an important special case of our more general result obtained Linear lndependence Any number of solutions of (1) oíthe form en* are linearly independent on an open interval I if and only if the corresponding ), are all dffirent. Simple Complex Roots If complex roots occur, they must occur in conjugate pairs since the coefficients of (1) arereal.Thus,if i: y* iaisasimpleroot of (2), soistheconjugate h: y- ia,and two corresponding linearly independent solutions are (as in Sec. 2.2, except for notation) lt : eY* cos )í, Simple Complex Roots. lnitial Value Problem Solve the initial value problem !2 : eY* sin arx. y"'-y"*100y/-100y:0, y(0) : 4, y'(0) : 11, y"(0) : -299. Solution. The characteristic equation is i3 - ^2 + 100^ - 100 : 0. It has the root 1, as can perhaps be seen by inspection. Then division by ), - l shows that the other roots are :l 10l. Hence a general solution and its derivatives (obtained by differentiation) are j: cIeI i Á cos 10x * B sin lOx, II !' : c " - l0Á sin 10x + 10B cos 10x, llí !" : c * - 100A cos 10x - 1008 sin 10x. From this and the initial conditions we obtain by setting x : 0 (a) c1,1 A: 4, (b) c1 * 10B : 1l, (c) c1 - 100Á : -299. We solve this system íbrthe unknowns A, B, cy Equation (a) minus Equation (c) gives 101Á : 303, Á : 3. Then c1 : l from (a) and B : 1from (b). The solution is (Fig. 72) ! : er i 3 cos 10x f sin 10x. This gives the solution curve, which oscillates about e" (dashed in Fig. 12 onp. ll4), I ll4 CHAP. 3 Higher Order Linear ODEs 10 4 o0' íQ.72. Solution in Example 2 Multiple Real Roots If a real double root occurs, say, ir : ,trr, then lt : lz in (3), and we take y, and xy1 as coffesponding linearly independent solutions. This is as in Sec.2.2. Ivtoie geneially, if ,\ is arealroot of order m,then /,? colTesponding linearlY indePendent solutions are We derive these solutions after the next example and indicate how to Prove their linear independence. EXAMPLE 3 Real Double andTriple Roots Solve the oDE yO - 3}i' + 3y"' - }" : 0. Solution.Thecharacteristicequation.tr5. 3^4+3^3-^2:0hastherootsi1 -lz:Oand is : i+ - is : 1, and the answer is (8) ) : c1 l c2xi (ca * c4x + c5x2)eÍ Deňvation oí (7). We write the left side of (1) as Líyl: y@) + ar_I!'n-l) +,,, * aoy. Let y : eOx. Then by performing the differentiations we have Llun"l : (^n l ar_lhn-l +,,, + ao)e^*. Now let i1 be a root of mth order of the polynomial on the right, where m a n. For m 1 nlet X,-11, . . , lnbe the other roots, all different from i1. Writing the polynomial in product form, we then have Llnn"l: (^ - h)*h(l)e^" withh(n): 1if /7l: fl,andh(n): (^ - i_*r) ",(^ - t)if m< n.Nowcomesthe key idea: We differentiate on both sides with respect to )", a . í, a . (9) ]: tlrn"f : -(t - ir)--1h(X)e^* + (^ _ ir)- ", |n(Deo"]. ó^ L- J - d^ y 20 (7) tr SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients The differentiations with respect to x and ,\ are independent and the occurring derivatives are continuous, so that we can interchange their order on the left: tl5 (10) * ilro,]dA : t[* ,*] : Llxen*l. The right side of (9) is zero for i : i, because of the factors i - i, (and m > 2 since we have a multiple root!). Hence L|xen"l: 0 by (9) and (10). This proves that xe^'* is a solution of (1). We can repeat this step and produ"" *'nn'*, , , , , x*-'r^'* by another m - 2 such differentiations with respect to ,tr. Going one step further would no longer give zero on the right because the lowest power of .\ - ),1 would then be (^ - .trr)o, multiplied by mlh(D and ň(,\1) * 0 because /z(n) has no factors n - nr, so we getprecisely the solutions in (7). We finally show that the solutions (7) are linearly independent. For a specific n this can be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrary m we can pull out the exponential functions from the Wronskian. This gives (nn")- - e^ln,, times a determinant which by "row operations" can be reduced to the Wronskian of 1 , x, , , , , x*-|. The latter is constant and different from zero (equal to Il2| , , , (m - 1)!). These functions are solutions of the ODE y(*) : 0, so that linear independence follows from Theroem 3 in Sec. 3.1. l Multiple Complex Roots In this case, real solutions are obtained as for complex simpleroots above. Consequently, if i: y-l ia isacomplexdoubleroot,soistheconjugate X: y- ia. Corresponding linearly independent solutions are (11) The filst and xe^* (l2) eY* Cos {DX, eY* Sin alx, XeY' Cos a)x, xeY* stn alx. y : eY'l(A, -l A2x) cos alí+ (Br * B2x) sin r,.lr]. For complex triple roots (which hardly ever occur in applications), one would obtain two more solutions x2e^* cos .rjr, x2eY* Sin rrlx, and so on. two of these result from e^' and ,i* u, before, and the second two from xe^' in the same fashion. Obviously, the corresponding general solution is E oDE FoR GIVEN BAsls Find an ODE (1) for which the given functions form a basis of solutions. 1, e*, e2*, e3* 3. e*, e-', cos,r, Sin x 2. e-'*, xe-2', 12r-2r 4. cos x, sin í,.tr cos t, x Sin x 5, I, x, cos 2.r, stn2x 6. "-'*, -', *, e2*, I @ GENERAL soluTloN Solve the given ODE, (Show the details of your work.) 7,y"'ty':0 8. y'" - 29y" + I00y : 3 9.y"'-|.u"-1,'_y:0 10. 16yi"-8y"*}:0 Ll.y"'-3y"-4y'*6y:6 12.y'"*3y"-4y:0 CHAP. 3 Higher-Order Linear ODEs @ INITIAL vALuE pRoBLEMs Solve by a CAS, giving a general solution and the particular solution and its graph. 13. yi" + 0.45y"'- 0.165y" + 0.0045y' - 0.00175y : Q, y(0) : I7.4, y' (0) : -2.82, y"(0) : 2.0485, y"'(0) - -1 .458615 14.4y"'+ 8y" + 4Iy' + 37y: 0, 1,(0) : 9, y'(0) : -6.5,y"(0) : -39.]5 |5. y"' + 3.2y" + 4.87y' :0, y(0) :3.4, y'(0) - -4.6, y"(0) : 9.9I 16. y'" l 4y :0, y(0) : L, y' (0) : -Z, y"(0) : Z, }"'(0) : -Z 17. y'" - 9y" - 400y: 0, },(0) : 0, }'(0) : 0, y"(0) : 4I, )"'(0) : 0 18. y"' + 7.5y" 1- l4.25yl - 9.125y : 0, y(0) : 10.05, y'(0) : -54-915, y"(0) : 251,5125 19. CAS PROJECT. Wronskians. Euler-Cauchy Equations of Higher Order. Although Euler-Cauchy equations have variable coefficients (powers of x), we include them here because they fit quite well into the present methods. (a) Write a program for calculating Wronskians, (b) Apply the program to some bases of third-order and fourth-order constant-coefficient ODEs. Compare the results with those obtained by the program most likely available for Wronskians in your CAS. (c) Extend the solution method in Sec. 2,5 to any order n, Solve ,"y"' * 2*'y" - 4,y' l 4y :0 and another ODE of your choice. In each case calculate the Wronskian. 20. PROJECT. Reduction of Order. This is of practical interest since a single solution of an ODE can often be guessed. For second order, see Example'7 in Sec,2,1, (a) How could you reduce the order of a linear constant-coefficient oDE if a solution is known? (b) Extend the method to a variable-coefficient ODE y"' + pr(x)y" + p7@)y' -l p6(x)y : 0. Assuming a solution y1 to be known, show that another solution is y2(x) : u(x)yl(x) with u(x) : I z(x) dx and z obtained by solving yrz" * (3yí + pzyt)z'+Gyi+2pry!,*pl|)z:O, (c) Reduce ,'y"' - 3r'y" + (6 - xz)xy' - (6 - x2)y : 0, using y, : x (perhaps obtainable by inspection), 21. CAS EXPERIMENT. Reduction of Order. Starting with a basis, find third-order ODEs with variable coefficients for which the reduction to second order turns out to be relatively simple. * p{x)y' + po@)y : r(x) : d.nyldxn as the first term, which is practical, and r(x) * 0. As for second-order general solution of (1) on an open interval 1of the _r-axis is of the form 3.3 Nonhomoteneous Linear ODEs We now turn from homogeneous to nonhomogeneous linear ODEs of nth order. We write them in standard form (1) with y(') ODEs, a .\J'' + pr_1(_r);,"r-1l 1 (2) Here y7r(x) : c t(x) + homogeneous ODE y(x):yn@)+yo(r). + cnjn(x) is a general solution of the corresponding (3) y@) + pn_t(x)y@-1)+ ", + p{x)y' *po(x)y:0 on 1. Also, }p is any solution of (1) on 1 containing no arbitrary constants. If (1) has continuous cóefficients and a continuous r(x) on 1, then a general solution of (1) exists and includes all solutions. Thus (1) has no singular solutions. lt6 117SEC. 3.3 Nonhomoseneous Linear ODEs An initial value problem for (1) consists of (1) anď n initial conditions y(ro) : Ko, y'(x : K1, ,, y'n-')(ro):Kn_l it has a unique solution. The ideas ofwith xo in 1. Under those continuity assumptions proof are the same as those for n : 2 in Sec. 2.7. Method of undetermined coefficients Equation (2) shows that for solving (1) we have to determine a particular solution of (1). For a constant-coefficient equation y(n) + ar,_ty@-I) l * ory' l ooy: r(x) (ao, . . . , an_l constant) and special r(x) as in Sec. 2.7, such a !p(x) can be determined by the method of undetermined coefficients, as in Sec. 2.7 , llsing the following rules. (A) Basic Rule as in Sec. 2.7. (B) Modification Rule. If a term in your choice for yo@) is a solution of the homogeneous equation (3), then multiply yo@) by ,o, where k is the smallest positive integer such that no term of xkyo@) is a solution of (3). (C) Sum Rule as in Sec. 2.7. The practical application of the method is the same as that in Sec. 2.]. It suffices to illustrate the typical steps of solving an initial value problem and, in particular, the new Modification Rule, which includes the old Modification Rule as a particular case (with F;$iT;:T:'oT:,:X*,r;,ff:Hfri::lil"":le same as for ft:2, perhaps except EXAMPLE 1 lnitial Value Problem. Modification Rule Solve the initial value problem (6) y"' + 3y" + 3,-' + y - 30e-', y(0):3, y'(o): -:, .r-"10): -4J. Solution. Sfup1. The characteristicequationis,tr3 +3^2 + 3^ + 1: (^+ t;3:0. Ithasthetripleroot ^ : -1. Hence a general solution of the homogeneous ODE is ),h- t,ť * + c2xe-Í + caxze-* : (cr + c2x l csxzle-Í. Step 2. If we try lp: Ce-',we 8et -C + 3C - 3C + C : 30, which has no solution. Try Cxe-* and Cx2e-'. The Modification Rule calls for Then ^3-r|p: Lx e In,l 23,-r )p:L(J,r -X ]e yi: C6, - 6x2 + xs e-*, !'i': Ct6 - l8x + 9x2 - *3lu-* (4) (5) 118 CHAP. 3 Higher-Order Linear ODEs Substitution of these expressions into (6) and omission of the common factor e-tr gives C6 - l8x * 9x2 - í3)+ 3C(6x - 6x2 + x3; + 3C(3x2 -,T3) + Cr3 : 30, The linear, quadratic, and cubic terms drop out, and 6C:3}.Hence C: 5.This gives l,p: 5x3e-*, Step 3. We now write down ! : ln t yo, the general solution of the given oDE, From it we find c1 by the first initial condition. we insert the value, differentiate, and determine c2 fiom the second initial condition, insert the value, and finally determine ca from y"10; and the third initial condition: ,}:)h+)p: (c1 l c2x+ c"x2le-* +5x3e-*, .v(0): ct:3 y' : [-3 ,l c2l (-cz-| 2cg)x * (15 - ca)x2 - 5.r3]e-,'. .y'(0) : -3 ,| c2 - -3, cz: 0 }": [3 -l 2cg-| (30 - 4cs).r + (-30 + c3)x2 + 5r3]e-", )"(0): 3-1 Zcg: -47, cs: -25, Hence the answer to our problem is (Fig, 73) ) : (3 - 25x2le-* + 5x3e-*. The curve of ,v begins at (0, 3) with a negative slope, as expected from the initial values, and approaches zero Ias _r --> oo. The dashed curve in Fig. 73 is yo. y 5 Fig. 73. y and )zp (dashed) in Exampte l Method of variation of parameters The method of variation of parameters (see Sec. 2.10) also extends to arbitrary order n. It gives a particular solution}p for the nonhomogeneous equation (1) (in standard form with y(') as the first term!) by the formula (7) yr,(x) :Ž,yu@) ÍW 1',) dx sW{x) l l I l t í-W"(x) ..l1 : )1(.r) J ffi rG) dx+,,, * y,(x) )iď rG\ d,r on an open interval 1on which the coefficients of (1) and r(x) are continuous. In (7) the functions jt, . .. , y," form a basis of the homogeneous ODE (3), with Wronskian W, and Wj(j - 1, . . . ,nj is obtained from Wby replacing the jth column of W bY the column tď 0 0 1]T. Thus, when fl:2, this becomes identical with (2)in Sec.2.10, W- 'r'r|' : !t. Ii *: l: !zl | : -yo. yi| l-vr n': |ri :l The proof of (7) uses an extension of the idea of the proof of (2) in Sec. 2.10 and can be found in Ref tA11] listed in App. 1. y and )zp (dashed) in Example 1 ll9SEC. 3.3 Nonhomogeneous Linear ODEs ExAMPLE-2 Variation of Parameters. Nonhomoteneous Euler-Cauchy Equation Solve the nonhomogeneous Euler-Cauchy equation ,')"" - 3x2l," + 6ry' - 6y : xa \n, (r > 0). Solution. Step 1. General solution of the homogeneous ODE. Substitution of y : x- anď the derivatives into the homogeneous ODE and deletion of the factor x- gives m(m - I)(m - 2) - 3mQn - l) + 6m - 6:0, The roots are 7,2,3 and give as a basis 2 |r : "l'. \'c : .\ , 3 }s:I Hence the corresponding general solution of the homogeneous ODE is lh: CIX + c2x2 + car3, Step 2. Detenninants needed in (7). These are ,l: ":ll ,"=11- lo ":l: 2 3lr-Yl 2x z*'| : zr" 2 6-1 2 3lrŤl ^ .rl 4 lX J,{ I: jr Z 6, l 0 x3| o 3x2| : -rr' l 6, 1 *2 0| 2x Ol : "2. 21l lp: x I: ,hx clx - -' I-u.x dx _ r Í+ xlnx dx ; (+ nx- +) *(:h,-+) - + Simplification gives ry : fxa (ln x -f;). Hence the answer is Step 3. Integration.In (7) we also need the right side r(x) of our ODE in standard form, obtained by division of the given equation by the coefficient .r3 of .y"'; thus, r(x) : 1xa ln x)lxs : x ln,r. In (7) we have the simPle quotients WI|W : xll, W2lW - - 1, WslW : Il(2x). Hence (7) becomes (xlnx-;). |: !n+ yo: clx * c2x2 + cgxs + árn 1tnx - 161). Figure 74 shows !p. Can you explain the shape of this curve? Its behavior near Jr : 0? The occurrence of a minimum? Its rapid increase? Why would the method of undetermined coefficients not have given the solution? l 120 CHAP.3 Higher-Order Linear ODEs yl 3O 20 10 0 -10 _20 F|.7a. Particular solution yp of the nonhomogeneous Euler-Cauchy equation in Example 2 Application: Elastic Beams Whereas second-order oDEs have various applications, some of the more imPortant ones we have seen, higher order oDEs occur much more rarely in engineering work, An ;;no5l *m:fi:i;.ff:erns the bending of elastic beams, such as wooden or iron Vibrations of beams will be considered in Sec, 12,3, ExAMPLE 3 Bending of an Elastic Beam under a Load we consider a beam B of length L anď constant (e.g., rectangular) cross section and homogeneous elastic material (e.g., steel); see Fig. 75. We assume that under its own weight the beam is bent so little that it is practically straight, If we apply a load to B in a vertical plane through the axis of symmetry (the x_axis in Fig. 75), B is bent. Its axis is curved into the so-called elastic curve C (or deflection curve), It is shown in elasticity theory that the bending moment M(x) isproportional to the curvature k(x) of c. we assume the bending to be small, so that the deflection y(x) and its derivative y'1"; 1determining the tangent direction of C) are small, Then, by calculus, 1r: ,"l1I + y'')'l': y", Hence (8) M(x) : EIy"(x), El is the constant of proportionality. E is Young's modulus of elasticity of the material of the beam, 1is the moment of inertia of the cross section about the (horizontal) z_axis in Fig. 75. Elasticity theory shows further that M"(x) : f(x), where í(x)is the load per unit length, Together, EIyIv : f(x). Fig. 75. Deformed beam under uniform load (simply supported) Elastic Beam )c SEC. 3.3 Nonhomogeneous Linear ODEs T2l The practically most important supports and corresponding boundary conditions are as follows (see Fig. 76). (9) (C) Clamped atx:0, free aíx: L y(0) : )'(0) : O,y"(L): y"'1L1 : O. Theboundaryconditiony:Omeansnodisplacementatthatpoint,}':0meansahorizontaltangent,y":O means no bending moment, anď y"' : 0 means no shear force. Let us apply this to the uniformly loaded simply supported beam in Fig. 75. The load is /(-r) : f g: const. Then (8) is ío y'u-Ř, k: _J ", El' This can be solved simply by calculus. Two integrations give y'':+x2 lcg+c2. y"(0) : 0 gives cz : O.Then y"(t; : LGkLf .r) : 0, cr : -kLl2(since L * 0). Hence ," : L(r' L.rl. Integrating this twice, we obtain k v: with ca : 0 from y(0) : 0. Then kL lts L3 \ L3 |tl: , (" - u *rr) 0. ,,": n. Inserting the expression for k, we obtain as our solution u: JL u4 - 2Lx3 + L3x)., 24EI Since the boundary conditions at both ends are the same, we expect the deflection y(x) to be "symmetric" with respect to Llz, that is, ,v(x) : y(L - x). Verify this directly or set J : tt * Ll2 anď show that y becomes an even function of u, ,:h(;-I*)(;-i*) From this we can see that the maximum deflection in the middl e at u :0 (x : Ll2) is 5f oL4l06 , 24EI). Recall that the positive direction points downward. :--...--: { _ rÁr Simp|ysupported r=O x=L (A) Simply supported (B) Clamped at both ends y:y":0atx:Oandt y:y':0atx:OandL (B) Clamped at both ends (C) Clamped at the left end, free at the right end (+-'-i;+.s*+.a) x=O x=L Fig.76. Supports of a Beam r=0 x:L ! 172 CHAP.3 Higher-Order Linear ODEs E ESolve GENERAL SOLUT|ON the following ODEs. (Show the details of your work,) !. y"' - 2y" - 4y' -l 8y : e,3' + 8x2 2. y"' + 3y" - 5y' - 39y : 30 cos x 3. y'" + 0.5y" + 0.0625y : e-* cos 0,5x 4. y"' + 2y" - 5y' - 6y:100e-3* -l 78e-* 5. "'y"' + 0.75xy' - 0.]5! : 9x5,5 6. (xDs + 4D2)y : 8e* 7. (Dn + IoDz + 9I)y : 13 cosh 2x 8. (D' - 2D2 - gD + 181)y : e2* E lNlTlAL vALuE pRoBLEMs Solve the following initial value problems, (Show the details.) 9. y"' - 9y" 1- 27y' - 21y: 54 sin 3x, y(0) : 3.5, y'(0) : 13.5, y"(0) : 38.5 10. y'" - t6y : I28 cosh 2x, y(0) : 1, y'(0) :24, y"(0) : 20, y"'(0) : -160 II. (x3D3 - x2D2 - ]xD + l6l)y : 9x In x, y(1) : 6, Dy(l) : 18, D'y(I) : 65 12. (D4 - 26D2 + 25I)y: 50(x + I)2, y(0) : 12,16, Dy(O) - -6, D'y(O) :34, D3y(0) : -130 13. (D3 + 4D2 + 85D)y : í35xe*, y(0) : 10,4, Dy(O) : -18.1, D'y(O) : -69L6 1,4, (2D3 - D2 - 8D + 4I)y : sin í, y(0) : 1, Dy(O) : 0, D2y(0) : 6 15. WRITING PROJECT. Comparison of Methods, write a report on the method of undetermined coefficients and the method of variation of parameters, discussing and comparing the advantages and disadvantages of each method. Illustrate your findings with typical examples, Try to show that the method of undetermined coefficients, say, for a third-order ODE with constant coefficients and an exponential function on the right, can be derived from the method of variation of parameters. 16. CAS EXPERIMENT. Undetermined Coefficients, Since variation of parameters is generally complicated, it seems worthwhile to try to extend the other method, Find out experimentally for what ODEs this is possible and for what not. Hint: Workbackward, solving ODEs with a CAS and then looking whether the solution could be obtained by undetermined coefficients, For example, consider y"' - I2y" + 48y' - 64y : ,|l2u4x and *"y"' + ,'y" - 6xy' + 6y: xlnx. 1. What is the superposition or linearity principle? For what nth-order ODEs does it hold? 2. List some other basic theorems that extend from second-order to nth-order ODEs. 3. If you know a general solution of a homogeneous linear ODE, what do you need to obtain from it a general solution of a corresponding nonhomogeneous linear oDE? 4. What is an initial value problem for an nth-order linear oDE? 5. What the wronskian? what is it used for? @ GENERAL soluTloN Solve the given ODE. (Show the details of your work,) 6. y"' + 6y" + 18y/ * 40y : 6 7. 4x2y"' -l l2xy" + 3y' : 0 8.y'"+10y"+9y:g 9. 8y"' + Izy" - 2y' - 3y : 0 10. (D3 + 3D2 + 3D 1- I)y : az II. (xDa + Ds)y : l50xa 12. (D4 - 2D3 - lDly : 1,6 cos 2x 13. (D3 + Dy - 9nrl2 14. (x3D3 - 3x2D2 * 6xD - 6I)y : 3Ox-2 15.(D3 -D2-D+I)y:"' @ INITIAL vALuE pRoBLEMs Solve the given problem. (Show the details.) 16. y"' - 2y" -l 4y' - 8y:0, y(0) : -1, y'(0) : 30, y"(0) : 282 17, x3y"' + 7x2y" - zry' - 10y : 0, y(1) : I, y'(1) - -J, y"(1) : 44 18. (D3 + 25D)y : 32 cos2 4x, y(0) : 0, Dy(O) : 0, D2y(0) : 6 Ig. (D4 + 40D2 - 4411)y: 8 coshx, y(0): 1,98, Dy(O) : 3, D'y(O) : -40,02, D3y(0) : 27 20. (x3D3 + 5x2D2 -l 2xD - ZI)y : 7x3l2, y(1) : 10.6, Dy(1) : -3,6, D'y(t) : 3L2 *retrne,nTl ONS AN D PROBLEMS Summary of Chapter 3 l23 Compare with the similar Summary of Chap. 2 (the case n - 2). Chapter 3 extends Chap. 2 from order n : 2 to arbitrary order n. An nth-order linear ODE is an ODE that can be written (1) y@) + pn_t(x)y@-1) + -| pt@)y' + po@)y : r(x) with y(') : dnyldxn as the first term; we again call this the standard form. Equation (1) is called homogeneous if r(x) : 0 on a given open interval I considered, nonhomogeneous ífr(x) * 0 on 1. For the homogeneous ODE (2) y@) + pn_t(x)y@-l) + + p1,@)y' * p6(x)y : 0 the superposition principle (Sec. 3.1) holds, just as in the case n:2. A basis or fundamental system of solutions of (2) on 1 consists of n linearly independent solutions !t . . . ,l,-of (2) on 1. A general solution of (2) on 1is a linear combination of these, }:Crhi * cnYn (ct,,,, crl.arbitrary constants). A general solution of the nonhomogeneous ODE (1) on 1is of the form j: jnl )'p (Sec. 3.3). Here, le is a particular solution of (1) and is obtained by two methods (undetermined coefficients or variation of parameters) explained in Sec. 3.3. An initial value problem for (1) or (2) consists of one of these ODEs and n initial conditions (Secs. 3.I, 3.3) (5) y(xo) : Ko, y'(.xo) : Kr, , y(n-"("o) : Kr_l with given 16 in I and given Ko, , , , , Kn_1.If po, , , , , pn_1,, r are continuous on 1, then general solutions of (1) and (2) on 1 exist, and initial value problems (1), (5) or (2), (5) have a unique solution. (3) (4) C .H.,.A,,,P,. T,,,E, R,__4 Systems of oDEs. Phase Plane. Óualitative Methods Systems of oDEs have various applications (see, for instance, Secs. 4.1 and 4.5)' Their theory is outlined in Sec. 4.2 anďincludes that of a single oDE. The PracticallY imPortant conversionofasinglerrth-orderoDEtoasystemisshowninSec.4.1. Linearsystems(Secs.4.3,4.4,4.6)arebesttreatedbytheuseofvectorsandmatrices, of which, however, only a few eleme naury facts will be needed here' as given in Sec' 4'0 and probably familiar to most students, Qualitative methods. In addition io actually solving systems (Sec, 4,3,4,6), which is often difficult or even impossible, we shall explain a tňliY different method' namelY' the powerful method of inveitigating ,t. g.n"rui behavior of whole families of solutions in the phase plane (Sec. 4.3). This approactr to systems of oDEs is called a qualitative method because it does not need ň"r solutions (in contrats to a "quantitative method" of actually solving a system), Thisphas, pbín mithod,as it is called, also gives information on stabilitY of solutions' which is of general importance in control th"o.y, circuit theory, population dynamics' and so on. Here, stability of a physical system*"án, that, roughly speaking, a small change at some instant causes only smalt .t á^g". in the behavior of the system at all later times' phase plane methods can be exteided to nonlinear systems, for which they are particularly usetul. We wil| show this in Sec, 4,5, which includes a discussion of the pendulum equation and the Lotka_Volterra population model, We finally discuss nonhomogeneous linear systems in Sec, 4,6, NoTATloN. Analogous to Chaps. 1-3, we continue to denote unknown functions by y;thus,y{t),yz|).Thisseem.p..f",ubletosuddenlyusingxforfunctions,xt(t),xz(t), u, i, ,ornetimes done in systems of ODEs, Prerequisite: Chap,2, References and Aiswers to Problems; App. 1 Part A, and App,2, 4.0 Basics of Matrices and Vectors In discuss ing linearsystems of oDEs we shall use matrices and vectors. This simPlifies formulas and clarifiesideas. But we sha|| need only a few elementary facts (by no means the bulk of material in Chaps. 7 anď8). These facts wil' very likelY be at the disPosal of most students. Hence this section is for reference only, Begin with sec, 4,íand consult 4.0 as needed. 124 - L l25 Most of our linear systems will consist of two ODEs in two unknown functions yt(t), YzG), I lt: al!t l anlz. (1) l yL: aztlt l azzjz, y'l,: -5yr * 2yz for example, yL: 13y1 + ly, (perhaps with additional given functions 8{t), g2(t) in the two ODEs on the right). Similarly, a linear system of n first-order ODEs in n unknown functions yr(r), , , , , y,(r) is of the form y|: atlyt t anlz + , , , l atnjn yL: aztyt* azzjz + ", * aznln (2) y'r: antltl anzlz + ", l annln (perhaps with an additional given function in each ODE on the right). some Definitions and Terms Matrices. In (1) the (constant or variable) coefficients form a 2 X 2 matrix A, that is, an aríay f o.,, anf [ -5 2] (3) A:[a;r] :| | forexample. A:I l_ |o^ orr) lvl l^qlrrťrvt L '' +_.l Similarly, the coefficients ín (2) form an n x n matrix f or, 0e or,1 I o^ azz or, l (4) A:la;r]:| | Il o.-) The a11, al2, . . . are called entries, the horizontal lines rows, and the vertical lines columns. Thus, in (3) the first row is [a11 ap], the second row is 1a21 a22], and the first and second columns are f orrf f orr1 L,,,_] and l,,,) In the "double subscript notation" for entries, the first subscript denotes the row and the second the columlz in which the entry stands. Similarly in (a). The main diagonal is the diagonal att azz ann in (4), hence att azz in (3). We shall need only square matrices, that is, matrices with the same number of rows and columns, as in (3) and (4). SEC. 4.0 Basics of Matrices and Vectors 126 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Vectors. A column vector x with n components x1,,,,, xnis of the form l-xr l I-.l *:|.'l, thusif n:2. l:l L-,] Similarly, a row vector v is of the form -: [;;] f b, brr1 and B:| l. l,b^ brr) o : [r, unf, thus if n : 2, then Y : [U1, uz]. calculations with Matrices and vectors Equality. Two lz X n matrtces are equal if and only if coffesponding entries are equal. Thus for n:2,Iet A: |-o" Lo^ Then A :'B if and only if '"'.,',f a11 -- b11, atz: btz a21: b21, azz: bzz. Two column vectors (or two row vectors) are equal if and only if theY both have components and coíTesponding components are equal. Thus, let lurl ["r-l ":L,;_] and -:L",_] , Then Addition is performed by adding corresponding must both be n x n, and vectors must both have for n -- 2, UI: X"j, v : x if and only if U2: X2, entries (or components); here, matrices the same number of components. Thus f a1 -l b1 a9 * blr] [r, (5) A+B:| l, v*x:| |a21 * b21 a22 * b22) Lu, Scalar multiplication (multiplication by a number c) is performed by entry (or component) by c. For example, if i xrl * ",_] multiplying each 19 3l A:I l. then |-z 0_] then r -63 -7A: I Lt4 |- 4l 10v : l l [ - t:o_] -2I1 .] If " : [::] SEC. 4.0 Basics of Matrices and Vectors 127 x n matrices j:I,"')n k:I,",,fl, j:l,"',n. (6) Matrix Multiplication. The product C : AB (in this order) of two n A: |aq]andB : |biu) isthe nX n matrix C: ycry"]withentries ,lL Cjk : ) ai*b*rc ln:1 CAUTION! Matrix multiplication is not commutative, AB example, For example, Ui [j:]] that is, multiply each entry in the7th row of A by the coffesponding entry in the lďh column of B and then add these n products. One says briefly that this is a "multiplication of rows into columns." For example, tg 3l [t -o-1 :[9.1+3.2 g.(-4)+3,5-1 l--, 0_] L, 5_] L-z, t + 0,2 (-2), (-4) + 0 , 5l I ts -2t]:|l. L-z 8] + BA in general. In our 1 .3 + (-4) .0l l 2.3 + 5.0 l ftl 3l_ll -L8 6] Multiplication of an n X n matňx A by a vector x with ,? components is defined by the same rule: v : Ax is the vector with the n components l: ;] l', ;] : |',',I',-r,'_,|,' Systems of ODEs as Vector Equations Differentiation. The derivative of components) is obtained by differenti [yrtOl f ,-"1 Y(/) : Ly,t,l_] : |,'",_] ' Using matrix multiplication and diffel . lyil fo,(7) y, : L y;): o, : |o^ : Ž,a3rnxrl [;,] : [j;: I,,-:,) a matrix (or vector) with variable entries (or ating each entry (or component). Thus, if then y,(t) :t;il] :|-l:,i] rentiation, we can now write (1) as ,.,..,,ft;;] , eg ,,: [,: 'lf[;;] l28 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Similarly for (2)by means of an nX nmatrixA and acolumnvectory with n components, namely, y' : Ay. The vector equation (7) is equivalent to two equations for the components, and these are precisely the two ODEs in (1). Some Further Operations and Terms Transposition is the operation of writing columns as rows and converselY and is indicated by T. Thus the transpose AT of the 2 X 2 matrix fo, anf [-5 21 o : |o,,, o,,,): | ,, ;] is The transpose of a column vector, say, f o-,, orr1 [ -5 'rlAT:l l:l l.A -|o,, ,,,)- | 2 +_] ": [:;] , is a row vector, o' : [r, uz], and conversely. Inverse of a Matrix. Then X nanitmatrix I is the n X n matrix with main diagonal I, I, . . ., I and all other entries zero. Iffor a given n X nmatrix A there is an n X n matrix B such that AB : BA : I, then A is called nonsingular and B is called the inverse of A and is denoted by A-1; thus (8) If A has no inverse, it is (9) where the determinant of A is AA-1 :A-lA:I. called singular. For n : 2, tr -1 _ _| [ azz -orrf .,* - d.tA l_-,o^ orr) (10) lal,t a el detA: I l:urtazz-(ttza2t. |o^ azzl (For general n, see Sec. 7.'I, b this will not be needed in this chapter.) Linear Independence. r given vectors y(1), , , v(') with n components are called a linearly indepenclent set or, more briefly, linearly independent, if (11) cry(')+",*crv('):0 implies that all scalars cL, . . . , c, must be zero; here, 0 denotes the zero vector, whose ,? components are all zero.If (11) also holds for scalars not aII zero (so that at leaSt one of these scalars is not zero), then these vectors are called alinearly dependent set or, briefly, linearly dependent, because then at least one of them can be expressed as l29 a linear combination of the others; that is, if, for instancQ, c1 * 0 in (11), then we can obtain .,(1) - l (rrr'rr+ . . . + crv(.)). C1 Eigenvalues, Ei genvectors Eigenvalues and eigenvectors will be very important in this chapter (and, as a matter of fact, throughout mathematics). Let A : |ait"] be an n X n matrix. Consider the equation Ax : .trx where ,tr is a scalar (a real or complex number) to be determined and x is a vector to be determined. Now for every A a solution is x : 0. A scalar ň such that (I2) holds for some vector x *'0 is called an eigenvalue of A, and this vector is called an eigenvector of A corresponding to this eigenvalue .[. We can write (I2) as Ax - .ňx : 0 or (13) (A - ,\I)x : 0. These ďíe n linear algebraic equations in the n unknowíls .í1, , , , , xr,. (the components of x). For these equations to have a solution x * 0, the determinant of the coefficient matrix A - ^I must be zero. This is proved as a basic fact in linear algebra (Theorem 4 in Sec. 7.7). In this chapter we need this only for n: 2. Then (13) is (l2) (I4) in components, (14*) |"-.^o ,,:,: ^f[;;] : [:] , (al-,\)x1 * at2xz -0 aztxt * (azz - i)xr: g. Now A - ^I is singular if and only if its determinant det (A * ^I), called the characteristic determinant of A (also for general n), is zero. This gives (15) |arr - ,\ an l det(A-il;:; I I ort azr-^l : (al - X)(azz - ^) - atzazt : X2 - (al -l ar))" l alla22 - al2a21 : 0. This quadratic equation in .tr is called the characteristic equation of A. Its solutions are the eigenvalues A1 and ,tr2 of A. First determine these. Then use (14*) with ,\ - ),, to determine an eigenvector x(1) of A corresponding to .tr1. Finally use (14*) with ,\ : .tr, to find an eigenvector x(2) of A corresponding to .tr2. Note that if x is an eigenvector of A, so is kx for any k + 0. SEC. 4.0 Basics of Matrices and Vectors t3o cHAp. 4 Systems of oDEs. Phase Plane. Qualitative Methods EXA M PLE l Eigenvalue Problem Find the eigenvalues and eigenvectors of the matrix o : [-o,o ool (16) ^-L-,.u 1.2) Solution. The characteristic equation is the quadratic equation |_^-\ A l det[A-^Il : I l:o'+2.8^+ 1,6:0. | -t.o 1.2-^l It has the solutions \: -2 and ,tr2 : -0,8, These are the eigenvalues of A, Eigenvector,u."obtuin"dfrom(14*).For),:i,r:-2wehavefrom(14*) (-4.0 + 2.0)x1 -| 4.0x2 - 0 -1.6x1 + 0.2-1 2.0)x2: 0. A solution of the first equation is x1 : 2, x2: 1. This also satisfies the second equation. (WhY?). Hence an eigenvector of A corresponding to ),1 : -2,0 is (I7) *" : [i] similarly, x(2) - [J_] is an eigenvector of A corresponding to iz : -0,8, as obtained from (14*) with ), : ,\z, Verify this, l 4.1 Systems of ODEs as Models We first illustrate with a few typical examples that systems of ODEs can serve as models in various applications. We further show that a higher order ODE (with the highest derivative standing alone on one side) can be reduced to a first-order sYstem. Both facts account for the practical importance of these systems. EXAMPLE 1 Mixing Problem lnvolvingTwoTanks A mixing problem involving a single tank is modeled by a single ODE, and you may first review the .o.."rponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two tanks, The model will be a system of two first-order ODEs, Tank71 anďT2inFig.77 containinitially 100galof watereach. InZlthewaterisPure,whereas 150 1bof fertilizer are dissolved itr rr.yy circulating liquid at a rate of 2 ga|lmin and stining (to keeP the mixture uniform) the amounts of fertilizer yrftl in Z1 and }z(r) in T2changewith time /. How long should we let the liquid circulate so that 71 will contain at least half as much fertilizer as there will be left in 12? Solution. Step 1. Setting up the model. As for a single tank, the time rate of change yi(r) or yl(r) equals inflowminusoutflow.SimilarlyfortankT2.FromFig.11weseethat 22 )i : Inflow/min - Outflow/min : 100 Y, - 100 Y' 22 yl : Inflow/min - Outflow/min : 100 Y' - 100 Y' Hence the mathematical model of our mixture problem is the system of first_order oDEs yi: -o.0zy, * 0.02y2 yL: 0.02y1 - 0.0Zy2 (Tank Q) (Tank Z2). (Tank Z1) (Tank 72). SEc.4.1 Systems of ODEs as Models l31 100 75 50 100 t Fig.77. Fertilizer content in Tanks T, (lower curve) and T, As a vector equation with column vector y : t;;] Step 2, General solution. As for a single equation, we try an exponential function of /, (1) y : xe^t. Then y' : lxe^' : Lxe^'. Dividing the last equation lxe^' : Axe^'by e^' and interchanging the left and right sides, we obtain Ax : .[x. We need nontrivial solutions (solutions that are not identically zero). Hence we have to look for eigenvalues and eigenvectors of A. The eigenvalues are the solutions of the characteristic equation | -0.02 - ^ 0.02 l (z) det(A- ^I): l |: (-0.02- ^)2- 0.022: ^(^+0.04):0.l o.oz -0.02 - ^ l We see that .tr1 : 0 (which can very well happen-don't get mixed up-it is eigenvectors that must not be zero) and .tr2 : -0.04. Eigenvectors are obtained from (14x) in Sec. 4.0 with i : 0 and ), : -0.04. For our present A this gives [we need only the first equation in (14i')] -0.02x1 -l 0.02x2: Q and (-0.02 + 0.04)11 -l 0.02x2: 0, respectively. Hence xI: x2 and x1 : -x), respectively, and we can take xl : x2: l and x1 : -x2: I. This gives two eigenvectors corresponding to ,tr1 : 0 and ,tr2 : -0.04, respectively, namely, and matrix A this becomes T -0.02 0.02l A:| l. L o.oz -0.02_] Y' : Ay, where *cr, _ ftl Lt_] and x,2,:[ l] From (1) and the superposition principle (which continues to hold for systems of homogeneous linear ODEs) we thus obtain a solution (3) y : ,t*G)"^'t + crx(2)"^2t - cl [l] - ,r|_'rf 'o*' where c1 and c2 dra arbitrary constants. Later we shall call this a general solution. Step 3. Use of initial conditions. The initial conditions are y1(0) : 0 (no fertilizer in tank [) and y2(0) : 150. From this and (3) with r : 0 we obtain y(0):.,[l] -" [_l] : [:l :',',f: [,:.] System of tanks y 2(t) 27,5 50 v(í) 150 2 gallmin T, T22 gallmin |_---=- -|-------=l ---- iv,(í)|" I I l32 CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods In components this is c1 l c2: 0, cr - c2: cz: -15. This gives the answer e-o.o4t, In components, y : 75x(1) _ 75x2)e-0,04t _ J5 -. -o.o4l _r't:/)-l)e -- -. -o.o4l |z -- l) Ť l)e yt: 15 - ]Se-o'oat : 50, e-o,o4t - !, Hence the fluid should circulate for at least about half an hour. lution is c1: 75, ['l - r, [ 'l Lr_] L-t] 150. The so Figure 77 shows the exponential increase of y1 and the exponential decrease of )2 to the common limit 75 lb, oid you expect this for physical reasons? Can you physically explain why the curves look "sYmmetric"? Would the limit change if 71 initially contained l00 lb of fertilizer and T2 contained 50 lb? Step 4. Answer. 71 contains half the t'ertilizer amount of T2 lf it contains 1i3 of the total amount, that is, 50 lb. Thus (Tank 71, lower curve) (Tank 72, upper curve). , : (ln 3)10.04 : 2'7.5. ExAM PLE 2 Electrical Network Find the currents 11(r) and I2Q) ínthe network in Fig. 78. Assume all currents and charges to be zero at l : 0, the instant when the switch is closed. L=lhenry C=0,25farad Switch t=O Rr = 4 ohms E = 12 volts..: c Rz = 6 ohms Fig. 78. Electrical network in Example 2 Solution. Step 1. Setting up the mathematical model. The model of this network is obtained from Kirchhoff's voltage law, as in Sec. 2.9 (where we considered single circuits). Letlr(l) ayd I2(t) be the currents in the left and right loops, respectively. In the left loop the voltage drops are LI| = /i tV] over the inductor and R1(11 - Iz) : 41It - 12) tV] over the resistor, the difference because 11 and 12 flow through the resistor in opposiie directions. By Kirchhoff's voltage law the sum of these drops equals the voltage of the batterY;that is, 1i + 4(Il * 12) : l2,hence t'r: -+t, + 4I2 + 12, ,}ňlV ! (4a) In the right loop the voltage drops are RzIz: 6I2[Yl and R1(12 - 11) (IICI 12 dt : 4 l I, dt [V] over the capacitor, and their sum is zero, 612+4(I2-11) + oÍrrclt:O or |0I2 Division by 10 and differentiation gives tL - O.+t', -l 0.412: Q. To simplify the solution process, we first get rid oíO.aI|, which l Substitution into the present ODE gives I;: 0.4I| - 0.412: 0.4(-4h + 4I2 + 12) - 0.412 : 4(.I2 - 11) [V] over the resistors and r -4I1 +oJ,rdt:0. by (4a) equals 0.4(-4It + 4I2 + l2). 133SEC. 4.1 Systems of ODEs as Models and by simplification (4b) Ii- -1.611 + 1.2I2+ 4.8. In matrix form, (4) is (we write J since I is the unit matrix) r 1,1 l- -4.0 4.0f T l2.0l (5) J':AJig. where J:|'I A:| | g:| | L,) L lo 1.2) - Los_] Step 2. Solving (5). Because of the vector g this is a nonhomogeneous system, and we try to proceed as for a single ODE, solving first the hotnog3neou., system J' : AJ (thus J' - AJ : 0) by substituting J : x"n'. This gives J':lxe^':Axe^', hence Ax : ),x. Hence to obtain a nontrivial solution, we again need the eigenvalues and eigenvectors. For the present matrix A they are derived in Example l in Sec. 4.0: trt - -2. -,,, : [i] , iz:-0.8. *,,,-['-l L0.8_] Hence a "general solution" of the homogeneous system is Jn: ctx(De-2t + ,zxQ)e-o,at, For a particular solution of the nonhomogeneous system (5), since g is constant, we try a constant column vector Jp-awithcomponentsrr1, a2.ThenJ;:0,andsubstitutioninto(5)givesAatg:O;incomponents, -4.0q*4.0a2 -F l2.0:0 -1.6q-l 1.2a2 * 4.8 : 0. The solution is c1 : 3. ct2:0: thus " - [;] . Hence (6) J: Jnl Jr: crx(l)r-zt l crx(2)"-o," + u; in components, 11: 2cp Iz: cle The initial conditions give 1r(0) : 2c1 -| c2-1 3 : 0 Iz(0): c1 *0,8c2 -0. Hence cI : -4 anď c2: 5. As the solution of our problem we thus obtain (7) J:-4x(De-2t+5x2)e-o,8'+a. In components (Fig. 79b), It: -8e-2t + se,o,Bt + 3 lz: -4e-2t 4ť-o,8t. Now comes an important idea, on which we shall elaborate further, beginning in Sec. 4.3. Figure 79a shows I1(t) and I2Q) as two separate curves. Figure 79b shows these two currents as a single curve [11(r),12(t)] in the I]2-plane, This is a parametric representation with time r as the parameter. It is often important to know in which sense such a curve is traced. This can be indicated by an arrow in the sense of increasing /, as is shown. The I1l2-plane is called the phase plane of our system (5), and the curve in Fig. 79b is called a trajectory. We shall see that such "phase plane representations" are far more important than graphs as in Fig. 79a because they will give a much better qualitative overall impression of the general behavior of whole families of solutions, not merely of one solution as in the present case. _2l -o.8r1 C2? ŤJ -2t + O.icre-o,Bt. tr /1(ť) l34 CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods I(t) 4 2 1 U Currents 1, (upper curve) and I, The first n - L of these n ODEs follow y'* : y(n) by (9), so that the last equation I2 1.5 1 0.5 ll (b) Trajectorv t1,(f), 1r(ť)ť in the /rlr-plane (the "phase plane") immediately from (9) by differentiation. in (10) results from the given ODE (8). 012 (a) Fig.79. Currents in Example 2 Conversion of an nth-Order ODE to a System We show that an nth-order oDE of the general form (8) (see Theorem 1) can be converted to a system of n first-order ODEs. This is practically and theoreticallY imPortantpractically because it permits the study and solution of single ODEs bY methods for -systems, ánd theoretically because it opens a way of including the theory of higher order ÓoB, into that of first-order systems. This conversion is another reason for the imPortance of systems, in addition to their use as models in various basic apPlications. The idea of the conversion is simple and straightforward, as follows. THEoREM l PRooF Also, l Conversion of an ODE An nth-order ODE (8) ,@) can be converted to a system (9) !t: !, lz This system is of the form : F(t,!,!',, , , ,y'n-D) of n first-order ODEs by setting : ]u', !3: !",' ",ln- r@-L) (10) lt: lz lz: ls : I !n-t: ln Y'. : F(t, Yt, jz,''', !n), SEC. 4.1 Systems of ODEs as Models EXAMPLE 3 MassonaSpring l35 To gain confidence in the conversion method, let us apply it to an old friend of ours, modeling the free motions of a mass on a spring (see Sec. 2.4) my"+cy' +lq:g or y":-!r' _Lr. mm For this ODE (8) the system (l0) is linear and homogeneous, Setting y : The characteristic equation is |-l l l llock detlA-^l):I k . l:i,+-^--:0. l; -;-o| m m It agrees with that in Sec. 2.4.For an illustrative computation,letm: 1, c :2, and : 0.75. Then ^2 + 2^ + 0.75: (^ + 0.5X^ + 1.5):6. This gives the eigenvalues ,\1 : -0.5 anď ),2: -1.5. Eigenvectors follow from the first equation in A - ^I:0,whichis -)-r1 * xz:0. Forilthisgives0.5-11 i xz:0, sá},Jr1 :2,xz - -1. Fori2: -1.5 it gives 1.5x1 -| xz: 0, say,,r1 : I, xz: -1.5. These eigenvectors Y't: Yz yL: - }r,- *r,. [Yrl l l . we get in matrix form Lyr_] ,,:o,:[ ; ]l ["l |-; -Á)Ll,) f 21 1-1l x(1)-l l. *'2':l I give L-l] L-r.s_j This vector solution has the first component ) :.Ir : 2'rn-o'" + c2e-|'5t which is the expected solution. The second component is its derivative Yz: Y' ': y' : _ clg-o'51 - L5c2e-1,5t [ 2-] -o5l I l l -r,Y:crL-,_] e +"l,_1.5_] " 3. Derive the eigenvectors in Example 1 without consulting this book. 4. In Example 1 find a "general solution" for any ratio a : (flow rate)l(tank size), tank sizes being equal. Comment on the result. 5. If you extend Example 1 by a tank 73 of the same size as the others and connected to T, by two tubes with 1. r MIXING pRoBLEMs Find out without calculation whether doubling the flow rate in Example 1 has the same effect as halfing the tank sizes. (Give a reason.) What happens in Example 1 if we replace Tzby a tank containing 500 gal of water and 150 lb of fertilizer dissolved in it? l CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods flow rates as between TlandT2,what system of ODEs will you get? 6. Find a "general solution" of the system in Prob, 5, @ ELEcTRlcAL NETwoRKs 7. Find the currents in ExampLe 2 if the initial currents are 0 and -3 A (minus meaning that 1r(0) flows against the direction of the arrow), 8. Find the cuments in Example 2 \f the resistance of R, and R, is doubled (general solution only), First, guess, 9. What are the limits of the cur:rents in Example 2? Explain them in terms of physics. 10. Find the currents in Example 2 if the capacitance is changed to C : Il5,4F (farad). EJšl coNvERsloN To sysTEMs Find a general solution of the given ODE (a) by first conveTting it to a system, (b), as given. (Show the details of your work.) ll.y" - 4y : O 12. y" -l 2y' - 24y : g 13. y" - y' : 0 14. y" + 15y' * 50y:6 '1,5. 64y" - 48y' - 7y : 0 16. TEAM PROJECT. Two Masses on Springs, (a) Set up the model for the (undamped) system in Fig, 80, (b) Solve the system of ODEs obtained, Hint, Try y : xe'' and set a2 : ).. Proceed as in Example 1 or 2, (c) Describe the influence of initial conditions on the possible kind of motions. hf 12 mt= I k,= 8 mz= I System in static equilibrium Fig. 80. Mechanical system in Team Project 16 -?Y,l L& t*.' change in -r-) spring length l \\_ = v^- v.'| -], *É--LW System in motion 4.2 Basic Theory of Systems of ODEs In this section we discuss some basic concepts and facts about sYstems of ODEs that are quite similar to those for single ODEs. The first-order systems in the last section Were Special cases of the more general system y'r -- í{t,!l, " , , !n) yi -- íz(t, yt, " , , !n) (1) Y'.: Ín(t,Yt, "' ,!n), We can write the system (1) as a vector equation by introducing the column vectors y : Lh y,rlT and f : [/, ír]'(rh.re T means transposition anď saves uS in" ,pu." that would be needed for writing y and f as columns). This gives (1) y' : f(r, y). This system (1) includes almost a1l cases of practical interest. For n : 1 it becomes y't: ír(t, y) or, simply, y' : í(t, y), well known to us from Chap, 1, Asolution of (1) on some interval a 1t < b is a setof n differentiablefunctions lt: htG), , ln: hn1) ------ SEC. 4.2 Basic Theory of Systems of ODEs on a 1 t < b that satisfy (1) throughout this interval. In vector form, introducing the "solution vector" h : lhí h.]T (a column vector!) we can write y : h(/). An initial value problem for (1) consists of (1) and n given initial conditions (2) yr(/o):Kr, yzG :K2, ",, yn(t :Kn, in vector form, y(/o) : K, where /g is a specified value of r in the interval considered and the components of K : [K, KnfT are given numbers. Sufficient conditions for the existence and uniqueness of a solution of an initial value problem (I), (2) are stated in the following theorem, which extends the theorems in Sec. 1 .7 for a single equation. (For a proof, see Ref. [A7].) THEoREM t Existence and Uniqueness Theorem Let f 1,. . , f nin(I) be continuousfunctions having continuous partial derivatives 0f1l6yy,,,, \ír/ó!n,,,,, afnlaln in some domain R oít!ůz,,, yn-space containing the point (to, Kt, , , , , Kn).Then (I) has a solution on some interval to- a Tl <__ T2 (Pure water) 64 gallmin -_+ + SEC.4.4 Criteria for Critical Points. Stability 147 in components, y'l,: anjl,* anlz yr: aztlt * azzlz. dy, : dyt yi dt (3) q : det A,: allar, - al2a21> A,: p2 - 4q. (6) Furthermore, the product representation of the equation gives ^2 - p^ + q: (^ - nrXn - b): 12 - (ir + ^2)^ + ^1^2. 4.4 Criteria for Critical Points. Stabitity We continue our discussion of homogeneous linear systems with constant coefficients (l) y' : Ay : [n" n"l ,, Lon azz_] From the examples in the last section we have seen that we can obtain an overview of families of solution curves if we represent them parametrically as y(/) : [yr(r) yzG)]T and graph them as curves in the yry2-plane, called the phase plane. Such a curve is called a trajectory of (1), and their totality is known as the phase portrait of (1). Now we have seen that solutions are of the form y(t): xe^t. Substitutioninto (1) gives y'(t): Xxe^t: Ay : Axe^t. Dropping the common factor e^t, wehaye (2) Ax : ,\x. Hence y(r) is a (nonzero) solution of (1) if ,\ is an eigenvalue of A and x a coíTesponding eigenvector. Our examples in the last section show that the general form of the phase portrait is determined to a large extent by the type of critical point of the system (1) defined as a point at which dy2tdyl becomes undetermined, 0/0; here [see (9) in Sec. 4.3] azlyt * arryz y: anlt * anjz We also recall from Sec. 4.3 that there are various types of critical points, and we shall now see how these types are related to the eigenvalues. The latter are solutions ), : .tr1 and .[, of the characteristic equation larr-tr atz l (4) det(A-^I): | |:tr"-(ar,-la22))" *detA:0. I o^ or,-X| This is a quadratic equation ^2 - p^ + q: O with coefficients p, q and discriminant A given by (5) p:attlorr, From calculus we know that the solutions of this equation are it: ž@+ VÁ), Xz: Žrp - VÁl. l48 DEFlNlTloNs (7) CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Hence p is the sum and q the product of the eigenvalues. Also ^1 - Together, ),2 : ÝT from (6). P--\lXr, 4: XtLz, A : (ňr - Xr)', for classifying critical points. A derivation will beThis gives the criteria in Table 4.1 indicated later in this section. Table 4.1 Eigenvalue Criteria for Critical Points (Derivation after Table a.2) Stabitity Critical points may also be classified in terms of their stability. Stability concepts are basic in engineering and other applications. They are suggested by physics, where stabilitY means, roughly speaking, that a small change (a small disturbance) of a physical system at some instant changes the behavior of the system only slightly at all future times /. For critical points, the following concepts are appropriate. Stable, Unstable, Stable and Attractive A critical point Po of (1) is called stablez if, roughly, all trajectories of (1) that at some instant are close to Po remain close to Pg at all future times; precisely: if for every disk D. of radius e ) 0 with center P6 there is a disk Du of radius 6 > 0 with center Pg such that every trajectory of (1) that has a point P, (corresponding to t : tr, say) in Du has all its points coíTesponding to t =- tttn D.. See Fig. 89. Po is called unstable if P6 is not stable. P6 is called stable and attractive (or asymptotically stable) if P0 is stable and every trajectory that has a point in Du approaches P6 as t ---> @. See Fig. 90. Classification criteria for critical points in terms of stability aíe given in Table 4.2. Both tables are summarizeďin the stability chart in Fig. 91. In this chart the region of instability is dark blue. 2In the sense of the Russian mathematician ALEXANDER MICHAILOVICH LJAPUNOV (l857-19l8), whose work was fundamental in stability theory fbr ODEs. This is perhaps the most appropriate definition of stability (and the only we shall use), but there are others, too. Name p: h1 l )"2 Q: Xtlz A:(i, -lr)' Comments on ,[1, .[2 (a) Node (b) Saddle point (c) Center (d) Spiral point p:0 p+0 q>0 q<0 q>0 ^=0 A<0 Rea1, same sign Real, opposite sign Pure imaginary Complex, not pure imaginary "------I SEC.4.4 Criteria for Critical Points. Stability Fig. 89. Stable critical point Po of (1) (The trajectory initiating at P' stays in the disk of radius e.) Table 4.2 Stability Criteria for Critical Points Fig. 90. Stable and attractive critical point Po of (1) 149 Fig. 9l. Stability chart of the system (l) with p, Q, L defined in (5). Stable and attractive: The second quadrant without the g-axis. Stability also on the positive g-axis (which corresponds to centers). Unstable: Dark blue region We indicate how the criteria in Tables 4.1 and 4.2 are obtained. If q: A1^2 > 0, both eigenvalues are positive or both are negative or complex conjugates. If also p : lt * Lz ( 0, both are negative or have a negative real part. Hence P6 is stable and attractive. The reasoning for the other two lines in Table 4.2 is similar. If A < 0, the eigenvalues are complex conjugates, say, ht : ot + iB anď hz: ot - iB. If also p : Ll t lz : 2a ( 0, this gives a spiral point that is stable and attractive. If p : 2d ) 0, this gives an unstable spiral point. If p:0, then Lz: -),1 and Q: htlz- -ll,'.If also qž 0, then ),12 - -q 10, so that .tr1, and thus ),2, must be pure imaginary. This gives periodic solutions, their trajectories being closed curves around P6, which is a center. EXAMPLE 1 Application of the Criteria in Tables 4.1 and 4.2 In Example l, Sec.4.3, we have y' : [-' '-l ,,, : -6, Q:8, L: 4,a node by Table 4.I(a),which L1-3] is stable and attractive by Table 4.2(a). l Type of Stability p:hllX2 Q: ltLz (a) Stable and attractive (b) Stable (c) Unstable p<0 l q>0 p=0 l qr} p>0 OR q<0 Spiral poi nt o,o/ o // ,O Node Spiral point Saddle point 150 ExAMPLE 2 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Free Motions of a Mass on a Spring What kind of critical point does my" + cy' + ky : 0 in Sec. 2,4 have? Solution. Division by ru gives ,l' : -(klm)y - (clm)y'. To get a system, set y1 : |, lz : y' (see Sec. 4.1). Then yL: y" : -(ktm)y1- @lm)y2. Hence '' : |-oo,* -',,*f ', |-l 1 l _ c k ctettA-.trI):l l:tr"+ -^+-:0.' |-rt. -ctn - tl m m Weseethatp:-clm,q:klm,6:1clm)2-4ktm.FromthisandTables4.1 and4.2weobtainthefollowing results. Note that in the last three cases the discriminant A plays an essential role. No damping. c : 0, p : 0, q } 0,a center. Ilnderdamping. c2 1 4mk, p 1 O, q > 0,A < 0, a stable and attractive spiral point, Critical damping. c2 : 4mk, p 1 0, q ž0,A : 0, a stable and attractive node. overdamping. 12 > 4-k, p < 0, q žO,A > 0, a stable and attractive node. l r TypE AND sTABlLlTy oF cRlTlcAL polNT Determine the type and stability of the critical point. Then find a real general solution and sketch or graph some of the trajectories in the phase plane. (Show the details of your work.) 1. Y', : 2Yz yL: 8y._ 3. yi :zyl + jz yL : lt ]- Zyz 5. y', : -4yl * lz !z: lt - 4!z 7. yr: -Zyz yL: 8y, 9. y', : yt * 2yz yL: 2yt * lz @ FoRM oF TRAJEcToRIEs What kind of cuíves are the trajectories of the following ODEs in the phase plane? ll.y" + 5y' :0 II.y"-k'y:0 12.y" + #y:0 13. (Damped oscillation) Solve y" + 4y' i 5y : 0. What kind of cuíves do you get as trajectories? 1,4. (Transformation of variable) What happens to the system (1) and its critical point if you introduce T : -t as a new independent variable? 15. (Types of critical points) Discuss the critical points in (10)-(14) in Sec. 4.3 by applying the criteria in Tables 4.I and4.2in this section. 16. (Perturbation of center) If a system has a center as its critical point, what happens if you replace the matrix A by Á : A * kI with any real number k + 0 (representing measurement elTors in the diagonal entries)? 17. (Perturbation) The system in Example 4 in Sec. 4.3 has a center as its critical point. Replace each a7, in Example 4, Sec. 4,3,by ayx -| á. Find values of b such that you get (a) a saddle point, (b) a stable and attractive node, (c) a stable and attractive spiral, (d) an unstable spiral, (e) an unstable node. 18. CAS EXPERIMENT. Phase Portraits. Graph phase portraits for the systems in Prob. 17 with the values of b suggested in the answer. Try to illustrate how the phase portrait changes "continuously" under a continuous change of á. 19. WRITING EXPERIMENT. Stability. Stability concepts are basic in physics and engineering. Write a two-part report of 3 pages each (A) on general applications in which stability plays a role (be as precise as you can), and (B) on material related to stability in this section. Use your own formulations and examples; do not copy. 20. (Stability chart) Locate the critical points of the systems (10)-(14) in Sec, 4.3 and of Probs. 1, 3, 5 in this problem set on the stability chart. 2. Y', : 4Yt ll jz: 5!z 4. Y', : Y, yL: -5y, -Zyz 6. y _ : y, -l 10y2 yL: ]yt - 8y, 8. yl : 3y1 * 5y, yL:-5yt-3y, SEC. 4.5 Qualitative Methods for Nonlinear Systems 151 (1) qualitative information on solutions are particularly valuable for systems impossible. This is the case for many y|: f {yt, yz) yL: fz(yt,y). y| : allt -l anlz -l h{yt, yz) yl, : aztlt * azzlz * hz(yl, y). is autonomous. One can prove the following App. 1). 4.5 Qualitative Methods for Nonlinear Systems Qualitative methods are methods of obtaining without actually solving a system. These methods whose solution by analytic methods is difficult or practically important nonlinear systems (2) y' : Ay + h(y), thus A is constant (independent of r) since (1) (proof in Ref. [A7], pp. 375-388, listed in y' : f(y), thus In this section we extend phase plane methods, as just discussed, from linear systems to nonlinear systems (1). We assume that (1) is autonomous, that is, the independent variable / does not occur explicitly. (A11 examples in the last section are autonomous.) We shall again exhibit entire families of solutions. This is an advantage over numeric methods, which give only one (approximate) solution at a time. Concepts needed from the last section are the phase plane (the yp2-plane), trajectories (solution curves of (1) in the phase plane), the phase portrait of (1) (the totality of these trajectories), and critical points of (1) (points (yr y) at which both /1(y1, y2) and f z(yt, yz) are zerc). Now (1) may have several critical points. Then we discuss one after another. As a technical convenience, each time we first move the critical point Pg:. (a, b) tobe considered to the origin (0, 0). This can be done by a translation l,: lt - a, z:lz-b which moves Po to (0, 0). Thus we can assume Po to be the origin (0, 0), and for simplicity we continue to write jy lz (instead of r, z). We also assume that P6 is isolated, that is, it is the only critical point of (1) within a (sufficiently small) disk with center at the origin. If (1) has only finitely many critical points, this is automatically true. (Explain!) Linearization of Nonlinear Systems How can we determine the kind and stability property of a critical point Pg: (0, 0) of (1)? In most cases this can be done by linearization of (1) near P6, writing (1) as y' : f(y) : Ay + h(y) and dropping h(y), as follows. Since Po is critical, /1(0, 0) : 0, f z(0,0) : 0, so that /1 and f 2have no constant terms and we can write l52 THEoREM l ExAMPLE t CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods Free Undamped Pendulum. Linearization Figure 92a shows a pendulum consisting of a body of mass m (the bob) and a rod of length L, Determine the locations and types oť the critical points, Assume itrat tne mass of the rod and air resistance are negligible, Solution. Step L Setting up the mathematical model. Let 0 denote the angular displacement, measured counterclockwise tiom the equilibrium position. The weight of the bob is mg (g the acceleration of gravitY), It callses a restoring fbrce nzg sin 0 tangent to the curve of motion (circular arc) of the bob. BY Newton's second 1aw, at each instant this 1brce is balanced by the íbrce of acceleration mL7" , where L0" is the acceleration; hence the resultant oť these two tbrces is zero, and we obtain as the mathematical model ttlLe"' lilgsiná:0. 0"+ksin0:0 when 0 is very small, we can approximate sin 0 rather accurately by 0 and obtain as an approximate sollltion A;;::{k, + b sin xTt,wutthe exact solution for any 0is not an elementarY function, Step 2. Critical points (0,0), !(2rrr0), +(4Ťrr0), , , , ,Linearization,To obtain a System of oDEs, We Set á ] r,r,0' : ),z.Then í}om (4) we obtain a nonlinear system (t) of the íbrm l" }r:Jr()r,jz'-lz l?,r jz: Jz(jt !z| : -k Sln )'1. Therightsidesarebothzerowhen'1',:0andsin1'':0.Thisgivesinfinitelymanycriticalpoints(nrr,0), where n:0, _|I, )_2,. . . . We consider (0,0). Since the Maclaurin series is siny1 : yr - áyr3 + - , , , :.}1, the linearized system at (0, 0) is ,Y't: Yz yL: -kyt ToapplyourcriteriainSec.4,4wecalculatep:at|lazz:O,q:detA:k:glL(>0),and L' :;Ž'_ +i : -+t .From this and Table 4.1(c) in Sec. 4.4 we conclude that (0, 0) is a center, which is alwaYs stable. Since sin 0 : siny1 is periodic with period 2rr,the critical Points (nrr,0), n: +2, !4,'", are all centerS. Step 3. Critical points i(rr,0), !(3Í,r0), =(5r,0), , , , ,Linearization"ý{e now consider the critical point (rr,, O), setting 0 - t: }1 and (0 - ,T)' : 0' : y2, Then in (4), sin 0: sin(,v1 * r): -sin_y1 : -J,1 + áyr'- + , ,'- -}1 Dividing this by ruL, we haye (4) (-: ;) (4") y':Ay:t: ;] ,, thus Linearization If f 1 and f z i, (1) are continuous and neighborhood of the criticctl point P6: (0, and stability of the critical point oí (|) system hctve continuous partial derivatives in a 0), and ií'detA + 0 in (2), then the kind are the same as those of the linearized (3) Y' : AY, thus y|: alyt YL : aztYt -l anjz -l azzlz, Exceptions occwr if A, has equal or pure imaginary eigenvalues; then (l) may have the iame kincl of critical point as (3) or a spiral point, SEC. 4.5 Qualitative Methods for Nonlinear Systems and the linearized system at (rr, 0) is now . 1-0 ll yi:y, v':Av-| lv. thus Lr 0] " yL: kyt. We see that p : O, q : -ft (< 0), and L : - 4q : 4k. Hence, by Table 4.1(b), this gives a saddle point. which is always unstable. Because of periodicity, the critical points (nn,0), n: + 1, -|3, , . ., are all saddle points. These results agree with the impression we get fiom Fig. 92b. l (a) Pendulum (b) Solution curves yr(y,) of (4) in the phase plane Fig. 92. Example l (C will be explained in Example 4.) EXAM PLE 2 Linearization of the Damped Pendulum Equation To gain further experience in investigating critical points, as another practically important case, let us see how Example 1 changes when we add a damping term c0' (clamping proportional to the angular velocity) to equation (4), so that it becomes (5) 0"+c0'+tsin0:0 wherek) 0andc > 0 (which includes ourpreviouscaseof nodamping, c:0). Setting 0: yt,0' : y2,as before, we obtain the nonlinear system (use 0" : ,!1 Y'l : Yz yi: -k sin y1 - cy2, We see that the critical points have the same locations as before, namely, (0,0), (*zr,0), (X.2r,0), . . .. We consider (0,0). Linearizing sin)1 :,vt as in Example 1, we get the linearized system at (0,0) , T 0 ll yi:y,(6) v:Av:| lv. thus L-t -c_] yL - -lcyt - c!z. This is identical with the system in Example 2 of Sec 4,4, except for the (positive!) íactor m (and except for the physical meaning of yr). Hence for c : 0 (no damping) we have a center (see Fig. 92b), for small damping we have a spiral point (see Fig. 93), and so on. We now consider the critical point (z,,0). We set 0 - r: !t, G - i)' : 0' : y2andlinearize sin 0 : sin (y1 + rr): -siny1 : -yt. This gives the new linearized system at (rr,0) |-0 l1 y':Ay:I lv. Lr -c,] 153 I lt: |z thus yL: kyt - c|z. (6") 154 Lotka-Volterra Population Model ExAMPLE 3 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods ForourcriteriainSec 4.4wecalculate p: a,ltl a22: -c,q: detA: -k,and L,- p2 - 4q:,2 + 4k, This gives the following results for the critical point at (n 0). No damping. c : 0, p : 0, q 1 0, A > 0, a saddle point. See Fig. 92b. Damping. c ) 0, p 10,q <0, A > 0, a saddlepoint. SeeFig.93. Since sin y1 is periodic with period 2rr, the critical points (*2t, 0), (.+4rr,0), , , , are of the same type aS (0,0), and the critical points (-rr, 0), (t3r,0), . , , are of the same type as (z,,0), so that our task is finished. Figure 93 shows the trajectories in the case of damping. What we see agrees with our physical intuition. Indeed, damping means loss of energy. Hence instead of the closed trajectories of periodic solutions in Fig. 92b we now have trajectories spiraling around one of the critical points (0, 0), (+2r,0), , , , . Even the wavY trajectories corresponding to whirly motions eventually spiral around one of these points. Furthermore, there are no more trajectories that connect critical points (as there were in the undamped case for the saddle points). l Fig. 93. Trajectories in the phase plane for the damped pendulum in Example 2 Predator-Prey Population Model3 This model concerns two species, say, rabbits and foxes, and the foxes prey on the rabbits. Step 1. Setting up the model. We assume the following. 1. Rabbits have unlimited food supply. Hence if there were no foxes, their number yl(r) would grow exponentially, yI : ayt, Actually, y1 is decreased because of the kill by foxes, say, at a rate proportional to y|yz, where y2(r) is the number of foxes. Hence y|: "yt - byl,yz, where a > 0 and Ď ) 0. If there were no rabbits, then y2Q) would exponentially decrease to zero, yL: -lyz. However, y2 is increased by a rate proportional to the number of encounters between predator and prey; together we have y!2 : - llz + kltlz,where k > 0 and l > 0. This gives the (nonlinear!) Lotka-Volterra system y|: í{yt, yz) : ayt - by z yL: f z(,yt, yz) : kyůz - lyz . 3lntroduced by ALFRED J. LOTKA (1880-1949), American biophysicist, and VITO VOLTERRA (1860-1940), Italian mathematician, the initiator of functional analysis (see [GR7] in App. 1). 2. 3. (7) SEC. 4.5 Qualitative Methods for Nonlinear Systems 155 Step 2. Critical poinl (0, 0), Linearizati.on. We see from (7) that the critical points are the solutions of (7*) "fr(yl, y): yt@ * by): o, ízOr y) : y2&yr - /) : 0. la The solutions are (y1, yz) : (0, 0,1 and (;, ; ). We consider (0, 0). Dropping -byůz anď lcy12 from (7) gives,kb the linearized system |" ol ,. L0 -l) Step 3. Cňtical point (llk, alb), Linearization. We set.}r : }r -l llk, y2: z + alb.Then the critical point (llk, alb) corresponds to (}r,Iz) : (0, 0). Since 'r: y'r, L: yL, we obtain from (7) [factorized as in (8)] ,i- (,,,-;) |"-,(o,- t)] : (,,- i) (-b ) ,;: (,,- ;) [-(o,- i) -,] : (o, - í)o,, Dropping the two nonlinear terms -b!1}2and l$1 2, we have the linearized system (7xx.. (a) (b) -l lb- }r:-TY, -l ak!z: , |tThe left side of (a) times the right side of (b) must equal the right side of (a) times the left side of (b), ak_ _, lb_ _, bYIYI: - n!z!z. By integration, ak_, lb_, Vr- t -Yo-: CollSl. DK This is a family ellipses, so that the critical point(llk, alb) of the linearized system (7*x) is a center (Fig. 9a). It can be shown by a complicated analysis that the nonlinear system (7) also has a center (rather than a spiral point) at (llk, alb) surrounded by closed trajectories (not ellipses). We see that the predators and prey have a cyclic variation about the critical point. Let us move counterclockwise around the ellipse, beginning at the right vertex, where the rabbits have a maximum number. Foxes are sharply increasing in number until they reach a maximum at the upper vertex, and the number of rabbits is then sharply decreasing until it reaches a minimum at the left vertex, and so on. Cyclic variations of this kind have been observed in nature, for example, for lynx and snowshoe hare near the Hudson Bay, with a cycle of about 10 years. For models of more complicated situations and a systematic discussion, see C. W. Clark, Mathematical Bioeconomics (Wiley, I9'7 6). yI Fig. 9a. Ecological equilibrium point and trajectory of the [inearized Lotka-Volterra system (7**) l L h 156 CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods Transformation to a First-order Equation in the phase plane Another phase plane method is based on the idea of transforming a second-order autonomóus oDE (an oDE in which r does not occur explicitly) F(y, y', )") : 0 to first order by taking ) : }r as the independent variable, setting y' : yzand transforming y" by the chain rule, ll I _} -J2 dy, : dy, dyt : dy, dt dy, dt dy, !z. Then the ODE becomes of first order, r(lr,lr, and can sometimes be solved or treated by direction fields. We illustrate this for the equation in Example l and shall gain much more insight into the behavior of solutions, ExAMPLE 4 An oDE (8) for the Free Undamped Pendulum If in (4) 0,, + ksin 0 : 0 we set 6 : ),r, 0, : yz(the angular velocity) and use #r,):o(8) 0,, : ,ly: _ dyz 9_L : bro We get * ,r: -k sin)t. dt clyt dt dyt " - tlyt Separation of variables gives y2 d,yz: -k sin yl dyt By integration, (9) Multiplying this by tttLz, we get Lyrr: kcosy1 t C (C constant). }m(Ly2S2 - mL2k cos y1 : mL2C, We see that these three terms are energies. Indeed, y2 is the angular velocity, so that I,y2 is the velocitY and the first term is the kinetic energy. The second term (including the minus sign) is the potential energY of the Pendulum, unÁ *ric is its total energy, which is constant, as expected fiom the law of conservation of energy, because there is no damping (no loss of energy). The type of motion clepends on the total energY, hence on C, as follows, Figure 92b on p. 153 shows trajectories íb.ua.ious values of C. These graphs continue periodically with pe.ioa 2rr to the left and to the right. We see that some of them are ellipse-like and closed, others are Wavy, and there are two trajectories (passing through the saddle pclints (nt, O), n : + 1, t3, , , ,) that Separate those two types of trajectories. From (9) we see that the smallest possible C is C - -k', then -Y2 : 0, and cos y1 : 1, so that the pendulum is at rest. The penclulum will change its direction of motion if there are Points oir,Ín'r"n y|r: o|i : o.Tr.nkcosy1 + C: Oby(9). If )l: n,thencos}1 : -l and C: k, Henceií _k k, then y2 : 0 is impossible and the pendulum makes a whirly motion that appears aS a WavY trajectorY in the yly2_plane. Finally, the value C : kcorresponds to the two "separating trajectories" in Fig. 92b connecting the saddle points. '- -.----"r'--- ' l The phase plane method of deriving a single first_order equation (8) may be of practical interest not ónly *h"n (8) can be solved (as in Example 4) but also when solution is not Possible and we have to utilize direction fields (Sec. 1.2). We illustrate this with a very famous examPle: SEC.4.5 Qualitative Methods for Nonlinear Systems 157 ExAMPLE 5 Self-Sustained Oscillations. Van der Pol Equation There are physical systems such that for snrall oscillations, energy is fed into the system, whereas for large oscillations, energy is taken from the system. In other words, large oscillations will be damped, whereas for small oscillations there is "negative damping" (feeding of energy into the system). For physical reasons we expect such a system to approach a periodic behavior, which will thus appear as a closed trajectory in the phase plane, called a limit cyc|e. A diííbrentialequation describing such vibratiorrs is the famous van der Pol equationa (10) )," - lr(l -._,,,2),),'*.r,:0 (p > 0,constant). It first occurred in the study of electrical circr"rits containing vacuum tubes. For LL : 0 this equation becomes y"+y:Oandweobtainharmonicoscillations.Letp)O.Thedampingtermhastheí'actor_ p(l _ y2). This is negative for small oscillations, when y2 ( 1, so that we have "negative damping"'is zero íbr v2 : l (no damping), and is positive if y2 > 1 (positive damping, loss of energy). If pl is small, we expect a limit cycle that is almost a circ|e because then our equation diíTers but little íiom ,y" i ,l, : 0. If p is large, the limit cycle will probably look different. Setting) : }r, y' : yz and using ,l| - (dy2ldl,r).yz as in (8), we have tiom (l0) (11) The isoclines in the )r_yz-plane (the phase plane) are the curves d1,2ldy1 : K : consl, that is, 92 : LL(l - y]| -tly l Solving algebraically for _y2, we see that the isoclines are given by )'r ,,) o p(l -.r,r') - K K- r / ,^ K=O _5 K=-I Fig. 95. Direction field for the van der Pol equation with p : 0.] in the phase plane, showing also the limit cycle and two trajectories. See also Fig. 8 in Sec.'1.2. agALtH.qsAR VAN DER PoL (l889-1959), Dutch physicist and engineer. ":U.Jz (Figs. 95, 96). Yr) 5 K=-5 5 Y1" K=l K=i -0 K=-l K=i K=I 158 CHAP. 4 Systems of oDEs. Phase Plane. Qualitative Methods Figure 95 shows some isoclines when p is small, p:0.1. the limit cycle (almost a circle), and two (blue) trajectories approaching it, one from the outside and the other from the inside, of which onlY the initial Portion, a small spiral, is shown. Due to this approach by trajectories, a limit cycle differs conceptually from a closed curve (a trajectory) surrounding a center, which is not approached by trajectories, For larger p the limit cycle no longer resembles a circle, and the trajectories approach it more rapidly than for smaller p, Figure 96 illustrates this for pL : 1. K=-I Fig. 96. Direction field for the van der Po[ equation with p : ] in the phase plane, showing also the limit cycle and two trajectories approaching it @ cilTlcAL polNTs, LlNEARlzATloN Determine the location and type of all critical points by linearization. In Probs.1_I2 first transform the oDE to a system. (Show the details of your work.) * yzz 2. y', : 4yt - !t2 yL: y, 4. y'r: -3yl i lz - jzz I yL: yl - 3y, !z2 6. y', : lz - !z2 yL:yt-!t2 8.y"t9y+y2:0 13. (Trajectories) What kind of curves are the trajectories of yy" l 2yl2 : Ql 14. (Trajectories) Write the ODE y" - 4y * y' :0 as a system, solve it for y2 as a function of )r, and sketch or graph some of the trajectories in the phase plane, 15. (Trajectories) What is the radius of a real general solution of y" + y : 0 in the phase plane? 16. (Trajectories) In Prob. 14 add a linear damping term to 8et y" + 2y' - 4y + y3 : 0, Using arguments from mechanics and a comparison with Prob, 14, as well as with Examples 1 and 2, guess the type of each critical point. Then determine these types by linearization, (Show all details of your work.) 9.y"*cosy:0 II.y"+4y-y3:0 10. y" + siny:0 12.y"+y'+2y-y':o I.Y'r:Yr, ljz : 3!t 3. y, : 4yz yL:2yt-ytz 5.yi :-|t-|lz- yL:-yt-yz 7.y"iy-4y':o SEC. 4.6 Nonhomogeneous Linear Systems of ODEs 17. (Pendulum) To what state (position, speed, direction of motion) do the four points of intersection of a closed trajectory with the axes in Fig.92b correspond? The point of intersection of a wavy curve with the y2-axis? 18. (Limit cycle) What is the essential difference between a limit cycle and a closed trajectory surrounding a center? 19. CAS EXPERIMENT. Deformation of Limit Cycle. Convert the van der Pol equation to a system. Graph the limit cycle and some approaching trajectories for LL: 0.2,0.4,0.6,0.8, 1.0, I.5,2.0. Try to observe how the limit cycle changes its form continuously if you vary p, continuously, Describe in words how the limit cycle is deformed with growing p. 20. TEAM PROJECT. Self-sustained oscillations. (a) Van der Pol Equation. Determine the type of the critical point at (0, 0) when p ž O, L,,: 0, p, { 0. 159 Show that if p ---> 0, the isoclines approach straight lines through the origin. Why is this to be expected? (b) Rayleigh equation. Show that the so-called Rayleigh equations y"-p(í-ly'')y' * :0 (p>0) also describes self-sustained oscillations and that by differentiating it and setting ! : Y' one obtains the van der Pol equation. (c) Duffing equation. The Duffing equation is y"+,o'y*Éy3:0 where usually |B| is small, thus characterizing a small deviation of the restoring force from linearity. B > 0 and B { 0 are called the cases of a hard spring and a soft spring, respectively. Find the equation of the traiectories in the phase plane. (Note that for B > 0 all these curves are closed.) 4.6 Nonhomoteneous Linear Systems of ODEs In this last section of Chap. 4 we discuss methods for solving nonhomogeneous linear systems of ODEs y' : Ay + g (see Sec. 4.2) where the vector g(r) is not identically zero. We assume g(/) and the entries of the n X n matrix A(r) to be continuous on some interval "/ of the /-axis. From a general solution y'n'(t) of the homogeneous system y' : Ay on J and a particular solution y(e)lr; of (1) on J |i.e., a solution of (1) containing no arbitrary constants], we get a solution of (1), y:y&)ly@) y is called a general solution of (1) on ,/ because it includes every solution of (1) on .I. This follows from Theorem 2 in Sec. 4.2 (see Prob. 1 of this section). Having studied homogeneous linear systems in Secs. 4.I-4.4, our present task will be to explain methods for obtaining particular solutions of (1). We discuss the method of undetermined coefficients and the method of the variation of parameters; these have counterparts for a single ODE, as we know from Secs. 2.] and 2.|0. 5LORD RAYLEIGH (JOHN WILLIAM STRUTT) (1842-1919), great English physicist and mathematician, professor at Cambridge and London, known by his important contributions to the theory of waves, elasticity theory, hydrodynamics, and various other branches of applied mathematics and theoretical physics. In 1904 he received the Nobel Prize in physics. (1) (2) (6) l CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Method of undetermined coefficients As for a single ODE, this method is suitable if the entries of A are constants and the components of g are constants, positive integer powers of /, exponential functions, or cosines and sines. In such a case a particular solution y(P) is assumed in a form similar to g; for instance, y(P) : u f v/ l wtz if g has components quadratic in /, with u, v, w to be determined by substitution into (1). This is similar to Sec. 2.7, except for the Modification Rule. It suffices to show this by an example. ExAMPLE l Method of Undetermined Coefficients. Modification Rule Find a general solution of [-: tl l--ol (3) y':Ay+g:l lv+l le-z'.' L l -3_] " L z) Solution. A general equation of the homogeneous system is (see Example 1 in Sec. 4.3) |-tl l- tl (4) y"":.,| |r-2'+r2| lr-+t. Lr_] 'L-t_] Since,tr : -2 is an eigenvalue of A, the function ,-2'onthe right also appears in y"', and we must apply the Modification Rule by setting to rc ,n ue suffi ondil than be s, ). cient ng to Ut-U2:-a_6 -Ul * uz: -a _l 2. Byaddition,O:-2a-4,a- -2,andthenuz:ut+4,say,ut:k,uz:k+4,thus,v:[t k+4]T We can simply choose r : 0. This gives íhe answer (5) y: y,h)+ y,o,:., ['-l "_ L1] For other fr we get other v; for instance, fr : (5*) y : cl |'l ,-" * ,, LlJ [l], _2t -i|- tl2tl1_4t"|-czl le-"-2 L-1] -2 gives v : |-2 2]T, |- tl l-tl l lr-a'-2l lrc L-t_] Lt] -2t [o-.l -2t* lo)' , so that the answer becomes |-:]"" etc. l Method of variation of parameters This method can be applied to nonhomogeneous linear systems y' : A(r)y + g(r) u-2v:Avi t :] thus [;] l:,,,,1: [-:;,_-,',,,f* [ ;] SEC. 4.6 Nonhomoteneous Linear Systems of ODEs l61 It yields a particular solution y(P) of (6) on some solution of the homogeneous system y' : A(r)y in terms of the previous example. e-2tlT andlr_at - e-4tfT .Hence with variable A : A(r) and general g(r). open interval J on the /-axis if a general on "I is known. We explain the method . EXAMPLE 2 Solution by the Method of Variation of Parameters Solve (3) in Example 1. Solution. A basis of solutions of the homogeneous system isfr-2t the general solution (4) of the homogenous system may be written (,]) y(h) -|",',', ",nn'uf[.;] : Y(r)c. Here, Y(t) : [y"' ,tzl]T is the fundamental matrix (see Sec. 4.2). As in Sec.2.10 we replace the constant vector c by a variable vector u(0 to obtain a particular solution y(Pl : Y(r)u(r). Substitution into (3) y' : Ay i g gives (B) Y'u + Yu' : AYu * g. Now since y(1) and yQ) are solutions of the homogeneous system, we have y(1)/ * Ay"', ,(2)| - lrrQ), thus y' : Ay. Hence Y'u : AYu, so that (B) reduces to Yu' : 9. The solution is u' : Y*lg; here we use that the inverse Y- 1 of Y (Sec. 4.0) exists because the determinant of Y is the Wronskian lV, which is not zero for a basis. Equation (9) in Sec.4.0 gives the form of Y-1, ,lrz-], I - -k-' We multiply this by g, obtaining f -st -st1 |-e -e 1_1 l ^_zt ^-zt l 2 L-e e -J ,fo2' o2,ff-oo-r'f l[-4 1 [-rlu' : Y-lg : 1|,o, _,n,)L ,n,,): ' L _8,,,): |_od,] f zt zt1 |',n -',n ) Integration is done componentwise fiust as differentiation) and gives r'f -2 l l- -2t l u(/) lo|_o,ri )o' : l__r,r, * r) (where i 2 comes from the lower limit of integration). From this and Y in (7) we obtain ",,: ["-rt ,-+tl f _z, l t-2te-2t - 2e-2t + 2e-all:l-r, - 11 ,_r, *l 2f n_n,. Lr_2, _r_n,)l_r.rr, * r): l__z,r-r, + 2e-2t - zr-n ): l--z, - zl L- z_J The last term on the right is a solution of the homogeneous system. Hence we can absorb it into y(b). We thus obtain as a general solution of the system (3), in agreement with (5x), (9) Y: cr [l] ,-" *., [ _'rf n* - r|',f ,ru -|-'r]r" 162 CHAP, 4 Systems of oDEs. Phase Plane. Quatitative Methods 1,. (General solution) Prove that (2) includes every solution of (1). @ GENERAL soluTloN Find a general solution. (Show the details of your work,) 14. y| : 3yt - 4y, * 20 cos r yL: yl - 2y, yr(0) : 0, y2(0) : 8 15. y'l : 4y2 l 3e3t yL :2y, - I5e-3t yr(O) : 2, y2(0) : 2 'r:4y, f 8y, * 2cost 2.y'r:lzlt yL: yt - 3t 3.yi:4yz*9t yL:-4y1 *5 4. y'r: jl * lz * 5 cos t 5. y't: y'r,:3yt- !z- 5sinr yL: 6. y|: -)r * 1.2 * e-zt , -rtV, : -|1 - Y2 - 7. y', : -I4:yt * 10y2 + 162 yL:*5y,.*y2-324t S. yi: 10y, - 6y, + 10(l - t - yL:6yr-l0y2*4-20tg. y't: -3)r - 4yz + 11í + 15 yL : 5.y1 + 6yz -| 3e-t - I5t 2y, ]- 2y2 + 12 5yr-}z-30 t2) 6t2 -20 10. CAS EXPERIMENT. Undetermined Coefficients, Find out experimentally how general you must choose y(o), in particular when the components of g have a different form (e.g., as in Prob. 9). Write a short report, covering also the situation in the case of the modification rule. @ lNlTlAL vALuE IRoBLEM Solve (showing details): 11. y't : -2y, * 4t yL:2yt - 2t yr(O) : 4, lz (0) : + 12. y't : 4y, _l 5et " - 2Oe-ty2 - ]l yr(0) : 1, y2(0) : 0 13. y|:yt*Zyz+e2t-2t "/ : _ ,, + l + /y2 - 12 yr(O) : 1, y2(0) - -4 !zL:6yr*Zyricos/yr(0) : 15, yz(0) : 13 17. (Network) Find the currents in Fig. 97 when R: 2,5 a, L : I H, C :0.04 F, E(t) : 845 sin / V, and {(0) : 0, 1z(0) : 0. (Show the details.) 18. (Network) Find the currents in Fig. 97 when R : 1 C), L : 1,0H, C : I.25 F, E(t) : 10 kV, and 11(0) : 0, /z(0) : 0. (Show the details,) Fig. 97. Network in Probs. 17, 18 19. (Network) Find the currents in Fig. 98 when R1: 2 O, Rz : 8 a, L : ll1', C : 0.5 F, E : 200 V, (Show the details.) Fig. 98. Network in Prob. ]9 20. WRITING PROJECT. Undetermined Coefficients, Write a short report in which you compare the application of the method of undetermined coefficients to a single ODE and to a system of two ODEs, using ODEs and systems of your choice. 16. }r - 16 sinr 14 sin r Chapter 4 Review Questions and Problems I63 1. State some applications that can be modeled by systems of ODEs. 2. What is population dynamics? Give examples. 3. How can you transform an ODE into a system of ODEs? 4. What are qualitative methods for systems? Why are they important? 5. What is the phase plane? The phase plane method? The phase portrait of a system of ODEs? 6. What is a critical point of a system of ODEs? How did we classify these points? 7. What are eigenvalues? What role did they play in this chapter? 8. What does stability mean in general? In connection with critical points? 9. What does linearization of a system mean? Give an example. 10. What is a limit cycle? When may it occuf in mechanics? FIJE GENERAL soluTloN. cRlTlcAL polNTs Find a general solution. Determine the kind and stability of the critical point. (Show the details of your work.) :4yt*3y2*2 : -6yt - 5y, l 4e-t :.}t - 2y, - sint :3yt-4yr-cos/ :}r*2y2-1 t2 :2yt*yr-t' 2ó. (Mixing problem) Tank T1 in Fig. 99 contains initially 200 gal of water in which 160 lb of salt are dissolved. TankTrcontains initially 100 gal of pure water. Liquid is pumped through the system as indicated, and the mixtures are kept uniform by stining, Find the amounts of salt yl(r) and yr(t) in 7, and T2, respectively. Fig. 99. Tanks in Problem 26 27. (Critical point) What kind of critical point does y' : Ay have if A has the eigenvalues -6 and 1? 28. (Network) Find the currents in Fig. 100, where Rt : 0.5 C), Rz : 0.7 O, Ll : 0-4 H, Lz: 0.5 H, E : I kV : 1000 V, and 1.(0) : 0, 12(0) : 0. LI L2 R2 Fig.l00. Network in Problem 28 (Network) Find the currents in Fig. 101 when R : 10 C), L : I.25 H, C :0.002 F, and 1r(0) : 12(0) : 3 A. Fig.1O1. Network in Problem 29 )r !z _Yr jz !t jz 25. 11, y', : 4yz Y', : I6Y, 12. y', : 9y,s, 13. 1,6.15. 18.17. , )r !z }r jz }r !z :lz : 6yl, - 5yz : 1.5yr - 6y, : -4.5yt 1- 3yz : 3yr -l 2yz : Zyt -f 3yz r a-,yl I Ly2 a-.Lll 12 :lz : 3yt - 3yz : 3y1 t 3y, : -3yt - Zyz : -Zyt - 3y, : 3yt ]- 5yz : -5yr - 3yz !z Ia. y', I !z jt !z }r !z 19. Y', !z @ NoNHoMoGENEoussysTEMs Find a general solution. (Show the details.) 29. 20. y't : 3y2 * 6t yL: I2y1 + I 22. y', : y, * y2 -l sin / yL : 4y, -| y, 2L. y', : yr, t 2y, + ezt .. l I.5e-2ty2 - 12 | Water, 10 gallmin M ixture, 10 gallmin TlONS AND PROBLEMS 6 gallmin T,T, + <- 16 gallmin ---.__+ .> 164 cHAp. 4 Systems of oDEs. Phase Plane. Qualitative Methods LINEAR|ZAT!oN Determine the location and kind of all critical points of the given nonlinear system by linearization. 32, y| : cos y, |:J' z : J.\'r 33. y', : y, - Zyr' yL:yt-Zyr' 30. Y', : Y, y'r:4yl,-yr3 31. y', : -9y, I !z: SlnY1 Whereas single electric circuits or single mass-spring systems are modeled bY single oDEs (Chap. 2), networks of several circuits, systems of several masses and sPrings, and other engineering problems lead to systems of ODEs, involving several unknown functions y{t), . . . , !nG). Of central interest are first-order systems (Sec. 4.2): y' : f(r, y), in components, Y'r -- í{t,!|, " ' , !n) : jn: Ín(t,Yt" ",!n), to which higher order ODEs and systems of ODEs can be reduced (Sec. 4.1). In this summary we let n : 2, so that (1) y' : f(r, y), in components, yi -- ítG, yt, yz) yL : ízG, !t, !z) Then we can represent solution curves as trajectories in the Phase Plane (the yly2-p1ane), investigate their totality [the "phase portrait" of (1)], and studY the kiná a"d stability of the critical points (points at which both f1 and f 2 are zero), and classify them as nodes, saddle points, centers, or spiral points (Secs. 4.3,4.4). These phase plane methods are qualitative; with their use we can discover various general properties of solutions without actually solving the system. TheY are primarily used for autonomous systems, that is, systems in which / does not occur explicitly. A linear system is of the form If g : 0, the system is called homogeneous and is of the form (3) y' : Ay. (2)y,:Aytg, where A- |"o','o oo',',7 r:[;] , 8:[;] of ODEs. Phase Plane. Qualitative Methods l65Summary of Chapter 4 Ifa1,...,az2areconstants,ithassolutionsy:xe^,wherer\isasolutionofthe quadratic equation larr-i atz l l l:(a1-L)(azz-^)-al2a21:0 I o1 azz-h| and x * 0 has components .T1, x2 determined up to a multiplicative constant by (al - i)x1 * apx2 -- 0. (These ň's are called the eigenvalues and these vectors x eigenvectors of the matrix A. Further explanation is given in Sec. 4.0.) A system (2) with g + 0 is called nonhomogeneous. Its general solution is of the form y : yn * yo, where yn is a general solution of (3) and yo a particular solution of (2). Methods of determining the latter are discussed in Sec. 4.6. The discussion of critical points of linear systems based on eigenvalues is summarized in Tables 4.1 and 4.2tn Sec.4.4. It also applies to nonlinear systems if the latter are first linearized. The key theorem for this is Theorem 1 in Sec. 4.5, which also includes three famous applications, namely the pendulum and van der Pol equations and the Lotka_Volterra predator-prey population model.