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Algeo is a graphing calculator. It has some notable features: calculate derivatives, definite integrals and Taylor-series of functions. What's New in 0.7.3: · added tangent What's New in This Release: · added function tracing Like it? Share with your friends! Other Android Freeware of Developer «Algeo»: Not found Similar Applications:Algeo graphing calculator Algeo is a graphing calculator with the ability to calculate derivatives. With it you can easily draw functions, find intersections in the trace mode and show a table of values Matrix Calculator This calculator performs all matrix, vector operations. You can add, subtract, find length, find dot and cross product, check if vectors are dependant. For every operation, calculator will generate a detailed explanation Ticker Tape Calculator Simple calculator, but keeps a ticker tape that saves all your calculations, like an accountant's calculator that spits out ticker tape	677.169	1
How to improve calculation speed during Examination The Vedic maths means, the way of doing mathematics in smarts and shorter way. Vedic comes from Vedas, In India the oldest and most knowledge scripture in the form of literature is available and that are known as Vedas which consists of practical lessons of life and high value teachings. Since from the ancient times Indians are well/god enough in maths and give numerous inventions and contribution to world mathematics. In today's education system, in India students not allowed to use calculators till 12th class (senior secondary schooling) But their syllabus which is mostly derive and having patterns of Europe and American education system where student can use calculators, graphers etc electronic machines after 5th – standard. No doubts foreign students are average has poor solving skills as compared to Indian students. But Indians students have to face Tough challenge while solving complex calculation especially in 9th & 10th class in physics numerical like etc. 11th & 12th class in mathematics, physics, chemistry or Accounts.	677.169	1
Differential Equations: I hate them. Yep. Taking the course now, and out of all my math courses (calc 1-3, matrix theory&linear algebra , abstract algebra I & II, linear programming, and real analysis) I'd have to say that it's the worst. Not because it's difficult, but because it's so boring. It seems to me, and this probably has a lot to do with my particular class, that the entire thing is set up as a recipe book. You look at the DE, decide what grouping it falls under, and apply repeated mechanical mindless steps to eventually get to a meaningless (and often rather ugly) answer. I'll conceed that they are useful, but mathematically speaking, especially at the procedural level I'm exposed to at the moment, they are no fun at all. the book i have by arnold is titled ordinary differential equations, mit press, copyright 1973, isbn 0-262-51018-9 paperback, and is much cheaper than the newer slightly expanded edition. the older book by witold hurewicz is also excellent but this is more modern and clearer, less technical. also my little post here titled trivial solution process is an attempt to introduce some logic and beauty into one case of solving de's, the easiest one. at least if you know some linear algebra, and how to find an inverse from a minimal polynomial it will speak to you.	677.169	1
Math.NET aims to provide a self contained clean framework for symbolic mathematical (Computer Algebra System) and numerical/scientific computations, including a parser and support for linear algebra, complex differential analysis, system solving and more	677.169	1
[center] Just in Time Math makes cramming for your test quick and easy with focused lessons and targeted practice. Topics covered include: fractions, decimals, ratios, absolute value, integers, percents, square roots, probability, and word problems. JIT Math covers everything you need to know about math basics that you can expect to see on college and graduate entrance exams, as well as on vocational and civil service tests. [center]	677.169	1
Asymptotes Matching PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 1.21 MB \| 5 pages PRODUCT DESCRIPTION COMMON CORE ALIGNED! Asymptotes Matching is an interactive and hands on way for students to practice finding asymptotes. Students find the horizontal and vertical asymptotes of six functions algebraically and match each to the correct graph of the function. This activity can be used in a variety of ways including as an interactive notebook page, class pairs activity, homework assignment, or individual assessment of student	677.169	1
Pre-Calculus: Trigonometry Pre-Calculus: Trigonometry Pre-Calculus: Trigonometry 加州大学尔湾分校关于此课程: This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles. This course is designed to help prepare students to enroll for a first semester course in single variable calculus. In this module, you will get an overview of the course and the foundations of trigonometry. We will also begin exploring angles and different systems for angle measure. 5 视频, 4 阅读材料已评分: Angles and their Measure WEEK 2 The Unit Circle and Solving Right Triangles In this module, we will explore circles and right triangles. We will see several special angles related to particular right triangles and we will learn how to find measurements of sides and angles in right triangles using trigonometric functions. 4 视频, 1 阅读材料已评分: Unit Circle and Solving Right Triangles WEEK 3 Properties of Trigonometric Functions In this module we will explore several properties of trigonometric functions and discover how to compute values of these functions given information about an angle or a unit circle point. 3 视频, 1 阅读材料已评分: Properties of Trigonometric Functions WEEK 4 Inverse Trigonometric Functions We will now explore the inverse trigonometric functions. These are useful to go backwards - we will seek to find the angles which produce a given value of a trigonometric function. 3 视频, 1 阅读材料已评分: Inverse Trigonometric Functions WEEK 5 Basic Trigonometric Identities I There are several useful trigonometric identities which allow us to simplify trigonometric expressions and find values for the trigonometric functions beyond the special angles. We will begin by exploring the sum and difference identities. Warning: Generally, ... 4 视频, 1 阅读材料已评分: Basic Trigonometric Identities I WEEK 6 Basic Trigonometric Identities II In this module, we continue our exploration of trigonometric function identities. We will begin by learning how to verify such identities. We will then talk about the double-angle and half-angle identities. 4 视频, 1 阅读材料已评分: Quiz: Basic Trigonometric Identities II WEEK 7 Trigonometric Equations In this module, we will focus on solving equations involving trigonometric functions. These are usually equations in which the variable appears inside of a trigonometric function and we must use a combination of algebra skills and trigonometry manipulation to ... 4 视频, 1 阅读材料已评分: Trigonometric Equations WEEK 8 Law of Sines and Law of Cosines The Law of Sines and the Law of Cosines give useful properties of the trigonometry functions that can help us solve for unknown angles and sides in oblique (non-right angle) triangles. We will focus on utilizing those laws in solving triangles, including those... 6 视频, 1 阅读材料已评分: Law of Sines and Law of Cosines WEEK 9 Trigonometry Final Exam We have completed the new content for the course. In this final module, you will review and practice the topics covered throughout the course. You will end by taking the comprehensive final exam. 已评分: Trigonometry 85 评分 RM A nice effort. good if more interactive lessons good review. Very useful course. I was able to use both The Trig and Functions courses to brush up on my math for my college assessment exams. That allowed me to skip all of the remedial courses that I would have been otherwise required to take. h Very clear and concise lessons. I learned so much. But I wish there were lessons for graphing trigonometric functions.	677.169	1
Solving word problems has never been easier than with Schaum's How to Solve Word Problems in Algebra ! This popular study guide shows students easy ways to solve what they struggle with most in algebra: word problems. How to Solve Word Problems in Algebra , Second Edition, is ideal for anyone who wants to master these skills. Completely updated,...	677.169	1
Mathematics in Games, Sports, and Gambling The Games People Play Along with exercises in every chapter, this text introduces students to probability, statistics, and elementary mathematics through the use of many motivating examples from games, sports, and gambling. It covers permutations in the two-deck matchingMore... Along with exercises in every chapter, this text introduces students to probability, statistics, and elementary mathematics through the use of many motivating examples from games, sports, and gambling. It covers permutations in the two-deck matching game so derangements can be counted, introduces graphs to find matches when looking at extensions of the five-card trick, and studies lexicographic orderings and ideas of encoding for card tricks. The text also explores linear equations and weighted equations in the section on the NFL passer rating formula and presents graphing to show how data can be compared or displayed. A solutions manual is available for qualifying	677.169	1
Product Description: Symplectic geometry is a central topic of current research in mathematics. Indeed, symplectic methods are key ingredients in the study of dynamical systems, differential equations, algebraic geometry, topology, mathematical physics and representations of Lie groups. This book is a true introduction to symplectic geometry, assuming only a general background in analysis and familiarity with linear algebra. It starts with the basics of the geometry of symplectic vector spaces. Then, symplectic manifolds are defined and explored. In addition to the essential classic results, such as Darboux's theorem, more recent results and ideas are also included here, such as symplectic capacity and pseudoholomorphic curves. These ideas have revolutionized the subject. The main examples of symplectic manifolds are given, including the cotangent bundle, Kahler manifolds, and coadjoint orbits. Further principal ideas are carefully examined, such as Hamiltonian vector fields, the Poisson bracket, and connections with contact manifolds. Berndt describes some of the close connections between symplectic geometry and mathematical physics in the last two chapters of the book. In particular, the moment map is defined and explored, both mathematically and in its relation to physics. He also introduces symplectic reduction, which is an important tool for reducing the number of variables in a physical system and for constructing new symplectic manifolds from old. The final chapter is on quantization, which uses symplectic methods to take classical mechanics to quantum mechanics. This section includes a discussion of the Heisenberg group and the Weil (or metaplectic) representation of the symplectic group. Several appendices provide background material on vector bundles, on cohomology, and on Lie groups and Lie algebras and their representations. Berndt's presentation of symplectic geometry is a clear and concise introduction to the major methods and applications of the subject, and requires only a minimum of prerequisites. This book would be an excellent text for a graduate course or as a source for anyone who wishes to learn about symplectic geometry. REVIEWS for An Introduction to Symplectic	677.169	1
MATH Algebra Documents Showing 1 to 1 of 1 Bronda Williams Math 126 Survey of Mathematical Methods Instructor: Dan Urbanski 5/13/2013 We were taught to develop our skills through talking, reading, and writing. The class am taking right now is a math class and this class I will learn how to speak, Showing 1 to 2 of 2 Math has always been a weakness of mine that I struggle with a lot, I have always been more gifted with reading and writing, so math I have always found rather difficult, no matter the subject. Course highlights: I was forced to push myself to extreme limits and create new boundaries while expanding my knowledge with the subject of math. The teacher created a fun learning environment while encouraging learning and providing positive encouragement. Hours per week: 3-5 hours Advice for students: Study and take advantage of free resources, you can never be too prepared, or study too much or too hard. Never not complete assignments, take advantage of every opportunity handed to you and run with it. Course Term:Summer 2015 Professor:Cortes Course Required?Yes Course Tags:Math-heavyParticipation Counts Jan 23, 2016 \| Would recommend. This class was tough. Course Overview: Algebra 2 isnt easy but its neccesary to do well if you want to excel at a higher level. Course highlights: The advancedments of algebra, the connection of graphing and formulas. Hours per week: 0-2 hours Advice for students: Try your best, it will be hard. IF you do not have a strong teacher SWITCH OUT. It is crucial that you are taught everything you need to know. Fall in love with it too, that helps.	677.169	1
Books By Author Clarence Raymond Wylie Geared toward students preparing to teach high school mathematics, this text explores the principles of Euclidean and non-Euclidean geometry and covers both generalities and specifics of the axiomatic method. 1964 edition. Author Biography - Clarence Raymond Wylie The late C. R. Wylie, Jr., taught for many years at the University of Utah, where he was Chairman of the Department of Mathematics. A Dover Original Clarence Raymond Wylie, Jr., had a long career as a writer of mathematics and engineering textbooks. His Advanced Engineering Mathematics was the standard text in that field starting in the 1950s and for many decades thereafter. He also wrote widely used textbooks on geometry directed at students preparing to become secondary school teachers, which also serve as very useful reviews for college undergraduates even today. Dover reprinted two of these books in recent years, Introduction to Projective Geometry in 2008 and Foundations of Geometry in 2009. The author is important to our program for another reason, as well. In 1957, when Dover was publishing very few original books of any kind, we published Wylie's original manuscript 101 Puzzles in Thought and Logic. The book is still going strong after 55 years, and the gap between its first appearance in 1957 and Introduction to Projective Geometry in 2008 may be the longest period of time between the publication of two books by the same author in the history of the Dover mathematics program. Wylie's 1957 book launched the Dover category of intriguing logic puzzles, which has seen the appearance of many books by some of the most popular authors in the field including Martin Gardner and, more recently, Raymond Smullyan. Here's a quick one from 101 Puzzles in Thought and Logic: If it takes twice as long for a passenger train to pass a freight train after it first overtakes it as it takes the two trains to pass when going in opposite directions, how many times faster than the freight train is the passenger train? Answer: The passenger train is three times as fast as the freight train	677.169	1
Summary and Info This book provides a pedagogical and comprehensive introduction to graph theory and its applications. It contains all the standard basic material and develops significant topics and applications, such as: colorings and the timetabling problem, matchings and the optimal assignment problem, and Hamiltonian cycles and the traveling salesman problem, to name but a few. Exercises at various levels are given at the end of each chapter, and a final chapter presents a few general problems with hints for solutions, thus providing the reader with the opportunity to test and refine their knowledge on the subject. An appendix outlines the basis of computational complexity theory, in particular the definition of NP-completeness, which is essential for algorithmic applications. More About the Author Jean-Claude Fournier (French: [fuʁnje]; born 21 May 1943, Paris), known simply as Fournier, is a French cartoonist best known as the comic book artist who handled Spirou et Fantasio in the years 1969-1979.	677.169	1
What is STEP? STEP is an additional mathematics examination, taken at the end of Y13, which forms part of conditional offers to applicants for mathematics and some related degrees at Cambridge and some other universities. STEP is designed to test candidates on questions that are similar in style to undergraduate mathematics. Although demanding, preparing for STEP gives you a chance to engage with interesting, challenging mathematics, to stretch yourself and to develop the mathematical reasoning, independence and confidence needed in an undergraduate mathematics degree. Free online resources to support STEP preparation Developed by the Faculty of Mathematics and NRICH, this new online STEP Support programme is designed to help potential university applicants develop their advanced problem-solving skills and prepare for sitting STEP Mathematics examinations. It builds on an earlier national pilot, led by Dr Stephen Siklos and funded by the DfE. The STEP Support programme includes online modules for individual additional study, starting in the summer of Y12, and an online discussion forum. The resources are free and open to everyone. Each module assignment starts with some warm-up exercises, followed by preparatory work leading to a STEP question. Supportive self-evaluation and development material will be published after each assignment, to help you assess your progress and identify areas that need more work. The programme also includes an online discussion forum to support students preparing for STEP. You can ask for help and hints if you're stuck, share what you have tried so far, offer advice to others and get guidance from Cambridge University students who've taken STEP themselves. Additional supporting resources These additional complementary resources are designed to help you develop your advanced problem solving skills and mathematical thinking. Free online Advanced Problem-Solving resources from NRICH The NRICH project, based at the University of Cambridge, has published a collection of free online advanced problem solving resources. The material has been carefully selected to provide an accessible and supportive introduction to advanced problem solving, and to help you build your confidence, fluency and speed. Advanced Problems in Core Mathematics book A book by Dr Stephen Siklos, analysing recent STEP questions selected to address the syllabus for Papers I and II, which is the A-level core (i.e. C1 to C4) with a few additions. Each question is followed by a comment and a full solution. The comments direct the reader's attention to key points and put the question in its true mathematical context. The solutions point students to the methodology required to address advanced mathematical problems critically and independently. The book can be downloaded as a free pdf; this revised and extended version can also be bought as a paperback.	677.169	1
Product Description: Knots and links are studied by mathematicians, and are also finding increasing application in chemistry and biology. Many naturally occurring questions are often simple to state, yet finding the answers may require ideas from the forefront of research. This readable and richly illustrated 2004 book explores selected topics in depth in a way that makes contemporary mathematics accessible to an undergraduate audience. It can be used for upper-division courses, and assumes only knowledge of basic algebra and elementary topology. Together with standard topics, the book explains: polygonal and smooth presentations; the surgery equivalence of surfaces; the behaviour of invariants under factorisation and the satellite construction; the arithmetic of Conway's rational tangles; arc presentations. Alongside the systematic development of the main theory, there are discussion sections that cover historical aspects, motivation, possible extensions, and applications. Many examples and exercises are included to show both the power and limitations of the techniques developed. Additional Details 21 Day Unconditional Guarantee REVIEWS for Knots and Links	677.169	1
Browse by ELEMENTARY DIFFERENTIAL EQUATIONS SlXTH EDitiON Ci I'!! I . . 1Y' ELEMENTARY DIFFERENTIAL EQ.Ui\TI(): NS stXTHmmoN I, 1 .,. I ~ '\_< ' lo ... EDWARDS & PENNEY EDWARDS & PENNEY ABOUT THE COVER. This image illustrates the trajectory of a moving point whose space coordinates satisfy (as functions of time) the Rossler system of differential equations that is discussed on page 553, and which originated in studies of oscillations in chemical reactions. In its motion along its trajectory the point may appear to spiral repeatedly around a set - the so-called Rossler bond - that somewhat resembles a (twisted) Mobius strip in space. To portray the progress of the moving point, we can regard its trajectory as a necklace string on which beads are placed to mark its successive positions at fixed increments of time (so the point is moving fa.stest where the spacing between beads is greatest). In order to aid the eye in following the moving point's progress, the color of the beads changes continuously with the passage of lime and motion along the trajectory. As the point travels around and around the band, it may be observed to drift radially back and forth across the band in an apparently unpredictable fashion . Two points that start from nearby initial positions may loop around and around the band somewhat in synchrony, while moving radially in quite different ways, so that their trajectories diverge appreciably with the passage of time. This illustrates the phenomenon of chaos, in which tiny differences in initial conditions can result in great differences in the resulting situations some time later. Further discussion of chaos associated with differential equations can be found in Section 7.6. Throughout this textbook computer-generated graphics are used to portray numerical and symbolic solutions of differential equations vividly and to provide additional insight. r- m m m C HENRY ~ EDWARDS & DAVID E. z ~ :::0 -< :::0 "'" m m "'" )> z c PENNEY -I rm /:) c: ~ 0 V" Upper Saddle River, NJ 07458 .r i)Tioof ·.t .L... ~~~~~ ISBN~13 87S.0.13·23t730·$ • ... SIXTH - z ELEMENTARY DIF·FERENT'IAL EQUATIONS ELEMENTARY Dl FFER ENTIAL EQUATIONS Sixth Edition C. Henry Edwards David E. Penney The University of Georgia with the assistance of he evolution of the present text in successive editions is based on experience teaching the introductory differential equations course with an emphasis on conceptual ideas and the use of applications and projects to involve students in active problem-solving experiences. At various points our approach reflects the widespread use of technical computing environments like Maple, Mathematica, and MATLAB for the graphical, numerical, or symbolic solution of differential equations. Nevertheless, we continue to believe that the traditional elementary analytical methods of solution are important for students to learn and use. One reason is that effective and reliable use of computer methods often requires preliminary analysis using standard symbolic techniques; the construction of a realistic computational model often is based on the study of a simpler analytical model. T While the successful features of preceding editions have been retained, the exposition has been significantly enhanced in every chapter and in most individual sections of the text. Both new graphics and new text have been inserted where needed for improved student understanding of key concepts. However, the solid class-tested chapter and section structure of the book is unchanged, so class notes and syllabi will not require revision for use of this new edition. The following examples of this revision illustrate the way the local structure of the text has been augmented and polished for this edition. Chapter 1: New Figures 1.3.9 and 1.3.10 showing direction fields that illustrate failure of existence and uniqueness of solutions (page 24); new Problems 34 and 35 showing that small changes in initial conditions can make big differences in results, but big changes in initial conditions may sometimes make only small differences in results (page 30); new Remarks 1 and 2 clarifying the concept of implicit solutions (page 35); new Remark clarifying the meaning of homogeneity for first-order equations (page 61); additional details inserted in the derivation of the rocket propulsion equation (page 95), and new Problem 5 inserted to investigate the liftoff pause of a rocket on the launch pad sometimes observed before blastoff (page 97). Chapter 2: New explanation of signs and directions of internal forces in mass-spring systems (page 101); new introduction of differential operators and clarification of the algebra of polynomial operators (page 127); new introduction and illustration of polar exponential forms of complex numbers (page 132); fuller explanation of method of undetermined coefficients in Examples 1 and 3 (page 149-150); new Remarks 1 and 2 introducing "shooting" terminology, and new Figures 2.8.1 and 2.8.2 illustrating why some endpoint vii viii Preface value problems have infinitely many solutions, while others have no solutions at all (page 181 ); new Figures 2.8.4 and 2.8.5 illustrating different types of eigenfunctions (pages 183-184). Chapter 3: New Problem 35 on determination of radii of convergence of power series solutions of differential equations (page 218); new Example 3 just before the subsection on logarithmic cases in the method of Frobenius, to illustrate first the reduction-of-order formula with a simple non-series problem (page 239). Chapter 4: New discussion clarifying functions of exponential order and existence of Laplace transforms (page 273); new Remark discussing the mechanics of partial-fraction decomposition (page 279); new much-expanded discussion of the proof of the Laplace-transform existence theorem and its extension to include the jump discontinuities that play an important role in many practical applications (page 286-287). Chapter 5: New Problems 20-23 for student exploration of three-railwaycars systems with different initial velocity conditions (page 392); new Remark illustrating the relation between matrix exponential methods and the generalized eigenvalue methods discussed previously (page 416); new exposition inserted at end of section to explain the connection between matrix variation of parameters here and (scalar) variation of parameters for second-order equations discussed previously in Chapter 3 (page 427). Chapter 6: New discussion with new Figures 6.3.11 and 6.3.12 clarifying the difference between rotating and non-rotating coordinate systems in moonearth orbit problems (page 473). Chapter 7: New remarks on phase plane portraits, autonomous systems, and critical points (page 488-490); new introduction of linearized systems (page 502); new 3-dimensional Figures 6.5.18 and 6.5.20 illustrating Lorenz and Rossler trajectories (page 552-553). Throughout the text, almost 550 computer-generated figures show students vivid pictures of direction fields, solution curves, and phase plane portraits that bring symbolic solutions of differential equations to life. About 15 application modules follow key sections throughout the text. Their purpose is to add concrete applied emphasis and to engage students is more extensive investigations than afforded by typical exercises and problems. A solid numerical emphasis provided where appropriate (as in Chapter 6 on Numerical Methods) by the inclusion of generic numerical algorithms and a limited number of illustrative graphing calculator, BASIC, and MATLAB routines. Organization and Content The traditional organization of this text still accommodates fresh new material and combinations of topics. For instance: • The final two sections of Chapter 1 (on populations and elementary mechanics) offer an early introduction to mathematical modeling with significant applications. • The final section of Chapter 2 offers unusually early exposure to endpoint Preface ix • • • • • problems and eigenvalues, with interesting applications to whirling strings and buckled beams. Chapter 3 combines a complete and solid treatment of infinite series methods with interesting applications of Bessel functions in its final section. Chapter 4 combines a complete and solid treatment of Laplace transform methods with brief coverage of delta functions and their applications in its final section. Chapter 5 provides an unusually flexible treatment of linear systems. Sections 5.1 and 5.2 offer an early, intuitive introduction to first-order systems and models. The chapter continues with a self-contained treatment of the necessary linear algebra, and then presents the eigenvalue approach to linear systems. It includes an unusual number of applications (ranging from brine tanks to railway cars) of all the various cases of the eigenvalue method. The coverage of exponential matrices in Section 5.7 is expanded from earlier editions. Chapter 6 on numerical methods begins in Section 6.1 with the elementary Euler method for single equations and ends in Section 6.4 with the RungeKutta method for systems and applications to orbits of comets and satellites. Chapter 7 on nonlinear systems and phenomena ranges from phase plane analysis to ecological and mechanical systems to an innovative concluding section on chaos and bifurcation in dynamical systems. Section 7.6 presents an elementary introduction to such contemporary topics as period-doubling in biological and mechanical systems, the pitchfork diagram, and the Lorenz strange attractor (all illustrated with vivid computer graphics). This book includes adequate material for different introductory courses varying in length from a single term to two quarters. The longer version, Elementary Differential Equations with Boundary Value Problems (0-13-600613-2), contains additional chapters on Fourier series methods and partial differential equations (including separation of variables and boundary value problems). To sample the range of applications in this text, take a look at the following questions: • What explains the commonly observed lag time between indoor and outdoor daily temperature oscillations? (Section 1.5) • What makes the difference between doomsday and extinction in alligator populations? (Section 1.7) • How do a unicycle and a two-axle car react differently to road bumps? (Sections 2.6 and 5.5) • Why are flagpoles hollow instead of solid? (Section 3.6) • If a mass on a spring is periodically struck with a hammer, how does the behavior of the mass depend on the frequency of the hammer blows? (Section 4.6) • If a moving train hits the rear end of a train of railway cars sitting at rest, how can it happen that just a single car is "popped" off the front end of the second train? (Section 5.5) X Preface • How can you predict the time of next perihelion passage of a newly observed comet? (Section 6.4) • What determines whether two species will live harmoniously together, or whether competition will result in the extinction of one of them and the survival of the other? (Section 7.4) • Why and when does non-linearity lead to chaos in biological and mechanical systems? (Section 7.6) The answer section has been expanded considerably to increase its value as a learning aid. It now includes the answers to most odd-numbered problems plus a good many even-numbered ones. The 605-page Instructor's Solutions Manual (0-13600614-0) accompanying this book provides worked-out solutions for most of the problems in the book, and the 345-page Student Solutions Manual (0-13-6006159) contains solutions for most of the odd-numbered problems. The approximately 15 application modules in the text contain additional problem and project material designed largely to engage students in the exploration and application of computational technology. These investigations are expanded considerably in the 320-pageApplications Manual (0-13-600679-5) that accompanies the text and supplements it with about 30 additional applications modules. Each section in this manual has parallel subsections "Using Maple," "Using Mathematica," and "Using MATLAB" that detail the applicable methods and techniques of each system, and will afford student users an opportunity to compare the merits and styles of different computational systems. In preparing this revision we profited greatly from the advice and assistance of the following very capable and perceptive reviewers: Raymond A. Claspadle, University of Memphis Semion Gutman, University of Oklahoma Miklos Bona, University of Florida It is a pleasure to (once again) credit Dennis Kletzing and his extraordinary TJ3Xpertise for the attractive presentation of both the text and the art in this book. Finally, but far from least, I am especially happy to acknowledge a new contributor to this effort, David Calvis, who assisted in every aspect of this revision and contributed tangibly to the improvement of every chapter in the book. C. H. E. h.edwards@rnindspring.corn : E· ····· M r EL . ' v ; ,: E·~···· ·· N 1 r· ·A ~- R; · · y~ . ' " ' ·- ~ . . ·, . D .r· I•f: ... F :. . . E ..-·.R ; -· E :_ .. . ...... . . ,. J . , ~-. N : ;r: ·~. A L :" . . . I! ' . .. · .·AT ..· . t. 0 . . N· .S EQ ' u · • 1 ; ! . : l l i . <>· - ', ''". . ·-·'' ; _. JI j-' » ,, ··~" i .. ' it is natural that equations involving derivatives are frequently used to describe the changing universe. 3. Algebra is sufficient to solve many static problems. 1 . To find-either exactly or approximately. An equation relating an unknown function and one or more of its derivatives is called a differential equation. Because the derivative dxjdt = f'(t) of the function f is the rate at which the quantity x = f (t) is changing with respect to the independent variable t. y of the independent variable x and the first two • The study of differential equations has three principal goals: 1. To discover the differential equation that describes a specified physical situation. Example 1 T The differential equation dx . The differential equation d 2y dy dx2 + 3 dx + 7 y = 0 involves the unknown function derivatives y' and y" of y.the appropriate solution of that equation. To interpret the solution that is found. 2. but the most interesting natural phenomena involve change and are described by equations that relate changing quantities.First-Order Differential Equations he laws of the universe are written in the language of mathematics.=x2 +t2 dt involves both the unknown function x(t) and its first derivative x'(t) = dxjdt. Eq.11x +41 = 0.4) it can be shown that every solution of the differential equation in (2) is of the form in Eq.=2xy dx for all x. (3) FIGURE 1. dT dt = -k(T . if possible. That is.A). then dTjdt < 0. where k is a positive constant. one for each choice of the arbitrary constant C.1). • -···· Example 4 Torricelli's law implies that the time rate of change of the volume V of water in a draining tank (Fig. Example 2 If C is a constant and y(x) = Cex . (4) . 2 (1) then ~~ = c (2xex2 ) = (2x) ( cex 2 ) = 2xy.1. we are challenged to find the unknown functions y = y (x) for which an identity such as y'(x) = 2xy(x)-that is.1. Eq. so that Tis increasing. in solving a differential equation. Example 3 Newton's law of cooling may be stated in this way: The time rate of change (the rate of change with respect to timet) of the temperature T(t) of a body is proportional to the difference between T and the temperature A of the surrounding medium (Fig. we will want to find all solutions of the differential equation. the differential equation - dy =2xy dx -holds on some interval of real numbers. By contrast. so the temperature is a decreasing function of t and the body is cooling. but we will see numerous examples in which some quantity other than time is the independent variable.1. (3). describes the cooling of a hot rock in water. In each of these examples the independent variable is time t. If we are given the values of k and A. Observe that if T > A. 1. Thus the physical law is translated into a differential equation. 1. we should be able to find an explicit formula for T(t). • Differential Equations and Mathematical Models The following three examples illustrate the process of translating scientific laws and principles into differential equations. In particular.1. Ordinarily. (1). and then-with the aid of this formula-we can predict the future temperature of the body. Newton's law of cooling. Thus every function y(x) of the form in Eq. But if T < A.2 Chapter 1 First-Order Differential Equations In algebra. (1) satisfies-and thus is a solution of-the differential equation dy (2) .2) is proportional to the square root of the depth y of water in the tank: dV dt = -ky'Y. By the method of separation of variables (Section 1. then dT jdt > 0. we typically seek the unknown numbers that satisfy an equation such as x 3 + 7x 2 . (1) defines an infinite family of different solutions of this differential equation. Because substitution of each function of the form given in (7) into Eq. then V = Ay. the predicted number of bacteria in the population after one and a half hours (when t = 1.ik. In this case Eq.! ch~ng~--~T~-p~p{. Torricelli's law of draining. If the tank is a cylinder with vertical sides and cross-sectional area A. (6) produces an identity. • dP dt = kP. We can use this particular solution to predict future populations of the bacteria colony.1.nst~tbirthandd.~~P-(t)~ith c.~-th_r _ atesi -s. This is typical of differential equations.!II]tflll!llr.!!!lljl!!ll!ii"-J.5) = 1000 · 2312 ~ 2828. Substitution of k = ln 2 and C = 1000 in Eq. • . Let us discuss Example 5 further. h) was 1000.ti~~ ~~. one for each choice of the "arbitrary" constant C. For instance. Suppose that P (t) = C ekt is the population of a colony of bacteria at time t.1 Differential Equations and Mathematical Models 3 where k is a constant. (6). Note first that each function of the form P(t) = Cek1 • (7) is a solution of the differential equation FIGURE 1.. so dVjdt =A· (dyjdt). because it may allow us to use additional information to select from among all these solutions a particular one that fits the situation under study.693147)P. That is. the differential equation dPjd t = k P has infinitely many different solutions of the form P (t) = C ekt. describes the draining of a water tank. dP -=kP dt in (6). Eq. 2000 = P(l) = Cek. It follows that C = 1000 and that ek = 2. This additional information about P(t) yields the following equations: = Ce 0 = C. (4) takes the form dy dt = -hy'y.693147. even if the value of the constant k is known.I.~!llfb. Thus. (7) yields the particular solution P(t) = 1000e(ln 2)1 = 1000(e1n2 ) 1 = 1000 · 21 (because e1n2 = 2) that satisfies the given conditions.. all such functions are solutions of Eq. We verify this assertion as follows: P'(t) = Ckekt = k (Cek1 ) = kP(t) for all real numbers t.5) is P(l.2. proportional to the size of the population. (4). (6) ll~!":l. that the population at timet = 0 (hours. and that the population doubled after 1 h. It is also fortunate.1. so k of k the differential equation in (6) is 1000 = P(O) = ln2 ~ 0. in many simple cases. where k is the constant of proportionality. With this value dP dt = (ln2)P ~ (0...-. (5) where h = kjA is a constant. Because exactly one solution passes through each such point. If.5) ~ 2828 bacteria after 1. answering the question originally posed." Figure 1. The analysis or solution of the resulting mathematical problem. The interpretation of the mathematical results in the context of the original real-world situation-for example. the real-world problem is that of determining the population at some future time. P(O) = 1000. describing the bacteria population of Example 6.4). 3.3 shows several different graphs of the form P (t) = C ek1 with k = ln 2. we see in this case that an initial condition P (0) = Po determines a unique solution agreeing with the given data. which involves the following: 1. P(O) = Po) that are known or are assumed to hold. that is. Finally. For instance. the selection of any one point Po on the P-axis amounts to a determination of P(O).4. As an example of this process. The mathematical analysis consists of solving these equations (here. 2. our prediction may be quite accurate.5 hours. Moreover. The formulation of a real-world problem in mathematical terms.1. The graphs of all the infinitely many solutions of dPjdt = kP in fact fill the entire two-dimensional plane. FIGURE 1.1. together with one or more equations relating these variables (dPjdt = kP . for no choice of the constants C and k does the solution P (t) = C ek1 in Eq. In the population example. for P as a function oft). A mathematical model consists of a list of variabies (P and t) that describe the given situation. The process of mathematical modeling. 6 4 2 -2 -4 -6 -8~~~~-r~~-7--~ The condition P(O) = 1000 in Example 6 is called an initial condition because we frequently write differential equations for which t = 0 is the "starting time. in which case we conclude that the mathematical model is quite adequate for studying this particular population.4 Chapter 1 First-Order Differential Equations C= 12 C= 6 C = 3 8.1.----. C=-12 -2 Mathematical Models Our brief discussion of population growth in Examples 5 and 6 illustrates the crucial process of mathematical modeling (Fig. FIGURE 1.3. 1. For an interpretation in terms of our real-world situation-the actual bacteria population-we substituted t = 1.----r---. On the other hand.1. Graphs of P(t) = Cekt withk = ln2. (7) accurately . Then our mathematical analysis there consisted of solving for the solution function P(t) = 1000eCinZ)t = 1000 · 21 as our mathematical result.--. and no two intersect.5 to obtain the predicted population of P(1. the construction of a mathematical model.------r--r---. the bacteria population is growing under ideal conditions of unlimited space and food supply.--. we apply these mathematical results to attempt to answer the original real-world question. it may turn out that no solution of the selected differential equation accurately fits the actual population we're studying. for instance. think of first formulating the mathematical model consisting of the equations dPjdt = kP. The remainder of this introductory section is devoted to simple examples and to standard terminology used in discussing differential equations and their solutions. . As indicated in Fig.x) 2 = y 2 if x ::j:: C . then dy dx 1 (C. (8) defines a one-parameter family of solutions of d yjdx = y 2 . With sufficient insight. Mathematical models are discussed throughout this book. Ways must be found to simplify the model mathematically without sacrificing essential features of the real-world situation. We must conclude that the differential equation dPjdt = kP is inadequate for modeling the world population---. This made the mathematical analysis quite simple. If the model is so detailed that it fully represents the physical situation.1.1. then the mathematical analysis may be too difficult to carry out. perhaps unrealistically so. we might formulate a new mathematical model including a perhaps more complicated differential equation.4 in a counterclockwise manner. yet it must be sufficiently simple to make the mathematical analysis practical. If we can solve the new differential equation. we get new solution functions to compare with the real-world population. The construction of a model that adequately bridges this gap between realism and feasibility is therefore the most crucial and delicate step in the process.3. But in Example 6 we simply ignored any complicating factors that might affect our bacteria population. the results may be so inaccurate as to be useless. Thus 1 y (x ) = -C---x defines a solution of the differential equation -=y (8) dy dx 2 (9) on any interval of real numbers not containing the point x = C. 1. Indeed. we may attempt to traverse once again the diagram of Fig. one for each value of the arbitrary constant or "parameter" C.x).x that satisfies the initial condition y (0) = 1. 1) but has a vertical asymptote at x = 1.oo.1.1 Differential Equations and Mathematical Models 5 describe the actual growth of the human population of the world over the past few centuries. With the formulation of this new mathematical model.5. Thus there is an inevitable tradeoff between what is physically realistic and what is mathematically possible. 1. Eq. a successful population analysis may require refining the mathematical model still further as it is repeatedly measured against real-world experience. one that that takes into account such factors as a limited food supply and the effect of increased population on birth and death rates. A satisfactory mathematical model is subject to two contradictory requirements: It must be sufficiently detailed to represent the real-world situation with relative accuracy. Actually. this solution • is continuous on the interval ( . With C = 1 we get the particular solution 1 y(x) = 1. If the model is too simple.-which in recent decades has "leveled off" as compared with the steeply climbing graphs in the upper half (P > 0) of Fig. 1. Examples and Terminology Example 7 If Cis a constant and y(x) = 1/(C.1. (13) where F is a specific real-valued function of n + 2 variables. For the sake of brevity. and = is a fourth-order equation. Our use of the word solution has been until now somewhat informal.1 12 Inx and y"(x) = ix.6 Chapter 1 First-Order Differential Equations Example 8 Verify that the function y (x) = 2x 112 - x 112 In x satisfies the differential equation (10) for all x > 0. This is why only a continuous function can qualify as a (differentiable) solution of a differential equation on an . . • (11) The fact that we can write a differential equation is not enough to guarantee that it has a solution. = u (x ) satisfies the Remark: Recall from elementary calculus that a differentiable function on ~~~. uCn) exist on I and ' U 11 .. For example. For a variation on this theme. ••• .3 12 • Then substitution into Eq.. The differential equation of Example 8 is of second order. it is clear that the differential equation (y') 2 +l = -1 has no (real-valued) solution. those in Examples 2 through 7 are first-order equations. because the sum of nonnegative numbers cannot be negative. The most general form of an nth-order differential equation with independent variable x and unknown function or dependent variable y = y(x) is ' y " . we may say that u differential equation in ( 13) on I. for all x in I.. U. Solution First we compute the derivatives y'(x) = -~x. U. (10) yields if x is positive. . u". an open interval is necessarily continuous there.y. . we say that the continuous function u = u (x) is a solution of the differential equation in (13) on the interval I provided that the derivatives u'. note that the equation (12) obviously has only the (real-valued) solution y(x) 0. U (n)) = 0 F (X ..y (n)) = 0 ' F (x. To be precise. . The order of a differential equation is the order of the highest derivative that appears in it. In our previous examples any differential equation having at least one solution indeed had infinitely many. . so the differential equation is satisfied for all x > 0.312 tnx.~x. y. 1. In addition. oo). 1 Differential Equations and Mathematical Models Example 7 Continued 7 Figure 1.5 shows the two "connected" branches of the graph y = 1/(1 .. In Chapters 1 through 7 we will be concerned only with ordinary differential equations and will refer to them simply as differential equations. . if they are. Figure 1. we will ordinarily assume that any differential equation under study can be solved explicitly for the highest derivative that appears. Eq. The solution of y' = y 2 defined by y(x) = 1/(1. If the dependent variable is a function of two or more independent variables. y).1. we will see that an nth-order differential equation ordinarily has an nparameter family of solutions-one involving n different arbitrary constants or parameters. (11) and (12). . The right-hand branch is the graph of a different solution of the differential equation that is defined (and continuous) on the different interval (1. All the differential equations we have mentioned so far are ordinary differential equations.6 shows the graphs of several such solutions. meaning that the unknown function (dependent variable) depends on only a single independent variable.1.9Bsin3x = -9y(x) for all x. = -9Acos3x...5. So the single formula y(x) = 1/(1 . then partial derivatives are likely to be involved.. The left-hand branch is the graph of a (continuous) solution of the differential equation y' = y 2 that is defined on the interval ( -oo.x) actually defines two different solutions (with different domains of definition) of the same • differential equation y' = y 2 • ~ .ax 2 ' where k is a constant (called the thermal diffusivity of the rod).x).1. t) of a long thin uniform rod at the point x at time t satisfies (under appropriate simple conditions) the partial differential equation ---k- au a2 u at . In this chapter we concentrate on first-order differential equations of the form dx = f(x.6.. that is.y 1 11 y (n) . 5 ~------~r-------~ on the whole real number line. . then two successive differentiations yield (14) y'(x) y"(x) = -3A sin 3x + 3B cos 3x. For example.x). we will always seek only real-valued solutions unless we warn the reader otherwise. 1). The three solutions y 1(x) = 3 cos 3x. • Although the differential equations in (11) and (12) are exceptions to the general rule. In both Eqs. Consequently. ···-· Example 9 If A and B are constants and y(x) =A cos 3x + B sin 3x. where G is a real-valued function of n + 1 variables.y. Y2(x) = 2 sin 3x. the temperature u = u (x. the equation is called a partial differential equation. (14) defines what it is natural to call a two-parameter family of solutions of the second-order differential equation -57---------~a-----~ ~ 0 5 y" + 9y = 0 (15) X FIGURE 1.y (n-1)) ' (16) -5~--------L-------~ -3 0 3 X FIGURE 1.y. For this reason.1. dy (17) . the appearance of y' as an implicitly defined function causes complications. that the equation can be written in the so-called normal form _ G (x. and y3(x) = -3 cos 3x + 2 sin 3x of the differential equation y" + 9y = 0. 28. 24. y(x) = cex. y) is the sum of x and y . y(x) = ln(x +C). How long does it take for the velocity of the boat to decrease to 1 m/s? To ro mls? When does the boat come to a stop? In Example 7 we saw that y(x) = 1/(C . y" + y = 0 43. 25. y(O) = 3 y' = y + 1. y) having the function g as its solution (or as one of its solutions).4y = 0 35. (b) Determine by inspection a solution of the initial value problem x' = kx 2 . xy' + y = 3x 40. y(rr) = 0 44. 32. 22. y" = 0 38. (a) Continuing Problem 43. the differential equation dyjdx = y 2 has exactly one solution y(x) satisfying the condition y (a) = b? = 10 23. Every straight line normal to the graph of g passes through the point (0. given any point (a. y(O) = 5 y' = X . 37. the time rate of change of the number N of those persons who have heard a certain rumor is proportional to the number of those who have not yet heard the rumor. 20. where Cis an arbitrary constant. y(O) = 0 dy x . (b) How would these solutions differ if the constant k were negative? Suppose a population P of rodents satisfies the differential equation dPjdt = kP 2 • Initially. 34. and vis decreasing at the rate of 1 m/s2 when v = 5 m/s. y(x) = ce. y(x) = ce-x +x.x. _ • _ 1 _ In x = 0.1 Differential Equations and Mathematical Models 9 11. x 2 y". 27.x) defines a one-parameter family of solutions of the differential equation dyjdx = y 2 • (a) Determine a value of C so that y(lO) = 10. 0). 1). 18. The time rate of change of the velocity v of a coasting motorboat is proportional"to the square of v.3y = x 3 . y(x) = (x +C) cosx. Then determine a value of the constant C so that y(x) satisfies the given initial condition. y 1 = x cos(lnx). (y') 2 + y 2 = 1 41. y2 = x sin(lnx) In Problems 13 through 16. 45. y' = y 2 39.in the manner of Eqs.x).2y =0 14. y) intersects the x-axis at the point (x/2. 21. 3y' = 2y 15. write. In Problems 32 through 36. In a city having a fixed population of P persons. y(x) = y(O) = 7 eYy' = 1. x). 31.+ 3y = 2x 5 . y(x) = tan(x 3 +C).3 . In Problems 27 through 31. 36. (a) If k is a constant. a function y = g(x) is described by some geometric property of its graph. y(l) = 17 y' = 3x 2 (y 2 + 1). y' + y = 0. How long will it take for this population to grow to a hundred rodents? To a thousand? What's happening here? Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dvjdt = kv 2 • The initial speed of the motorboat is v(O) = 10 meters per second (m/s). first verify that y(x) satisfies the given differential equation. y(x) = ix 5 + cx. The acceleration dvjdt of a Lamborghini is proportional to the difference between 250 kmlh and the velocity of the car.. use your knowledge of derivatives to make an intelligent guess. show that a general (one-parameter) solution of the differential equation dx 2 -=kx dt is given by x(t) = 1/(C. Does it appear that these solution curves fill the entire xyplane? Can you conclude that. (3) through (6) of this section-a differential equation that is a mathematical model of the situation described. 3y" + 3y'. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation.1. y(O) ce-x 3 .2 X X 12. y' + y = ex 42. Write a differential equation of the form d y jdx = f (x. x 2 y +5xy +4y -O. 13. 33. In Problems 37 through 42. 4y" = y 16. y(x) = x 3 (C + lnx). b) in the plane. .xy' + 2y . Then test your hypothesis. y) passes through the point ( -y.yt. and their number is increasing at the rate of dPjdt = 1 rodent per month when there are P = 10 rodents. The time rate of change of a population P is proportional to the square root of P. and highlight the one that satisfies the given initial condition.y. The line tangent to the graph of g at the point (x. determine by inspection at least one solution of the given differential equation. 47. and then sketch graphs of solutions of x' = kx 2 with several typical positive values of x(O). The slope of the graph of g at the point (x.2·Y2. y' + 3x 2 y = 0.8 shows typical graphs of solutions ofthe form y(x ) = 1/(C. 29. 26.1. The line tangent to the graph of g at (x. 17. assume that k is positive. In Problems 17 through 26. . the time rate of change of the number N of those persons infected with a certain contagious disease is proportional to the product of the number who have the disease and the number who do not. Can you guess what the graph of such a function g might look like? 30. y(O) = 2 y' = 2y. (b) Is there a value of C such that y(O) = 0? Can you nevertheless find by inspection a solution of d yjdx = y 2 such that y(O) = 0? (c) Figure 1. substitute y = e'x into the given differential equation to determine all values of the constant r for which y = e'x is a solution of the equation.kt). y(x) = Ce2x. That is. y(O) = 1 y' + ytanx = cosx. 46. y(2) = 1 dx xy'. x(O) = 0.1. there are P(O) = 2 rodents. 19. In a city with a fixed population of P persons. y" + y' . The graph of g is normal to every curve of the form y = x 2 + k (k is a constant) where they meet.1. ---. The graph y = Cx 4 for various values of C.2.. (c) Given any two real numbers a and b. if G' (x) f (x )-then = y(x) = G(x ) +C. meaning that it involves an arbitrary constant C.1..1. (a) Show that y(x) = Cx 4 defines a one-parameter family of differentiable solutions of the differential equation xy' = 4y (Fig.2.--.. (1) In this special case we need only integrate both sides of Eq.10 Chapter 1 First-Order Differential Equations 3~~~~~~~~~ C=-2 C=-1 C=O C= 1 C=2 C=3 100 80 60 40 20 . we obtain the particular solution of Eq. 1. The first-order equation dyjdx = f(x. but is not of the form y(x) = Cx 4 . If G (x) is a particular antiderivative of f-that is.----. Graphs of solutions of the equation dyfdx = y 2 • FIGURE 1. There we see that the constant C is geometrically the vertical distance between the two curves y(x) = G(x ) and y(x ) = G(x) +C.---r---y-~.G(xo).8. so that C = Yo .. so dy dx = f(x). With this choice of C. (1). we need only substitute x = x o and y = Yo into Eq. (1) to obtain y(x) = J f(x) dx +C. explain why-in contrast to the situation in part (c) of Problem 47-there exist infinitely many differentiable solutions of x y' = 4 y that all satisfy the condition y(a) =b. (3) The graphs of any two such solutions YI (x) G(x) + C1 and Y2(x) = G (x) + C2 on the same interval I are "parallel" in the sense illustrated by Figs. 1.2.---. . y (xo) =YO· . To satisfy an initial condition y(x0 ) = y 0 .9. 48. and for every choice of C it is a solution of the differential equation in ( 1). (1) satisfying the initial value problem dy dx = f(x). (b) Show that y(x) = ~-x 4 x4 if x < 0. (3) to obtain Yo = G(xo) + C.9). if x ~ 0 defines a differentiable solution of xy' = 4y for all x. y) takes an especially simple form if the right-hand-side function f does not actually involve the dependent variable y.---r--1 2 C=4 0 j---J~~~t-H~~~y -2 -20 -40 -60 -80 -1oo~~~~~~~~~ -5-4-3-2-1 0 1 2 3 4 5 X FIGURE 1.1.1 and 1. (2) This is a general solution of Eq. 2. we call it the general solution of the differential equation.1. (2) immediately yields the general solution y(x) 2 0 = J (2x + 3) dx = x 2 + 3x +C. Solution 4 dy dx y(l) = 2. A natural question is whether a given general solution contains every particular solution of the differential equation. When this is known to be true. a general solution of a first-order differential equation is simply a one-parameter family of solutions. It follows that C = -2.2 Integrals as General and Particular Solutions 4rT~~~~~~~~-. 11 6 3 2 C=-1 C=-2 4 2 1 ~ 0 1--\-~~____::::::. Thus Eq. • Example 1 Solve the initial value problem .. Solution curves for the differential equation in Example 1.2.3.. (1) is of the form in (2)..2. it follows that every solution of Eq.3 shows the graph y = x 2 + 3x + C for various values of C..10 -6 0 2 4 y(l) = (1) 2 + 3. thereby satisfying the initial condition -4 -2 X . Graphs of y = sin x + C for various values of C. • .. Graphs of y = ~ x 2 + C for various values of C. FIGURE 1. We will see that this is the typical pattern for solutions of first -order differential equations. by appropriate choice of C. so the desired particular solution is y(x) = x 2 + 3x . The particular solution we seek corresponds to the curve that passes through the point (1. -2 ~ -4 -6 -8 Figure 1.. Integration of both sides of the differential equation as in Eq.= 2x + 3.1. 2). (2) serves to define the general solution of (1). (1) + c = 2.2. we will first find a general solution involving an arbitrary constant C.. because any two antiderivatives of the same function f (x) can differ only by a constant.t. FIGURE 1.2.. We can then attempt to obtain.2. a particular solution satisfying a given initial condition y(xo) = YO· Remark: As the term is used in the previous paragraph. For example.:::::---:~HY -1 -2 -2 -3 -4 3 X 4 0 X 2 4 6 FIGURE 1. Ordinarily. The motion of a particle along a straight line (the x-axis) is described by its position function X= j(t) (5) giving its x-coordinate at time t. dx V=-. dx Velocity and Acceleration Direct integration is sufficient to allow us to solve a number of important problems concerning the motion of a particle (or mass point) in terms of the forces acting on it. The velocity of the particle is defined to be v(t) = f' (t). We simply integrate once to obtain ~. The observation that the special first-order equation d y jdx = f (x) is readily solvable (provided that an antiderivative off can be found) extends to second-order differential equations of the special form (4) in which the function g on the right-hand side involves neither the dependent variable y nor its derivative dyjdx. where Cz is a second arbitrary constant.dt 2 • dv d 2x (7) J v(t) dt or in the definite integral form Equation (6) is sometimes applied either in the indefinite integral form x(t) = x(t) = x(t0 ) +it to v(s) ds. In effect. a ---. that is. which you should recognize as a statement of the fundamental theorem of calculus (precisely because d x jdt = v). dt (6) Its acceleration a (t) is a(t) = v'(t) = x"(t). = j y"(x) dx = j g(x) dx = G(x) + Ct.dt . Then another integration yields y(x) = J y' (x) dx = J [G(x) + Ct] dx = J G(x) dx + C 1x + C 2 . the second-order differential equation in (4) is one that can be solved by solving successively the first-order equations dv dx = g(x) and dy = v(x). in Leibniz notation. where G is an antiderivative of g and C 1 is an arbitrary constant. .12 Chapter 1 First-Order Differential Equations Second-order equations. So dx v(t) = .2 Integrals as General and Particular Solutions 13 Newton's second law of motion says that if a force F(t) acts on the particle and is directed along its line of motion. Its retrorockets. with Eq. suppose that the force F. of the particle at any time t in terms of its constant acceleration a.5 meters per second per second (mjs 2 ) (the gravitational acceleration produced by the moon is assumed to be included in the given deceleration). its initial velocity v0 . and substitution of this information into the preceding equation yields the fact that C 1 = v0 . (8) where m is the mass of the particle. Constant acceleration. These two arbitrary constants are frequently determined by the initial position x 0 = x(O) and the initial velocity v0 = v(O) of the particle.= at dt + vo . (11) Thus. and its initial position x 0 • Example 2 A lunar lander is falling freely toward the surface of the moon at a speed of 450 meters per second (mjs). x (t) = ~at 2 + vot + xo. Therefore. Then we begin with the equation dv = a dt (a is a constant) - (9) and integrate both sides to obtain v(t)= Jadt=at+C 1 • We know that v = v0 when t = 0. For instance. (10) A second integration gives X(t) = J V(t) dt = J (at+ Vo) dt = ~at 2 + Vot + C2.1. then ma(t) = F(t). (10) we can find the velocity. when fired. then the equation x"(t) = F(t)jm can be integrated twice to find the position function x(t) in terms of two constants of integration. that is. If the force F is known. and the substitution t = 0. and with Eq. X = Xo gives c2 = xo. At what height above the lunar surface should the retrorockets be activated to ensure a "soft touchdown" (v = 0 at impact)? . (11) the position. F = ma. provide a constant deceleration of 2. and therefore the acceleration a = F jm. are constant. more precise values are 9.8 m/s2 The last line of this table gives values for the gravitational acceleration g at the surface of the earth.2. negative because the height x(t) is decreasing). where xo is the height of the lander above the lunar surface at the time t = 0 when the retrorockets should be activated. the foot-pound-second (fps) and meter-kilogram-second (mks) unit systems are used more generally in scientific and engineering problems.5 = 180 s (that is.500 meters-that is.7805 m/s2 and 32. However.5t .4.2. (13) yields x0 = 0. Thus the retrorockets should be activated when the lunar lander is 40.) . • Physical Units Numerical work requires units for the measurement of physical quantities such as distance and time. Although these approximate values will suffice for most examples and problems. We let t = 0 denote the time at which the retrorockets should be fired. From Eq. 1.25)(180) 2 + 450(180) = 40.450t + xo. Thus 1 N is (by definition) the force required to impart an acceleration of 1 mjs2 to a mass of 1 kg.5 kilometers above the surface of the moon. as indicated in Fig. Both systems are compatible with Newton's second law F = ma. Then v0 = -450 (m/s. Similarly. while mks units constitute the standard international system of scientific units. (12) we see that v = 0 (soft touchdown) occurs when t = 450/2. 1 slug is (by definition) the mass that experiences an acceleration of 1 ft/s 2 under a force of 1 lb. Force Mass Distance Time g pound (lb) slug foot (ft) second (s) 32 ft/s 2 newton (N) kilogram (kg) meter (m) second (s) 9.450 (12) (13) and Lunar surface x(t) = 1. and a = +2.4. x = 0 into Eq.14 Chapter 1 First-Order Differential Equations Solution We denote by x (t) the height of the lunar lander above the surface. FIGURE 1. and it will touch down softly on the lunar surface after 3 minutes of decelerating descent. (10) and (11) become v(t) = 2.088 ft/s 2 (at sea level at the equator). because an upward thrust increases the velocity v (although it decreases the speed Ivi).(1. 3 minutes). We sometimes use ad hoc units-such as distance in miles or kilometers and time in hours-in special situations (such as in a problem involving an auto trip). then substitution oft = 180. xo = 40. The lunar lander of Example 2. fps units are commonly used only in the United States (and a few other countries). Then Eqs.25t 2 .5 km ~ 25t miles. (We will use mks units in all problems requiring mass units and thus will rarely need slugs to measure mass. In fact.5. 54-.8 N 2 9. Yo denotes the initial (t = 0) height of the body and vo its initial velocity.S. For conversions between fps and mks units it helps to remember that 1 in. x 2.48 em.4 kg.448 Njlb) = 444. Successive integrations (as in Eqs. speed limit of 50 mi/h means that-in international terms-the legal speed limit is about 50 x 1. 1 m1 = em 12 in. so its mass is m =.= g W 444. a mass m weighing 100 pounds has mks weight W = (100 lb)(4.45 kmjh. a mass of m = 20 kg has a weight of W = (20 kg)(9. To discuss vertical motion it is natural to choose the y-axis as the coordinate system for position.609 km.8 mjs ~ 45.54 em (exactly) and 1 lb ~ 4.gt + vo and y(t) = . then the effect of gravity on a vertically moving body is to decrease its height and also to decrease its velocity v = dyjdt . Vertical Motion with Gravitational Acceleration The weight W of a body is the force exerted on the body by gravity. (17) . = 30. = 2.4 em~ 1.8 mjs2 ). Substitution of a= g and F =Win Newton's second law F = ma gives W=mg (14) for the weight W of the mass mat the surface of the earth (where g ~ 32 ftjs 2 ~ 9. dt This acceleration equation provides a starting point in many problems involving vertical motion. Similarly.48- em ft = 160934. For instance. Thus a posted U .2 Integrals as General and Particular Solutions 15 Inches and centimeters (as well as miles and kilometers) also are commonly used in describing distances.1. (10) and (11)) yield the velocity and height formulas (16) v(t) = .m. if we ignore air resistance.448 N. Consequently. frequently with y = 0 corresponding to "ground level." If we choose the upward direction as the positive direction. For instance. = 5280 ft x 30.609 ~ 80. then the acceleration a = dvjdt of the body is given by dv (15) -=-g.8 N.8 mjs2 ) = 196 N. 1 ft and it follows that .tgt 2 + Vot + YO· Here. Hence the maximum height that the ball attains is y(3) = -~ . substitution using (18) gives the differential equation (19) for the swimmer's trajectory y = y(x) as he crosses the river. ~~~~----·-~-~-~--~--------. 3 + 0 = 144 (ft) and thus when t = 3 s. his velocity vector (relative to the riverbed) has horizontal component Vs and vertical component VR.= 3(1.5 shows a northward-flowing river of width w = 2a. (16)) is zero. then it returns to the ground when y(t) = -~.9)t( -t + 10) = 0. (17)).----·--- - - Integration yields y(x) = J (3 .2. 32 (with the aid of Eq. then Eq.--·--~--.16 Chapter 1 First-Order Differential Equations Example 3 (a) Suppose that a ball is thrown straight upward from the ground (y 0 = 0) with initial velocity v0 = 96 (ftjs.------~---~-~ Example 4 Suppose that the river is 1 mile wide and that its midstream velocity is vo = 9 mijh.12x 2 )dx = 3x . (19) takes the form dy 2 . (9. + 96. so we use g = 32 ftjs 2 in fps units). 1. Suppose that a swimmer starts at the point (-a. Hence the swimmer's direction angle a is given by tana = Vs VR FIGURE 1.8)t 2 + 49t = (4. 32. The lines x =±a represent the banks of the river and the y-axis its center.5. where VR = vo.2.8 mjs2 in mks units). Because tan a = d y jdx. so we use g = 9. and thus after 10 s in the air. (18) to verify that the water does flow the fastest at the center. 0) on the west bank and swims due east (relative to the water) with constant speed v5 .4x ). A swimmer's problem (Example 4). Then it reaches its maximum height when its velocity (Eq. Suppose that the velocity v R at which the water flows increases as one approaches the center of the river. and that VR = 0 at each riverbank.4x 3 +C . and indeed is given in terms of distance x from the center by (18) You can use Eq. v(t) = -32t + 96 = 0.5.--~--. If the swimmer's velocity is v 5 = 3 mi/h. dx ·---H•·----- ~~. • A Swimmer's Problem y-axis Figure 1. (b) If an arrow is shot straight upward from the ground with initial velocity vo = 49 (m/s. As indicated in Fig.2. How high could this person throw the ball on the planet Gzyx of Problem 29? 35.5 mi and v0 = 9 mijh as in Example 4. Chapter 1 First-Order Differential Equations 10 . The car in question is known to have a constant deceleration of 20 m/s2 under these conditions. 35 miles away. How far does the car travel before coming to a stop? 26. a projectile is fired straight upward toward the bomb. 5) (7. At noon a car starts from rest at point A and proceeds at constant acceleration along a straight road toward point B. It strikes the ground with a speed of 60 mjs. a ball dropped from a height of 20 ft hits the ground in 2 s. 30. and Vs = 3 mi/h as in Example 4.2. when fired. If the car starts from rest (x 0 = 0. If a ball is dropped from the top of a 200-ft-tall building on Gzyx. The skid marks made by an automobile indicated that its brakes were fully applied for a distance of 75 m before it came to a stop. (18). v 0 = 0). 36. What is the maximum height attained by the arrow of part (b) of Example 3? 24. what is the distance from A to B? 38. (c) its total time in the air. A projectile is fired straight upward with an initial velocity of 100 mjs from the top of a building 20m high and falls to the ground at the base of the building. (b) when it passes the top of the building.was the car traveling when the brakes were first applied? rather than the quadratic function in Eq. A baseball is thrown straight downward with an initial speed of 40 ft/s from the top of the Washington Monument (555 ft high). how far will it skid if it is moving at 100 km/h when the brakes are applied? 33. From the ground directly beneath the helicopter. If a = 0. how long will it take to hit the ground? With what speed will it hit? 34.J'Iih.in km/h. 23. Find (a) its maximum height above the ground.000 mi/h2 • At what height above the lunar surface should the astronauts fire the retrorockets to insure a soft touchdown? (As in Example 2. 27. but that the velocity of the river is given by the fourth-degree function . With what initial velocity should the projectile be fired. How long does it take to reach the ground? With what speed does the ball strike the ground? 25. Show that the speed with which it strikes the ground is v = . At noon a car starts from rest at point A and proceeds with constant acceleration along a straight road toward point C.5 mi. A ball is thrown straight downward from the top of a tall building. How long does it take to reach the ground. provide a constant deceleration of 20. ignore the moon's gravitational field. If the car reaches B at 12:50 P.---. (3.M. How fast.18 22. 41. 8 6 . Now find how far downstream the swimmer drifts as he crosses the river. v0 = 9 mijh. 5) 4 6 8 10 FIGURE 1. Suppose a woman has enough "spring" in her legs to jump (on earth) from the ground to a height of 2. A stone is dropped from rest at an initial height h above the surface of the earth. Graph of the velocity function v(t) of Problem 22.6)t (ft/s 2 ). exactly 2 seconds after the bomb is released. On the planet Gzyx. A ball is dropped from the top of a building 400 ft high.= dv dt (0. what is that deceleration? For how long does the skid continue? 31. Suppose that a car skids 15 m if it is moving at 50 km/h when the brakes are applied. If the constantly accelerated car arrives at C with a velocity of 60 mijh. what must the swimmer's speed v 5 be in order that he drifts only l mile downstream as he crosses the river? 40. in order to hit the bomb at an altitude of exactly 400 feet? 42. Its retrorockets.-~. The initial speed of the ball is 10 mjs. Suppose that a = 0. A car traveling at 60 mi/h (88 ftj s) skids 176 ft after its brakes are suddenly applied. A bomb is dropped from a helicopter hovering at an altitude of 800 feet above the ground. A spacecraft is in free fall toward the surface of the moon at a speed of 1000 mph (mi/h). How tall is the building? 28.25 feet. find the distance it has traveled at the end of the first 10 s and its velocity at that time. Under the assumption that the braking system provides constant deceleration. A diesel car gradually speeds up so that for the first 10 s its acceleration is given by 32. at what time does it arrive at C? 39. The brakes of a car are applied when it is moving at 100 km/h and provide a constant deceleration of 10 meters per second per second (mjs 2) . Assuming that the car has the same constant deceleration.3 ft/s 2-how high above the surface will she rise? 37.---. If she jumps straight upward with the same initial velocity on the moon-where the surface gravitational acceleration is (approximately) 5. with a velocity of 60 mijh.-~---.9. and with what speed does the baseball strike the ground? 29.12)t 2 + (0. A person can throw a ball straight upward from the surface of the earth to a maximum height of 144 ft.) . and therefore cannot be evaluated explicitly. 19 and how far will it have traveled by then? 44. y) of the . y(x)) through which it passes-that is. y) (1) y where the right-hand function f (x. A solution curve for the differential equation y' = x . y). they found that when its brakes were applied at 25 mph. and hence write y(x) = J f(x.0098 rn/s 2 • Suppose this spacecraft starts from rest at time t = 0 and simultaneously fires a projectile (straight ahead in the same direction) that travels at one-tenth of the speed c = 3 x 108 rn/s of light. y) involves both the independent variable x and the dependent variable y. . the car skidded only 45 feet before coming to a stop. At each point (x. the graphical and numerical methods of this and later sections can be used to construct approximate solutions of differential equations that suffice for many practical purposes.---- Example 1 - ···-····· . How long will it take the spacecraft to catch up with the projectile. y2).1 shows a solution curve of the differential equation y' = x . the value of f(x.y 1 at the point (xJ. y 1) . But the driver's skid marks at the accident scene measured 210 feet.2 (a)-(d) show slope fields and solution curves for the differential equation d y =ky dx (2) with the values k = 2.y 3 at the point (x3. y(x)) dx +C. the solutions of such a simple-looking differential equation as y' = x 2 + y 2 cannot be expressed in terms of the ordinary elementary functions studied in calculus textbooks.xy-plane whose tangent line at each point (x. y(x)). .y together with tangent lines having • slope m 1 = x 1 . a spacecraft propelled by the solar wind.1. y'(x) = f(x. A driver involved in an accident claims he was going only 25 mph.OOlg = 0. Slope Fields and Graphical Solutions There is a simple geometric way to think about solutions of a given differential equation y ' = f(x . Assuming the same (constant) deceleration.3. y) . ·· · ~···· FIGURE 1. and • slope m 3 = x 3 . Indeed. Fig. Its aluminized sail provides it with a constant acceleration of O.. - Figures 1. We might think of integrating both sides in (1) with respect to x. This geometric viewpoint suggests a graphical method for constructing approximate solutions of the differential equation y' = f(x . Nevertheless. y) determines a slope m = f (x.1. -1.3. Thus a solution curve of the differential equation y' = f (x. . 1.3 Slope Fields and Solution Curves 43. y3). Note that each slope field yields important qualitative information about the set of all solutions . • slope m2 = x 2 . Actually. Arthur Clarke's The Wind from the Sun (1963) describes Diana. y).xy-plane.. there exists no straightforward procedure by which a general differential equation can be solved explicitly. because the indicated integral involves the unknown function y(x) itself. When police tested his car. and -3 of the parameter k in Eq. All these line segments constitute a slope field (or a direction field) for the equation y ' = f(x. y) . this approach does not lead to a solution of the differential equation. y) has slope m = f(x. 0. determine the speed he was actually traveling just prior to the accident.Y2 at the point (x2. For instance. (2). A solution of the differential equation is simply a differentiable function whose graph y = y(x) has this "correct slope" at each point (x. --·· ··--. E Sl~pe Fields and Solution Curves Consider a differential equation of the form dy dx = f(x. However. Through each of a representative collection of points (x .5.-·····--. y) in the plane we draw a short line segment having the proper slope m = f (x .. y )-the graph of a solution of the equation-is simply a curve in the .3. y).y together with its tangent lines at three typical points. y). The numerical slope m = x . All this is apparent from slope fields like those in Fig. Through each point a solution curve should proceed in such a direction that its tangent line is nearly parallel to the nearby line segments of the slope field..2.. 4).. 1.5)y..l_~~~3ill~i!!!i!l~ -1 2 .3.n-Tr-nr>o-Tr-n~.2(a) Slope field and solution curves for y' = 2y.2(a) and (b) suggest that each solution y(x) approaches ±oo as x ~ +oo if k > 0.2(c) Slope field and solution curves for y' = .. 1. (2) is given explicitly by y (x) = C ekx. although the sign of k determines the direction of increase or decrease of y(x). 1.3. its absolute value lkl appears to determine the rate of change of y(x). Moreover. even without knowing that the general solution of Eq.3.-or-.~~~~ 4--~~~~~~~~--70 3 3 2 2 -1 -1 -2 -3 -2 0 X 2 3 4 X FIGURE 1. 1.r-v--r-.3.y appears at the intersection of the horizontal x-row and the vertical y-column of the table. we can attempt to sketch freehand an approximate solution curve that threads its way through the slope field. of the differential equation.3.3 shows a table of slopes for the given equation.2(b) Slope field and solution curves for y' = (0. (Of Solution .y. • A slope field suggests visually the general shapes of solution curves of the differential equation. FIGURE 1. 3 2 .-. Starting at any initial point (a.2(d) Slope field and solution curves for y' = -3y . 4 3 4 .:. Figs. 0 ~. you can see that it was easily and quickly constructed.3..2(c) and (d) suggest that y (x) ~ 0 as x ~ +oo if k < 0. Example 2 Construct a slope field for the differential equation y' = x . b). If you inspect the pattern of upper-left to lower-right diagonals in this table.. For instance. 0 -1 I I -2 -2 -3 X X FIGURE 1.20 Chapter 1 First-Order Differential Equations 4. whereas Figs.:.3. Solution Fig..3.4. following the visible line segments as closely as possible. . FIGURE 1.y and use it to sketch an approximate solution curve that passes through the point (. y for -4 -2 -1 0 1 x. together with typical solution curves treading through this slope field. 4).y of Example 2..4. FIGURE 1.3.3. Although a spreadsheet program (for instance) readily constructs a table of slopes as in Fig.6. such commands are illustrated in the application material for this section..1.5 shows an approximate solution curve sketched through the point (-4. If you look closely at Fig.3.j.':-7f-\-J -1 -2 -3 -4 -4 . This inference illustrates the fact that a slope field can suggest tangible information about solutions that is not at all evident from the differential equation itself.. 0 J--i--i\-i--+--C--:'-:. 1. a more complicated function f(x.6. 1.:>.3. and Fig. you can verify that the linear function y = x . At each point it appears to proceed in the direction indicated by the nearby • line segments of the slope field.r:. ..y:.4 shows the corresponding slope field. FIGURE 1.3.3.3. 4) so as to follow as this slope field as closely as possible. the more accurately solution curves can be visualized and sketched. Values of the slope y' 5 =x \ \ I I I \ \ .y corresponding to the table of slopes in Fig..y.3..5.3 Slope Fields and Solution Curves 21 x\Y -4 -3 -4 0 1 -3 -1 0 1 -2 -2 -1 0 -1 -3 0 1 -5 2 -6 -5 3 -4 -3 -7 -6 -5 -2 -1 -4 -3 4 -8 -7 -6 -5 -2 -1 0 1 2 3 -2 -1 0 1 -4 -3 2 3 4 5 6 1 2 3 0 1 -2 -1 -4 -3 4 5 6 2 3 0 -2 -1 0 -4 -3 2 3 4 5 6 2 3 4 7 8 4 5 1 2 3 . \ \ \ \ \ \ \ \ \ -/ \ " "- \ \ " / 5~~~---r------~ / I I I 4 3 2 I 0 \ \ \ "--/ "I I I 0 X "" I I O~r~~-r~+-+-~ ".--'j-!.3.1 is a solution of the equation y' = x . 1.3. most computer algebra systems include commands for quick and ready construction of slope fields with as many line segments as desired.) Figure 1. y ~ 7 4 1 2 ~ -2 -1 0 4.~. by tracing the 4 3 I .3. and it appears likely that the other solution curves approach this straight line as an asymptote as x ~ +oo. 1.>. you may spot a solution curve that appears to be a straight line! Indeed.3. Can you. Figure 1. Slope field for y' = x . The more line segments are constructed. y) on the right-hand side of the differential equation would necessitate more complicated calculations. Slope field and typical solution curves for y' = X .3. course.3. The solution curve through ( -4.6 shows a "finer" slope field for the differential equation y' = x .3 -2 -1 0 X I 2 3 4 FIGURE 1./ -5 -5 / I I I I I I I I I I I I I I 5 -1 -2 -3 -4 / -5~------~------~ -5 0 X 5 FIGURE 1. However. it can be quite tedious to plot by hand a sufficient number of slope segments as in Fig. 1.4..>.y. This implies that-however you throw it-the baseball should approach the limiting velocity v = 200 ft/s instead of accelerating indefinitely (as it would in the absence of any air resistance).0.22 Chapter 1 First-Order Differential Equations appropriate solution curve in this figure.16v. y(-4) = 4? ~ 2 for the solution y(x) of Applications of Slope fields The next two examples illustrate the use of slope fields to glean useful information in physical situations that are modeled by differential equations. infer that y(3) the initial value problemy'= x. then the initial acceleration given by Eq. h Perhaps a catcher accustomed to 100 milh fastballs would have some chance of fielding this speeding ball. (4) is negative.kv. Note that all these solution curves appear to approach the horizontal line v = 200 as an asymptote. because of air resistance. The handy fact that 60 milh = 88 ft/s yields V ft = 200s X 60 milh 88 ft/s ~ mi 136.in the absence of air resistance. 1. then Eq.(0. (3) A typical value of the air resistance proportionality constant might be k = 0. Its velocity therefore remains unchanged.3. If the y-axis is directed downward.7. It therefore seems quite reasonable that.0.36 . you can use your laptop's computer algebra system to construct a slope field for the differential equation dv . • Comment If the ball's initial velocity is v(O) = 200. while its acceleration due to air resistance is negative. then the ball's velocity v = dyjdt and its gravitational acceleration g = 32 ft/s 2 are both positive.16)(200) = 0. But if the initial velocity is less than 200.16. You might like to verify that. Example 3 is based on the fact that a baseball moving through the air at a moderate speed v (less than about 300 ft/s) encounters air resistance that is approximately proportional to v.y. then it experiences both the downward acceleration of gravity and an upward acceleration of air resistance. Slope field and typical solution curves for v' = 32. Hence its total acceleration is of the form dv dt = g. dt (4) 5 10 15 20 25 FIGURE 1. the baseball will approach a limiting velocity of 200 ft/s. (4) gives v' (0) = 32 . so the ball slows down as it falls.= 32. and hence v(t) 200 is a constant "equilib- = rium solution" of the differential equation. If the baseball is thrown straight downward from the top of a tall building or from a hovering helicopter.16v. Example 3 Suppose you throw a baseball straight downward from a helicopter hovering at an altitude of 3000 feet. so the ball speeds up as it falls. If the initial velocity is greater than 200.7. The result is shown in Fig.whatever initial velocity it starts with.. together with a number of solution curves corresponding to different values of the initial velocity v(O) with which you might throw the baseball downward. • . then the initial acceleration given by (4) is positive. You wonder whether someone standing on the ground below could conceivably catch it. so the ball experiences no initial acceleration.3.this ball would hit the ground at over 300 milh. In order to estimate the speed with which the ball will land. = 0. then the initial rate of change given by (6) is positive. We may also want to know whether there is only one solution of the equation satisfying a given initial condition-that is. X 1 y(O) = 0 (7) has no solution. dt 250 200 (6) ~150~~~~~~~~~ 100 50 00~--~----~--~--~ 25 50 t 75 100 The positive term 0.06P.0004P 2 • Comment If the initial population is P(O) = 150.3. We see this graphically in Fig.0.3.8 shows a slope field for Eq. (6) gives P' (0) = 0.06P.0004?(150. then the initial rate of change given by (6) is negative. This implies that-whatever the initial population-the population P(t) approaches the limiting population P = 150 as t ---+ oo. so the population experiences no initial (instantaneous) change. and hence P(t) 150 is a constant "equilibrium solution" of the differential equation. because no solution y(x) = j(ljx) dx =In lxl + C of the differential equation is defined at x = 0. then Eq. for instance). it is wise to know that solutions actually exist. Figure 1.8.3 Slope Fields and Solution Curves 23 In Section 1. Example 5 (a) [Failure of existence] The initial value problem y I = -. Note that all these solution curves appear to approach the horizontal line P = 150 as an asymptote. This means that M is the maximum population that this environment can sustain on a long-term basis (in terms of the maximum available food.0004? 2 .0004? 2 represents the inhibition of growth due to limited resources in the environment. It therefore remains unchanged.0. 1.150) = 0.1. 7 we will discuss in detail the logistic differential equation dP = kP(M.P) = 0. But if the initial population is less than 150. Slope field and typical solution curves for P' = 0. 0).0004 and M = 150. so the population immediately begins to decrease.0004(150)(150. so the population immediately begins to increase.P) dt (5) that often is used to model a population P (t) that inhabits an environment with carrying capacity M. whether its solutions are unique. It therefore seems quite reasonable to conclude that the population will approach a limiting value of 150-whatever the (positive) initial population. which shows a direction field and some typical solution curves for the equation y' = ljx. (6). then the logistic equation in (5) takes the form dP .06P on the right in (6) corresponds to natural growth at a 6% annual rate (with timet measured in years). It is apparent that the indicated direction field "forces" all solution curves near the y-axis to plunge downward so that none can pass through the point (0. If the initial population is greater than 150. . • FIGURE 1. Example 4 If we take k = 0. • = Existence and Uniqueness of Solutions Before one spends much time attempting to solve a given differential equation. together with a number of solution curves corresponding to possible different values of the initial population P(O).9. The negative term -0.3. Figure 1.24 Chapter 1 First-Order Differential Equations I I I I I I I I I I I I I I I I I I I I I I I I / I I / I I / I / I / I I / / / . the solution interval I may not be as "wide" as the original rectangle R of continuity./.) .y. The rectangle R and x-interval I of Theorem I. you can readily verify that the initial value problem (8) y' = 2y/y.9. see Remark 3 below. and the solution curve y = y(x) through the point (a./.11. This raises an immediate question as to whether the mathematical model adequately represents the physical system. The theorem stated below implies that the initial value problemy' = f(x. Direction field and typical solution curves for the equation y' = ljx. for some open interval I containing the point a./. Methods of proving existence and uniqueness theorems are discussed in the Appendix. 0) 0 X FIGURE 1. We see that the curve y 1 (x) = x 2 threads its way through the indicated direction field./. but that our proposed mathematical model involves a differential equation not having a unique solution satisfying those conditions. Questions of existence and uniqueness of solutions also bear on the process of mathematical modeling.1 0 shows a direction field and these two different solution curves for the initial value problem in (8)..3. y) are continuous on some rectangle R in the xy-plane that contains the point (a. FIGURE 1./. Then. y(O) = 0. Suppose that we are studying a physical system whose behavior is completely determined by certain initial conditions. the initial value problem X dy dx = f(x. y).9. has one and only one solution that is defined on the interval I. y(a) = b has one and only one solution defined near the point x =a on the x-axis./ 0~~~----~~----~ Y2(x) =0 (0. Direction field and two different solution curves for the initial value problemy' = 2-JY. whereas the differential equation y' = 2-JY specifies slope y' = 0 along the x-axis y 2 (x) = 0. y(O) = 0 has the two different solutions y 1 (x) = x 2 and y 2 (x) = 0 (see Problem 27). before we can speak of "the" solution of an initial value problem. (As illustrated in Fig.(x)=x2 /.3 .10. we need to know that it has one and only one solution.a y are continuous near the point (a. y) and its partial derivative D y f(x. • Example 5 illustrates the fact that. y(a) = b (9) FIGURE 1. y R THEOREM 1 Existence and Uniqueness of Solutions b Suppose that both the function f(x. (b) [Failure of uniqueness] On the other hand. / / / / / / .3./.11. b) in its interior.3. 1. provided that both the function f and its partial derivative a f . y). b) in the xy-plane. b)./.3. y) and/or its partial derivative 8ff8y fail to satisfy the continuity hypothesis of Theorem 1. b)..1 we examined the especially simple differential equation dyjdx = y 2 • Here we have f(x. each of which satisfies the initial condition solutions y 1 (x) = x 2 and y2 (x) y(O) = 0. (0.x) is discontinuous at x = 1. Both of these functions are continuous everywhere in the xy-plane.3. dx dy (11) Applying Theorem 1 with f(x. y) = -2...3 Slope Fields and Solution Curves 25 Remark 1: In the case of the differential equation dyfdx = -y of Example 1 and Fig.._ •····· -·-··~··~·· . (11) must have a unique solution near any point in the xy-plane where x =f. .JY of Example 5(b) and Eq. each solution y(x) = ce-x actually is defined for all x.12). so our unique continuous solution does not exist on the entire interval -2 < x < 2. and hence at the point (0. Theorem 1 guarantees a unique solution-necessarily a continuous function-of the initial value problem 6 Remark 3: dy dx 4 y = 1/(l -x) =y 2 ' y (0) =1 (10) ~ 2 r---- l __. 1) leaves the rectangle R before it reaches the right side of R. Example 6 Consider the first-order differential equation x ... 1) lies in the interior of this rectangle. Although the theorem ensures existence only on some open interval containing x =a. (8). The solution curve through the initial point (0. 0).= 2y. and in particular on the rectangle -2 < x < 2. if the function f (x. the function f(x. 0 < y < 2... 0. ·······-···•· . but the partial derivative 8fj8y = 1/..3.JY is continuous wherever y > 0. so Theorem 1 implies the existence of a unique solution for any initial data (a. Remark 2: In the case of the differential equation dyjdx = -2.. we see immediately by substitution in (11) that y(x) = Cx 2 (12) .2(c).3. 1. both the function f(x. y) = y 2 and ajjay = 2y. This is why it is possible for there to exist two different 0. y) = -y and the partial derivative 8fj8y = -1 are continuous everywhere. the reason is that the solution curve provided by the theorem may leave the rectangle-wherein solutions of the differential equation are guaranteed to exist-before it reaches the one or both ends of the interval (see Fig. that we discussed in Example 7.. Indeed. Indeed this is the solution 1 y(x) = -1- IR : I -x -4 -2 0 X 2 4 FIGURE 1. But y(x) = 1/(1 . Geometrically. l) on some open x-interval containing a = 0.1.. then the initial value problem in (9) may have either no solution or many-even infinitely many-solutions. y) = 2yjx and 8ff8y 2/x .12.JY is discontinuous when y = 0. Because the point (0. • The following example shows that. 1. 0 -2 I I i --. Thus the solution interval I of Theorem 1 may not be as wide as the rectangle R where f and 8ff8 y are continuous. = In Example 7 of Section 1. .. we conclude that Eq. Then for any value of the constant C the function defined by X x2 y(x) = { FIGURE 1. but none of them passes through any other point on the y-axis. then the initial value problem dy xdx = 2y. • no solution if a = 0 but b =1= 0. We therefore see that Theorem 1 (if its hypotheses are satisfied) guarantees uniqueness of the solution near the initial point (a.3.3.) Observe that all these parabolas pass through the origin (0. the solution y (x) = x 2 of the initial value problem in (17) exists on the whole xaxis. -1).26 Chapter 1 First-Order Differential Equations (0. if a =I= 0. • infinitely many solutions if a = b = 0. y(a) =b (15) has a unique solution on any interval that contains the point x = a but not the origin x = 0 In summary.13. Finally. It follows that the initial value problem in (13) has infinitely many solutions. but this solution is unique only on the negative x -axis .14. 0) satisfies Eq. if X> 0 (16) Still more can be said about the initial value problem in (15). 1.3. Cx 2 is continuous and satisfies the initial value problem dy xdx=2y. but the initial value problem dy xdx=2y. There are infinitely many solution curves through the point ( 1. whose solution curves are the parabolas y = Cx 2 illustrated in Fig. has no solution if b =I= 0.3. b) may eventually branch elsewhere so that uniqueness is lost. Thus a solution may exist on a larger interval than one on which the solution is unique. Thus the unique solution curve near ( -1. 0).14. 1. b) if a =I= 0. (In case C = 0 the "parabola" is actually the x-axis y = 0. (17) For a particular value of C.13. note that through any point off the y-axis there passes only one of the parabolas y = C x 2 . b) (0. 1) indicated in Fig.14. There are infinitely many solution curves through the point (0. y(O)=b (14) FIGURE 1. the solution curve defined by ( 16) consists of the left half of the parabola y = x 2 and the right half of the parabola y = C x 2 . (11) for any value of the constant C and for all values of the variable x. but no solution curves through the point (0. the initial value problem dy xdx =2y. b) if b :f: 0. 1) branches at the origin into the infinitely many solution curves illustrated in Fig. 1.3. y(-1)=1. 0). For instance.oo < x < 0. Consider a typical initial point off the y-axis-for instance the point ( -1. the initial value problem in (15) has • a unique solution near (a. . b). In particular. • if X S 0. Hence. but a solution curve through (a. y(O)=O (13) X has infinitely many different solutions. A suggestion for Problem 29. tial value problem y' = 31. +Jf=Y2.3 Slope Fields and Solution Curves Problems 23 and 24 are like Problems 21 and 22. 30. You bail out of the helicopter of Example 3 and pull the ripcord of your parachute. Carry out an investigation similar to that in Problem 30. 1. or infinitely many solutions. (ii) a unique solution that is defined for all x? 28. 24.) Sketch a variety of such solution curves. construct a slope field for this differential equation and sketch the appropriate solution curve. Verify that if c is a constant.0225 p . Then determine (in terms of a and b) how many different solutions the initial value problem y' = 4x-JY.1. none. Verify that if k is a constant.l. Then determine (in terms of a and b) how many different solutions the iniy (a) = b has. y(a) = b has--one.c) { 3 for x ~ c. then the function y(x ) = kx satisfies the differential equation xy' = y for all x . lfyou wish (and know how). then the function defined piecewise by y(x ) ~ { ~x. how long will it take the number of deer to double? What will be the limiting deer population? The next seven problems illustrate the fact that. if X~ C + 7r 27. y (a) = b has.= 0.3. y(2) =? 29 29. 23. Verify that if c > 0. FIGURE 1.I. (a) Verify that if c is a constant. for x > c 2? 25. or infinitely many. y' = x 2 + y 2 . if the hypotheses of Theorem 1 are not satisfied. finitely many solutions. y(O) = 0. b) of the xy-plane such that the initial value problemy' = 3y 213 .) Sketch a variety of such solution curves. Is there a point (a.25. Sketch a variety of such solution curves for different values of c. you can check your manually sketched solution curve by plotting it with the computer. except with the differential equation y' = Does it suffice simply to replace cos(x. y(O) = b have (i) no solution.c) with sin(x. Then determine (in terms of a and b) how many different solutions the initial value problem x y' = y. if x 2 > c satisfies the differential equation y' = 4x -JY for all x . Can you also use the "left half" of the cubic y = (x . for x > c satisfies the differential equation y' = 2-JY for all x (including the point x = c). (b) For what values of b does the initial value problemy' = 2-JY. then the function defined piecewise by JI=Y2 -JI=Y2. c. y(x) = { 0 (x 2 - c) 2 if x 2 ::s.25.c) in piecing together a solution that is defined for all x ? 32. What will your limiting velocity be? Will a strategically located haystack do any good? How long will it take you to reach 95% of your limiting velocity? 26.c) 3 in piecing together a solution curve of the differential equation? (See Fig. y(a) = b has either no solution or a unique solution that is defined for all x? Reconcile your answer with Theorem 1. y ). but now use a computer algebra system to plot and print out a slope fieldfor the given differential equation. y' =X+ y(-2) = 0. v(O) = 0. satisfies the differential equation y' = for all x . y (a) = b may have either no solutions. Construct a figure illustrating the fact that the initial value problemy' = 2-JY. Verify that if c is a constant. then the function defined piecewise by y(x) ~ ~x . Suppose the deer population P(t)in a small forest satisfies the logistic equation dP . (3). then the function defined piecewise by y(x) = 1 +1 -1 cos(x. satisfies the differential equation y' = 3y 213 for all x.0003 P 2 • dt c Construct a slope field and appropriate solution curve to answer the following questions: If there are 25 deer at time t = 0 and t is measured in months. then the initial value problem y' = f (x.0. y(O) = 0 has infinitely many different solutions.c) 2 for x ~ c . Now k = 1.6 in Eq. (Perhaps a preliminary sketch with c = 0 will be helpful. so your downward velocity satisfies the initial value problem dv dt = 32.c) ifx~c.6v. ifc < x < c +n. Construct a slope field and several of these straight line solution curves. y In order to investigate your chances of survival. . y(2) =? 1 2.3. 5 .2. Mathematica™. b = -3.x + 1 with initial values y(-1) = -1. 1. 1..1).. " " .. '"" "\ II II \ _. Slope field and solution curves for the differential equation - dy dx = sm(x .8.2 and y(-1) = -0. y(a) = b has.5.27. -2. b). as do some graphing calculators (see Fig. (a) Use the direction field of Problem 6 to estimate the values at x = 2 of the two solutions of the differential equation y' = x .01 and y( -3) = -2. (b) Use a computer algebra system to estimate the val- ues at x = 3 of the two solutions of this differential equation with initial values y( -3) = -3. determine (in terms of a and b) how many different solutions the initial value problem x 2 y' + y 2 = 0.5 :S x . Sketch a variety of such solution curves for different values of c.. -1. verify that the function defined by y(x) = xj(cx. 35.x) =sin(x._ .3. y =. .y(x)).3. 4 and window . If c Chapter 1 First-Order Differential Equations "I 0. Also.. '- ______ .2 andy( -3) = +0. 2.3. with initial points (0. The applications manual accompanying this textbook includes discussion of Maple™.27). \ \ " " " "" I \ \ I \ \ I I \ "" (b) Use a computer algebra system to estimate the val'- \ \ I \ \ '- "" ues at x = 3 of the two solutions of this differential equation with initial values y(. 5). FIGURE 1. 1. Slope field for x 2 y' + y 2 = 0 and graph of a solution y(x) = x j(cx . FIGURE 1.. \ \ I \ ___ .30 33.3.1) (with graph illustrated in Fig.y ) .26..26) satisfies the differential equation x 2 y' + y 2 = 0 if x "I 1jc. The lesson of this problem is that big changes in initial conditions may make only small differences in results. and MATLAB™ resources for the investigation of differential equations. The lesson of this problem is that small changes in initial conditions can make big differences in results. For instance.. 34.99. y(x). 0. x=-5 .3) = -0.5 and y( -3) = +0. the Maple command with(DEtools): DEplot(diff(y(x). note the constant-valued function y(x) = 0 that does not result from any choice of the constant c. 5.y + 1 with initial values y( -3) = -0. Finally.3 Application Widely available computer algebra systems and technical computing environments include facilities to automate the construction of slope fields and solution curves. y ::: 5 on a TI-89 graphing calculator. (a) Use the direction field of Problem 5 to estimate the values at x = 1 of the two solutions of the differential equation y' = y . 28. Upper Saddle River. a number of apparent straight line solution curves should be visible.3. Use a graphing calculator or computer system in the following investigations.math. {y. produce slope fields similar to the one shown in Fig. 1.3 Slope Fields and Solution Curves and the Mathematica command << Graphics\PlotField.y).edul-dfield). FIGURE 1. 0 l-7::i~~~~~~~~'-! -1 31 -2 -3 -4 X FIGURE 1.29). -10 ~ y ~ 10. but with a larger window than that of Fig.2- x.3.3. Sin[x-y]}.3.2 tan.29.y). -5.rice. you can (with mouse button clicks) plot both a slope field and the solution curve (or curves) through any desired point (or points).m PlotVectorField[{l.n /2 corresponding to the initial condition y(n/2) = 0? Are there any values of C for which the corresponding solution curves lie close to this straight line solution curve? .3.28.3.28 itself was generated with the MATLAB program dfield [John Polking and David Arnold. Ordinary Differential Equations Using MATLAB. INVESTIGATION A: Plot a slope field and typical solution curves for the differential equation dyjdx = sin(x. 5}] .28. for instance.. MATLAB dfield setup to construct slope field and solution curves for y' = sin(x . You might warm up by generating the slope fields and some solution curves for Problems 1 through 10 in this section. (a) Substitute y = ax + b in the differential equation to determine what the coefficients a and b must be in order to get a solution. When a differential equation is entered in the dfield setup menu (Fig. 5}.edu/iode)... 1.y). NJ: Prentice Hall. Can you see that no value of C yields the linear solution y = x .1. 1. -5. 2nd edition. {x. 1.. Figure 1.uiuc. (b) A computer algebra system gives the general solution y (x) = x . Computergenerated slope field and solution curves for the differential equation y' = sin(x ..3.C c) • Plot this solution with selected values of the constant C to compare the resulting solution curves with those indicated in Fig. With -10 ~ x ~ 10.28. Another freely available and user-friendly MATLAB-based ODE package with impressive graphical capabilities is lode (www. 1999] that is freely available for educational use (math.1 ( x . (b) Then generate a slope field for this differential equation. . dx n (a) First investigate (as in part (a) oflnvestigation A) the possibility of straight line solutions.ny). State your inference as plainly as you can. Can you make a connection between this symbolic solution and your graphically generated solution curves (straight lines or otherwise)? The first-order differential equation dy dx = H(x. y) (1) is called separable provided that H (x . try to predict (perhaps in terms of y 0 ) how y(x) behaves as x --+ +oo.32 Chapter 1 First-Order Differential Equations INVESTIGATION B: For your own personal investigation. In this case the variables x and y can be separated-isolated on opposite sides of an equation-by writing informally the equation f(y) dy = g(x ) dx . which we understand to be concise notation for the differential equation dy f( y ) dx = g(x ). and consider the differential equation dy 1 = cos(x. (c) A computer algebra system gives the general solution y(x) = ~ [ x + 2 tan- 1 (x ~C) J. plus a sufficient number of nonlinear solution curves that you can formulate a conjecture about what happens to y(x) as x --+ +oo. (2) It is easy to solve this special type of differential equation simply by integrating both sides with respect to x: I f( y (x)) ~~ dx = I g(x ) d x + C. y) can be written as the product of a function of x and a function of y: dy dx = g(x)h(y) = . with the viewing window chosen so that you can picture some of these straight lines. Given the initial value y (0) = y0 . let n be the smallest digit in your student ID number that is greater than 1. f(y) g(x ) where h (y) = 1/f (y ).. (2) and (3) are equivalent. • . Hence ld: = l(-6x)dx. (0. This is the upper emphasized solution curve shown in Fig.4. we divide both sides of the differential equation by y and multiply each side by dx to get dy -y = -6xdx. The condition y(O) = 7 yields A = 7. y(O) = 7. (3) All that is required is that the antiderivatives I 1 f(y) dy G(x) =I g(x) dx can be found. (4) because two functions have the same derivative on an interval if and only if they differ by a constant on that interval. Solution Informally.3x2' = ec. I F(y) = f(y)dy =I and g(x)dx +C. 7) ln IYI = -3x 2 +C.4. Example 1 Solve the initial value problem dy dx = -6xy.1. so the desired solution is y(x) X FIGURE 1. We see from the initial condition y(O) = 7 that y(x) is positive near x may delete the absolute value symbols: lny = -3x 2 8 I I I I I I 6 = 0.1. -4 and hence where A -6 -8 -~2--~~---L~~~--~2 = e-3x2+C = e-3x2ec = Ae. 1.1.4 Separable Equations and Applications 33 equivalently. To see that Eqs. note the following consequence of the chain rule: Dx[F(y(x))] =F (y(x))y (x) 1 = dy f(y)dx = g(x) = Dx[G(x)].6xy in Example 1. so we 4 2 -2 "' 0 f-~~E------1f------3:~~-l + C. which in tum is equivalent to F(y(x)) = G(x) + C. Slope field and solution curves for y' = . 5)..J4 . and Singular Solutions The equation K (x. differentiation of x 2 + y 2 = 4 yields dy x + y dx = 0. 1. ~ 0 -5 -10 -15 -20 Figure 1. It appears that this particular solution through (1.J4 .= -x(y .2x) dy dx (9) .x 2 and y = .2.x 2 of the differential equation x + y y' = 0. with the initial condition y(1) = -2 we get the lower contour curve in Fig.4. satisfies the initial condition y (0) = 2 (Fig.8552 of the particular solution in Example 3. With a graphing calculator we can solve for the single real root y :::::::: 2.4 Separable Equations and Applications 20 15 35 Remark I: When a specific value of x is substituted in Eq. For instance. then we get the new implicit solution (y. if we multiply the implicit solution x 2 + y 2 . 4) but not on the interval ( -1. Remark 2: If the initial condition in (7) is replaced with the condition y(l) = 0. consider the differential equation (y .4 = 0 by the factor (y. (8) for x-values at desired increments from x = -1 to x = 5 (for instance). y) = 0 may or may not satisfy a given initial condition. 1.4.2. x = 4 yields the equation 10 5 f (y) = l - 5y . But note that a particular solution y = y (x) of K (x. Such a table of values serves effectively as a "numerical solution" of the initial value problem in (7). • Implicit. 1.2x).x 2 FIGURE 1.3.4). 0) is defined on the interval (0. then the resulting particular solution of the differential equation in (5) lies on the lower "half" of the oval contour curve in Fig.9.9 = 0.2x ) y .4. Remark 2: Similarly.1. Thus the initial condition can determine whether a particular solution is defined on the whole real line or only on some bounded interval.5y. General. For instance. Slope field and solution curves for y' = -xjy.x 2 and y(x) = -)4.4) = 0 that yields (or "contains") not only the previously noted explicit solutions y +.8552. On the other hand. so x 2 + y 2 = 4 is an implicit solution of the differential equation x only the first of the two explicit solutions X + yy' = 0.4. Remark I: You should not assume that every possible algebraic solution y = y (x) of an implicit solution satisfies the same differential equation. but also the additional function y = 2x that does not satisfy this differential equation. For instance. we can attempt to solve numerically for y. But y(x) = +)4. y) = 0 is commonly called an implicit solution of a differential equation if it is satisfied (on some interval) by some solution y = y(x) of the differential equation.4.2x)(x 2 +i .4. solutions of a given differential equation can be either gained or lost when it is multiplied or divided by an algebraic factor. For example. -4 -2 0 y -6 2 4 6 FIGURE 1. This yields the value y(4) :::::::: 2. With a computer algebra system one can readily calculate a table of values of the y-solutions of Eq.3 shows the graph of f. Graph of f(y) = y 3 .4. (8). This particular solution is defined for all x. The argument preceding Example 1 actually suffices to show that every particular solution of the differential equation f(y)y' = g(x) in (2) satisfies the equation F(y(x)) = G(x) + C in (4). By contrast. 1. whereas negative values yield those that dip below it.4. any particular choice of a specific value for C yields a single particular solution of the equation. then we get the previously discussed differential equation dy ydx=-x.1) 213 1 dy = J 2x dx . we "gain" this new solution when we multiply Eq. Positive values of the arbitrary constant C give the solution curves in Fig. (10) of which y = 2x is not a solution.5. • A solution of a differential equation that contains an "arbitrary constant" (like the constant C in the solution of Examples 1 and 2) is commonly called a general solution of the differential equation.6 that lie above the line y = 1.2x).6. but the general solution of (2). dx -15 -10 -5 0 X 5 10 15 FIGURE 1. but the possibility of loss or gain of such "extraneous solutions" should be kept in mind. • = . and to observe that the constant-valued function y(x) = 0 is a singular solution that cannot be obtained from the general solution by any choice of the arbitrary constant C . Thus we "lose" the solution y = 2x of Eq.1)1/3 = xz + C. Example 4 Solution Separation of variables gives f 3(y.= 6x(y. it is appropriate to call (4) not merely a general solution of (2). In Section 1.1) 3 both satisfy the initial condition y(l) = 1. (y . Note that the two different solutions y(x) 1 and y(x) = 1 + (x 2 . y (x) = 1 + (x 2 + C) 3 • -2 -1 0 X 2 FIGURE 1.C) 2 and the singular solution curve y = 0 of the differential equation (y')2 = 4y. In Problem 30 we ask you to show that the general solution of the differential equation (y') 2 = 4y yields the family of parabolas y = (x .5 we shall see that every particular solution of a linear first-order differential equation is contained in its general solution. These exceptional solutions are frequently called singular solutions. (9) upon its division by the factor (y . y) = 6x (y .C) 2 illustrated in Fig. The value C = 0 gives the solution y (x ) = 1 + x 6 . or x + y dx dy = 0. General and singular solution curves for y' = 6x (y .2x).4.4. The general solution curves y = (x . (10) by (y .4. alternatively.1)213 is not differentiable. 1. Find all solutions of the differential equation dy . Indeed. Such elementary algebraic operations to simplify a given differential equation before attempting to solve it are common in practice.5.1) 213 . but no value of C gives the singular solution y(x) = 1 that was lost when the variables were separated.2x). the whole singular solution curve y = 1 consists of points where the solution is not unique and where the function f (x.1)2/ 3. But if we divide both sides by the common factor (y. Consequently.36 Chapter 1 First-Order Differential Equations having the obvious solution y = 2x. it is common for a nonlinear first-order differential equation to have both a general solution involving an arbitrary constant C and one or several particular solutions that cannot be obtained by selecting a value for C. 10.= l i m . we use Eq. and with f3 = 0.t---+0 b. (Note that 10% annual interest means that r = 0.t (12) where k = f3. It has been observed that a constant fraction of those radioactive atoms will spontaneously decay (into atoms of another element or into another isotope of the same element) during each unit of time. This same ratio permeates all life.P ~ ({3.8)P(t) b. Then. This proportion remains constant because the fraction of 14 C in the atmosphere remains almost constant. and living matter is continuously taking up carbon from the air or is consuming other living matter containing the same constant ratio of 14 C atoms to ordinary 12C atoms.4 Separable Equations and Applications 37 Natural Growth and Decay The differential equation dx =kx dt (k a constant) (11) serves as a mathematical model for a remarkably wide range of natural phenomenaany involving a quantity whose time rate of change is proportional to its current size. because organic processes seem to make no distinction between the two isotopes.t deaths occur.A = r A(t) b. and therefore dP b. with k > 0 in place of 8. POPULATION GROWTH: Suppose that P(t) is the number of individuals in a population (of humans. We thus get the differential equation dN =-kN. approximately f3P(t) b. dt fl.t. To write a model for N(t). Consequently. or insects.8.t ---+ 0 b. Here are some examples.=rA. Let A (t) be the number of dollars in a savings account at time t (in years). The key to the method of radiocarbon dating is that a constant proportion of the carbon atoms in any living creature is made up of the radioactive isotope 14 C of carbon.t births and 8P(t) b. dt fl.) Continuous compounding means that during a short time interval b.P -=lim -=kP.A . so that COMPOUND INTEREST: dA b. dt (14) The value of k depends on the particular radioactive isotope.1. during a short time interval b. (12) with N in place of P.t.t. so the change in P(t) is given approximately by b. . and suppose that the interest is compounded continuously at an annual interest rater. the amount of interest added to the account is approximately b. the sample behaves exactly like a population with a constant death rate and no births.t (13) RADIOACTIVE DECAY: Consider a sample of material that contains N(t) atoms of a certain radioactive isotope at time t.t. or bacteria) having constant birth and death rates f3 and 8 (in births or deaths per individual per unit of time). 38 Chapter 1 First-Order Differential Equations The ratio of 14C to normal carbon remains constant in the atmosphere because.4.8. (Matters are not as simple as we have made them appear. (16) Because of the presence of the natural exponential function in its solution. 7 shows a typical graph of x (t) in the case k > 0.) In many cases the amount A(t) of a certain drug in the bloodstream. measured by the excess over the natural level of the drug. Then we solve for x : Because C is a constant. will decline at a rate proportional to the current excess amount. For 14 C. it is known that k ~ 0. the differential equation dx (17) -=kx dt is often called the exponential or natural growth equation. although 14C is radioactive and slowly decays. That is. Over the long history of the planet. and consequently the ratio of 14C to normal carbon begins to drop. so the particular solution of Eq. the cosmic ray levels apparently have not been constant. researchers in this area have compiled tables of correction factors to enhance the accuracy of this process. The Natural Growth Equation The prototype differential equation dxjdt = kx with x(t) > 0 and k a constant (either negative or positive) is readily solved by separating the variables and integrating: lnx = kt +C. the amount of time elapsed since the death of the organism can be estimated.4. DRUG ELIMINATION: dA = -AA. it ceases its metabolism of carbon and the process of radioactive decay begins to deplete its 14C content. Figure 1. There is no replenishment of this 14C. the amount is continuously replenished through the conversion of 14 N (ordinary nitrogen) to 14C by cosmic rays bombarding the upper atmosphere. In addition.0001216 if tis measured in years. this decay and replenishment process has come into nearly steady state. . the case k < 0 is illustrated in Fig. (11) with the initial condition x(O) = x 0 is simply x (t) = xoekt. It is also clear that A = x (0) = x 0 . so is A = ec . dt (15) where A > 0. In applying the technique of radiocarbon dating. 1. By using independent methods of dating samples. extreme care must be taken to avoid contaminating the sample with organic matter or even with ordinary fresh air. so the ratio of 14C in the atmosphere has varied over the past centuries. The parameter A is called the elimination constant of the drug. Of course. when a living organism dies. By measuring this ratio. For such purposes it is necessary to measure the decay constant k. ~ ~ 0. Natural decay. This value of k gives the world population function P(t) = 6eO. when t = In 10 0. P(O) 6 t = 0 we now Thus the world population was growing at the rate of about 1. FIGURE 1.07743 k = -. and thus in the year 2177.1. so Po = 6. and was then increasing at the rate of about 212 thousand persons each day.29% annually in 1999.OI 29t .0129.000212 billion.0129 ~ 178.8. or 0.25) ~ 0. • . Natural growth. The fact that P is increasing by 212. From the natural growth equation P' = k P with obtain P' (O) 0. that is.000212)(365. persons per day at timet = 0 means that P' (O) = (0.4. (b) With t = 51 we obtain the prediction 0 ·01 29)(S l) ~ 11.07743 Solution billion per year.4 Separable Equations and Applications X X 39 x =xo ekt (k > 0) FIGURE 1.gov. the world's total population reached 6 billion persons in mid-1999. We take t = 0 to correspond to (mid) 1999.OJ 29t .58 (billion) P(51) = 6e< for the world population in mid-2050 (so the population will almost have doubled in the just over a half-century since 1999). we want to answer these questions: (a) What is the annual growth rate k? (b) What will be the world population at the middle of the 21st century? (c) How long will it take the world population to increase tenfold-thereby reaching the 60 billion that some demographers believe to be the maximum for which the planet can provide adequate food supplies? (a) We measure the world population P(t) in billions and measure time in years.census. Assuming that natural population growth at this rate continues. (c) The world population should reach 60 billion when 60 = 6 eo.7. Example 5 According to data listed at 1 billion.63)N0 now. What is the age of the sample? We take t = 0 as the time of the death of the tree from which the Stonehenge charcoal was made and No as the number of 14C atoms that the Stonehenge sample contained then. because this parameter is more convenient. The decay constant of a radioactive isotope is often specified in terms of another empirical constant.63) 0_ 0001216 ~ 3800 (years).T.40 Chapter 1 First-Order Differential Equations Note: Actually. (3) of Section 1. Thus the sample is about 3800 years old. so we solve the equation (0. To find the relationship between k and T . approximately 5700 Example 6 A specimen of charcoal found at Stonehenge turns out to contain 63% as much 14C as a sample of present-day charcoal of equal mass. ···-······· . initially at 50°F. the time rate of change of the temperature T(t) of a body immersed in a medium of constant temperature A is proportional to the difference A .T) . or templewhichever it may be-dates from 1800 B. We are given that N = (0.M.63)No = N 0 e-kt with the value k = 0. • Cooling and Heating According to Newton's law of cooling (Eq. is placed in a 375°F oven at 5:00 P. A simple mathematical model cannot be expected to mirror precisely the complexity of the real world. the rate of growth of the world population is expected to slow somewhat during the next half-century. we set t = T and N = 4 in the equation N (t) = N 0 ekt.1 ). and the best current prediction for the 2050 population is "only" 9. the half-life of years. or earlier. After 75 minutes it is found that the temperature T(t) of the roast is 125°F. we find that (18) For example. Thus we find that t =- Solution ln(0. so that 4 No = N 0ekr: .. monument. That is. When will the roast be 150°F (medium rare)? . The half-life T of a radioactive isotope is the time required for half of No it to decay. = k(A .C. _ _ ____ _ Example 7 A 4-lb roast. When we solve for r. dt (20) It includes the exponential equation as a special case (b = 0) and is also easy to solve by separation of variables. This is an instance of the linear first-order differential equation with constant coefficients: dx -=ax +b.0001216. 14C is r ~ (In 2)/(0. our computations suggest that this observatory. the half-life of the isotope. If it has any connection with the builders of Stonehenge. dt dT (19) where k is a positive constant.0001216). the total cooking time required.4 Separable Equations and Applications Solution 41 We take timet in minutes. We have T(t) < A = 375.T dT=fkdt· ' -ln(375. Denote by y(t) the depth of water in the tank at timet. As a consequence ofEq. from which water is leaking.oo3s)t for t = -[ln(225j325)]/(0. and T(75) = 125. (22b) This is a statement of Torricelli 's law for a draining tank. under ideal conditionsthat the velocity of water exiting through the hole is v = fiiY. (21) which is the velocity a drop of water would acquire in falling freely from the surface of the water to the hole (see Problem 35 of Section 1.T) = kt + C. 375. Let A (y) denote the horizontal cross-sectional area of the tank at height y. Hence we finally solve the equation 150 = 375 . f 375. One can derive this formula beginning with the assumption that the sum of the kinetic and potential energy of the system remains constant. (21). Substitution of these values in the preceding equation yields k = .0035.M.6 for a small continuous stream of water). We also know that T = 125 when t = 75.J'liY.T). so T(t) = 375. dt (22a) equivalently. we have dV = -av = -afiiY.. Because the roast was placed in the oven at 5:00P. applied to a thin horizontal .325e(-o. it should be removed at about 6:45P. v = c.fii. = -kJy dt dV where k= a. Under real conditions.T = Be-k1 • Now T(O) =50 implies that B = 325. For simplicity we take c = 1 in the following discussion. T(O) = 50.71 5 ln (~~~) >=:::: 0. • Torricelli's Law Suppose that a water tank has a hole with area a at its bottom.1.325e-kr. Then. Hence 1 dT dt = k(375.2).M. with t = 0 corresponding to 5:00P. and by V(t) the volume of water in the tank then. taking into account the constriction of a water jet from an orifice. We also assume (somewhat unrealistically) that at any instant the temperature T(t) of the roast is uniform throughout.0035) >=:::: 105 (min). It is plausible-and true. where cis an empirical constant between 0 and 1 (usually about 0.M. 44 Chapter 1 First-Order Differential Equations weight. (a) At what depth is the intensity half the intensity / 0 at the surface (where x = 0)? (b) What is the intensity at a depth of 10 m (as a fraction of / 0 )? (c) At what depth will the intensity be 1% of that at the surface? 46. (b) Without prior conditioning.51 x 109 years for 238 U and 7. of mercury. there were equal amounts of the two uranium isotopes 235 U and 238 U at the creation of the universe in the "big bang. Suppose that the temperature of the buttermilk has dropped to l5 °C after 20 min. (a) If the initial amount is 10 pu (pollutant units). What is the age of the rock. There are now about 3300 different human "language families" in the whole world. and 5 months later it is still 10 su.will it be until A = 1 su. A cake is removed from an oven at 210oF and left to cool at room temperature. it initially contains water to a depth of 9 ft.5 years. how long will this take? 40.000 ft and again at 30. An accident at a nuclear power plant has left the surrounding area polluted with radioactive material that decays naturally. A certain moon rock was found to contain equal numbers of potassium and argon atoms. After 1 h the depth of the water has dropped to 4 ft. A pitcher of buttermilk initially at 25°C is to be cooled by setting it on the front porch.10 x 108 years for 235 U.2) p. How . few people can survive when the pressure drops to less than 15 in." Assume (as in Problem 52) that the number of these language families has been multiplied by 1. (b) What amount of radioactive material will remain after 8 months? (c) How long-total number of months or fraction thereof.000 people had heard this rumor. with a half-life of 5 h. write a formula for A(t) giving the amount (in pu) present after t years.kA (k > 0)." At present there are 137. The half-life of radioactive cobalt is 5.92. Suppose that a mineral body formed in an ancient cataclysm-perhaps the formation of the earth itselforiginally contained the uranium isotope 238 U (which has a half-life of 4.4)/.5 language families every 6 thousand years. How long does it take for all the water to drain from the tank? 55. The initial amount of radioactive material present is 15 su (safe units). About how long ago was the single original human language spoken? 53. How long will it be until half the population of the city has heard the rumor? 51. How high is that? 47. The intensity I of light at a depth of x meters below the surface of a lake satisfies the differential equation dljdx = (-1.51 x 109 years) but no lead. measured from the time it contained only potassium? 43. p(O) = 29. The barometric pressure p (in inches of mercury) at an altitude x miles above sea level satisfies the initial value problem dpjdx = (-0. When will it be at 5°C? 44.27 years. If today the ratio of 238 U atoms to lead atoms in the mineral body is 0. Assume that the rate of increase of the number who have heard the rumor is proportional to the number who have not yet heard it. If 25% of the sugar dissolves after 1 min.000 ft. (a) Calculate the barometric pressure at 10. how long does it take for half of the sugar to dissolve? 45. when did the cataclysm occur? 42. (b) What will be the amount (in pu) of pollutants present in the valley atmosphere after 5 years? (c) If it will be dangerous to stay in the valley when the amount of pollutants reaches 100 pu. Using the half-lives 4. About when did the ancestors of today's Native Americans arrive? 54. and that a language family develops into 1. A tank is shaped like a vertical cylinder. and a bottom plug is removed at timet = 0 (hours). calculate the age of the universe.7 atoms of 238 U for each atom of 235 U. Within a week. the end product of the radioactive decay of 238 U. Thousands of years ago ancestors of the Native Americans crossed the Bering Strait from Asia and entered the western hemisphere.) 41.28 x 109 years) and that one of every nine potassium atom disintegrations yields an argon atom. There are now 150 Native American language families in the western hemisphere. 10. Suppose also that sodium pentobarbital is eliminated exponentially from the dog's bloodstream. A certain piece of dubious information about phenylethylamine in the drinking water began to spread one day in a city with a population of 100. Suppose that the tank of Problem 48 has a radius of 3 ft and that its bottom hole is circular with radius 1 in. 49. According to one cosmological theory. Since then. What single dose should be administered in order to anesthetize a 50-kg dog for 1 h? 48. the amount A that remains undissolved after t minutes satisfies 'the differential equation dAjdt = . where the temperature is ooc. When will it be 100° F? 50. which is 70°F. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. When sugar is dissolved in water. Assume that all these are derived from a single original language. so it is safe for people to return to the area? 52.5 every 6000 years.9. The single language that the original Native Americans spoke has since split into many Indian "language families. (a) Write a formula giving the amount A(t) of radioactive material (in su) remaining after t months. they have fanned out across North and South America. How long will it be until the region is again habitable? (Ignore the probable presence of other radioactive isotopes. The amount A(t) of atmospheric pollutants in a certain mountain valley grows naturally and is tripling every 7. After 30 min the temperature of the cake is 140° F. Assume that all the argon is the result of radioactive decay of potassium (its half-life is about 1.000. show that dx 1 k-=dt t where k is a constant.4. it had moved an additional 3 miles. What time did it start snowing? This is a more difficult snowplow problem because now a transcendental equation must be solved numerically to find the value of k.M. John Bernoulli proposed this problem as a public challenge. it had traveled 2 miles. how long wi11 it take for the liquid to drain completely? 61. (The clepsydra.the curve of minimal descent time is an . When will the tank be empty? 57. In June of 1696. (b) What time did it start snowing? (Answer: 6 A. received Bernoulli's challenge on January 29. A plug at the bottom is removed at 12 noon. (a) Find the depth y(t) of water remaining after t hours.M. A snowplow sets off at 7 A./h)? 65. What should be this curve. What was the time of death? 66.M.M.J28Y" (taking constriction into account) to determine when the tank will be empty. At 12 noon the temperature of the body is 80°F and at 1 P.4.M. it had traveled 4 miles and that by 9 A. but it took two more hours (until 10 A. and what should be the radius of the circular bottom hole. After 1 h the water in the tank is 9 ft deep.1 0. A cylindrical tank with length 5 ft and radius 3 ft is situated with its axis horizontal. How long will be required for all the gasoline to drain from the tank? 62. Suppose that a cylindrical tank initially containing V0 gallons of water drains (through a bottom hole) in T minutes.M. shaped like the surface obtained by revolving the curve y = f (x) around the y-axis. what is the radius of the circular hole in the bottom? 60.M.M. 58. (Answer: 4:27A. in order that the water level wi11 fall at the constant rate of 4 inches per hour (in.M. If a circular bottom hole with a radius of 1 in. (b) When will the tank be empty? (c) Ifthe initial radius of the top surface of the water is 2 ft. or water clock) A 12-h water clock is to be designed with the dimensions shown in Fig. Assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour. circular bottom hole with radius 1 in. The very next day he communicated his own solution. Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of 70°F. when will the tank be empty? 63.M. when a circular plug in the bottom of the tank is removed. 1. A water tank has the shape obtained by revolving the parabola x 2 = by around the y-axis. The clepsydra.6)rrr 2 . the depth of the water is 1 ft. At 7 A. At timet = 0 the bottom plug (at the vertex) of a full conical water tank 16 ft high is removed. it is 75°F.) 67. the bottom hole is opened and at 1:30 P.10.6°F and that it has cooled in accord with Newton's law. At 1 P. the depth of water in the tank is 2ft. The brachistochrone problem asks what shape the wire should be in order to minimize the bead's time of descent from P to Q. At 1 P. (a) Lett = 0 when it began to snow and let x denote the distance traveled by the snowplow at time t. By 8 A.11 shows a bead sliding down a frictionless wire from point P to point Q. a snowplow set off to clear a road. Early one morning it began to snow at a constant rate. Suppose that an initially full hemispherical water tank of radius 1 m has its flat side as its bottom.M. say). If this bottom hole is opened at 1 P. except that the radius r of its circular bottom hole is now unknown.£ T minutes is V = V0 [1 -(tiT) f .M. Suppose now that by 8 A. (b) What is the radius of the bottom hole? 64. Consider the initially full hemispherical water tank of Example 8. the depth of the water is 6 ft.1. is opened. The water depth is 4 ft at 12 noon.M.) 68.. At 1 P. with a 6-month deadline (later extended to Easter 1697 at George Leibniz's request). as in Problem 66.) for the snowplow to go an additional 2 miles. Use Torricelli's law to show that the volume of water in the tank after t . A water tank has the shape obtained by revolving the curve y = x 413 around the y-axis. Assume that the temperature of the body at the time of death was 98. It has a bottom hole of radius 1 em.M. is opened and the tank is initially half full of xylene. (a) Use Torricelli's law in the form dV/dt = -(0. A spherical tank of radius 4 ft is full of gasoline when a y = f(x) or X FIGURE 1. Isaac Newton.4 Separable Equations and Applications long will it take the water (initially 9 ft deep) to drain completely? 45 56. 1697. Figure 1.4. When will the tank be empty? 59. then retired from academic life and serving as Warden of the Mint in London. when the depth of water in the tank is 12ft. . Principles of physics can be used to show that the shape y = y (x) of the hanging cable satisfies the differential equation where the constant a = TIp is the ratio of the cable's tension T at its lowest point x = 0 (where y' (0) = 0 ) and its (constant) linear density p. d vjdx = d 2 yjdx 2 in this secondorder differential equation.4. H) where a is an appropriate positive constant.11. y = a(l -cos 2t) (iii) Yo X for which t = y = 0 when x = 0.· . which . 7th edition (Upper Saddle River.] . y (L. For a modem derivation of this result.4 we saw how to solve a separable differential equation by integrating after multiplying both sides by an appropriate factor.= 2xy (y > 0).4. dy = 4a sin t cost dt in (ii) to derive the solution x = a(2t . y dx that is. (i) the differential equation Solve this differential equation for y'(x) = v(x) sinh(x/a). Then integrate to get the shape function y (x) = a cosh ( ~) + C dy= dx V----y- ~ (ii) of the hanging cable. the substitution of e = 2a in (iii) yields the standard parametric equations X = a(e -sin e). This curve is called a catenary. H) at equal heights located symmetrically on either side of the x-axis (Fig.= 2x. from the Latin word for chain.J2iY is the bead's velocity when it has descended a distance y vertically (from KE = ~mv 2 mgy = -PE).12. to solve the equation dy (1) . Finally.12). For instance. all that remains are two simple integrations. we get the first-order equation dv a-= dx Q v 1 + v2 • ~ FIGURE 1. 1. Suppose a uniform flexible cable is suspended between two points (±L.4. The catenary.= constant.4 of Edwards and Penney. NJ: Prentice Hall. [See Example 5 in Section 9.sin 2t) . (a) First derive from Eq. dx we multiply both sides by the factor 1/y to get 1 dy . If we substitute v = dymyslashdx . suppose the bead starts from rest at the origin P and let y = y(x) be the equation of the desired curve in a coordinate system with the y-axis pointing downward.46 Chapter 1 First-Order Differential Equations arc of an inverted cycloid-to the Royal Society of London. A bead sliding down a wire-the brachistochrone problem. y = a(l -COS e) FIGURE 1. Calculus: Early Transcendentals. p 69. Dx (ln y) = Dx (x). 2 (2) Because each side of the equation in (2) is recognizable as a derivative (with respect to the independent variable x ). 2008). (b) Substitute y = 2a sin2 t. E Linear First-Order Equations In Section 1. v sin a (i) where a denotes the angle of deflection (from the vertical) of the tangent line to the curve-so cot a = y' (x) (why?)-and v = . Then a mechanical analogue of Snell's law in optics implies that of the cycloid that is generated by a point on the rim of a circular wheel of radius a as it rolls along the xaxis. + P(x)y = dy Q(x) (3) on an interval on which the coefficient functions P(x) and Q(x) are continuous. (6) This formula should not be memorized. Next. (3) with the coefficient functions P(x) and Q(x) displayed explicitly. 2. . Begin by calculating the integrating factor p (x) = ef P(x ) dx. the function p (y) = 1/y is called an integrating factor for the original equation in (1). the left-hand side is the derivative of the product y(x) · ef P(x) dx. For this reason. ef P(x)dx J = Q(x)ef P(x)dx. there is a standard technique for solving the linear first-order equation dx . METHOD: SOLUTION OF FIRST-ORDER EQUATIONS 1. = Q(x)ef P(x)dx. in order to solve an equation that can be written in the form in Eq. solving for y. 3. you should attempt to carry out the following steps. With the aid of the appropriate integrating factor. An integrating factor for a differential equation is a function p (x. y) such that the multiplication of each side of the differential equation by p(x. (3) by the integrating factor p(x) = ef P(x)dx. We multiply each side in Eq. recognize the left-hand side of the resulting equation as the derivative of a product: D x [p(x)y(x)] = p(x) Q(x).5 Linear First-Order Equations 47 yield ln y = x 2 + C. (5) is equivalent to Dx [y(x). Then multiply both sides of the differential equation by p (x). y) yields an equation in which each side is recognizable as a derivative. That is. we obtain the general solution of the linear first-order equation in (3): y(x) = e. In a specific problem it generally is simpler to use the method by which we developed the formula.J P(x)dx [! ( Q(x)ef P(x) dx) dx + C J. Finally. Integration of both sides of this equation gives y(x)ef P(x)dx = J( Q(x)ef P(x)dx ) dx +C. (4) (5) The result is ef P(x)dx dy dx + P(x)ef P(x)dx y Because Dx [! P(x) dx J= P(x).1. so Eq. then y(x) is given by the formula in Eq. Conversely. because y(O) = 2 implies that y(x) = 2 for all x (and thus the value of the solution remains forever where it starts). there is-as previously noted-a unique value of C such that y(xo) = Yo· Consequently. f xo . Note that substitution of y' = 0 in the differential equation (9) yields 3xy = 6x. (11)-can be selected "automatically" by writing p(x )=exp(1: P(t)dt) . for which it follows that y'(x) 0. (12) y(x ) = - p (x ) 1 . Remark 2: Theorem 1 tells us that every solution of Eq. Then the anti derivatives J P(x) dx and J( Q(x)ef P(x)dx) dx exist on I. Remark 1: Theorem 1 gives a solution on the entire interval I for a linear differential equation.3. Our derivation of Eq. Finally. in contrast with Theorem 1 of Section 1.as needed to solve the initial value problem in Eq. (6). given by the formula in Eq. Thus a linear first-order differential equation has no singular solutions. we have proved the following existence-uniqueness theorem. as seems visually obvious in Fig. (6) of the linear first-order equation y' + P y = Q bears closer examination. • = = = A Closer Look at the Method The preceding derivation of the solution in Eq.2. then the initial value problem dy dx + P(x)y = Q(x).50 Chapter 1 First-Order Differential Equations solution can be described as an equilibrium solution of the differential equation. Remark 3: The appropriate value of the constant C in Eq.[Yo + { x p(t)Q(t) dt]. you may verify by direct substitution (Problem 31) that the function y(x) given in Eq. (3)." so by an equilibrium solution of a differential equation is meant a constant solution y(x) c. (3) is included in the general solution given in Eq. (6) for some choice of the constant C. (6) with an appropriate value of C. given a point xo of I and any number y 0 . Hence we see that y (x) 2 is the only equilibrium solution of this differential equation. 1. so it follows that y = 2 if x :j= 0. the word "equilibrium" connotes "unchanging. (6) satisfies Eq. (3) on I. which guarantees only a solution on a possibly smaller interval. y(xo) = y 0 (11) has a unique solution y(x) on I. More generally. Suppose that the coefficient functions P (x) and Q (x) are continuous on the (possibly unbounded) open interval I. (6) shows that if y = y(x) is a solution of Eq.5. (6). THEOREM 1 The Linear First-Order Equation If the functions P(x) and Q(x) are continuous on the open interval I containing the point x 0 . Withx0 = 1 theintegratingfactorin(12) is p(x) = exp (lx ~ dt) = exp(ln x) = x. (6) that guarantees in advance that p(x0 ) = 1 and that y(x0 ) = y0 (as you can verify directly by substituting x = x 0 in Eqs.1..5.5 Linear First-Order Equations 51 The indicated limits x 0 and x effect a choice of indefinite integrals in Eq. (14) reduces to y(x) ( 1.when a solution of a differential equation can be expressed in terms of elementary functions. S1(x)S1(1)]..3) -I 1 [ Yo+ =X 1x 0 sin t dt t 1 1 sin t dt t -2 -3 0 J= -[Yo+ 1 . for instanceto find the value y(x) of the solution at x. We will study various devices for obtaining good approximations to the values of the nonelementary functions we encounter. (12)). It appears that on each solution curve. an integral such as the one in Eq. • Comment: In general.3. Typical solution curves defined by Eq.. ···· ··· ~« ' . (15).. y (x ) ~ 0 as x ~ + oo.. ... . . . Then the particular solution in Eq. The sine integral function is available in most scientific computing systems and can be used to plot typical solution curves defined by Eq. In Chapter 6 we will discuss numerical integration of differential equations in some detail. (14) would (for given x) need to be approximated numerically-using Simpson's rule. • In the sequel we will see that it is the exception-rather than the rule.3 shows a selection of solution curves with initial values y (l) = y0 ranging from y0 = -3 to y0 = 3. '"" " '"""'~ •-· ··. 1965).· Example 3 Solve the initial value problem x dx 2dy + xy = smx. and this is in fact true because the sine integral function is bounded.5. X (15) 0 5 10 X 15 FIGURE 1... In this case. . • . -t- J (14) In accord with Theorem 1. however. Handbook of Mathematical Functions (New York: Dover. A good set of tables of special functions is Abramowitz and Stegun. (15).. this solution is defined on the whole positive x-axis. Figure 1. y( ) =YO· 1 (13) Solution Division by x 2 gives the linear first-order equation -+-y=2 dx x x dy 1 sinx with P(x) = 1/x and Q(x) = (sinx)jx 2 . 1 0 t 3 2 which appears with sufficient frequency in applications that its values have been tabulated.. we have the sine integral function x sin t Si(x) = -dt.. so the desired particular solution is given by 1 [ Yo+ y(x) = ~ lx 1 sint dt . {grams output}~ rici ~t: ~X ~t- r0 c0 ~t. but c0 denotes the variable concentration x(t) C0 (t) = V (t) (17) of solute in the tank at time t.c. Suppose that solution with a concentration of ci grams of solute per liter of solution flows into the tank at the constant rate of ri liters per second.examination of the behavior of the system over a short time interval [t . Important: Equation ( 18) need not be committed to memory. = r0 ) of solution in the tank at time t. dt v (18) If V0 = V(O). where V(t) denotes the volume (not constant unless r. note how the cancellation of dimensions checks our computations: Amount x(t) Volume V(t) Concentration c0 (t) Output: r0 Lis. then the error in this approximation also approaches zero. we estimate the change ~x in x during the brief time interval [t. given the amount x (0) = x 0 at time t = 0. The amount of solute that flows out of the tank during the same time interval depends on the concentration c0 (t) of solute in the solution at time t. then V(t) = V0 + (r.c. Remark: It was convenient for us to use giL mass/volume units in deriving Eq.roCo . In the following example we measure both in • cubic kilometers. liters ) ( grams) c. There is both inflow and outflow. ( 18).= rici ..r0 C0 • Finally.5. because it is a very useful tool for obtaining all sorts of differential equations.x . . and we obtain the differential equation . we take the limit as ~t --+ 0.that you should strive to understand. dt dx (16) in which ri. (18) is a linear first-order differential equation for the amount x (t) of solute in the tank at time t. The single-tank mixture problem.4. t + ~t]. yields a quantity measured in grams. It is the process we used to obtain that equation.4. c. Then ~x ={grams input}. so Eq. 1.. c 0 (t) = x(t)jV(t).5. The amount of solute that flows into the tank during ~t seconds is r. C0 giL = f.. But as noted in Fig. ( r. To check this. Thus the amount x (t) of solute in the tank satisfies the differential equation dx ro . and we want to compute the amount x(t) of solute in the tank at time t.52 Chapter 1 First-Order Differential Equations Mixture Problems As a first application of linear first-order equations.=r. hter second (~t seconds) FIGURE 1. . We now divide by - ~t ~ riCi . . we consider a tank containing a solution-a mixture of solute and solvent-such as salt dissolved in water.. and that the solution in the tank-kept thoroughly mixed by stirring-flows out at the constant rate of r 0 liters per second. and r 0 are constants. if all the functions involved are continuous and x (t) is differentiable. ~t grams. t + ~t ].r0 )t. To set up a differential equation for x(t). But any other consistent system of units can be used to measure amounts of solute and volumes of solution. 3 ( 90 + t x ) .t.6. and the well-stirred mixture flows out of the tank at the rate of 3 galjmin. due to the differing rates of inflow and outflow.+ px =q dt dx (20) with constant coefficients p = r jV.-x dt v .x in the amount x of salt in the tank from time t to time t + . Eq. Brine containing 2lbjgal of salt flows into the tank at the rate of 4 galj min. and the question is this: When is x (t) = 2c V? With this notation.~ (ert/V- 1) J.6. (21) x(t) = cV +4cve-rr. If the outflow henceforth is perfectly mixed lake water.v_ To find when x (t) = 2c V.901 (years) .1.v [5cV + . we therefore need only solve the equation cV + 4cve.5 Linear First-Order Equations 53 Example 4 Assume th~t L~e Erie has a ~~lume of 480 km3 and that its rate of inflow (from Lake Huron) and outflow (to Lake Ontario) are both 350 km3 per year.pt [xo +~ (eP 1 .t (minutes) is given by .x Solution ~ (4)(2) .= rc. Suppose that at the timet = 0 (years). . The change . = ro = r = 350 (km3 jyr). How much salt does the tank contain when it is full? The interesting feature of this example is that.rt/V = 2cV for t = -ln4 r v = 480 -ln4 350 ~ 1.6. the volume of brine in the tank increases steadily with V (t) = 90 + t gallons. the pollutant concentration of Lake Erie-caused by past industrial pollution that has now been ordered to cease-is five times that of Lake Huron.I) J = e-rr. how long will it take to reduce the pollution concentration in Lake Erie to twice that of Lake Huron? Here we have Solution V = 480 (km3 ). q = rc.t. and integrating factor p = eP1 • You can either solve this equation directly or apply the formula in (12). and xo = x(O) = 5cV. (19) which we rewrite in the linear first-order form .6. ri ci = c (the pollutant concentration of Lake Huron). The latter gives x(t) = e-pt [xo + 1 1 qeP1 dt] = e . (18) is the separable equation dx r .6. • Example 5 A 120-gallon (gal) tank initially contains 90 lb of salt dissolved in 90 gal of water. 5. (b) Find the maximum amount of ethanol ever in tank 2. A cascade of two tanks. 33. 1. Show that y(x) = Yc(x) + Yp(x) is a general solution of dyjdx + P(x)y = Q(x).1.5. with V 1 = 100 (gal) and V2 = 200 (gal) the volumes of brine in the two tanks. A multiple cascade is shown in Fig. 35. The three flow rates indicated in the figure are each 5 gal/min. 39. 1. The only differences are that this lake has a volume of 1640 km3 and an inflow-outflow rate of 410 km3 /year. is a particular solution of dyjdx + P(x)y = Q(x). Pure water flows into tank 1 at 10 galjmin. and the well-mixed brine in the tank flows out at the rate of 3 galjs. all the remaining tanks contain 2 gal of pure water each. find the maximum amount of salt ever in tank 2. and the other two flow rates are also 10 gal/min. = e. Pure water is pumped into tank 0 at 1 galjmin. There is a daily inflow of 500 million ft3 of water with a pollutant concentration of 0. A tank initially contains 60 gal of pure water. How long will it take to reduce the pollutant concentration in the reservoir to 0. Each tank also initially contains 50 lb of salt. FIGURE 1. 55 time t. tank 0 contains 1 gal of ethanol and 1 gal of water. 1. Pure water is pumped into the tank at the rate of 5 Ljs. How much salt will the tank contain when it is full of brine? Consider the cascade of two tanks shown in Fig.200' 31. (b) Suppose that y (t) is the amount of salt in tank 2 at FIGURE 1. A multiple cascade. with pure water flowing into tank 1.5.10%? Rework Example 4 for the case of Lake Ontario. 40. 37. (b) What is the maximum amount of salt ever in the tank? A 400-gal tank initially contains 100 gal of brine containing 50 lb of salt. (c) Suppose that yc(x) is any general solution of dyjdx + P (x) y = 0 and that Yp (x) is any particular solution of dyjdx + P(x)y = Q(x). y(O) = 1.5.5. Suppose that in the cascade shown in Fig.6. Lawrence River and receives inflow from Lake Erie (via the Niagara River). .is pumped out at the same rate.f P(x)dx [! ( Q(x)ef P(x)dx) dx J 32. Brine containing 1 lb of salt per gallon enters the tank at the rate of 5 galj s.05% and an equal daily outflow of the well-mixed water in the reservoir. Show first that dy dt 5x 5y 100. How long will it be until only 10 kg of salt remains in the tank? Consider a reservoir with a volume of 8 billion cubic feet (ft3 ) and an initial pollutant concentration of 0.5 Linear First-Order Equations Problems 31 and 32 illustrate-for the special case of firstorder linear equations-techniques that will be important when we study higher-order linear equations in Chapter 3. A tank contains 1000 liters (L) of a solution consisting of 100 kg of salt dissolved in water. 36. (a) Find constants A and B such that Yp(x) = A sinx + B cosx is a solution of dyjdx + y = 2 sinx.5. (a) Show that Yc(X) = ce. and the mixture-kept uniform by stirring. 34. (c) Finally.25%.5. thus the tank is empty after exactly 1 h. (b) Use the result of part (a) and the method of Problem 31 to find the general solution of dyjdx + y = 2sinx. (a) Find the amount of salt in the tank after t minutes. (a) Find the amount x(t) of salt in tank 1 at time t. (c) Solve the initial value problem dyjdx + y = 2 sin x. (a) Find the amounts x(t) and y(t) of ethanol in the two tanks at time t ~ 0. Brine containing 1 lb of salt per gallon enters the tank at 2 galjmin. (b) Show and then solve for y(t). using the function x(t) found in part (a). tank 1 initially contains 100 gal of pure ethanol and tank 2 initially contains 100 gal of pure water. and the (perfectly mixed) solution leaves the tank at 3 galjmin.6. 38. which empties into the St. At time t = 0.5.f P(x)dx is a general solution of dyjdx that Yp(X) + P(x)y = 0. Meanwhile.5.5. the amount available for her retirement at age 70.9. which accumulates interest at a continuous annual rate of 6%. Then show that dvjdt = g/4.112 • (b) Show by induction on n that tne-t/2 Xn(t) = . 10 8 6 4 2 :>-. How long does it take the pollutant concentration in the reservoir to reach 5 Llm 3 ? 25 X 20 15 10 5 ~~ 10 20 Probl<m 46 Problem 45 30 40 50 60 FIGURE 1.. = mg. 46. Slope field and solution curves for y' = x . Slope field and solution curves for y' = x + y. 12% of her salary is deposited continuously in a retirement account. Suppose that a falling hailstone with density 8 = 1 starts from rest with negligible radius r = 0. The well-mixed water in the reservoir flows out at the same rate.y. (a) Show that every solution curve approaches the straight line y = -x . Verify that the graph of x(t) resembles the steadily rising curve in Fig.9.998. Let Xn (t) denote the amount of ethanol in tank n at time t.7. that the mixtures are kept perfectly uniform by stirring.I asx---+ -oo. The incoming water has a pollutant concentration of c(t) = 10 liters per cubic meter (L/m3 ) . (a) Estimate L'lA in terms of L'lt to derive the differential equation satisfied by the amount A(t) in her retirement account after t years.5.J2irii that Mn ~ (2rrn). 5.999. Assume. 43. Figure 1.002.n.1 as x ---+ +oo. and the positive y-axis points downward.56 Chapter 1 First-Order Differential Equations 44. where m is the variable mass of the hailstone. (b) Foreachofthefivevaluesy 1 = -10. n! 2n (c) Show that the maximum value of Xn(t) for n > 0 is Mn = Xn(2n) = nne-njn!. Thereafter its radius is r = kt (k is a constant) as it grows by accretion during its fall. (a) Show that every solution curve approaches the straight line y = x . Does it seem predictable that the lake's polutant content should ultimately oscillate periodically about an average level of 20 million liters? . (b) For each of the five values y 1 = 3. 4. with an average concentration of 10 L/m 3 and a period of oscillation of slightly over 6 ± months. 0 -2 -4 -6 -8 -10 X FIGURE 1.5. Thus the hailstone falls as though it were under one-fourth the influence of gravity.000 per year.112 • 41. You r first task is to find the amount x (t) of pollutant (in millions of liters) in the reservoir after t months. determine the initial value y0 (accurate to five decimal places) such that y(5) = y 1 for the solution satisfying the initial condition y( -5) =Yo· 10 and the varying mixture in each tank is pumped into the one below it at the same rate.8. 42. 0 -2 -4 -6 -8 -10 X 45. (b) Compute A(40). 4. The incoming water has pollutant concentration c (t) 10(1 + cos t) L/m3 that varies between 0 and 20. Initially it is filled with fresh water. which approaches asymptotically the graph of the equilibrium solution x(t) = 20 that corresponds to the reservoir's long-term pollutant content. as usual.001. v = dyjdt is its velocity. (d) Conclude from Stirling's approximation n! ~ nne. (a) Show that x 0 (t) = e.8 shows a slope field and typical solution curves for the equation y' = x + y. and 4. but at time t = 0 water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month.y . Figure 1.000.5) = Yo· FIGURE 1. Her salary S(t) increases exponentially. Use Newton's second law-according to which the net force F acting on a possibly variable mass m equals the time rate of change dpjdt of its momentum p = m v-to set up and solve the initial value problem d -(mv) dt 8 6 4 2 :>-. Problems 45 and 46 deal with a shallow rese rvoir that has a one square kilometer water suiface and an average water depth of 2 meters. -5. A 30-year-old woman accepts an engineering position with a starting salary of $30. 1. v(O) = 0.7 shows a slope field and typical solution curves for the equation y' = x . 3.5.5. determine the initial value y0 (accurate to four decimal places) such that y(5) = y 1 for the solution satisfying the initial condition y( . with S(t) = 30e 1120 thousand dollars after t years. 0. and I 0.for n > 0. Graphs of solutions in Problems 45 and 46. If we write Newton's law of cooling (Eq. For instance. GA with a minimum temperature of 70°F when t = 4 (4 A.M. but with the outside temperature A(t) given by Eq. then these oscillations have a period of 24 hours (so that the cycle of outdoor temperatures repeats itself daily) and Eq. and we cannot afford to have it repaired until payday at the end of the month.M.5v'3 sin wt . Suppose that our air conditioner fails at time to = 0 one midnight.kt + c1 cos wt + d1 sinwt.2 for a wellinsulated building with tightly sealed windows). (1) instead of a constant ambient temperature A.1. we would take A(t) = 80. Typical values of the proportionality constant k range from 0.2 to 0. or less than 0.5 for a poorly insulated building with open windows. indeed. (1). (4) .5 cos wt. du .4) = 80.5. (1) provides a realistic model for the temperature outside a house on a day when no change in the overall day-to-day weather pattern is occurring. consider indoor temperature oscillations that are driven by outdoor temperature oscillations of the form A(t) = a0 + a 1 coswt + b 1 sinwt.9. You may want to use the integral formulas in 49 and 50 of the endpapers. resemble the oscillatory curve shown in Fig. You should get the solution SCENARIO: u(t) = ao + coe.5 Application For an interesting applied problem that involves the solution of a linear differential equation.5 Linear First-Order Equations Verify that the graph of x(t) does.{3) = cos a cos f3 +sin a sin f3 to get ao = 80. Begin your investigation by solving Eq.+ ku = k(ao + a 1 coswt + dt b1 sinwt) (3) with coefficient functions P(t) = k and Q(t) = kA(t).). we get the linear first-order differential equation dt = that is. We therefore want to investigate the resulting indoor temperatures that we must endure for the next several days. or possibly a computer algebra system. du -k(u. and b1 = -5v'3 in Eq. (1) If w = n /12.5 (although k might be greater than 0. (2) by using the identity cos(a . (3) with the initial condition u(O) = u 0 (the indoor temperature at the time of the failure of the air conditioner). 1. (2) We derived Eq.1) for the corresponding indoor temperature u(t) at timet. a 1 = -5.) and a maximum of 90°F when t = 16 (4 P.A(t)). How long does it 57 take the pollutant concentration in the reservoir to reach 5 L/m 3 ? 1. (3) of Section 1.10 cosw(t. for a typical July day in Athens. 5.-----. 71.. Solution curves given by Eq..--..3351) cos 12 ...the indoor temperature "settles down" within about 18 hours to a periodic daily oscillation. Comparison of indoor and outdoor temperature oscillations. Indeed. b 1 = -5. = rr/12. 90 :S 80 ::! eo 85 75 = 80.. 1.--. (What is the winter-summer difference for the indoor temperature problem?) You may wish to explore the use of available technology both to solve the differential equation and to graph its solution for the indoor temperature in comparison with the outdoor temperature. JTt (5) 90 :3 80 ::! eo 85 as t ~ Observe first that the "damped" exponential term in Eq. and k = 0. But the amplitude of temperature variation is less indoors than outdoors. comparison of Eqs.(5.11. '92. .----.J3 (as in Eq.5 hours.6036) sin 12 . so the hottest part of the day inside is early evening rather than late afternoon (as outside).5.5082-4 ~ 3. a 1 = -5. Consequently.--.(6.M.---.2 JTt u(t) = 80 + e-r. With a0 = 80.-----.0707) cos 12 (t - 7T (7) 70 65 600 FIGURE 1..5082). Do you see that this implies that the indoor temperature varies between a minimum of about 74 oF and a maximum of about 86 oF? Finally.---. this solution reduces (approximately) to 1()() .s (u 0 - 95 82.11. (6) can be rewritten (verify this!) as l 00 . Eq.5.10.(5... leaving the long-term "steady periodic" solution JTt JTt 75 70 65 u 5p(t) = 80 + (2.-----.. the long-term indoor temperatures oscillate every 24 hours around the same average temperature 80° F as the average outdoor temperature. (5) approaches zero +oo.---r----. For a personal problem to investigate.-. (5) with u 0 = 65.3351) + (2. each evening.-.3351) cos-.6036) sin. (2)).(6.. Thus the temperature inside the house continues to rise until about 7:30 P.5. Figure 1. 68.---. w (for instance).9656) 7. Outdoor 95 u(t) = 80. You might also consider a winter day instead of a summer day..58 Chapter 1 First-Order Differential Equations where dl = kwa1 + k 2 b1 k2 + w2 with w = rr/12. carry out a similar analysis using average July daily maximum/minimum figures for your own locale and a value of k appropriate to your own home. (2) and (7) indicates that the indoor temperature lags behind the outdoor temperature by about 7. . as illustrated in Fig. Observe that-whatever the initial temperature. using the trigonometric identity mentioned earlier.-----.0707) cos ( 7~ - 1.10 shows a number of solution curves corresponding to possible initial temperatures u 0 ranging from 65°F to 95°F. 95. 12 12 (6) 600 FIGURE 1. y). (5) is one we can solve. (5). If the substitution relation in Eq. V) and a{Jjav = fJv(X. If this new equation is either separable or linear. v). If we substitute the right-hand side in (4) for d y jdx in Eq. Example 1 Solve the differential equation . (1) with dependent variable y and independent variable x. v(x )) will be a solution of the original Eq. But many applications involve differential equations that are neither separable nor linear. V) are known functions of x and v.6 Substitution Methods and Exact Equations 59 - Substitution Methods and Exact Equations The first-order differential equations we have solved in the previous sections have all been either separable or linear. v ) (5) with new dependent variable v. (2). the result is a new differential equation of the form dv dx = g(x . the differential equation dy dx = f(x. The trick is to select a substitution such that the transformed Eq. then we can apply the methods of preceding sections to solve it. (1). For instance. Thus the differential equation dy dx = (x + y + 3)2 practically demands the substitution v = x + y + 3 of the form in Eq. If v = v(x) is a solution of Eq. y) (2) of x and y that suggests itself as a new independent variable v. it may require a fair amount of ingenuity or trial and error. Even when possible. (1) and then solve for dvjdx .1. then y = fJ(x. may contain a conspicuous combination v = a(x . this is not always easy. (2) can be solved for y = fJ(x. In this section we illustrate (mainly with examples) substitution methods that sometimes can be used to transform a given differential equation into one that we already know how to solve. (3) then application of the chain rule-regarding v as an (unknown) function of xyields (4) where the partial derivatives a{Jjax = fJx (X. We see that. dy dv -= v + x -.1 v +C. dx dx (8) . This situation is fairly typical of nonlinear differential equations. y = v . in contrast with linear differential equations. whose solutions are continuous wherever the coefficient functions in the equation are continuous.x . dx x If we make the substitutions (7) v= . and we have no difficulty in obtaining its solution 10 8 6 1111111 I I I I I I dv Ill I I II I I I I I x = I I I I I I I I I I 11111111111 1111111 Ill I dv = tan.. 1+v 2 4 2 . n/2). although the function f (x. 0 H-il"7'::::::-=-':-':--7-t!77-f+++H-.x .60 Chapter 1 First-Order Differential Equations Solution As indicated earlier.3. each solution is continuous only on a bounded interval. y(x) = tan(x. • Example 1 illustrates the fact that any differential equation of the form - dy dx = F(ax +by +c) (6) can be transformed into a separable equation by use of the substitution v = ax + by + c (see Problem 55). the general solution of the original equation dyjdx = (x + y + 3) 2 is x + y + 3 = tan(x. y) = (x+ y+3) 2 is continuously differentiable for all x andy.6. Homogeneous Equations A homogeneous first-order differential equation is one that can be written in the form dy = F (~). The paragraphs that follow deal with other classes of first-order equations for which there are standard substitutions that are known to succeed. Slope field and solution curves for y' = (x + y + 3) 2 • Remark: Figure 1.C) . • -5 X 5 FIGURE 1. C -n/2 < x < C + n j2. because the tangent function is continuous on the open interval ( -n/2.:-H -2 -4 -6 -8 I I I -JO I I I I I I So v = tan(x . that is..C < n j2. Because v = x + y + 3. that is. let's try the substitution V =X+ y + 3.1. X y y = vx .3. dx dx so the transformed equation is = 1 + v 2• dx This is a separable equation. In particular.1 shows a slope field and typical solution curves for the differential equation of Example 1.C) .. - that is. the particular solution with arbitrary constant value C is continuous on the interval where -n /2 < x..6. Then dy dv = --1.C). y ) y ' = Q(x . a differential equation of the form P (x . and so forth-is the equation dy _ (y)q A --B + C(y)s ( y)n x dx X X which evidently can be written (by another division) in the form of Eq.xm+n = (yfxt. but we recognize it as a homogeneous equation by writing it in the form dy dx 4x2 + 3y2 = 2 2xy (~) + ~ (~). then the result-because xmyn." Consider a differential equation of the form whose polynomial coefficient functions are "homogeneous" in the sense that each of their terms has the same total degree.= 4x 2 + 3l. y ) with polynomial coefficients P and Q is homogeneous if the terms in these polynomials all have the same total degree K. If we divide each side of() by xK. dy dx dv dx y v = -.6 Substitution Methods and Exact Equations then Eq. X and 1 X v y v+ x and hence dv x dx dv 2 =dx v + -v .1.dv = v2 +4 J 2_ dx. dx Solution This equation is neither separable nor linear. (7) is transformed into the separable equation dv x . m + n = p + q = r + s = K.= F(v). v v2 + 4 2v ' 3 2 = ~+2= 2 f --:- 2-v. ··········· ·······- Example 2 Solve the differential equation dy 2xy. The differential equation in the following example is of this form with K = 2.v. Remark: A dictionary definition of "homogeneous" is "of a similar kind or nature. More generally. -=v+x-. . x ln(v 2 + 4) = ln lxl + ln C. y 2 x The substitutions in (8) then take the form y = These yield VX. (7). dx 61 Thus every homogeneous first-order differential equation can be reduced to an integration problem by means of the substitutions in (8). we finally obtain _a y(x)-- 2 [(x)l-k . The components of the velocity vector of the airplane. ( 16) and then solving for v is (18) Because y = xv. The plane travels with constant speed vo relative to the wind.VoY. As indicated in Fig. (12) takes the homogeneous form (14) The substitution y = xv. Figure 1.1. (16) = 0 yields (17) C=klna. y' = v + xv' then leads routinely to J.k In x + C.5.wJx2 dx dxjdt v0 x If we set X k= . As we ask you to show in Problem 68.::=::::::::=::::: dt )x2 + yz' - dy . 0) located due east of its intended destination-an airport located at the origin (0. (12) dy = dyjdt 1 ( . They are (a. = -vo sm () dt +w = - VoY )x2 + y2 + w. the ratio of the windspeed to the plane's airspeed.6. then Eq. 0). Hence the trajectory y - = f (x) of the plane satisfies the differential equation + y2 ) .= .o= .J1 + In ( v and the initial condition v(a) ---. we assume that the plane's pilot maintains its heading directly toward the origin.--. vo w (13) FIGURE 1. 1.4.5 helps us derive the plane's velocity components relative to the ground. The airplane headed for the origin.6. the result of substituting ( 17) in Eq. we find that + J1 + v2) = y(a)ja = .(x)l+k] a a (19) for the equation of the plane's trajectory.v2 - J~ x dx · (15) By trigonometric substitution.6. - dx vox = . which is blowing due north with constant speed w.6 Substitution Methods and Exact Equations 65 Flight Trajectories y y =f(x) Suppose that an airplane departs from the point (a.vo cos () = .6. 0) x FIGURE 1. or by consulting a table for the integral on the left. .4.== d v===.. Eq. y ). On the other hand. If w = v 0 (so that k = 1). so the plane's trajectory approaches the point (0._) 1/5] 200 5 200 ' and we readily solve the equation y ' (x) = 0 to obtain (x /200) 215 = ~. ax ay d x that is. (19) pass through the origin. aj2) rather than (0. Provided that Eq. The three cases w < v0 (plane velocity exceeds wind velocity). (19) yields y(x) = 100 [ ( 200 X )4/5. y ) and N(x . y) = Fy(x. w = v0 (equal velocities).::.6. (20) yields dy = dx 2 ! [~ 5 (..6. 0) X Ymax = 100 2) . The vertical scale there is exaggerated by a factor of 4. FIGURE 1.6.) • Exact Differential Equations We have seen that a general solution y(x) of a first-order differential equation is often defined implicitly by an equation of the form F(x.. so that the plane reaches its destination. (19) that y ---+ +oo as x ---+ 0.6. With these values. The situation is even worse if w > v0 (so k > 1)-in this case it follows from Eq.::_)-1/5. w < v0 ) does the curve in Eq. and w > v0 (wind is greater). y ) + N(x.6.=0. (21) where C is a constant.= 0. y ) = Fx(x . 1.66 Chapter 1 First-Order Differential Equations Note that only in the case k < 1 (that is. 0). 0). v0 = 500 mijh. y(x)) = C. (The graph of the function in Eq. y). 1. (19) takes the form y(x) = ~a(l . this gives the original differential equation in the form aF aF dy + . what is the maximum value of y(x) for 0 ~X~ 200? Solution y Differentiation of the function in Eq. dx (22) where M (x.(200 )6/5] · X (20) Now suppose that we want to find the maximum amount by which the plane is blown off course during its trip.~ (. Example 7 If a= 200 mi. (20) is the one used to construct Fig. then k = wjv0 = ~'so the plane will succeed in reaching the airport at (0.(2) 3 [( 3 2 3 ] = 400 27 ~ 14.4. then Eq. given the identity in (21). . (21) implicitly defines y as a differentiable function of x . Hence (a.x 2ja 2 ). we can recover the original differential equation by differentiating each side with respect to x. The three cases are illustrated in Fig. That is.. and w = 100 mijh. Thus the plane is blown almost 15 mi north at one point during its westward journey. dy M(x .81... y) such that aF -=M ax then the equation F(x. y) dx + N(x. then the differential equation in question is not exact. aN ax aM ay aN ax (24) is a necessary condition that the differential equation M dx + N dy = 0 be exact. Eq. The general first-order differential equation y' = f (x. The preceding discussion shows that. (23). if My =F Nx. Natural questions are these: How can we determine whether the differential equation in (23) is exact? And if it is exact. if you prefer. (22) in the more symmetric form M(x. it then follows that Thus the equation aM ay = Fxy = Fyx = -.113 • • . (23) is called an exact differential equation-the differential dF = Fxdx + Fydy of F(x.1. (23) is exact and M and N have continuous partial derivatives.6 Substitution Methods and Exact Equations 67 It is sometimes convenient to rewrite Eq. if there exists a function F (x. Example 8 The differential equation y 3 dx + 3xl dy = 0 (25) is exact because we can immediately see that the function F (x. (23) called its differential form. implicitly defines a general solution of Eq. y) such that Fx = M and Fy = N-there is no such function. y) = C and -=N aF ay . y) dy = 0. That is. so we need not attempt to find a function F (x. y) = x y 3 has the property that Fx = y 3 and Fy = 3xy 2 . y) and N = -1. In this case. (25) is xi =C. how can we find the function F such that Fx = M and Fy = N? To answer the first question. then they are equal: Fxy = Fyx· If Eq. let us recall that if the mixed second-order partial derivatives Fxy and Fyx are continuous on an open set in the xy-plane. y) is exactly M dx + N dy. y(x) = kx . Thus a general solution of Eq. y) can be written in this form with M = f (x. y)= J M(x. (24) holds on R. y) defined on R with aF jax = M and aF jay= N if and only ifEq. In other words. y)dy = 0 (23) is exact in R if and only if aM ay aN ax (24) at each point of R.) We plan to choose g (y) so that aF = N = ay (a J ay ay M(x.68 Chapter 1 First-Order Differential Equations But suppose that we divide each term of the differential equation in Example 8 by y 2 to obtain + 3xdy = 0. whether a given differential equation is exact or not is related to the precise form M dx + N dy = 0 in which it is written. (24) is not satisfied. That is. y) with respect to x. y)dx + N(x. (27). the notation J M(x. The differential equations in (25) and (26) are essentially equivalent. ay ax (26) Hence the necessary condition in Eq. then the differential equation M dx + N dy = 0 is exact. for any function g(y). y) and N (x. so that g I (y) = N. Theorem 1 tells us that (subject to differentiability conditions usually satisfied in practice) the necessary condition in (24) is also a sufficient condition for exactness. and they have exactly the same solutions. y) dx denotes an antiderivative of M (x. Then the differential equation M(x. (28) . (In Eq. we must show that if Eq. Proof: We have seen already that it is necessary for Eq. c < y < d. with M = y and N = 3x.y)dx ) I +g(y) as well. (24) to hold if Eq. We are confronted with a curious situation here. then we can construct a function F(x. if My = Nx. In brief. yet one is exact and the other is not. y) such that aF. y) are continuous and have continuous first-order partial derivatives in the open rectangle R: a < x < b. there exists a function F (x. that is.ax = M and aFjay = N. To prove the converse. This equation is not exact because. Note first that. we have ydx aM aN -=1#3=-. (24) holds. (23) is to be exact.y)dx+g(y) (27) satisfies the condition aFjax = M. the function F(x.- a/ M(x. y)dx. THEOREM 1 Criterion for Exactness Suppose that the functions M (x. 6 Substitution Methods and Exact Equations 69 To see that there is such a function of y. y)dx =---=0 by hypothesis.. First we integrate M (x . (27) to obtain F(x.l ) dx = 3x 2 y . ay ax Integrating a F . Then we differentiate with respect toy and set aFjay = N(x. This yields a general solution in the implicit form F(x. we get F(x. indeed. y) = J M(x. y) dx + g(y). y) with respect to x and write F(x._!_ ~ ax ax ay aN a a = -. This yields aF = 3x 2 ay 3xl + g'(y) = 4y + 3x 2 - 3xl. Then we determine g (y) by imposing the condition that aFjay = N(x. it suffices to show that its derivative with respect to x is identically zero. (28). y). y) = 6xy. y) =C. (30) Solution Let M(x.ax ay ax aN ax aM ay J J M (x. We substitute this result in Eq. y) with respect to x. y). .1. y) dx M(x. We can then find g(y) by integrating with respect toy. But _!_ ax (N - _!_ ay J M (x. (28) is a function of y alone.aay J M(x. and hence on an interval as a function of x.xl + g(y).3xy 2 • The given equation is exact because aM aN 2 =6x-3y = . (29). Because the right-hand side in Eq. y) = J - (6xy . y) = 4y + 3x 2 .ax = M (x. y) dx + J( N(x. (27) and (28). y). Example 9 Solve the differential equation (6xy . So we can. • Instead of memorizing Eq. find the desired function g(y) by integrating Eq. y) = J M(x. (28) is defined on a rectangle. it is usually better to solve an exact equation M dx + N d y = 0 by carrying out the process indicated by Eqs. y) dx) = aN . y) dx) dy (29) as the desired function with Fx = M and Fy = N. it suffices to show that the right-hand side in Eq.l ) dx + (4y + 3x 2 - 3xl) dy = 0.y 3 and N(x. thinking of the function g(y) as an "arbitrary constant of integration" as far as the variable x is concerned. and thus has the general form F(x . Hence g(y) = 2y 2 + C 1 . a general solution of the differential equation is defined implicitly by the equation 5 4 3 2 (31) (we have absorbed the constant C 1 into the constant C)..6.75. 2. then it is easily reduced by a simple substitution to a firstorder equation that may be solvable by the methods of this chapter. y'. y") = 0. If y is missing. p. If we can solve this equation for a general solution p(x. The other two special points in the figure-at (0..7. (32) takes the form F(x. (32) If either the dependent variable y or the independent variable x is missing from a second-order equation.12)-are ones where both coefficient functions in Eq. C. y. (30) vanish. y'. The solution satisfying a given initial condition y(xo) = Yo is defined implicitly by Eq. Dependent variable y missing.70 Chapter 1 First-Order Differential Equations and it follows that g'(y) = 4y. and thus Therefore. Slope field and solution curves for the exact equation in Example 9.7 shows a rather complicated structure of solution curves for the differential equation of Example 9. C 1 ) involving an arbitrary constant C 1 . Reducible Second-Order Equations A second-order differential equation involves the second derivative of the unknown function y(x). (33) that involves two arbitrary constants C 1 and C2 (as is to be expected in the case of a second-order differential equation).)dx + Cz to get a solution of Eq. (33) Then the substitution I dy p = y = dx' II y dp dx (34) results in the first-order differential equation F(x. (31). p') = 0. the particular solution satisfying y(O) = 1 is defined implicitly by the equation 3x 2 y . For instance. 0 ~ -~~~~~~~~~ -I • -2 -3 -4 -5UL~L-~~LL-L~~~ X Remark: Figure 1. . with C determined by substituting x = x 0 and y = y 0 in the equation.3 does not guarantee a unique solution. then we need only write y(x) = J y'(x)dx = J p(x . . 0) and near (0. so the theorem of Section 1. • FIGURE 1.6.:.xy 3 + 2y2 = 2. y") = 0. then Eq. Solutions of the Clairaut equation of Problem 67. suppose that a = 100 mi. (c) If w = 6 ftjs. which of course changes only by integral increments-that is. substitute p = y' to derive a general solution of the second-order differential equation (with r constant) in the form y+ Jx + y 2 2 = 2(200x 9 ) 1110 • Thus a circle of radius r (or a part thereof) is the only plane curve with constant curvature 1/r. by one birth or death at a time. always heading toward a tree at (0. . (16) and (17). 0) on the west bank directly across from the dog's starting point.4 we introduced the exponential differential equation dPjd t = k P. • 8 (t) is the number of deaths per unit of population per unit of time at time t. the parabola at the point iC C!C. one learns that the curvature K of the curve y = y(x) at the point (x. A dog starts at (100. 7th edition (Upper Saddle River. (18) in this section from Eqs. y) is given by ly''(x)l K = ---'----:----:-:-::(1 + y'(x)2]3/2.10. and w = 40 mijh.6. t + ~t] of length ~tis ~p = {births} - {deaths} ~ {J(t) · P(t) · ~t - o(t) · P(t) · ~t. iC 68. In the calculus of plane curves. 0) and swims at v0 = 4 ftjs. (a) If w = 2 ftjs. Derive Eq. In the situation of Example 7. The "typical" straight 2 is tangent to line with equation y = Cx2). It is customary to track the growth or decline of a population in terms of its birth rate and death rate functions defined as follows: • f3 (t) is the number of births per unit of population per unit of time at time t. A river 100ft wide is flowing north at w feet per second. with solution P (t) = P0 ekr. t + ~t] is given (approximately) by deaths: 8(t) · P(t) · ~t. show that the dog reaches instead the point on the west bank 50ft north of the tree. As in the text discussion. as a mathematical model for natural population growth that occurs as a result of constant birth and death rates.] Conversely. [t. 2008). 69. Here we present a more general population model that accommodates birth and death rates that are not necessarily constant.6 of Edwards and Penney. 150). Suppose that the population changes only by the occurrence of births and deaths-there is no immigration or emigration from outside the country or environment under consideration. Calculus: Early Transcendentals. and the plane begins at the point (200. Hence the change ~ P in the population during the time interval [t. v0 = 400 mijh. FIGURE 1. our population function P(t) will be a continuous approximation to the actual population. show that the dog never reaches the west bank. As before. show that the dog reaches the tree. Now how far northward does the wind blow the airplane? 70. (b) If w = 4 ft/s. NJ: Prentice Hall. D P'?pulation Models I n Section 1. show that its trajectory is described by and that the curvature of a circle of radius r is K = 1j r. however. 72. Then the numbers of births and deaths that occur during the time interval births: {J(t) · P(t) · ~t.74 Chapter 1 First-Order Differential Equations 71. suppose that an airplane maintains a heading toward an airport at the origin. If v0 = 500 mijh and w =50 mijh (with the wind blowing due north). [See Example 3 in Section 11. then Eq. Eq.0005) P 2 in Example 1.B = .B = (0.. Slope field and solution curves for the equation dP/dt = (0.1. .·~iii·g. 1 . 0 10 20 30 40 50 FIGURE 1. But it also includes the possibility that f3 and 8 are variable functions of t. dt P(O) = 100 (with t in years). the direction field and solution curves shown in Fig.8. Then upon separating the variables we get f. Equation (1) is the general population equation.7.0005) P-and thus increases as the population does. If the birth rate is .0005)? .7 Population Models 75 so !1P !1t ~ [. p = 0. and then we readily solve for P(t) = 20.finite period of time. In particular. 1.B and 8 are constant.t.ti~~-~~~b~~. and P = P(t) for brevity. Indeed. 8 = 8(t). If .= (0. they may well depend on the unknown function P(t). it appears that the • population always becomes unbounded in a . But we see that P -+ +oo as t -+ 20. so after 10 years the alligator population has doubled.B(t) . (1) in which we write .th. The birth and death rates need not be known in advance.7. Substitution oft 2 dP = f (0. ( 1) gives the initial value problem dP 2 .8(t)] P(t).1..1 indicate that a population explosion always occurs. so-taking the limit-we get the differential equation Tt dP = (... P (1 0) = 2000 j 10 = 200. (1) reduces to the natural growth equation with k = . For instance.t~~p~p~i. lllt'·'l·'·'tsl' -··s~pp~.~ ·th~t· .~at~-is 8 = 0 (so none of the alligators is dying).8.:· ~~ctrh~tits d~.B .8)P.= (0.0005) =2000 dt. The error in this approximation should approach zero as !1t -+ 0.0005)t +C. P = 100 gives C 1/100. so a real "population explosion" occurs in 20 years. whatever the size of the (positive) initial population P(O) = P0 .--1ooi~iti~ii.B(t). .f3t P .o6r ___ = ±ec eo.Po).{3 1 P.0004P(150. (2) that is.0004P 2 .~" · '•·-· . it is useful to rewrite the logistic equation in the form dt where k m"''''""" " .PI= 0. • -· . ..p 150.0.P). (2) is called the logistic equation. so that f3 = {30 .0._._. this equation is easy to solve for the population 150Po P(t) = .Po Finally. .. Hence [where B = ±ec ].· •··•• ""' "' •··. 0. then Eq. we find that Poe0.~ dP = kP(M.76 Chapter 1 First-Order Differential Equations Bounded Populations and the Logistic Equation In situations as diverse as the human population of a nation and a fruit fly population in a closed container.p p f f # 0. .-..h. dt (4) To solve this differential equation symbolically.Po )e. We get 1 150 f (_!__ + p f dP = P(150.8o)P. then Eq. = Po p 150 into this last equation. Suppose.06t 150. that the birth rate f3 is a linear decreasing function of the population size P. In IPI -In 1150. --• ··• •.u.=aP-bP dt ' where a= f3o. we separate the variables and integrate.•. ·-•.8o and b = f3t· If the coefficients a and b are both positive. " " ' . where f3o and {3 1 are positive constants.0004dt .. . ." """ .. ~ ... The reasons may range from increased scientific or cultural sophistication to a limited food supply.P) = 0._. Example 2 In Example 4 of Section 1.0004dt [partial fractions] . If the death rate 8 = 80 remains constant.3 we explored graphically a population that is modeled by the logistic equation - dP = 0.Po+ (150 ..o6r = If we substitute t = 0 and P B = Po/ (150 . for example. For the purpose of relating the behavior of the population P(t) to the values of the parameters in the equation. dP 2 .p + C. (1) takes the form dP dt = (f3o.. it is often observed that the birth rate decreases as the population itself increases.P) 1 ) dP = 150. (3) = b and M = ajb are constants.06r (5) .06r 150. B eo.06P.. it approaches the finite limiting population M as t ~ +oo.< .0004P 2 • The finite limiting population noted in Example 2 is characteristic of logistic populations. Typical solution curves for the logistic equation P' = kP(M. if Po > M. If Po = M.Po)e-kMr --------------.0. Po + 0 (8) FIGURE 1. but steadily decreases and approaches M from above if Po > M.2 shows a number of solution curves corresponding to different values of the initial population ranging from Po = 20 to Po = 300. Typical solution curves for the logistic equation P' = 0. Po However. Indeed. the behavior of a logistic population depends on whether 0 < Po < M or Po> M. whatever the initial value Po > 0. P(O) = Po (6) P(t) = MP0 Po + (M .7 Population Models p 77 300 240 1801 150r 120 60 :::=§3111~!!!!!!!~====-- at time t in terms of the initial population Po = P (0). Example 3 . Assume that this population satisfies the logistic equation.Po)e-kMt (7) Actual animal populations are positive valued. you should be able to see directly from Eq. then we see from (6) and (7) that P' < 0 and P(t) = p MPo Po + (M . 7.3. It follows that lim P(t) = t --+ +oo MPo = M.> MPo Po + {neg. Note that all these solution curves appear to approach the horizontal line P = 150 as an asymptote. In Problem 32 we ask you to use the method of solution of Example 2 to show that the solution of the logistic initial value problem dt IS dP = kP(M. the population P(t) steadily increases and approaches M from below ifO < Po< M.) Thus a population that satisfies the logistic equation does not grow without bound like a naturally growing population modeled by the exponential equation P' = k P.P). Sometimes M is called the carrying capacity of the environment. IfO < Po< M.06P. Suppose that in 1885 the population of a certain country was 50 million and was growing at the rate of 750.2. As illustrated by the typical logistic solution curves in Fig. 1.7. (See Problem 34. • 20 Limiting Populations and Carrying Capacity 25 50 -r--~----~--~----~1 75 100 FIGURE 1. (5) that lim1--+oo P(t) = 150.- Po + MPo {pos. Suppose also that in 1940 its population was 100 million and was then growing at the rate of 1 million per year.Po)e-kMt -------------. then (7) reduces to the unchanging (constant-valued) "equilibrium population" P(t) M. Each solution curve that starts below the line P = M/2 has an inflection point on this line.3. considering it to be the maximum population that the environment can support on a long-term basis.. Instead. Figure 1.7. Otherwise. Determine both the limiting population M and the predicted population for the year 2000. number} ----M MPo Po - · In either case.7.1.000 people per year at that time.P). the "positive number" or "negative number" in the denominator has absolute value less than Po and--because of the exponential factor-approaches 0 as t ---+ +oo. number} MPo =M. then we see from (6) and (7) that P' > 0 and = P(t) = MPo Po+ (M. 78 Chapter 1 First-Order Differential Equations Solution We substitute the two given pairs of data in Eq. If we take Po = 5.212 (for 1900) in the logistic model formula in Eq.212 = 5..oooi)(200)(60) ' 100. Verhulst as a possible model for human population growth. P = 76.192 million.308)e(0. (7). we find that 76. population census data.121.S.03I55I )r (10) .192 (for 1850) and t = 100.308 + (M. .50). population during the 19th century is P(t) = (5.308 and substitute the data pairs t =50. F.7% per year. . (7)-the population in the year 2000 will be P(60) = ----------:-::-:=--=-=-=:-:-:-::::100 + (200.000167716.100)e-< o. (7) yields the logistic model 998.026643)r (9) (with t in years and P in millions). (3) and find that 0. and with t = 0 corresponding to the year 1940 (in which Po = 100).(0. We solve simultaneously forM = 200 and k = 0. Thus the limiting population of the country in question is 200 million. Nonlinear systems like this ordinarily are solved numerically using an appropriate computer system.308e 100r. Thus our natural growth model for the U. M = 188. • Example 5 The U. Example 4 The U.75 = 50k(M.308 ~ 0.5.308)e-IOOkM in the two unknowns k and M.100).00 = lOOk(M.S. With these values of M and k. we find thataccording to Eq. Because e 0·026643 ~ 1.308)e .02700. Substitution of these values in Eq.308 million and in 1900 was 76. But with the right algebraic trick (Problem 36 in this section) the equations in (1 0) can be solved manually for k = 0.813)e.026643. P = 76. population in 1800 was 5.212 5.0001.308 (with t = 0 in 1800) in the natural growth model P(t) = Poerr and substitute t = 100. • Historical Note The logistic equation was introduced (around 1840) by the Belgian mathematician and demographer P. then compare projections for the 20th century.7 million people.212 r = ln 100 5. 1. If we take Po= 5.192 5.S.546 (11) P(t) = · 5. In the next two examples we compare natural growth and logistic model fits to the 19th-century U.308)M = 76.308 + (M. population in 1850 was 23.212 million.S. the average population growth between 1800 and 1900 was about 2.212. so 1 76. P = 23. we get the two equations (5308 )M = 23.308 + (182.50kM ' (5.200 about 153.5. 805 15.638 12.S.492 221.253 34.113 26. it is customary to define the average error (in the model) as the square root of the average of the squares of the individual errors (the latter appearing in the fourth and sixth columns of the table in Fig.849 169.000 0.4).237 288.593 0.972 25. ·.000 1.212 92.387 76. By the end of the 20th century the exponential model vastly overestimates the actual U.S.326 50.620 -288.354 154.721 6.437 0.394 0. In order to measure the extent to which a given model fits actual data.079 5.052 161.308 7. the exponential error is considerably larger during the 19th century and literally goes off the chart during the first half of the 20th century.202 9.001 1.710 281.316 173.4 compares the actual 1800-1990 U.251 -23.203 132.·.252 177.192 30.323 203.501 23.545 -0. Both agree well with the 19th-century figures.7. .240 9. while the average error in the logistic model is only 0. -.459 4. But the exponential model diverges appreciably from the census data in the early decades of the 20th century.• Exponential .943 492.038 Logistic Error 0.271 41.422 5.834 132.4.730 58.5.192 31.155 0.594 1. Consequently.452. 1. population-predicting over a billion in the year 2000-whereas the logistic model somewhat underestimates it.'.7. ' odeli .780 376. even in 1900 we might well have anticipated that the logistic model would predict the U.818 Logistic Model 5.990 168.312 58.023 642.929 9.308 1094.861 17.228 106. Comparison of exponential growth and logistic models with U. where plots of their respective errors-as a percentage of the actual population-are shown for the 1800-1950 period.290 5.827 -46.598 -812.694 -589.165 151.289 -89.212 99.226 76.S. census population figures with those predicted by the exponential growth model in (9) and the logistic model in (11).000 -7. '.479 129.022 123.834 105.213 90.980 76.056 1.~ . • .0.369 . We see that the logistic model tracks the actual population reasonably well throughout this 150-year period.097 -0.7.612 119. 1.044 11.384 Error 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 5.443 38.034 62.409 20.410 3.7 Population Models 79 The table in Fig.'.735 13.308 6. However.038 -0.458 104.886 144.072 .435 76.162 for the average error in the exponential model.<::. The two models are compared in Fig.~~~~btlaJ(.S.064 23. FIGURE 1. census populations (in millions). population growth during the 20th century more accurately than the exponential model.655 3.095 17.234 -0.189 62.236 838. 1.768 0.558 50.268 44.311 0.721 -415.454 -197.1. Using only the 1800-1900 data. whereas the logistic model remains accurate until 1940.000 0.5.:t:-:-~-:::-:: '.326 179.7. Error 0.137.542 248.308 7.7.190 4.240 FIGURE 1. • ·.. this definition gives 3.038 -0. Percentage errors in the exponential and logistic population models for 1800-1950.302 226.405 39. Limited environment situation. Then P' (t) should be proportional to the product of the number P of individuals having the disease and the number M . in which all deaths result from chance encounters between individuals. 3.P of those not having it. then . and therefore dPjdt = kP(M . when will 80% of the city's population have heard the rumor? Substituting Po= 10 and M = 100 (thousand) in Eq. Then f3 . It is then reasonable to expect the growth rate f3 . The mathematical description of the spread of a rumor in a population of M individuals is identical. we get 1000 P(t) = 10 + 90e. If the birth rate f3 is constant but the death rate 8 is proportional to P. Assuming that P(t) satisfies a logistic equation.P). Then substitution oft (12) Solution = 1. Of course.P as the potential for further expansion. More Applications of the Logistic Equation We next describe some situations that illustrate the varied circumstances in which the logistic equation is a satisfactory mathematical model.80 Chapter 1 First-Order Differential Equations The moral of Examples 4 and 5 is simply that one should not expect too much of models that are based on severely limited information (such as just a pair of data points). The classic example of a limited environment situation is a fruit fly population in a closed container.8 (the combined birth and death rates) to be proportional to M . Competition situation.P). because we may think of M . so that dP dt = ({3. 10 thousand people in a city with population M = 100 thousand people have heard a certain rumor. Again we discover that the mathematical model is the logistic equation. A certain environment can support a population of at most M individuals.JOOkr. The disease is spread by chance encounters. competition between individuals is not usually so deadly. Example 6 Suppose that at timet = 0. so that 8 = a P.P). Let P(t) denote the number of individuals in a constant-size susceptible population M who are infected with a certain contagious and incurable disease. nor its effects so immediate and decisive. P = 20 gives the equation 20 = 1000 10 + 90e.8)P = kP(M.8 = k(M . (7). Much of the science of statistics is devoted to the analysis of large "data sets" to formulate useful (and perhaps reliable) mathematical models. 2. This might be a reasonable working hypothesis in a study of a cannibalistic population. 1. Joint proportion situation.= dP dt ({3.aP)P = kP(M- P).100k .P. After 1 week the number P(t) of those who have heard it has increased to P(1) = 20 thousand. With P(t) = 80.06P. p ___ = ±ece.. lOOk In 4 thus after about 4 weeks and 3 days. We will see that the constant M is now a threshold population.150) (~p 1 p.150) = 0.. (14) We want to find P(t) if (a) P(O) = 200.9 ~ 4.42.0004dt .0004P(P. In IPI-In IP.150 ) dP =J 0.1.-~-- . We get f 1 -.150 [where B = ±ec ].06t + C. Example 7 Consider an animal population P(t) that is modeled by the equation dP dt = 0. Note that the right-hand side in Eq. with k constant). (b) P(O) = 100.0.0. (13) is the negative of the right-hand side in the logistic equation in (3). then the general population equation in (1) yields the differential equation dP 2 . we separate the variables and integrate. (12) takes the form 1000 80=----10 + 90e-lOOkt' which we solve for e-lOOkt = 3~. with the way the population behaves in the future depending critically on whether the initial population Po is less than or greater than M.7 Population Models 81 that is readily solved for e-IOOk = ~.= kP .M) dt (13) (where M = 8/k > 0) as a mathematical model of the population. Solution To solve the equation in (14). Eq.= . • Doomsday versus Extinction Consider a population P(t) of unsophisticated animals in which females rely solely on chance encounters to meet males for reproductive purposes.06r p .. If the death rate 8 is constant. (15) . · -~ .J 150 dP = J0.0004dt [partial fractions]. hence at a rate proportional to P 2 • We therefore assume that births occur at the rate k P 2 (per unit time.06r = Be.8P = kP(P.008109.1501 = -0. P(P. so k = Ibo In ~ ~ 0. The birth rate (births/ time/population) is then given by f3 = kP. It is reasonable to expect such encounters to occur at a rate that is proportional to the product of the number P/2 of males and the number P /2 of females.0004P 2 - 0. It follows that 80% of the population has heard the rumor when In 36 In 36 t = . as t increases and approaches T = ln(4)/0. depending on whether or not its initial size exceeds the threshold population M = 150. the positive denominator on the right in (16) approaches +oo.13). Use either the exact solution or a computer-generated slope field to sketch the graphs ofseveral solutions of the given differential equation.x 2 . - 1 Note that.4x 2 . x(O) = 2 dx 7. as t increases without bound. dt = 3x(5.06t · (17) Note that. dt dx = x. P(t) --+ 0 as t --+ +oo. the positive denominator on the right in (16) decreases and approaches 0. - dx dt = ?x(x. (15) for 30oe-0·061 300 P(t).x 2 . With this value of B we solve Eq. Consequently. this equilibrium situation is very unstable.x 2 x(O) = 3 4. Figure 1. (13). (b) If Po = 100 and after 6 months there are 169 fish in the lake.x).= 1. then the population remains constant. This is a doomsday situation-a real population explosion. and highlight the indicated particular solution.6. how many will there be after 1 year? . (15) for 4e.105. Consequently P(t) --+ +oo as t --+ r-. Thus the population in Example 7 either explodes or is an endangered species threatened with extinction. If Po exceeds M (even slightly). how long did it take all the fish in the lake to die? 11. If Po = M (exactly!). . x(O) = 17 9. This is an (eventual) extinction situation. x(O) = 2 2.2 + e0.2e-0.0·061 oo61 . with the result that the fish cease to reproduce (so that the birth rate is f3 = 0) and the death rate 8 (deaths per week per fish) is thereafter proportional to 1/v'P. the birth and death rates f3 and 8 are both inversely proportional to v'P.-~. At time t = 0 (months) where k is a constant.82 Chapter 1 First-Order Differential Equations (a) Substitution oft = 0 and P = 200 into (15) gives B = 4. such as the alligator population in certain areas of the southern United States. dt ' dt dx 5. However. dt=4x(7-x). whereas if the initial (positive) population is less than M (however slightly).-. x(O) = 0 = 3x(x - 5).7. x(O) = 1 dx dx 3.M).7. See Problem 33. An approximation to this phenomenon is sometimes observed with animal populations. (a) Show that Separate variables and use partial fractions to solve the initial value problems in Problems 1-8. (b) Substitution oft = 0 and P = 100 into (15) gives B = -2. How many rabbits will there be one year later? 10. then it decreases (more gradually) toward zero as t --+ +oo.06t + 1 . dt = dx lOx.6 shows typical solution curves that illustrate the two possibilities for a population P(t) satisfying Eq. • p FIGURE 1.· 60oe. If there were initially 900 fish in the lake and 441 were left after 6 weeks. dt = 9. Typical solution curves for the explosion/extinction equation P' = kP(P . Suppose that the fish population P(t) in a lake is attacked by a disease at time t = 0. 1.x(0)=11 8. x(O) = 8 dx 6. B Problems the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. The time rate of change of a rabbit population P is proportional to the square root of P. P(t) = With this value of B (16) we solve Eq. then P(t) rapidly increases without bound. Suppose that when a certain lake is stocked with fish.06 >:::::: 23. Consider an alligator population P (t) satisfying the extinction/explosion equation as in Problem 18. Eq. and that the number of those who have heard it is then increasing at the rate of 1000 persons per day. 19. the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation dxjdt = 0.1. If the initial population is P (0) = P0 . Predict this country's population for the year 2000. The swamp contained a dozen alligators in 1988. Consider a rabbit population P (t) satisfying the logistic equation as in Problem 15.7. show that the threshold population isM = DoP0/B 0 .h) 2h with h = 1 to estimate the values of P'(t) when P = 25. where B = aP is the time rate at which births occur and D = b P 2 is the rate at which deaths occur. At time t = 0 the number N (t) of people who have developed Michaud's syndrome is 5000 and is increasing at the rate of 500 per day. Suppose that a community contains 15.04 FIGURE 1. The time rate of change of an alligator population P in a swamp is proportional to the square of P. Consider an alligator population P(t) satisfying the extinction/explosion equation as in Problem 18. (b) Suppose that P0 = 6 and that there are nine rabbits after ten months. f3 and 8. Consider a prolific breed of rabbits whose birth and death rates. How long will it take for this rumor to spread to 80% of the population? (Suggestion: Find the value of k by substituting P(O) and P' (0) in the logistic equation. If P(O) = 100 and P' (O) = 8. how many months does it take for P(t) to reach 10 times the threshold population M? 20. 26. are each proportional to the rabbit population P = P(t). 1.P(t.38 1974 1975 1976 47.0.00 25. Ifthe initial population is 120 rabbits and there are 8 births per month and 6 deaths per month occurring at time t = 0. (Suggestion: Take t = 0 to correspond to the year 1925.) (b) Use the values of k and M found in part (a) to determine when P = 75.000 persons have heard a certain rumor. Ifthe initial population is 110 alligators and there are 11 births per month and 12 deaths per month occurring at time t = 0. Assume that N ' (t) is proportional to the product of the numbers of those who have caught the disease and of those who have not. When does doomsday occur? 14.) 23. a contagious disease. 16.bP 2 .7.P) with k constant. and B0 births per month and Do deaths per month are occurring at time t = 0. Consider a rabbit population P (t) satisfying the logistic equation as in Problem 15. How long will it take for another 5000 people to develop Michaud's syndrome? 25. If the initial population is P (0) = Po and B0 births per month and Do deaths per month are occurring at time t = 0.00 and when P = 47. how many months does it take for P (t) to reach 10% of the threshold population M? 21. show that the limiting population is M = BoPo/Do. Its population in 1940 was 100 million and was then (a) What is the maximum amount of the salt that will ever dissolve in the methanol? (b) If x = 50 when t = 0. two dozen in 1998.bP. Repeat part (a) of Problem 13 in the case f3 < 8. (a) Show that 83 growing at the rate of 1 million per year. Note that P(t) ---+ +oo as t ---+ 1j (kP0 ). Consider a population P(t) satisfying the logistic equation dP/dt = aP. with f3 > 8. A population P(t) of small rodents has birth rate f3 (O. (3).000 people who are susceptible to Michaud's syndrome.OOI)P (births per month per rodent) and constant death rate 8.63 25.54.8x . Population data for Problem 25. The data in the table in Fig. how long (in . If the initial population is 100 alligators and there are 10 births per month and 9 deaths per months occurring at time t = 0.) Year 1924 1925 1926 P (millions) 24.004x 2 • P (t) - R 0 1-kPot' k constant.54 48. Consider a population P(t) satisfying the extinctionexplosion equation dPjdt = aP 2 . When will there be four dozen alligators in the swamp? What happens thereafter? 13. how many months does it take for P (t) to reach 105% of the limiting population M? 18. Suppose that the population P(t) of a country satisfies the differential equation dP/dt = kP(200 . (a) What is the limiting population M? (Suggestion: Use the approximation p (t) ~ I P(t +h). half of a "logistic" population of 100. What now happens to the rabbit population in the long run? 15.7 are given for a certain population P(t) that satisfies the logistic equation in (3).04 47.7 Population Models 12. how many months does it take for P (t) to reach 95% of the limiting population M? 17. 22. If the initial population is 240 rabbits and there are 9 births per month and 12 deaths per month occurring at time t = 0. Suppose that at time t = 0. where B = aP 2 is the time rate at which births occur and D = b P is the rate at which deaths occur.7. As the salt KN0 3 dissolves in methanol. This is doomsday. Then substitute these values in the logistic equation and solve fork and M . how long will it take for an additional 50 g of salt to dissolve? 24. symbolically by analyzing the solution given in Eq.M). and 1960. or numerically by substituting successive values of t. Use the method of Problem 36 to fit the logistic equation to the actual U. and hence dP dt = f3oe . (3).7.2 million.9 million to 123. 38. 1930. population through the year 1930 (long after his own death. With M now known.= 0. it follows that any solution curve that crosses the line P = ~ M has an inflection point where it crosses that line.S. Birth and death rates of animal populations typically are not constant. Make it clear how your derivation depends on whether 0 < P0 < M or Po > M . population P(t) (tin years) grew from 3. P1. A tumor may be regarded as a population of multiplying cells. solve this differential equation to determine what happens to the alligator population in the long run. 29.0001x 2 .0001489P 2 . population data (Fig. and 1950.JOOkM.e. and 2000. Upon equating the two resulting expressions for x 2 in terms of M . Consider two population functions P1(t) and P2 (t). except with 150 alligators initially. 1900. Observe that P (t) approaches the finite limiting population P0 exp ({30/a) as t ---+ + oo. (a) Derive the solution p (t) months) will it take this population to double to 200 rodents? (Suggestion: First find the value of 8. 1. (b) Repeat part (a). the U. P(O) = P0 • = (b) How does the behavior of P(t) as t increases depend on whether 0 < Po < M or Po > M ? 34.01 (deaths per animal per month) and with birth rate f3 proportional to P.7. P (t) remained close to the solution of the initial value problem = MPo -R-o_+_(_M_-_::__Po _)_ ek :--:-M-r of the extinction-explosion initial value problem P' kP(P. and Pz corresponding to equally spaced times to = 0. suppose that at time t = 0 there are Po = 106 cells and that P(t) is then increasing at the rate of 3 x 105 cells per month.0. 35.) 27. and P" > 0 if P > M. dt (a) What 1930 population does this logistic equation predict? (b) What limiting population does it predict? (c) Has this logistic equation continued since 1930 to accurately model the U. Solve the resulting logistic equation. This technique can be used to "fit" the logistic equation to any three population values P0 .M). dP . t 1.4) for the years 1850. 32. both of which satisfy the logistic equation with the same limiting population M but with different values k 1 and k 2 of the constant k in Eq. (7). 31. Solve this initial value problem for Conclude that P" > 0 if 0 < P < ~M.S. For the tumor of Problem 30. so that f3(t) = f30 e-ar (where a and {3 0 are positive constants).P). dP . Derive the solution P(t) = Po + (M - MPo Po)e-kMt of the logistic initial value problem P' = kP(M .03135P.0.9. begin by solving the first equation for the quantity x = e-SO kM and the second equation for x 2 = e. Suppose that the number x(t) (with tin months) of alligators in a swamp satisfies the differential equation dx jdt = 0. If P(t) satisfies the logistic equation in (3). P(O) = Po. they vary periodically with the passage of seasons.S. population data (Fig.01x. population? [This problem is based on a computation by Verhulst. Assume that k 1 < k2 • Which population approaches M the most rapidly? You can reason geometrically by examining slope fields (especially if appropriate software is available).4) for the years 1900.3. Find P (t) if the population P satisfies the differential equation P(t) = P0 exp ( ~0 (1 . During the period from 1790 to 1930.S. P(O) = 3. 36.S.= (k dt + b cos27Tt)P. .84 Chapter 1 First-Order Differential Equations 33. and therefore resembles one of the lowerS-shaped curves in Fig. 1990.] 30. (a) If initially there are 25 alligators in the swamp.S. P(O) = P0 . Solve the resulting logistic equation and compare the predicted and actual populations for the years 1990 and 2000.7. Suppose that P (0) = 200 and P' (0) = 2. Consider an animal population P(t) with constant death rate 8 = 0. Throughout this period. who in 1845 used the 1790-1840 U. either of the original equations is readily solved fork. 39. use the chain rule to show that P"(t) = 2e P(P- ~M)(P. and t2 = 2t1 • 37.ar P . 1. you get an equation that is readily solved for M . P" = 0 if P = ~M. instead. of course). It is found empirically that the "birth rate" of the cells in a tumor decreases exponentially with time. To solve the two equations in ( 10) for the values of k and M. and then find the limiting population of the tumor. Fit the logistic equation to the actual U. population data to predict accurately the U. Solve numerically for a. In particular. After 6 months the tumor has doubled (in size and in number of cells). then compare the predicted and actual populations for the years 1980. P" < 0 if ~M < P < M.ar)) . (a) When is P = 1000? (b) When does doomsday occur? 28. 1. + 49 = 0.8.· -·- . How would the two populations compare after the passage of many years? In Section 1. p = 1 for relatively . (1) with g = 9. g ~ 32 ftjs 2 in fps units). The bolt returns to the ground when y seconds aloft.= -9.8)t + 49] dt = -(4. Generally speaking.8 mjs2 (in mks units.8)t + 49.. The force F R exerted by air resistance on the moving mass m must be added in Eq.8 gives ------- --··· . = -(4. Thus the growth-rate function r(t) = k + b cos 2rrt varies periodically about its mean value k.8 Acceleration-Velocity Models where t is in years and k and b are positive constants..9)t 2 + 49t + y 0 = -(4.9)(5 2 ) + (49)(5) = 122.10) = 0. (1).9)t(t.2 we discussed vertical motion of a mass m near the surface of the earth under the influence of constant gravitational acceleration. Thus its maximum height is Ymax = y(5) = -(4.8)t + v0 = -(9.8)t t = 5 (s). ---·····- "" ------·· ·· - ------ -- ------ -···---- ----- - ----------···· . and thus after 10 • Now we want to take account of air resistance in a problem like Example 1.=Fe+ FR. dt (2) Newton showed in his Principia Mathematica that certain simple physical assumptions imply that F R is proportional to the square of the velocity: F R = kv 2 • But empirical investigations indicate that the actual dependence of air resistance on velocity can be quite complicated.e dt ' (1) where Fe = -mg is the (downward-directed) force of gravity. Hence the bolt's height function y(t) is given by y(t) = J[ -(9. Then Eq.9)t2 + 49t. where the gravitational acceleration is g ~ 9. Suppose that a crossbow bolt is shot straight upward from the ground (y0 = 0) with initial velocity v0 = 49 (m/s). If we neglect any effects of air resistance. dv dt so v(t) = -(9. then Newton's second law (F = ma) implies that the velocity v of the mass m satisfies the equation dv m. . so now dv m.5 (m). hence when The bolt reaches its maximum height when v = -(9.1. For many purposes it suffices to assume that where 1 ~ p ~ 2 and the value of k depends on the size and shape of the body. --- Example 1 ·--.= F. as well as the density and viscosity of the air. Construct a graph that contrasts the growth of this population with one that has 85 the same initial value Po but satisfies the natural growth equation P' = kP (same constant k). it approaches afinite limiting speed. the net force acting on the body is then F = FR +Fa= -kv. = lim v(t) = _ §__ t~oo p (6) Thus the speed of a body falling with air resistance does not increase indefinitely. subject to two forces: a downward gravitational force Fa and a force FR of air resistance that is proportional to velocity (so that p = 1) and of course directed opposite the direction of motion of the body. (3) makes FR positive (an upward force) if the body is falling (vis negative) and makes FR negative (a downward force) if the body is rising (vis positive). Thus air resistance is a complicated physical phenomenon. Vertical motion with air resistance. As indicated in Fig.1.) L.mg. 1. then Fa= -mg and FR = -kv. (3) m m (Note: F R acts upward when the body is falling. You should verify for yourself that if the positive y-axis were directed downward. (4) would take the form dvjdt = -pv +g. fo><oF • F. or terminal speed. If we set up a coordinate system with the positive y-direction upward and with y = 0 at ground level. Note that the minus sign in Eq. where k is a positive constant and v = dyfdt is the velocity of the body. dt (4) where p = kf m > 0. and Newton's law of motion F = m(dvjdt) yields the equation dv mdt = -kv -mg. and its solution is v(t) = (vo + ~) e. Note that v. yields a tractable mathematical model that exhibits the most important qualitative features of motion with resistance.1.~· (5) Here.pt. But how slow "low speed" and how fast "high speed" are depend on the same factors that determine the value of the coefficient k.8. instead. then Eq.g. whereas 1 < p < 2 for intermediate speeds.. Thus - dv = -pv.8.+ Fa Ground level FIGURE 1. But the simplifying assumption that FR is exactly of the form given here. Resistance Proportional to Velocity Let us first consider the vertical motion of a body with mass m near the surface of the earth.86 Chapter 1 First-Order Differential Equations low speeds and p = 2 for high speeds. (7) . Equation (4) is a separable first-order differential equation. with either p = 1 or p = 2. v0 = v(O) is the initial velocity of the body. 5 meters without air resistance). y(t) = 7350.245 = 0 for tm = 25ln(294/245) : : : : 4. We substitute y0 obtain Solution = 0. We now rewrite Eq. We . (5) and (9). v0 = 49. + v.tf25 . I : : : : 21. Thus we find that C =Yo+ (vo. we must solve the equation y(t) = 7350.5. The two equations also involve the terminal velocity v.04 in Eq.7350e-tf 25 • To find the time required for the bolt to reach its maximum height (when v = 0). Example 2 We again consider a bolt shot straight upward with initial velocity v0 = 49 mj s from a crossbow at ground level.245t. With an unbuttoned overcoat flapping in the wind in place of a parachute. The formulas depend on the initial height y0 of the body.v. it even helps explain the occasional survival of people who fall without parachutes from high-flying airplanes. pr (9) Equations (8) and (9) give the velocity v and height y of a body moving vertically under the influence of gravity and air resistance.( vo .3 ftjs. about 44 mijh. To find when the bolt strikes the ground. an unlucky skydiver might increase p to perhaps as much as 0.1125 = 0.v. which corresponds to a terminal speed of Iv.5. We ask how the resulting maximum height and time aloft compare with the values found in Example 1. and v. its initial velocity v0 .y(tn)/y' (tn) to generate successive approximations to the root. or about 14. But now we take air resistance into account. we solve the equation v(t) = 294e. (5) in the form dy -pr . and v(t) = 294e-tf25 - 245. (8) Integration gives 1 y(t) = --(vo. the constant such that the acceleration due to air resistance is aR = -pv. Its maximum height is then Ymax = v(tm) : : : : 108.8 Acceleration-Velocity Models 87 This fact is what makes a parachute a practical invention. )e dt + v. Or we can simply use the Solve command on a calculator or computer.)e-P' p We substitute 0 fort and let y 0 = y(O) denote the initial height of the body. See Problems 10 and 11 for some parachute-jump computations.245t. (4). For a person descending with the aid of a parachute.t 1 + -(vov. with p = 0. )(1 p e.v. we can begin with the initial guess to = 10 and carry out the iteration tn+l = tn . Using Newton's method.)jp. defined in Eq. a typical value of p is 1. = -gjp = -245 in Eqs. and so y(t) =Yo+ v.7350e. which gives a terminal speed of Iv..5 mi/h. (6). and the drag coefficient p..280 meters (as opposed to 122.1.558 (s). I : : : : 65 ft/ s.t +C.). ln I cos u I + C .tm ~ 4.g ( 1 + ~ v2 ) • (12) In Problem 13 we ask you to make the substitution u = v v'P7i and apply the familiar integral 1 du = tan.853 s) than in ascent Ctm ~ 4. Then Eq./Pi) cos Ct (14) .. Then Newton's second law gives dv mdt (10') = Fa+ FR = -mg.88 Chapter 1 First-Order Differential Equations find that the bolt is in the air for tt ~ 9. (11) with v > 0 gives the differential equation ~~ = . Thus the effect of air resistance is to decrease the bolt's maximum height. Taking the positive y-direction as upward. Note also that the bolt now spends • more time in descent (tf.g - p v2 = .kvlvl.227 mjs (as opposed to its initial velocity of 49 mjs). and its final impact speed. It hits the ground with a reduced speed of lv(tf)l ~ 43. Resistance Proportional to Square of Velocity Now we assume that the force of air resistance is proportional to the square of the velocity: (10) with k > 0. (11) where p = kfm > 0. so we can rewrite Eq.1 u + C 2 l+u to derive the projectile's velocity function f v(t) =/!tan (C 1 - tv'iJi) with C1 = tan. The choice of signs here depends on the direction of motion. the total time spent aloft. Thus the sign of FR is always opposite that of v.1 ( vo/f) .411 seconds (as opposed to 10 seconds without air resistance). FR < 0 for upward motion (when v > 0) while FR > 0 for downward motion (when v < 0).558 s). that is. UPWARD MOTION: Suppose that a projectile is launched straight upward from the initial position y 0 with initial velocity v0 > 0. (13) Because J tan u d u = . dt = dv -g . a second integration (see Problem 14) yields the position function y (t) = Yo + -1 ln p --'-----'-~ COS ( C 1 - t .pvlvl. (10) as FR = -kvlvl. which the force of resistance always opposes. We must discuss the cases of upward and downward motion separately. provide a deceleration of T = 4 rn/s 2 • At what height above the lunar surface should the retrorockets be activated to ensure a "soft touchdown" ( v = 0 at impact)? Let r(t) denote the lander's distance from the center of the moon at time t (Fig. the difference is hardly visible.74 x 106 meters (or 1740 km. which has a radius of R = 1.2.8.90 Chapter 1 First-Order Differential Equations y 120 100 80 60 40 20 FIGURE 1. a little over a quarter of the earth's radius). but now we take account of lunar gravity.8. However. the atmospheric reentry and descent of a space vehicle. The following example is similar to Example 2 in Section 1.2. and Example 3 (with air resistance proportional to the square of the velocity) are all plotted. the distance r is measured between the centers of the spheres. Example 2 (with linear air resistance). and at an altitude of 53 kilometers above the lunar surface its downward velocity is measured at 1477 kmlh. The formula is also valid if either or both of the two masses are homogeneous spheres. The height functions in Example 1 (without air resistance).6726 x w. where the corresponding height functions are graphed.35 x 1022 (kg) is the mass of the moon.11 N · (m/kg)2 in mks units). the gravitational acceleration acting on it is not constant. And in Fig.3). in this case. when fired in free space. Its retrorockets. the difference between linear and nonlinear resistance can be significant in more complex situations-such as. . Comparison of the last two lines of data here indicates little difference-for the motion of our crossbow bolt-between linear air resistance and air resistance proportional to the square of the velocity.8. The graphs of the latter two are visually indistinguishable. When we combine the (constant) thrust acceleration T and the (negative) lunar acceleration F/m = GMjr 2 of Eq. we get the (acceleration) differential equation (20) Solution where M = 7. for instance. (19).2. 1. " " Example 4 A lunar lander is free-falling toward the moon. 1. • Variable Gravitational Acceleration Unless a projectile in vertical motion remains in the immediate vicinity of the earth's surface. According to Newton's law of gravitation. the gravitational force of attraction between two point masses M and m located at a distance r apart is given by F=-~ GMm r2 (19) where G is a certain empirical constant ( G ~ 6. A mass mat a great distance from the earth.680 ftjs. (20).378 x 106 (m). With r(t) denoting the projectile's distance from the earth's center at time t (Fig. so it continues forever to move away from the earth. Remark: Equation (24) gives the escape velocity for any other (spherical) planetary body when we use its mass and radius.180 (m/s) (about 36. Therefore.. then r (t) ---* oo as t ---* oo. about 6. Similarly. In particular. 1. when we use the mass M and radius R for the moon given in Example 4. R 2GM so v will remain positive provided that v5 ~ 2GMjR . this gives vo ~ 11.8. which has an equatorial radius of R = 6. • . but with T = 0 (no thrust) and with M = 5. v >v0 2 2 - -. we find that escape velocity from the lunar surface is v 0 ~ 2375 mjs..4). 2 r Now v = v0 and r = R when t = 0. With the given values of G and the earth's mass M and radius R./2G M j R.95 mijs. This will be so if its velocity v = drjdt remains positive for all t > 0. if the projectile's initial velocity exceeds .975 x 1024 (kg) denoting the mass of the earth. indeed. For instance.GMjR. (23) This implicit solution ofEq. a fact that greatly facilitates the return trip ("From the Moon to the Earth"). Substitution of the chain rule expression dvjdt = v(dvjdr) as in Example 4 gives GM r2 Then integration of both sides with respect to r yields v-=--. similar to Eq. "escape" from the earth. soC= !v5. about 25. and hence solution for v 2 gives 2 v 2 = v0 + 2GM ~ .8.000 mi/h).92 Chapter 1 First-Order Differential Equations Escape Velocity lr Velocity v(t) In his novel From the Earth to the Moon (1865). This is just over one-fifth of the escape velocity from the earth's surface. we have the equation dv d 2r dt = dt 2 = GM r2 ' (22) FIGURE 1. (22) determines the projectile's velocity vas a function of its distance r from the earth's center.R 1 2 GM (1 1) . we can ask what initial velocity vo is necessary for the projectile to escape from the earth altogether. the escape velocity from the earth is given by Vo = J2~M· (24) In Problem 27 we ask you to show that. dv dr -v =-+C.4. Jules Verne raised the question of the initial velocity necesary for a projectile fired from the surface of the earth to reach the moon. so it does. so that dvjdt = -kv 2 • Show that v(t) and that = --1 + v0 kt Vo x(t) = xo + k ln(l + v0 kt). Suppose that a body moves through a resisting medium with resistance proportional to its velocity v. A woman bails out of an airplane at an altitude of 10. and that 10 s later the boat has slowed to 20 ft/ s. Then dv 1000. Suppose that a car starts from rest. (17) with initial condition y(O) = Yo· .5 with the parachute. 15. Assume that a body moving with velocity v encounters resistance of the form dvjdt = -kv 312 • Show that m- dv dt = -W +B+ FR. (13) to obtain the upward-motion position function given in Eq. so that dvjdt = -kv. with the sole difference that the deceleration due to air resistance now is (0. Conclude that under a ~-power resistance a body coasts only a finite distance before coming to a stop. Assume. Rework both parts of Problem 7. dt If the boat starts from rest.000 ft. while air resistance provides 0. The acceleration of a Maserati is proportional to the difference between 250 km/h and the velocity of this sports car.= 5000. a paratrooper survived a training jump from 1200 ft when his parachute failed to open but provided some resistance by flapping unopened in the wind. Which offers less resistance when the body is moving fairly slowly-the medium in this problem or the one in Problem 2? Does your answer seem consistent with the observed behaviors of x(t) as t --* oo? 5. Assume that the water resistance is 100 pounds for each foot per second of the speed v of the boat.JVO +2) x(t)=xo+~. It is proposed to dispose of nuclear wastes-in drums with weight W = 640 lb and volume 8 ft3-by dropping them into the ocean ( v0 = 0). x (t) --* +oo as t --* +oo.fii!i to obtain the downward-motion velocity function given in Eq. According to a newspaper account. (12) and substitute u = v.) 11. Test the accuracy of this account. where the buoyant force B is equal to the weight (at 62. (a) Find the car's maximum possible (limiting) velocity. How far will the boat coast in all? 4. (16) with initial condition v(O) = v0 • Integrate the velocity function in Eq. 1 Note that.JV0+2 14. and find that distance. (Suggestion: Find p in Eq. what is the maximum velocity that it can attain? 10.15 without the parachute and p = 1.jpjg to obtain the upward-motion velocity function given in Eq. 7.000 lb and its motor provides a thrust of 5000 lb. 1 ftjs 2 of deceleration for each foot per second of the car's velocity. as in Problem 2. If the drums are likely to burst upon an impact of more than 75 ftjs. 2 )· kt. taking p = 0. Assuming resistance proportional to the square of the velocity (as in Problem 4). How long will it take her to reach the ground? Assume linear air resistance pv ftjs 2 . Then calculate the time required to fall 1200 ft. If this machine can accelerate from rest to 100 km/h in 10 s. what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting? Separate variables in Eq. (b) Find how long it takes the car to attain 90% of its limiting velocity. in contrast with the result of Problem 2.kr and x(t) = Xo + (~0 ) (1. The force equation for a drum falling through water is 1. 16. v(t) and that = 4v0 2 (kt. Suppose that a motorboat is moving at 40 ftjs when its motor suddenly quits. (14) with initial condition y(O) = Yo· Separate variables in Eq. (4) by assuming a terminal velocity of 100 mi/h. (b) Conclude that the body travels only a finite distance. and how far it travels while doing so. A motorboat weighs 32. (13) with initial condition v(O) = v0 • Integrate the velocity function in Eq. 9. Allegedly he hit the ground at 100 mi/h after falling for 8 s.e-kr). how far does the motorboat of Problem 3 coast in the first minute after its motor quits? 6.5 lb/ft3 ) of the volume of water displaced by the drum (Archimedes' principle) and FR is the force of water resistance.8 Acceleration-Velocity Models 93 E Problems 8. (15) and substitute u = v.JV0(1k 13. found empirically to be 1 lb for each foot per second of the velocity of a drum.) 12. then opens her parachute. its engine providing an acceleration of 10 ft/ s2 . (a) Show that its velocity and position at time t are given by v(t) = voe. (Suggestion: First determine her height above the ground and velocity when the parachute opens. falls freely for 20 s. 3. how long will it take for the car to accelerate from rest to 200 km/h? 2. Consider a body that moves horizontally through a medium whose resistance is proportional to the square of the velocity v.100v. that the resistance it encounters while coasting is proportional to its velocity.001)v2 ft/s 2 when the car's velocity is v feet per second.1. (16) to obtain the downward-motion position function given in Eq. Consider the crossbow bolt of Example 3. Suppose that you are stranded-your rocket engine has failed-on an asteroid of diameter 3 miles..R(R 2GMy + y) for the velocity v of the projectile at height y. (a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity v0 less than escape velocity J2GMJR.80 slater with an impact speed of about 43.1 2GM V~ Jr). Take g = 9." . the maximum possible speed of the boat). (b) If the projectile is launched vertically with initial velocity v0 > J2GM/R. g 22. suppose that the bolt is now dropped (v0 = 0) from a height of y0 = 108. respectively. In Jules Verne's original problem. (b) With what initial velocity v0 must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth.-------~ max . If you =-+ (S. (a) Suppose that a body is dropped ( v0 = 0) from a distance r 0 > R from the earth's center. can you blast off from this asteroid using leg power alone? 27. Then use Eqs. r(O)=vo 1 where Me and Mm denote the masses of the earth and the moon. r y'r Why does it again follow that r(t) --+ oo as t --+ oo? 20. How long will it now take him to reach the ground? 24. how long does it take to fall and with what speed will it strike the earth's surface? 29.94 Chapter 1 First-Order Differential Equations upward from the ground (y = 0) at time t = 0 with initial velocity v0 = 49 mjs. Continuing Problem 17. respectively. expressed in terms of earth radii. so its acceleration is dvjdt = -GMjr 2 • Ignoring air resistance. An arrow is shot straight upward from the ground with an initial velocity of 160 ftjs. (15) for a paratrooper falling with parachute open. attained by a projectile launched from the surface of the earth with 90% of escape velocity.Rv 2 ' 0 (Suggestion: Substitute r = r0 cos 2 e to evaluate J r j(r0 . (23) in this section) explicitly to deduce that r(t) --+ oo as t --+ 00.49 mjs.47 min about 4. the projectile launched from the surface of the earth is attracted by both the earth and the moon. (a) To what radius (in meters) would the earth have to be compressed in order for it to become a black hole-the escape velocity from its surface equal to the velocity c = 3 x 108 m/s of light? (b) Repeat part (a) with the sun in place of the earth. R is the radius of the earth and S = 384.0011 in Eq. Then use Eqs.61 s. 19. y(O) = 0.000 ft and opens his parachute immediately. (16) and (17) to show that it hits the ground about 4.00075 before opening his parachute.!.. y' (0) = vo.8 mjs 2 and p = 0.400 km is the distance between the centers of the earth and the moon. and falls from there to the lunar surface.. shot straight 18.r) d r.320 times that of the earth and its radius is 109 times the radius of the earth. 25. If he jumps from an altitude of 10._. with g = 32 ftjs 2) in Eq. deduce from Eq. Suppose that a projectile is fired straight upward from the surface of the earth with initial velocity v0 < J2GMJR . It experiences both the deceleration of gravity and deceleration v 2j800 due to air resistance. 28.r) 2 . what will be his terminal speed? How long will it take him to reach the ground? 23. Then its height y(t) above the surface satisfies the initial value problem J (y GM + R)2.47 m. Then solve the differential equation drjdt = kjy'r (from Eq. To reach the moon. so its distance r(t) from the center of the earth satisfies the initial value problem where M and R are the mass and radius of the earth. dt 2 r2 d 2r GMe GMm r(O)=R. 26. deduce that dr = dt Jk2 +a>. have enough spring in your legs to jump 4 feet straight up on earth while wearing your space suit. Suppose that p = 0. the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. 1 ( = 2 p In 1 + pv~) .075 (in fps units. with density equal to that of the earth with radius 3960 miles. Thereafter it is "under the control" of the moon.r 2 + rocos. show that it reaches the height r < r0 at time 21. ( 14) that the maximum height it attains is Ymax 17. Suppose that the paratrooper of Problem 22 falls freely for 30 s with p = 0. (a) Suppose a projectile is launched vertically from the surface r = R of the earth with initial velocity v0 = J2GM/R so vJ = k 2 fR wherek 2 = 2GM. (13) and (14) to show that the bolt reaches its maximum height of about 108. A motorboat starts from rest (initial velocity v(O) = v0 = 0).) (b) If a body is dropped from a height of 1000 km above the earth's surface and air resistance is neglected. then the maximum distance from the center of the earth attained by the projectile is 2GMR r . The mass of the sun is 329. Substitute dvfdt = v(dvjdy) and then integrate to obtain vz = v6 . How high in the air does it go? If a ball is projected upward from the ground with initial velocity v 0 and resistance proportional to v 2 . Its motor provides a constant acceleration of 4 ftjs 2 . (12). and also find the limiting velocity as t --+ +oo (that is. but water resistance causes a deceleration of v 2 j 400 ft/ s2 • Find v when t = 10 s. Find the minima/launch velocity v0 that suffices for the projectile to make it "From the Earth to the Moon.2GM . t = J ro (Jrro. What maximum altitude does it reach if its initial velocity is 1 kmfs? 30. 8. where Fa = -mg is a constant force of gravity and FR = -kv is a force of air resistance proportional to velocity. Then the change in the momentum of the rocket itself is 11P ~ (m + 11m)(v + 11v). But here m is not constant. Hence the total change in momentum during the time interval 11t is 11P ~ = + v 11m+ 11m 11v) + (-11m)(v.). To derive the equation of motion of the rocket.mv = m 11v + v 11m+ 11m 11v. m(tJ) = mr. Because of the combustion of its fuel. m(t) = mo. we use Newton's second law in the form dP (1) -=F dt where P is momentum (the product of mass and velocity) and F denotes net external force (gravity. 1.1.c) m 11v + c 11m+ 11m !1v. with mass -!:lm and approximate velocity v -c. during which time the mass of the rocket decreases from m 0 to m 1• Thus m(O) = mo. But the system also includes the exhaust gases expelled during this time interval. so 11m ~ 0. (1) gives F= d(mv) dv dm dv =m-+-v=mdt dt dt dt' FIGURE 1. dm = -{3 dt fort ::S tr.8 Application Rocket Propulsion y F l Suppose that the rocket of Fig. t 1]. for instance-then Eq. Suppose m changes to m + 11m and v to v + 11 v during the short time interval from t to t + 11t. An ascending rocket. The rocket is propelled by exhaust gases that exit (rearward) with constant speed c (relative to the rocket).5 blasts off straight upward from the surface of the earth at timet = 0. (m 11v Now we divide by 11t and take the limit as 11t ~ 0.8. The substitution of the resulting expression for dPjdt in (1) yields the rocket propulsion equation dv dm m-+c-=F. dm (3) Constant Thrust Now suppose that the rocket fuel is consumed at the constant "bum rate" fJ during the time interval [0. air resistance. (2) finally gives dv m dt + c dt = -mg. the mass m = m(t) of the rocket is variable.5. (4) with burnout occurring at time t = t 1• .8 Acceleration-Velocity Models 95 1.{Jt. Ifthe mass m of the rocket is constant so m'(t) = 0-when its rockets are turned off or burned out. We want to calculate its height y and velocity v = dyjdt at time t. which (with dvjdt =a) is the more familiar form F = ma of Newton's second law.kv. then Eq. etc. assuming continuity of m(t). dt dt (2) IfF = Fa + FR. {Jt)g.-gt 1 --ln-...f3 mo.{Jt) ln . .{Jt)- dv dt + kv = {Jc.{Jt 1 = m 1.{Jt (7) Because mo. (5) Solve this linear equation for (6) where vo = v(O) and M = _m_(t_) = _m_o_-. it follows that the velocity of the rocket at burnout (t = t 1) IS (8) PROBLEM 3 Start with Eq. set k = 0 in Eq. (5) and integrate to obtain v(t) = vo.gt + cln .850 kg. Assume it encounters air resistance of 1.-gt 2 y(t) = (vo c mo -(mo. No Resistance PROBLEM 2 For the case of no air resistance.. of which 68.45 N per rn/s of velocity.{J_t mo mo denotes the rocket's fractional mass at timet.{Jt (9) It follows that the rocket's altitude at burnout is 1 2 cm1 mo + c)t1.96 Chapter 1 First-Order Differential Equations PROBLEM 1 Substitute the expressions in (4) into Eq.mo -mo. 2 YI = y(tJ) = (vo f3 (10) m1 PROBLEM 4 The V-2 rocket that was used to attack London in World War II had an initial mass of 12.5% was fuel. (3) to obtain the differential equation (m. This fuel burned uniformly for 70 seconds with an exhaust velocity of 2 km/s.(mo..:. (7) and integrate to obtain 1 2 + c)t. Then find the velocity and altitude of the V-2 at burnout under the assumption that it is launched vertically upward from rest on the ground... 4).. if the rocket blasts off from rest (vo = 0) and c = 5 km/s and m 0 fmi = 20.8 Acceleration-Velocity Models PROBLEM 97 5 Actually. and that the time t 8 which then elapses before actual blastoff is given by ts = mog.6). our basic differential equation in (3) applies without qualification only when the rocket is already in motion.{3c . its increase in velocity is ~ v = VI - mo v0 = c In . linear. when a rocket is sitting on its launch pad stand and its engines are turned on initially. but does not depend on the bum rate {3. {3g Free Space Suppose finally that the rocket is accelerating in free space. With the notation introduced to described the constant-thrust case. show that the rocket initially just "sits there" if the exhaust velocity c is less than mog/{3. In this chapter we have discussed applications of and solution methods for several important types of first-order differential equations. Given a differential equation f(x . y)] = 0. But the rocket does not descend into the ground. mi (11) Note that ~ v depends only on the exhaust gas speed c and the initial-to-final mass ratio m 0 jm I. The reason is that if v = 0 in (3). Thus if a rocket initially consists predominantly of fuel. including those that are separable (Section 1.6 we also discussed substitution techniques that can sometimes be used to transform a given first-order differential equation into one that is either separable. With g = 0 in Eq. In Section 1. it just "sits there" while (because m is decreasing) this calculated acceleration increases until it reaches 0 and (thereafter) positive values so the rocket can begin to ascend. linear (Section 1. it is observed that a certain time interval passes before the rocket actually "blasts off" and begins to ascend. (1) we attempt to write it in the form d x [G(x. then it can attain velocities significantly greater than the (relative) velocity of its exhaust gases. For example. y. (8) we see that. so g = k = 0. Lest it appear that these methods constitute a "grab bag" of special and unrelated techniques. as the mass of the rocket decreases from m 0 to m I. it is important to note that they are all versions of a single idea. or exact.1. then the resulting initial acceleration -=---g dv dt c dm m dt of the rocket may be negative. or exact (Section 1. then its velocity at burnout is VI = 5ln 20 ~ 15 km/s. However.5). d (2) . where there is neither gravity nor resistance. y') = 0. y'. B(x). We now tum to equations of higher order n ~ 2. 100 . By contrast. beginning in this chapter with equations that are linear. which we outline in this initial section. y.Linear Equations of Higher Order BJ Introduction: Second-Order Linear Equations n Chapter 1 we investigated first-order differential equations. Recall that a second-order differential equation in the (unknown) function y (x) is one of the form I G(x. and F(x) are continuous on some open interval I (perhaps unbounded) on which we wish to solve this differential equation.! X is linear because the dependent variable y and its derivatives y' and y" appear linearly. Thus a linear second-order equation takes (or can be written in) the form A(x)y" + B(x)y' + C(x)y = F(x). (2) Unless otherwise noted. C(x). we will always assume that the (known) coefficient functions A(x). The general theory of linear differential equations parallels the second-order case (n = 2). (1) This differential equation is said to be linear provided that G is linear in the dependent variable y and its derivatives y' and y". but we do not require that they be linear functions of x. Thus the differential equation ex y" + (COSX)y' + (1 + Vx)Y =tan. the equations y" = yy' and y" + 3(y') 2 + 4l = 0 are not linear because products and powers of y or its derivatives appear. y") = 0. We assume that the dashpot force FR is proportional to the velocity v = dxjdt of the mass and acts opposite to the direction of motion. we will return to this problem in detail in Section 2.1. In general. Of course.+c.2. it is nonhomogeneous. then we call Eq. dx FR = -cv = . (2) is A(x)y" + B(x)y' + C(x)y = 0. Then X. 2. (2) a homogeneous linear equation. the second-order equation x 2 y" + 2xy' + 3y = cosx is nonhomogeneous. the homogeneous linear equation associated with Eq. (3) In case the differential equation in (2) models a physical system. dt 2 dt (5) Thus we have a differential equation satisfied by the position function x(t) of the mass m. then Newton's law F = ma gives (4) that is.+kx = 0.1 ).6).1 Introduction: Second-Order Linear Equations 101 If the function F(x) on the right-hand side of Eq. (2) vanishes identically on I. This homogeneous second-order linear equation governs the free vibrations of the mass. suppose that a mass m is attached both to a spring that exerts on it a force Fs and to a dashpot (shock absorber) that exerts a force FR on the mass (Fig. . the nonhomogeneous term F (x) frequently corresponds to some external influence on the system. If FR and Fs are the only forces acting on the mass m and its resulting acceleration is a = dvjdt. For example.1. For example. Directions of the forces acting on m.1. d 2x dx m . Assume that the restoring force Fs of the spring is proportional to the displacement x of the mass from its equilibrium position and acts opposite to the direction of displacement.2. V > 0 FIGURE 2. otherwise. A Typical Application Spring Mass Dashpot ~ -==~ x=O x>O Equilibrium position Linear differential equations frequently appear as mathematical models of mechanical systems and electrical circuits.4. it is not unusual-either in mathematics or in the English language more generally-for the same word to have different meanings in different contexts.c dt (with c > 0) so FR < 0 if v > 0 (motion to the right) while FR > 0 if v < 0 (motion to the left). its associated homogeneous equation is x 2 y" + 2xy' + 3y = 0. A massspring-dashpot system. Then FIGURE 2. Fs = -kx (with k > 0) so Fs < 0 if x > 0 (spring stretched) while Fs > 0 if x < 0 (spring compressed).1. Remark: Note that the meaning of the term "homogeneous" for a secondorder linear differential equation is quite different from its meaning for a first-order differential equation (as in Section 1. If c 1 and c2 are constants. + p(x)y' + q(x)y = 0. 0 at each point of I. (6) This nonhomogeneous linear differential equation governs the forced vibrations of the mass under the influence of the external force F (t). (7) by A(x) and write it in the form y" + p(x)y' + q(x)y = f(x). so we can divide each term in Eq. in addition to Fs and FR. (8) We will discuss first the associated homogeneous equation y" (9) A particularly useful property of this homogeneous linear equation is the fact that the sum of any two solutions of Eq. (9) is again a solution. THEOREM 1 Principle of Superposition for Homogeneous Equations Let y 1 and y 2 be two solutions of the homogeneous linear equation in (9) on the interval I.102 Chapter 2 Linear Equations of Higher Order If. then the linear combination (10) is also a solution of Eq. (7) where the coefficient functions A. Here we assume in addition that A (x) =J=.+ kx = dt F(t). B. . (4)-the resulting equation is d 2x mdt 2 dx +c . C. as is any constant multiple of a solution. (9) on I. This is the central idea of the following theorem. the mass m is acted on by an external force F(t)-which must then be added to the right-hand side in Eq. which gives Then y" + py' + qy = (CJYl + c2y2)" + p(CIYI + c2y 2)' +q(CIYI + c2y2) = (CJY~ + c2y~) + p(CJY~ + c2y~) + q(CJYI + C2Y2) = CJ (y~ + PY~ + qyi) + c2(y~ + PY~ + qy2) = CJ · 0 + C2 · 0 = 0 because Yl and Y2 are solutions. Proof: The conclusion follows almost immediately from the linearity of the operation of differentiation. Homogeneous Second-Order Linear Equations Consider the general second-order linear equation A(x)y" + B(x)y' + C(x)y = F(x). Thus y = c1 Y1 + c2Y2 is also a solution. and F are continuous on the open interval I. 4. FIGURE 2. Solutions y(x) = c 1 cosx of y" + y = 0.5.. We will see later that.1. (6) ought to have a unique solution satisfying these initial conditions.. 2. More generally. Solutions of y" + y = 0 with c 1 and c2 both nonzero.1 .. every solution of y" + y = 0 is a linear combination of these two particular solutions y 1 and Y2· Thus a general solution of y" + y = 0 is given by y(x) = c 1 cos x + Cz sinx. Earlier in this section we gave the linear equation mx" +ex'+ kx = F(t) as a mathematical model of the motion of the mass shown in Fig.1.1 . Physical considerations suggest that the motion of the mass should be determined by its initial position and initial velocity.>. FIGURE 2.1 Introduction: Second-Order Linear Equations ~------- 103 Example 1 We can see by inspection that Yt (x) = cos x and Y2 (x) = sin x are two solutions of the equation y II +y= 0. 8 6 4 2 .. It is important to understand that this single formula for the general solution encom- passes a "twofold infinity" of particular solutions. in order to be a "good" mathematical model of a deterministic physical situation. Hence. Theorem 1 tells us that any linear combination of these solutions.3 through 2. a differential equation must have unique solutions satisfying any appropriate initial conditions.>.2 sin x y = 3 cos x + 4 sin x I 0 0 0 -2 -2 -4 -6 -8 -:rt 0 X :l't -4 -6 2:rt -8 -:rt 0 X :l't 2:rt -2 -4 -6 -8 -10 -:rc 0 :l't X 2:rc 3n: FIGURE 2.3. Solutions y(x) = c2 sin x of y" + y = 0. because the two coefficients c 1 and c2 can be selected independently. Figures 2. given any preassigned values of x(O) and x' (0). y = 6 cos x . is also a solution. . • 8 6 4 2 . / c. The following existence and uniqueness theorem (proved in the Appendix) gives us this assurance for the general second-order equation.5 illustrate some of the possibilities. 8 = 10 5 6 4 2 . such as y(x) = 3yt(x).2.>.1.1. with either c 1 or c2 set equal to zero. or with both nonzero. conversely.2y2(x) = 3cosx..2sinx.1.. Eq. 7. y'(a) = b 1• That is.. Theorem 2 tells us that any such initial value problem has a unique solution on the whole interval I where the coefficient functions in (8) are continuous. given any two numbers bo and b 1.6. Example 1 continued c2y2(a) = bo.3 that a nonlinear differential equation generally has a unique solution on only a smaller interval. (11) FIGURE 2. every nonvertical straight line through (a. one for each (real number) value of the initial slope y' (a) = b1• That is. bo) is tangent to some solution curve of Eq. (8). Figure 2. First. also as guaranteed by Theorem 2. Solutions of y" + 3y' + 2y = 0 with the same initial slope y' (0) = 1 but different initial values. we might actually find the solution y(x) whose existence is assured by Theorem 2. we find two "essentially different" solutions y 1 and y2. The application at the end of this section suggests how to construct such families of solution curves for a given homogeneous second-order linear differential equation. the equation . More generally. b).7 shows a number of solution curves all having the same initial slope y' (0) = 1. b0 )-namely. It has the initial values y(O) = 3. Then. • We saw in the first part of Example 1 that y (x) = 3 cos x .2 sin x is a solution (on the entire real line) of y" + y = 0.6 shows a number of solution curves of the equation y" + 3y' + 2y = 0 all having the same initial value y (0) = 1. y'(a) = bt . we attempt to solve the CtYi (a)+ 2 . 0 -1 y" + p(x)y' + q(x)y = f(x) (8) -2 " 0 y'(O) = -6 has a unique (that is.first-order differential equation dyjdx = F(x. 5 4 3 y(O) = 3 Remark I: Equation (8) and the conditions in (11) constitute a secondorder linear initial value problem. and f are continuous on the open interval I containing the point a.104 Chapter 2 Linear Equations of Higher Order THEOREM 2 Existence and Uniqueness for Linear Equations Suppose that the functions p. given a homogeneous second-order linear equation.1. Solutions of y" + 3y' + 2y = 0 with the same initial value y(O) = 1 but different initial slopes. this illustrates the existence of such a solution. the solution y(x) = bo cosx + b1 sinx satisfies the arbitrary initial conditions y(O) = b 0 . 2. instead of there being only one line through (a. y) generally admits only a single solution curve y = y(x) passing through a given initial point (a. . 0 -1 -2 -3 -4 -5 -2 -1 0 X 2 3 4 FIGURE 2. Theorem 2 tells us that this is the only solution with these initial values. • Example 1 suggests how. one and only one) solution on the entire interval I that satisfies the initial conditions 4 -1 2 X 3 5 y(a) = bo.1. Recall from Section 1. while Fig.. second.1. Remark 2: Whereas a.>. y' (0) = b 1 . bo) tangent to a solution curve. CtY~ (a)+ c2y~(a) = b1 (13) for the coefficients c 1 and c 2 .>. we attempt to impose on the general solution (12) the initial conditions y(a) simultaneous equations bo. y'(O) = -2. q. Theorem 2 implies that the second-order equation in (8) has infinitely many solution curves passing through the point (a.1 . The following definition tells precisely how different the two functions y 1 and Y2 must be. c2 = -2. =C) = -4 -6 -8 -10 -2 The resulting solution is c 1 value problem is -1 0 X = 3. no matter what the initial conditions bo and b 1 might be. we omit it. x2· x+ l X lxl.. Hence the solution of the original initial y(x) = 3ex . e. Two functions are said to be linearly dependent on an open interval provided that they are not linearly independent there. y' (O) = 1.1. one of them is a constant multiple of the other.8 shows several additional solutions of y". Example 3 Thus it is clear that the following pairs of functions are linearly independent on the entire real line: sinx ex ex and and and and and cos x.8. and then find a solution satisfying the initial conditions y(O) = 3. 10 8 6 4 2 .1 Introduction: Second-Order Linear Equations 105 Example 2 Verify that the functions Yl (x) =ex and Y2(x) = xex are solutions of the differential equation y".2y' +y = 0. ' xex. that is. We can always determine whether two given functions f and g are linearly dependent on an interval I by noting at a glance whether either of the two quotients f jg or gjf is a constant-valued function on I.. Figure 2. • In order for the procedure of Example 2 to succeed.2. to obtain the simultaneous equations y(O) 0 -2 3. Solution The verification is routine. We impose the given initial conditions on the general solution for which y'(x) = (cl +c2)ex +c2xex. y'(O) = c1 +c2 = 1.2xex. the two solutions y 1 and Y2 must have the elusive property that the equations in (13) can always be solved for c 1 and c2 .2y' + y = 0 with the same initial value y(O) = 3. all having the same initial value y(O) = 3. FIGURE 2. DEFINITION Linear Independence of Two Functions Two functions defined on an open interval I are said to be linearly independent on I provided that neither is a constant multiple of the other. Different solutions y (x) = 3ex + c2 x ex of y" . ' .1.2y' + y = 0.2x.. 2 + sm x = 1 and W( e x . x smx = I ex ex sm x cosx I = cos xe x 2 x . W (cos x. y~(a) = 1.ex is a constant-valued function. But the identically zero function j(x) 0 and any other function g are linearly dependent on every interval. . y2. depending on whether we wish to emphasize the two functions or the point x at which the Wronskian is to be evaluated. It is then impossible that either y 1 = ky2 or Y2 = ky 1 because k · 0 =!= 1 for any constant k. the determination of the constants c 1 and c2 in (12) depends on a certain 2 x 2 determinant of values of y 1 . and so forth. Given two functions f and g . the functions f (x) = sin 2x and g (x) = sin x cos x = are linearly dependent on any interval because f(x) = 2g(x) is the familiar trigonometric identity sin 2x = 2 sin x cos x. Also. that given any two linearly independent solutions y 1 and Y2 of the homogeneous equation y"(x) + p(x)y'(x) + q(x)y(x) = 0. finally.f 'g . • General Solutions But does the homogeneous equation y" + py' + qy = 0 always have two linearly independent solutions? Theorem 2 says yes! We need only choose y 1 and Y2 so that Yt (a) = 1. (9)-it provides all possible solutions of the differential equation. and in greater detail beginning in Section 2. These are examples of linearly independent pairs of solutions of differential equations (see Examples 1 and 2).2x = e 3x nor e-2x. (9) every solution y of Eq. neither sinxjcosx = tanx nor cosxjsinx = cotx is a constant-valued function. We want to show. Yi (a) =0 and Y2(a) = 0. Theorem 2 tells us that two such linearly independent solutions exist. As suggested by the equations in (13). the Wronskian of f and g is the determinant W= f g ! ' g' = fg'.cos . We write either W(f.106 Chapter 2 Linear Equations of Higher Order That is. because 0 · g(x) = 0 = j(x). g) or W(x ). neither ex je. Note that in both cases the Wronskian is everywhere nonzero. sm x) . actually finding them is a crucial matter that we will discuss briefly at the end of this section. and their derivatives. = I. (9) can be expressed as a linear combination (12) of Yt and Y2· This means that the function in ( 12) is a general solution of Eq. For example.3. xe x) ex X~ + I = e 2x . In Section 2. then there exist numbers c 1 and c2 such that . (9)) ·.:_---.J. then W(f. By ._ --. If y is any solution whatsoever ofEq. The determinant of the coefficients in this system of linear equations in the unknowns c1 and c2 is simply the Wronskian W(y 1. y 2) evaluated at x = a. with f I.2 we will prove that. (9) if Yt and Y2 are linearly independent solutions. (14) CtYi (a)+ c2y~(a) = Y'(a). Y2) ":f 0 at each point of I.. the Wronskian is never zero if the solutions are linearly independent. . y" + p(x)y' + q(x)y =0 on an opeJ. " · · . The latter fact is what we need to show that y = Ct Yt + c2Y2 is the general solution of Eq. .-: . THEOREt.A4 General Solutions of Homogeneous Equations Letyrandy2 '6 e·twolinearlyindependent solutions of the homogeneous equation {Eq. y2) 0 on I.. if the two functions Yt and Y2 are solutions of a homogeneous second-order linear equation.kg'g = 0. Y(x) = CtYt (x) + c2y2(x) for allx in L In essence.intei"Vali on which p and q are continuous..___ -· . We therefore call the linear combination Y = c1 YI + c2Y2 a general solution of the differential equation.atld are Hneady independent. :' Suppo~~ttl~f)it an<. then w(YI. then we have found all of its solutions. (9)on I.1<ly2 arelinearly dependent. then the strong converse stated in part (b) of Theorem 3 holds. '' .ly2are two solutions of the homogeneous second-order linear equation(Eq.1 Introduction: Second-Order Linear Equations 107 = On the other hand. Yi = Thus. . and consider the simultaneous equations CtYl (a)+ c2y2(a) = Y(a). then W(y 1.(9)) y"+p(x)y' +q(x)y = 0 with _p andq'i¢bntil1UOUS on the open interval I.. Thus the Wronskian of two linearly dependent functions is identically zero. given two solutions of Eq. g)= f and g are linearly dependent. THEOREM 3 Wronskians of Solutions ·-: ..l. (9). Theorem 4 tells us that when we have found two linearly independent solutions of the second-order homogeneous equation in (9)... there are just two possibilities: The Wronskian W is identically zero if the solutions are linearly dependent.. (a}Ifyra. Proof of Theorem 4: Choose a point a of I. '. " (b) :Ifyt .2. I= kgg'. if the functions kg (for example). because cosh 2x = ~e 2x + ~e. (9).2 (cosh2x) = -(2sinh2x) = 4cosh2x dx dx and." • .2x and sinh 2x = ~e 2x - d2 d ~e. Example 4 y~ = (2)(2)e 2x = 4e2x = 4yl and y~ = ( -2)( -2)e-2x = 4e-2x = 4Y2· Therefore. (15).4y = 0. sinh x are two different pairs of linearly independent solutions of the equation y" . so do Y' and G'. With these values of c 1 and c2. • Remark: Because e 2x.2x by the definitions of the hyperbolic cosine and hyperbolic sine. because . Theorem 4 implies that every particular solution Y (x) of this equation can be written both in the form and in the form Y(x) = a coshx + b sinhx. so by elementary algebra it follows that the equations in (14) can be solved for c 1 and c2. this is no surprise. (sinh 2x )" = 4 sinh 2x. we define the solution of Eq. By the uniqueness of a solution determined by such initial values (Theorem 2). as desired. Of course. y 1 and Y2 are linearly independent solutions of y".2x and cosh x. this determinant is nonzero. Hence each of these two linear combinations is a general solution of the equation. Thus these two different linear combinations (with arbitrary constant coefficients) provide two different descriptions of the set of all solutions of the same differential equation y" .108 Chapter 2 Linear Equations of Higher Order Theorem 3. then G(a) = CJYI (a)+ c2y2(a) = Y(a) and G'(a) = CJYi (a)+ c2y~(a) = Y'(a). (15) But y3(x) =cosh 2x and y4 (x) =sinh 2x are also solutions of Eq. It therefore follows from Theorem 4 that the functions cosh 2x and sinh 2x can be expressed as linear combinations of YI (x) = e 2x and y2(x) = e.4 y = 0. it follows that Y and G agree on I. Thus we see that Y(x) = G(x) = CJYl (x) + c2y2(x). likewise. Thus the two solutions Y and G have the same initial values at a.4y = 0 in (15). Indeed. similarly.2x. e. this is why it is accurate to refer to a specific such linear combination as "a general solution" rather than as "the general solution. (18) has two distinct (unequal) roots r 1 and r 2 . we actually have found two solutions: y 1 (x) = e 2x and y2 (x) = e3x. then each term would be a constant multiple of erx. We first look for a single solution ofEq. b. if we substitute y = erx in the equation y". and c. Hence. (18) This quadratic equation is called the characteristic equation of the homogeneous linear differential equation ay" + by' + cy = 0.3)erx = 0. (16). then y = erx will be a solution ofEq. . b. THEOREM 5 Distinct Real Roots If the roots r 1 and r 2 of the characteristic equation in ( 18) are real and distinct. Hence y = erx will be a solution if either r = 2 orr = 3. For example. This suggests that we try to find a value of r so that these multiples of erx will have sum zero. we discuss the homogeneous second-order linear differential equation ay" + by' + cy = 0 (16) with constant coefficients a. we find the result to be Because erx is never zero. we substitute y = erx in Eq. then (19) is a general solution ofEq. (16) and begin with the observation that (17) so any derivative of erx is a constant multiple of erx.1 Introduction: Second-Order Linear Equations 109 Linear Second-Order Equations with Constant Coefficients As an illustration of the general theory introduced in this section.2. we conclude that y(x) = erx will satisfy the differential equation in ( 16) precisely when r is a root of the algebraic equation ar 2 + br + c = 0. (16). with the constant coefficients dependent on r and the coefficients a. (16). we obtain Thus (r 2 - 5r + 6)erx = 0. So. (16) If Eq. With the aid of the equations in (17). (16). (r . if we substituted y = erx in Eq. (Why?) This gives the following result.5y' + 6y = 0. If we succeed. To carry out this procedure in the general case. in searching for a single solution. then the corresponding solutions YI(x) = er 1x and y 2 (x) = erzx of(16) are linearly independent.2) (r . and c. • = Remark: Note that Theorem 5 changes a problem involving a differential equation into one involving only the solution of an algebraic equation. we get the general Figure 2. (16). Solution We can solve the characteristic equation 2r 2 -7r +3 = 0 by factoring: (2r . so Theorem 5 yields the general • Example 6 The differential equation y" + 2y' = 0 has characteristic equation r2 ····~- . all appearing to approach the solution curve y(x) 1 (with c2 = 0) as x ---+ +oo. (20). Because e0 ·x solution = 1. The problem in this case is to produce the "missing" second solution of the differential equation. A double root r = r 1 will occur precisely when the characteristic equation is a constant multiple of the equation Any differential equation with this characteristic equation is equivalent to (20) But it is easy to verify by direct substitution that y = xer'x is a second solution of Eq.x of Eq. It is Clear (but you should verify) that are linearly independent functions.9 shows several different solution curves with c 1 = 1. we get (at first) only the single solution y 1(x) = er. + 2r = r(r + 2) = 0 with distinct real roots r 1 = 0 and r2 = -2. so the general solution of the differential equation in (20) is .1.2x of y" + 2y' with different values of c2 • =0 If the characteristic equation in (18) has equal roots r 1 = r2 . The roots r 1 solution = ~ and r 2 = 3 are real and distinct.1.7y' + 3y = 0.. Solutions y(x) = I+ c 2 e.1)(r.9.3) = 0. • X FIGURE 2. -.11 0 Chapter 2 Linear Equations of Higher Order Example 5 Find the general solution of 2y". . Yn be n solutions of the homogeneous linear equation in (3) on the interval I. we will always assume that the coefficient functions Pi (x) and F (x) are continuous on some open interval I (perhaps unbounded) where we wish to solve the equation.3x . A second-order Euler equation is one of the form ax 2 y" 113 + bxy' + cy = 0 (22) conclude that a general solution of the Euler equation in (22) is y (x) = c 1x'~ + c 2x'2 • where a. + 8xy'.12y = 0 55. we can divide each term in Eq.2. and y 3(x) = sin 2x . Under the additional assumption that P0 (x) =!= 0 at each point of I. 54. 52. x 2 y" + 2xy'. (22) into the constant-coefficient linear equation Make the substitution v = ln x of Problem 51 to find general solutions (for x > 0) of the Euler equations in Problems 5256. c are constants. (1) by Po(x) to obtain an equation with leading coefficient 1.y d2y a dvl + (b. c2 . The proof of the following theorem is essentially the same-a routine verification-as that of Theorem 1 of Section 2. a homogeneous nth-order linear differential equation has the valuable property that any superposition. en are constants. (3) on I. If c 1 . or linear combination.(x)y<n-l) + · · · + Pn-I(x)y' + Pn(x)y = F(x). then the linear combination Y = CJYI + C2Y2 + · · · + CnYn (4) is also a solution of Eq.3xy' + 4y = 0 We now show that our discussion in Section 2. 4x 2 y" 56. THEOREM 1 Principle of Superposition for Homogeneous Equations Let y.I) + · · · + Pn-1 (x)y' + Pn (x)y = f(x). then the substitution v = In x transforms Eq. b.a) dv + cy = dy 0 (23) =0 53. of solutions of the equation is again a solution. y 2 (x) = cos2x..3y = 0 2 x y" . + Pn-J(x)y' + Pn(x)y = 0. x 2 y" + xy' = 0 with independent variable v. (1) Unless otherwise noted.1. (a) Show that if x > 0. (b) If the roots r 1 and r2 of the characteristic equation ofEq... Example 1 It is easy to verify that the three functions y 1 (x) = e. (23) are real and distinct. x 2 y" + xy'. (2) is yen)+ PI(x)y(n-!) + .. (3) Just as in the second-order case. ••.. . of the form yen) +PI (x)y(n . y2.2 General Solutions of Linear Equations 51. . (2) The homogeneous linear equation associated with Eq.1 of second-order linear equations generalizes in a very natural way to the general nth-order linear differential equation of the form Po(x)y(n) + P. 2. Theorem 1 tells us that any linear combination of these solutions. ••• . It tells us nothing.3 we will see how to construct explicit solutions of initial value problems in the constan. a particular solution of an nth-order linear differential equation is determined by n initial conditions. y' (0) = 5.. and y 3 • Thus a general solution is given by y(x) = c 1e. 2. conversely.3e.1) looks periodic on the right. y'(a)=bt.2. we see that y (x) ~ 3 cos 2x . We saw earlier that y(x) = -3e. The particular solution y(x) = .2 sin 2x.2 sin 2x for large positive x.tcoefficient case that occurs often in applications. Pn. (2)) y<n) +PI (x)y(n-l) + · · · + Pn-1 (x)y' + Pn(x)y = f(x) has a unique (that is. We will see that.3x + 3 cos 2x - 2 sin 2x. ' (5) 0 2 4 . Theorem 2 tells us that any such initial value problem has a unique solution on the whole interval I where the coefficient functions in (2) are continuous . . In Section 2. however. proved in the Appendix. .X 6 10 FIGURE 2. and Theorem 2 implies that there is no other solution with these same initial values.114 Chapter 2 Linear Equations of Higher Order are all solutions of the homogeneous third-order equation y(3) + 3y" + 4y' + 12y = 0 on the entire real line. is also a solution on the entire real line. Equation (2) and the conditions in (5) constitute an nth-order initial value problem. and f are continuous on the open interval I containing the point a.3x Example 1 Continued + 3 cos 2x - 2 sin 2x 0 is a solution of y<3 ) + 3y" + 4y' + 12y = on the entire real line.3x + 3 cos 2x . is the natural generalization of Theorem 2 of Section 2. b 1 . because of the negative exponent. Note that its graph (in Fig. THEOREM 2 Existence and Uniqueness for Linear Equations Suppose that the functions p 1 . about how to find this solution. This particular solution has initial values y(O) = 0. such as y(x) = -3y 1 (x) + 3yz(x) - 2y3 (x) = -3e. one and only one) solution on the entire interval I that satisfies the n initial conditions y(a)=bo.1. given n numbers b0 .1.1 that a particular solution of a second-order linear differential equation is determined by two initial conditions. Indeed. every solution of the differential equation of this example is a linear combination of the three particular solutions y 1 . Similarly. The following theorem. p 2 . and y"(O) = -39. • .3x + Cz cos 2x + C3 sin 2x. • Existence and Uniqueness of Solutions We saw in Section 2. y 2 . bn-l• the nth-order linear equation (Eq.. ••. Then. If we write these equations as (l)!J + (-k)h = 0 or (k)j. we say that n functions f 1. c2 . . (6). if c 1 and c2 are not both zero.. . DEFINITION Linear Dependence of Functions The n functions !I. Yn are particular solutions of Eq. . h . c2 . ••. Cn not all zero such that (7) on I.. But these n particular solutions must be "sufficiently independent" that we can always choose the coefficients ci.. . . ••. .. if either !I = kh or h = kfi for some constant k. we anticipate that a general solution of the homogeneous nth-order linear equation y<n) + P1 (x)y(n-I) + · · · + Pn .I (x)y' + Pn(x)y = 0 (3) will be a linear combination (4) where YI. that is. then clearly we can solve for at least one of the functions as a linear combination of the others.. (7) are zero. c 2 . .116 Chapter 2 Linear Equations of Higher Order Linearly Independent Solutions On the basis of our knowledge of general solutions of second-order linear equations. fn are said to be linearly dependent on the interval I provided that there exist constants CI. + (-l)h = 0. that is.. Y2 • . Thus the functions !I.. (6) certainly implies that f 1 and h are linearly dependent. In analogy with Eq.. h. we see that the linear dependence of f 1 and c 1 and c2 not both zero such that h implies that there exist two constants (6) Conversely. Cn are zero (although some of them may be zero). (3). fn are linearly dependent provided that some nontrivial linear combination of them vanishes identically.. The question is this: What should be meant by independence of three or more functions? Recall that two functions !I and h are linearly dependent if one is a constant multiple of the other.. then Eq. . h . . f n are linearly dependent if and only if at least one of them is a linear combination of the others. . If not all the coefficients in Eq. Cn in (4) to satisfy arbitrary initial conditions of the form in (5). nontrivial means that not all of the coefficients c 1. for all x in /. . . and conversely. . Then their Wronskian is the n x n determinant ft W= fn f{ ~~ (8) We write W(j1 . fn are called linearly independent on the interval I provided that they are not linearly dependent there. h . . obtaining the n equations cifi(x ) + c2h(x) CJf{(x) + c2j. and j)(x) =ex are linearly dependent on the real line because (l)JI + (-2)h + (O)j) = 0 (by the familiar trigonometric identity sin 2x = 2 sin x cos x ). Wronski (1778-1853). M. We then differentiate this equation n . But in order to show that n given functions are linearly independent. h. . nontrivial values of the coefficients so that Eq. More generally. To prove this. the functions f 1 . . • (7) Then functions f 1. h(x) = sinx cos x. that is. h • .(x) + ··· + +···+ Cnfn(X) = 0. in the case of n solutions of a homogeneous nth-order linear equation. .. We saw in Section 2.1. Cn not all zero. .. they are linearly independent on I provided that the identity Ctfi + c2h + · · · + Cnfn = = C2 0 holds on I only in the trivial case Ct = · · · = Cn = 0. no nontrivial linear combination of these functions vanishes on I. . (7) holds on the interval I for some choice of the constants c 1 ... we must prove that nontrivial values of the coefficients cannot be found. there is a tool that makes the determination of their linear dependence or independence a routine matter in many examples. fn) or W(x). c2 . . . Suppose that the n functions !I. The Wronskian is named after the Polish mathematician J... as in Example 3.1 times differentiable. fn is identically zero. fn are each n . Fortunately. H. This tool is the Wronskian determinant. Put yet another way. . h . the Wronskian of n linearly dependent functions fi.. Cnf~(x) = 0.1 times in succession. assume that Eq. which we introduced (for the case n = 2) in Section 2. h ...1 that the Wronskian of two linearly dependent functions vanishes identically.2 General Solutions of Linear Equations 117 Example 3 The functions !1 (x) =sin 2x. depending on whether we wish to emphasize the functions or the point x at which their Wronskian is to be evaluated.2. Equivalently. (9) .. fn are linearly independent if no one of them is a linear combination of the others. (7) holds. .. . and this is seldom easy to do in any direct or obvious manner. (Why?) Sometimes one can show that n given functions are linearly dependent by finding. en and the determinant of coefficients is simply the Wronskian W (JI . fn) evaluated at the typical point x of I. When we compute the Wronskian of the three given solutions. to show that the functions !I. 0 X 2 . . y' (1) = 2.. We recall from linear algebra that a system of n linear homogeneous equations in n unknowns has a nontrivial solution if and only if the determinant of coefficients vanishes. Yz(x)=xlnx. In Eq. y 2 (x) Example 1) are linearly independent. and y3(x) sin 2x (of Solution sin2x 2cos2x -4 sin 2x W = -3e. (11) Solution Note that for x > 0. y 2 . it suffices to show that their Wronskian is nonzero at just one point of I.3x cos2x sin2x -4 sin 2x -4 cos 2x cos2x = 26e. (9) the unknowns are the constants c 1 . .:.3x i= 0.3x -2sin2x -4 cos 2x 2cos2x -4 sin 2x sin 2x 2 cos 2x + 3e.x 2 y" + 2xy'.Jx-. fn are linearly independent on the interval I. . it follows that y 1 . . Then find a particular solution of Eq.. it follows that W (x) 0. . -2 sin 2x Because W i= 0 everywhere. Their Wronskian is cos2x cos 2x. h. Because we know that the ci are not all zero.118 Chapter 2 Linear Equations of Higher Order which hold for all x in I. (10) that satisfies the initial conditions y(l) = 3. h. Example 5 Show first that the three solutions YI(x)=x.3x -2 sin 2x -4 cos 2x 9e. and y 3 are linearly independent • on any open interval (including the entire real line). .2y = 0 (10) are linearly independent on the open interval x > 0. we could divide each term in ( 10) by x 3 to obtain a homogeneous linear equation of the standard form in (3). and Y3(x)=x 2 of the third-order equation x 3 y( 3) . y" (1) = 1. = Example 4 Show that the functions y 1 (x) ~. ••. . .. as we wanted to prove. c2 . Therefore. we find that X W= x lnx 1 + lnx 1 x2 1 2x =X. It can be shown that the Wronskian of n solutions y 1.x 2)y" + 2xy'. but we said little about how actually to find even a single solution. n linearly independent solutions of a given nth-order linear equation if it has constant coefficients. 39. Use the method of reduction of order as in Problem 37 to find a second linearly independent solution Yz· W' = y. Yt (x) = ex/2 x 2 y" .4y' 0 (x > 0). z cos x is one solution (for x > 0) of Bessel's equation of order ~. y 1(x) = x First note that y 1 (x) = x is one solution of Legendre's equation of order 1. Thus. Thence solve for v(x) = C ln x.2 we saw that a general solution of an nth-order homogeneous linear equation is a linear combination of n linearly independent particular solutions. a differential equation and one solution y 1 are given.(x + 2)y' + y = 0 (x > . + (2y. According to Problem 32 of Section 2. Prove this for the case n = 3 as follows: (a) The derivative of a determinant of functions is the sum of the determinants obtained by separately differentiating the rows of the original determinant. 40..x 2)y" .1 12 sinx.2xy' + p(x)y' + q(x)y = 0 (18) + 2y = 0..124 Chapter 2 Linear Equations of Higher Order 37.2y = 0 (-1 < x < 1). First verify by substitution that y 1(x) = x -t. use the fact that y 1(x) is a solution to deduce that Yt v" Then use the method of reduction of order to derive the second solution Yz(x) = 1. Then derive by reduction of order the second solution Y2(x) = x . Conclude that YI substitute y = vx 3 and deduce that xv" + v' = 0. The . 43.l n . and yj3 ) from the equation y< 3) +PI y" + P2y' + P3Y = 0. the equation must first be written in the form of ( 18) with leading coefficient 1 in order to correctly determine the coefficient function p(x). . explicitly and in a rather straightforward way. The method of reduction of order consists of substituting y 2(x) = v(x)y 1(x) in (18) and attempting to determine the function v(x) so that y 2(x) is a second linearly independent solution of ( 18). In each of Problems 38 through 42. 36.2 1. y 1(x) = x (x + 1)y". y 1(x) = ex (1.)v' = 0.. Y2. x 2y " +xy'. Y1 (3) y~ Y2 (3) y~ Y3 (3) (b) Substitute for y} 3). . (19) with a given homogeneous second-order linear differential equation and a known solution y 1 (x). (1. (x) dx) for some constant K.p 1 W. Suppose that one solution y 1(x) of the homogeneous second-order linear differential equation y" 38.1). starting with the readily verified solution y 1 (x) = x 3 of the equation x 2y".1. + py. But we can now show how to find. Yt(X) = x 3 + y = 0. is known (on an interval I where p and q are continuous functions). Yn of the nth-order equation y<n) +PI (x)y(n-l) + · · · + Pn-1 (x)y' + Pn(X)y = 0 + 9y = 0 (x > 0).X x 1 +x (for-1 < x < 1). satisfies the same identity. and thereby obtain (with C = 1) the second solution y 2 (x) = x 3 In x. yi3 ). If y 1 (x) is known. and then show that W' = . Integration now gives Abel's formula. 42. (18). Frequently it is more convenient to simply substitute y = v (x) y 1(x) in the given differential equation and then proceed directly to find v(x).x(x + 2)y' + (x + 2)y = 0 (x > 0).9y = 4y" . (19) 44. Before applying Eq. lfiJ Homogeneous Eq]JatiOJ:lS with Constant Coef¥ic~ents In Section 2. After substituting y = v(x)y 1(x) in Eq. Integration of v' (x) then gives the desired (nonconstant) function v(x). the Wronskian W(y 1. 41. y 2 ) of two solutions of the second-order equation y" +PI (x)y' + P2(x)y = 0 is given by Abel's's formula W(x) = K exp (- j p. then ( 19) is a separable equation that is readily solved for the derivative v' (x) of v(x).Sxy' 35. The solution of a linear differential equation with variable coefficients ordinarily requires numerical methods (Chapter 6) or infinite series methods (Chapter 3). but for equations of higher degree we may need either to spot a fortuitous factorization or to apply a numerical technique such as Newton's method (or use a calculator j computer solve command). . . (1) precisely (3) This equation is called the characteristic equation or auxiliary equation of the differential equation in (1).1 we substituted y = erx in the second-order equation ay" + by'+ cy = 0 to derive the characteristic equation ar 2 + br + c = 0 that r must satisfy. Finding the exact values of these zeros may be difficult or even impossible. the quadratic formula is sufficient for second-degree equations.. For example. in Section 2. every nth-degree polynomial. This suggests that we try to find r so that all these multiples of erx will have sum zero. we substitute y = and with the aid of Eq. that is. if we substituted y = erx in Eq. ( 1).3 Homogeneous Equations with Constant Coefficients 125 general such equation may be written in the form (1) where the coefficients a0. is 'reduced to the solution of this purely algebraic equation. . 0.. (1). + a2r 2 + azr + ao) = 0. with the constant coefficients depending on r and the coefficients ak. (3). a2. erx Because erx is never zero. The Characteristic Equation We first look for a single solution of Eq. To carry out this technique in the general case. According to the fundamental theorem of algebra. Our problem. in which case y = erx will be a solution of Eq. (1). Hence. erx (anrn + an-lrn-l + . (2) we find the result to be erx in Eq. though not necessarily distinct and not necessarily real.2. (1).such as the one in Eq. . then. each term would be a constant multiple of erx. an are real constants with an f. we see that y = when r is a root of the equation will be a solution ofEq.has n zeros. and begin with the observation that (2) so any derivative of erx is a constant multiple of erx. a 1 . . Consequently.a2 ) • • • (x .a)(D ..b)(y' .b)(D . then.by).a)(D .ay.b) = D 2 . it can be shown by induction on the number of factors that an operator product of the form (D .a) y. The proof of the formula in (7) is (D.(a + b)D + ab.a)(D ..by) = y" . A first-degree polynomial operator with leading coefficient 1 has the form D .b(y'.ay = y' . 2. Similarly. The problem. . We also denote by D differentiation with respect to x . the algebra of polynomial differential operators closely resembles the algebra of ordinary real polynomials.2.a 1)(x . We see here also that (D .a)y = Dy. + a2 / 2) + a1Y 1 + aoy of y and its first n derivatives. is to produce the missing linearly independent solutions.by) = D(y' .in the same way as does an ordinary product (x . For example.a) y (7) for any twice differentiable function y the following computation: = y(x). it is a polynomial differential operator. It operates on a function y = y (x ) to produce (D.3 Homogeneous Equations with Constant Coefficients 127 Polynomial Differential Operators If the roots of the characteristic equation in (3) are not distinct-there are repeated roots-then we cannot produce n linearly independent solutions of Eq. the operator Lin (5) may be written (6) and we will find it useful to think of the right-hand side in Eq. In terms of D.(a + b)y' + bay = D(y' . it is convenient to adopt "operator notation" and write Eq.an) of linear factors. where a is a real number.b)y = (D. we obtain only the two functions ex and e 2 x. 2. (1) in the form Ly = 0. where the operator operates on the n-times differentiable function y(x) to produce the linear combination Ly = an/n) + an-! yen-!) + .a)(y' .ay ) = (D . The important fact about such operators is that any two of them commute: (D.a(y'.ay) = (D . (6) as a (formal) nthdegree polynomial in the "variable" D.b)(D. and 2.a .a2 ) · · • (D -an) expands-by multiplying out and collecting coefficients. so that = djdx the operation of and so on.a 1)(D.b)y = (D . (1) by the method of Theorem 1. with x denoting a real variable. For this purpose.(b + a)y' + aby = y" . if the roots are 1.ay) . r 1)ky ] = 0.1 > 1. For example. .. But this is so if and only if U(X ) = CJ + C2X + c 3x 2 + · · · + Ck Xk-I.r1) [ ue' 1x ] = (Du)e' 1x + u(r1 e' 1x ) - r1 (ue' 1x ) = (Du) e' 1x. xk-l e'lx of the original differential equation Ly = 0.1 solutions. Consequently. because the equation is of order k + 1. suppose that Eq. 1 (11) where u (x) is a function yet to be determined. (1 0) is y (x ) = U e ' 1x = ( CJ In particular. ro of multiplicity 1 and r 1 of multiplicity k = n . however.128 Chapter 2 Linear Equations of Higher Order Repeated Real Roots Let us now consider the possibility that the characteristic equation (3) has repeated roots. This is. we need k + 1 linearly independent solutions in order to construct a general solution. To find the missing k .. we see here the additional solutions x e'l x. Hence y = ue' 1x will be a solution of Eq. a polynomial of degree at most k . + C2X + C3X 2 + · · · + Ck Xk l) e ' 1x . The fact that y 1 = e'I x is one solution of Eq. Observe that (D .ro)[(D. (10) suggests that we try the substitution y(x) = u(x )y 1(x) = u(x)e' x . the corresponding operator Lin (6) can be written as (9) the order of the factors making no difference because of the formula in (7). (3) has only two distinct roots. it follows that (13) for any sufficiently differentiable function u(x ). (12) Upon k applications of this fact. Two solutions of the differential equation Ly = 0 are certainly Yo = e'0 x and y 1 = e'Ix.r 1 replaced with an arbitrary polynomial operator. not sufficient. (3) can be rewritten in the form (8) Similarly. . Then (after dividing by an) Eq. Hence our desired solution of Eq. The preceding analysis can be carried out with the operator D .1. When this is done. we note that Ly = (D. (10) if and only if Dku = u(k) = 0. x 2 e'l x . every solution of the kth-order equation (10) will also be a solution of the original equation L y = 0. Hence our problem is reduced to that of finding a general solution of the differential equation in (10). . the result is a proof of the following theorem. A complex-valued function F of the real variable x associates with each real number x (in its domain of definition) the complex number F(x) = f(x) + ig(x). The proof of this assertion is a straightforward computation based on the definitions and formulas given earlier: Dx(erx) = D x (eax cosbx) + iDx (eax sinbx) = (aeax cosbx.beax sinbx) + i (aeax sinbx + beax cosbx) = (a+ bi)(eax cosbx + ieax sinbx) = rerx. we define the derivative F' of F by F'(x) = J'(x) + ig'(x). we define the exponential function eZ. (17) The real-valued functions f and g are called the real and imaginary parts. for z = x + iy an arbitrary complex number. to be (16) Thus it appears that complex roots of the characteristic equation will lead to complex-valued solutions of the differential equation. where r = a ± bi. If they are differentiable. respectively. We note from Euler's formula that (19a) and (19b) The most important property of erx is that (20) if r is a complex number. (15) This result is known as Euler's formula. Because of it. . this implies that e 10 =cost/+ i sine. We say that the complex-valued function F(x) satisfies the homogeneous linear differential equation L[F(x)] = 0 provided that its real and imaginary parts in (17) separately satisfy this equation-so L[F(x)] = L[f(x)] + iL[g(x)] = 0. respectively. of F. The particular complex-valued functions of interest here are of the form F (x) = er x. (18) Thus we simply differentiate the real and imaginary parts ofF separately.130 Chapter 2 Linear Equations of Higher Order Because the two real series in the last line are the Taylor series for cost/ and sine. It follows that c2 = 3.1. y) + iy = reiB of the complex number z. in terms of the modulus r = x 2 + y2 > 0 of the number z and its argument e indicated in Fig. we find that and ~= (2e i3rr/2)1 /2 = -Jle i3rr / 4 = -J2 ( cos 3: + 3:) i sin = -1 + i.3. Similarly. J Example 5 Solution 4S + 4 y = 0.1 ± i . Thus the four (distinct) roots of the characteristic equation are r = ± (± 1+ i ). This form follows from Euler's formula upon writing z= r X (~ + i~) = r(cose + i sine)= rei 8 FIGURE 2. 1 ± i and .132 Chapter 2 Linear Equations of Higher Order Then so the initial conditions give y(O) = c1 = 1 and y'(O) = 2ct + c2 = 5. give a general solution of the differential equation yC4 ) + 4y = 0. Modulus and argument of the complex number X+ iy. 2.Jr denotes (as usual for a positive real number) the positive square root of the modulus of z. -i = e 3rr12 . so i = eirr/ 2 . Find a general solution of y < The characteristic equation is and its four roots are ±-J±2T. • .1. For instance. the imaginary number i has modulus 1 and argument n j2. and so the desired particular solution is y(x) = e2x(cosx + 3 sinx).3. • (22) In Example 5 below we employ the polar form y z =x (x. These two pairs of complex conjugate roots. Another consequence is the fact that the nonzero complex number z = reie has the two square roots (23) where . Since i = eirr/ 2 and -i = ei 3rrl 2 . 2. and ± 10 of the constant term 10. sin bx).. (24) are linearly independent.: k- 1 that appear in Eq. The three roots we The roots of this quotient are the complex conjugates -1 have found now yield the general solution • . (24) It can be shown that the 2k functions xPeax cosbx. then the corresponding part of the general solution has the form (AI+ Azx + . and division of the former into the latter produces as quotient the quadratic polynomial r 2 + 2r +5 = (r + 1)2 + 4.lOy= 0 + r10 = 0. + Bkxk-l)e(a. we see that the characteristic equation has as its roots the conjugate pair -3 ± 2i of multiplicity k = 2.. By trial and error (if not by inspection) we discover the root 2. cosbx p=O + d. ±5.3 Homogeneous Equations with Constant Coefficients 133 Repeated Complex Roots Theorem 2 holds for repeated complex roots. ±2. 0. Often the most difficult part of solving a homogeneous linear equation is finding the roots of its characteristic equation. + Akxk-l)e(a+bi)x + (BI + Bzx + . Hence Eq..: p. The factor theorem of elementary algebra implies that r . the only possible rational roots are the factors ± 1.. If the conjugate pair a ± bi has multiplicity k.. ± 2i .bi)x k-1 = L:xpeax(c.10. Example 7 illustrates an approach that may succeed when a root of the characteristic equation can be found by inspection. Example 6 Solution Find a general solution of (D 2 + 6D + 13)2 y = 0. xPeax sinbx. (24) gives the general solution In applications we are seldom presented in advance with a factorization as convenient as the one in Example 6. By completing the square.. Example 7 The characteristic equation of the differential equation / 3) + y'. is the cubic equation r3 By a standard theorem of elementary algebra.2 is a factor of r 3 + r . 57. 2. x 2 y" + xy' + 9y = 0 x 2 y" + 7xy' + 25y = 0 x 3 y"' + 6x 2 y" + 4xy' = 0 x 3 y"'. 51. (0) = 0. A massspring-dashpot system. .3x 2 y" + xy' = 0 x 3 y"' + 6x 2 y" + 7xy' + y = 0 Mechanical Vibrations The motion of a mass attached to a spring serves as a relatively simple example of the vibrations that occur in more complex mechanical systems.2.) 49. 54. We take x > 0 when the spring is stretched. so that it can move only back and forth as the spring compresses and stretches. 52.1. Because this is the same as the displacement x of the mass m from its m c I I Equilibrium position FIGURE 2.3.2. Solve the initial value problem FIGURE 2. a . We consider a body of mass m attached to one end of an ordinary spring that resists compression as well as stretching. Y1 (0) = y~(O) = 1 and Yz(O) = y.4. According to Hooke's law. (25) nevertheless has two linearly independent solutions Y1 (x) and Yz (x) defined for all x such that • • • Make the substitution v = ln x of Problem 51 to find general solutions (for x > 0) of the Euler equations in Problems 52 through 58. the substitution v = ln x (x > 0) transforms the second-order Euler equation ax 2 y" + bxy' + cy = 0 to a constant-coefficient homogeneous linear equation. For many such systems. The differential equation y" + (sgnx)y =0 ax 3 y"' + bx 2 y" + cxy' + dy = 0 has the discontinuous coefficient function (where a. the other end of the spring is attached to a fixed wall. Graphs of y 1 (x) and Yz(x) in Problem 50.+ dy = dv ~ 0.1 = 0. and thus x < 0 when it is compressed.x 2 y" + xy' = 0 x 3 y"' + 3x 2 y" + xy' = 0 x 3 y"'.) The graphs of these two solutions are shown in Fig. if X < 0.+ (b. 56. the analysis of these vibrations is a problem in the solution of linear differential equations with constant coefficients.1. (Suggestion: Each Yi (x) will be defined by one formula for x < 0 and by another for x ~ 0.4.2. c. as shown in Fig. 58.b + 2a). Show that Eq. 53. (25) at each point x =1= 0. d are constants) into the constantcoefficient equation sgnx = 1-1 +1 if X > 0.4 Mechanical Vibrations where a and f3 are the complex conjugate roots of r 3 . the restorative force Fs that the spring exerts on the mass is proportional to the distance x that the spring has been stretched or compressed. y(O) = y' (0) = y" (0) = 0.3a)dv 3 dv 2 ~y ~y + (c. Each satisfies Eq. to discover that 135 1 ( ex+ 2e-xfZ cos xv-'3) y(x) = 3 2is a solution. b. Denote by x the distance of the body from its equilibrium positionits position when the spring is unstretched.3. 55. Show similarly that this same substitution transforms the third-order Euler equation (25) y<4) = y<3) + y" + y' + 2y. 2. Each has a continuous derivative at x = 0. Assume that the body rests on a frictionless horizontal plane. 50. According to Problem 51 in Section 2. 2y<3 \0) = 30.1. We refer to the motion as free in this case and forced in the case F(t) "/:. (3). the mass is subjected to a given external force FE = F(t).2.=mx ' dt 2 we obtain the second-order linear differential equation mx" +ex'+ kx = F(t) (3) that governs the motion of the mass.2. That is. it follows that Fs = -kx.1 shows the mass attached to a dashpot-a device. 0. For an alternative example. . we might attach the mass to the lower end of a spring that is suspended vertically from a fixed support. If there is no dashpot (and we ignore all frictional forces).136 Chapter 2 Linear Equations of Higher Order equilibrium position. Using Newton's law d2x " F=ma = m . then we ask you to show in Problem 9 that y satisfies Eq. we replace F(t) with 0 in Eq. specifically. (1) The positive constant of proportionality k is called the spring constant. then we set c = 0 in Eq. dx FR = -cv =-c-. Note that Fs and x have opposite signs: Fs < 0 when x > 0.4.6. (1) with Fs = -Wand x = s0 . 2.4. We will defer discussion of forced motion until Section 2. Figure 2. dt (2) The positive constant c is the damping constant of the dashpot. that my"+ cy' + ky = F(t) (5) if we include damping and external forces (meaning those other than gravity). so that so = mgjk.4. like a shock absorber. A mass suspended vertically from a spring. then the total force acting on the mass is F = Fs + FR + F£. mg = ks0 . Fs > 0 when x < 0. we may regard Eq. If there is no external force. describes free motion of a mass on a spring with dashpot but with no external forces applied. measured downward from its static equilibrium position. This gives the static equilibrium position of the mass. (2) as specifying frictional forces in our system (including air resistance to the motion of m). More generally. If y denotes the displacement of the mass in motion. in addition to the forces Fs and FR. as in Fig. that provides a force directed opposite to the instantaneous direction of motion of the mass m. that is. In this case the weight W = mg of the mass would stretch the spring a distance s0 determined by Eq. it is damped motion if c > 0. (3). If. Thus the homogeneous equation mx" + ex' + kx = 0 (4) FIGURE 2. We assume the dashpot is so designed that this force FR is proportional to the velocity v = dxjdt of the mass. (3) and call the motion undamped. ~mv2 = ~m (ds)2 = ~mL2 (d8)2 2 2 dt 2 dt We next choose as reference point the lowest point 0 reached by the mass (see Fig. The distance along the circular arc from 0 to m is s = L8.3.3. (7) . The fact that the sum of T and V is a constant C therefore gives 2 (d8) 1 -mL +mgL(l-cos8)=C.4. However. In fact. the result is an equation in the form of Eq. a simple pendulum consists of a mass m swinging back and forth on the end of a string (or better.cos8) . It therefore seems reasonable to simplify our mathematical model of the simple pendulum by replacing sin 8 with e in Eq. sin 8 and 8 agree to two decimal places when 181 is at most n/12 (that is. (6). and it may be instructive to see the energy method in a simpler application like the pendulum. 2. The simple pendulum. then sin 8 ~ 8 (this approximation obtained by retaining just the first term in the Taylor series for sin 8).4. so V = mgL(l. Now recall that if 8 is small. We may specify the position of the mass at time t by giving the counterclockwise angle 8 = 8(t) that the string or rod makes with the vertical at timet. derivations of differential equations based on conservation of energy are often seen in more complex situations where Newton's law is not so directly applicable. 2 dt 2 We differentiate both sides of this identity with respect to t to obtain so (6) after removal of the common factor mL 2 (d8jdt). we will apply the law of the conservation of mechanical energy. In a typical pendulum clock.4. so the velocity of the mass is v = dsjdt = L(d8jdt). 2. and therefore its kinetic energy is T = FIGURE 2. If we also insert a term c8' to account for the frictional resistance of the surrounding medium. according to which the sum of the kinetic energy and the potential energy of m remains constant.3). as shown in Fig. (3) and (5) stems from the fact that it describes the motion of many other simple mechanical systems. (4): e" + c8' + k8 = 0. This differential equation can be derived in a seemingly more elementary manner using the familiar second law F = rna of Newton (applied to tangential components of the acceleration of the mass and the force acting on it). 8 would never exceed 15°. To analyze the motion of the mass m.4 Mechanical Vibrations 137 The Simple Pendulum The importance of the differential equation that appears in Eqs. for example. For example. Then its potential energy V is the product of its weight mg and its vertical height h = L (1 .cos 8) above 0.2. 15°). a massless rod) of length L. (12) Thus the mass oscillates to and from about its equilibrium position with . In any event. Free Undamped Motion If we have only a mass on a spring. then Eq. Instead. although tan a= BjA. n/2) given by a calculator or computer. we choose constants C and a so that cosa = c' A and sma = . (8) as x II 2x = 0 . We might. where either A or B or both may be negative. .4. (7) will probably not describe accurately the actual motion of the pendulum over a long period of time. (8) It is convenient to define wo=ff (9) and rewrite Eq. expect the effects of the discrepancy between () and sin() to accumulate over a period of time. tan.4. 2. we first analyze free undamped motion and then free damped motion. we find that x(t) = C cos(wot -a). if A > 0. c' B (II) as indicated in Fig. Note that. (8') is x(t) = Acoswot + Bsinwot.n/2 < x < n/2). however. B > 0 (first quadrant). The angle a. With the aid of the cosine addition formula. (10) To analyze the motion described by this solution.n/2. so that Eq.138 Chapter 2 Linear Equations of Higher Order where k = gjL. if A < 0 (second or third quadrant). a is the angle between 0 and 2n whose cosine and sine have the signs given in (11).4. the angle a is not given by the principal branch of the inverse tangent function (which gives values only in the interval . In the remainder of this section. with neither damping nor external force. Note that this equation is independent of the mass m on the end of the rod. ) +C smwot = C(cosa coswot .1 (8/A) 2n + tan.1 (8/A) is the angle in ( . (3) takes the simpler form mx" + kx = 0. Thus A FIGURE 2.1 (B/A) if A > 0. where tan.1 (B/A) a= { n + tan.4. ) + sma smwot . from (I 0) and ( 11) we get A x(t ) = C ( C cos wot B . B < 0 (fourth quadrant). +w0 (8' ) The general solution of Eq. cvo.= T 2n 2rr ~ 1. (10). that is. period of oscillation. and the time lag a o=cvo are indicated.6283 s and with frequency 1 cvo 10 v= .4.4. then find the amplitude C and phase angle a by carrying out the transformation of x(t) to the form in Eq. and Such motion is called simple harmonic motion. If the initial position x (0) = x 0 and initial velocity x' (0) = v0 of the mass are given. The period of the motion is the time required for the system to complete one full oscillation.. Solution The spring constant is k SOx = 0. which measures the number of complete cycles per second. Amplitude 139 2. as indicated previously. (8) yields ~x" x" + + 100x = 0. 0 )) = C cos(cvo(t. Note that frequency is measured in cycles per second.= . If time t is measured in seconds. its frequency is 1 cvo V=-=T 2rr (14) in hertz (Hz). whereas circular frequency has the dimensions of radians per second.) Find the position function of the body as well as the amplitude. Hence it will oscillate with period 2rr 2rr T = cvo = lO ~ 0. we first determine the values of the coefficients A and Bin Eq. the circular frequency cv0 has dimensions of radians per second (radjs). Consequently. frequency. Circular frequency 3. Phase angle C. = (100 N)/(2 m) = 50 (N/m).5. a. the circular frequency of the resulting simple harmonic motion of the body will be cv0 = .5915 Hz.5. so is given by T=X 2rr cvo (13) seconds.JIOO = 10 (rad/s).a)= C cos ( cvo (t. .2. is shown in Fig. so Eq. (Note that these initial conditions indicate that the body is displaced to the right and is moving to the left at time t = 0.4 Mechanical Vibrations 1.8)) FIGURE 2. Example 1 A body with mass m = ~ kilogram (kg) is attached to the end of a spring that is stretched 2 meters (m) by a force of 100 newtons (N). A typical graph of a simple harmonic position function 1---T ----1 x(t) = C cos(cvot. Simple harmonic motion. where the geometric significance of the amplitude C. It is set in motion with initial position x 0 = 1 (m) and initial velocity v0 = -5 (rnls). (12). and time lag of its motion. 2. the period T. ~. the position function of the body takes the form x(t) ~ ~Jscos(lOt.8195).6. we write x (t) = v'5(2 1 ) 2 v'5 cos lOt.. = 1 and B = .5820 s.140 Chapter 2 Linear Equations of Higher Order We now impose the initial conditions x (0) function x(t) =A cos lOt+ B sin lOt It follows readily that A = 1 and x' (0) = -5 on the position with x'(t) =-lOA sin lOt+ lOB cos lOt. 2. Hence its amplitude of motion is To find the time lag. where the phase angle a satisfies 2 .< 0. 1 cosa = v'5 > 0 and sma = . • . With the amplitude and approximate phase angle shown explicitly.5.4. v'5 Hence a is the fourth-quadrant angle and the time lag of the motion is a 0= wo ~ 0. and its graph is shown in Fig. 1 .v'5 sin lOt = v'5 2 cos(lOt - a) . so the position function of the body is x(t) =cos lOt - 2 sm lOt. 118.5 c .J4kiri.JK7iii.2. Graph of the position function = C cos(w0 t . r2 + w6 = 0 of Eq. Solution curves are graphed with the same initial position x 0 and different initial velocities.628.a) in Example 1.4.5 2.6. (15) .7 shows some typical graphs of the position function for the overdamped case.4. both of which are negative. (16) The characteristic equation r 2 + 2pr r1.4. It is easy to see that x (t) ---+ 0 as t ---+ +oo and that the body settles to its equilibrium position without any oscillations (Problem 29). we chose xo a fixed positive number and illustrated the effects of changing the initial velocity v0 .5 3 -0. Then ( 17) gives distinct real roots r 1 and r2 . the differential equation we have been studying takes the form mx" + ex' + kx = 0. we are dealing with a strong resistance in comparison with a relatively weak spring or a small mass. Free Damped Motion With damping but no external force. x" where w0 = + 2px' + w5x c 2m = 0. Overdamped motion: x(t) = c 1er 11 + c2 e'2 1 with r 1 < 0 and r2 < 0.4 Mechanical Vibrations X 141 T 0. alternatively.7. with amplitude C ~ 1. period T ~ 0. (19) FIGURE 2. and we distinguish three cases.582. Because cis relatively large in this case. The position function has the form 0 OVERDAMPED CASE: c > Ccr(c2 > 4km). In every case the would-be oscillations are damped out. . and time lag 8 ~ 0. (15) has roots = -p ± (p 2 - w6) 112 (17) that depend on the sign of p-w=---0 2 2 c2 4m 2 k m c2 4km 4m 2 - (18) The critical damping Ccr is given by Ccr = according as c > Ccn c = Ccr• or c < Ccr· .5 -1 x(t) FIGURE 2. Figure 2.> 0. is the corresponding undamped circular frequency and p = . 4. The motion is not actually periodic. Most of these quantities are shown in the typical graph of underdamped motion in Fig. but it is nevertheless useful to call cv 1 its circular frequency (more properly.4. In the critically damped case. Solution curves are graphed with the same initial position x 0 and different initial velocities.p ± i . 2. and they resemble those of the overdamped case (Fig. in accord with the time-varying FIGURE 2.4. the body passes through its equilibrium position at most once.142 Chapter 2 Linear Equations of Higher Order CRITICALLY DAMPED CASE: c Ccr(c 2 4km).P 1 an integral multiple of rr . and C e . we may rewrite Eq.pt its time-varying amplitude. 2. 2.8. that is. (20) as so x(t) = ce-pr cos(cv1t. Thus the action of the dashpot has at least two effects: 1.4. 2. its pseudofrequency). The graph of x(t) lies between the "amplitude 1 and touches them when cv t . but even a slight reduction in resistance will bring us to the remaining case. . Underdamped oscillations: X(t) = Ce-pt COS(Wtt - a).jw6 - The characteristic equation now p 2 . .P1 and x = ce.a) (23) where cosa .j4km .a is envelope" curves x = -ce.c 2 2m (22) Using the cosine addition formula as in the derivation of Eq. ---/ X= + Ce-Pt 0 The solution in (22) represents exponentially damped oscillations of the body around its equilibrium position.a) '/ / '-.7). (17) gives equal roots r 1 = r 2 = . It exponentially damps the oscillations. (21) that in this case cv 1 is less than the undamped circular frequency cv0 . = c· B a I x = Ce-P1 cos(w 1t . It slows the motion. = -. T1 = 2rr 1cv 1 its pseudoperiod of oscillation. and the general solution is (21) FIGURE 2. the one that shows the most dramatic behavior. (12). the resistance of the dashpot is just large enough to damp out any oscillations. so T1 is larger than the period T of oscillation of the same mass without damping on the same spring.9. so the general solution is = = (20) Because e .8. UNDERDAMPED CASE: c < Ccr(c2 < 4km). 0 has two complex conjugate roots . Critically damped motion: x(t) = (ct + c2t)e-pt with p > 0. Note from Eq. Some graphs of the motion in the critically damped case appear in Fig. the dashpot decreases the frequency of the motion. c A and sma .pt > 0 and c 1 + c2 t has at most one positive zero.p of the characteristic equation. amplitude.4.9. In this case.w. where CVJ- _J w0 2 - p2 - _ . and it is clear that x(t) --+ 0 as t --+ +oo. 1 (cos 59 t. • ~ ·- '" " '••~'o. damping typically also delays the motion further-that is. It follows that x(O) =A= 1 whence we find that A the body is and x'(O) =-A+ B59 = -5. Recall that m = and k = 50.5915 > v 1 in Example 1).. = 1 and B = -4/ . · -- Example 2 The mass and spring of Example 1 are now attached also to a dashpot that provides 1 N of resistance for each meter per second of velocity.5836 Hz (as compared with T ~ 0. The characteristic equation r 2 + 2r + 100 = (r r2 = -1 ± . The new (pseudo)period and frequency are 2rr T1 = . its new frequency and pseudoperiod of motion. The mass is set in motion with the same initial position x(O) = 1 and initial velocity x'(O) = -5 as in Example 1.1 (A cos -J99 t + 1) 2 + 99 0 has roots r 1. its new time lag.1 (A cos -J99 t + B sin -J99 t) + -J99 e. so the general solution is x(t) = e.2. Thus the new position function of x(t) = e. (4) is ~x" + x' +SOx = 0. that is. increases the time lag-as compared with undamped motion with the same initial conditions. Hence Eq.•N. Now find the position function of the mass.J99 ~ 9. the new circular (pseudo)frequency is w 1 = . and the times of its first four passages through the initial position x = 0. 1 v99 IN'\ ~ 0. + B sin -J99 t).6315 s and v1 = - T1 = - WI 2rr = -- _J99 2rr ~ 1.6283 < T 1 and v ~ 1.4 Mechanical Vibrations 143 As the following example illustrates. we are now given c = 1 in mks units.9499 (as cornpared with wo = 10 in Example 1).1 (-A sin -J99 t + B cos J99 t).J99 i. it is better practice in a particular case to set up the differential equation and then solve it directly. We now impose the initial conditions x (0) = 1 and x' (0) = -5 on the position function in (23) and the resulting velocity function x' (t) = -e. o.= 2rr w.J99.~ sin 59 t) . . (24) Consequently.. Hence its time-varying amplitude of motion is . Solution 1 x" + 2x' + 100x = 0. Rather than memorizing the various formulas given in the preceding discussion. X • x (t) = C cos (w0 t . Determine the period and frequency of the simple harmonic motion of a body of mass 0.. 8o+-. A body with mass 250 g is attached to the end of a spring that is stretched 25 ern by a force of 9 N. and thus when Jr Jr 3n 2' 2' 2 . period. 8o+-. t 4 we calculate for the 1 0.1107 0. A mass of 3 kg is attached to the end of a spring that is stretched 20 em by a force of 15 N. . that is. 2.a 1) = 0. 1. respectively.1195 2 0.11.0667 Accordingly.--. 3n 2w1 We see similarly that the undamped mass of Example 1 passes through equilibrium when Jr 3n 3n Jr t = 80 ...11 (where only the first three equilibrium passages are shown) we see the damped oscillations lagging slightly behind the undamped ones .4 Mechanical Vibrations 145 From (24) we see that the mass passes through its equilibrium position x = 0 when cos(w 1t ..a) -1 FIGURE 2. and frequency of the resulting motion. Determine the period and frequency of the simple harmonic motion of a 4-kg mass on the end of a spring with spring constant 16 N j m.4352 3 0. n tn (undamped) tn (damped) t 1.0532 1. when t =01.2. t3. t2. It is set in motion with initial position x 0 = 0 and initial velocity v0 = -10 mj s.7509 4 1...8 illustrating the additional delay associated with damping. 3. 4. At time t = 0 . 8o. in Fig.4. Graphs on the interval 0 ~ t ~ 0. 2.7390 0. Find the amplitude.4. .75 kg on the end of a spring with spring constant 48 N j m. 2wo 2wo 2wo 2wo The following table compares the first four values undamped and damped cases.4249 0. when located at the respective distances R 1 and R 2 from the center of the earth-have periods p 1 and p 2 .x 0 in this differential equation. with mass M and radius R = 3960 (mi).4. with a period of about 84 min.2 ft j s2 .5 (recall that the density of water is 1 gj cm3 ). 7..146 Chapter 2 Linear Equations of Higher Order the body is pulled 1 m to the right.4. and uniform density p . A mass m falling down a hole through the center of the earth (Problem 12). (Suggestion: First denote by x(t) the displacement of the mass below the unstretched position of the spring. has the same period as does a pendulum of length 100. (a) Show that F. Most grandfather clocks have pendulums with adjustable lengths. where M. Conclude from Newton's second law and part (a) that r " (t) = -k 2 r(t). 2. Conclude that the buoy undergoes simple harmonic motion around its equilibrium position X e = ph with period p = 2:rr ~· Compute p and the amplitude of the motion if p = 0. How do you explain the coincidence? Or is it a coincidence? (e) With what speed (in miles . where the radius of the earth is R = 3956 (mi). (b) Now suppose that a small hole is drilled straight through the center of the earth. Two pendulums are of lengths L 1 and L 2 and. The buoy of Problem 10.00 in. compare with the result in part (c). 8. A certain pendulum keeps perfect time in Paris.. M denotes the mass of the earth).10 in. Show that FIGURE 2. (5) describing the motion of a mass attached to the bottom of a vertically suspended spring. Assume that friction is negligible. 2. But this clock loses 2 min 40 s per day at a location on the equator. where k 2 = GMjR 3 = gjR. and g = 980 cmj s2 • FIGURE 2.13). PI Pz 6. and set in motion with an initial velocity of 5 mjs to the left. 11. A pendulum of length 100. = -GMmrjR 3 . (a) Find x(t) in the form C cos(w0 t . set up the differential equation for x .) 10. The buoy is initially suspended at rest with its bottom at the top surface of the water and is released at time t = 0. Find the radius of the buoy. h = 200 em. Use the result of Problem 5 to find the amount of the equatorial bulge of the earth.5 g j cm 3 . assume that the differential equation of a simple pendulum of length L is L()" + g() = 0. When depressed slightly and released. Consider a floating cylindrical buoy with radius r. Derive Eq. (d) Look up (or derive) the period of a satellite that just skims the surface of the earth. is the mass of the part of the earth within a sphere of radius r. Let a particle of mass m be dropped at time t = 0 into this hole with initial speed zero. One such clock loses 10 min per day when the length of its pendulum is 30 in.12). atop a nearby mountain. A cylindrical buoy weighing 100 lb (thus of mass m = 3. Thereafter it is acted on by two forces: a downward gravitational force equal to its weight mg = :rr r 2 hg and (by Archimedes' principle of buoyancy) an upward force equal to the weight :rr r 2 xg of water displaced. For a particle of mass m within the earth at distance r from the center of the earth. Use the result of Problem 5 to find the height of the mountain. height h. With what length pendulum will this clock keep perfect time? 9. (c) Take g = 32. and conclude from part (b) that the particle undergoes simple harmonic motion back and forth between the ends of the hole. where x = x(t) is the depth of the bottom of the buoy beneath the surface at time t (Fig. it oscillates up and down four times every 10 s. 0. stretching the spring. Then substitute y = x .4. Assume that the earth is a solid sphere of uniform density. located at a point at sea level where the radius of the earth is R = 3960 (mi). the gravitational force attracting m toward the center is F.mjr 2 . thus connecting two antipodal points on its surface.4.a). where g = G M / R 2 is the gravitational acceleration at the location of the pendulum (at distance R from the center of the earth.12. In Problems 5 through 8. = -GM.13. and let r(t) be its distance from the center of the earth at timet (Fig. (b) Find the amplitude and period of motion of the body. 5.125 slugs in ft-lb-s (fps) units) floats in water with its axis vertical (as in Problem 10). 12. {a) Find the position function x(t) and show that its graph looks as indicated in Fig. findthe undampedposition 25. c = 30. write the position function in the f orm x(t) = C1 e. Also. 24. If it is underdamped. and wf = w5 . (Critically damped) Deduce from Problem 24 that the mass passes through x = 0 at some instant t > 0 if and only if x 0 and v0 + px 0 have opposite signs. vo = 4 m = 1. k = 40.15. How do you explain the coincidence? Or is it a coincidence? 13. and phase angle of the motion.a 1) . 0 5 10 15 20 FIGURE 2.4. and k = 226 is set in motion with x(O) = 20 and x' (0) = 41. or underdamped. v0 = 16 m = 2. The position function x(t) of Problem 13. and the resulting damped vibrations have a frequency of 78 cycles/min. (a) If the weight is pulled down 1 ft below its static equilibrium position and then released from rest at timet = 0.14. as specified in each problem. x 0 = 2. 18. and to a dashpot that provides 3 lb of resistance for every foot per second of velocity. 20. (Overdamped) Show in this case that . 15. Vo =50 A 12-lb weight (mass m = 0. Suppose that the mass in a mass-spring-dashpot system with m = 25. The system is critically damped. time-varying amplitude. c = 9. After how long will the time-varying amplitude be 1% of its initial value? t. xo = 6. (a) Find the stiffness coefficient k of the spring if the car undergoes free vibrations at 80 cycles per minute (cycles/ min) when its shock absorbers are disconnected.4. c = 8. compare with the result in part (e). 2. Suppose that the mass in a mass-spring-dashpot system with m = 10.2. x 0 = 5. Vo = -10 m = 2. x 0 = 0. (b) Find how far the mass moves to the right before starting back toward the origin. k = 50. find its position function x(t). but with the dashpot disconnected (so c = 0). the car is set into vibration by driving it over a bump. 5 4 147 function u(t) =Co cos(wot. The remaining problems in this section deal with free damped motion. 17. 27. 14. c = 16. c = 12.14. v0 = 2 m = 1. overdamped.4. ' ' 10 ~ 0 -10 / Problems 24 through 34 deal with a mass-spring-dashpot system having position fun ction x (t) satisfying Eq. In Problems 15 through 21. The mass is set in motion with initial position Xo and initial velocity v0 • Find the position fun ction x(t) and determine whether the motion is overdamped. 22. critically damped. c = 20. (4 ).4 Mechanical Vibrations per hour) does the particle pass through the center of the earth? (f) Look up (or derive) the orbital velocity of a satellite that just skims the surface of the earth. xo = 2. k = 169. w5 = k j m . k = 16. 2. c = 10. k = 125.pt cos(w 1t . Finally. k = 4. x 0 = 4. We write x 0 = x(O) and v0 = x' (O) and recall that p = cj(2m) . and k = 2 is set in motion with x(O) = 0 and x ' (O) = 5. 16. Xo = 5. (b) With the shock absorbers connected. construct a figure that illustrates the effect of damping by comparing the graphs of x(t) and u(t). 20 m = c = 3. vo = -8 m = 4.p 2 . c = 10. 21. Assume that the suspension system acts like a single spring and its shock absorbers like a single dashpot.4. (Critically damped) Deduce from Problem 24 that x (t) has a local maximum or minimum at some instant t > 0 if and only if v0 and v0 + px 0 have the same sign. 3 2 ~ 0 -I -2 23. (a) Find the position function x(t) and show that its graph looks as indicated in Fig. 19. (b) Find the frequency. (Critically damped) Show in this case that 5 10 / -20 0 I 15 20 x (t) = (xo + vot + p xot)e .15. The position function x(t) of Problem 14.ao) that would result if the mass on the spring were set in motion with the same initial position and velocity. a mass m is attached to both a spring (with given spring constant k) and a dashpot (with given damping constant c). (b) Find the pseudoperiod ofthe oscillations and the equations of the "envelope curves" that are dashed in the figure. k = 63. (4) with appropriate values of the coefficients. so that its vertical vibrations satisfy Eq.375 slugs in fps units) is attached both to a vertically suspended spring that it stretches 6 in. or underdamped. This problem deals with a highly simplified model of a car of weight 3200 lb (mass m = 100 slugs in fps units). v0 = 0 m = 3.P1 • FIGURE 2. 26. . Thus the future motion of an ideal mass-spring-dashpot system is completely determined by the differential equation and the initial conditions. Suppose that m = 1 and c = 2 but k = I . D Nonhomoge~eous Eq~~~ions and Undetermined Coefficients We learned in Section 2. y x(t) 29. = 2rr pjw 1 is called the logarithmic decrement of the oscillation.. c. 30. Wt has a unique solution for t ~ 0 satisfying given initial conditions x (0) = x 0 . + a!y' + aoy = f(x). a spherical body of radius a moving at a (relatively slow) speed through a fluid of viscosity fL experiences a resistive force FR = 6rr fLaV. (1) . Differential Equations and Determinism Given a mass m. Of course in a real physical system it is impossible to measure the parameters m.. = 1. Note also that c = mw 1 ll.17 s.4. Conclude that on a given finite time interval the three solutions are in "practical" agreement if n is sufficiently large. respectively. which is an important parameter in fluid dynamics but is not easy to measure directly. 33. (Underdamped) Show that in this case 34. The final formula in Problem 33 then gives c and hence the viscosity of the fluid. 8mk 32.2n.JSmk.4.. but we saw in Section 2. (26).4 that an external force in a simple mechanical system contributes a nonhomogeneous term to its differential equation. 37. The frequency Wt and logarithmic decrement ll.!Y(n . Note: The result of Problem 33 provides an accurate method for measuring the viscosity of a fluid. Suppose that m = 1 and c = 2 but that k = 1 + 10.125 slugs in fps units) is oscillating attached to a spring and a dashpot. Whereas the graphs of x 1(t) and x 2 (t) resemble those shown in Figs. (26) with x(O) = 0 and x' (0) = 1 is x2(t) = lOne.1) + .~). The general nonhomogeneous nth-order linear equation with constant coefficients has the form GnY(n) + Gn . a dashpot constant c. According to Stokes's drag law. Show that the solution of Eq. 2.34 s and 1.8.4. and k precisely.a ) = . Compute the damping constant (in pound-seconds per foot) and spring constant (in pounds per foot). c = 2. :::::o wo(l . Thus if a spherical mass on a spring is immersed in the fluid and set in motion. 35.73 in. this drag resistance damps its oscillations with damping constant c = 6rr a fL.7 and 2. (Underdamped) If the damping constant cis small in comparison with .46 in. 2.148 Chapter 2 Linear Equations of Higher Order 28. the graph of x 3 (t) exhibits damped oscillations like those illustrated in Fig.1 implies that the equation 31. Deduce from the result of Problem 32 that lnx' = 2rrp x2 w. x' (0) = v0 . Its first two maximum displacements of 6.1o-2 n.t 1 = 2rrjw 1 if two consecutive maxima occur at times t 1 and t2. (Underdamped) A body weighing 100 lb (mass m = 3. show that for each fixed t > 0 it is true that n---+oo lim X2(t) = lim x 3 (t) n~ oo = X t (t). Show that the solution with x (0) = 0 and x' (0) = 1 is x 1 (t) = te- 1• 36. and 1. are observed to occur at times 0. (Underdamped) Show that the local maxima and minima of occur where p tan(wtt. Suppose that m Conclude that t 2 . apply the binomial series to show that mx" + ex' + kx = 0 (26) w. The constant ll. and k = 1 in Eq. (Overdamped) Prove that in this case the mass can pass through its equilibrium position x = 0 at most once. (26) with x(O) = 0 and x'(O) = 1 is 38.9. but with a very long pseudoperiod. of the oscillations can be measured by direct observation./rr because p = cj(2m). Nevertheless. (Underdamped) Let x 1 and x 2 be two consecutive local maximum values of x(t).1 sinh 10-nt. and a spring constant k. (Overdamped) If x 0 = 0.3 how to solve homogeneous linear equations with constant coefficients. Theorem 2 of Section 2. deduce from Problem 27 that = vo e-pt sinh yt. Show that the solution of Eq. Problems 35 through 38 explore the resulting uncertainty in predicting the future behavior of a physical system. suppose that f (x) = a cos kx + b sin kx. we illustrate it with several preliminary examples. (1) is sufficiently simple that we can make an intelligent guess as to the general form of Yp. Example 1 Find a particular solution of y" + 3 y' + 4 y = 3x + 2. Then. a linear combination with undetermined coefficients A and B . Then it is reasonable to expect a particular solution of the same form: Yp(x) =A cos kx + B sin kx. Similarly. that is. a general solution ofEq. substitute this expression for Yp into Eq. so Yp will satisfy the differential equation provided that (0) + 3(A) + 4(Ax +B) = 3x + 2. (1). Here f(x) = 3x Solution + 2 is a polynomial of degree 1. but with as yet undetermined coefficients. Thus our remaining task is to find Yp· The method of undetermined coefficients is a straightforward way of doing this when the given function f(x) in Eq. We may therefore substitute this form of Yp in Eq. be a particular solution of Eq. ••• .2. (1) has the form Y = Yc + Yp (2) where the complementary function Yc(x) is a general solution of the associated homogeneous equation (3) and Yp(x) is a particular solution of Eq. A 1 . We may. For example. suppose that f (x) is a polynomial of degree m. (4A )x + (3A + 4 B) = 3x +2 . (1).!Xm-! + · · · + A1x + Ao that is also a polynomial of degree m. Then y~ = A and y~ = 0. Before describing the method in full generality. be a particular solution.5 Nonhomogeneous Equations and Undetermined Coefficients 149 By Theorem 5 of Section 2. Am so that Yp will. It turns out that this approach does succeed whenever all the derivatives of f (x ) have the same form as f (x) itself. it is reasonable to suspect a particular solution Yp(X) = AmXm + Am. The reason is that any derivative of such a linear combination of cos kx and sin kx has the same form. (1). indeed. because the derivatives of a polynomial are themselves polynomials of lower degree.2. so our guess is that Yp(x) =Ax+ B. (1). and then-by equating coefficients of like powers of x on the two sides of the resulting equation-attempt to determine the coefficients A 0 . indeed. and then-by equating coefficients of cos kx and sin k x on both sides of the resulting equation-attempt to determine the coefficients A and B so that y P will. therefore. 2(A cosx + B sinx) = 2cosx. So we try Yp(x) = y~(x) = y~(x) Solution A cos x + B sinx.1~. -A. • Example 3 Find a particular solution of 3 y" + y' . (-SA+ B) cosx +(-A.B sinx) + (-A sinx + B cosx).~~- • Example 2 Solution Any derivative of e 3x is a constant multiple of e3x. -A sinx + B cosx. Thus our particular solution is Yp (x) = ~ e 3x. that is (collecting coefficients on the left). A first guess might be Yp(x) =A cosx.SB = 0 with readily found solution A =- 5 13 . which superficially resembles Example 2.2 y = 2 cos x. 5 Yp ( x ) = .150 Chapter 2 Linear Equations of Higher Order for all x.SB) sinx = 2cosx. B = /3 • Hence a particular solution is I · + 13 smx. This will be true if the x-terms and constant terms on the two sides of this equation agree. This will be true for all x provided that the cosine and sine terms on the two sides of this equation agree. but the presence of y' on the left-hand side signals that we probably need a term involving sin x as well. It therefore suffices for A and B to satisfy the two linear equations -SA+ B = 2. Then y~ = 9Ae 3x. indicates that the method of undetermined coefficients is not always quite so simple as we have made it appear. Thus we have found the particular solution Yp(X) =~X. SA = 2. Example 4 . It therefore suffices for A and B to satisfy the two linear equations 4A = 3 and 3A + 4B = 2 that we readily solve for A= ~and B = . so the given differential equation will be satisfied provided that that is. so it is reasonable to try Yp(x) = Ae3x. so that A = ~.B sinx. =-A cosx . Then substitution of Yp and its derivatives into the given differential equation yields 3( -A cosx.13 cosx • The following example. 4Ae2 x = 0 =/= 2e 2x. we find that y~. (1) is a linear combination of (finite) products of functions of the following three types: 1. so that The terms involving xe2x obligingly cancel. . In fact. The method of undetermined coefficients applies whenever the function f (x) in Eq. Then take as a trial solution for Yp a linear combination of all linearly independent such terms and their derivatives.5 Nonhomogeneous Equations and Undetermined Coefficients 151 Solution If we try Yp (x) = Ae2X. and Rule 2 tells what to do when we do have it. Therefore.4yp = 4Ae 2 x . Thus. no matter how A is chosen. • The General Approach Our initial difficulty in Example 4 resulted from the fact that f(x) = 2e 2x satisfies the associated homogeneous equation. Rule 1. An exponential function erx. cos kx or sin kx. 2e 2x. leaving only 4Ae2x A = Consequently. RULE 1 Method of Undetermined Coefficients Suppose that no term appearing either in f (x) or in any of its derivatives satisfies the associated homogeneous equation Ly = 0. 3. A reasonable guess is Yp(x) = Axe2 x. a particular solution is !. the preceding computation shows that Ae 2x satisfies instead the associated homogeneous equation. A polynomial in x. Substitution into the original differential equation yields (4Ae 2x + 4Axe2x) - 4(Axe 2 x) = 2e 2 x. given shortly. Any such function-for example. tells what to do when we do not have this difficulty.2. we should begin with a trial function Yp(x) whose derivative involves both e 2x and something else that can cancel upon substitution into the differential equation to leave the e2x term that we need. Ae 2x cannot satisfy the given nonhomogeneous equation. Then determine the coefficients by substitution of this trial solution into the nonhomogeneous equation Ly = f(x). In Rules 1 and 2 we assume that Ly = f(x) is a nonhomogeneous linear equation with constant coefficients and that f (x) is a function of this kind. 2. for which y~(x) = Ae2x + 2Axe2x and y~(x) = 4Ae2x + 4Axe2x. (4) has the crucial property that only finitely many linearly independent functions appear as terms (summands) in f(x) and its derivatives of all orders. To see how to amend our first guess. x 2 e 2x . and r = 3i. the trial solution takes the form Yp(x) =A cosx + B sinx + Cx cosx + Dx sinx + Ee2x + Fxe 2x + Gx 2 e 2x. (8) is (10) so substitution of (9) in the left-hand side of (8) would yield zero rather than (2x. F. we observe that by Eq. suppose that we want to find a particular solution of the differential equation (8) Proceeding as in Rule l. ' X The derivatives of the right-hand side in Eq. we get seven equations determining the seven coefficients A.154 Chapter 2 Linear Equations of Higher Order Solution The characteristic equation r 3 + 9r the complementary function is Yc(x) = 0 has roots r = 0. D. sinx. If y(x ) is any solution of Eq.3)erx . e 2x x cosx. our first guess would be (9) This form of Yp (x) will not be adequate because the complementary function of Eq. we see that y (x) is also a solution ofthe equation (D . and G. e 2x and Because there is no duplication with the terms of the complementary function. (8) and we apply the operator (D. Upon substituting Yp in Eq. + Cz cos 3x + c3 sin3x. (13) of Section 2. r = -3i. E.r) 2 to both sides. (7) and equating coefficients oflike terms. • The Case of Duplication Now we turn our attention to the situation in which Rule 1 does not apply: Some of the terms involved in f (x ) and its derivatives satisfy the associated homogeneous equation. C .3 . ' x sinx. The general solution of this homogeneous equation can be written as Thus every solution of our original equation in (8) is the sum of a complementary function and a particular solution of the form (11) .r ) 5 y = 0. So = c. For instance. B. (7) involve the terms cosx. The table in Fig. where f (x) is a linear combination of products of the elementary functions listed in (4 ). (12) it suffices to find separately particular solutions Y1 (x) and Y 2 (x) of the two equations Ly = ft (x) and Ly = h(x). it is common to have j(x ) =!I (x) + h(x) . (12). Thus f (x ) can be written as a sum of terms each of the form Pm(x)erx cos kx or Pm (x)erx sin kx. we observe that to find a particular solution of the nonhomogeneous linear differential equation L y = j. corresponding to the possibilities m = 0. (11) can be obtained by multiplying each term of our first guess in (9) by the least positive integral power of x (in this case.5 Nonhomogeneous Equations and Undetermined Coefficients 155 Note that the right-hand side in Eq. (14) where Pm(x ) is a polynomial in x of degree m. In practice we seldom need to deal with a function f (x) exhibiting the full generality in (14). RULE 2 Method of Undetermined Coefficients Ifthe function f(x) is of either form in (14). (15) by substituting Yp into the nonhomogeneous equation.5. The procedure by which we arrived earlier at the particular solution in (11) of Eq. (8) can be generalized to show that the following procedure is always successful. This procedure is illustrated in Examples 8 through 10. This procedure succeeds in the general case. (This is a type of "superposition principle" for nonhomogeneous linear equations. choosing s separately for each part to eliminate duplication with the complementary function. where !1 (x) and h (x) are different functions of the sort listed in the table in Fig. To simplify the general statement of Rule 2. For linearity then gives and therefore Yp = Y1 + Y 2 is a particular solution of Eq. Note that any derivative of such a term is of the same form but with both sines and cosines appearing. 2. r = 0. On the other hand.x + B zx 2 + · · · + Bmxm)erx sinkx] .5.(x) + h(x) . x + A zx 2 + · · · + Am x m)erx coskx + (Bo + B. (15) where s is the smallest nonnegative integer such that such that no term in Yp duplicates a term in the complementary function Yc · Then determine the coefficients in Eq. In this event we take as Yp the sum of the trial solutions for j 1(x ) and h(x). (13) respectively. . and k = 0.) Now our problem is to find a particular solution of the equation Ly = f(x) .1 lists the form of Yp in various common cases.2.1. take as the trial solution Yp(x ) = xs[(Ao + A. 2. x 3 ) that suffices to eliminate duplication between the terms of the resulting trial solution Yp(x) and the complementary function Yc(x) given in (10). The part Aex corresponding to 3ex does not duplicate any part of the complementary function, but the part B + Cx + Dx 2 must be multiplied by x 2 to eliminate duplication. Hence we take Yp = Aex + Bx 2 + Cx 3 + Dx 4 , To eliminate duplication with terms of Yc (x), the first part-corresponding to x 2e2x must be multiplied by x 3 , and the second part-corresponding to x sin 3x-must be multiplied by x. Hence we would take Variation of Parameters Finally, let us point out the kind of situation in which the method of undetermined coefficients cannot be used. Consider, for example, the equation y" +y = tan x, (17) which at first glance may appear similar to those considered in the preceding examples. Not so; the function f (x ) = tan x has infinitely many linearly independent derivatives sec2 x, 2 sec2 x tan x, 4 sec 2 x tan 2 x + 2 sec4 x, Therefore, we do not have available a finite linear combination to use as a trial solution. 158 Chapter 2 Linear Equations of Higher Order We discuss here the method of variation of parameters, which-in principle (that is, if the integrals that appear can be evaluated)-can always be used to find a particular solution of the nonhomogeneous linear differential equation y(n) + Pn-1 (x)yCn-l) +· · ·+PI (x)y + po(x)y = 1 f(x), (18) provided that we already know the general solution Yc = CJYI + CzYz + · · · + CnYn (19) of the associated homogeneous equation y(n) + Pn-1 (x)yCn-I) + · · · + Pl (x)y + po(x)y = 1 0. (20) Here, in brief, is the basic idea of the method of variation of parameters. Suppose that we replace the constants, or parameters, c 1, c2 , ..• , Cn in the complementary function in Eq. (19) with variables: functions u 1, u 2 , ... , un of x. We ask whether it is possible to choose these functions in such a way that the combination Yp(x) = Ut(X)YI (x) + uz(x)yz(x) + · · · + Un(X)Yn(x) (21) is a particular solution of the nonhomogeneous equation in ( 18). It turns out that this is always possible. The method is essentially the same for all orders n ~ 2, but we will describe it in detail only for the case n = 2. So we begin with the second-order nonhomogeneous equation L[y] = y" + P(x)y + Q(x )y = 1 f(x) (22) with complementary function Yc(x) = CtYI(x) + CzYz(x) (23) on some open interval I where the functions P and Q are continuous. We want to find functions u 1 and u 2 such that Yp(x) = u1 (x)yi (x) + uz(x)yz(x) (24) is a particular solution of Eq. (22). One condition on the two functions u 1 and u 2 is that L[yp] = j(x). Because two conditions are required to determine two functions, we are free to impose an additional condition of our choice. We will do so in a way that simplifies the computations as much as possible. But first, to impose the condition L[yp ] = f(x), we must compute the derivatives y~ and y~. The product rule gives Yp = I ( UJ YJ + UzYz I I ) + ( U1Y1 I + UzYz · I ) To avoid the appearance of the second derivatives u 1{ and u ~ , the additional condition that we now impose is that the second sum here must vanish: (25) 2.5 Nonhomogeneous Equations and Undetermined Coefficients 159 Then (26) and the product rule gives (27) But both Yt and Y2 satisfy the homogeneous equation y" + Py' + Qy = 0 associated with the nonhomogeneous equation in (22), so I Q Yi YiII = - p Yi- (28) for i = 1, 2. It therefore follows from Eq. (27) that In view of Eqs. (24) and (26), this means that hence (29) The requirement that Yp satisfy the nonhomogeneous equation in (22)-that is, that L[yp] = f(x)-therefore implies that (30) Finally, Eqs. (25) and (30) determine the functions u 1 and u2 that we need. Collecting these equations, we obtain a system (31) of two linear equations in the two derivatives u~ and u;. Note that the determinant of coefficients in (31) is simply the Wronskian W (y 1 , y2 ). Once we have solved the equations in (31) for the derivatives u'1 and we integrate each to obtain the functions u 1 and u 2 such that u;, (32) is the desired particular solution of Eq. (22). In Problem 63 we ask you to carry out this process explicitly and thereby verify the formula for Yp(x) in the following theorem. The complementary function is Yc (x) = c 1 cos x + c2 sin x, and we could simply substitute directly in Eq. (33). But it is more instructive to set up the equations in (31) and solve for u1 1 and u;, so we begin with Yt = cosx, y 1 = - smx, I • In Problems 58 through 62, a nonhomogeneous second-order linear equation and a complementary function Yc are given. Apply the method of Problem 57 to find a particular solution of the equation. 58. x 2 y" - 4xy' 42. Find the solution of the initial value problem consisting of the differential equation of Problem 41 and the initial conditions y(O) Forced Oscillations and Resonance In Section 2.4 we derived the differential equation mx" +ex'+ kx = F(t) (1) that governs the one-dimensional motion of a mass m that is attached to a spring (with constant k) and a dashpot (with constant c) and is also acted on by an external force F(t). Machines with rotating components commonly involve mass-spring systems (or their equivalents) in which the external force is simple harmonic: Equilibrium F(t) = F0 cos wt or F(t) = F0 sin wt, (2) FIGURE 2.6.1. The cartwith-flywheel system. where the constant F0 is the amplitude of the periodic force and w is its circular frequency. For an example of how a rotating machine component can provide a simple harmonic force, consider the cart with a rotating vertical flywheel shown in Fig. 2.6.1. The cart has mass m - m 0 , not including the flywheel of mass m 0 • The centroid of the flywheel is off center at a distance a from its center, and its angular speed is w radians per second. The cart is attached to a spring (with constant k) as shown. Assume that the centroid of the cart itself is directly beneath the center of the flywheel, and denote by x(t) its displacement from its equilibrium position (where the spring is unstretched). Figure 2.6.1 helps us to see that the displacement x of the centroid of the combined cart plus flywheel is given by _ X = (m - mo)x + mo(x +a cos wt) m moa = X + -COS Wt. m Let us ignore friction and apply Newton's second law mx" = - kx, because the force exerted by the spring is -kx. We substitute for x in the last equation to obtain mx" - m 0 aw 2 cos wt = -kx; that is, mx" + kx = m 0 aw2 cos wt. (3) Thus the cart with its rotating flywheel acts like a mass on a spring under the influence of a simple harmonic external force with amplitude F0 = m 0 aw2 . Such a system is a reasonable model of a front-loading washing machine with the clothes being washed loaded off center. This illustrates the practical importance of analyzing solutions of Eq. (1) with external forces as in (2). Undamped Forced Oscillations To study undamped oscillations under the influence of the external force F(t) F0 cos wt, we set c = 0 in Eq. (1), and thereby begin with the equation mx" + kx = Fo cos wt (4) whose complementary function is Xc = c 1 cos w 0 t + c2 sin w0 t. Here wo=/f 2.6 Forced Oscillations and Resonance 163 (as in Eq. (9) of Section 2.4) is the (circular) natural frequency of the mass-spring system. The fact that the angle w 0 t is measured in (dimensionless) radians reminds us that if t is measured in seconds (s), then w0 is measured in radians per secondthat is, in inverse seconds (s- 1). Also recall from Eq. (14) in Section 2.4 that division of a circular frequency w by the number 2n of radians in a cycle gives the corresponding (ordinary) frequency v = wj2n in Hz (hertz= cycles per second). Let us assume initially that the external and natural frequencies are unequal: w =!= w0 . We substitute Xp = A cos wt in Eq. (4) to find a particular solution. (No sine term is needed in Xp because there is no term involving x' on the left-hand side in Eq. (4).) This gives -mw2 A cos wt A rapid oscillation with a (comparatively) slowly varying periodic amplitude exhibits the phenomenon of beats. For example, if two horns not exactly attuned to one another simultaneously play their middle C, one at w0 j(2n) = 258 Hz and the other at wj(2n) = 254 Hz, then one hears a beat- an audible variation in the amplitude of the combined sound-with a frequency of (wo - w)/2 • FIGURE 2.6.3. The phenomenon of beats. 2JT 258 - 254 = 2 (Hz). 2 2.6 Forced Oscillations and Resonance 165 Resonance Looking at Eq. (6), we see that the amplitude A of Xp is large when the natural and external frequencies w0 and w are approximately equal. It is sometimes useful to rewrite Eq. (5) in the form A=------,k-mw2 Fo __h_o_/k_--=- _ ± _Ph_o 1 - (wfwo) 2 - k ' (11) where Folk is the static displacement of a spring with constant k due to a constant force Fo, and the amplification factor p is defined to be 1 p- ------.,.-- 11- (wfwo) 2 1. (12) It is clear that p ---+ +oo as w ---+ w0 . This is the phenomenon of resonancethe increase without bound (as w ---+ w 0 ) in the amplitude of oscillations of an undamped system with natural frequency w0 in response to an external force with frequency w ~ w 0 . We have been assuming that w i= w 0 . What sort of catastrophe should one expect if wand wo are.precisely equal? Then Eq. (4), upon division of each term by m, becomes x" 2 + w0 x =-cos w0 t. m Fo (13) 1.5 1.0 0.5 "< ffX = I I I .-' Because cos wot is a term of the complementary function, the method of undetermined coefficients calls for us to try .-' __ .-- t------- .-' Xp(t) = t(Acoswot + Bsinw0 t). Fo/(2mwo). 0.0 -0.5 -1.0 -1.5 fktl_/} ·x ....v.... ff- v l_1f_f - --~ I We substitute this in Eq. (13) and thereby find that A = 0 and B Hence the particular solution is ' ' ... ' ' --~ I x p(t) = - - t Fo x f- = I t sin SOt 2mwo smw0 t. . (14) o.oo 0.25 o.so ' 0.75 1.00 1.25 1.50 I I FIGURE 2.6.4. The phenomenon of resonance. The graph of xp(t) in Fig. 2.6.4 (in which m = 1, F0 = 100, and w 0 = 50) shows vividly how the amplitude of the oscillation theoretically would increase without bound in this case of pure resonance, w = w 0 . We may interpret this phenomenon as reinforcement of the natural vibrations of the system by externally impressed vibrations at the same frequency. Suppose that m = 5 kg and k = 500 Njm in the cart with the flywheel of Fig. 2.6.1. Then the natural frequency is w 0 = ,Jkliii = 10 radjs; that is, 10/ (2n) ~ 1.59 Hz. We would therefore expect oscillations of very large amplitude to occur if the flywheel revolves at about (1.59)(60) ~ 95 revolutions per minute (rpm). • In practice, a mechanical system with very little damping can be destroyed by resonance vibrations. A spectacular example can occur when a column of soldiers marches in step over a bridge. Any complicated structure such as a bridge has many natural frequencies of vibration. If the frequency of the soldiers' cadence is approximately equal to one of the natural frequencies of the structure, then-just as in our simple example of a mass on a spring- resonance will occur. Indeed, the resulting Example 3 166 Chapter 2 Linear Equations of Higher Order resonance vibrations can be of such large amplitude that the bridge will collapse. This has actually happened-for example, the collapse of Broughton Bridge near Manchester, England, in 1831-and it is the reason for the now-standard practice of breaking cadence when crossing a bridge. Resonance may have been involved in the 1981 Kansas City disaster in which a hotel balcony (called a skywalk) collapsed with dancers on it. The collapse of a building in an earthquake is sometimes due to resonance vibrations caused by the ground oscillating at one of the natural frequencies of the structure; this happened to many buildings in the Mexico City earthquake of September 19, 1985. On occasion an airplane has crashed because of resonant wing oscillations caused by vibrations of the engines. It is reported that for some of the first commercial jet aircraft, the natural frequency of the vertical vibrations of the airplane during turbulence was almost exactly that of the massspring system consisting of the pilot's head (mass) and spine (spring). Resonance occurred, causing pilots to have difficulty in reading the instruments. Large modem commercial jets have different natural frequencies, so that this resonance problem no longer occurs. Modeling Mechanical Systems The avoidance of destructive resonance vibrations is an ever-present consideration in the design of mechanical structures and systems of all types. Often the most important step in determining the natural frequency of vibration of a system is the formulation of its differential equation. In addition to Newton's law F = rna, the principle of conservation of energy is sometimes useful for this purpose (as in the derivation of the pendulum equation in Section 2.4). The following kinetic and potential energy formulas are often useful. 1. Kinetic energy: T = ~mv 2 for translation of a mass m with velocity v; 2. Kinetic energy: T = ~I ui for rotation of a body of a moment of inertia I with angular velocity w; 3. Potential energy: V = 4kx 2 for a spring with constant k stretched or compressed a distance x; 4. Potential energy: V = mgh for the gravitational potential energy of a mass m at height h above the reference level (the level at which V = 0), provided that g may be regarded as essentially constant. Example 4 Find the natural frequency of a mass m on a spring with constant k if, instead of sliding without friction, it is a uniform disk of radius a that rolls without slipping, as shown in Fig 2.6.5. With the preceding notation, the principle of conservation of energy gives Solution Equilibrium _2lmv2 + _ 2 1 I-.,2 ""' + _21kx2 = E where E is a constant (the total mechanical energy of the system). We note that v = aw and recall that I = ma 2j2 for a uniform circular disk. Then we may simplify the last equation to ~mv 2 + 4kx 2 =E. x = o Because the right-hand side of this equation is constant, differentiation with respect to t (with v = X 1 and V 1 = x ") now gives 3 I II k I 0 2mx x + xx = . FI Gu RE 26 . .5. The rolling disk. 2.6 Forced Oscillations and Resonance 16 7 We divide each term by ~mx' to obtain II 2k x +-x =0. 3m Thus the natural frequency of horizontal back-and-forth oscillation of our rolling disk is J2kj3m, which is ~ ~ 0.8165 times the familiar natural frequency .Jk7iii of a mass on a spring that is sliding without friction rather than rolling without sliding. It is interesting (and perhaps surprising) that this natural frequency does not depend on the radius of the disk. It could be either a dime or a large disk with a radius of one meter (but of the same mass). • Example 5 Suppose that a car oscillates vertically as if it were a mass m = 800 kg on a single spring (with constant k = 7 x 104 Njm), attached to a single dashpot (with constant c = 3000 N·s/m). Suppose that this car with the dashpot disconnected is driven along a washboard road surface with an amplitude of 5 em and a wavelength of L = 10 m (Fig. 2.6.6). At what car speed will resonance vibrations occur? m 2ns y - a cos L s=O position FIGURE 2.6.6. The washboard road surface of Example 5. FIGURE 2.6.7. The "unicycle model" of a car. Solution We think of the car as a unicycle, as pictured in Fig. 2.6.7. Let x(t) denote the upward displacement of the mass m from its equilibrium position; we ignore the force of gravity, because it merely displaces the equilibrium position as in Problem 9 of Section 2.4. We write the equation of the road surface as y = a cos L 2ns (a = 0.05 m, L = 10m). (15) When the car is in motion, the spring is stretched by the amount x - y , so Newton's second law, F = ma, gives mx" = - k(x - y); that is, mx" + kx = ky (16) Ifthe velocity of the car is v, then s = vt in Eq. (15), so Eq. (16) takes the form mx " + kx = 2nvt ka cos£· (16') 168 Chapter 2 Linear Equations of Higher Order This is the differential equation that governs the vertical oscillations of the car. In comparing it with Eq. (4), we see that we have forced oscillations with circular frequency w = 2nvjL. Resonance will occur when w = w0 = ,Jkliii. We use our numerical data to find the speed of the car at resonance: v =- 2n Lff - = -10~x104 m 2n 800 ~ 14.89 (m/s); • that is, about 33.3 mijh (using the conversion factor of 2.237 mi/h per mjs). Damped Forced Oscillations In real physical systems there is always some damping, from frictional effects if nothing else. The complementary function Xc of the equation mx" + ex' + kx = Fo cos wt (17) is given by Eq. (19), (20), or (21) of Section 2.4, depending on whether c > Ccr = ,J4kfli., c = Ccr. or c < Ccr· The specific form is not important here. What is important is that, in any case, these formulas show that xc(t) --+ 0 as t --+ +oo. Thus Xc is a transient solution of Eq. (17)-one that dies out with the passage of time, leaving only the particular solution xp. The method of undetermined coefficients indicates that we should substitute x(t) = A cos wt + B sin wt in Eq. (17). When we do this, collect terms, and equate coefficients of cos wt and sin wt, we obtain the two equations -cwA + (k- mw2 )B = 0 as usual, we see that the resulting steady periodic oscillation Xp(t) = C cos(wt - a) (20) has amplitude (21) Now (19) implies that sin a = BjC > 0, so it follows that the phase angle a lies in the first or second quadrant. Thus tan a ==kA B cw mw2 with 0 <a < n , (22) 2.6 Forced Oscillations and Resonance 169 so tan- 1 a = { ;r cw k- mw 2 cw + tan- 1 k- mw2 ifk > mw2 , if k < mw2 (whereas a = rr /2 if k = mw2 ). Note that if c > 0, then the "forced amplitude"-defined as a function C(w) by (21)-always remains finite, in contrast with the case of resonance in the undamped case when the forcing frequency w equals the critical frequency wo = ,Jkfm. But the forced amplitude may attain a maximum for some value of w, in which case we speak of practical resonance. To see if and when practical resonance occurs, we need only graph C as a function of w and look for a global maximum. It can be shown (Problem 27) that C is a steadily decreasing function of w if c ~ v'?JWi. But if c < v'?JWi, then the amplitude of C attains a maximum value-and so practical resonance occurs-at some value of w less than wo, and then approaches zero as w ~ +oo. It follows that an underdamped system typically will undergo forced oscillations whose amplitude is • Large if w is close to the critical resonance frequency; • Close to F0 jk if w is very small; • Very small if w is very large. Example 6 Find the transient motion and steady periodic oscillations of a damped mass-andspring system with m = 1, c = 2, and k = 26 under the influence of an external force F (t) = 82 cos 4t with x (0) = 6 and x' (0) = 0. Also investigate the possibility of practical resonance for this system. The resulting motion x(t) problem Solution = Xtr(t) + Xsp(t) of the mass satisfies the initial value x " + 2x' + 26x = 82cos4t; x(O) = 6, x'(O) = 0. (23) Instead of applying the general formulas derived earlier in this section, it is better in a concrete problem to work it directly. The roots of the characteristic equation r 2 + 2r + 26 = (r + 1)2 + 25 =0 are r = -1 ± 5i , so the complementary function is When we substitute the trial solution x (t) = A cos 4t + B sin4t in the given equation, collect like terms, and equate coefficients of cos 4t and sin 4t, we get the equations lOA+ 8B = 82, - 8A + lOB = 0 170 Chapter 2 Linear Equations of Higher Order with solution A = 5, B = 4. Hence the general solution of the equation in (23) is x(t) = e-t (c 1 cos 5t + c2 sin 5t) + 5 cos 4t + 4 sin 4t. At this point we impose the initial conditions x (0) = 6, x' (0) = 0 and find that c1 = 1 and c 2 = -3. Therefore, the transient motion and the steady periodic oscillation of the mass are given by X1r(t) = e-t (cos 5t - 3 sin 5t) for the different values x 0 = -20, -10, 0, 10, and 20 of the initial position. Here we see clearly what it means for the transient solution x 1r(t) to "die out with the passage of time," leaving only the steady periodic motion Xsp(t). Indeed, because X~r(t) --+ 0 exponentially, within a very few cycles the full solution x(t) and the steady periodic solution Xsp(t) are virtually indistinguishable (whatever the initial position xo). X To investigate the possibility of practical resonance in the given system, we substitute the values m = 1, c = 2, and k = 26 in (21) and find that the forced amplitude at frequency w is C(w) = 82 .J676- 48w2 + w4 . 5 4 3 2 I The graph of C(w) is shown in Fig. 2.6.9. The maximum amplitude occurs when 5 Problems Each of Problems 15 through 18 gives the parameters for a forced mass- spring-dashpot system with equation mx" +ex'+ kx = Fo cos wt. Investigate the possibility of practical resonance of this system. In particular, find the amplitude C(w) of steady periodic forced oscillations with frequency w. Sketch the graph of C (w) and find the practical resonance frequency w (if any). In Problems 1 through 6, express the solution of the given initial value problem as a sum of two oscillations as in Eq. (8 ). Throughout, primes denote derivatives with respect to time t. In Problems 1-4, graph the solution function x(t ) in such a way that you can identify and label (as in Fig. 2.6.2) its period. In each of Problems 11 through 14, find and plot both the steady periodic solution Xsp(t) = C cos(wt -a) of the given differential equation and the transient solution x1r(t) that satisfies the given initial conditions. 11. 12. 13. 14. Show that there are two values of w for which resonance occurs..172 Chapter 2 Linear Equations of Higher Order In particular.a).6..10. k )(k.:. (21). The first floor is attached rigidly to the ground.) 27. Assume that in an earthquake the ground oscillates horizontally with amplitude Ao and circular frequency w. show that C attains a maximum value (practical resonance) when w = Wm = J~ . except with sin(wt -a) in place of cos(wt.6. 2. The pendulumand-spring system of Problem 21. ( 17)) is given by (b) Suppose that e 2 < 2mk. it requires a horizontal force of 5 tons to displace the second floor a distance of 1 ft. The mass. a.Z 0 2 FIGURE 2. as shown in Fig. A building consists of two floors.000 lb).Jkfiii. 22. and g. Find the natural circular frequency of the system in terms of m. FIGURE 2. (Suggestion: Add the steady periodic solutions separately corresponding to E 0 cos wt and F0 sin wt (see Problem 25). (a) Show that the amplitude of the steady periodic solution of the differential equation mx" +ex' + kx = mAu} cos wt (with a forcing term similar to that in Eq. A mass m hangs on the end of a cord around a pulley of radius a and moment of inertia I.2~2 < Wo = [f. The elastic frame of the building behaves as a spring that resists horizontal displacements of the second floor. (a) What is the natural frequency (in hertz) of oscillations of the second floor? (b) If the ground undergoes one oscillation every 2. k. Thus the resonance frequency in this case is larger (in contrast with the result of Problem 27) than the natural frequency w0 = ..J2. ·~ ·~ ·~ ' k -with both cosine and sine forcing terms-derive the steady periodic solution Xsp(t) = j E0 + F. .4iWi. According to Eq.. Given the differential equation L mx" + ex' + kx = Eo cos wt + Fo sin wt . and find both..25 s with an amplitude of 3 in.{3). what is the amplitude of the resulting forced oscillations of the second floor? 24.'. Assume small oscillations so that the spring remains essentially horizontal and neglect friction. the amplitude of forced steady periodic oscillations for the system mx" + ex' + kx = F0 cos wt is given by (a) If e . .11. show that it is what one would expect-the same as the formula in (20) with the same values of C and w. 23. where a is defined in Eq. ecr/. and the second floor is of mass m = 1000 slugs (fps units) and weighs 16 tons (32.) . Derive the steady periodic solution of mx" +ex'+ kx 28. (Suggestion: Maximize the square of C. Show that the maximum amplitude occurs at the frequency Wm given by Wm = k ( 2mk ) m 2mk.. 25.springpulley system of Problem 22.a . I. 26. resulting in an external horizontal force F(t) = mA 0 w 2 sin wt on the second floor. A mass on a spring without damping is acted on by the external force F (t) = F0 cos 3 wt. The rim of the pulley is attached to a spring (with constant k). As indicated by the cart-with-flywheel example discussed in this section.e 2 • = Fo sinwt.1(F0 jE0 ). an unbalanced rotating machine part typically results in a force having amplitude proportional to the square of the frequency w.11..6. (22) and {3 = tan.mw 2 ) 2 + (cw) 2 cos(wt. where ecr = steadily decreases as w increases. show that C (b) If e < ecr/h. this gives C as a function of v.2.7.7. Maximum resonance vibrations with amplitude about 14 em occur around 32 mij h.1.7.1 satisfy the basic circuit equation Ldt FIGURE 2. As a consequence.1. dt (1) R FIGURE 2. but then subside to more tolerable levels at high speeds. (2) . Verify these graphically based conclusions by analyzing the function C(w). • • •. and A capacitor with a capacitance of C farads in series with a source of electromotive force (such as a battery or a generator) that supplies a voltage of E (t) volts at time t. it initially oscillates with 3 Velocity (mi/h) FIGURE 2. The series RLC circuit..7 Electrical Circuits 173 Automobile Vibrations Problems 29 and 30 deal further with the car of Example 5.6. . 7.1 is closed.. As shown in Fig. dl + RI + -1 Q = c E (t) . Table of voltage drops. An inductor with an inductance of L henries.1 with the aid of this table and one of Kirchhoff's laws: The (algebraic) sum of the voltage drops across the elements in a simple loop of an electrical circuit is equal to the applied voltage.E -o 11) = cwa and Fo = ka.2. 2.7. as the car accelerates gradually from rest. 15 12 + kx = Eo cos wt + Fo sin wt 8 . Figure 2. the voltage drops across the three circuit elements are those shown in the table in Fig. it consists of L A resistor with a resistance of R ohms. 7. this differential equation becomes mx" + ex' where Eo amplitude slightly over 5 em. 2. in which time is measured in seconds. 6 E <t: Because w = 2nvjL when the car is moving with velocity v. find the practical resonance frequency and the corresponding amplitude. The relation between the functions Q and I is dQ =l(t). 2.2. CIr(!Uit.12 shows the graph of the amplitude function C(w) using the numerical data given in Example 5 (including c = 3000 N·s/m). . With y = a sin wt for the road surface.12. Here we examine the RLC circuit that is a basic building block in more complicated electrical circuits and networks. Amplitude of vibrations of the car on a washboard surface. Element · Inductor Resistor Capacitor L1 d! dt Rl CQ We will always use rnks electric units. the current and charge in the simple RLC circuit of Fig. 2.7. 30. It indicates that. this results in a current of /(t) amperes in the circuit and a charge of Q(t) coulombs on the capacitor at time t . ~ 29. Its upward displacement function satisfies the equation mx" + ex' + kx = cy' + ky when the shock absorber is connected (so that c > 0 ).6. According to elementary principles of electricity. In particular. If the switch shown in the circuit of Fig. 2. Apply the result of Problem 26 to show that the amplitude C of the resulting steady periodic oscillation for the car is given by 9 0. We can analyze the behavior of the series circuit of Fig. All we need to know about these constants and functions is that they satisfy Eqs.6 for mechanical systems can be applied at once to electrical circuits. Mechanical. When the switch is open. (3) and substitute I for Q' to obtain LI" + RI' +-I= E ' (t). This idea is the basis of analog computers--electrical models of mechanical systems. or even dangerous. The table in Fig. More concretely. The Mechanical-Electrical Analogy It is striking that Eqs. under the assumption that the voltage E(t) is known. there is a current I (t) in the circuit and a charge Q(t) on the capacitor. our mathematical model for the RLC circuit. 2 . no current flows in the circuit.174 Chapter 2 Linear Equations of Higher Order If we substitute (1) in Eq. We can then learn a good deal about electricity by studying this mathematical model. and capacitor in an electrical circuit as "black boxes" that are calibrated by the constants R.3 details this important mechanical-electrical analogy. when the switch is closed. Massm Damping constant c Spring constant k Position x Force F Inductance L Resistance R Reciprocal capacitance 1I C Charge Q (using (3) (or current I using (4))) Electromotive force E (or its derivative E' ) FIGURE 2. In most practical problems it is the current I rather than the charge Q that is of primary interest. (1) through (4). we get the second-order linear differential equation LQ" + RQ' + -Q = c 1 E(t) (3) for the charge Q(t). It suffices to regard the resistor. 2. .7. A battery or generator is described by the voltage E(t) that it supplies.7. The performance of the mechanical system can then be predicted by means of accurate but simple measurements in the electrical model. most of the results derived in Section 2. 1 c (4) We do not assume here a prior familiarity with electrical circuits. inaccurate.electrical analogies.3 can be used to construct an electrical model of a given mechanical system. so we differentiate both sides of Eq. L. This is especially useful when the actual mechanical system would be expensive to construct or when measurements of displacements and velocities would be inconvenient. The fact that the same differential equation serves as a mathematical model for such different physical systems is a powerful illustration of the unifying role of mathematics in the investigation of natural phenomena. Analog computers modeled the first nuclear reactors for commercial power and submarine propulsion before the reactors themselves were built. As a consequence. the correspondences in Fig. inductor.7. and C. using inexpensive and readily available circuit elements.3. (2). (3) and (4) have precisely the same form as the equation mx" +ex'+ kx = F(t) (5) of a mass-spring-dashpot system with external force F(t). 7 Electrical Circuits 175 In the typical case of an alternating current voltage E(t) =Eo sinwt.LCw o ::s. k-mw 2 0 ~a~ rr. and a steady periodic current / 5p. Then the steady periodic current /5 p(t) = Eo z cos(wt. Eq. (5) with F(t) = Fo cos wt is X 5 p(t) = Fo cos(wt. z= J R 2 + ( wL- ~C J (ohms).a) j(k. 1/ C fork. If we make the substitutions L form.a) (11) has amplitude lo= reminiscent of Ohm's law. thus (7) Recall from Section 2. so the roots of the characteristic equation have negative real parts). (19) through (22) there) that the steady periodic solution of Eq. the solution of Eq. a ::. R for c.2.6 (Eqs. we get the steady periodic current (8) with the phase angle a=tan _1 wRC 2' 1. JT. (6) As in a mass-spring-dashpot system with a simple harmonic external force. (6) is the sum of a transient current 11r that approaches zero as t --* +oo (under the assumption that the coefficients in Eq. I = EI R. (9) Reactance and Impedance The quantity in the denominator in (8).mw2 ) 2 + (cw)l . (6) are all positive. (4) takes the form 1 LI" + RI' + CI = wE0 coswt. where a= tan -1 cw . z' Eo (12) . (10) is called the impedance of the circuit. and wE0 for F0 . 5). and voltage. Eo . and C 5 x 10-4 farad (F). The differential equation in (6) takes the form (O. Isp(t) = .sm(wt .7. Example 1 Consider an RLC circuit with R = 50 ohms (Q).J R 2 + S2 . To do so. (14) z where S LCw2 . Find the current in the circuit and the time lag of the steady periodic current behind the voltage. Initial Value Problems FIGURE 2. When we want to find the transient current. (9) that a is as in Fig.7.= tan. approximately 377 radjs.4.1 8 = tan. Solution . Reactance and delay angle. So we must first find I' (0). Time lag of current behind imposed voltage..sin 8 cos wt).7. So we take E(t) = 110 sin 377t and use equality in place of the symbol for approximate equality in this discussion.1 henry (H). we first introduce the reactance 1 S = wL. To convert Isp to a sine function..5.176 Chapter 2 Linear Equations of Higher Order Equation (11) gives the steady periodic current as a cosine function. wC -S (13) Then Z = .8). the circuit is connected to a 110-V.1 . charge. 2. A frequency of60 Hz means thatw = (2n)(60) radjs.1 . 2. Equation (11) now yields Isp(t) = ~0 (cos a cos wt +sin a sin wt) S cos wt + Z R sin wt ) Z R Eo ( =Z - = Eo (cos 8 sin wt . with delay angle 8 =a -1n. when both I (0) and Q(O) are zero.7. At time t = 0. 60-Hz alternating current generator.l)I" + 501' + 2000I = (377)(110) cos 377t. z Therefore. whereas the input voltage E(t) = Eo sin wt was a sine function. (2) to obtain the equation LI'(O) + RI (0) + C Q(O) = E(O) 1 (16) to determine I' (0) in terms of the initial values of current. and we see from Eq.4. L = 0. FIGURE 2.-. we substitute t = 0 in Eq.R wRC (15) This finally gives the time lag 8jw (in seconds) of the steady periodic current Isp behind the input voltage (Fig. we are usually given the initial values I (0) and Q(O). Therefore. L. In Section 2.6. • Electrical Resonance Consider again the current differential equation in (6) corresponding to a sinusoidal input voltage E (t) = Eo sin cvt. Suppose that we wanted to pick up a particular radio station that is broadcasting at frequency cv. The radio is a familiar example.1)I" + 50I' + 2000I = E ' (t) = 0. so Eq. but its capacitance C is varied as one operates the tuning dial. I (t) = (2. L 0.670. For typical values of the constants R. . and thereby (in effect) provides an input voltage E(t) = Eo sin cvt to the tuning circuit of the radio. Its inductance L and resistance R are constant.1 and the differential equation is (0. the graph of Io as a function of w resembles the one shown in Fig.JLC and then approaches zero as cv ---+ +oo. I'(O) = -44c 1 .670)(e-44r e-456t) . Note that I (t) ---+ 0 as t ---+ +oo even though the voltage is constant.178 Chapter 2 Linear Equations of Higher Order Solution We now have E(t) =110.456c2 = 1100 for c1 = -c2 = 2. By contrast.7. many common electrical devices could not function properly without taking advantage of the phenomenon of resonance.6. It reaches a maximum value at cvm = 1/. The resulting steady periodic current Isp in the tuning circuit drives its amplifier. with the volume of sound we hear roughly proportional to the amplitude I 0 of Isp · To hear our preferred station (of frequency cv) the loudest-and simultaneously tune out stations broadcasting at other frequencies-we therefore want to choose C to maximize Io. Its general solution is the complementary function we found in Example 1: We solve the equations I (0) = C) + C2 = 0.6 we emphasized the importance of avoiding resonance in most mechanical systems (the cello is an example of a mechanical system in which resonance is sought). 2. (16) gives I (0) = 1 £(0) 110 = = 1100 (A/s).7. The effect of frequency on lo. A highly simplified model of its tuning circuit is the RLC circuit we have discussed. and in tum its loudspeaker. We have seen that the amplitude of its steady periodic current is (17) FIGURE 2. and E0 . the critical frequency Wm is the resonance frequency of the circuit. C. so that a steady current of 4 A is flowing in the circuit.7. (b) Write the solution in the form l sp(t) = C cos(wt .Q= c 1 E(t) for the charge Q = Q(t) on the capacitor at timet. and I (0) = 0. 3. the switch is thrown to position 2. Note that I(t) = Q'(t). when c= 1 --2. In the circuit of Fig. Given the same circuit as in Problem 1. Problems 7 through 10 deal with the RC circuit in Fig.7. Suppose that the battery in Problem 2 is replaced with an alternating-current generator that supplies a voltage of E (t) = 100 cos 60t volts. wC 1 that is. suppose that the switch is initially in position 2. the second controls the frequency of a signal that the radio itself generates. but modem AM radios have a more sophisticated design.8. This is the way that old crystal radios worked. 2.=0. a series circuit containing an inductor with an inductance of L henries. . known as the intermediate frequency. Lw (18) So we merely turn the dial to set the capacitance to this value.. Find I (t) and show that I (t) --+ 4 as t --+ +oo. a resistor with a resistance of R ohms. Find the maximum current in the circuit for t ~ 0. so that I(O) = 0 and E = 100 fort ~ 0. a capacitor (C farads). 2. but is thrown to position 1 at timet = 0. 2. and I (0) = 0. take L = 1. R = 20. (a) Substitute l sp(t) = A cos 60t + B sin 60t and then determine A and B to find the steady-state current l sp in the circuit. 1. containing a resistor ( R ohms).2. R = 40. with the switch in position 1. E(t) = WOe .7. This technique has the advantage that the several RLC circuits used in the amplification stages easily can be designed to resonate at 455 kHz and reject other frequencies. We see at a glance-no calculus required-that / 0 is maximal when w L .a). 2. Suppose also that the switch has been in position 1 for a long time. With everything else the same. suppose that E(t) = WOe . A pair of variable capacitors are used.7. now find I (t). (2) reduces to the linear first-order equation LI' + RI = E(t). resulting in far more selectivity of the receiver as well as better amplification of the desired signal. but no inductor.7. In the circuit of Fig. and the source E of emf is a battery supplying 100 V to the circuit. 2. In the circuit of Fig. The circuit for Problems I through 6. 6. with the switch in position 1. thinking of w as a constant with C the only variable.7. 7.101 cos60t.7. is then amplified in several stages. 4. The resulting beat frequency of 455 kHz. and a source of electromotive force (emf). 2. Find I (t). and E(t) = 30cos60t + 40sin60t. ( 3) gives the linear first-order differential equation dQ dt R- +.7. but no capacitor. L R FIGURE 2. 7. L = 2. R = 10. The first controls the frequency selected as described earlier. 2. a switch. 5. suppose that L = 2. kept close to 455 kilohertz (kHz) above the desired frequency. In this case Eq. so that I (0) = 4 and E = 0 for t ~ 0.101 .7.7.7. Problems 1 through 6 deal with the RL circuit of Fig. In the circuit of Fig. with the switch in position 1.7 Electrical Circuits 179 But examine Eq. R = 25 Q . a source of emf. At time t = 0. Find I (t). suppose that L = 5 H. Substitution of L = 0 in Eq. (17). 2.2. y(a) = 0.J3 with B =I= 0 hits the desired target value y = 0 when x = n.2. Thus the endpoint value problem in (4) has infinitely many different nontrivial solutions. so the condition y(O) = 0 implies that A = 0. ·''' """' ""~"'"'''''•"'~""'. .2. Examples 1 and 2 illustrate the sorts of complications that can arise in endpoint problems. Now y(O) = A. • Jr.8 Endpoint Problems and Eigenvalues 181 In this section we will see that the situation is radically different for a problem such as y" + p(x)y' + q(x)y = 0.1 illustrates the fact that no possible solution y (x) = B sin x . In (2) we are to find a solution of the differential equation on the interval (a. so the condition y(O) = 0 implies that A = 0. Various possible solutions y(x) = B sinxv'3 ofthe endpoint value problem in Example 1. Example 1 Consider the endpoint problem y" + 3y = 0. No matter what the coefficient B is. Perhaps this does seem a bit surprising.74588. y(b) = 0. Graphically.8. But then FIGURE 2. y(rr) = 0. b) that satisfies the conditions y(a) = 0 and y(b) = 0 at the endpoints of the interval. every possible solution y(x) = B sin 2x hits automatically the desired target value y = 0 when x = n (whatever the value of B). Therefore the only possible solutions are of the form y(x) = B sin 2x. y(O) =A. the solution automatically hits the target value y = 0 for X = Again. Various possible solutions y (x) = B sin 2x of the endoint value problem in Example 2. 2. Thus the only solution of the endpoint value problem in (3) is the trivial solution y(x) = 0 (which probably is no surprise). Therefore the only possible solutions are of the form y (x) = B sin x . so the other condition y(n) = 0 requires that B = 0 also. FIGURE 2.J3. But now y(n) = B sin 2n = 0 no matter what the value of the coefficient B is. y(n) = B sin nJ3 ~ -0.8. with the coefficient 3 in one replaced by the coefficient 4 in the other. We consider a possible solution which starts at the left endpoint value and ask whether it hits the "target" specified by the right endpoint value. In mathematics as elsewhere. Remark 1: Note that the big difference in the results of Examples 1 and 2 stems from the seemingly small difference between the differential equations in (3) and (4 )."''"'"' • " "' Example 2 Consider the endpoint problem y" + 4y = 0.8. Such a problem is called an endpoint or boundary value problem. (4) The general solution of the differential equation is y(x) =A cos 2x 1t X + B sin 2x. (3) The general solution of the differential equation is y(x) =A cosxJ3 + B sinxJ3.1. y(n) = 0. as illustrated graphically in Fig. y(O) = 0." Remark 2: The "shooting" terminology used in Examples 1 and 2 is often useful in discussing endpoint value problems.8. Fig. For no B =I= 0 does the solution hit the target value y = 0 for x = rr. y(O) = 0. (2) The difference between the problems in Eqs. (1) and (2) is that in (2) the two conditions are imposed at two different points a and b with (say) a < b. Hence. sometimes "big doors tum on small hinges. One might think of such a value as a "proper" value of A for which there exist proper (nonzero) solutions of the problem. (5) with p(x) = 0. Thus we saw in Example 2 that A = 4 is an eigenvalue of the endpoint problem y" + Ay = 0. so eigenvalues are sometimes called proper values (or characteristic values). If we take A = 3. y(O) = 0. (7) Solution We must consider all possible (real) values of A-positive. note that the problem in (5) is homogeneous in the sense that any constant multiple of an eigenfunction is again an eigenfunction-indeed. we get the equations in (3). That is. the prefix eigen is a German word that (in some contexts) may be translated as the English word proper. The number A is a parameter in the problem (nothing to do with the parameters that were varied in Section 2. Suppose that A* is an eigenvalue of the problem in (5) and that y(x) is a nontrivial solution of the endpoint problem that results when the parameter A in (5) is replaced by the specific numerical value A. q(x) = 1. Indeed. If A = 0. Thus we saw in Example 2 that y* (x) = sin 2x is an eigenfunction associated with the eigenvalue A* = 4.e. a = 0. so Then we call y* an eigenfunction associated with the eigenvalue A. y(O) = 0. and negative. Examples 1 and 2 illustrate the typical situation for an endpoint problem as in (2): It may have no nontrivial solutions. and b = JT. More generally. then the equation is simply y" = 0 and its general solution is y (x) = Ax + B. as is any constant multiple of sin 2x. one associated with the same eigenvalue. y(rr) = 0. y(b) = 0.. nonzero) solution of the endpoint value problem? Such a value of A is called an eigenvalue of the problem. y(a) = 0. (6) whereas Example 1 shows that A = 3 is not an eigenvalue of this problem. Note that the problems in (3) and (4) both can be written in the form y" + p(x)y' + Aq(x)y = 0.182 Chapter 2 Linear Equations of Higher Order Eigenvalue Problems Rather than being the exceptional cases. Examples 1 and 2 show that the situation in an endpoint problem containing a parameter can (and generally will) depend strongly on the specific numerical value of the parameter. or it may have infinitely many nontrivial solutions. The question we ask in an eigenvalue problem is this: For what values of the parameter A does there exist a nontrivial (i. with A = 4.5). then so does any constant multiple cy(x). y(L) =0 (L > 0). if y = y(x) satisfies the problem in (5) with A = A. An endpoint value problem such as the problem in (5)-one that contains an unspecified parameter A-is called an eigenvalue problem. we obtain the equations in (4). Example 3 Determine the eigenvalues and associated eigenfunctions for the endpoint problem y" + Ay = 0. zero. It can be proved (under mild restrictions on the coefficient functions p and q) that any two eigenfunctions associated with the same eigenvalue must be linearly dependent. . 8 Endpoint Problems and Eigenvalues 183 Then the endpoint conditions y(O) = 0 = y(L) immediately imply that A = B = 0. (Recall that cosh ax = (eax that sinh ax = (eax . ' nn.2. nnx L n = 1. • . 9n 2 L2.3. f . If A < 0. This implies that B = 0. if n2 A =a 2 = L 2' y 4n 2 L2 . and its general solution is where A = c 1 + c2 and B = c 1 . (9) FIGURE 2. The hyperbolic sine and cosine graphs.. B sin a x .4 shows graphs of the first several of these eigenfunctions. and we may therefore conclude that the problem has no negative eigenvalues.c2 . . let us then write A = -a 2 (with a > 0) to be specific. •• 0' n2n2 L2 ' Thus we have discovered that the problem in (7) has an infinite sequence of positive eigenvalues (8) L X With B = 1.L forn = 1. 0? Yes. A= 0 is not an eigenvalue of the problem in (7). so y(x) y(L) = 0 then gives y(L) = B sinaL= 0. but only provided that aL is a (positive) integral multiple of n: aL =n. Then the differential equation takes the form = II y . The condition y(O) = 0 implies that A = 0.a2 y =0 . gives y(L) = B sinhaL = 0. the eigenfunction associated with the eigenvalue An is Yn(X) = sin . and sinhx = 0 only for x = 0 (examine the graphs of y = sinhx andy = coshx in Fig. 3. 2n..8.8. The only remaining possibility is that A = a 2 > 0 with a > 0. y(L) = 0. FIGURE 2. Figure 2. But now the second endpoint condition..8. + e-ax)/2 and X so that y(x) = B sinh ax. 2. because a f. 3. Therefore.3..) Thus the only solution of the problem in (7) in the case A < 0 is the trivial solution y 0. . nnx etgen unctwns Yn (x ) = sin .8. 0. The . 3n. The condition Can this occur if B f.) The condition y(O) = 0 then gives y(O) =A coshO + B sinhO =A= 0. We see visibly how the endpoint conditions y(O) = y(L) = 0 serve to select just those sine functions that start a period at x = 0 and wind up at x = L precisely at the end of a half-period..e-ax)/2. .4. 2. 4. 2. •• 0 ' that is. In this case the differential equation is = with general solution y(x) =A cos ax+ B sin ax. so the only solution in this case is the trivial function y(x) 0.. a1y(a) + a2y'(a) = 0..8. The second endpoint condition y' (L) = 0 now gives y' (L) = BacosaL = 0.1)2rr 2 9rr 2 A=. The eigenfunctions . y(O) = 0. we get the problem of Example 4 (in which p(x) 0 and q(x) 1. With a 1 = 1 = b2 and a2 = 0 = b1. Then the differential equation is with general solution y(x ) = A cos ax+ B sin ax. in which the endpoint conditions involve values of the derivative y' as well as values of y: + p(x)y' + Aq(x)y = 0. (2n..8. so y(x) = B sin ax and y'(x) = Ba cos ax. so we take A = a 2 > 0 (a > 0) to be specific. b1y(b) + b2y'(b) = y" 0. According to a theorem whose precise statement we will defer until Section 9.. under the assumption that q (x) > 0 on the interval [a. y' (L) = 0. as in the previous example).. • .l)rrx Yn(x ) = sm ----'---2L for n = 1. 3. (11) Solution Virtually the same argument as that used in Example 3 shows that the only possible eigenvalues are positive. = = Example 4 Determine the eigenvalues and eigenfunctions of the problem y" + AY = 0.1)rr 7r 3rr aL=. (10) where a1..5. ' 2' 2 ' 2 that is. y This will hold with B =1= 0 provided that a L is an odd positive integral multiple of rrj2: (2n .. 2. We see visibly how the endpoint conditions y(O) = y' (L) = 0 serve to select just those sine functions that start a period at x = 0 but wind up at x = L precisely in the middle of a half-period. This is also true of the following more general type of eigenvalue problem. b 1. 4. .l)rr x 2L (1 2) FIGURE 2.. if rr 2 (2n .. for n = 1. any eigenvalue of the form in (5) has a divergent increasing sequence )q < A2 < A3 < · · · < An < · · · ---+ +oo of eigenvalues. b ].5 shows graphs of the first several of these eigenfunctions. a2.184 Chapter 2 Linear Equations of Higher Order Example 3 illustrates the general situation. Figure 2. The condition y(O) = 0 immediately gives A = 0. each with an associated eigenfunction. ' . and b2 are given constants. 2.1. ' 4L 2 ' 4L 2 ' 4L 2 Thus the nth eigenvalue An and associated eigenfunction of the problem in (11) are given by -I and Yn(x ) = sin (2n . 3. . JIL) + B ·0 = + B sin(. The remainder of this section is devoted to three such applications.Jix). A) because y 1 and y 2 will depend on A. (as we saw in Example 3). A)+ By2(X. When we collect coefficients of A and B in the resulting pair of equations. Most of the interest in eigenvalue problems is due to their very diverse physical applications. We assume that the portion of the string to one side of any point exerts a constant tension force T on the portion of the string to the other . We first write the general solution of the differential equation in the form y = Ayt(X. (13) Now A is an eigenvalue if and only if the system in (13) has a nontrivial solution (one with A and B not both zero). 3. 0 (in the unknowns A and B) which correspond to the equations in (13). . We write Yi (x. But such a homogeneous system of linear equations has a nontrivial solution if and only if the determinant of its coefficients vanishes. which implies that .. The Whirling String Who of us has not wondered about the shape of a quickly spinning jump rope? Let us consider the shape assumed by a tightly stretched flexible string of length L and constant linear density p (mass per unit length) if it is rotated or whirled (like a jump rope) with constant angular speed w (in radians per second) around its equilibrium position along the x-axis. as in Examples 3 and 4. in which y 1 (x) = cosax = cosx~ and y 2 (x) = sinax = sinx~. (14) in a concrete problem. and hence also linear in A and B.JIL = nn. Then we impose the two endpoint conditions. If A > 0.JIL) = 0. let's revisit the eigenvalue problem of Example 3. 0. then the differential equation y" + AY = 0 has the general solution y(x) = A cos(.. noting that each is linear in y and y'. The endpoint conditions y(O) = 0 and y(L) = 0 then yield the equations y(O) =A· 1 y(L) =A cos(. We therefore conclude that the eigenvalues of the problem in (10) are the (real) solutions of the equation (14) To illustrate Eq.2.8 Endpoint Problems and Eigenvalues 185 A general procedure for determining the eigenvalues of the problem in ( 10) can be outlined as follows.)B = at (A)A 0. The determinant equation D(A) = 0 corresponding to (14) is then simply the equation sin(.JIL) = 0. Numerous additional applications are included in Chapters 8 and 9 (on partial differential equations and boundary value problems).. 2. the solution of the equation D(A) = 0 in (14) may present formidable difficulties and require a numerical approximation method (such as Newton's method) or recourse to a computer algebra system.Jix) + B sin(. we therefore get a system of the form + f3t (A)B = a2(A)A + f32(J. or A= n 2 n 2 jL 2 for n = 1. A). For more general problems. 6(c). substitution of this and ( 15) in F = ma yields Ty'(x + 6.8. (16) x If we write FIGURE 2.Ty'(x). then the only solution of the problem . 2. x + 6. But what does all this mean in terms of the whirling string? It means that unless A in ( 17) is one of the eigenvalues in (8). Here we have r = y. each point moves in a circle centered at that point's equilibrium position on the x-axis.7 we see that the net vertical force in the positive y-direction is F = T sin(() x=O Whirling string (b) x=L + 6.8. (15) -- 8 --' Vertical component: 1 Tsin e 1 String (c) FIGURE 2.8. T (17) and impose the condition that the ends of the string are fixed. Next we recall from elementary calculus or physics the formula a = ro} for the (inward) centripetal acceleration of a body in uniform circular motion (r is the radius of the circle and w is the angular velocity of the body). we assume that the deflection of the string is so slight that sin() ~ tan() = y' (x) in Fig. t ) Equilibrium position (a) X side of the point.x string: / / 0 to get the differential equation of motion of the Ty" x . the minus sign because the inward direction is the negative y-direction.x).8) - T tan(). so the vertical acceleration of our piece of string is a = -w2 y. We further assume that. Forces on a short segment of the whirling string.x ---+ ~ -pw 2 y 6. Because m = p 6. From Fig. so that as it whirls it also stretches to assume a curved shape. y so that T T · y'(x y'(x) ~ .7.6.x). so that F ~ Ty'(x + 6. we finally get the eigenvalue problem y" + AY = 0.pw y.x. We plan to derive a differential equation for y(x) by application of Newton's law F = ma to the piece of string of mass p 6. We found there that the eigenvalues of the problem in (7) are (8) with the eigenfunction Yn(x) = sin(mrx/L) associated with An.8. y(L) =0 (7) that we considered in Example 3. 2. y(O) = 0. Denote by y(x) the displacement of the string from the point x on the axis of rotation. Thus the string is elastic.186 Chapter 2 Linear Equations of Higher Order T ' I l .x.8) - T sin() ~ T tan(() + 6. 2 We now take the limit as 6. Finally. with the direction of T tangential to the string at that point.X x + L1x + pw2 y pwz A=- = 0.x ]. The whirling string.x) 6.Ty'(x) + 6. as the string whirls around the x-axis. The only forces acting on this piece are the tension forces at its two ends.x corresponding to the interval [x. We want to investigate the shape y = y(x) of this curve. But when w = w1. And when w is increased still further.4.. Suppose that we start the string rotating at speed = then gradually increase its speed of rotation. A consequence of the theory of elasticity is that for relatively small deflections of such a beam (so small that [y' (x)] 2 is negligible in comparison with unity). . with the positive y-axis directed downward. 2.8. the string pops back into its undeflected position along the axis of rotation! = The Deflection of a Uniform Beam We include now an example of the use of a relatively simple endpoint value problem to explain a complicated physical phenomenon-the shape of a horizontal beam on which a vertical force is acting. Consider the horizontal beam shown in Fig.. In this case the string remains in its equilibrium position with zero deflection. where • • E is a constant known as the Young 's modulus of the material of the beam. I denotes the moment of inertia of the cross section of the beam around a horizontal line through the centroid of the cross section. and • F(x) denotes the density of downward force acting vertically on the beam at the point x .2. So long as w < w 1. so the numerical value of en would necessarily be significant! y smaller than 1. then the force of its own weight distorts its longitudinal axis of symmetry into the curve shown as a dashed line in the figure.8.9. 2. 2.8. At angular speed w it assumes a shape of the form Yn = en sin(mr xj L) illustrated in Fig. we get a sequence of critical speeds of angular rotation.9. an adequate mathematical model of the deflection curve is the fourth-order differential equation EI/4 ) = F(x). 2. (19) FIGURE 2. 2. the string pops into a whirling position y = c 1 sin(n xj L). but it assumes much smaller deflections than those observed in Fig.8. The deflection curve.4 (where Cn 1).8. 3.8.8.8.8 Endpoint Problems and Eigenvalues 187 in (7) is the trivial solution y (x) 0. = (18) for n = 1. . the string remains in its undeflected position y 0. Distortion of a horizontal beam. L -----------X Positive y-values FIGURE 2. But if we equate (17) and (8) and solve for the value Wn corresponding to An. . Only at these critical angular speeds can the string whirl up out of its equilibrium position. uniform both in cross section and in material. Our mathematical model is not sufficiently complete (or realistic) to determine the coefficient en. If it is supported only at its ends. the deflection curve of the beam_ We will use the coordinate system indicated in Fig. which depends only on the material of the beam. where A. which depends only on the shape of the cross section of the beam. w pounds per foot.10. For instance. The following table shows the boundary or endpoint conditions corresponding to the three most common cases. Eq. Figure 2. x + ~x] of the beam is approximately F(x) ~x. Thus we obtain a solution of Eq. (21) FIGURE 2. C3 . C2 . Values of the Young's modulus E of various materials can be found in handbooks of physical constants. B. (20) Note: We assume no previous familiarity with the theory of elasticity or with Eq. We will consider here the case in which the only force distributed along the beam is its own weight. (20) yields a second yields another yields Ely'= iwx 3 a final integration gives x-0 + iC1x 2 + Czx + C3. (20) is a fourth-order differential equation. two constants: E. although a discussion here of their origin would take us too far afield.188 Chapter 2 Linear Equations of Higher Order Density of force? Yes. This proportionality involves.8. A beam might also be supported one way at one end but another way at the other end. X=L Simply supported or hinged where C 1 .. 2. its solution involves only the solution of simple first-order equations by successive simple integrations. I. Fig.x 2 + Ax 3 24£/ w + Bx 2 + Cx + D. such as pounds per foot. .10 shows two common types of support. (20) of the form x=L x=O I ----Built in I! y(x) = . (19) takes the form Ely<4 l = w where E. (19) or (20) here. and w are all constant.8. and C4 are arbitrary constants. I = ~ n a 4 for a circular cross section of radius a. this means that the force acting downward on a very short segment [x. and Dare constants resulting from the four integrations. Two ways of supporting a beam. These last four constants are determined by the way in which the beam is supported at its ends. thus we develop an understanding of the equation by examining its solutions. so that F(x) = w. Although Eq. We will see that these conditions are applied readily in beam problems. Observe that. Then Eq. and I. where x = 0 and x = L. It is important to be able to begin with a differential equation that arises in a specific applied discipline and then analyze its implications.11 shows a cantilever-a beam firmly fastened at x = 0 but free (no support whatsoever) at x = L. in essence. C. One integration of Eq. (20) implies that the fourth derivative y<4 l is proportional to the weight density w. however.8. The units of F (x) are those of force per unit length. . .. ~ ... .27 ~) (980 c~) ~ 38484..190 Chapter 2 Linear Equations of Higher Order as the shape of the simply supported beam..60) 4 ~ 16_ 96 em (384)(2 X 1012 )(2.75 gjcm 3 and that its Young's modulus is E = 2 x 10 12 gjcm-s 2 . .68 in. so it will be more convenient to work in cgs units.54 ~m) = 1. . .) = 609. It is interesting to note that Ymax is proportional to L 4. . so I = = 4n(l. . . (25) that the same reduction in maximum deflection • could be achieved by doubling the radius a of the rod. ..· - v . ..48 c. .04) ' Ymax about 6. 5wL 4 = 384 EI · Ymax (25) • -- . It is apparent from symmetry (see also Problem 17) that the maximum deflection Ymax of the beam occurs at its midpoint x = L/2. ._. suppose that we want to calculate the maximum deflection of a simply supported steel rod 20ft long with a circular cross section I in.._.6 dyn. (25) yields ~ (5)(38484. _ · · · -· · ·· Example 6 For instance.. its mass per unit length) is so w = pg = (39. .) ( 2. Its linear mass density (that is.·~~· . its maximum deflection would be only one-sixteenth as much.-. . _ . Because I = ~na\ we see from Eq.. 4 Therefore Eq. . ··~···'<· '"' ' 4"~· • -~ ~~··· --~-· . . and thus has the value L) y(2 Ymax = = 24 E I w ( 1 16 L - 4 2 4 1 4) S L + 2L .27) 1 4 ~ 2. Thus our rod has length: and radius: a L = (20ft) ( 30.6)(609.. that is.. as the maximum deflection of the rod at its midpoint.42 in. in diameter. 2 m...60 em = (~ in..04 em . em s em The area moment of inertia of a circular disk of radius a around a diameter is I ~na4.only about 0. so if the rod were only 10ft long.. From a handbook we find that typical steel has density 8 = 7.-~ .27 em. . that has been "buckled" by an axial force of compression P applied at one end. y(O) = y(L) = 0 (7) that we considered in Example 3.E I. .. Only when the compressive force P is one of these critical forces should the rod "buckle" out of its straight (undeflected) shape. y(O) = y(L) = 0 (26) is used to model the actual (nonlinear) behavior of the rod. hinged at each end. (In practice a rod may fail at a significantly smaller force due to a contribution of factors not taken into account by the mathematical model discussed here. 3.8 Endpoint Problems and Eigenvalues 191 The Buckled Rod y y = y(x) p I p X x=O lx=L Figure 2. The forces n = 1. it is the upper bound for those compressive forces to which the rod can safely be subjected without buckling. recall from Eq. The buckled rod.12 shows a uniform rod of length L. If we write p 'A=EJ' (27) then the problem in (26) becomes the eigenvalue problem y" + A. (27) that P = A. E denotes the Young's modulus of the material of the beam and I denotes the moment of inertia of each cross section of the beam around a horizontal line through its centroid.y = 0.£ L. We found that its eigenvalues {'An} are given by (8) with the eigenfunction Yn = sin(mrx/L) associated with 'An.) . The smallest compressive force for which this occurs is (29) This smallest critical force P1 is called the Euler buckling force for the rod.8.8. In the theory of elasticity the linear endpoint boundary value problem FIGURE 2. Ely"+ Py = 0.12.2.. We assume this buckling to be so slight that the deflection curve y = y(x) of the rod may be regarded as defined on the interval 0 . (28) are the critical buckling forces of the rod.) To interpret this result in terms of the buckled rod. (Thus whirling strings and buckled rods lead to the same eigenvalues and eigenfunctions.£ x . 2. As in our discussion of the deflection of a uniform beam. y" + AY = 0. 12.'"·-·····-- Example 7 For instance.. respectively.rr) = y'(rr) . • The eigenvalues in Problems 1 through 5 are all nonnegative.448 x 105 dynjlb. y(O) = 0. (c) Draw a sketch indicating the roots {an}f as the points of intersection of the curves y = tan z and y = .48 c. y'(rr) = 0 y" + AY = 0. (a) Show that Ao = 0 is an eigenvalue with associated . y(rr) = 0 y" + AY = 0. (b) Show that y = A cos ax + B sin ax satisfies the endpoint conditions if and only if B = 0 and a is a positive root of the equation tan z = 1/z.192 Chapter 2 Linear Equations of Higher Order . (a) Show that A = 0 is not an eigenvalue. (10ft) ( 30. (b) Show that the eigenfunctions are the functions {sinanx}f. all its eigenvalues are nonnegative. y' (. Use a method similar to that suggested in Problem 10 to show that the eigenvalue problem in Problem 6 has no negative eigenvalues. All the eigenvalues are nonnegative. y(-2) = 0. y'(O) = 0. which is not of the type in (1 0) because the two endpoint conditions are not "separated" between the two endpoints. 8.54 4 ~m) ~ 2. y'(O) = 0. Deduce from this sketch that f3n ~ (2n + l)rr /2 when n is large. y'(O) = 0. Deduce from this sketch that a n ~ (2n . in diameter. Consider the eigenvalue problem y" +AY FIGURE 2. (Suggestion: Show graphically that the only root of the equation tanh z = . Draw a sketch showing these roots. y(l) = 0 y" + AY = 0. 2.. where f3n is the nth positive root of the equation tan z = z. where a n is the nth positive root of the equation tan z = .) = 304. y" +AY = 0. ln. 4. . In cgs units we have E L =2x = 10 12 gjcm·s 2 . First determine whether A = 0 is an eigenvalue.. 1. --· .l)rr /2 when n is large.. 3.z. y (I) = y'(l).. (b) Show that the remaining eigenfunctions are given by Yn(x) = sin f3nx . Consider the eigenvalue problem = 0. y (l) + y' (l) = 0 has no negative eigenvalues. 2.04 cm4 .34 x 108 dyn ~ 976 lb.13.. y'(2) = 0 Consider the eigenvalue problem y" + AY all its eigenvalues are nonnegative. as indicated in Fig..) (2.13. Prove that the eigenvalue problem of Example 4 has no negative eigenvalues. and I= rr [co.) 11.. (a) Show that A = 0 is an eigenvalue with associated eigenfunction Yo(x) = x. Prove that the eigenvalue problem y" + AY = 0. y(. y (l) + y'(l) = 0. ·--~··. 5. = 0. These roots {an }f are the abscissas of the points of intersection of the curves y = tan z andy= ljz. using the conversion factor 4.8 em. y(O) = 0.z. suppose that we want to compute the Euler buckling force for a steel rod 10 ft long having a circular cross section 1 in. (29) we find that the critical force for this rod is P 1 ~ 4.5 in. then find the positive eigenvalues and associated eigenfunctions. y'(-rr) = 0. y'(rr) = 0 y" + AY = 0. so write A = a 2 where a ~ 0. Consider the eigenvalue problem y" + AY = 0. 10.8.8. 9. (a) Show that A = 0 is not an eigenvalue.. y(O) = 0. The eigenvalues are determined by the intersections of the graphs of y = tan z and y = l j z (Problem 6). y(l) + y'(I) = 0.rr ) = y(rr). y(-rr) = 0. r Upon substituting these values in Eq. 6. Thus the eigenvalues and eigenfunctions of this problem are the numbers {a~}f and the functions {cosanx}f. 7.z is z z = 0. y'(l) = 0. cos nx and sin nx.( x4 w (b) Show that its maximum deflection occurs where x = ( 15 . 14. Show that its deflection curve IS y(x) = . y(O) = 0. For the simply supported beam whose deflection curve is given by Eq.6% of the maximum deflection that would occur if the beam were simply supported at each end. 24£/ (b) Show that y' (x) = 0 only at x = 0. and thus that it follows (why?) that the maximum deflection of the cantilever is Ymax = y(L) = wL 4/(8£ /). Show that its shape is given by = y" + 2y' + AY = 0. (c) Show that the nth positive eigenvalue is n 2 and that it has two linearly independent associated eigenfunctions. Consider the eigenvalue problem (a) Show that A (b) Show that the roots of y'(x) = 0 are x = 0. y(O) = y(l) = 0. show that the only root of y'(x) = 0 in [0. (b) Show that there are no negative eigenvalues. so it follows (why?) that the maximum deflection of the beam is Ymax = Y( 2 L) = 384£/ ' wL 4 y" + 2y' + AY = 0. (b) Show that there is no eigenvalue A such that A < 1. and x = Lj2. (a) Suppose that a beam is fixed at its ends x x = L. 17. 15.. (a) A beam is fixed at its left end x = 0 but is simply supported at the other end x = L. (25). 13. (c) Show that the nth positive eigenvalue is A11 = n 2 rr 2 + 1. 18. . where an is the nth positive root of tan z = z. Show that its shape is given by . (24).4Lx 3 + 6L 2 x 2 ).. one-fifth that of a beam with simply supported ends. (a) A uniform cantilever beam is fixed at x = 0 and free at its other end. Consider the eigenvalue problem 193 0 and = 16. so it follows (why?) that the maximum deflection is indeed that given in Eq. Show that the eigenvalues are all positive and that the nth positive eigenvalue is An = a~ + I with associated eigenfunction Yn(x) =e-x sinanx.8 Endpoint Problems and Eigenvalues eigenfunction y0 (x) 1. where x = L. L] is x = L/2.J33) Lj16 and is about 41 . = 1 is not an eigenvalue. with associated eigenfunction Yn(x) = e-x sin nrr x.2. x = L . at+··· .3 we saw that solving a homogeneous linear differential equation with constant coefficients can be reduced to the algebraic problem of finding the roots of its characteristic equation.2xy' -+. state (without proof) several theorems that constitute a review of the basic facts about power series. These techniques suffice for many of the nonelementary differential equations that appear most frequently in applications. n=O 00 (2) 194 . It has the form (1 .Power Series Methods - Introduction and Review of Power Series I n Section 2.a is an infinite series of the form L Cn(X n=O 00 at =co+ c1 (x. Perhaps the most important (because of its applications in such areas as acoustics. There is no similar procedure for solving linear differential equations with variable coefficients.a) 2 + · · · + Cn(X . such as the occasional equation that can be solved by inspection. heat flow. (1) If a = 0.x 2 )y" . at least not routinely and in finitely many steps. With the exception of special types.a)+ Cz(x. Recall first that a power series in (powers of) x . In this section we introduce the power series method in its simplest form and.n(n + l) y = 0. and electromagnetic radiation) is Bessel's equation of order n: Legendre's equation of order n is important in many applications. this is a power series in x : L CnXn = Co+ C1 X + CzX 2 + · · · + CnXn + · · · . linear equations with variable coefficients generally require the power series techniques of this chapter. along the way. =L 1-x and (1 1 00 xn = 1 + x + x 2 + x 3 + ··· . 3! 5! x2 · ' (7) cosh x = n=O x 2n x4 L -= 1 +.1)x 2 2! + a(a.a by replacement of x with x. is the binomial series.+ .- 2'. In this case the sum 00 f(x) = LCnXn n=O (4) is defined on I...1 Introduction and Review of Power Series 195 We will confine our review mainly to power series in x.+ . including x = 0.+-. The following power series representations of elementary functions should be familiar to you from introductory calculus: (5) oo COS X (-l)nx2n x2 = "'\"" ~ n=O (2n)'' = 1. The power series in (2) converges on the interval I provided that the limit (3) exists for all x in I. · · (2n)! 2! 4! ' oo (8) sinh x = L x2n+l n=O (2n + 1)! = x + . and we call the series L CnXn a power series representation of the function f on I. + ..·· · ' 4'. (What if lx I = 1?) The series in (11) is the geometric series.a. with a an arbitrary real number.2)x 3 3! + ··· . smx = n=O L oo oo ( -l)nx2n+1 (2n + 1)! = x . Otherwise. If a is a nonnegative integer n. ' x3 xs x4 (6) .3. (11) n=O + x) a = 1 + ax + a(a. In contrast.+ . the series in ( 10) and ( 11) converge if lx I < 1 but diverge if lx I > 1. The series in (5) through (9) converge to the indicated functions for all x . but every general property of power series in x can be converted to a general property of power series in x .1)(a.. then the series in (1 2) terminates and the binomial series reduces to a polynomial of degree n which converges for all x.. · · 3! 5! ' x3 xs (9) (10) .+. . The series in (12). we observe the usual conventions that 0! = 1 and that x 0 = 1 for all x. its behavior for lx I = 1 depends on the value of a.. the series is actually infinite and it converges if lxI < 1 and diverges if lxI > 1. (12) In compact summation notation.. power series may be manipulated algebraically in much the same way as polynomials. then the series in (13) is the Maclaurin series 00 j(n) (0) L n=O n! xn = f(O) + j'(O)x + .( x . (13) are all defined). suppose that f (x) = ex. f (x) for all x in some open f is analytic at x = a. If a = 0. (13') reduces to the exponential series in (5). Power Series Operations If the Taylor series of the function f converges to interval containing a. if 00 00 f(x) = LanXn n=O and g(x) = L bnxn. It is rather awkward to compute the Taylor series of the tangent function using Eq. The Taylor series with center x = a of the function f is the power series >- oo f(nl(a) L n=O (x. (6) and (7)). and hence f(n) (0) = 1 • for all n ~ 0. (13) because of the way in which its successive derivatives grow in complexity (try it!).196 Chapter 3 Power Series Methods Remark: Power series such as those listed in (5) through (12) are often derived as Taylor series.a)+ . · (13) in powers of x -a. • every rational function is analytic wherever its denominator is nonzero. For instance.. if the two functions f and g are both analytic at x = a. Fortunately. then we say that the function example. n=O (14) then 00 f(x) + g(x) = L(an + bn)Xn (15) n=O and 00 f(x)g(x) = L CnXn n=O ..a)n = f(a) n! + f"(a) j'(a)(x. Then f(n) (x) = eX. For • every polynomial function is analytic everywhere. then so are their sum f + g and their product f · g. • more generally. the function h (x) = tan x = (sin x) 1(cos x) is analytic at x = 0 because cos 0 = 1 f 0 and the sine and cosine functions are analytic (by virtue of their convergent power series representations in Eqs. For example. In this case Eq..x 2 + 2! J"(O) f(3) (0) 3! x3 + · · ·. (13') For example. as is their quotient fIg if g(a) f 0. under the hypothesis that f is infinitely differentiable at x =a (so that the coefficients in Eq.a) 2 2! + . Similarly... The Power Series Method The power series method for solving a differential equation consists of substituting the power series (18) in the differential equation and then attempting to determine what the coefficients co.1. The series in (15) is the result of term wise addition and the series in ( 16) is the result of formal multiplication-multiplying each term of the first series by each term of the second and then collecting coefficients of like powers of x. but when it is we obtain an infinite series representation of a solution..1. we must first know what to substitute for the derivatives y'. This method is not always successful. .3. 3. on any open interval on which both the series in (14) converge.2 1 X4 + 24 ••. This division of the Taylor series for cos x into that for sin x yields the first few terms of the series tanx = x + 3x 3 + 15 x 5 + 315 x 7 + · · · . y".x 6 120 =2 1[ (2x) 3 (2x). 6 2 24 12 120 4 3 16 5 =x. in contrast to the "closed form" solutions that our previous methods have yielded.) The series in (15) and (16) converge to f(x) + g(x) and f(x)g(x).1 Introduction and Review of Power Series 197 where cn = aobn + a 1bn -I + · · · + an bo. the sine and cosine series converge for all x.~) x3 + (_!_ + _1 + _1_) x5 + .. This is reminiscent of the method of undetermined coefficients. cz. but the tangent series in (17) converges only if lx I < rr /2. must be in order that the power series will satisfy the differential equation. the quotient of two power series can be computed by long division.-x + . For example. . I 2 17 (17) Division of power series is more treacherous than multiplication. respectively.~ (2x) 1 . Before we can substitute the power series in (18) in a differential equation. + -----s!· · · J= 2 sm 2x 5 for all x.. X COS X Slll = (X - 1X 3 + 1 X 5 120 6 • · · ) (1 1X 2 . . as illustrated by the computation shown in Fig. . . . ) = x + (-~ . (Thus the processes strongly resemble addition and multiplication of ordinary polynomials. but now we have infinitely many coefficients somehow to determine. c. . The following theorem (stated without proof) tells us that the derivative y' of y = L cnxn is obtained by the simple procedure of writing the sum of the derivatives of the individual terms in the series for y. For example. the series thus obtained for fIg may fail to converge at some points where the series for f and g both converge. that is. we shift the index of summation in the first sum. then an = bn for all n 0. Example 1 Solution Solve the equation y' + 2y = 0. then they are the same series. Thus we can replace n with n + 1 if.l' n=l (21) 00 L ncnXn. . In particular. we need the general term in each sum to be the term containing xn.1 Introduction and Review of Power Series 199 gives 1 -------o- (1 . The result of making this shift in Eq. we start counting one step lower.3. To accomplish this. This theorem-also stated without proof-tells us that if two power series represent the same function. In particular. (21) is the identity 00 :L)n n=O + 1)cn+1Xn + 2 L CnXn = 0.x)2 =L 00 n=l nxn-l = 1 + 2x + 3x 2 + 4x 3 + · · · . at n = 0 rather than at n = 1. it follows from Theorem 2 that an = 0 for all n ~ 0. at the same time.a) that represents the function f. n=O 00 that is. THEOREM 2 If Identity Principle :L: anxn = :L: bnxn n=O n=O ~ 00 00 for every point x in some open interval I. note that 00 00 'L:ncn xn-l = c1 + 2c2x + 3c3x 2 + · · · = L(n n=l n=O + 1)cn+lxn . The process of determining the coefficients in the series y = L CnXn so that it will satisfy a given differential equation depends also on Theorem 2. 00 :L)Cn + 1)cn+1 + 2cn]Xn = n=O 0. the Taylor series in (13) is the only power series (in powers of x. if L anxn = 0 for all x in some open interval. This is a shift of + 1 in the index of summation. To see how to do this. n=l n=O 00 00 To compare coefficients here. We substitute the series and and obtain y' = ~ L ncnXn.l + 2 L CnXn = 0. consequently..x n =co L n=O n=O · n=O · 2x n' n' .) Consequently.X = coe y(x) = L CnX n = L (-1) n . thereby obtaining the series on the right. (22) for all n 2: 0. 3 23 co = .' n n ~ l. 1 With n = 1. In the final step we have used the familiar exponential series in Eq. our solution takes the form oo oo 2n oo ( 2 )n co . we will have n 2nco c = ( . n. Eq.=--. (22) gives 2co c. in terms of c0 . then it follows from the identity principle that (n + l)cn+t + 2cn = 0 for all n ~ 0. 2. (22) gives C3 2cz 23co = . With n = 0. Eq.--.200 Chapter 3 Power Series Methods If this equation holds on some interval. from n = 1 to n = 0.1 ) I. the latter will tum out to be the arbitrary constant that we expect to find in a general solution of a first-order differential equation. (22) gives With n = 2. c2 .=- 3 1 .l = n=l 00 00 L(n n=O + 1)cn+tXn (23) by shifting the index of summation by + 1 in the series on the left. •. (This is easy to prove by induction on n. • Shift of Index of Summation In the solution of Example 1 we wrote L ncnxn . Eq. c3 . That is. n --+ n + 1) and decreased the starting point by 1. This procedure is valid because each infinite series in (23) is simply a compact notation for the single series (24) .. 3! By now it should be clear that after n such steps. (5) to identify our power series solution as the same solution y(x ) = c0 e .2x we could have obtained by the method of separation of variables. we simultaneously increased the index of summation by 1 (replacing n with n + 1. Equation (22) is a recurrence relation from which we can successively compute c 1 . (b) If 0 < p < oo. then L CnXn converges if lx I < p and diverges if lx I > p. then the series diverges for all x =!= 0. More commonly. a power series solution is not recognizable in terms of the familiar elementary functions. Then (a) If p = 0. then the series converges for all x. THEOREM 3 Radius of Convergence Given the power series L cnxn. Hence this formal process is justified only if in the end we can show that the power series we obtain converges on some open interval. The following theorem (which we state without proof) may be used for this purpose. n=l 00 00 n=3 we have decreased the index of summation by 2 but increased the starting point by 2. (25) will be quite sufficient for computing the radius of convergence. suppose that the limit p = hm Cn n-oo Cn+l . Eq. After all. we interpret a "decrease by k" as an increase by -k = shift by -2 (n ---+ n . The procedure illustrated in Example 1 for finding the coefficients {en} is merely a formal process and may or may not be valid. Its validity-in applying Theorem 1 to compute y ' and applying Theorem 2 to obtain a recurrence relation for the coefficientsdepends on the convergence of the initially unknown series y = L cnxn. I (-1t2n cofn! I= h . we have p = hm L .2) in the index of summation yields lkl. y = L CnXn is merely an assumed form of the solution.l)n+12n+lco/(n + 1)! n-OO 2 ' and consequently the series we obtained in Example 1 converges for all x. there always will be a number p such that exactly one of the three alternatives in Theorem 3 holds. The number p in (25) is called the radius of convergence of the power series CnXn. If so. Even if the limit in (25) fails to exist. I I (25) exists (p is finite) or is infinite (in this case we will write p = oo ).m n+1 .2Xn-3. it then represents a solution of the differential equation on that interval. (c) If p = oo. Thus a L nCnXn-1 = L(n. we can shift the index of summation by k in an infinite series by simultaneously increasing the summation index by k (n ---+ n+k) and decreasing the starting point by k. You should check that the summation on the right is merely another representation of the series in (24). For instance. for the power series obtained in Example 1.2)Cn. from n = 1 ton = 3.1 Introduction and Review of Power Series 201 More generally. a shift by +2 (n ---+ n + 2) yields If k is negative. we need a way of finding where it converges.3.. We know that the power series obtained in Example 1 converges for all x because it is an exponential series. For instance. This number may be difficult to find.= oo n-oo (. When we get an unfamiliar power series solution. . but for the power series we will consider in this chapter. 1 (replace n = 1 with n = 2 and n with n . c0 + c 1x. An elementary solution (obtained by separation of variables) of our differential equation is y = 1j (3 .1 Introduction and Review of Power Series 203 Thus the series in (26) converges if -3 < x < 3 and diverges if lxl > 3.X+ Co+ C]X + L n=2 C11 X11 • Because the left-hand side contains neither a constant term nor a term containing x to the first power.I X11 n=2 = -1. which yield 11 x2 LnCnXn-l = -1. in general. Thus we obtain the power series 00 y(x ) = 1 +x + L(n.x)2 on the interval -3 < x < 3. It follows that C2 = 1 · CJ = 1!. Thus this series (with the arbitrary constant c0 appropriately chosen) represents the solution 1 y(x) = (3. and the singularity at x = 3 is the reason why the • radius of convergence of the power series solution turned out to be p = 3.x on the right-hand side.J.X+ LCnX 11 n=i n=O 00 so that 00 L:nc11 xn+l = -1-x + LCnX 11 • n=i n=O Because of the presence of the two terms -1 and .l)! x n=2 11 • . of the series on the right for comparison. C3 = 2 · C2 = 2! . c 1 = 1. the identity principle now yields co = 1. C4 = 3 · C3 = 3! .1)! for n ~ 2.1)cn-J for n ~ 2.x) 2 • If we differentiate termwise the geometric series we get a constant multiple of the series in (26). we need to split off the first two terms. we get co co L(n. Example 3 Solution We make the usual substitutions y = 00 Lc 11 x 11 and y' = 00 L nc xn.1). and Cn = (n . In this particular example we can explain why.1)Cn.3. If we also shift the index of summation on the left by . that C11 = (n . and. . 2.. we get Cz =- co 2! ... When we apply the recurrence relation in (27) with n = 0... CJ 7! ...204 Chapter 3 Power Series Methods But the radius of convergence of this series is p = n->oo hm .Z + L CnXn = n=O 00 0.. We shift the index of summation in the first sum by +2 (replace n = 2 with n = 0 and n with n + 2). and 4 in turn. and thus we obtain the recurrence relation c n+Z - .-~·. 3.(n + 1)(n + 2) Cn (27) for n ~ 0.· · . C4 = 4! . (n.. we find that 00 00 y' = Z:ncnXn-l n=l and y 11 = ' ""' { ~n n- 1) CnX n. CJ co and C6 = --.. The identity (n + 2)(n + 1)cn+Z + Cn = 0 now follows from the identity principle.. n=2 Substitution for y and y" in the differential equation then yields L n=2 00 n(n - l)cnxn . c0 and c 1 are not predetermined and thus will be the two arbitrary constants we expect to find in a general solution of a second-order equation.= 0. n 1 so the series converges only for x = 0. This gives 00 L(n n=O + 2)(n + 1)cn+2Xn + L n=O 00 CnXn = 0. This example serves as a warning that the simple act of writing y = L CnXn involves an assumption that may be false. • . If we assume a solution of the form - ...5!' and C7 = .1)! n! = n->oo lim ... What does this mean? Simply that the given differential equation does not have a (convergent) power series solution of the assumed form y = L cnxn. Example 4 Solution Solve the equation y" + y = 0. It is evident that this formula will determine the coefficients Cn with even subscript in terms of co and those of odd subscript in terms of c 1 . _ _____ . co 6! Taking n = 1. we find that cs .-.. and 5 in turn.2 . using only their initial values (at x = 0) and the derivatives in (30). and that y = S (x) is the unique solution that satisfies the initial conditions y(O) = 0 and y' (0) = 1.1 Introduction and Review of Power Series Again.3. the pattern is clear. the Taylor series for the sine and cosine functions converge for ali x. Indeed. • The solution of Example 4 can bear further comment. (Can you use the series in (28) and (29) to show that [C(x)] 2 + [S(x)f = 1 for all x?) This demonstrates that The cosine and sine functions are fully determined by the differential equation y" + y = 0 of which they are the two natural linearly independent solutions. Both of these power series converge for all x. the ratio test in Theorem 3 implies convergence for all z of the series I:< -l)nzn/(2n)! obtained from (28) by writing z = x 2 . Suppose that we had never heard of the sine and cosine functions. x2 x4 x6 x3 xs x 7 ) + · · · ) + CJ ( X--+---+··· . . and-recognizing the importance of the differential equation y" + y = O--we can agree to call C the cosine function and S the sine function. We would then have discovered the two power series solutions (28) and (29) of the differential equation y" + y = 0. c 2k+l - ( -1)kcl ---(2k + 1)!.. we have discovered that y = C (x) is the unique solution of y" +y = 0 that satisfies the initial conditions y (0) = 1 and y' (0) = 0. and termwise differentiation of the two series in (28) and (29) yields C' (x) =- S(x) and S' (x) = C(x). Hence it follows that (28) itself converges for all x. It is clear that C(O) = 1 and S(O) = 0. It follows that C(x) and S(x) are linearly independent. as does (by a similar ploy) the series in (29).+--2! 4! 6! 205 ~ 1. C' (0) = 0 and S' (0) = 1. Note that we have no problem with the radius of convergence here. y (x) = c0 cos x + c 1 sin x. 3! 5! 7! ' that is. Thus with the aid of the power series method (all the while knowing nothing about the sine and cosine functions). we leave it for you to show (by induction) that fork and Thus we get the power series solution y(x) =Co ( 1 . (30) Consequently. For instance. let alone their Taylor series. there is no need to refer to triangles or even to angles. all the usual properties of these two functions can be established. identify the particular solution in terms offamiliar elementary functions. y(O) = 0.1. 21. y (l) = l . FIGURE 3. 6.l)y' + 2y = 0 2. x 3 y' = 2y has no power series solution of the form y = L CnX 11 • 24.2. y" = 4y +y = X Show (as in Example 3) that the power series method fails to yield a power series solution of the form y = L C11 X 11 for the differential equations in Problems 15 through 18.2y' + y = 0.. y" 14. 15. 10. use the method ofExample 4 to find two linearly independent power series solutions of the given differential equation. Establish the binomial series in ( 12) by means of the following steps. Taylor polynomial approximations to sinx.2)y' + y = 0 2(x + l)y' = y 2(x . y" + 4y = 0. y" +9y = 0 12. y'(O) y" + y' . y" = y 13. y n=8 n = 16 y n = 24 n=5 n= 13 n=21 n=6 n= 14 n = 22 n=7 n = 15 n = 23 FIGURE 3. y'(O) Show that the equation 1. y' = 4y y' + 2xy = 0 (x. finally. Taylor polynomial approximations to cosx. and identify the general solution in terms offamiliar elementary functions. y'(O) = 0 y". In Problems 1 through 10. Determine the radius of convergence of each series. y(O) = 2. 23. (Of course. This is by no means an uncommon situation. 22. 4. y' =y 2y' + 3y = 0 y' = x 2 y (2x -l)y' +2y = 0 (x. 19. y(O) = l. xy' + y = 0 17. 9.1.4y = 0. (c) Explain why the validity of the binomial series given in (12) follows from parts (a) and (b). Next determine Cn (in terms of n. (a) Show that y = (1 + x )C/ satisfies the initial value problem (1 +x)y' = ay. 7.1. find a power series solution of the given differential equation. x 2 y' + y = 0 16. and that this series converges if lxl < 1. first derive a recurrence relation giving C11 for n ~ 2 in terms of co or c 1 (or both).1. y (O) = 1. 3. 11. y'(O) = 3 y" .2 and 3. 8. (5) through (12) to identify the series solution in terms of familiar elementary functions. Many important special functions of mathematics occur in the first instance as power series solutions of differential equations and thus are in practice defined by means of these power series. 25. 5. Then apply the given initial conditions to find the values of co and c.3. For the initial value problem y" = y' + y. y(O) = 0. 20. as in the text) and. In the remaining sections of this chapter we will see numerous examples of such functions. Determine the radius of convergence of the resulting series. and use the series in Eqs.206 Chapter 3 Power Series Methods Figures 3. 2xy' = y 18.3 show how the geometric character of the graphs of cos x and sin x is revealed by the graphs of the Taylor polynomial approximations that we get by truncating the infinite series in (28) and (29).l)y' = 3y =1 = -2 In Problems 11 through 14.2y = 0. no one can prevent you from checking your work by also solving the equations by the methods of earlier chapters!) In Problems 19 through 22. (b) Show that the power series method gives the binomial series in (12) as the solution of the initial value problem in part (a). y(O) = 0. differential equations can sometimes be used to sum infinite series. consider the infinite series 1 1 1 1 1 1+---+-+---+···· 1! 2! 3! 4! 5! ' note the + + .3. 13. F1 = 1.x 5 + . = 2c7 + 2c3 cs. Conversely. We saw in Example 3 of Section 3. see Problem 48 of Section 2. 5. but its most important applications are to homogeneous second-order linear differential equations of the form A(x)y" + B(x)y' + C(x)y = 0. Given these facts. This section introduces the use of infinite series to solve differential equations. = 2cg + 2c3 c 7 + (c5 ) 2 . and C are analytic functions of x. Substitute this series in y' = 1 + y 2 and equate like powers of x to derive the following relations: 3c3 7c7 1lc11 = = 1. (1) where the coefficients A. and with P = B I A and Q = CIA. . (3) ._l for n > 1. 26.· · · pattern of signs superimposed on the terms of the series for the number e. in most applications these coefficient functions are simple polynomials. 3. 27.. Note that P(x) and Q(x) will generally fail to be analytic at points where A(x) vanishes. y(O) = y'(O) = 1. (a) It's possible to show that the power series given here converges for all x and that termwise differentiation is valid. The power series method introduced in Section 3. y(O) is y(x) = tanx.-x + · · · 2! 3! 4! 5! ' because the sum of the numerical series in question is simply f(l). 2. (1) in the form y" + P(x)y' + Q(x)y = 0 (2) with leading coefficient 1. To discover when it does succeed.2 Series Solutions Near Ordinary Points derive the power series solution 207 y(x) = L _!!_xn n=l n! ()() F. 1.x 7 3 15 315 (b) Solve this initial value problem to show that 62 g 1382 II + 2835x + 155925x + · · · · (d) Would you prefer to use the Maclaurin series formula in (13) to derive the tangent series in part (c)? Think about it! For a suggestion. (b) Because tanx is an odd function with y' (0) = 1. Fn = Fn-2 + F. y " (O) = -1.3. B. (c) Evaluate f (1) to find the sum of the numerical series given here. 1.1 that the series method does not always yield a series solution.-x + -x + -x . we rewrite Eq. 2cs 5cs 9c9 + (c3 ) 2 . of Fibonacci numbers defined by F0 = 0. tanx = 2c3. (a) Show that the solution of the initial value problem y' = 1 + l. 8. =x 1 2 17 + -x 3 + . For example.. (c) Conclude that yC3l = y. Indeed. For instance. its Taylor series is of the form =0 y(x) = Y =X + c3 x 3 + c5 x 5 + c7 x 7 + · · · . show that f (x) satisfies the initial value problem where {F~~}~0 is the sequence 0.+ + .1 can be applied to linear equations of any order (as well as to certain nonlinear equations). We could evaluate this series if we could obtain a formula for the function 12131415 f(x) = 1 + x. consider the equation xy" + y' +xy = 0. • Theorem 2 of Section 2. A proof of the following theorem can be found in Chapter 3 of Coddington. 1 implies that Eq. It follows that. If A(x). (1) with analytic coefficients. these solutions will be power series in powers of x .) • ····-···· Example 3 The point x = 0 is an ordinary point of the equation because the coefficient functions A(x ). Thus the only singular point of Eqs. x = a is a singular point. Example 1 The point x = 0 is an ordinary point of the equation xy" + (sinx)y' + x 2y = 0. (2) has two linearly independent solutions on any open interval where the coefficient functions P(x) and Q(x) are continuous. . and C(x) are polynomials with no common factors. (Theorem 1 of Section 3.: Prentice Hall. (2)-and of the equivalent Eq. Recall that a quotient of analytic functions is analytic wherever the denominator is nonzero. then x = a is an ordinary point.a.1 implies that an analytic function must be differentiable. and C(x) are polynomials with A(O) -::1 0. N. An Introduction to Ordinary Diff erential Equations (Englewood Cliffs. • Example 2 The point x = 0 is not an ordinary point of the equation For while P(x ) = x 2 is analytic at the origin. B(x) . Otherwise. 1961). if A(a) -::1 0 in Eq. despite the fact that A (x) = x vanishes at x = 0. The basic fact for our present purpose is that near an ordinary point a. then x = a is an ordinary point if and only if A(a) -::1 0. But in the form of (2) it is the equation 1 y" + . B(x). The point x = a is called an ordinary point of Eq. (3) and (4) is x = 0.y' X + y =0 (4) with P (x) = 1I x not analytic at x = 0. (1 )-provided that the functions P (x) and Q (x) are both analytic at x = a.J. Q(x ) = x 112 is not.208 Chapter 3 Power Series Methods The coefficient functions in (3) are continuous everywhere. The reason is that is nevertheless analytic at x = 0 because the division by x yields a convergent power series. The reason is that Q(x) is not differentiable at x = 0 and hence is not analytic there. a)n. Solution This example illustrates the fact that we must take into account complex singular points as well as real ones. The distance (in the complex plane) of each of these from 0 is 3. n=O (5) The radius of convergence of any such series solution is at least as large as the distance from a to the nearest (real or complex) singular point of Eq. The distance of each singular point from 4 is 5. Radius of convergence as distance to nearest singularity. . The coefficients in the series in (5) can be determined by its substitution in Eq. Then Eq.2. so a (x . so a series solution of the form L cnxn has radius of convergence at least 3. (1) that is. (1) has two linearly independent solutions.3.4. 3.4t has radius of convergence at least 5 (see series solution of the form L en Fig. Example 4 Determine the radius of convergence guaranteed by Theorem 1 of a series solution of (6) in powers of x.2 Series Solutions Near Ordinary Points 209 THEOREM 1 Solutions Near an Ordinary Point Suppose that a is an ordinary point of the equation A(x)y" + B(x)y' + C(x)y = 0. (6) are ±3i. (1). • y FIGURE 3. Repeat for a series in powers of x .2.1).1. the functions P = B/A and Q = CjA are analytic at x =a. (1). each of the form 00 y(x) = Lcn(X. Because X P(x) = x2 +9 and the only singular points of Eq. . dy dx and d 2y dt 2 - [!!_ (dy)] dx dx dx dt - !!_ I _ II dx (y ) .3) dt2 + 3(t. we would have needed the general solution in the form 00 y(x) = LCn(X- at. For only with a solution of the form in (13) is it true that the initial conditions y(a) =co and y 1 (a) = c1 determine the arbitrary constants c0 and c 1 in terms of the initial values of y and y 1 • Consequently.1) 2 + 6(t. Example 6 Solve the initial value problem 2 d2y (t . . so that we wind up looking for a series of the form L CnXn after all. We substitute t .1) dt + y = dy 0. it simplifies the computations if we first make the substitution x = t . y 1 (1) = -1.Y ' where primes denote differentiation with respect to x. (14) into one with the new independent variable x. Hence we transform Eq.a rather than in powers of x.212 Chapter 3 Power Series Methods Translated Series Solutions If in Example 5 we had sought a particular solution with given initial values y(a) and y 1 (a). we note that t2 - 2t .. To transform Eq. we need a series expansion of the general solution centered at the point where the initial conditions are specified.1 for x in Eq. -=--=-=y .4)y" + 3xy 1 + y = 0 with initial conditions y = 4 and y 1 = 1 at x = 0 (corresponding tot = 1). y(l) = 4.1)n. (14) into (x 2 .3 = (x dy dt + 1)2 dy dx dx dt 2(x + 1) 1 3 = x2 - 4. But instead of substituting this series in (14) to determine the coefficients. so the particular solution in (12) is available.1) 3 1) 5 1 + 32 (t - 3 1) 4 + 30 (t - 1 + . in powers of x . to solve an initial value problem. (12) and thereby obtain the desired particular solution 1 y(t) = 4 + (t.1.1) + 2(t. .2t. (13) n=O that is. (14) Solution We need a general solution of the form Len (t. This is the initial value problem we solved in Example 5. A manyterm recurrence relation expresses each coefficient in the series in terms of two or more preceding coefficients.2 = -ncn --(n + 2)(n + 1) (16) . (Why?) A series such as this can be used to estimate numerical values of the solution.LnCnXn .8) ~ 3. n=2 n= l n=O We can start the second sum at n = 0 without changing anything else.8188.. so we must separate the terms corresponding to n = 0 and n = 1 in the first two sums before collecting coefficients of xn. • The last computation in Example 6 illustrates the fact that series solutions of differential equations are useful not only for establishing general properties of a solution.2 . Example 7 Find two linearly independent solutions of y If . n=2 n=O The common range of these three summations is n . that c3 = ic 1 . The identity principle now implies that 2c2 = 0. and the three-term recurrence relation Cn+2 +Cn .2] xn = 0.. it is generally inconvenient or even impossible to find a formula that gives the typical coefficient en in terms of n.2) 3 1 1 + 332 ( -0. (15) Solution We make the usual substitution of the power series y = equation 00 00 00 L cnxn. (8) is an example of a two-term recurrence relation.2).Cn. These shifts yield 00 00 00 L(n n=O + 2)(n + 1)cn+2Xn - L ncnxn .2)4 + 310 ( -0.2) 2 + 6(-0.2Xn = 0. The next example shows what we sometimes can do with a three-term recurrence relation. we shift the index of summation in the first sum by +2 (replace n with n + 2). 2. it expresses each coefficient in the series in terms of one of the preceding coefficients. but also for numerical computations when an expression of the solution in terms of elementary functions is unavailable. For instance. To make each term include xn in its general term.1)CnXn. Types of Recurrence Relations The formula in Eq. ' so that y(0.2) + 2(-0. In the case of a many-term recurrence relation.2 Series Solutions Near Ordinary Points 213 This series converges if -1 < t < 3.2)5 + . This results in the Ln(n.ncn.c1x + L n=2 [ (n + 2)(n + 1)cn+2..3.8) = 4 + ( -0. This gives 00 2c2 + 6c3x .L Cn..x 2 y = 0 . and we shift it by -2 in the third sum (replace n with n . y(0.LCnXn+ 2 = 0.xy I . To get our first solution y 1 of Eq. so that c2 = 0 and c 3 = ~. The only singular points of the Legendre equation are at + 1 and -1. • The Legendre Equation The Legendre equation of order a is the second-order linear differential equation (1 . cs = C7 3c3 + c1 20 5cs + c3 = 42 (17) Thus all values of Cn for n ~ 4 are given in terms of the arbitrary constants c0 and = 0 and c3 = ~cr. Then the formulas in ( 17) yield 3 cs = 40' so that C6 = 0.x + ··· . (20) where the real number a satisfies the inequality a > . (15). so it has two linearly independent solutions that can be expressed as power series in . so that c2 = c3 = 0. we choose c0 = 1 and c 1 = 0. we take co = 0 and c 1 = 1.x + . A general solution of Eq. it is clear from Eq.2xy' + a(a + 1)y = 0..1. (15).x + . (16) that this series contains only terms of even degree.x 2 )y". This differential equation has extensive applications. 12 90 1120 (18) Because c 1 = c 3 = 0.1120' 3 thus Yr (x) = 1 1 4 1 6 3 8 +x + .· . Then the formulas in (17) yield cr because c2 C4 = 12 . The solutions y 1(x) and y 2 (x) are linearly independent because YI (0) = 1 and y~ (0) = 0. In particular. C6 = 1 90 . 6 40 1008 (19) Because co = c2 = 0. 1 Cs = 0. C7 = 0. it is clear from Eq.x + ··· . so the power series representing y 1 (x) and y 2 (x) converge fur~lx.. whereas y 2 (0) = 0 and y~ (0) = 1.. 13 C? = 1008' Y2(x) = x 1 3 3 5 13 7 + -x + . (16) that this series contains only terms of odd degree. c8 . (15) is a linear combination of the power series in (18) and (19).. Equation (15) has no singular points.214 Chapter 3 Power Series Methods for n ~ 2. To obtain a second linearly independent solution y2 ofEq. ranging from numerical integration formulas (such as Gaussian quadrature) to the problem of determining the steady-state temperature within a solid spherical ball when the temperatures at points of its boundary are known. If a = n is even. The substitution y Eq. we see from Eq.2 Series Solutions Near Ordinary Points 215 powers of x with radius of convergence at least 1. y 1 (x ) is a polynomial of degree n and y 2 is a (nonterminating) infinite series. c5 = (a.4)· · ·(a.3) · · · (a . In terms of the arbitrary constants c0 and c 1.1)(a + 2) 3! 4! CJ. (23) that a 2m+ l = 0 when 2m + 1 > n . y 2(x) is a polynomial of degree n and y 1 is a (nonterminating) infinite series. one of the two solutions in (24) is a polynomial and the other is a nonterminating series. = a(a. If a = n is an odd positive integer. Now suppose that a = n.2)(a.1)(a. (20) leads (see Problem 31) to the recurrence relation (a.m)(a + m + 1) (m + 1)(m + 2) Cm = L cmxm in (21) Cm+2 =- form ~ 0. Thus in either case. Eq.3)(a + 2)(a + 4) 5! c 1• We can show without much trouble that form > 0. In this case. we get two linearly independent power series solutions 00 Yt (x) =CoL( -l)ma2mX 2m m =O and Y2(x) = Ct I:<-1)ma2m+tX2m+l m =O 00 (24) of Legendre's equation of order a. C2m = ( -1) ma(a. In this case.2m + 1) (a + 2) (a ( 2m+ 1)! + 4) · · · (a + 2m) C I· (23) Alternatively.2)(a + 1)(a + 3) Co.1) (2m)! Co (22) and C - 2m+l- (-1)m (a . We are using m as the index of summation because we have another role for n to play. a nonnegative integer. respectively. where a2m and a 2m+ I denote the fractions in Eqs. 2! C3=C4 (a. With this notation.3. (21) yields C2 = - a(a + 1) Co. .2m+ 2)(a + 1)(a + 3)· ··(a+ 2m. (22) that a 2m = 0 when 2m > n.1) (a . we see from Eq. (22) and (23). y. y~(O) = 1.3... B l(X)- . write the series of terms of even degree in Eq. has the two singular points x = -1 and x = 1. B. Then verify that your results agree with the formulas Yl (x) = 1 + f=( ~ 1 . whereas the Legendre equation of order n..3 exhibit trigonometric-like oscillatory behavior for x < 0.3 116r(~) ----o- Y1 (x) + Y2(x) 3 -lt6r(~) define the standard Airy functions that appear in mathematical tables and computer algebra systems. respectively..1) (3k + 1)! x3k+l FIGURE 3. Their graphs shown in Fig. For instance.. and C are polynomials having no common factors.218 Chapter 3 Power Series Methods y serves to introduce two new special functions that appear in applications ranging from radio waves to molecular vibrations.2xy' + n(n + l)y = 0. .32f3r(~) Yz(x) 31 /3 r(~) Yo(x) = 1 + LCznX 2n = 1 + Lanzn n=l n=l and . (3k . Y1 (x) A 1(x)----o.. where an = c2n and z = x2 • Then apply the recurrence relation in Eq. The Airy function graphs y = Ai(x) andy= Bi(x). k=l ~ 2 .. Recall that if the functions A.. (1 . It turns out that some of the features of the solutions of such equations of the most importance for applications are largely determined by their behavior near their singular points. (1) are simply those points where A (x) = 0. (11) in the form to show similarly that its radius of convergence (as a power series in x) is also 2. 3. whereas Ai (x) decreases exponentially and Bi(x) increases exponentially as x -+ +oo. Derive the first three or four terms of two different power series solutions of the Airy equation..2) (3k)! x3k and Y2(x) = x + L. x = 0 is the only singular point of the Bessel equation of order n..1 to show that the radius of convergence of the series in z is 4. (0) = 0 and y 2 (0) = 0.2.3 (which is based on high-precision approximations to the Airy functions). (3k . How does this corroborate Theorem 1 in this section? (b) Write the series of terms of odd degree in Eq. (8) and Theorem 3 in Section 3.2.x 2 )y" . Hence the radius of convergence of the series in x is 2. It is interesting to use a computer algebra system to investigate how many terms must be retained in the y 1 . then the singular points of Eq. 35.n 2 )y = 0. (11) in the form 00 00 for the solutions that satisfy the initial conditions y 1 (0) = 1.and y2 -series above to produce a figure that is visually indistinguishable from Fig. 5 . 3.2. 4 . IEIJ Regular Singular Points We now investigate the solution of the homogeneous second-order linear equation A(x) y" + B(x)y' + C(x)y = 0 (1) near a singular point. x 2 y" + xy' + (x 2 . (a) To determine the radius of convergence of the series solution in Example 5. The special combinations . Now it can be proved that each of the functions P(x) and Q(x) either is analytic or approaches ±oo as x ---+ 0. This is a way of saying that P(x) and Q(x) have only "weak" singularities at x = 0. This new equation has the singular point t = 0 corresponding to x = 1 in the original equation. we assume that Eq. A differential equation having x = a as a singular point is easily transformed by the substitution t = x -a into one having a corresponding singular point at 0.t 2 . x = 0 is a singular point of Eq. (1). To investigate the form that a solution of such an equation might take. we get the equation 2(t -t(t + 2dt )2 - d2y + 1)+ n(n + 1)y = dt dy 0. x2 (3) where p(x) = x P(x) and q(x) = x 2 Q(x). we rewrite Eq. as x ----+ 0. if we rewrite the Bessel equation of order n in the form y II 1 + + -y x I ( n-) y = 0. Consequently.2 fails in this case. Types of Singular Points A differential equation having a singular point at 0 ordinarily will not have power series solutions of the form y(x) = L cnxn.(t + 1)2 = -2t. To state this more precisely. (2) where P = BjA and Q = CjA. if P (x) and Q(x) have convergent power series expansions in powers of x on some open interval containing x = 0. let us substitute t = x. Recall that x = 0 is an ordinary point (as opposed to a singular point) of Eq. (2) if the functions P(x) and Q(x) are analytic at x = 0. (1) has analytic coefficient functions and rewrite it in the standard form y'' + P(x)y' + Q(x)y = 0. 1x2 2 we see that P(x) = ljx and Q(x) = 1 . (4) . and Q(x) no more rapidly than 1jx 2 .1 into the Legendre equation of order n. For example. Because and 1 .x 2 = 1 . so the straightforward method of Section 3.(njx) 2 both approach infinity as x ---+ 0. that is.3. (2). (2) in the form y" + p(x) y' x + q(x) Y = O. it has also the singular point t = -2 corresponding to x = -1.3 Regular Singular Points 219 We will restrict our attention to the case in which x = 0 is a singular point of Eq. provided that P (x) approaches infinity no more rapidly than 1/x. We will see presently that the power series method can be generalized to apply near the singular point x = 0 of Eq. (2) provided that either P(x) or Q(x) (or both) approaches ±oo as x ---+ 0. For instance. . Alternatively. (1) and rewrite it in the form in (3). so the singular point x = 0 is regular.· H - H OH o '<'-H oO. 0 = (6) If Po x 2 y 11 + xp(x)y' + q(x)y = = 0 = q 0 .. • If either limit fails to exist or is infinite. then x 0 may be an ordinary point of the differential equation 0 in (3).. Then l'H6pital's rule gives the values Po = xlim .. . we see that x = 0 is not an ordinary point.0 1 lim sin x cosx 1 1.• • H•ON H H Example 2 To investigate the nature of the point x = 0 for the differential equation x 4 y 11 + (x 2 sinx)y + (1.HO. we could write qo and = .= x ---+ 0 x2 x ---+ 0 2x 2 for the limits in (5) and (6).3..= lim . H<U• N " H- -- ---- 000 . ••~ ---· --H"Ooo R..3 Regular Singular Points 221 It may happen that when we begin with a differential equation in the general form in Eq.- "''''" '"N H H ...= ---+ 0 X x ---. Since they are not both zero.-..·· --~~HoO • "·~Ooo h>OO<"" -N H - . In this case p 0 = p (0) and qo = q(O) are simply the constant terms of these polynomials.0 (5) and qo = q{O) = x---+0 lim q(x) = lim x 2 Q(x).. In this case the situation is determined by the limits Po= p(O) = lim p(x) = lim x P(x) x---. .. Otherwise: • If both the limits in (5) and (6) exist and are finite. Remark: The most common case in applications.cosx sinx 1 = lim .. the functions p(x) and q(x) as given in (4) are indeterminate forms at x = 0.cosx)jx 2 X 2 y = 0. x x2 (3) is that the functions p (x) and q (x) are polynomials. Oo.cos x)y = 1 0.. But both limits are finite. then x = 0 is a regular singular point. for the differential equation written in the form y II+ p(x) I+ q(x) 0 -y --y= .. then x = 0 is an irregular singular point.0 x---. (5) and (6). we first write it in the form in (3): y 11 + (sinx)/x X y 1 + (1. x ---. so there is no need • to evaluate the limits in Eqs. (9) -the product of xr and a power series. in which p(x) and q(x) are power series rather than constants.x2 x2 2! x .x6 ++ · ·· ) ] 4 4! 6! These (convergent) power series show explicitly that p(x) and q(x) are analytic and moreover that p 0 = p(O) = 1 and q0 = q (0) = ~.222 Chapter 3 Power Series Methods and 2 x q(x) = 1-cosx = .1 [ 1 . An infinite series of the form in (9) is called a Frobenius series. This turns out to be a very fruitful conjecture. To investigate the possible existence of Frobenius series solutions. named for the German mathematician Georg Frobenius ( 1848-1917). In this case we can verify by direct substitution that the simple power function y(x) = xr is a solution of Eq. it is a reasonable conjecture that our differential equation might have a solution of the form y(x) = xr L n=O 00 CnXn =L n=O 00 CnXn+r = coxr + C!Xr+l + C2Xr+2 +.( 1 . + qo = (8) In the general case.. For instance. • The Method of Frobenius We now approach the task of actually finding solutions of a second-order linear differential equation near the regular singular point x = 0. according to Theorem 1 (soon to be stated formally). indeed. thereby verifying directly that x = 0 is a regular singular point.~ the series in (9) takes the form it is not a series in integral powers of x. (3) reduces when p(x) p 0 and q(x) q0 are constants. Note that a Frobenius series is generally not a power series. have at least one such solution. with r = . (7) if and only if r is a root of the quadratic equation r(r.. every equation of the form in (1) having x = 0 as a regular singular point does.1) + por = = 0. This fact is the basis for the method of Frobenius. we begin with the equation x 2 y" + xp(x)y' + q(x)y = 0 (10) . who discovered the method in the 1870s. The simplest such equation is the constant-coefficient equidimensional equation x 2 y II + poxy'+ qoy = 0 (7) to which Eq. (11) Suppose that Eq. (16) shows that if the Frobenius series y = xr L cnxn is to be a solution of the differential equation in ( 10). then the exponent r must be one of the roots r 1 and r2 of the indicia! equation in ( 16). so it follows that r must satisfy the quadratic equation r(r. it follows that there are two possible Frobenius series solutions. r2.3 Regular Singular Points 223 obtained by multiplying the equation in (3) by x 2 • If x = 0 is a regular singular point. (15) is co[r(r -1) + por +q0 ]xr. (10) now yields [r(r . n=O (12) We may (and always do) assume that c0 ::f. then the coefficient of this term (as well as those of the higher-degree terms) must vanish. But we are assuming that c0 ::f. (12) leads to 00 y' = L Cn(n n=O + r)xn+r. The exponents r 1 and r2 in the possible Frobenius series solutions are determined (using the indicia! equation) by the values Po = p(O) and qo = q(O) that we have discussed.1) + por + qo = 0 (16) of precisely the same form as that obtained with the equidimensional equation in (7).1)coxr + (r + 1)rcl x r+I + · · ·] +[pox+ PIX 2 + · · ·] · [rcoxr. Equation (16) is called the indicia! equation of the differential equation in (10). Our derivation ofEq.3. particularly when the coefficients in the differential equation . we see that the lowest-degree term in Eq. and its two roots (possibly equal) are the exponents of the differential equation (at the regular singular point x = 0). 0 because the series must have a first nonzero term. Termwise differentiation in Eq. IfEq.1 (13) and 00 y" = L Cn(n n=O + r)(n + r- l)xn+r. q(x) = qo + q1x + q2x 2 + q3x 3 + · · · . whereas if r 1 = r2 there is only one possible Frobenius series solution. (14) Substitution of the series in Eqs.I + (r + 1)cix r + · · ·] 0. (10) has the Frobenius series solution 00 Y = LCnXn+r.2. (11) through (14) in Eq. 0. then p(x) and q(x) are analytic at x = 0. so p(x) = Po+ PIX+ P2x 2 + p3x 3 + · · · . If r 1 ::f. (15) + [qo + q1x + · · ·] · [coxr + c1xr+I + · · ·] = Upon multiplying initial terms of the two products on the left-hand side here and then collecting coefficients of x r. the second solution cannot be a Frobenius series. In practice. (15) is to be satisfied identically. (12) through (14) in the differential equation.224 Chapter 3 Power Series Methods in the original form in (I) are polynomials. with roots r 1 = ~ and r 2 = -1.~) = 0. .~ . we need only replace x r 1 with lx lr1 to obtain a solution for x < 0. We also will seek solutions only for x > 0. essentially the same method as was used to determine coefficients in power series solutions in Section 3. If the exponents r 1 and r 2 are complex conjugates. Hence the indicia] equation is r(r . n=O 00 • Frobenius Series Solutions Once the exponents r 1 and r2 are known. We will restrict our attention here to the case in which r 1 and r2 are both real.x) 2 x2 y = 0. The two possible Frobenius series solutions are then of the forms Y1 (x) = x 112 L anxn n=O 00 and Y2(x) = x - 1 L bnxn.2.~ = (r + l)(r. and thus see that Po = ~ and qo = .1) + ~r. then there always exist two linearly independent Frobenius series solutions. the simplest way of finding p 0 and q0 is often to write the equation in the form Then inspection of the series that appear in the two numerators reveals the constants Po and qo. Once such a solution has been found. Example 3 Find the exponents in the possible Frobenius series solutions of the equation Solution We divide each term by 2x 2 (1 + x) to recast the differential equation in the form y' y" + 1(1 2 + 2x + x 2 ) x + -l(l . The following theorem is proved in Chapter 4 of Coddington's An Introduction to Ordinary Differential Equations. the coefficients in a Frobenius series solution are determined by substitution of the series in Eqs.~ = r 2 + ~r. then there exists a second linearly independent solution for x > 0 of the form yz(x) = xr2 L bnxn n=O 00 (bo =1.r 2 is a positive integer. Example 4 Find the Frobenius series solutions of 2x 2 y" + 3xy' 3 (x 2 + 1)y = 0. and that p 0 = ~ and q0 = . It turns out that. (10) Let p > 0 denote the minimum of the radii of convergence of the power series 00 00 p(x) = LPnXn n=O and ~ q(x) = Lqnxn .4. then there can exist only one Frobenius series solution. of the indicia! equation r(r.1) + (a) For x > 0.r 2 is neither zero nor a positive integer. Examples 4 through 6 illustrate the process of determining the coefficients in those Frobenius series solutions that are guaranteed by Theorem 1. the Frobenius series we = -!- . (18) and (19) are each at least p. The coefficients in these series can be determined by substituting the series in the differential equation x 2 y" + xp(x)y' + q(x)y = 0. These exceptional cases are discussed in Section 3. (1 0) of the form 00 YI(X) = Xq LanXn (ao =f. n=O Let r 1 and r2 be the (real) roots.3. with r 1 por + qo = 0. there exists a solution of Eq. Then r2 . (20) Solution First we divide each term by 2x 2 to put the equation in the form in (17): y" + ly' + x _l _ lx2 2 x2 2 y = 0. if r 1 . Because p (x ) ~ and q(x ) = ~x 2 are polynomials.~ .0) (19) corresponding to the smaller root r 2 • The radii of convergence of the power series in Eqs.3 Regular Singular Points 225 THEOREM 1 Frobenius Series Solutions Suppose that x = 0 is a regular singular point of the equation x 2 y" + xp(x)y' + q(x)y = 0. ( 19) corresponding to the smaller root r2 . (21) We now see that x = 0 is a regular singular point. 0) (18) n=O corresponding to the larger root r 1• (b) If r 1 . We have already seen that if r 1 = r2 . there may or may not exist a second Frobenius series solution of the form in Eq. 1)cnxn+r + 3 L(n + r)CnXn+r n=O 00 00 . whereas with r 2 = -1 it becomes a recurrence relation for the series for y 2 • When we substitute 00 00 Y = LCnXn+r. In this example. n=O y' = L(n n=O + r)cnxn+r-l' and 00 y" = L)n + r)(n + rn=O 1)cnxn+r-2 in Eq.~ ) co = 0. A good standard practice is to shift indices so that each exponent will be the same as the smallest one present. rather than Eq. Following our standard practice.226 Chapter 3 Power Series Methods obtain will converge for all x > 0. This gives 2 I)n n=O 00 00 + r)(n + r. n=O n=O (22) At this stage there are several ways to proceed.L CnXn+r+2 .L Cn Xn+r = 0. so we must treat n = 0 and n = 1 separately. They do not differ by an integer. n=2 n=O (23) The common range of summation is n ~ 2. (20). so Theorem 1 guarantees the existence of two linearly independent Frobenius series solutions.L CnXn+r = 0. We will then get a recurrence relation that depends on r. the terms corresponding to n = 0 will always give the indicia! equation [2r(r .~ = (r- D(r + 1) = 0. it is more efficient to begin by substituting y = xr L CnXn. we shift the index of summation in the third sum by -2 to reduce its exponent from n + r + 2 to n + r. so the exponents are r 1 = ~ and r2 = -1. .l)cnxn+r + 3 L(n + r)cnxn+r n=O 00 00 .1) + ~r. The indicia} equation is r(r .1) + 3r - 1]co = 2 (r 2 + ~ r .L Cn. (21)-we get 00 00 2 L(n n=O + r)(n + r. Rather than separately substituting 00 00 Yt = x 112 Lanxn n=O and Y2 = x-l L bnxn n=O in Eq.2Xn+r . With the value r 1 = ~ it becomes a recurrence relation for the series for y 1. (20)-the original differential equation. 1tco = . it is in fact a power series. Eq. denoted by lo(x). the indicia! equation is r(r. 22. it will not be a Frobenius series. 00 Thus we obtain only the single exponent r = 0. Substituting n 6 in Eq. Because p 0 = 1 and q 0 = 0. the Bessel function of order zero of the first kind.4. We will derive that solution in Section 3. and the term for xn yields the recurrence relation Cn. (28).1) = + r = r 2 = 0. (28) Solution In the form of ( 17). and so there is only one Frobenius series solution y(x) = x0 L CnX n n=O of Eq. Thus (30) In this example we have not been able to find a second linearly independent solution of Bessel's equation of order zero. (29). 4. 42 . we get = 2. the pattern is ~n = - ( -l)nc0 ( . we see that Cn = 0 whenever n is odd. so our series will converge for all x > 0. Hence x = 0 is a regular singular point with p(x) 1 and q(x) = x 2 . The term corresponding to x 0 gives 0 = 0: no information... (2n)2 22n(n!)2 The choice c0 = 1 gives us one of the most important special functions in mathematics.n2 for n 2:: 2.228 Chapter 3 Power Series Methods Example 5 Find a Frobenius solution of Bessel's equation of order zero. the result is 00 00 00 Ln(n.2Xn = 0.. (29) Because c 1 = 0. • . Thus we substitute y = L CnXn in (28). and Evidently. The term corresponding to x 1 gives c 1 = 0. We combine the first two sums and shift the index of summation in the third by -2 to obtain 00 00 L n 2cnXn n=O +L n=2 Cn . (28) becomes y II + -y + -2y = x x 1 I X2 0.1)cnxn n~ + L:ncnxn + LCnXn+2 = n~ n~ 0.2 Cn = . which differ by an integer.3.r2 is a positive integer. we had obtained an equation such as 0 · c 1 = 3. The r (r . (31).2)cnxn. our computations will simultaneously yield both Frobenius series solutions. if r 1 .2 + Cn-2Xn. then Theorem 1 guarantees only the existence of the Frobenius series solution corresponding to the larger exponent r1• Example 6 illustrates the fortunate case in which the series method nevertheless yields a second Frobenius series solution. we begin anew with the larger exponent r = r 1 to obtain the one Frobenius series solution guaranteed by Theorem 1. If the second solution does not exist. 2 and q0 so we see that x indicia} equation 0 is a regular singular point with Po = 0.3 Regular Singular Points 229 When r 1 . .1) + 2r = r (r + 1) = 0 has roots r 1 = 0 and r2 = -1.l)(n. this would have told us that no second Frobenius series solution could exist. which can be satisfied for no choice of c 1 . (31) Solution In standard form the equation becomes II y + -y +y= x x2 2 I X2 0. Example 6 Find the Frobenius series solutions of xy" + 2y' + xy = 0. n=O n= 2 L 00 (32) = 0 and n = 1 reduce to 0 · co =0 and 0 · c 1 = 0. Hence we have two arbitrary constants c 0 and c 1 and therefore can expect to find a general solution incorporating two linearly independent Frobenius series solutions. In this case when r 1 . for n = 1.2 + Lcnxn = 0. As you will see. If. the recurrence relation will then tell us whether or not a second Frobenius series solution exists. The case in which the second solution is not a Frobenius series will be discussed in Section 3.r2 Is an Integer Recall that.l)cnxn. This gives 00 L(n.l)cnXn.2 + 2 L(n. n~ n~ n~ We combine the first two sums and shift the index by -2 in the third to obtain L The cases n 00 n(n .4.r 2 is an integer. it is better to depart from the standard procedure of Example 4 and begin our work with the smaller exponent. Hence we begin by substituting y = x-1 L CnXn = L CnXn-1 n=O n=O 00 00 00 00 in Eq.2 = 0. If it does exist. Show that (1) F(l. 2. But we saw in Example 6 of Section 3.4 Method of Frobenius: The Exceptional Cases with c0 =I= 0.3. (d) The series in (36) is known as the hypergeometric series and is commonly denoted by F(a. For applications.3 guarantees that Eq. this is the most important second-order linear differential equation with variable coefficients. 1.3 we derived the indicial equation by substituting the power series p (x) = L PnXn and q (x) = L qnxn and the Frobenius series y(x) = xr L CnXn = L CnXn+r n=O n=O 00 00 (co =I= 0) (3) in the differential equation in the form x 2y" + xp(x)y' + q(x)y = 0. with N a positive integer. x) = . (c) Conclude that with c0 part (b) is = 1 the series in 1 1-x (the geometnc series). The Nonlogarithmic Case with r 1 = r 2 +N In Section 3. If the roots r 1 and r2 of the indicia! equation </J(r) = r(r- 1) + por + qo = 0 (2) do not differ by an integer. and fJn and Yn are defined similarly.0 for n ~ 0. 1. (thebinomialseries). and thus there can be only one Frobenius series solution. ~. If r 1 = r 2. 1. then it is possible that a second Frobenius series solution exists. We consider now the more complex situation in which r 1 . F(-k. 1. 1. and x = 0 is a regular singular point. (iii) xF (iv) (k. 0 (ii) xF(l. then Theorem 1 of Section 3. after collection of the coefficients of like powers of x. {3.-x) = (l+x)k Method of Frobenius: The Exceptional Cases We continue our discussion of the equation y" + p(x) y' + q(x) Y = x2 x 0 (1) where p(x) and q(x) are analytic at x = 0. We will also see that it is possible that such a solution does not exist. (4) The result of this substitution. (1) has two linearly independent Frobenius series solutions. then there is only one exponent available. In fact.r 2 is an integer. the second solution involves ln x when it is not a Frobenius series. 1. Show that the recurrence relation for this series is (a+ n)(fJ + n) c n+l - 233 - (y + n)(1 + n) cn where an = a(a + 1)(a + 2) ···(a+ n . is an equation of the form L Fn(r)xn+r = 0 n=O 00 (5) . As you will see in Examples 3 and 4.3 that if r 1 = r 2 + N. -x) (36) = ln(l + x). -x 2 ) = tan.1) for n ~ 1.1 x. y . x). these exceptional cases occur in the solution of Bessel's equation. . If it happens that then we can choose eN arbitrarily and continue to determine the remaining coefficients in a second Frobenius series solution. For in this case ¢(r2 + N) = 0. . .. Cn . (9) for en and continue to compute successive coefficients in the Frobenius series solution corresponding to the exponent r 1 • But when we use the smaller exponent r2 . . . (10) is not satisfied with any choice of eN. c1. so the indicia! equation is ¢ (r) = r (r - 1) + 6r = r 2 + 5r = 0 (12) . Cn-l· Although the exact formula is not necessary for our purposes.234 Chapter 3 Power Series Methods in which the coefficients depend on r.1) + por + qoJco = ::1 ¢(r)co. . ••. it follows that the exponent r and the coefficients c 0 . But if it happens that then Eq. there is a potential difficulty in computing eN. ••• . in this case there cannot exist a second Frobenius series solution corresponding to the smaller exponent r2 • Examples 1 and 2 illustrate these two possibilities. (7) Here Ln is a certain linear combination of c0. then the coefficient ¢(r 1 + n) of en will be nonzero for every n ~ 1 because ¢ (r) = 0 only when r = r 1 and when r = r2 < r 1. also. en must satisfy the equation ¢(r + n)cn + Ln(r. It turns out that the coefficient of xr is Fo(r) = [r(r. we therefore can solve Eq.J have been determined. (9) becomes (10) At this stage co. Cn . c 1 . Example 1 Consider the equation (11) Here Po = 6 and q 0 = 0. c 1..1· Now suppose that r 1 = r2 + N with N a positive integer. Cn-1) = 0. (6) which gives the indicia! equation because c0 the coefficient of xn+r has the form 0 by assumption. for n > 1. (9). ••• . •. Co.. c 1. c1. c 1.. (9) This is a recurrence relation for Cn in terms of co. (4 ). CN-I have already been determined. If we use the larger exponent r 1 in Eq. Once c0 . so Eq. . it happens that n-1 Ln = :L)Cr + k)Pn-k + qn-kkk· k=O (8) Because all the coefficients in (5) must vanish for the Frobenius series to be a solution of Eq. The radii of convergence of the power series of this theorem are all at least p. Theorem 1 (next) is a summation of the preceding discussion and also tells where the series in (33) and (34) converge.r2 = N.242 Chapter 3 Power Series Methods In our derivation of Eqs. then Eq. of the indicia! equation r(r. a positive integer. with r 1 ~ r2. (4) Let p > 0 denote the minimum of the radii of convergence of the power series 00 00 p(x) = LPnXn n=O and q(x) = Lqnxn. respectively-we have said nothing about the radii of convergence of the various power series that appear. so the logarithmic term may or may not actually be present in this case. THEOREM 1 The Exceptional Cases Suppose that x = 0 is a regular singular point of the equation x 2y" + xp(x)y' + q(x)y = 0. (33) and (34)-which exhibit the general form of the second solution in the cases ri = r2 and ri .1) + por + qo = 0. then Eq. (4) has two solutions y 1 and yz of the forms 00 YI(x) = Xr Lanxn n=O 1 (ao =I 0) (35a) and 00 Y2(x) = YI(x )lnx +xq+l Lbnxn. (36b).(x)lnx +xr Lbnxn. . The coefficients in these series (and the constant C in Eq. n=O Let ri and r2 be the roots. (a) If ri = r2. we restrict our attention to solutions for x > 0. n=O 2 (36b) In Eq.r2 = N > 0.3. As in Theorem 1 of Section 3. (36b)) may be determined by direct substitution of the series in the differential equation in (4). n=O (35b) (b) If r1 . (4) has two solutions YI and Y2 of the forms 00 YI(X) = x rl L:anXn n=O (ao =I 0) (36a) and 00 Y2(X) = c y. bo =I 0 but C may be either zero or nonzero. Tf . The power series in (45) converges for all x. 1 1 1 1 c6=c4+ =1+-+3 2 3' Cg = C6 + 4 = 1 + 2 + 3 + 4. Noting also the alternation of sign.4 x2 128 + . Finally. Substitution of (43) in (42) gives ( -2)( -2t(2n) 22n(n!)2 which boils down to the extremely simple recurrence relation C2n = C2n . Evidently. We note the presence of 22n (n !) 2 on the righthand side in (42). with n = 1 in (43)..2 + -. and their solution often requires a bit of ingenuity. Note the "nonhomogeneous" term (not involving the unknown coefficients) on the right-hand side in (42). 1 1 1 (44) where we denote by Hn the nth partial sum of the harmonic series L (1 I n). we get c2 = 1.. we make the substitution (43) in the expectation that the recurrence relation for c2n will be simpler than the one for bzn· We chose ( -l)n+l rather than ( -l)n because b2 = ~ > 0. The most commonly used linearly independent [of J0 (x)] second solution is Yo(x) = .··· 13824 11x 6 (45) of Bessel's equation of order zero. in conjunction with the coefficient (2n) 2 on the left-hand side. we substitute (43) and (44) in (39) to obtain the second solution y 2 (x) = J0 (x) lnx +"' n=l oo ( -l)n+l Hnx2n ~ 22n(n!) 2 3x 4 = Jo(x) lnx +. Such nonhomogeneous recurrence relations are typical of the exceptional cases of the method of Frobenius. n 1 1 Thus C4 = C2 + 2 = 1 + 2. 1 1 1 1 and so on.244 Chapter 3 Power Series Methods from (40).(y .ln2)yi Tf 2 +- 2 y2. we are induced to think of b 2n as something divided by 2 2n (n !) 2 .. c2n =1+-+-+···+-=H 2 3 n n. The usual strategy depends on detecting the most conspicuous dependence of b2n on n. keeping in mind that the coefficients of odd subscript are all zero. If we substitute y = 2. The equation itself appears in a 1764 article by Euler on the vibrations of a circular drumhead. 2nd ed.248 Chapter 3 Power Series Methods in Eq. The Case r = p formula > 0 If we use r = p and write am in place of em. (2) is left to the reader (Problem 6). (d) Take b 2 = ~ and substitute ( -1)"Czn 22"(n. 26 . 3(p + l)(p + 2)(p + 3) . The first few even coefficients are a2 - ao - 2(2p + 2) . (3) Because a 1 = 0.1] b2n+2 + bzn = (~ 1)"(2n + 1) 2 "(n + l)!n! b _ 2" - (63) C2n+2.. 1. (2) yields the recursion am =- am. The verification of Eq. Such functions first appeared in the 1730s in the work of Daniel Bernoulli and Euler on the oscillations of a vertically suspended chain. The standard source of information on Bessel functions is G.:. (63) to obtain 1 (b) Deduce from Eq.C2n = -- n+1 + -. then Eq.- - ao 22 (p + 1) .2 m(2p + m) . and Fourier used Bessel functions in his classical treatise on heat (1822). (62) that C = -1 and that bn = 0 for n odd. (1) Its solutions are now called Bessel functions of order p. with roots r = ±p. then b 2n is determined for all n > 1.: cm x m+r in Eq. who was investigating the motion of planets. N. Its 36 pages of references. a2 ao a 4 = . which cover only the period up to 1922. Note that if b2 is chosen arbitrarily. (c) Next deduce the recurrence relation [ (2n + If. n 1 for n . Watson's A Treatise on the Theory of Bessel Functions. (Cambridge: Cambridge University Press.-2-(p_+_1_)(_p_+_2_)' 4 a6 = a4 6(2p + 6) = - ao .1)!n! (e) Note that c2 = 1 = H 1 + H 0 and deduce that liD Bessel's Equation We have already seen several cases of Bessel's equation of order p ~ 0.4(2p + 4) = -2 -. (1). Bessel ( 1784-1846). it follows that am = 0 for all odd values of m. give some idea of the vast literature of this subject. But their general properties were first studied systematically in an 1824 memoir by the German astronomer and mathematician Friedrich W.p 2 = 0. 1944). Bessel's equation in (1) has indicia! equation r 2 . 2 . we find in the usual manner that Ct = 0 and that (2) form ~ 2. · (p + m) = 00 ( -l)m x2m+p (4) If p 0 this is the only Frobenius series solution. it is the The Case r = -p 0 If we user = . we take bm = 0 form odd and define the coefficients of even subscript in terms of bo by means of the recursion formula bm. k(k. then no value of bm can satisfy this equation.. (5) is simply 0 · bm + bm_2 = 0. Hence if pis not a positive integer. Thus in this case we obtain the second solution 00 ( -l)m x 2m. .p Y2(x ) = bo "\" £=o 22mm! (-p + 1)(-p + 2) .. (7) The series in (4) and (7) converge for all x > 0 because x = 0 is the only singular point of Bessel's equation.2p) bm = - m ::::: 2. Eq. For suppose that p = k/2 where k is an odd positive integer. · (-p + m) . whereas the leading term in y2 is b0 x .5 Bessel's Equation 249 The general pattern is ( -l)mao 2 2mm! (p a2m = + I)(p + 2) · · · (p + m) . but y 2 (x) ----+ ±oo as x ----+ 0. 22mm! (p + I)(p + 2) . Thus if bm-2 =I= 0. whereas b 1 = 0. If p > 0. We see that there is a potential difficulty if it happens that 2 p is a positive integer-that is. i. if p is either a positive integer or an odd positive integral multiple of i· For then when m = 2p. except with p replaced with . < = 1 as well.k)bk + bk-2 = 0.p. we see that (6) will lead to the same result as that in (4). Eq. and this equation will hold because bk = bk. The crucial step is the kth step.3. so it is clear that y 1 and y 2 are linearly independent solutions of Bessel's equation of order p > 0.2 ' m(m .2p)bm + bm-2 = 0 (5) for m ~ 2. Then we need only choose bm = 0 for all odd values of m . so with the larger root r = p we get the solution Yt (x) = ao £=o "\" .P . (6) In comparing (6) with (3). (2) takes the form m(m. with a0 function J0 (x) we have seen before. Hence y 1 (0) = 0.p and write bm in place of Cm. then the leading term in y 1 is a 0 xP. But if p is an odd positive integral multiple of we can circumvent this difficulty.2 = 0. (9) and (10). r(2) = 2! . for instance. we note first that f(1) = 1 0 00 e-t dt = lim b-'>00 [-e-r] b 0 = 1. X . [See. (11) An important special value of the gamma function is (12) where we have substituted u 2 fort in the first integral. r(x + 1) = xr(x) . Calculus: Early Transcendentals. (10) This is the most important property of the gamma function. To see the way in which r(x) is a generalization of n!. r(4) = 3. r(3) = 3!. (9) Then we integrate by parts with u = tx and dv = e-r dt: =X (lim rb e-t tx-l dt) . b-+ oo lo that is. If we combine Eqs. and in general that r(n + 1) = n! for n ~ 0 an integer. which is defined only if n is a nonnegative integer. but is far from obvious. 2008). the fact that 1 o oo e . NJ: Prentice Hall.4 of Edwards and Penney. then r(x ) = r(x + 1).] Although r (x ) is defined in (8) only for x > 0.u2 d ~ u= 2 is known.250 Chapter 3 Power Series Methods The Gamma Function The formulas in (4) and (7) can be simplified by use of the gamma function r (x). If . 7th edition (Upper Saddle River. r(l) = 1!. Example 5 in Section 13. r(3) = 2. The gamma function is a generalization for x > 0 of the factorial function n!.1 < x < 0. we see that r(2) = 1 . which is defined for x > 0 by (8) It is not difficult to show that this improper integral converges for each x > 0. we can use the recursion formula in (10) to define r(x) whenever x is neither zero nor a negative integer. The graph of the extended gamma function. If pis not an integer. respectively (see Problem 27). -2). (13) through (15) to get the correct solutions for x < 0. Bessel Functions of the First Kind FIGURE 3. (p + 2)(p + l)r(p + 1) by repeated application of Eq. this is one of the finest expositions in the entire literature of mathematics.1. if p > 0 is not an integer. Indeed. the zeros of the . As suggested by Fig.5. The same formula may then be used to extend the definition of r (x) to the open interval (-2.2. where p (p > 0. xP must be replaced with lx IP in Eqs. 3. then to the open interval ( -3. Thus and The graphs of J0 (x) and 11 (x) are shown in Fig. (13) gives ln(x ) .3. (-l)m (X )2m+n m=O m! (m + n)! (16) 2 for the Bessel functions of the first kind of integral order. 3. (10). The student who would like to pursue this fascinating topic further should consult Artin's The Gamma Function (New York: Holt.. If p = n . 1964)..5. If we choose a 0 r(p = 1/[2Pr(p + 1)] in (4). 3. and so on. In only 39 pages. In a general way they resemble damped cosine and sine oscillations. if you examine the series in (17). and note that + m + 1) = + m)(p + m.p + m + 1) 2 (X )2m-p (14) of Bessel's equation of order p. -1). a nonnegative integer.~ _ f--.5 Bessel's Equation 251 the right-hand side is defined because 0 < x + 1 < 1. you can see part of the reason why J0 (x) and cos x might be similar-only minor changes in the denominators in (17) are needed to produce the Taylor series for cosx.1. Rinehart and Winston.5. then Eq. we have the general solution (15) for x > 0.5. we choose bo obtain the linearly independent second solution = 1/[2-Pr(-p + 1)] in (7) to J_p(x) = ~ oo ( -l)m m! r(.2.1) . we can write the Bessel function of the first kind of order p very concisely with the aid of the gamma function: -l)m lp(x) = ~ m!r(p + m 00 ( + 1) 2 (X )2m+p · (13) Similarly. The graph of the gamma function thus extended is shown in Fig. (19) Bessel Functions of the Second Kind The methods of Section 3. It turns out that J P (x) is an elementary function if the order p is half an odd integer.. the nth zero of 1 1(x) is approximately (n + !} 7T.!} rr.1735 13.1)! m_!_x_n----:2-m-- with the notation used there.7915 14.9270 7. FIGURE 3. For n large.2(x) = {2 cosx. the nth zero of J0 (x) is approximately (n. You can see the way the accuracy of these approximations increases with increasing n by rounding the entries in the table in Fig.6537 11.4706 (n + D1r 3.3.3237 16. For instance.6537.5.2(x) = y-. (13) and (14).6394 11. Yn (x) is called the Bessel function of the second kind of integral order n 2:: 0.252 Chapter 3 Power Series Methods y ·n. Here.9226 nth Zero of Jt(x) 3. 3.2. Thus the interval between consecutive zeros of either lo(x) or J 1(x) is approximately rr-another similarity with cosx and sinx.4048 5. y-. If n = 0 then the first sum in (20) is taken to be zero.5201 8. 8.. 2.. and 11. functions J0 (x) and 11(x) are interlaced-between any two consecutive zeros of lo(x) there is precisely one zero of 1 1(x) (Problem 26) and vice versa. 5.8317 7. (21) It is important to note that Yn(x) --+ -oo as x --+ 0 (Fig. A very complicated generalization of Example 3 in that section gives the formula Yn(X) = 2 ( 7T y +In 2 ln(X)- X) 1 7T L __ m=O n-l 2n . Thus if y(x) is a continuous solution of Bessel's equation of order n.4978 8.0156 10.7915.4 must be used to find linearly independent second solutions of integral order.5. Zeros of J0 (x) and J 1 (x).0686 10.5. The graphs of the Bessel functions J0 (x) and 1 1 (x) . The first four zeros of J0 (x) are approximately 2. the results can be recognized (Problem 2) as l1. it follows that y(x) = cln(x) .5201.3562 5. 3.3 to two decimal places..3518 16.2102 13. respectively.4934 X 1 2 3 4 5 2.m(n - 2 m.9309 FIGURE 3.4048. {2 sinx and l-1.':""'41r ( ·" 'i) . The general solution of Bessel's equation of integral order n is y(x) = c1ln(x) + c2Yn(x). on substitution of p = ~and p = -~in Eqs. Hence c2 = 0 in Eq. (21) if y(x) is continuous at x = 0.7810 14.4).5. Subtraction of Eq.lp-1 (x). The identities in Eqs.lo(x)+C. (26). Using first p = 2 and then p = 1 in Eq. X 2p (27) it can be used to express Bessel functions of large negative order in terms of Bessel functions of numerically smaller negative orders. (22) gives J xJ0 (x)dx = x11 (x) +C. Eq. X 2p (26) which can be used to express Bessel functions of high order in terms of Bessel functions of lower orders. Example 3 To antidifferentiate x h (x ). In the form lp-1 (x) = -lp(x). (22) through (27) hold wherever they are meaningfulthat is. they hold for all nonintegral values of p. (25) from Eq. (23) with p = 1.lp+ l (x). • Jl1(x) .1h(x)dx = .lo(x) so that J3(x) = -~lo(x) + (:2 - 1) l1(x ). (23) gives J Example 2 J 1 (x)dx= .l1(x) = X X 4 4[2 X -J1(x). we first note that J x .1 J 1 (x) +C by Eq. Eq. we get J3(x) = -h(x). with p = 0. With similar manipulations every Bessel function of positive integral order can be • expressed in terms of J0 (x) and 1 1 (x).x . In particular. We therefore write . whenever no Bessel functions of negative integral order appear. Similarly.254 Chapter 3 Power Series Methods and (25) Thus we may express the derivatives of Bessel functions in terms of Bessel functions themselves. (24) gives the recursion formula lp+l (x) = -Jp(x). Example 1 With p = 0. (see Fig.2 and hence has an infinite sequence of positive zeros Yni· Yn2• Yn 3• . y(L) = 0 (31) on the interval [0. Because Yn(x)---+ -oo as x---+ 0 but ln(O) is finite. du = 2xdx. -xJI(x) and v = -x-I JI (x). A Stegun. 0 (29) with general solution y(t) = c 1 ln(t) Eq. (28). This gives J xh(x)dx = +2 J JI(x)dx = -xJI(x). (32) for n ~ 8 and k ~ 20 are tabulated in Table 9. L] . 3. ln (x) oscillates rather like JI (x) in Fig.5 of M. 3.3. As we will see in Chapter 9. then the differential equation in (31) is that in Eq. . (28) is Hence the general solution of (30) Now consider the eigenvalue problem x 2y" + x y' + (Ax 2 - n 2 ) = 0. L].2J0 (x) + C.. • (28) The Parametric Bessel Equation The parametric Bessel equation of order n is where a is a positive parameter. If we write A = a 2 ..5. this equation appears in the solution of Laplace's equation in polar coordinates. Abramowitz and I. It is easy to see (Problem 9) that the substitution t = ax transforms Eq. It follows that the kth positive eigenvalue of the problem in (31) is Ak = (ak) 2 = (y2~) 2 . with the aid of the second result of Example 1. (28) into the (standard) Bessel equation d 2y dy t 2+t- dt 2 dt + (t 2 -n 2)y = + c2Yn(t) . We seek the positive values of A for which there exists a nontrivial solution of (31) that is continuous on [0. (32) For n > 1. (30). (33) and that its associated eigenfunction is (34) The roots Ynk of Eq.5 Bessel's Equation 255 and integrate by parts with dv =x-I h(x) dx. Handbook of Mathematical Functions (New York: Dover. so its general solution is given in Eq. The endpoint condition y(L) = 0 now implies that z = a L must be a (positive) root of the equation ln(Z) = 0. 1965). Thus y(x) = ciln(ax). the continuity of y (x) requires that c2 = 0.5.6). ~ y-. 25.. and x large. 24. Use Eqs.~. {2 cos [x.3. liD Applications _?f B~ssel F~nctions The importance of Bessel functions stems not only from the frequent appearance of Bessel's equation in applications. It can be shown that ln (x) = - (b) If X is so large that (p 2 jx 2 is negligible./'I1i[ and a = (2n + 1)n I 4 in (35) yield the best approximation to ln (x) for x large: ln(x) = -1 12n 0 eixsine 2n dO. (22) and (23) and Rolle's theorem to prove that between any two consecutive zeros of ln (x) there is precisely one zero of ln+l (x). but also from the fact that the solutions of many other second-order linear differential equations can be expressed in terms of Bessel functions.112 cos(x -a) with C and a constants. Explain why this constitutes a proof. Yn(x) ~ y-. {2 sin [x. Deduce from Problem 22 that lo(x) Asymptotic Approximations It is known that the choices C = . {2 cos (x. Prove that 11(x) 257 yields =- 11n 0 cos(O . 12n 0 eixsin& de= 1n 0 (eixsin& + e-ixsine) de. Use a computer algebra system to construct a figure illustrating this fact with n = 10 (for instance). ][ = cx. These are asymptotic approximations in that the ratio of the two sides in each approximation approaches unity as x-+ +oo. Explain why this does not suffice to prove the preceding assertion. J 0 (x) and Y0 (x) ~ {f sin (x .~(2n + l)n J.01 y. With n ~ 2. (a) Show that the substitution y = x.x sin B) dO z" + ( 1 - p'x~ l) n z= 0 ][ by showing that the right-hand side satisfies Bessel's equation of order 1 and that its derivative has the value J{ (0) when x = 0.112 z in Bessel's equation of order p.. (2) Then a routine but somewhat tedious transformation (Problem 14) of Eq.x sin 0) dO. (0). To see how this comes about.. we begin with Bessel's equation of order p in the form (1) and substitute w = x . (36) (Suggestion: Show first that Similarly. z= kx/3.. Explain why this suggests (without proving it) that if y(x) is a solution of Bessel's equation..112 (A cosx 11n 0 + B sinx) (35) cos(ne . ( 1) yields .rr) if x is large.. ~ y-. 27.~(2n + l)n J. then y(x) ~ x. show that the right-hand side satisfies Bessel's equation of order n and also agrees with the values ln (0) and 1.) 26..~n) (37) then use Euler's formula. then the latter equation reduces to z" + z ~ 0..6 Applications of Bessel Functions 23. In particular. 1/ 2 [ CJ J1. (4) It is a simple matter to solve the equations in (4) for a=-- 1-A 2 ' q (3 = 2' j(l. This establishes the following result. i· = i. and p= Under the assumption that the square roots in (5) are real. C a= (3 = 2.A) 2 q (5) 4B k=2~. C = {Pk 2.~. C. (3) is Solutions in Bessel Functions 0.2 (z) = {2' sin z v~ and 1.312 (A cos~+ B sin : .258 Chapter 3 Power Series Methods that is. (7) can be written in the elementary form y(x) = x .5 that J1 . then J_p is to be replaced with Yp. v~ 2 ). (7) is y(x) = x. • If we recall from Eq. Then the equations in (5) give Thus the general solution in (6) ofEq. q and q = 2(3. (3) where the constants A. (3) is where w(z) = CJJp(Z) + c2Y-p(Z) (assuming that p is not an integer) is the general solution of the Bessel equation in (1). and q = 4.2 {ix 2) + c2l.2a. (3. B = . If p is an integer.r. and q are given by A= 1. q Eq. Example 1 Solve the equation 4x 2 y" + 8xy' + (x 4 - 3)y = 0. (7) with Eq.A) 2 ~ 4B. (3). it follows that the general solution of Eq. then the general solution (for x > 0) of i= (6) where a. i. k = and p = -i. B = a2 - {Pp 2. {2' cos z. (19) of Section 3. we rewrite the former as x 2 y" + 2xy' + ( -~ + ix 4 ) y = 0 and see that A = 2. and p are given by the equations in (5). THEOREM 1 If C > 0. (7) Solution To compare Eq.1. and (1 . k.2 (Z) = we see that a general solution of Eq.2 {ix 2)]. B. perhaps. and g is gravitational acceleration. The buckling column. and q = 3. being nudged laterally just a bit by a passing breeze).y .6. we now consider the problem of determining when a uniform vertical column will buckle under its own weight (after. (8) Solution First we rewrite the given equation in the form This is the special case of Eq. where E is the Young's modulus of the material of the column.6. we assume that the bottom is rigidly imbedded in the ground.1.6 Applications of Bessel Functions ~· ····· 259 Example 2 Solve the Airy equation y" + 9xy = 0. B(L) = 0. perhaps in concrete. (9) FIGURE 3. We take x = 0 at the free top end of the column and x = L > 0 at its bottom. B(L) = 0. fJ = ~. p is the linear density of the column. 3. and p = ~. fJ = ~. C = y 2 . The equations in (5) give a = ~.. k = ~ y. (3) with A = B = 0. (10) We will accept (9) and ( 10) as an appropriate statement of the problem and attempt to solve it in this form. Thus the general solution of Eq. and q = 3. C = 9. k = 2. Denote the angular deflection of the column at the point x by B(x).EI' (11) we have the eigenvalue problem 8" + y 2 xB = 0. It follows from the equations in (5) that a = ~. I is its crosssectional moment of inertia.3.gp . The differential equation in (12) is an Airy equation similar to the one in Example 2. see Fig. (12) The column can buckle only if there is a nontrivial solution of (12).ected vertical position. From the theory of elasticity it follows that EIdx 2 d2 B + gpxB = 0. 8'(0) = 0.1. So the general solution is (1 3) . It has the form of Eq. otherwise the column wiii remain in its undefl. (8) is • Buckling of a Vertical Column For a practical application. With A.2 . and p = ~. (3) with A = B = 0. For physical reasons-no bending at the free top of the column and no deflection at its imbedded bottom-the boundary conditions are B' (0) = 0. so (14) The endpoint condition ()(L) = 0 now gives (15) Thus the column will buckle only if z = ~ y L 312 is a root of the equation l-I f3 (Z) = 0.5) is shown in Fig. From this it is clear that the endpoint condition ()' (0) = 0 implies that c 1 = 0.. For instance.. The graph of J-l j3 (Z). (see Problem 3 of Section 3.x)=O.• 2) FindRoot[BesselJ[-1/J. Most technical computing systems can find roots like this one. {x. 2) (Maple) (Mathematica) (MATLAB) yield the value z1 = 1. each of the computer system commands fsolve(BesselJ(-1/J.2}] fzero ( 'besselj ( -1/3.2. we finally get EI ) goA (17) . ) 180 .6..86635 and p = oA. where 8 is the volumetric density of the material of the column and A is its cross-sectional area. where we see that the smallest positive zero z 1 is a bit less than 2. we substitute p = ±~ in lp(x) =I.260 Chapter 3 Power Series Methods In order to apply the initial conditions.x]==O. oo ( -l)m m! f(p + m + 1) 2 (X )2m+p ' and find after some simplifications that + y I/ 3 f cz31 /3 n) (1 - y2x3 6 + y4x6 . The shortest length L 1 for which the column will buckle under its own weight is Ll ~ c~r ~ [3~~ L1 ~ (1 .6. x. The graph of (16) 5 10 15 z FIGURE 3. 3.86635 (rounded accurate to five decimal places). 1 .2.986) ( - (!:rr 1 3 / If we substitute z 1 ~ 1. x)'. 18. and Hobson in the March 1985 issue of Mathematics Magazine (Vol.= A(x )l dx + B(x)y + C(x). Many Riccati equations like the ones listed next can be solved explicitly in terms of Bessel functions.6. Assume that its length is increasing linearly with time. y(a) = y(b) = 0. 3. + A.262 Chapter 3 Power Series Methods Note that if a = b. (28) of Section 2." see the article by Borrelli. 3. where A. about a foot in length from horn to horn. y(a) = y(b) = 0.6.6. = fJ} = P b4 j E I0 • (a) Apply the theorem of this section to show that the general solution of x 4 y" + JJ} y = 0 is y(x) =x (A cos~+ Bsin~).6. Here. this result reduces to Eq. (1) (2) . the moment of inertia I cross section at x is given by I(x) = I (x) of the = ~rr(kx) 4 = 10 · (~) 4 . and the under edge as keen as that of a razor .5. L (t) = a + bt. . . subject to an axial force P of compression. (b) Conclude that the nth eigenvalue is given by /-Ln nrrabjL.6.8. As in Section 2. Figure 3.6 Application A Riccati equation is one of the form dy . down and still down it came") of Edgar Allan Poe's macabre classic "The Pit and the Pendulum.y = 0. where L = b -a is the length of the rod. where I0 = I (b). Coleman. It can be shown that the oscillations of this pendulum satisfy the differential equation LB" 17. .6. the horns upward. and hence that the nth buckling force is m FIGURE 3. (l8) + 2L'()' + g() = 0 under the usual condition that () is so small that sin() is very well approximated by (): () ~ sin e. and the whole hissed as it swung through the air . 58. the value of I at x =b.5 shows a linearly tapered rod with circular cross section. however. The tapered rod of Problem 17. Consider a variable-length pendulum as indicated in Fig.8. pp. Substitute L = a + bt to derive the general solution X x=a x=b FIGURE 3. 78-83). its deflection curve y = y(x) satisfies the endpoint value problem Ely"+ Py = 0. Substitution of I (x) in the differential equation in ( 18) yields the eigenvalue problem x 4 y" For the application of this solution to a discussion of the steadily descending pendulum ("its nether extremity was formed of a crescent of glittering steel. A variable-length pendulum. . (7) are involved. y(x)) or the Mathematica command DSolve[ y'[x) == x~2 y[x]~2.y2. dx (4) (5) (6) For example. dx dy 2 . x) agrees with Eq. y[x].4. the general solution of Eq. dx ' dy .c/1 .c I -2/3 (2 3x 3/ 2) 112 I 2/3 ( 3x L1 . such as the Maple command dsolve(diff(y(x). (5) is given for x > 0 by y(x) =x 2 3/2) .=y -X. Next. 3 3x 312) (2 ) (2 ' (9) . If Bessel functions other than those appearing in Eq.= X .6 Applications of Bessel Functions 263 (3) dy dx = l-x2. Problem 15 in this section says that the general solution of Eq. 3.5 to transform the computer's "answer" to (7).6. Each has a general solution of the same general form in (7)-a quotient of linear combinations of Bessel functions.3. you may need to apply the identities in (26) and (27) of Section 3. dy . 3 3x 3l 2 . Then see whether your system can take the limit as x ~ 0 in (7) to show that the arbitrary constant c is given in terms of the initial value y(O) by c=---:-::-'-'-'y(O)r (t) 2r (D (8) Now you should be able to use built-in Bessel functions to plot typical solution curves like those shown in Fig. (7).=x+y2. (1) is given by (7) See whether the symbolic DE solver command in your computer algebra system.x) = + x~2 + y(x)~2. In addition to lp(x) and Yp(x). these solutions may involve the modified Bessel functions and Kp(x) = li . For instance. investigate similarly one of the other equations in (2) through (6).p [ lp(ix) n + Yp(ix)] 0 that satisfy the modified Bessel equation x 2y" + xy' - (x 2 + p 2)y = of order p. .+ . . G) (10) 4 2 >-..6. = y2 - x. (6)-we see a funnel near y = +Jx and a spout near y = -Jx. (1) through (6) have interesting solutions which. 1956)..7 shows some typical solution curves. . For instance. The second-order differential equations of the form y" = f (x.X (11) with the same initial value y(O) = 0 but different slopes y'(O) = -3. an equation that arose historically in the classification of nonlinear second-order differential equations in terms of their critical points (see Chapter 14 of E. L. Nevertheless..8 shows solution curves of the second-order equation y" = y2.5.6. -3. .3.1.3. however. y'(O) = . . . y) with the same right-hand sides as in Eqs. but the power series of the modified function In (x) is simply that of the unmodified function In (x) except without the alternating minus signs. Equation (11) is a form of the first Painleve transcendant.6. The first Painleve transcendant y" y(O) = 0. . . Solution curves dy of-= x.3.8.. the interesting pattern in Fig. 3. New York: Dover Publications. 0.0 -2 Figure 3.6.3. The Bessel functions with imaginary argument that appear in the definitions of IP (x) and K P (x) may look exotic.7. (17) and (18) in Section 3.31/3f (~). Figure 3. Ordinary Differential Equations.y 2 • dx and x x3 h (x) = 2 + 16 + 384 + 18432 + · · · · Check these power series expansions using your computer algebra system-look at Bessell in either Maple or Mathematica-and compare them with Eqs. together with the parabola y 2 = x that appears to bear an interesting relation to Eq. Io(x) = 1 + 4 X -4 x2 + .1.7.264 Chapter 3 Power Series Methods where y(O)r c =. Ince.6. they can be investigated using an ODE plotter.7. 64 2304 x7 x5 x4 x6 FIGURE 3. 2 0 -2 -4 X FIGURE 3. cannot be expressed in terms of elementary functions and/or "known" special functions such as Bessel functions. 0. .+ . For instance.8 was suggested by an article by Anne Noonburg containing a similar figure in the Spring 1993 issue of the C• ODE•E Newsletter. 1. Can you explain where the rational function part comes from. more exotic special functions are involved. where r is an even integer. or at least verify it? For further examples of this sort. and/or replace the x on the right-hand side with x 5 . here's a related example that was inspired by a Maple demonstration package. (13) Show that Theorem 1 in this section explains the "Bessel part" of the alleged solution in Eq. (13) with r 2 .1 (c1 lw(x) + c2Yw(x)) (12) + x. (With parities other than these.11 (1857945600 + 51609600x 2 + 806400x 4 + 9600x 6 + 100x8 + x 10 ) of the nonhomogeneous second-order equation x 2y" + 3xy' + (x 2 - 99)y = x. (12).6 Applications of Bessel Functions 265 Finally. where s is an odd integer. The Maple dsol ve command yields the general solution y(x) = x.3.) . you can replace the coefficient 99 in Eq. we will see in Section 4. In this case the methods of Chapter 2 can be quite awkward.1.2 that the Laplace transform converts a differential equation in the unknown function f(t) into an algebraic equation in F(s). when the voltage supplied to an electrical circuit is turned off and on periodically. and the Laplace transform method is more convenient..for example.1. corresponding to a mass-spring-dashpot system and a series RLC circuit. has discontinuities. The Laplace transformation £ involves the operation of integration and yields the new function £{f(t)} = F(s) of a new independent variable s .1. It often happens in practice that the forcing term. The situation is diagrammed in Fig. recall the differential equations m x " +ex'+ kx = F(t) and LI" 1 + RI' + -1 = c E'(t) D {f(t)) = f'(t) f (t)--. yields the new function D{f(t)} = J'(t) . '. The differentiation operator D can be viewed as a transformation which. I n Chapter 2 we saw that linear differential equations with constant coefficients have numerous applications and can be solved systematically. 266 . when applied to the function f(t). After learning in this section how to compute the Laplace transform F (s) of a function f(t). this is one method that simplifies the problem of finding the solution f(t). Transformation of a function: £ in analogy with D.i'{f(t) ) = F (s ) FIGURE 4. 4.Laplace Transform Methods Laplace Transforms and Inverse Transforms f(t)--.. in which the alternative methods of this chapter are preferable. respectively. There are common situations. F(t) or E'(t).1. For example. Because algebraic equations are generally easier to solve than differential equations. however. e -(s-a)rJoo o o s.1 Laplace Transforms and Inverse Transforms 267 DEFINITION The Laplace Transform ~ Given a function f(t) defined for all t function F defined as follows: F(s) 0. 1 a 00 g(t) dt = lim b---'> oo 1b a g(t) dt.st dt = [-~e-s 1 ] = lim [-~e-bs + ~]' }0 s b---'>oo s s 0 and therefore d.in problems as well as in examples. the definition of the Laplace transform in (1) gives = 00 r)() e. Note that the integrand of the improper integral in (1) contains the parameters in addition to the variable of integration t. This is typical of Laplace transforms. Recall that an improper integral over an infinite interval is defined as a limit of integrals over bounded intervals.a r=o . Therefore. it diverges or fails to exist. otherwise. • Example 2 ·····-·····-····· · With f(t) =eat fort oC{eat} ~ 0. it is typical for the improper integral in the definition of oC {f (t)} to converge for some values of s and diverge for others. b---'>oo a (4) • Remark: The limit we computed in Example 1 would not exist if s < 0. then we say that the improper integral converges. the domain of a transform is normally of the forms >a for some number a. Also. but to afunction F of s. we obtain . in this computation we have used the common abbreviation [ g(t)J 00 a = lim [g(t)Jb.. the Laplace transform off is the = oC{f(t)} = 1 00 e-st f(t) dt (1) for all values of s for which the improper integral converges.4. when the integral in (1) converges. Hence oC{1} is defined only for s > 0. As in the following examples.- p{ 1} - ~ s <: lOr s > 0. (3) As in (3). Example 1 With f(t) = 1 fort oC{1} ~ 0. it's good practice to specify the domain of the Laplace transform.st eat dt = 1 oo e-<s-a)t dt = [. = 1 oo e. it converges not merely to a number. that is. (2) If the limit in (2) exists. for then (1js)e-bs would become unbounded as b-+ +oo. 1)1(n . then e-(s-a)t --+ 0 as t --+ +oo. (5) Note here that the improper integral giving £{ea 1 } diverges if s ~ a .1) · (n. which is defined for x > 0 by the formula (6) For an elementary discussion of 1 (x) . and dt = dujs in this integral. agrees with the factorial function for x = n. so it follows that _1_ c 10r d. thus 1(n + 1) = n! (9) if n is a positive integer.268 Chapter 4 Laplace Transform Methods If s -a > 0. e-(s-a)t = eifJt e-(s-a)t --+ 0 as t --+ +oo.5.st ta dt. the function 1(x + 1). Therefore. see the subsection on the gamma function in Section 3. which is defined and continuous for all x > -1 . For then.2) = n(n. It is worth noting also that the formula in (5) holds if a is a complex number. then 1(n + 1) = n1(n) (8) = n · (n . It then follows that if n is a positive integer.2)1(n . • The Laplace transform £{ta} of a power function is most conveniently expressed in terms of the gamma function 1(x). where it is shown that 1(1) = 1 (7) and that 1(x + 1) = x1(x) for x > 0.. Example 3 Suppose that f (t) = t a where a is real and a > -1. provided that s > a = Re[a]. Then £ {ta} = ··- 100 e .l)(n . with a =a+ if3.2) · · · 2 · 1(2) = n(n . a positive integer.1) = n · (n . recall that eifJt = cos f3t + i sin f3t.- P{eat} _- s-a s > a.1)(n.2) · · · 2 · 1 · 1(1). we get (10) . If we substitute u = st. t = ujs. .........4... Once we know the Laplace transforms of several functions. For instance.. (11) For instance........st f(t) dt) c--+oo lo r lo e... The computation of £{tnl 2 } is based on the known special value (13) _.. ....1 Laplace Transforms and Inverse Transforms 269 for all s > 0 (so that u = st > 0)....... Example 4 -·... .. .. . ....... The proof of Theorem 1 follows immediately from the linearity of the operations of taking limits and of integration: oC{af(t) + bg(t)} = 1 00 e ... ... THEOREM 1 Linearity of the Laplace Transform If a and b are constants.... ....... ~ r (~) = ~0T 2 2 2 2 2 2 4.. • Linearity of Transforms It is not necessary for us to proceed much further in the computation of Laplace transforms directly from the definition. we see that oC{tn} = n! n+I + 1) = n! if n is a nonnegative s for s > 0........st [af(t) + bg(t)] dt = lim { c e-st [af(t) c--+oo lo + bg(t)] dt + b (lim c---+ oo = a (lim re.. without the use of the gamma function.... ... .. ........st g(t) dt) = aoC{f(t)} + boC{g(t)} .. ..... 1 oC{t} = 2' s oC{t } = 3' s 2 2 and oC{t } = 4 . then oC{af(t) + bg(t)} = aoC{f(t)} + boC{g(t)} (12) for all s such that the Laplace transforms of the functions f and g both exist. The reason is that the Laplace transformation is a linear operation... ... ... .. .. . these formulas can be derived immediately from the definition. it follows that r (~) = ~r (~) = ~ . . Because r(n integer..... of the gamma function.. ...---···-······· ... s 3 6 As in Problems 1 and 2..... we can combine them to obtain transforms of other functions....... and 5 we see that -1 { £ s3 1} = 1 2 2t ' £-1 {-1 } s+2 = e-2r ' and so on.1 Laplace Transforms and Inverse Transforms 271 Inverse Transforms According to Theorem 3 of this section. then we call f(t) the inverse Laplace transform of F(s) and write f(t) = £. f(t) has a finite limit as t approaches each endpoint of each subinterval from its interior.E) . This observation allows us to make the following definition: If F(s) = £{f(t)}. We say that f is piecewise continuous for t ~ 0 if it is piecewise continuous on every bounded subinterval of [0.k2 k s2. no two different functions that are both continuous for all t ~ 0 can have the same Laplace transform.2. (18) Example 7 Using the Laplace transforms derived in Examples 2.a) s s2 + kz k f is continuous in the interior of each of these subintervals. The table in Fig. A short table of Laplace transforms.1. s-a coskt sin kt cosh kt sinh kt u(t.2 lists the transforms derived in this section.1. Throughout this chapter we denote functions oft by lowercase letters.4. A table of Laplace transforms serves a purpose similar to that of a table of integrals. many additional transforms can be derived from these few.4. whose graph appears in Fig. using various general properties of the Laplace transformation (which we will discuss in subsequent sections).1 . 4. fort ~ 0. The jump in f(t) at the point c is defined to be f(c+). then f(t) is uniquely determined. 3.as s and f(c-) = lim f(c. 4. and sZ + k2 2. 4. Thus F(s) is the Laplace transform of f (t) and x (t) is the inverse Laplace transform of X (s).1 {F(s)}.3. E-+0+ FIGURE 4. Thus a piecewise continuous function has only simple discontinuities (if any) and only at isolated points.k2 e . Perhaps the simplest piecewise continuous (but discontinuous) function is the unit step function. The transform of a function will al- 1 s s2 tn (n ~ (s > 0) (s > 0) (s > 0) 0) n! 5 n+l ways be denoted by that same letter capitalized. as indicated in Fig.1. The function f(t) is said to be piecewise continuous on the bounded interval a ~ t ~ b provided that [a.f(c-). (19) . +oo). where f(c+) = lim f(c +E) E-+0+ s s z . It is defined as follows: u(t) = {1 0 fort < 0. At such points the value of the function experiences a finite jump. b] can be subdivided into finitely many abutting subintervals in such a way that 1. Thus if F(s) is the transform of some continuous function f(t). • f(t) NOTATION: FUNCTIONS AND THEIR TRANSFORMS. ta (a > -1) ear f(a + 1) (s > 0) 5 a+l (s > 0) (s > 0) (s > 0) (s > lkl) (s > lkl) (s > 0) Piecewise Continuous Functions As we remarked at the beginning of this section. we need to be able to handle certain types of discontinuous functions. More precisely. we see that the improper integral f000 e-stet2 dt that would define £{e12 } does not exist (for any s). then its Laplace transform F(s) = £{f(t)} exists. consider the function f(t) = et 2 = exp(t 2 ). Proof: First we note that we can take T = 0 in (23). and T are not so important. c. this is true (with c = 0) if f(t) itself is bounded. The particular values of M. The function 273 f is said to be of exponential order as t ---+ +oo if there exist nonnegative constants M. To do this. lf(t)l is bounded on [0. = t-+oo hm e 1 2_ et = +oo because t 2 . For an example of an elementary function that is continuous and therefore bounded on every (finite) interval. What is important is that some such values exist so that the condition in (23) is satisfied. it then follows that IJ(t)l ~ Meet for all t ~ 0. and T such that lf(t)l :£Meet fort 2: T.ct ---+ +oo as t ---+ +oo. THEOREM 2 Existence of Laplace Transforms If the function f is piecewise continuous for t ~ 0 and is of exponential order as t ---+ +oo. we see that (t) hm . The condition in (23) merely says that f(t)/eet lies between -M and M and is therefore bounded in value fort sufficiently large. we can therefore assume that If (t) I :£ M if 0 :£ t :£ T. Because ec1 ~ 1 for t ~ 0. Thus every bounded function.t-+ oo eet . f . so we conclude that the function f(t) = e12 is not of exponential order. Whatever the value of c.4. (23) Thus a function is of exponential order provided that it grows no more rapidly (as t ---+ +oo) than a constant multiple of some exponential function with a linear exponent. By a standard theorem on convergence of improper integrals-the fact that absolute convergence implies convergence-it suffices for us to prove that the integral exists for s > c. because e-st e 12 ---+ +oo as t ---+ +oo. In particular. then the familiar fact that p(t)e-t ---+ 0 as t ---+ +oo implies that (23) holds (for T sufficiently large) with M = c = 1. c. if f is piecewise continuous and satisfies the condition in (23).such as coskt or sin kt-is of exponential order. Hence the condition in (23) cannot hold for any (finite) value M. If p(t) is a polynomial. T]. Increasing Min (23) if necessary. and therefore that the function e 12 does not have a Laplace transform. Thus every polynomial function is of exponential order. Similarly. then F(s) exists for all s >c. it suffices in tum to show that the value of the integral . For by piecewise continuity. The following theorem guarantees that piecewise functions of exponential order do have Laplace transforms. et = t-+oo hm eet 2 .1 Laplace Transforms and Inverse Transforms t ---+ +oo. but nevertheless is not of exponential order. 274 Chapter 4 Laplace Transform Methods remains bounded as b ---+ that +oo. the function f(t) = 1/. The remainder of this chapter is devoted largely to techniques for solving a differential equation by first finding the Laplace transform of its solution.) = £{t-1 /2 } = _2_ sl/2 ~ s both exists and violates the condition in (24 ). that the function of s we have found has only one inverse Laplace transform that could be the desired solution. (New York: McGrawHill. we get the following result. COROLLARY F(s) for s Large If f (t) satisfies the hypotheses of Theorem 2. that IF(s)l ~ 1 0 le-st f(t)l dt ~ -- M s-c (24) if s > c. the function G (s) = s j (s + 1) cannot be the Laplace transform of any "reasonable" function because its limit as s ---+ +oo is 1. then f(t) = g(t) wherever on [0. But the fact that If (t) I ~ M ect for all t ~ 0 implies ~ M 1 o oo oo e-(s-c)t dt = . We have shown. which would imply that sF (s) remains bounded as s ---+ +oo. The following theorem is proved in Chapter 6 of Churchill's Operational Mathematics. as we shall see) a Laplace transform only if the degree of its numerator is less than that of its denominator. More generally. not 0. On the other hand.- M s-c if s > c. It is then vital for us to know that this uniquely determines the solution of the differential equation. (25) The condition in (25) severely limits the functions that can be Laplace transforms. +oo) both f and g are continuous. the hypotheses of Theorem 2 are sufficient. but nevertheless (Example 3 with a=-~> -1) its Laplace transform r (1. THEOREM 3 Uniqueness of Inverse Laplace Transforms Suppose that the functions f(t) and g(t) satisfy the hypotheses of Theorem 2. conditions for existence of the Laplace transform of f(t). 1972). . that is. This proves Theorem 2. so that their Laplace transforms F(s) and G(s) both exist. If F(s) = G(s) for all s > c (for some c). For instance. 3rd ed. a rational function-a quotient of two polynomials-can be (and is. moreover. but not necessary. then lim F(s) S--+00 = 0. When we take limits ass ---+ +oo. For example./i fails to be piecewise continuous (at t = 0). Use the transforms in Fig. f (t) = sinh2 3t .1. Historical Remark: Laplace transforms have an interesting history. 13. The so-called operational methods for solving differential equations. 0) ~(1. f(t) = t 2. and around the beginning of the twentieth century their validity was the subject of considerable controversy. It is customary in mathematics to name a technique or theorem for the next person after Euler to discover it (else there would be several hundred different examples of "Euler's theorem").) Apply the definition in (I) to find directly the Laplace transforms of the functions described (by formula or graph) in Problems I through 10. f(t) = sin 3t cos 3t 20.IOr = t 312 - 16.1. These techniques were successfully and widely applied before they had been rigorously justified. f(t) = -Ji + 3t f(t) = t . 4. and hence must be the same solution if they have the same Laplace transform. 15. f(t) = e3r+I 5.4. 21. 10.6.1 Laplace Transforms and Inverse Transforms 275 Thus two piecewise continuous functions of exponential order with the same Laplace transform can differ only at their isolated points of discontinuity.1)~ (1. One reason is that Heaviside blithely assumed the existence of functions whose Laplace transforms contradict the condition that F (s) --j> 0 ass --j> 0.l) I FIGURE 4.8. f(t) = sin 2 t 3. f(t) =sin 2t +cos 2t 18. who employed such integrals in his work on probability theory. The integral in the definition of the Laplace transform probably appeared first in the work of Euler. 1) j I ~ • c 11. so we may regard inverse Laplace transforms as being essentially unique. f(t) = cost 6. were not exploited by Laplace. FIGURE 4. two solutions of a differential equation must both be continuous.7.2 to find the Laplace transforms of the functions in Problems II through 22. thereby raising questions as to the nature and role of numbers in mathematics. k:(l. f(t) = (1 + t) 3 = te 1 FIGURE 4. (0. the next person was the French mathematician Pierre Simon de Laplace (1749-1827). thereby raising questions as to the meaning and nature of functions in mathematics. f(t) = 3t 512 14. A preliminary integration by parts may be necessary. f(t) f(t) = cos2 2t 19.1. 17. they were discovered and popularized by practicing engineers-notably the English electrical engineer Oliver Heaviside (1850-1925). 9. (1.1) FIGURE 4.2e31 f(t) = 1 +cosh 5t 12. In this case.I. 1. which are based on Laplace transforms. In particular. Indeed. 8. f(t) = t cos 2t 22. f(t) = sinht 7. f(t) - 4t 3 e . 1) (2. This is of no importance in most practical applications.9. f(t) = t 2 4.1. (This is reminiscent of the way Leibniz two centuries earlier had obtained correct results in calculus using "infinitely small" real numbers. Then express f in terms of unit step functions to show that£{/(!)}= s. n=1. Express g in terms of the function f of Problem 40 and hence deduce that to obtain £{cos kt} directly from the definition of the Laplace transform. Show that f can be written in the form 00 27. First. 38. 50 FIGURE 4. First. 4.:.2 to find the inverse Laplace transforms of the functions in Problems 23 through 32. The graph of the function of Problem 41.1. Derive the transform of f(t) =sin kt by the method used in the text to derive the formula in ( 16).e-s) =-s 1 2 s 512 26. ..__ .. Given that 0 < a < b. F(s) = s 2 + 4 9+s 30. 36. F(s) = 4 s 25.n). 4. F(s) 24.:. let f(t) = 1 if 0 . n=O 5.276 Chapter 4 Laplace Transform Methods Apply the geometric series to obtain the result £{f(t)} Use the transforms in Fig.. Derive the transform of f(t) =sinh kt by the method used in the text to derive the formula in (14). F(s) = s..1.:.1.1(1. F(s) = -s-4 3 3s + 1 28. Given constants a and b. f(t) = 0 if t ~ a. F(s) (b) Use the method of Problem 39 to show that = lOs. Given a > 0. 1 33.. 1f n . . Then express f in terms of unit step functions to show that£{!(!)}= s .1 .10.--. 39. Show that the function f(t) = sin(e12 ) is of exponential order as t ~ +oo but that its derivative is not. sketch the graph of the function f.2. Use the tabulated integral 0 4 • 5 06 - f eax eax cosbxdx= 2 2 (acosbx +b sinbx)+C a +b FIGURE 4. t < b.1 :::.312 1 = 1 s(l.11. 41. The graph of the square-wave function g(t) is shown in Fig. 42. F(s) = -s+5 40. (c) Assume that the Laplace transform of the infinite series in part (b) can be taken term wise (it can). The unit staircase function is defined as follows: f(t)=n I L ~I . 3 23.e .1(e . t < a. F(s) = s 2 + 9 31. =L n=O 00 u(t...bs ). . f(t) = 0 if either t < a or t ~ b. t < n and n IS even. making clear its value at t = a.__ . = a + be-s s(l + e-s) ..F(s)= 4 _s 2 f(t) = L(-l)nu(t.--.3 25. (a) Sketch the graph off to see why its name is appropriate.--.1. making clear its values at t = a and t = b..10. (a) The graph of the function f is shown in Fig... define h(t) fort ~ 0 by if n-I .3s 29..s2 £{f(t)} = s(l + e-s) .1. . let f(t) = 1 if a .-a 4 .-a 34. (b) Show that f(t) h(t) = I a b if n ..-a 2 3 .. t < n and n is odd. sketch the graph of the function f.e-as) . 37. 4. The graph of the function of Problem 40.3.. .n) Sketch the graph of h and apply one of the preceding problems to show that £{h(t)} for all t ~ 0.11. t < n .as ..:. 35. We may assign arbitrary values to f(t) at the isolated points at which f is not differentiable. the integrated term e . (3) b X Then £{f'(t)} exists for s > c. and its value at the lower limit t = 0 contributes .1. 'b X a ~ Piecewise continuous derivative FIGURE 4. c. lo Because of (3). Then £{f' (t)} exists when s > c. and T such that lf(t)l ~Meet a I : I I I I I I I I I I I I I I I I fort~ T.1 indicates how "comers" on the graph of f correspond to discontinuities in its derivative f './(0). such as ax"(t) + bx'(t) + cx(t) = f(t). we get £{!' (t)} = r>O e .2 Transformation of Initial Value Problems 277 II!J Transformation of Initial Value Problems We now discuss the application of Laplace transforms to solve a linear differential equation with constant coefficients.st f(t)]oo t=O + s {00 e. y THEOREM 1 Transforms of Derivatives Suppose that the function f(t) is continuous and piecewise smooth fort ~ 0 and is of exponential order as t --+ +oo.4. The function f is called piecewise smooth on the bounded interval [a. We will defer the case in which f'(t) has isolated discontinuities to the end of this section.st!' (t) dt = lo [e . The main idea of the proof of Theorem 1 is exhibited best by the case in which f' (t) is continuous (not merely piecewise continuous) fort ~ 0.2. The integral that remains is simply £{f(t) }. (4) I I I I I I : ~ I . b] and differentiable except at finitely many points.f(O) to the evaluation of the preceding expression. . which tells us how to express the transform of the derivative of a function in terms of the transform of the function itself. beginning with the definition of £ {f' (t)} and integrating by parts. (1) with given initial conditions x (0) = x 0 and x' (0) = xb.2.1. Figure 4. We say that f is piecewise smooth for t ~ 0 if it is piecewise smooth on every bounded subinterval of [0.f(O) = sF(s). Then. b] if it is piecewise continuous on [a. with f'(t) being piecewise continuous on [a. by Theorem 2 of Section 4. so that there exist nonnegative constants M. b]. The transformed equation is a£{x"(t)} + b£{x'(t)} + c£{x(t)} = £{f(t)}. (1) by separately taking the Laplace transform of each term in the equation.st f(t) dt.st f(t) approaches zero (when s > c) as t --+ +oo. +oo). (4). we can transform Eq. The discontinuities of f' correspond to "corners" on the graph of f. and its value is that given in Eq. and £{/'(t)} I I y' N I I I = s£{f(t)}. (2) it involves the transforms of the derivatives x' and x" of the unknown function x (t). The key to the method is Theorem 1. the integral converges when s > c. By the linearity of the Laplace transformation. substitution of s 3 7 -2 shows that X(s) = £{x(t)} = ___l_ s-3 + ___l_.3) (s + 2) s. it follows that is the solution of the original initial value problem. (s .4.a)}= eat.3).6 = (s . 4.2 Transformation of Initial Value Problems 279 By the method of partial fractions (of integral calculus). 2A . Then upon equating coefficients of terms of like degree. In lieu of any such shortcut. If we substitute s B =~ .3 = A(s + 2) + B(s. we find that A = ~. Note that we did not first find the general solution of the differential equation.3)(s + 2) into the equation 2s. 2s . x(O) = x' (0) = 0.3 = A(s + 2) + B(s.3B = .3) (s + 2) yields the identity 2s.3 = (A+ B)s + (2A. automatically taking into accountvia Theorem 1 and its corollary-the given initial conditions.s .3 which are readily solved for the same values A = ~ and B = ~- • Example 2 Solve the initial value problem x" + 4x = sin 3t.1{1/(s. we get the linear equations A+ B = 2.2. there exist constants A and B such that 2s-3 A B -----=--+--.3).3) that resulted from clearing fractions. Such a problem arises in the motion of a mass-and-spring system with external force.3 s+2 and multiplication of both sides of this equation by (s . • Remark: In Example 1 we found the values of the partial-fraction coefficients A and B by the "trick" of separately substituting the roots s = 3 and s = -2 of the original denominator s 2 .Hence 3. The Laplace transform method directly yields the desired particular solution. the "sure-fire" method is to collect coefficients of powers of s on the right-hand side. as shown in Fig. .2. s+2 Because £. . A mass-andspring system satisfying the initial value problem in Example 2. and D. + 4) (s 2 + 9) · The method of partial fractions calls for 3 (s 2 + 4)(s 2 + 9) As+ B s2 + 4 Cs + D s2 + 9 · + The sure-fire approach would be to clear fractions by multiplying both sides by the common denominator. it follows that sin 2t- ! sin 3t. 4. Figure 4.4. Because £{sin 2t} = 2/(s 2 + 4) and £{sin 3t} = x(t) = 3 10 3/(s 2 + 9) . Note that the Laplace transform method again gives the solution directly.-----=-----::--(s 2 FIGURE 4. Equating coefficients of like powers on the two sides of the resulting equation would then yield four linear equations that we could solve for A. The mass is initially at rest in its equilibrium position. because neither the numerator nor the denominator on the left involves any odd powers of s. here we can anticipate that A = C = 0. Eq.1) and thereby get the transformed equation 3 s 2 X (s) + 4X (s) = 2 9 s + Therefore. When we equate coefficients of like powers of s we get the linear equations B+D =0. which are readily solved forB= ~and D =-~. FIGURE 4. However. X sZ 2 1 +4. Thus nonhomogeneous equations • are solved in exactly the same manner as are homogeneous equations.2.Hence 3 X(s) = oC{x(t)} = 10. The position function x(t) in Example 2.3 shows the graph of this period 2n position function of the mass. 98 +4D = 3. 3 X (s) .3. 4.2.280 Chapter 4 Laplace Transform Methods Solution Because both initial values are zero.2.1.2 (Section 4.2. whereas nonzero values for A or C would lead to odd-degree terms on the right.5. without the necessity of first finding the complementary function and a particular solution of the original nonhomogeneous differential equation. B. So we replace A and C with zero before clearing fractions. Examples 1 and 2 illustrate the solution procedure that is outlined in Fig. C.2. The result is the identity 3 = B(s 2 + 9) + D(s 2 + 4) = (B + D)s 2 + (9B + 4D) . (5) yields oC{x" (t)} = s 2 X (s). sZ 3 +9. We read the transform of sin 3t from the table in Fig. and then collect coefficients of powers of s on the right-hand side. Then the initial conditions in (9) and £ {y"(t)} £{x" (t)} = s 2 X (s) = s 2Y(s). ..and. y(t) . Using the Laplace transform to solve an initial value problem. A mass.2. ..2 Transformation of Initial Value Problems Differential equation in x(t) Solution x(t) of differential 281 Algebraic equation in X(s) Solution X(s) of algebraic equation FIGURE 4.2. y" = 2x.4.2. the Laplace transform reduces such a linear system of differential equations to a linear system of algebraic equations in which the unknowns are the transforms of the solution functions. Solution We write X(s ) imply that = £{x(t)} and Y(s) = £{y(t)} . of the independent variable t .4. As Example 3 illustrates.spring system satisfying the initial value problem in Example 3. FIGURE 4. assuming that the force f(t) = 40 sin 3t is suddenly applied to the second mass at the time t = 0 when both masses are at rest in their equilibrium positions.5. the technique for a system is essentially the same as for a single linear differential equation with constant coefficients. (8) (9) This initial value problem determines the indicated displacement functions x(t) and y (t) of the two masses shown in Fig. Linear Systems Laplace transforms are frequently used in engineering problems to solve a system of two or more constant-coefficient linear differential equations involving two or more unknown functions x(t). Both masses are initially at rest in their equilibrium positions. When initial conditions are specified. 4.5. Example 3 Solve the system 2x" = -6x + 2y . subject to the initial conditions x(O) = x'(O) = y(O) = y' (O) = 0.2y + 40sin3t. (13) for X (s). I (s) (14) where Z(s) = ms 2 + cs + k and I (s) = mx(O)s + mx'(O) Note that Z(s) depends only on the physical system itself. Much of the remainder of this chapter is devoted to finding Laplace transforms and inverse transforms.7. 4. m [s 2 X(s). a linear equation-in the "unknown" X (s). It follows from the linearity of the transform that . (14). we get F(s) X (s) = Z(s) + Z(s)' + cx(O). we seek those methods that are sufficiently powerful to enable us to solve problems that-unlike those in Examples 1 and 2.x(O)] + kX(s) = F(s). (13) Note that Eq.2 Transformation of Initial Value Problems 283 The Transform Perspective Let us regard the general constant-coefficient second-order equation as the equation of motion mx" +ex'+ kx = f(t) of the familiar mass-spring-dashpot system (Fig. respectively.Z(s) of the steady periodic solution and the transient solution. Hence Theorem I gives cl{eat +ateat} = cl{f'(t)} = scl{f(t)} = scl{tea1 }. In particular. Additional Transform Techniques Example 4 Show that Solution If f(t) =teat. In the case of an underdamped system. then f(O) = 0 and J'(t) =eat+ at eat.x'(O)] + c [sX(s). (13) is an algebraic equation-indeed. (14) presents X (s) = cl {x (t)} as the surn of a term depending only on the external force and one depending only on the initial conditions.7). these two terms are the transforms cl{X5p(t)} = F(s) Z(s) and I (s) cl{Xrr(t)} = . If we solve Eq. A massspring-dashpot system with external force f(t).cannot be solved readily by the methods of Chapter 2.2.sx(O). This is the source of the power of the Laplace transform method: Linear differential equations are transformed into readily solved algebraic equations.2. The only potential difficulty in finding these solutions is in finding the inverse Laplace transform of the right-hand side in Eq.4. Thus Eq. Then the transformed equation is FIGURE 4. . . in terms of the factorization of the denominator Q(s) into linear factors and irreducible quadratic factors corresponding to the real and complex zeros. and Bn are constants. First we must find the partial fraction decomposition of R(s). then .1{R(s)} is based on the same method of partial fractions that we use in elementary calculus to integrate rational functions. THEOREM 1 Translation on the s-Axis If F(s) = .4. The following two rules describe the partial fraction decomposition of R(s).£{/(t)} exists for s > c. Finding . B2..a in the transform corresponds to multiplication of the original function oft by eat..a ) 2 + b 2 of multiplicity n is a sum of n partial fractions.. of Q(s).. s-a (s -a) 2 where A1. An.. and .. . .. Thus the translation s -+ s .£{eat f(t)} exists for s >a+ c. The latter step is based on the following elementary property of Laplace transforms. . respectively.£-I {R (s)} involves two steps..a).£.a of multiplicity n is a sum of n partial fractions. and An are constants. and then we must find the inverse Laplace transform of each of the individual partial fractions of the types that appear in (2) and (3). A2. having the form where A1. B 1. . (4) (5) Equivalently. The technique for finding . . .£{eat f(t)} = F(s. . RULE 2 Quadratic Factor Partial Fractions The portion of the partial fraction decomposition corresponding to the irreducible quadratic factor (s . RULE 1 Linear Factor Partial Fractions The portion of the partial fraction decomposition of R (s) corresponding to the linear factor s .3 Translation and Partial Fractions 289 IIIJ Translation and Partial Fractions As illustrated by Examples 1 and 2 of Section 4. A2.2. the solution of a linear differential equation with constant coefficients can often be reduced to the matter of finding the inverse Laplace transform of a rational function of the form R(s) = P(s) Q(s) (1) where the degree of P(s) is less than that of Q(s). having the form A1 Az An (2) --+ +···+ (s -a)n . r ) d r = 1° f(t .r) dr (1) has the desired property that £{h(t)} = H(s) = F(s) · G(s). Thus the convolution is commutative: f g = g f .u)g(u)( .4. (3) We will also write f(t) g (t) when convenient. In terms of the convolution product.du) g(u)f(t. we get · 2 1 = cl {cos t} · cl {sin t}. Theorem 1 of this section will tell us that the function h(t) = lr f(r)g(t . DEFINITION The Convolution of Two Functions The convolution f * g of the piecewise continuous functions for t ~ 0 as follows: (f f and g is defined * g)(t) = 1 1 f(r)g(t. when we transform the initial value problem x" + x =cost. But obviously x(t) is not simply the product of cost and sin t. It is denoted by f * g. x(O) = x ' (O) = 0. Thus cl {cos t sin t} =1= £ {cos t} · £ {sin t}. . because X (s) = (s 2 + 1)2 s = £ {cos t sin t} = £ Bsin 2t } = s 2 +4 1 =I= (s 2 + 1)2 s . we see that f (t) * g (t) = lt = lt f (r )g (t .r) dr. Integrals. For example. Theorem 1 of this section says that cl{f * g} = cl{f}. and Products of Transforms 297 The Laplace transform of the (initially unknown) solution of a differential equation is sometimes recognizable as the product of the transforms of two known functions. (2) The new function oft defined as the integral in (1) depends only on f and g and is called the convolution of f and g. the idea being that it is a new type of product of f and g.r in the integral in (3). £{g }. 2s s +1 s +1 This strongly suggests that there ought to be a way of combining the two functions sin t and cost to obtain a function x(t) whose transform is the product of their transforms. so tailored that its transform is the product of the transforms of f and g. If we make the substitution u = t .u) du = g(t) * f(t).4 Derivatives. . Theorem 1 tells us that this fact.1 we saw that if a .1 {e .a) dt. . when properly interpreted.st f(t . then £{u(t. Then u ( t . if t ~ a.5. so it does not start arriving until time t = a.a)}= . Translation of f(t) a units to the right.a) f (t . Note that u(t. (2) s Because £{u(t)} = ljs. (2) implies that multiplication of the transform ofu(t) bye-as corresponds to the translation t --+ t -a in the original independent variable.. 4.as F(s) = 1 00 e . Note that the part (if any) of the graph of f(t) to the left oft = 0 is "cut off" and is not translated (Fig. 0. then e-as £{u(t. The notation Ua (t) indicates succinctly where the unit upward step in value takes place (Fig. In some applications the function f(t) describes an incoming signal that starts arriving at time t = 0.a) if t < a.1). a is the translation by a units to the right of the graph of f (t) for t . we get a FIGURE 4.2). Recall that the unit step function at t = a is defined by X Ua(t) = u(t -a) = a 0 1 1 if t <a.a)f(t.5 Periodic and Piecewise Continuous Input Functions 305 Section 4.a)f(t . 0..1 {e-as F(s)} is the function whose graph fort.1...a) connotes the sometimes useful idea of a "time delay" a before the step is made.a) (3b) for s > c +a.. (1) FIGURE 4.as F(s)} = u(t. > 1ft_ a.5.4.5.a) = IOf(t.5. (4) Thus Theorem 1 implies that £ . Proof of Theorem 1: From the definition of £{f(t)}.a) denotes a signal ofthe same "shape" but with a time delay of a.2.1. Eq. The substitution t = r + a then yields e .a)}= e-as F(s) (3a) and £ . In Example 8 of Section 4.a)f(t.. The graph of the unit step function at t =a. is a general property of the Laplace transformation. . THEOREM 1 Translation on the t-Axis If £{f(t)} exists for s > c. whereas u(t. 4. 2t = 4 by Eq.2n)] sin 2t.4. 4.6. x(O) = x'(O) = 0. but at timet = 2n this force is turned off (abruptly discontinued) and the mass is allowed to continue its motion unimpeded. The graph of the function x(t) of Example 4. As indicated by the graph of x(t) shown in Fig. but we see that its effect ceases immediately at the moment it is turned off.5.2n) sin2(t.2n) because of the periodicity of the cosine function. the mass oscillates with circular frequency w = 2 and with linearly increasing amplitude until the force is removed at time t = 2n.2n)] cos 2t =cos 2t . Hence Theorem 1 gives cl(f(t)} = cl{cos2t} -e.6.u(t. Beginning at time t = 0 (seconds). light spring that is stretched 1 ft by a force of 4 lb (k = 4lb/ft).2n) = ~ [t .5.2rrscl{cos2t} = s(l _ e-2rrs) .5 Periodic and Piecewise Continuous Input Functions Solution 307 We note first that f(t) = [1 . s Because oC -1 { (s 2 + 4) 2 S } I t sm . The transformed equation is (s 2 + 4)X(s) so = F(s) = s(l _ e-2rrs) . Find the resulting position function x(t) of the mass. The force F(t) = cos 2t would produce pure resonance if continued indefinitely.u(t. Solution where f(t) is the function of Example 3. The mass is initially at rest in its equilibrium position. the mass continues to oscillate with the same frequency but with constant amplitude n/2. we find that the position function may x(t) = { jt sin 2t ~n sin 2t if t < 2n.2n) cos 2(t. 2 s +4 s X(s) = (s2 + 4)2 - e . FIGURE 4. If we separate the cases t < 2n and t be written in the form X=-rr:/2 ~ 2n. • .u(t. (16) of Section 4. 2 s +4 • Example 4 A mass that weighs 32 lb (mass m = 1 slug) is attached to the free end of a long. if t 2:: 2n.2n) · j(t. Thereafter. it follows from Theorem 1 that x(t) = ~t sin2t. We need to solve the initial value problem x" + 4x = f(t).2n) · (t .u(t.3.2rrs (s2 + 4)2. an external force f(t) =cos 2t is applied to the mass. Thus e'(t) appears to have an infinite discontinuity at t = 1.5. and a battery supplying Eo = 90 V. With the given circuit elements.u(t .7 our strategy was to differentiate both sides of Eq. I Here we do not use that method. corresponding to the opening and closing of the switch. 1 e(t). (8) We substitute Eq. (6) FIGURE 4.7 the basic series circuit equation di Ldt Solution + Rl + -q = c . with R = 110 Q. For now. Example 5 Consider the RLC circuit shown in Fig. (5). + R+l = dt 2 dt c e (t). The Laplace transform method works well with such an equation. Eq. C = 0.001 F. where e(t) = 90[1 . (7) yield.£ t < 2n and then solve a new problem with different initial conditions for the interval t ~ 2n. 4. . q(t) =loti c 0 (r) dr . it involves both the integral and the derivative of the unknown function i (t). and voltage and reserve uppercase letters for their transforms. This phenomenon will be discussed in Section 4. At time t = 1 it is opened and left open thereafter. upon integration. The series RLC circuit of Example 5. whereas the jump from e(t) = 90 when t < 1 to e(t) = 0 when t > 1 would seem to require that e'(l) = -oo. (5) we use lowercase letters for current. In Section 2. we observe that the initial value q(O) = 0 and Eq. =- dq dt (7) to obtain the second-order equation d2i Ldi 1. L = 1 H. At time t = 0 the switch is closed and left closed for 1 second.7. charge. (8) in Eq. we would need to solve one problem for the interval 0 . (5) is R + llOi + 1000q = dt di e(t). (9) This is the integrodifferential equation of a series RLC circuit.7. (5) to obtain di Ldt 1 + Ri +- 1t i(r) dr = e(t). we will simply note that it is an odd situation and circumvent it rather than attempt to deal with it here.1)]. We recall from Section 2.308 Chapter 4 Laplace Transform Methods If we were to attack Example 4 with the methods of Chapter 2.5. Find the resulting current in the circuit. then apply the relation l .6. because e' (t) = 0 except at t = 1. In such a situation the Laplace transform method enjoys the distinct advantage of not requiring the solution of different problems on different intervals. Initially there is no current in the circuit and no charge on the capacitor. To avoid the possible problem at t = 1. This is a function of t. Our strategy for handling such a situation is to set up a reasonable mathematical model in which the unknown force f(t) is replaced with a simple and explicit force that has the same impulse. In the case of a force f(t) that acts on a particle of mass m in linear motion. then we see (Fig. This is fortunate. Suppose for simplicity that f(t) has impulse 1 and acts during some brief time interval beginning at time t = a ~ 0. with a and E being parameters that specify the time interval [a. we need know neither the precise function f (t) nor even the precise time interval during which it acts. (3) I 1 I I I I I I 1 ii I I I I I I otherwise. a + E]. (2) Thus the impulse of the force is equal to the change in momentum of the particle. The number pin Eq.E (f) df = la ra+E! df E = 1. b] is a a +e P = FIGURE 4.E(t)dt = 1. b]. with f(t) = 0 outside this interval. integration of Newton's law f(t) = mv' (t) = - d dt [mv(t)] yields p = 1a bd dt [mv(t)] dt = mv(b) - mv(a).6. Thus da .E has a unit impulse.1) that the impulse of da. 4. for instance) is an analogous electrical phenomenon. we are unlikely to have such detailed information about the impulsive force that acts on the ball. Then we can select a fixed number E > 0 that approximates the length of this time interval and replace f (t) with the specific function X f-. ! (t).1. (4) .6. lb a da.. In such a situation it often happens that the principal effect of the force depends only on the value of the integral p = 1b f(t)dt (1) and does not depend otherwise on precisely how f(t) varies with timet. A quick surge of voltage (resulting from a lightning bolt. because in a situation such as that of a batted ball. (1) is called the impulse of the force j(t) over the interval [a.316 Chapter 4 Laplace Transform Methods liD !mpulses and Delta Functions Consider a force f (t) that acts only during a very short time interval a ~ t ~ b. (t) ={ ~ if a~ t < a+ E. whatever the number E may be. If b ~ a+ E. So if change in momentum is the only effect with which we are concerned. Essentially the same computation gives 1 00 da. The graph of the impulse function da. we need know only the impulse of the force.e -----j r I I I ITI I Area= I da . A typical example would be the impulsive force of a bat striking a ball-the impact is almost instantaneous.E over [a . We might try to model such an instantaneous unit impulse by taking the limit as E ~ 0. a+ E). then the mean value theorem for integrals implies that 1a+Eg(t) dt = Eg (t) for some point tin [a. Although we call it the delta function. Obviously.when applied to a continuous function g(t). It follows that lim E~O 1 0 oo g(t)da. it is not a genuine function. then it would follow that 1 But the limit in Eq. we therefore could conclude that Function g(l)~ 1 00 g(t)8a(t) dt = g(a). instead. and if we could interchange the limit and the integral in Eq. (4). (8). (5) gives Oa(t) = 00 Oa(t) dt = 1. (6) +oo {O if t =a. (9) as the definition(!) of the symbol 8a(t). Note that we will use the symbol 8a(t) only in the context of integrals such as that in Eq. This idea is shown schematically in Fig. If g(t) is continuous function.2.6. (9). it specifies the operation FIGURE 4.4. (6) and (7). However interpreted. 4.M.E(t)dt =lim 1a+E a E~O 1 _ g(t) · -dt = limg (t) = g(a) E E ~O (8) by continuity of g at t = a. then its integral is not 1 but zero. A.6. who in the early 1930s introduced a "function" allegedly enjoying the properties in Eqs.sifts out or selects the value g(a) of this function at the point a ~ 0. the symbol 8a(t) is very useful. A diagram illustrating how the delta function "sifts out" the value g(a). it is tempting to think of an instantaneous impulse that occurs precisely at the instant t =a.6 Impulses and Delta Functions 317 Because the precise time interval during which the force acts seems unimportant. . which. no actual function can satisfy both (6) and (7)-if a function is zero except at a single point. Delta Functions as Operators The following computation motivates the meaning that we will attach here to the symbol 8a(t). it is called the Dirac delta function at a after the British theoretical physicist P.2. if t -=/= (7) a. thereby defining (5) where a ~ 0. (9) We take Eq. Dirac (1902-1984). Nevertheless. If 8a(t) were a function in the strict sense of the definition. or when it will appear subsequently in such an integral. If we could also take the limit under the integral sign in Eq. (13) Note that if 8 (t) were an actual function satisfying the usual conditions for existence of its Laplace transform. (17). we get the equation . 8(t) is not a function. E---+ 0 (16) where x" (t) is a solution of Ax"+ Bx' + Cx = da. (15) provided that x(t) = lim xE(t). (9). Delta Function Inputs Now. (14) To investigate the response of this system to a unit impulse at the instant t = a. But there is no problem here. (17) makes sense.1. the result is (10) We therefore define the Laplace transform of the delta function to be £{8a(t)} = e-as (a ~ 0). and Eq. When we transform Eq. then Eq. it seems reasonable to replace f(t) with 8a(t) and begin with the equation Ax"+ Bx' + Cx = 8a(t). (15) But what is meant by the solution of such an equation? We will call x (t) a solution of Eq. suppose that we are given a mechanical system whose response x(t) to the external force f(t) is determined by the differential equation Ax" + Bx' + Cx = f(t). (11) If we write 8(t) = 8o(t) and 8(t -a) = 8a(t). (13) would contradict the corollary to Theorem 2 of Section 4.318 Chapter 4 Laplace Transform Methods For instance. For simplicity suppose the initial conditions to be x(O) = x'(O) = 0. writing XE = oC{xE}. (12) then (11) with a = 0 gives £{8(t)} = 1. Because (17) (18) is an ordinary function. finally. (13) is our definition of £ {8 (t)}. E(t). if we take g(t) = e-sr in Eq. Eq. 3. we consider it beyond the scope of the present discussion.2rrs.5). (19) Note that this is precisely the same result that we would obtain if we transformed Eq.3s + 4X(s) = 8e.2rr) = 3 cos 2t + 8e.2rrs 2 • + 4u2rr (t) sin 2t. Note that the impulse at t = 2rr results in a visible discontinuity in the velocity at t = 2rr.a) with a= tan.6 Impulses and Delta Functions 319 If we take the limit in the last equation as E 1- ~ 0.9273) if t ~ 2rr. The resulting motion is shown in Fig.. separation of the cases t < 2rr and t :.2rr) sin 2(t ..6. According to Problem 15. • . 4. It is important to verify that the solution so obtained agrees with the one defined in Eq. and note that = 1 e-sf lim E--+0 SE by l'Hopital's rule.----. we need to solve the initial value problem x" + 4x = 882rr (t). At the instant t = 2rr the mass is struck with a hammer. we see that the inverse transform is x(t) = 3 cos 2t + 4u(t . providing an impulse p = 8. we get the equation (As 2 + Bs +C) X (s) = if X(s) =lim X~'(x).. using the fact that £{8a(t)} = e-as. (16). x'(O) = 0. (15) directly.: 2rr gives x (t ) ~ { 3 cos 2t 5 cos(2t . The formal method is valid in all the examples of this section and will produce correct results in the subsequent problem set. there is no dashpot. E--+ 0 e-as.?:: 2rr. On this basis it is reasonable to solve a differential equation involving a delta function by employing the Laplace transform method exactly as if 8a (t) were an ordinary function.------------------.- --~-- Solution We apply the Laplace transform to get s 2X(s). - A mass m = 1 is attached to a spring with constant k = 4.--·-. Determine the motion of the mass. Because 3cos2t + 4sin2t = 5cos(2t.0. ~~~ Example 1 ·-·"----·---- ·-------. but this depends on a highly technical analysis of the limiting procedures involved. x(O) = 3. so X(s) = 2 3s s +4 s +4 Recalling the transforms of sine and cosine. as it instantaneously increases the amplitude of the oscillations of the mass from 3 to 5.-----~·-··'-·~~~---. .1(4/3) ~ 0. if t . as well as the theorem on translations on the t-axis (Theorem 1 of Section 4.4. The mass is released from rest with x(O) = 3.9273. 4.5. c 1 In this example we have e(t) = 90 . with R = 110 Q.90ui (t) . . df E--->0 df ' E--->0 . dt ' ' Because a a+E FIGURE 4. Find the resulting current i (t) in the circuit. d . Example 2 We return to the RLC circuit of Example 5 of Section 4. At time t = 0 the switch is closed and at timet = 1 it is opened and left open.6. L = 1 H.ua(t) dt d = 8a(t) = 8(t.3.< (t) .4. we may begin with the ordinary circuit equation Li" + Ri' Solution + -i = e' (t).E (t) to Ua (t) shown in Fig. Delta Functions and Step Functions X It is useful to regard the delta function 8a (t) as the derivative of the unit step function ua (t). The motion of the mass of Example 1.90u(t . . 6 - -- - - x=5 X= -3 X= -6 0 2n -5 4n t 6n FIGURE 4.6. C = 0. We readily verify that d -Ua E(t) = da E(t).320 Chapter 4 Laplace Transform Methods t = 2Jr. consider the continuous approximation ua. Approximation ofu0 (t) by Ua. (20) We may regard this as the formal definition of the derivative of the unit step function.ua(t) = hm -Ua E(t) = hm da E(t). and therefore . Suppose that the circuit is initially passive-no current and no charge.4.6.5 we circumvented the discontinuity in the voltage by employing the integrodifferential form of the circuit equation. and a battery supplying e0 = 90 V. although Ua (t) is not differentiable in the ordinary sense at t = a. Now that delta functions are available. an interchange of limits and derivatives yields d .a) . To see why this is reasonable.001 F.1) = 90 . In Section 4. The mass-spring-dashpot system and the series RLC circuit are familiar examples of this general situation. Once w (t) has been determined. (23) The function 1 W(s)=---as2 + bs + c (24) is called the transfer function of the system. From Eq. via the convolution integral-Duhamel's principle reduces the problem of finding a system's outputs for all possible inputs to calculation of the single inverse Laplace transform in (25) that is needed to find its weight function. (24) we see by convolution that x(t) = 1 1 w(r)f(t . For simplicity we assume that the system is initially passive: x(O) = x' (0) = 0.322 Chapter 4 Laplace Transform Methods Systems Analysis and Duhamel's Principle Consider a physical system in which the output or response x(t) to the input function f (t) is described by the differential equation ax"+ bx' +ex = j(t).1{W(s)} (25) is called the weight function of the system. for instance) the response x (t) given by (26) automatically whenever a desired force function f (t ) is input. Then the transform of Eq. the integral in (26) gives the response of the system to an arbitrary input function j(t). Thus the transform of the response to the input j(t) is the product of W(s) and the transform F(s). b. so X (s) = F(s) as 2 + bs + c = W(s)F(s).r) dr. (22) is as 2X(s) + bsX(s) + cX(s) = F(s).spring-dashpot system described by (22) can be constructed in the form of a "black box" that is hard-wired to calculate (and then tabulate or graph. In engineering practice. (22) where the constant coefficients a. so their behaviors can be studied without need for expensive or time-consuming experimentation. In principle-that is. (26) This formula is Duhamel's principle for the system. Hence. The function w(t) = £. a computational analogue for a physical mass. all manner of physical systems are "modeled" in this manner. What is important is that the weight function w(t) is determined completely by the parameters of the system. . and care determined by the physical parameters of the system and are independent off (t). ..r) dr .. (23) that the transform of h(t) is H(s) = W(s).. or unit impulse response. either numerically or graphically. --·-. and measure the response h(t) with an ammeter..... 1 1 h'(r)e' (t ...-···------.. Then Duhamel's principle implies that the response x ( t) to the force f (t) is x(t) = 1' e._... • Note that W(s) = as 2 1 £{8(t)} + bs + c as 2 + bs + c · Consequently. APPLICATIONS: To describe a typical application of Eq... What if the coefficients a. Substitution of (27) in Duhamel's principle gives x(t) = 1 1 h' (t)f(t..__ _ ---._ ___ _ Example 4 Consider a mass-spring-dashpot system (initially passive) that responds to the external force f(t) in accord with the equation x" + 6x' + lOx = f(t)..·---· 323 . . Because £{u(t)} = 1/s..--.... we see from Eq.4. h (t) is the unit step response.. ..6 Impulses and Delta Functions ~. b.··-· ... _________ _.---.... (28).(s + 3) 2 + 1 ' 1 so the weight function is w(t) = e.. the output current i (t) corresponding to the input voltage e(t) will be given by i (t) = (using the fact that f(t) = e' (t )). but with i in place of x. For this reason w(t) is sometimes called the unit impulse response. Then according to Eq..------10 .r) dr (28) for the response of the system to the input f (t).. resistors. Assume that its circuit equation is a linear equation of the form in (22). We then compute the derivative h'(t). Then W (s) - 1 ..r) dr. .s 2 + 6s + ....3 ' (sin r)j(t.3t sin t. and c are unknown. it follows from Eq.------·-· .. .... and capacitors. s It follows from the formula for transforms of integrals that h(t) = 1 1 w(r) dr.. is the derivative of the unit step response... so that w(t) = h' (t) ... (23) that the weight function is simply the response of the system to the delta function input 8(t) . A response that is usually easier to measure in practice is the response h (t) to the unit step function u (t).. perhaps only because they are too difficult to compute? We would still want to know the current i(t) corresponding to any input j(t) = e'(t). so that f(t) = e'(t) = 1 = u(t) . . suppose that we are given a complex series circuit containing many inductors....... We connect the circuit to a linearly increasing voltage e(t) = t.....----------··-¥ .... (28). (27) Thus the weight function._.... then x(t) = (n + 1) sint. Thus resonance occurs because the mass is struck each time it passes through the origin moving to the right-in contrast with Example 3. Thus a resonance phenomenon occurs (see Fig.6. 4rr. 21.. 2rr /10. Consider an RLC circuit in series with a battery. Show that the position function x(t) ofthe mass satisfies the initial value problem x" +X= L n=O 00 o(t. Show that i (t) satisfies the initial value problem i" + 60i' + 1000i . 2rr.8). . Thus the current in alternate cycles of length rr/5 first executes a sine oscillation during one cycle. (a) Show that i (t) satisfies the initial value problem i"+lOOi = 10 ~)-lto(t.. n=O 10 = (n + 1) sin lOt if mr (n <t < + l)rr 10 10 i(O) = i'(O) = 0. . x(O) = x ' (O) = 0. in which the mass was struck each time it returned to the origin. Repeat Problem 19. n=O 10 00 i(O) = i'(O):::: 0. 2rr /10. nn/5. FIGURE 4.6. Suppose that each hammer blow imparts an impulse of+ 1.. rr/5. -10 FIGURE 4. Consider the LC circuit of Problem 18(b). construct a figure showing the graph of this position function.2nn). and so on (see Fig. i(t) (b) Solve this problem to show that if -<t<--nJT (n + 1)rr lO 10 10 then i(t) = e3mr +3rr e3rr - 1 1 e. Consider a mass m = 1 on a spring with constant k = 1. initially at rest. but struck with a hammer at each of the instants t = 0. . except suppose that the switch is alternately closed and opened at times t = 0. (b) Solve this initial value problem to show that i(t) = oo ( 10 L(-l)no t -nn) . rr /10. C = 10-3 F. .6.8. (a) Suppose that the switch is alternately closed and opened at times t = 0. Finally.4..6. The current function of Problem 20.6. if n is odd. 19. . then is dormant during the next cycle. 22. Now show that if nrr (n + l)rr <t<---- Construct a figure showing the graph of this current function.. 2rr/5. .6 Impulses and Delta Functions i(t) i(t) 1\ 1\ 325 In 1\ A -1 v v v v v FIGURE 4. with L = 1 H. The current function of Problem 19. and e0 = 10 V..301 sin lOt. . except suppose that the switch is alternate! y closed and opened at times t = 0. Solve this problem to show that if 2nn < t < 2(n + 1)rr. 4.. 4.7. R = 60 Q.. . The current function of Problem 18. .6.. 20. . . rr/10. .7).nrr). 5 5 then if n is even. . y. the position (x(t). y. x. y 1) = 0. y 1 (t). Applying Newton's law rna= F componentwise. require the use of two or more dependent variables. y 1) = 0.. each a function of a single independent variable (typically time). x. y I . II F2t. z. z . however. x 1 .z II 326 . We will usually denote the independent variable by t and the dependent variables (the unknown functions oft) by x 1. I I I) m z = F3t. Primes will indicate derivatives with respect to t. . x I . y. x . or by x. y. x 2. For an example of a second-order system. A solution of this system is a pair x(t). y. a system of two first-order equations in the dependent variables x and y has the general form f(t . .y . and its velocity (x 1(t) . Z1 (t)). Many applications. X 1 . x . consider a particle of mass m that moves in space under the influence of a force field F that depends on time t. x 3.zI ) ....Linear Systems of Differential Equations Ill First-Order ~ystems and Ap_plications n the preceding chapters we have discussed methods for solving an ordinary differential equation that involves only one dependent variable. we get the system mx II = F 1 (t. g(t .x. For instance. z I ) .y. (2) ( x . (1) where the functions f and g are given. y(t).z. Such a problem leads naturally to a system of simultaneous ordinary differential equations.x.. y. I We will restrict our attention to systems in which the number of equations is the same as the number of dependent variables (unknown functions). ( I I my= z . y (t) of functions of t that satisfy both equations identically over some interval of values oft. z(t)) of the particle. Consider the system of two masses and two springs shown in Fig. . The two brine tanks of Example 2. . .. ..1.30 · + 10 · = ... ..1.. .. m 2y" = -k2(y.1. 20y' = 6x..1. In addition. fresh water flows into tank 1 at 20 galjmin..x units... The "free of differential equations that the position functions x(t) and y(t) must satisfy...-x + . F3 are the components of the vector-valued function F. In the configuration in Fig..5.1.. We denote by x(t) the displacement (to the right) of the mass m 1 from its static equilibrium position (when the system is motionless and in equilibrium and f(t) = 0) and by y(t) the displacement of the mass m 2 from its static position.. and brine is pumped from each tank to the other at the rates indicated in Fig. 5. + 2y. ifm 1 = 2.. .... k 2 = 2. and the brine in tank 2 flows out at 20 gal/min (so the total volume of brine in the two tanks remains constant). . we therefore get the system of differential equations that x(t) and y(t) must satisfy: 1 X Y 3 1 x = ..y 100 200 10 20 ' 1 X Y Y 3 3 y = 30 · .. 5. .3.1...x). Initial Applications Examples 1 through 3 further illustrate how systems of differential equations arise naturally in scientific problems.3. . The brine in each tank is kept uniform by stirring. respectively._t) FIGURE 5. we thereby obtain the system m1x" = -k1x + k2(Y..1. 20x' = -6x + y... The salt concentrations in the two tanks are x 1100 pounds per gallon and y /200 pounds per gallon.1 First-Order Systems and Applications 327 x.. . m 2 = 1.. ' ~ ... then the system in (3) reduces to 2x" = . k 1 = 4. (4) • Example 2 20 gal/min Fresh water 20 gal/min FIGURE 5. 5.. the three right-hand-side functions F 1. The mass- and-spring system of Example I.3.2. " "'" ' '' '" .1.. Thus the two springs are neither stretched nor compressed when x and y are zero. Consider two brine tanks connected as shown in Fig... For instance... . Tank 1 contains x(t) pounds of salt in 100 gal of brine and tank 2 contains y(t) pounds of salt in 200 gal of brine. and f(t) = 40sin3t in appropriate physical units. 5.x) + f(t) (3) k~(y - x) [ii:'t-__ f~.... the first spring is stretched x units and the second by y ... F2 .2.1.6x y" = 2x - body diagrams" for the system of Example 1. z. 5 . When we compute the rates of change of the amount of salt in the two tanks.. . (5) • .1.......10 · .. with a given external force f(t) acting on the right-hand mass m 2 .. We apply Newton's law of motion to the two "free body diagrams" shown in Fig.x .. ' ~··· Example 1 Equilibrium positions FIGURE 5.3y..20 · = . of three second-order equations with independent variable t and dependent variables y.1. 2y + 40 sin 3t.y· 100 200 200 10 20 ' that is. . (6) FIGURE 5.-. 7.=. differentiation of each side of Eq.h in the direction indicated.· ·· .5. where / 1(t) denotes the current in the indicated direction through the inductor L and h (t) denotes the current through the resistor R 2 .-¥--. 5.1. because the voltage drop from the negative to the positive pole of the battery is -100. Voltage drops across common circuit elements.4. We recall Kirchhoff's voltage law to the effect that the (algebraic) sum of the voltage drops around any closed loop of such a network is zero. (7) yields 125/2 dlz d/1 + 75-50. respectively..008 farads £0 : .1. Element :~~rcui't >.~-- "' "·--·· _ _. ~ Example 3 --- ---· . 2 .100 = 0.-. 5.= dt dt 0.~- ~--·.1.·.. 100 volts -_ +G 11 R1 : 50 ohms FIGURE 5. we get the system dh + 25h dt d/1 25/z = 50. The current through the resistor R 1 is I = It . the voltage drops across the three types of circuit elements are those shown in Fig.-.y ·-·¥~-----·~¥.4.1. We apply Kirchhoff's law to the left-hand loop of the network to obtain CQ 2dt dh + 50(/t - /2) .3-5/z =0 dt dt of differential equations that the currents It (t) and / 2 (t) must satisfy.328 Chapter 5 Linear Systems of Differential Equations L: 2 henries C: 0. (6) and (8) by the factors 2 and -25. The right-hand loop yields the equation (7) where Q 2 (t) is the charge on the capacitor. (8) After dividing Eqs. The electrical network of Example 3.. 'Voltage DtQP Ldt RI 1 Inductor Resistor Capacitor dl Consider the electrical network shown in Fig. dh (9) • . As in Section 2. Because dQ 2 jdt = ]z .5.~ -~- · .. as explicit functions of t and lower-order derivatives of the dependent variables.5.x. .. . .. Example 4 The third-order equation xC 3l + 3x " + 2x 1 - Sx = sin 2t is of the form in ( 11) with f (t ... • . X3 = X = X2 t yield the system X I = X2 . II . I X3 =X . . Evidently. x . X].. Hence the substitution of (l2). To describe how such a transformation is accomplished. x 2 . (11) We introduce the dependent variables x 1. (11) yields the system = x" = x 3 .2x 1 - 3x" + sin 2t. in the case of a system of two second-order equations. X 1 . X2 =X = t I X I.3x 3 + sin2t of three first-order equations. .in (13) I X n-1 = Xn. our assumption is that it can be written in the form (10) It is of both practical and theoretical importance that any such higher-order system can be transformed into an equivalent system of first-order equations.2x2 .. For instance. Xn defined as follows: Xn =X. X2. X2 =X . Xn (t) defined in (12) satisfy the system of equations in (13). I I x2 = X 3.1 First-Order Systems and Applications 329 First-Order Systems Consider a system of differential equations that can be solved for the highest-order derivatives of the dependent variables that appear. (12) Note that x ~ = X 1 = x 2 . If Hence the substitutions Xi = X. Xn) X~= j(t. x " ) = Sx .. of n first-order equations. we consider first the "system" consisting of the single nth-order equation _ j( 1 x (n) t. and so on.x.. ' = X(n. . .. in the sense that x(t) is a solution of Eq. X] . (11) if and only if the functions x 1 (t). x 2 (t). x (n -1)) . this system is equivalent to the original nthorder equation in (11). . .1). x~ = Sx1 . x~ Eq. Solution Motivated by the equations in (12). + 40sin3t 2x. large systems of higher-order differential equations typically are solved numerically with the aid of the computer. The corresponding first-order system is (14) and we will see in Section 6. I (17) .py. From a practical viewpoint. Transform this system into an equivalent first-order system. x2=x =x1 . But suppose that we were confronted with the nonlinear equation x" = x 3 + (x 1 ) 3 . x 2.4 that there exist effective numerical techniques for approximating the solution of essentially any first-order system.qx . y~ = 2xi .330 Chapter 5 Linear Systems of Differential Equations It may appear that the first-order system obtained in Example 4 offers little advantage because we could use the methods of Chapter 2 to solve the original (linear) third-order equation. we define XJ=X. y 1. and the first step is to transform such a system into a first-order system for which a standard computer program is available. So in this case the transformation to a first-order system is advantageous. and Y2· Simple Two-Dimensional Systems The linear second-order differential equation x" + px + qx = 1 0 (16) (with constant coefficients and independent variable t) transforms via the substitutions x 1 = y. to which none of our earlier methods can be applied. Example 5 The system 2x" = -6x y" = + 2y.2y1 I (15) + 40sin 3t • of four first-order equations in the dependent variables x 1. Y2=Y =Y1· I I Then the system in (4) yields the system Y1 = Y2. y = . I I Y1=y.2y (4) of second-order equations was derived in Example 1. x" = y 1 into the two-dimensional linear system xl = y. .. Because the moving point (x(t)._. Thus the trajectories of the system in (18) are the ellipses of Fig.1. Example 6 To solve the two-dimensional system x' = -2y. y(t)) has velocity vector (x'(t). Additional information can be shown in the separate graphs of x(t) and y(t) as functions oft. for each value oft.L.1 First-Order Systems and Applications 331 Conversely. A solution (x(t). the direction field plotted in Fig.can be plotted. The choice of an initial point (x(O). Direction field and solution curves for the system x' = -2y. this direction field indicates the point's direction of motion along its trajectory.1. with semiaxes C and C /2. Figure 5.. This gives the single second-order equation x" x (t) = A cos t 5 4 3 2 I "' 0 -1 1-'t__. The picture showing a system's trajectories in the xy-plane-its so-called phase plane portrait-fails to reveal precisely how the point (x(t)..a).6 shows several such ellipses in the xy-plane .a) where A= C cos a and B = C sin a. I ! (18) we begin with the observation that x" = -2y' = -2 Gx) = -x. we can solve this system in ( 17) by solving the familiar single equation in (16).. the point (x(t).:.1.5. x. y(O)) determines which one of these trajectories a particular solution parametrizes.._"'t-". y(t)) lies on the ellipse FIGURE 5.6 indicates that each such point moves counterclockwise around its elliptical trajectory.y) • may be regarded as a parametrization of a solution curve or trajectory of the system in the xy-plane. y(t)) of a two-dimensional system x' = f(t.tr. 5...x. y'=g(t. y) and y' = g (x . -2 -3 -4 -5 X The identity cos 2 () + sin2 () = 1 therefore implies that. y = 2X. then a direction field-showing typical arrows representing vectors with components (proportional to) the derivatives x' = f (x.. y'(t)). y ).6. y' = ~x of Example 6.. If the functions f and g do not involve the independent variable t. . 5..1.. y(t)) moves along its trajectory. Then y(t) = -~x'(t) = -~(-Asint + Bcost) = !C sin(t..H +x = 0 with general solution + B sin t = C cos (t . For instance._~-7--:-'-..6. y).. . then the system has a unique solution satisfying given initial conditions. the system in (24) has a unique solution on the entire interval I that satisfies the n initial conditions . . Xn (t) that (on some interval) identically satisfy each of the equations in (24 ).. given then numbers b 1. Typical initial values would be the initial positions x (0) and y (0) and the initial velocities x' (0) and y' (0). fn are continuous on the open interval I containing the point a.. It tells us that if the coefficient functions PiJ and /j in (24) are continuous. fn are all identically zero. x2(t). .... Thus the linear system in (5) is homogeneous. we found that the amounts x (t) and y (t) of salt in the two tanks of Example 2 are described by the system 20x' = . h .6x + y. y"= 2x-2y+40sin3t. We will see that the general theory of a system of n linear first-order equations shares many similarities with the general theory of a single nth-order linear differential equation. . we saw in Example 5 that the second-order linear system 2x" = . . otherwise. The system in ( 14) is nonlinear because the right -hand side of the second equation is not a linear function of the dependent variables x 1 and x2. Then. Theorem 1 (proved in the Appendix) is analogous to Theorem 2 of Section 2. .3y of two first-order linear equations. (25) Thus n initial conditions are needed to determine a solution of a system of n linear first-order equations. . it is nonhomogeneous. ... .. bn. and we therefore expect a general solution of such a system to involve n arbitrary constants. h. whereas the linear system in (15) is nonhomogeneous.334 Chapter 5 Linear Systems of Differential Equations We say that this system is homogeneous if the functions f 1. . we often must transform it into an equivalent first-order system to discover how many initial conditions are needed to determine a unique solution. is equivalent to the system of four first-order linear equations in (15). For instance. THEOREM 1 Existence and Uniqueness for Linear Systems Suppose that the functions Pll· PI2· . which describes the position functions x(t) and y(t) of Example 1.6x + 2y. On the other hand.2. Pnn and the functions ft. 20y' = 6x. A solution of the system in (24) is ann-tuple of functions x 1 (t).. Hence the two initial values x(O) and y(O) should suffice to determine the solution. Hence four initial conditions would be needed to determine the subsequent motions of the two masses in Example 1. Theorem 1 tells us that the number of such conditions is precisely the same as the number of equations in the equivalent first-order system.. ••• . Given a higher-order system. b2 . -r3 X FIGURE 5. Kepler concluded that the motion of the planets around the sun is described by the following three propositions. Repeat Problem 27. which shows an inductor. Show that the equations of motion of the projectile are 27. A particle of mass m moves in the plane with coordinates (x(t). The square of the planet's period of revolution is proportional to the cube of the major semiaxis of its elliptical orbit. In his Principia Mathematica ( 1687) Isaac Newton deduced the inverse-square law of gravitation from Kepler's laws. 2.14.1. now known as Kepler's laws of planetary motion: 1.14. Suppose that a projectile of mass m moves in a vertical plane in the atmosphere near the surface of the earth under the influence of two forces: a downward gravitational r Jx mx" = +qBy'.15. 3. 5. The orbit of each planet is an ellipse with the sun at one focus. my"= -kvy' .336 Chapter 5 Linear Systems of Differential Equations force of magnitude mg. Johannes Kepler analyzed a lifetime of planetary observations by the astronomer Tycho Brahe.15). 28. The radius vector from the sun to each planet sweeps out area at a constant rate. 29. Then the inverse-square law of gravitation implies (Problem 29) that the . y v '' FIGURE 5. (1) where i = (1. The trajectory of the projectile of Problem 30. y(t)) = x i+ yj.andydirections. except with the generator replaced with a battery supplying an emf of 100 V and with the inductor replaced with a 1-millifarad (mF) capacitor. so the force on the particle is F = qv x B if its velocity is v. see Fig. In this application we lead you (in the opposite direction) through a derivation of Kepler's first two laws from Newton's law of gravitation. Suppose that a particle with mass m and electrical charge q moves in the xy-plane under the influence of the magnetic field B = Bk (thus a uniform field parallel to the z-axis).1. = J (x')2 + (y')2. 5. where = 30.1. Set up a system of first-order differential equations for the indicated currents / 1 and [z in the electrical circuit of Fig. 31. and write the position vector of the planet in the form r(t) = (x(t). 5. Assume that the sun is located at the origin in the plane of motion of a planet. y(t)) under the influence of a force that is directed toward the origin and has magnitude kj(x 2 + y 2)-an inverse-square central force field. Show that II kx mx = . and a resistive force F R that is directed opposite to the velocity vector v and has magnitude kv 2 (where v = lvl is the speed of the projectile.mg. Show that the equations of motion of the particle are and II ky my = . 0) and j = (0. 2 + y2 .1 Application Gravitation and Kepler's Laws of Planetary Motion Around the tum of the 17th century. and a generator which supplies an alternating voltage drop of E(t) = l00sin60t V in the direction of the current / 1• mx" where v = -kvx'. two resistors. my"= -qBx' .r3 .1. 1) denote the unit vectors in the positive x . The electrical circuit of Problem 27. Equating the transverse components-that is.= h dt ' FIGURE 5. Because the polar-coordinate area element-for computation of the area A(t) in Fig. (8) implies that the derivative A' (t) is constant.1. Eq. 5. Area swept out by the radius vector. so Ur = r j r. STEP 1: Differentiate the equations in (3) componentwise to show that dt du.1.5.+r-ue. (8) where h is a constant. which is a statement of Kepler's second law.16 are given by u.1. the coefficients ofue-we get (7) so it follows that 2d8 r .= u. 8 (t)).1. then the radial and transverse unit vectors shown in Fig. If the polar coordinates of the planet at time t are (r (t). The radial and transverse unit vectors Ur and u 11 • The radial unit vector Ur (when located at the planet's position) always points directly away from the origin.17. (3) FIGURE 5. de = ue dt and -=-Ur-· due dt dB dt (4) STEP 2: Use the equations in (4) to differentiate the planet's position vector r = ru. dt dt dt (5) Differentiate again to show that the planet's acceleration vector a dvjdt is given by STEP 3: a= [d r _ r (d8) ] dt 2 dt 2 2 Ur + [~~ (r2d8)]ue.17-is given by dA = ~r 2 d8.1 First-Order Systems and Applications 337 acceleration vector r" (t) of the planet is given by y kr r" =- r 3' (2) where r = J x 2 + y 2 is the distance from the sun to the planet. 5: Equate radial components in (2) and (6) and then use the result in (8) to show that the planet's radial coordinate function r(t) satisfies the second-order differential equation STEP (9) . and thereby show that its velocity vector is given by dr dr dB v = . and the transverse unit vector ue is obtained from Ur by a 90° counterclockwise rotation..16. X = icose +jsin8 and ue = -isin8 +jcose. (2) and (6) must agree. 5. r dt dt (6) STEP 4: The radial and transverse components on the right-hand sides in Eqs. (8) to show that if r = ljz. and first use the chain rule and Eq.1.. it can be transformed to a linear equation by means of a simple substitution. semilatus rectum L (Fig. In rectangular coordinates you can write x(t) = r(t)cost. a parabola if e = 1. (9) that the function z(()) = 1/r(()) satisfies the second-order equation (10) STEP 7: Show that the general solution ofEq.a) FIGURE 5. so close to zero that the orbit looks nearly circular (though with the sun off center). and the eccentricities of the other planetary orbits range from 0.e).-. As indicated in Fig. The object of this procedure is to eliminate dependent variables in succession until there remains only a single equation containing only one dependent variable.2486 for Pluto. C cos a= A. After its solution has been found.1.1. The elliptical orbit r= - L 1 + ecos(B.1 ecos((). + Bcos() + h2 .0068 for Venus and 0. BJ The Method of Elimination The most elementary approach to linear systems of differential equations involves the elimination of dependent variables by appropriately combining pairs of equations.1. sizes. Planetary orbits are bounded and therefore are ellipses with eccentricity e < 1. C sin a = B. y(t) = r(t)sint . (10) is z(()) = Asm() . the other dependent variables can be found in tum. 5.2056 for Mercury and 0. and L = h 2 jk.1. 0 ::::. The shape of the orbit of Halley's comet. using either the original differential equations or those that have appeared in the elimination process. the major axis of the ellipse lies along the radial line() = a. This remaining equation will usually be a linear equation of high order and can frequently be solved by the methods of Chapter 2.338 Chapter 5 Linear Systems of Differential Equations STEP 6: Although the differential equation in (9) is nonlinear.18.18. and rotation angle a.97 (Fig. k (l1) STEP 8: Finally. (12) is a conic section of eccentricity e-an ellipse if 0 ~ e < 1. like Halley's comet withe ~ 0. withe= Ch 2 jk. deduce from Eq. 5 . For this purpose.0167. STEP 9: Plot some typical elliptical orbits as described by (12) with different eccentricities. 5.a) ----- L + (12) with perihelion distance r 1 = Lj(l +e) and aphelion distance r 2 = Lj(l . The eccentricity of the earth's orbit is e ~ 0. assume that the orbit can be written in the polar-coordinate form r = r(()). (11) that r(()) = 1/z(()) is given by r(()) = . But many comets have highly eccentric orbits. The polar-coordinate graph of Eq. 2n FIGURE 5. and a hyperbola if e > 1-with focus at the origin. .19). then dr dt y -hd() dz Differentiate again to deduce from Eq.18). to plot an elliptical orbit with eccentricity e.0933 for Mars to 0.19. and orientations. t::::. as well as for theoretical discussion. that satisfies the initial conditions x(O) Solution y 1 = 6x. -. this yields 1 If 7 I 4([ 7 ) 6Y + 6Y = 6Y I + 6Y . (4) and (5) constitute the general solution of the system in (1).3y.10 = (r.2 The Method of Elimination 339 The method of elimination for linear differential systems is similar to the solution of a linear system of algebraic equations by a process of eliminating the unknowns one at a time until only a single equation with a single unknown remains. (5) Thus Eqs. For such systems the method of elimination provides a simple and concrete approach that requires little preliminary theory or formal machinery. . But for larger systems of differential equations. (3) We then substitute these expressions for x and x 1 in the first equation of the system in (1).-. the matrix methods of later sections in this chapter are preferable.2)(r + 5) = 0. we get (2) so that X - 1 _1 6Y"+7 6Y I . It is most convenient in the case of manageably small systems: those containing no more than two or three equations. which we simplify to y" + 3Y 1 - lOy = 0.3y. If we solve the second equation in ( 1) for x.5. The given initial conditions imply that and that y (O) = CJ + cz = . substitution of (4) in (2) gives that is. This second-order linear equation has characteristic equation r 2 + 3r. so its general solution is (4) Next. .7y (1) = 2.1. y(O) = -1.~ ·· Example 1 Find the particular solution of the system x 1 = 4x. • Remark: The general solution defined by Eqs. if L 1 = D + a and L 2 = D + b. 3 = c I (le2t e2t) 2 ' This expression presents the general solution of the system in ( 1) as a linear combination of the two particular solutions • Polynomial Differential Operators In Example 1 we used an ad hoc procedure to eliminate one of the independent variables by expressing it in terms of the other. y(t)). Direction field and solution curves for the system x' = 4x. By contrast. We now describe a systematic elimination procedure. Operator notation is most convenient for these purposes.St) . e.340 Chapter 5 Linear Systems of Differential Equations these equations are readily solved for c 1 = 2 and c2 = -3. (x(t). Recalling the componentwise addition of vectors (and multiplication of vectors by scalars).2. we can write the general solution in (4) and (5) in the form FIGURE 5. Recall from Section 2.3e-St. D(Dx This illustrates the fact that two polynomial operators with constant coefficients can be multiplied as if they were ordinary polynomials in the "variable" D. . Figure 5.St .1 shows this and other typical solution curves parametrized by the equations x(t) = ~c 1 e 2 t + ~c2 e-sr.7y of Example 1. then = L 1 L2[x] = (D + a)[(D + b)x] = + bx) + a(Dx + bx) [D 2 +(a+ b)D + ab]x.3 that a polynomial differential operator is one of the form (6) where D denotes differentiation with respect to the independent variable t. it follows that (8) if the necessary derivatives of x(t) exist. (4) and (5) may be regarded as the pair or vector (x(t). y' = 6x.1. Because the multiplication of such polynomials is commutative. y(t)) = (~cie2t + ~c2e-sr. this property of commutativity generally fails for polynomial operators with variable coefficients-see Problems 21 and 22. y(t) = c 1e 2t + c2e-sr with different values of the arbitrary constants c 1 and c2 • We see two families of curves resembling hyperbolas sharing the same pair of (oblique) asymptotes. If L 1 and L 2 are two such operators.3y. y(t) = 2e2t . cle2t + c2e-st) + c2 (le . Hence the desired solution is x(t) = 3e2t -e-St.2. then their product L 1 L 2 is defined this way: (7) For instance. Either process is especially simple if the system is homogeneous (f1(t) 0 and fz(t) 0). After solving for y = y(t) we can substitute the result into either of the original equations in (9) and then solve for x = x(t). L 2. = = . and (15) are zero. (11) Subtraction of the first from the second of these equations yields the single equation (12) in the single dependent variable y. Note that the same operator L 1 L 4 . we would get the equation (1 3) which can now be solved for x = x(t). For instance. (12) and (13) can be rewritten as Lz L 3 L4 L1 L1 L3 lx =I Lz L4 ly= I L3 L1 f1 (t) Lz fz(t) L4 f1 (t) fz(t) I. we operate with L 3 on the first equation and with L 1 on the second. 0. This is the operational determinant (14) of the system in (9). you can solve a system of two linear differential equations either by carrying out the systematic elimination procedure described here or by directly employing the determinant notation in ( 15). L 3.4)x 3y = + -6x + (D + 7)y = 0. (9) where L 1 . (13). because in this case the right-hand sides of the equations in (12). (6).2 The Method of Elimination 341 Any system of two linear differential equations with constant coefficients can be written in the form L 1x + Lzy = !1 (t). L 2 = 3. and L 4 = D + 7. L 3 = -6. If so. and f 1 (t) and fz(t) are given functions. (12) and Eq. Thus we obtain the system L3L1x + L 3L2y = L 3f1 (t). Indeed.5. we could eliminate in like manner the dependent variable y from the original system in (9). In determinant notation Eqs. L1L3x + L1L4y = Lifz(t).4. (15) I· It is important to note that the determinants on the right-hand side in (15) are evaluated by means of the operators operating on the functions. The equations in ( 15) are strongly reminiscent of Cramer's rule for the solution of two linear equations in two (algebraic) variables and are thereby easy to remember. (10) with L 1 = D. and L 4 are polynomial differential operators (perhaps of different orders) as in Eq. (1 3). the system in (1) (Example 1) can be written in the form (D. Alternatively. To eliminate the dependent variable x from the system in (9). L3x + L4y = fz(t).L 2L 3 appears on the left-hand side in both Eq. lOx = 0.4)x + 3y = 0. (13) and (12) are x" + 3x'.342 Chapter 5 Linear Systems of Differential Equations Example 2 Find a general solution of the system (D.5a2e-5t). • As illustrated by Example 2. but actually are not independent.10 = (r. (16) Hence Eqs. The "extra" constants must then be eliminated by substitution of the proposed general solution into one or more of the original differential equations. Solution (10) The operational determinant of this system is (D. This apparent difficulty demands a resolution. the elimination procedure used to solve a linear system frequently will introduce a number of interdependent constants that may appear to be arbitrary. and b2.2)(r + 5) = 0. so it follows that a 1 = ~b 1 and a2 = ~b2 . -6x+(D+7)y=0. Therefore. we get 0 = x ' . We can discover them by substituting the solutions in (17) into either of the original equations in (10). so their (separate) general solutions are x(t) = a1e 2t y(t) = + a 2e. the desired general solution is given by Note that this result is in accord with the general solution (Eqs.1 that the general solution of a system of two first-order equations involves only two arbitrary constants. b1e 2t + b2e-St.st are linearly independent functions.5t. (17) At this point we appear to have four arbitrary constants a 1. (4) and (5)) that we obtained by a different method in Example 1.lOy = 0. a 2. But e 21 and e.4x + 3y = (2ale2t.4)(D + 7) - 3 · ( -6) = D 2 + 3D. The explanation is simple: There must be some hidden relations among our four constants. On substitution in the first equation. that is. b 1. 0 = (-2al + 3b!)e2t + (-9a2 + 3b2)e-st. The characteristic equation of each is r2 + 3r.4(ale2t + a2e-5t) + 3(b!e2t + b2e-5t).10. The appropriate number of arbitrary constants in a general solution of a linear system is determined by the following proposition: . But it follows from Theorem 1 in Section 5. y" + 3y'. On the other hand. A degenerate system may have either no solution or infinitely many independent solutions. its degree as a polynomial in D. 2Dx. For the system L11x L21x L 31X + L12y + L13Z = f1 (t). however. the equations Dx 2Dx + Dy = + 2Dy = t.2Dy = 1 with operational determinant zero are obviously inconsistent and thus have no solutions. then the system is said to be degenerate. Although the aforementioned procedures and results are described for the case of a system of two equations.L. For instance. because its operational determinant D 2 +3D. If the operational determinant is identically zero. we can substitute any (continuously differentiable) function for x (t) and then integrate to obtain y (t). see pages 144-150 of E. every degenerate system is equivalent to either an inconsistent system or a redundant system. the equations Dx. the dependent variable x (t) satisfies the single linear equation L11 L12 L13 L21 L22 L23 X = L 31 L32 L 33 !1 (t) L12 L13 h(t) L22 L23 !J(t) L 32 L33 (19) with analogous equations for y = y (t) and z = z (t). Roughly speaking.Dy = 0.10 is of order 2. they can be generalized readily to systems of three or more equations.) Thus the general solution of the system in (1 0) of Example 2 involves two arbitrary constants. + L22y + L23Z = h(t).5. Example 3 illustrates this approach. The methods of this section often can be applied to analyze the "natural modes of oscillation" of a given system.2 The Method of Elimination If the operational determinant in (15) is not identically zero. then the number of independent arbitrary constants in a general solution of the system in (9) is equal to the order of its operational determinant-that is. 1956). Ince's Ordinary Differential Equations (New York: Dover. 2t with operational determinant zero are obviously redundant. 343 (For a proof of this fact. For most systems of more than three equations. the method of operational determinants is too tedious to be practical. Mechanical Vibrations A mechanical system typically vibrates or oscillates periodically in one or more specific ways. + L 32Y + L 33Z = j 3(t) (18) of three linear equations. . .5. bn ] ' it is understood that a 1 . aij + bij = bij + aij for all i and j because addition of real numbers is commutative. such as c = [ 5 17 0 -3].3 Matrices and Linear Systems 349 We denote ( -1)A by -A and define subtraction of matrices as follows: A.•• . (9) Each of these properties is readily verified by elementwise application of a corresponding property of the real numbers.one having only a single row. we will frequently write a column vector as the transpose of a row vector. The transpose AT of them x n matrix A = [aiJ] is then x m (note!) matrix whose jth column is the jth row of A (and consequently. Consequently. and am are the row vectors of the matrix A and that b1. For aesthetic and typographical reasons. (5) The matrix operations just defined have the following properties. . (c+d)A =cA+dA. a row vector is a 1 x n matrix. a 2 . Thus AT = [ aJi although this is not notationally perfect. you must remember that AT will not have the same shape as A unless A is a square matrix-that is. . or simply a vector. A + B = [ aij + biJ J = [ biJ + aij J= B + A. each of which is analogous to a familiar algebraic property of the real number system: A+O=O+A=A.B =A+ (-B). Thus if we write and B = [ b1 . . (distributivity) (6) (7) (8) A + (B + C) = (A + B) + C c(A +B) = cA + cB. (commutativity). . (associativity). An m x 1 matrix-one having only a single column-is called a column vector. for example. and bn are the column vectors of the matrix B. whose ith row is the ith column of A). unless m = n. as in J. . b2.. For example. A+B=B+A A-A=O. We often denote column vectors by boldface lowercase letters. . or x= Similarly. the two preceding column vectors may be written in the forms b= [3 -7 0 r and X = [X] X2 x mr- Sometimes it is convenient to describe an m x n matrix in terms of either its m row vectors or its n column vectors. The product AB of two matrices is defined only if the number of columns of A is equal to the number of rows of B. It may help to think of "pouring the rows of A down the columns of B. (12) For purposes of hand computation.350 Chapter 5 Linear Systems of Differential Equations Matrix Multiplication The properties listed in Eqs. The first surprises in the realm of matrix arithmetic come with multiplication. Example 2 Check your understanding of the definition of matrix multiplication by verifying that if . (6) through (9) are quite natural and expected. ai -----+ bpi bp2 am! a m2 a mp f?pJ bpn t b· J which shows that one forms the dot product of the row vector ai with the column vector b 1 to obtain the element ciJ in the ith row and the jth column of AB. a 21 a12 a22 a1p a2p bll b2! b12 b 22 btj bi n b2n hz · ·. (11) can be ciJ = L k= I aikbkJ. If A is an m x p matrix and B is a p x n matrix. each having the same number p of elements. (11) and (12) is easy to remember by visualizing the picture a. then their product AB is the m x n matrix C = [ ciJ ] . the definition in Eqs. We define first the scalar product of a row vector a and a column vector b. Thus C = AB = [ ai · b1 ] . and B (11) In terms of the individual entries of A recast in the form = [ aiJ] p = [ biJ ]. (10) exactly as in the scalar or dot product of two vectors-a familiar topic from elementary calculus. where ciJ is the scalar product of the ith row vector ai of A and the jth column vector b1 of B.1." This also reminds us that the number of columns of A must be equal to the number of rows of B. If then a · b is defined as follows: p a· b = Lakbk = a 1b 1 k=I + a2b2 + · · · + apbp. .. Eq.c-: . Inverse Matrices A square n x n matrix is said to have order n. if A and B are both n x n matrices (so that both the products AB and BA are defined and have the same dimensions-n x n ). it can happen that AB = 0 even though A =f. AB Moreover. 0.3 Matrices and Linear Systems 351 then AB = [ Similarly.2z 6x -7y 15 23 3J 7-1] 1 3 5 .5. (15) =f. 0 and B =f. But matrix multiplication is not commutative. that is. although you can easily construct your own examples using 2 x 2 matrices with small integral elements. then. The identity matrix of order n is the square matrix I= 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 (17) 0 0 . verify that i -. (16) Examples illustrating the phenomena in (15) and (16) may be found in the problems. A(BC) = (AB)C and A(B + C) = AB + AC. • It can be shown by direct (though lengthy) computation based on its definition that matrix multiplication is associative and is also distributive with respect to matrix addition. ] [ ~ ~ J 1 = [ 1 ~ ~~ l ] -3 5 -7 and that -2 0 1] [x] [ y = z 2x+z 4x + 53y y . in general. That is. BA. (14) (13) provided that the matrices are of such sizes that the indicated multiplications and additions are possible. 1• It is clear that some square matrices do not have inverses-consider any square zero matrix. In linear algebra it is proved that A . If A = [ aij J is ann x n matrix. then this inverse is unique. It is also easy to show that if A -I exists. the results are the same in all 2n cases.1 exists and (A -I ). let Aij denote the (n .1 exists if and only the determinant det(A) of the square matrix A is nonzero. then an inverse of A is a square matrix B of the same order as A such that both AB = I and BA = I . The expansion of the determinant IAI along its ith row is given by n IAI = L(-l)i+jaijiAijl j=1 (i fixed). we may speak of the inverse of A. (20a) and its expansion along its jth column is given by n IAI = L(-l)i+jaijiAijl i= 1 (j fixed). (20b) It is shown in linear algebra that whichever row we use in Eq.1 =A. and we will denote it by A.1) x (n . Determinants We assume that the student has computed 2 x 2 and 3 x 3 determinants in earlier courses. the matrix A is said to be nonsingular. Consequently. If A [ aij J is a 2 x 2 matrix.1 • Thus AA. then A is called a singular matrix.352 Chapter 5 Linear Systems of Differential Equations for which each entry on the principal diagonal is 1 and all off-diagonal entries are zero. It is not difficult to show that if the matrix A has an inverse. If A is a square matrix.1A . . Hence IAI is well defined by these formulas. (20a) and whichever column we use in Eq.1 =I= A.1). if det(A) = 0. If so. then its determinant det(A) = IAI is defined as Determinants of higher order may be defined by induction. (20b). It is quite easy to verify that AI= A= lA (18) for every square matrix A of the same order as I. as follows. (19) given the existence of A . then (A .1) matrix obtained from A by deleting its ith row and its jth column. A solution of Eq. x~ = 6x 1 . especially those computations that would be burdensome in scalar notation. The matrix notation used in Eq.7xz can be written as the single matrix equation To verify that the vector functions Xt (t) 3e2t = [ 2e2t J and x2 (t) = [ -5t 3: _ 51 J . X~ = P2I (t)Xt + P22(t)xz + · · · + P2n (t)Xn + /z(t). (28) on the open interval I is a column vector function x(t) = [Xi (t) J such that the component functions of x satisfy the system in (27) identically on I.5. We discuss here the general system of n first-order linear equations + Ptz(t)xz + · · · + Pin(t)Xn + /J(t). Example 6 The first -order system x~ = 4xt .3xz. (28) not only emphasizes this analogy. then the system in (27) takes the form of a single matrix equation - dx dt = P(t)x + f(t). but also saves a great deal of space. Our main use for matrix notation will be the simplification of computations with systems of differential equations. X~= P3t(t)Xt + P3z(t)xz + · · · + P3n(t)xn + /3(t).1 guarantees the existence on I of a unique solution x(t) satisfying preassigned initial conditions x(a) = b. but it is readily assimilated with practice. X~= Pll(t)Xt (27) X~= Pni(t)XI + Pnz(t)xz + · · · + Pnn(t)Xn + fn(t). then Theorem 1 of Section 5. (28) We will see that the general theory of the linear system in (27) closely parallels that of a single nth-order equation. P(t) = ( Pij(t)] If we introduce the coefficient matrix and the column vectors x= [ Xi ] and f(t) = [ fi (t) ] . If the functions Pij (t) and /i (t) are all continuous on I.3 Matrices and Linear Systems 355 First-Order Linear Systems The notation and terminology of matrices and vectors may seem rather elaborate when first encountered. . c11 are constants. then the linear combination (31) is also a solution of Eq. and such that every solution of Eq.2. x2 . (29) on I.. so it follows = = + c2x. We expect it to have n solutions x 1 . (28). x' = P(t)x. . Theorem 1 is analogous to Theorem 1 of Section 2. THEOREM 1 Principle of Superposition Let x 1. but with f(t) 0. + · · · + CnX~ CJP(t)Xt + c2P(t)x2 + · · · + CnP(t)X P(t)(CJXJ + C2X2 + · · · + C X 11 11 ). Xn that are independent in some appropriate sense. The remarkable simplicity of this proof demonstrates clearly one advantage of matrix notation. dt (29) which has the form shown in Eq. let us write Xtj(t) = (30) Thus Xij (t) denotes the ith component of the vector xj (t). .356 Chapter 5 Linear Systems of Differential Equations are both solutions of the matrix differential equation with coefficient matrix P. c2 .. . we consider first the associated homogeneous eq nation dx . . (29). x 2 .= P(t)x. If c 1 . . x2. as desired. . (28). 11 That is. so the second subscript refers to the vector function Xj (t). . Xn of Eq.A. . . we need only calculate Px I =[4 6 -3 ] [ 3e21 -7 2e2t ] =[ 6e 21 4e2r ] = x1 1 and • To investigate the general nature of the solutions of Eq. whereas the first subscript refers to a component of this function.. Proof: We know that x. Xn ben solutions of the homogeneous linear equation in (29) on the open interval I. •. (29) is a linear combination of these n particular solutions.. •. Given n solutions x 1. immediately that x' = CJX~ P(t)xi for each i (1 < < n). If x 1.2.. its proof is essentially the same. en not all zero such that C[XJ (t) + CzXz(t) + · · · + CnXn (t) = 0 (32) for all t in I.51. Xz. We may write either W(t) or W(x 1 ..2).5. . Xz(t) + 3c2e. x 2 . . they are linearly independent.. In scalar form with x = [x 1 X! V.51 ] is also a solution. Xn.•• . there is a Wronskian determinant that tells us whether or not n given solutions of the homogeneous equation in (29) are lir:tearly dependent. For instance. .e 21 + c 2e. x 2 . with the definition of W (x. [ 2ez1 x2 3e21 ] + cz [ 3e-51 e . Just as in the case of a single nth-order equation. Xn) in Eq.. neither is a scalar multiple of the other. . . . .2. Equivalently. Otherwise.51. e21 = 2c. The vector-valued functions x 1. x2 . . (33) substituted for the definition of the Wronskian of n solutions of a single nth-order equation (see Problems 42 through 44). Xn are such solutions. . •• • .. then the linear combination x(t) = c 1x 1 (t) + czxz(t) = c. . this gives the solution (t) = 3c. Xn are linearly dependent on the interval I provided that there exist constants c 1. then their Wronskian is the n x n determinant X11 (t) XJ2(t) X in (t) x21(t) xzz(t) Xzn(t) W(t) = X n! (t) Xnz(t) (33) using the notation in (30) for the components of the solutions. . x2 .3 Matrices and Linear Systems 357 Example 6 Continued If x 1 and x2 are the two solutions of dx _ [ 4 dt 6 -3] -7 X discussed in Example 6. c2 . Moreover. Xn).. the two solutions x 1 and x 2 of Example 6 are linearly independent because. . they are linearly independent provided that no one of them is a linear combination of the others. Note that W is the determinant of the matrix that has as its column vectors the solutions x 1 . Independence and General Solutions Linear independence is defined in the same way for vector-valued functions as for real-valued functions (Section 2. which is equivalent to the general solution we found by the method of elimination • in Example 2 of Section 5. clearly.. Theorem 2 is analogous to Theorem 3 of Section 2. . en such that (35) for all t in I. Xn Suppose that XI. and x3 are • Theorem 3 is analogous to Theorem 4 of Section 2. • If XI. . c2. . .2. x 2 .358 Chapter 5 Linear Systems of Differential Equations THEOREM 2 Wronskians of Solutions . which is never zero. . or W = 0 at no point of I. Xn be n linearly independent solutions of the homogeneous linear equation x' = P(t)x on an open interval I . .•.. Xn. . Thus there are only two possibilities for solutions of homogeneous systems: Either W = 0 at every point of I. are n solutions of the homogeneous linear equation x' = P(t)x on an open interval I. then W = 0 at every point of I. It says that a general solution of the homogeneous n x n system x' = P(t)x is a linear combination (35) of any n given linearly independent solutions XI. XI .. -e3r are solutions of the equation (34) The Wronskian of these solutions is W= 2et 2e 31 2e1 0 et -e3r 2 2 2 2 0 -2 1 -1 1 = -16e91 .. . then there exist numbers CI. x 2 . Suppose also that P(t) is continuous on I. x 2 . x2. . Hence Theorem 2 implies that the solutions linearly independent (on any open interval). x 2 .. THEOREM 3 General Solutions of Homogeneous Systems Let xI.••" " . where P(t) is continuous. Xn are linearly dependent on I . . < - N Oo <<•. .--~''"- "'''"''' '"' W Example 7 It is readily verified (as in Example 6) that X2(t) =[ 2e 03r ] . If x(t) is any solution whatsoever of the equation x' = P(t )x on I . x2.•• .. then W f= 0 at each point of I.' "' """ ' ' " " ' '' '" o . Xn are linearly independent on I. .• . Let Then: • If XI. note that the given solution x(t) and the solution y(t) of Eq. This establishes Eq.1 that x(t) = y(t) for all t in I. x2 . It suffices to choose for Xj (t) the unique solution such that 0 0 0 Xj(a) = 0 1 +--.. Xn.& Remark: Every n x n system x' = P(t )x with continuous coefficient matrix does have a set of n linearly independent solutions x 1 . Finally.1x(a) satisfies Eq. as desired. that is. . x2 .. We show first that there exist numbers c 1. . . . .5.. .1 • Therefore the vector c = X(a). c2.. such that (37) Let X(t) be then x n matrix with column vectors x 1. . It follows from the existence-uniqueness theorem of Section 5. (36)with the values of ci determined by the equation c = X(a). Xn are linearly independent. x2 . . . • . and let c be the column vector with components c 1. . en. (37) may be written in the form (38) X(a)c = x(a). (In other words. Xj(a) is merely the jth column of the identity matrix.. Cn such that the solution (36) has the same initial values at t =a as does the given solution x(t). (38). •.. . Xn are linearly independent by Theorem 2.3 Matrices and Linear Systems 359 Proof: Let a be a fixed point of I.. Then Eq.. c2 . (35).1x(a)-have the same initial values (at t = a). Xn as in the hypotheses of Theorem 3.position j 0 0 -that is. Hence the matrix X(a) has an inverse matrix X(a) . How actually to find these solutions explicitly is another matter-one that we address in Section 5. (35) of the homogeneous linear system x' = P(t)x can be written in the form x(t) = X(t)c. . The Wronskian determinant W (a) = IX (a) I is nonzero because the solutions x1.. (39) . the column vector with all elements zero except for a 1 in row j. • Initial Value Problems and Elementary Row Operations The general solution in Eq. x2 .) Then so the solutions x 1 .4 (for the case of constant coefficient matrices). it suffices to solve the system · · · bn r is given. until the first equation is finally solved for x r. constant vector b = [bi ]. First the last equation in (44) is solved for Xn. 2. n). The transformed system is then easily solved by the process of back substitution. (44) in which only the unknowns Xj. Xn appear explicitly in the jth equation (j = 1. Xj+I · . .. arrxr (43) with nonsingular coefficient matrix A = [aij ]. The transformation of the system in (43) to upper triangular form is most easily described in terms of elementary row operations on the augmented coefficient matrix all ar2 (A!h] = a2r a22 an2 arn br a2n b2 ann bn (45) ani that is obtained by adjoining the vector b to the matrix A as an additional column.I. The admissible elementary row operations are of the following three types: . . . a2IXI + a22X2 + · · · + a2nXn = b2. .•• . x 2 . Xn. x(a) = b. Cn in Eq. The basic idea is to transform the system in (43) into the simpler upper triangular form aiiXI + a 12X2 + · · · + arnXn = br. c2 . (39). according to (42) X(a)c = b to find the coefficients c 1 . and unknowns x 1 . then the next-to-last is solved for Xn. We therefore review briefly the elementary technique of row reduction to solve an n x n algebraic linear system + ar2x2 + · · · + arnXn = br. Xn. and so forth.360 Chapter 5 Linear Systems of Differential Equations where X(t) = [ x 1 (t) X2(t) Xn (t)] (40) is then x n matrix whose column vectors are the linearly independent solutions x 1. a22X2 + '' ' + a2nXn = b2.. (41) where the initial vector b = [ br b2 Eq.• .... . . (35).•• . and C = [ Ct C2 Cn is the vector of coefficients in the linear combination r (35) Suppose now that we wish to solve the initial value problem dx dt = Px. X2. Then. Multiply any (single) row of the matrix by a nonzero constant. 3.1 -1 1 : 6 I : i OJ 1 . The process of transforming [ A ! b Jis carried out one column at a time. In scalar form.C 2 + C3 = 6 with augmented coefficient matrix 2 2 2 ! 0 -2 2 [ 2 1 . This upper triangular augmented coefficient matrix then corresponds to an upper triangular system as in (44).5. Example 8 Use the solution vectors given in Example 7 to solve the initial value problem dx dt Solution = [ -~ 0 -1 -2 OJ 3 -2 3 X. from left to right. Ct .1 1 : 6 Multiplication of each of the first two rows by i OJ .2c3 = 2. X3(0) = 6. = X3(t) = Cte1 - We seek the particular solution satisfying the initial conditions X! (0) = 0. we get the algebraic linear system 2ct + 2c2 + 2c3 = 0. . this gives the general solution Xt(t) = 2cte 1 x 2(t) 2cte 1 + 2c2e31 + 2c 3e51 .2c3e51 . ! gives [i 1 1 0 . as in the next example. Interchange any two rows of the matrix. x(O) = [H (46) It follows from Theorem 3 that the linear combination is a general solution of the 3 x 3 linear system in (46). . in tum) to transform [A ! b] into an upper triangular matrix. one that has only zeros beneath its principal diagonal. The goal is to use a sequence of such operations (one by one. C2e 31 + c3e51 . X2(0) = 2. 2ct . 361 2. When we substitute these values in the three preceding scalar equations. Subtract a constant multiple of one row from any other row.3 Matrices and Linear Systems 1. Now we multiply the second row by -1.. . (47) The following theorem is analogous to Theorem 5 of Section 2. If x(t) is any solution whatsoever of Eq. We finally solve in tum for c3 = 1. (47) has the form x(t) = Xc(t) + Xp(t). 4c3 = 4. (47) and the complementary function Xc(t) is a general solution of the associated homogeneous equation x' = P(t)x.. and c 1 = 2. x2. (48) where xp(t) is a single particular solution ofEq. Theorem 4 means that the general solution of Eq. . In brief. . Let x 1. . c2.2. Xn be linearly independent solutions of the associated homogeneous equation on I. . en such that (49) for all tin I. . (t) - 3x2(t) + x 3(t) = 4e1 [ 4e 1 2er - 6e 31 + 2e 5' 3e3r + esr - + 2e 51 ] • • Nonhomogeneous Solutions We finally tum our attention to a nonhomogeneous linear system of the form dx dt = P(t)x + f(t). then add twice the result to the third row.. then there exist numbers c1. THEOREM 4 Solutions of Nonhomogeneous Systems Let Xp be a particular solution of the nonhomogeneous linear equation in (47) on an open interval I on which the functions P(t) and f(t) are continuous. c 2 = -3.2 and is proved in precisely the same way. Thereby we get the upper triangular augmented coefficient matrix [~ CI 1 1 0 2 1: OJ !-1 I 4 : 4 that corresponds to the transformed system + C2 + C3 = c2 + 2c3 = Q. -1. Thus the desired particular solution is given by x(t) = 2x.362 Chapter 5 Linear Systems of Differential Equations then subtraction of the first row both from the second row and from the third row gives the matrix 1 1 1 [ 0 -1 -2 0 -2 0 The first column of this matrix now has the desired form. substituting the preceding theorems in this section for the analogous theorems of Section 2. (4 7) on I. In Example 7 we saw that a general solution of the associated homogeneous linear system dx dt is given by [-! -23 -2 OJ -1 3 X and we can verify by substitution that the function (found using a computer algebra system. xz(t) = 2c.15.e1 + 2c2 e 31 + 2c3e 51 + 3t. 2. Finding the general solution Xc(t) of the associated homogeneous system. by (t) = 2c.e' - Cze 3' + c3e 51 + 2t.5.8) is a particular solution of the original nonhomogeneous system. XJ x3(t) = c. Finding a single particular solution Xp (t) of the nonhomogeneous system.e1 . The sum x(t) = Xc(t)+xp(t) will then be a general solution of the nonhomogeneous system.2c 3 e 51 + 5. . Example 9 The nonhomogeneous linear system xi = 3x. Consequently. - - xz + 3x3 - 6t + 7 is of the form in (47) with OJ P(t)=[ -2 -1 3 3 -2 -! ' f(t) = [ -9t + 13 7t.15 -6t + 7 J. Theorem 4 implies that a general solution of the nonhomogeneous system is given by x(t) = Xc(t) + Xp(t). that is. • . 2x3 + 7t .2x2 x~ = -x 1 + 3x2 x~ = - 9t + 13 . or perhaps by a human being using a method discussed in Section 5.3 Matrices and Linear Systems 363 Thus finding a general solution of a nonhomogeneous linear system involves two separate steps: 1. the linear combination (2) with arbitrary coefficients will then be a general solution of the system in (1). the Mathematica command C = Inverse[A]. . . VI. Vn in Eq.B).t V3eAt VI vz V3 eAt x(t) = X3 = veAt (3) Xn VneAI Vn where A. . (1) By Theorem 3 of Section 5. we know that it suffices to find n linearly independent solution vectors XI. . we proceed by analogy with the characteristic root method for solving a single homogeneous equation with constant coefficients (Section 2. To search for the n needed linearly independent solution vectors. indeed.366 Chapter 5 Linear Systems of Differential Equations C := multiply(inverse(A). It is reasonable to anticipate solution vectors of the form XI xz VI e AI v 2 eA. n) in (1 ).B or the MATLAB commands C = inv(A)B Use your own calculator or available computer algebra system to solve "automatically" Problems 31 through 40 in this section. v 2 ... Vn are appropriate scalar constants. .. 2.. For if we substitute (i = 1.we can hope to solve for values of the coefficients v 1. . (3) so that x(t) = ve At is.. This will leave us with n linear equations which.3).• . We now introduce a powerful alternative to the method of elimination for constructing the general solution of a homogeneous first-order linear system with constant coefficients. •. a solution of the system in (1). vz.. then each term in the resulting equations will have the factor eAt. Xz. x~ = x~ = a1IXJ aziXI + aizXz + · · · + a1nXn.for appropriate values of A. To investigate this possibility. so we can cancel it throughout. xn... it is more efficient to write the system in ( 1) in the matrix form x' = Ax (4) .3. + azzXz + · · · + a znXn. that is.A [A.4 The Eigenvalue Method for Homogeneous Systems where A = [ aij ]. For this reason. •. (5) in the form (A. (6) by c =!= 0.AI[= 0. (5) holds.. then so is any nonzero constant scalar multiple cv of v-this follows upon multiplication of each side in Eq. When we substitute the trial solution x x' = AVeM in Eq. The prefix eigen is a German word with the approximate translation characteristic in this context.1 with derivative (5) This means that x = veM will be a nontrivial solution of Eq. (7) An eigenvector associated with the eigenvalue A is a nonzero vector v such that Av = AV. 367 = ve". (7) In its simplest formulation. the eigenvalue method for solving the system x' = Ax consists of finding A so that Eq. v2 . Vn. . By a standard theorem of linear algebra. the terms characteristic value and characteristic vector are in common use. (6) Given A. (6) Note that if vis an eigenvector associated with the eigenvalue A.5. The question now is this: How do we find v and A? To answer this question.Al)v = 0. so that (A. it has a nontrivial solution if and only if the determinant of its coefficient matrix vanishes. DEFINITION Eigenvalues and Eigenvectors The number A (either zero or nonzero) is called an eigenvalue of the n x n matrix A provided that [A . (4).A (8) .. The name of the method comes from the following definition. the matrix product Avis a scalar multiple of the vector v. Vn· Then x =veAl will be a solution vector. this is a system of n homogeneous linear equations in the unknowns v 1. v2 . the result is AveM = Ave"-1 • We cancel the nonzero scalar factor e"-1 to get Av = AV. the equation a11a 21 A a12 a in a2n a22. . (6) with this value of A to obtain v 1 .Al)v = 0.AI) = 0. if and only if [A . that is.AI[= ani an2 =0 ann . (7) holds and next solving Eq. . (4) provided that vis a nonzero vector and A is a constant such that Eq. we rewrite Eq..AI[ = det(A . The case of repeated eigenvalues. (4) through (7) provides a proof of the following theorem. We will discuss separately the various cases that can occur. Our discussion of Eqs. if any).will be deferred to Section 5. possibly some are repeated-and thus an n x n matrix has n eigenvalues (counting repetitions. . . . V11 associated with these eigenvalues.. we get n linearly independent solutions (10) In this case the general solution of x' = Ax is a linear combination of these n solutions.6.368 Chapter 5 Linear Systems of Differential Equations is called the characteristic equation of the matrix A.. dt If v is an eigenvector a~sociated with A. 3. An of the matrix A. A2 . THEOREM 1 Eigenvalue Solutions of x' =Ax Let A be an eigenvalue of th~ [constant] coefficient matrix A of the first-order linear system dx -=Ax. 2. Next we attempt to find n linearly independent eigenvectors v 1. we allow the possibility of complex eigenvalues and complex-valued eigenvectors. which is the basis for the eigenvalue method of solving a first-order linear system with constant coefficients. depending on whether the eigenvalues are distinct or repeated. real or complex. . its roots are the eigenvalues of A. but when it is. this method for solving then x n homogeneous constant-coefficient system x' = Ax proceeds as follows: 1. .multiple roots of the characteristic equation. Although we assume that the elements of A are real numbers. this equation has n roots-possibly some are complex. The Eigenvalue Method In outline. Step 2 is not always possible.. Upon expanding the determinant in (8). then x(t) = ve'At is a nontrivial solution of the system. We first solve the characteristic equation in (8) for the eigenvalues A1. v2 . we evidently get an nth-degree polynomial of the form (9) By the fundamental theorem of algebra. . .. (11) Solution The matrix form of the system in ( 11) is (12) The characteristic equation of the coefficient matrix is 2 1=(4-A)(-1-A)-6 -1. the two scalar equations 6a 3a + 2b = + b= 0. 1988).5. (13) yields that is. Example 1 Find a general solution of the system xi = 4xl + 2x 2 . For the coefficient matrix A in Eq.. NJ: Prentice Hall. .2 in Eq. for the associated eigenvector v = [ a CASE 1: AJ the system = -2.Al)v = 0 takes the form (13) b ]7 .4 The Eigenvalue Method for Homogeneous Systems 369 Distinct Real Eigenvalues If the eigenvalues A1 .we can choose a arbitrarily (but nonzero) and then solve for b.) In any particular example such linear independence can always be verified by using the Wronskian determinant of Section 5. Elementary Linear Algebra (Englewood Cliffs. (For instance. then we substitute each of them in turn in Eq. see Section 6. Therefore. x2 = 3xl. (14) has infinitely many nonzero solutions.5) = 0.A = A2 - 3A .10 =(A + 2)(A .3..2 of Edwards and Penney. . A2 . .x2.. In this case it can be proved that the particular solution vectors given in (10) are always linearly independent.3. An are real and distinct. Vn . v2 . . The following example illustrates the procedure. Substitution of the first eigenvalue AJ = . so we have the distinct real eigenvalues A1 = -2 and A2 = 5. (6) and solve for the associated eigenvectors v 1 . (14) In contrast with the nonsingular (algebraic) linear systems whose solutions we discussed in Section 5. the homogeneous linear system in (14) is singular-the two scalar equations obviously are equivalent (each being a multiple of the other). Eq. ( 12) the eigenvector equation (A. 0. 370 Chapter 5 Linear Systems of Differential Equations Substitution of an eigenvalue A in the eigenvector equation (A . Hence a general solution of the system in (11) is X(t ) 1- = CtXt (t) + C2X2(t) = Ct X t (t) 1 ] e[ _3 21 + C2 [ 2 l Je 51. x2(t) = -3ct e. They are linearly independent because their Wronskian 2est 7 3t e sr . 3a.Al)v = 0 always yields a singular homogeneous linear system. so is an eigenvector associated with A2 = 5.3. we had made another choice a = c =I= 0. b = c =I= 0 would merely give a [constant] multiple of v 2 . = Cte- 21 + 2c2 e51 . the choice a = 1 yields b = -3. Looking at the second equation in (14). A different choice a = 2c. and thus is an eigenvector associated with At = -2 (as is any nonzero constant multiple of vi). b = . These two eigenvalues and associated eigenvectors yield the two solutions Xt(t) = [ -~ Je-21 and x2 (t) =[ i] e5 r. any choice we make leads to (a constant multiple of) the same solution CASE 2: A2 = 5. and among its infinity of solutions we generally seek a "simple" solution with small integer values (if possible). in scalar form. b = -3c. we would have obtained the eigenvector = [ Vt -3~ J= C [ -~ J· Because this is a constant multiple of our previous result. With b 0.6b = 0 (15) = 1 in the first equation we get a = 2. Substitution of the second eigenvalue A = 5 in (13) yields the pair -a+ 2b = of equivalent scalar equations.21 + c2 e 51• . Remark: If instead of the "simplest" choice a = 1.e is nonzero. .- . We see two families of hyperbolas sharing the same pair of asymptotes: the line x 1 = 2x2 obtained from the general solution with c 1 = 0. 2. .-. ..1.. If each flow rate is r gallons per minute..··-···""""""" ..x 2 of Example 1. . then a simple accounting of salt concentrations... Xn denote the components of a single vector-valued solution x.-------· . -.. and out of tank 3.5. and then the whole physical system is modeled by a system of differential equations. . yields the firstorder system x. ____ --~--. (17) Example 2 ·-... If VI = 20. • "" r (gal/min) Compartmental Analysis Frequently a complex process or system can be broken down into simpler subsystems or "compartments" that can be analyzed separately...3x 1. x~ = 3x 1 . xz(O) = x3(0) = 0.~-vi· i = 1. . x 2 .1 shows some typical solution curves of the system in (11). • FIGURE 5..4.4. . and the line x 2 = -3xi obtained with cz = 0.. 3. it is convenient when discussing a linear system =Ax to use vectors x 1..•. and 3. -. X1 Remark: As in Example 1.. The whole system can then be modeled by describing the interactions between the various compartments.4. then Xi (t) and xz(t) both tend to +oo as t --+ +oo.1.. . Given initial values x 1 (0) = b 1 .-~----~ ·· -·· ... then xi (t) and x 2 (t) both tend to -oo as t --+ +oo. = -k 1x 1. --. . - _. while mixed brine flows from tank 1 into tank 2. 5. and the initial amounts of salt in the three brine tanks. The three brine tanks of Example 2.4. b 2 ) lies to the right of the line x 2 = -3x 1. ~ find the amount of salt in each tank at time t 0. in pounds..2 shows three brine tanks containing VI. . Thus a chemical plant may consist of a succession of separate stages (or even physical compartments) in which various reactants and products combine or are mixed.. = 4x 1 + 2x2... kiXJ .2. whereas the scalars x 1 ... x 2 . 2. xz(O) = hz.. Fig. Let xi (t) denote the amount (in pounds) of salt in tank i at time t for i = 1. respectively. Fresh water flows into tank 1.. and V3 gallons of brine..k3X3.. x2 = I I Tink{ · . :..kzX2. : Vr (gal) ~. b 2) lies to the left of the line x 2 = . It may happen that a single differential equation describes each compartment of the system.... • If (b 1 . . V3 = 50. V2 .r· tl ~ ~v FIGURE 5. are Xi (0) = 15..-.·· ·····~.._. Xn to denote different vector-valued solutions of the system. As a simple example of a three-stage system. Direction field and solution curves for the linear system x. (16) x3 = where r k· ... it is apparent from the figure that • If (b 1 .. kzx2 . . r = 10 (galjmin).4 The Eigenvalue Method for Homogeneous Systems 371 Figure 5. V2 = 40. as in Example 2 of Section 5. from tank 2 into tank 3.. if v= [a! +.AI)v = 0.. The complex-valued solution associated with A and vis then x(t) = veA. The only complication is that the eigenvectors associated with complex eigenvalues are ordinarily complex valued. ] a2 b2z .AI)v = o since A =A and I= I (these matrices being real) and the conjugate of a complex product is the product of the conjugates of the factors. x(t) = eP1 (acosqt .. Thus the conjugate v of v is an eigenvector associated with I. (21) Because the real and imaginary parts of a complex-valued solution are also solutions. we thus get the two real-valued solutions x 1 (t) = Re[x(t)] = eP1 (acosqt. so long as they are distinct the method described previously still yields n linearly independent solutions. If v is an eigenvector associated with A. we note that-because we are assuming that the matrix A has only real entries-the coefficients in the characteristic equation in (8) will all be real. it is preferable in a specific example to proceed as follows: • First find explicitly a single complex-valued solution x(t) associated with the complex eigenvalue A. ] [bl. so that (A. Rather than memorizing the formulas in (22).bi. so we will have complex-valued solutions. i = a + bi. . then taking complex conjugates in this equation yields (A .bsinqt) + i eP1 (bcosqt + asinqt). that is. Of course the conjugate of a vector is defined componentwise. Consequently any complex eigenvalues must appear in complex conjugate pairs. x2(t) = lm[x(t)] = eP1 (bcosqt + asinqt) (22) associated with the complex conjugate eigenvalues p ± qi.. bl~] [a!.t = ve<p+qi)r =(a+ bi)eP1 (cos qt + i sinqt). Suppose then that A = p + qi and I= p . and I. + b2 .qi are such a pair of eigenvalues. It is easy to check that the same two real-valued solutions result from taking real and imaginary parts of veir. = a2 . To obtain real-valued solutions.bsinqt). (20) an+ bni an bn then v =a.374 Chapter 5 Linear Systems of Differential Equations Complex Eigenvalues Even if some of the eigenvalues are complex. • Then find the real and imaginary parts x 1 (t) and x2 (t) to get two independent real-valued solutions corresponding to the two complex conjugate eigenvalues A. 4.'. 5. _. It provides a sample of the more technical applications of eigenvalues to physics and engineering problems. 0 (2) • This optional section may be omitted without loss of continuity. A plot like Fig.3 and 5. In this section we apply the matrix methods of Sections 5. ?: · . suppose that the system of n tanks is closed as in Fig. X2(0) = X3(0) = · · · = Xn(O) = 0.k 1x 1..~ ~ Second-Order Systems and Mechanical Applications ~' .. _.__. "' k3 .~. We assume that the masses slide without friction and that each spring obeys Hooke's law.- •M ¥0N O. Graph the solution functions and estimate graphically the maximum amount of salt that each tank ever contains.5.5. they actually follow similarly from Hooke's and Newton's laws. and x 3 of the three masses (from their respective equilibrium positions) are all positive.4.. whatever the signs of these displacements. The fourth spring is compressed the distance x 3 ..i k4. then • • • • The first spring is stretched the distance x 1 ..o¥--> > N N - . Although we assumed in writing these equations that the displacements of the masses are all positive.X2) .. in this closed system.5. (1) m 3 X~ = .~~------N>"~"'''>. For an alternative investigation.···. .5. ¥ •••• •••• •¥ ~ ..1 shows three masses connected to each other and to two walls by the four indicated springs..·i k 2 .'!!: .-- N • ¥ " ' "'''-"'"'"'-"·-"" - · .k3(X3.x 1 .>~• ~ VO~-·. Then the first equation should be replaced with x.~ x1 x2 x3 FIGURE 5. as t ---+ +oo the salt originally in tank 1 distributes itself with constant density throughout the various tanks. .6 should make this fairly obvious.Xt) + k3(X3 - X2)._. = knxn.x 2 . x 2 . Apply the eigenvalue method to solve this system with initial Xt (0) = 10.4 to investigate the oscillations of typical mass-and-spring systems having two or more degrees of freedom.-· A-' ------.__ . application of Newton's law F = ma to the three masses (as in Example 1 of Section 5... so that tank 1 receives as inflow the outflow from tank n (rather than fresh water). The third spring is stretched the distance x 3 .. ~--· ~. 5.k4X3.its extension or compression x and force F of reaction are related by the formula F = -kx.1. '? z._ • ~~ ~-'""''-''"'--~···-• ' >N m _ _. Three spring-coupled masses. _ _ _ _ > < ' .. If the rightward displacements x 1. Therefore.1) yields their equations of motion: m2X~ = -k2(X2._. Figure 5. Now show that. The second spring is stretched the distance x 2 .5 Second-Order Systems and Mechanical Applications 381 where ki conditions r /Vi.~-.hN·. the mass matrix In terms of the displacement vector X = [ XJ X2 X3 r. Our examples are chosen to illustrate phenomena that are generally characteristic of complex mechanical systems. . (4).4 for a first-order system) a trial solution of the form x(t) = v eal' (8) where v is a constant vector. Hence multiplication of each side in Eq..5.5.: The notation in Eqs.2. M= [~' . and a stiffness matrix K satisfying Eq.1 yields the homogeneous second-order system x" =Ax. Then x" = a 2 vear. (7) where A = M . There is a wide variety of frictionless mechanical systems for which a displacement or position vector x.j> x. 0 mz 0 and -(kt 0 IJ 0 0 0 0 -(kn. so substitution of Eq. !i //:( . (1) through (4) generalizes in a natural way to the system of n spring-coupled masses shown in Fig.1 +kn) kn -(kn kn (5) + kz) 0 kz -(kz 0 k3 -(k3 kz + k3) k3 + k4) K= 0 0 k4 (6) 0 0 0 0 + kn+d for the mass and stiffness matrices in Eq. .382 Chapter 5 Linear Systems of Differential Equations and the stiffness matrix (3) the system in ( 1) takes the matrix form Mx" = Kx. (4) ~~ ( !. '•• ':. x2 xn-l xn FIGURE 5. 5. . (7). .~1/~': .. . a nonsingular mass matrix M. We need only write '-h:c•s:pyr:ry:!0p1fr>.·-~·:"' .___-_-_ . A system of n spring-coupled masses. (4) by M. Solution of Second-Order Systems To seek a solution ofEq.. to get its inverse M.2. we substitute (as in Section 5.1 we need only replace each diagonal element with its reciprocal.1 K. (4) can be defined. (8) in (7) gives . The diagonal matrix M is obviously nonsingular. . (11) with ai and bi arbitrary constants. then a general solution of x" =Ax is given by n x(t) = L(ai cos w. -w~ with associated (real] eigenvectors v 1 . If x" = Ax models a mechanical system.5. xo(t) = (ao + bot)vo (12) is the corresponding part of the general solution. then x" = 0 · vo = (ao + bot) · 0 = (ao + bot)· (Avo) =Ax. . . If x(t) = (ao + bot)vo. v2 . ( 12).o with associated eigenvector v0 . In this case the solution given by Eq. THEOREM 1 Second-Order Homogeneous Linear Systems If the n x n matrix A has distinct negative eigenvalues -wf..t i= ! + bi sin w.. thus verifying the form in Eq. (8) is x(t) = The real and imaginary parts x 1(t) = vcoswt and x2 (t) = vsinwt (10) v e i wr = v( cos wt + i sin wt). Remark: The nonzero vector v0 is an eigenvector corresponding to Ao = 0 provided that Av0 = 0. -wi. . . In the special case of a nonrepeated zero eigenvalue A.t)v.. • . .5 Second-Order Systems and Mechanical Applications 383 which implies that (9) Therefore x(t) = vear is a solution of x" =Ax if and only if a 2 =A. of x(t) are then linearly independent real-valued solutions of the system. and v is an associated eigenvector. then it is typical that the eigenvalues of A are negative real numbers. This analysis leads to the following theorem. an eigenvalue of the matrix A. Vn. If then a = ±wi. 5. k 2 =50. The case in which w is a natural frequency corresponds to the phenomenon of resonance discussed in Section 2.5 (b) 1. W 11 of the system. Position functions of the three railway cars of Example 2. Then with m 1 = 2. m 2 = 1. an eigenvalue of A. Example 3 Suppose that the second mass in Example I is subjected to the external periodic force 50coswt.5.9. k 1 = 100. (31) exists provided that the external forcing frequency does not equal one of the natural frequencies w 1.. (29) takes the form 11 X = [ -75 50 (33) FIGURE 5. We then anticipate a periodic particular solution Xp(t) = ccoswt (31) with the known external frequency w and with a coefficient vector c yet to be determined. 5. Thus a periodic particular solution of the form in Eq. F2 tiplication by M . followed by cancellation of the common factor cos wt.Fo (32) to be solved for c. FIGURE 5.5 FIGURE 5.5 Second-Order Systems and Mechanical Applications 389 50 (a) 25 0.5. Because x~ = -w 2 c cos wt.5 2. (b) after. w2 .8. substitution of (30) and (31) in (29).7. The forced mass-and-spring system of Example 3.9.5..0 2.1 yields Fn r is the external force vector for the system.5. Mul- x" =Ax+ f (29) where f is the external force vector per unit mass. and the substitution x = c cos wt leads to the equation (34) .6. (32) can be solved for c-unless -w2 = /. Observe that the matrix A + w2 1 is nonsingular-in which case Eq. (a) Before. We are especially interested in the case of a periodic external force f(t) = F 0 coswt (30) (where F 0 is a constant vector).0 1. where F = [ F. gives the linear system (A + w 2 1)c = . ••• . Eq. and F0 =50 in Fig. (w2.25)(w2 .. (t) = ~cos wt.E "0 . The resulting forced periodic oscillation is described by X! 0. (38) Hence Xc(t) is a transient solution that depends only on the initial conditions. then (35) yields c 1 = ~.100)' -. c 2 = -1. leaving the steady periodic solution Xp(t) resulting from the external driving force: x(t) ~ Xp(t) as t ~ +oo.-----m-----. Periodic and Transient Solutions It follows from Theorem 4 of Section 5. and now the two masses oscillate in synchrony in opposite directions. It is typical for the effects of frictional resistance in mechanical systems to damp out the complementary function solution Xc(t)..-----. The peaks • at w 2 = 5 and w 2 = 10 exhibit visually the phenomenon of resonance. xz(t) = -coswt. Figure 5.5.. The resulting forced periodic oscillation is described by XJ(t) = -coswt. This system is readily solved for cz =50(w2 . (39) As a practical matter. every physical system includes frictional resistance (however small) that damps out transient solutions in this manner. If the external squared frequency is w 2 = 125. then (35) yields -1. but with the amplitude of motion of m 2 twice that of m 1 • It is evident from the denominators in (35) that c 1 and c2 approach +oo as w approaches either of the two natural frequencies w 1 = 5 and w2 = 10 (found in Example 1). 15.----- c1 = For instance.. Frequencyamplitude plot for Example 3..10.390 Chapter 5 Linear Systems of Differential Equations for the coefficient vector c = [ c 1 c2 JT.. c2 = -1. (37) where Xp(t) is a particular solution of the nonhomogeneous system and Xc (t) is a solution of the corresponding homogeneous system.1 0 shows a plot of the amplitude J cf + ci of the forced periodic solution x(t) = c cos wt as a function of the forced frequency w. . <t: 5 OOL-----~----~--~~ Forced frequency 5 10 15 FIGURE 5. (w2. so that Xc(t) ~ 0 as t ~ +oo. if the external squared frequency is w 2 = 50. xz(t) = -cos wt... .----------:... Io s Thus the two masses oscillate in synchrony with equal amplitudes and in the same direction.25)(w2.3 that a particular solution of the forced system x" = Ax + F o cos wt will be of the form x(t) = Xc(t) (36) + Xp (t).100) (35) CJ - 1250 . it dies out with time..5 .75) . let x(t) denote the vertical displacement of the center of mass of the car from equilibrium. x3(0) = . 27. 10008". k 1 = k2 = 2000 = = 100. and that x. We assume that the car body acts as would a solid bar of mass m and length L = L 1 + L2. x.5.6 and discussed in Example 2. (a) Find the two natural frequencies of oscillation (in hertz). x~(O) = vo. (t ). I= 800. L 2 = 3 ft (it's a rear-engine car). 19. x~(O) = -2vo x. Then Newton's laws of and an angular oscillation with frequency W2 = )k£2j(21). (b) Assume that this car is driven along a sinusoidal washboard surface with a wavelength of 40ft.8000x + 116.14 (the symmetric situation). 23. show that the two springs are compressed until t = 1r /2 and that FIGURE 5. respectively.(t) = -~v0 and x~(t) = x~(t) = +~v0 thereafter. x~ (t ). k 1 = 1000. and x~ (t) of the three cars for t > 1r /2.5. I= 800. 20.5. Resonance occurs when with w = w 1 or w = w2 . and k = 1500 lbjft. In Problems 27 through 29. The car has front and back suspension springs with Hooke's constants k 1 and k 2 . k 1 = k 2 = 2000 lbjft. that car 2 weighs 8 tons.5. Suppose that m = 75 slugs (the car weighs 2400 lb). L 1 = 6. Thus both cars continue in the original direction of motion. x~(O) = 0.vo x. Determine the values of these constant final velocities x. If cars 1 and 2 weigh 8 and 16 tons.14 is taken as a model for an undamped car with the given parameters in fps units. L 2 = 4. 26. If cars 1 and 2 weigh 24 and 8 tons. x~(O) = 2vo. Find the two critical speeds. the system of Fig.392 Chapter 5 Linear Systems of Differential Equations motion for linear and angular acceleration can be used to derive the equations 18. k2 = 2000 . (0) = vo. m 28. and that each spring constant is 4 tons/ft.6. Model of the two-axle automobile. 24. v x. which is at distance L 1 from the front of the car.14. (t) = +~vo and x~(t) = +~vo Equilibrium position thereafter. respectively. but cars 2 and 3 continue with the same velocity.5. (0) = 3vo. Problems 20 through 23 deal with the same system of three railway cars (same masses) and two buffer springs (same spring constants) as shown in Fig. x3(0) = 2vo In the three-railway-car system of Fig. k 1 = k2 = 2000 100. respectively. show that the cars separate after 1r /2 seconds. Show that the railway cars remain engaged until t = 1r /2 (s ). Then show that every free oscillation is a combination of a vertical oscillation with frequency The Two-Axle Automobile In Example 4 of Section 2. I= 1000. and I = 1000 ft·lb·s 2 • Then the equations in (40) take the form 75x" + 4000x - 8oooe = o. 5. Figure 5. (0) = 2v0 . m = 100.5. after which time they proceed in their respective ways with constant velocities.(O) = v0 . (a) Find the two natural frequencies w 1 and w 2 of the car. Find the corresponding two critical speeds of the car (in feet per second and in miles per hour). L 1 = L 2 = 5. Suppose that k 1 = k 2 = k and L 1 = L 2 = ~L in Fig. but with different velocities.vo x. 5. The cars engage at time t = 0 with x 1 (0) = x 2 (0) = x 3 (0) = 0 and with the given initial velocities (where v0 = 48ftls). let e(t) denote its angular displacement (in radians) from the horizontal. and k = 3000 lb/ft.6 we investigated the vertical oscillations of a one-axle car-actually a unicycle. 5. L 1 = 7 ft. Thus the two cars rebound in opposite directions.oooe = o. Thus car 1 rebounds. 22.(t) = -~vo and x~(t) = +~v0 (40) thereafter. per second along a washboard surface shaped like a sine curve with a wavelength of 40 ft. (b) Now suppose that the car is driven at a speed of v feet x. If (0) = 0 and x~ (0) = x3 (0) = 0. When the car is in motion. £ 1 = L2 = 5. It has moment of inertia I about its center of mass C. The result is a periodic force on the car with frequency w = 2nvj40 = nv/20. 21. show that the two cars separate after 1r /3 seconds. x~ (0) = 0. In each problem you should find (as in Example 2) that the first and third railway cars exchange behaviors in some appropriate sense. 25. m 29. 5.14 represents the suspension system of such a car. and that x. Now we can analyze a more realistic model: a car with two axles and with separate front and rear suspension systems. x3 (0) = . suppose that cars 1 and 3 each weigh 32 tons. )(.. Example 1 Find a general solution of the system 4 -1 (5) 4 Solution The characteristic equation of the coefficient matrix A in Eq. then a general solution of the system -=Ax dt is given by dx (l ) (2) with arbitrary constants c 1.8A +A 2 ) = (5. as needed to form the general solution in (2). An with respective associated eigenvectors v 1. If every eigenvalue of the matrix A is complete..)+ 24] = (3. so there is at least one eigenvector associated with A.A. A2 . Vn associated with the eigenvalues AJ..6 Multiple Eigenvalue Solutions 393 Multiple Eigenvalue Solutions ·· m••••«•~ In Section 5. the eigenvector equation (A. c2 . Let us call an eigenvalue of multiplicity k complete if it has k linearly independent associated eigenvectors. In this section we discuss the situation when lA.All= 0 (3) does not have n distinct roots.Al)v = 0 (4) has at least one nonzero solution v. Vn . and thus has at least one repeated root. For each eigenvalue A.All= -6 6 0 0 3-A 4 = (3 . But an eigenvalue of multiplicity k > 1 may have fewer than k linearly independent associated eigenvectors.A)(15. In this case a general solution of x' = Ax is still given by the usual combination in (2). . . . .4 we saw that if the n x n matrix A has n distinct (real or complex) eigenvalues AJ. . An eigenvalue is of multiplicity k if it is a k-fold root of Eq. . .5. en. . . (3).. (5) is 9-A 4 -1-A lA .. .A)[(9 . the characteristic equation •. v2. A2. v2. . Thus A has the distinct eigenvalue AJ = 5 and the repeated eigenvalue A2 = 3 of multiplicity k = 2...A)(3 . then-because eigenvectors associated with different eigenvalues are linearly independent-it follows that A does have a complete set of n linearly independent eigenvectors v 1 .A. An (each repeated with its multiplicity).A) 2 = 0..1 . In this case we are unable to find a "complete set" of n linearly independent eigenvectors of A. A. and the new general solution X(t) = CJ VJ eSt+ C2V2e 3t + c 3 ve 3t would be equivalent to the one in Eq. 3a + 3b = 0.. Thus we need not worry about making the "right" choice of independent eigenvectors associated with a multiple eigenvalue.5. and hence is incomplete. we could choose v rather than v3 as our third eigenvector. we generally make the simplest one we can. The fact that b = -~a for any eigenvector associated with .4) 2 = 0.A 1 = 4 of multiplicity 2. • r. given a and c not both zero.A 2 = 3 means that any such eigenvector can be written as and thus is a linear combination of v2 and v3 . (7). The eigenvector equation (A.A)+ 9 = .All = 11 -3 . Defective Eigenvalues The following example shows that-unfortunately-not all multiple eigenvalues are complete.3 7 .. Therefore any eigen- vector associated with AI = 4 is a nonzero multiple of v = [ I -1 Thus the multiplicity 2 eigenvalue .6 Multiple Eigenvalue Solutions 395 Remark: Our choice in Example I of the two eigenvectors V2 = [0 0 ] r and V3 = [2 -3 0 r associated with the repeated eigenvalue .3a . Example 2 The matrix (8) has characteristic equation lA . Therefore.A 1 = 4 has only one independent eigenvector.A)(7.41)v = [ -3 3 then amounts to the equivalent scalar equations .A . .3b Hence b = = 0.A = (1 .A 2 - 8.A + 16 =(. • Any choice will do..A 2 = 3 bears comment. Thus A has the single eigenvalue . -a if v = [a b JT is to be an eigenvector of A. we might hope to find a second solution of the form (11) When we substitute x = v 2 teJ.t in x' =Ax. and hence that x2 (t) = 0. Thus the defective eigenvalue AJ = 4 in Example 2 has multiplicity k = 2 and defect d = 1. then the eigenvalue method as yet described will produce fewer than the needed n linearly independent solutions of the system x' = Ax.. we get the equation But because the coefficients of both eJ. The Case of Multiplicity k = 2 Let us begin with the case k = 2. and suppose that we have found (as in Example 2) that there is only a single eigenvector v 1 associated with the defective eigenvalue A.396 Chapter 5 Linear Systems of Differential Equations An eigenvalue A of multiplicity k > 1 is called defective if it is not complete. we get the equation v 1te>. This means that-contrary to our hope.t (13) We equate coefficients of eJ.t here.t must balance. We therefore need to discover how to find the "missing solutions" corresponding to a defective eigenvalue A of multiplicity k > 1. .. let us extend it slightly and replace v2 t with v 1t + v2 • Thus we explore the possibility of a second solution of the form (12) where v 1 and v2 are nonzero constant vectors. By analogy with the case of a repeated characteristic root for a single linear differential equation (Section 2.the system x' =Ax does not have a nontrivial solution of the form assumed in (11). If the eigenvalues of the n x n matrix A are not all complete. then the number d=k-p (9) of "missing" eigenvectors is called the defect of the defective eigenvalue A... it follows that v2 = 0.3). When we substitute x = in x' = Ax.H)vi =0 (14) and (15) that the vectors v 1 and v2 must satisfy in order for (12) to give a solution of x' = Ax. If A has only p < k linearly independent eigenvectors.t and teJ. because we saw that it has only p = 1 associated eigenvector..t and teJ. and thereby obtain the two equations (A. (11). Then at this point we have found only the single solution (10) of x' = Ax. Instead of simply giving up on the idea behind Eq.r +v2 e>. Al)v2 is nonzero. If . the procedure described in the following algorithm always succeeds in finding two independent solutions associated with such an eigenvalue. in order to solve simultaneously the two equations in (14) and (15). X1 = [ 1 3 -3] 7 X. (15) says that the vector v2 satisfies the equation It follows that. Consequently. it could happen that (A .5. it suffices to find a solution v2 of the single equation (A . (16) is [~ ~ J v2 = 0.6 Multiple Eigenvalue Solutions 397 Note that Eq. and therefore is satisfied by any choice of v2 . (14) merely confirms that v 1 is an eigenvector of A associated with the eigenvalue A. It turns out that this is always possible if the defective eigenvalue A of A is of multiplicity 2.41)v2 is nonzero (as desired) for some choices of v 2 though not for others.U) 2 v2 = 0 such that the resulting vector v 1 = (A. Then form the two independent solutions (18) and (19) of x' =Ax corresponding to A. 2. We therefore begin by calculating Hence Eq. and therefore is an eigenvector v 1 associated with A. First find a nonzero solution of the equation (16) such that (17) is nonzero. (20) has the defective eigenvalue A = 4 of multiplicity 2. ALGORITHM Defective Multiplicity 2 Eigenvalues v2 1. (20) Solution In Example 2 we found that the coefficient matrix A in Eq. In principle. Then Eq. These n generalized eigenvectors may be arranged in chains.t. (27) ) A. vd of generalized eigenvectors (with v 1 an ordinary eigenvector associated with A) determines a set of k independent solutions of x' = Ax corresponding to the eigenvalue A: Xt(t) = VJeAt. Note that. • Two length 1 chains and a length 2 chain (defect 1). ALGORITHM Chains of Generalized Eigenvectors Begin with a nonzero solution u 1 of Eq. or • A length 4 chain (defect 3).Jt + vk e . . we can begin with the equation >- (A. For instance.1)! + ··· + Vk.l)uk-1 = Uk -:j:. x 3(t) = (~v1t 2 + v2t + v 3) eA. But the structure of these chains depends on the defect of A. • A length 1 chain and a length 3 chain (defect 2). • Two length 2 chains (defect 2). . where d is the defect of the eigenvalue. X2(t) = (Vtf + V2) eA. If (A . Consequently..t + vk. (26) and successively multiply by the matrix A . a multiplicity 4 eigenvalue can correspond to • Four length 1 chains (defect 0).t.Al)uk = 0.6 Multiple Eigenvalue Solutions 401 The General Case A fundamental theorem of linear algebra states that every n x n matrix A has n linearly independent generalized eigenvectors. but (A .5. 0.Al)d+lu = 0 (26) to start building the chains of generalized eigenvectors associated with A.. in each of these cases. xk(t)= ( VJ t k -1 (k . and can be quite complicated.. once we have found all the ordinary eigenvectors associated with a multiple eigenvalue A.2t 2 2! . then the vectors (listed in reverse order of their appearance) form a length k chain of generalized eigenvectors based on the (ordinary) eigenvector v 1• Each length k chain {v 1./. the length of the longest chain is at most d + 1. v2 . and therefore know the defect d of A. with the sum of the lengths of the chains associated with a given eigenvalue A equal to the multiplicity of A.AI until the zero vector is obtained. = k(XJ .21 .2.c(x. of magnitude c(x. • If two chains of generalized eigenvectors are based on linearly independent eigenvectors.x. Then A. then the union of these two chains is a linearly independent set of vectors (whether the two base eigenvectors are associated with different eigenvalues or with the same eigenvalue). whereas A. The railway cars of Example 6. x1) . respectively.C(X~ .402 Chapter 5 Linear Systems of Differential Equations Note that (27) reduces to Eqs. X2(t) (28) In terms of the position vector x(t) = [ X!(t) written in the matrix form r. . x3(t) = (v1t + v2)e-21. 1 = -2 and A. • An Application Figure 5. and therefore produce a complete set of n linearly independent solutions of x' = Ax when we amalgamate all the "chains of solutions" corresponding to different chains of generalized eigenvectors.c1xi . X2(t) = v 1e. Example 5 Suppose that the 6 x 6 matrix A has two multiplicity 3 eigenvalues A. v 2 } of generalized eigenvectors (with the eigenvectors u1 and VJ being linearly independent).2 shows two railway cars that are connected with a spring (permanently attached to both cars) and with a damper that exerts opposite forces on the two cars.C2X~ . To ensure that we obtain n generalized eigenvectors of the n x n matrix A that are actually linearly independent. and w 3 are then linearly independent and yield the following six independent solutions of x' = Ax: XJ (t) = u 1e-21 .x~ ) .X4. The six generalized eigenvectors u 1. these equations can be (29) Mx" = Kx + Rx' . (iw1t 2 + w2t + w 3 ) e 31• As Example 5 illustrates. . (18) through (19) and (24) in the cases k = 2 and k = 3. w2.). v 1 . 1 must have an associated eigenvector 01 and a length 2 chain {v 1 . xs(t) = (w1t x6(t) = + w2)e 31 . we may rely on the following two facts: • Any chain of generalized eigenvectors constitutes a linearly independent set of vectors. The determination of the chain structure associated with a given multiple eigenvalue can be more interesting (as in Example 6).6. . the computation of independent solutions corresponding to different eigenvalues and chains of generalized eigenvalues is a routine matter. respectively.x~) proportional to their relative velocity. . w 1. The two cars are also subject to frictional resistance forces c 1xi and c 2x~ proportional to their respective velocities.6.X2 ) . An application of Newton's law rna = F (as in Example 1 of Section 5.1) yields the equations of motion m1x~' = k(x2 m 2X~ FIGURE 5. w3} of generalized eigenvectors based on its single eigenvector w 1.(t) = w 1e 31 .2 = 3 with defects 1 and 2. v2 .2 must have a length 3 chain {w 1. w2 . Then then x n matrix (2) having these solution vectors as its column vectors. that is. (4 ). Then the [unique] solution of the initial value problem x' = Ax.. (1). x2 (t). (1). so W(t) is a fundamental matrix for the system in (1 ). If lJI(t) is any other fundamental matrix for (1). Xn(t) are n linearly independent solutions of Eq. the general solution (3) of the system x' = Ax can be written in the form x(t) = ci>(t)c (4) where c = [c 1 c2 cn]T is an arbitrary constant vector. Fundamental Matrix Solutions Because the column vector x = x j (t) of the fundamental matrix ci> (t) in (2) satisfies the differential equation x' = Ax. any nonsingular matrix solution 'II (t) of Eq. In order that the solution x(t) in (3) satisfy a given initial condition x(O) = xo. THEOREM 1 Fundamental Matrix Solutions Let ci>(t) be a fundamental matrix for the homogeneous linear system x' = Ax. . (5) it suffices that the coefficient vector c in (4) be such that ci>(O)c = x0 . it follows (from the definition of matrix multiplication) that the matrix X = cl>(t) itself satisfies the matrix differential equation X' = AX. is called a fundamental matrix for the system in (1). . x(O) = Xo (7) is given by (8) . and therefore has an inverse matrix ci>(t).1 • Conversely. we get the conclusion of the following theorem. (1). (4) that 'll(t) = ci>(t)C (4' ) for some n x n matrix C of constants.408 Chapter 5 Linear Systems of Differential Equations associated with Eq. In terms of the fundamental matrix ci>(t) in (2). it also follows that the fundamental matrix cl»(t) is nonsingular. Suppose that x 1 (t). (1') has linearly independent column vectors that satisfy Eq. then each column vector of W(t) is a linear combination of the column vectors of ci>(t). that (6) When we substitute (6) in Eq. Because its column vectors are linearly independent.. so it follows from Eq. A. . Therefore.) The formula in (12) is also valid upon generalization to n x n matrices.4 tells us how to find a fundamental matrix for the system x' =Ax (9) with constant n x n coefficient matrix A.y . Xn as column vectors is a fundamental matrix for the system x' = Ax. 0. . Vn associated with the (not necessarily distinct) eigenvalues )q. (12) where ~ = det(A) f. then x n matrix (10) having the solutions x 1. (8).•. The inverse of the nonsingular 3 x 3 matrix A -1 = +A11 1 ~ [ -Azi + A31 -A12 +A13] +A22 . respectively.!. y' = 3x. n. we must be able to compute the inverse matrix «Jl(0). . Example 1 Find a fundamental matrix for the system x' = 4x + 2y . 0 and AiJ denotes the determinant of the 2 x 2 submatrix of A obtained by deleting the ith row and jth column of A... 2. v2 . In order to apply Eq. In this event the corresponding solution vectors of Eq. . An. . but in practice inverses of larger matrices are usually computed instead by row reduction methods (see any linear algebra text) or by using a calculator or computer algebra system. (12). 2 .5. ..7 Matrix Exponentials and Linear Systems 409 Section 5. (Do not overlook the symbol T for transpose in Eq.. . ••• ._ [ ~ -c d -b a J' (11) f.A32 -A23 + A 33 T . x2 . y (O) = .1. at least in the case where A has a complete set of n linearly independent eigenvectors v1 . The inverse of the nonsingular 2 x 2 matrix is A -I where ~ = det(A) = ad -be A = [aiJ] is given by = _. (9) are given by fori = 1. (13) then use it to find the solution of (13) that satisfies the initial conditions x(O) = 1.1. k An) n! (18) where A0 =I. A2 = AA. and so on.k -? oo ?. 2! 3! n! z2 z3 zn (16) Similarly.7 Matrix Exponentials and Linear Systems 411 Exponential Matrices We now discuss the possibility of constructing a fundamental matrix for the constantcoefficient linear system x' = Ax directly from the coefficient matrix A-that is. The exponential ez of the complex number z may be defined (as in Section 2. ranging from the scalar equation x' = kx with solution x(t) = x 0ekt to the vector solution x(t) = veJ..5. A3 = AA 2 . Example 2 Consider the 2 x 2 diagonal matrix Then it is apparent that . An+! = AAn ifn ~ 0.+ · · · + .+ . if A is an n x n matrix. the exponential matrix eA is defined (by Eq. 00 An ( --lim n! . That is. with associated eigenvector v.3) by means of the exponential series ez = 1+ Z + . The meaning of the infinite series on the right in (17) is given by ?. then the exponential matrix eA is the n x n matrix defined by the series eA=I+A+-+···+-+··· 2! n! A2 An (17) where I is the identity matrix. We now define exponentials of matrices in such a way that X(t) =eAr is a matrix solution of the matrix differential equation X'=AX with n x n coefficient matrix A-in analogy with the fact that the ordinary exponential function x(t) =eat is a scalar solution of the first-order differential equation x' =ax. (17)) for every square matrix A. We have seen that exponential functions play a central role in the solution of linear differential equations and systems.+ · · · . without first applying the methods of earlier sections to find a linearly independent set of solution vectors. inductively.r of the linear system x' = Ax whose coefficient matrix A has eigenvalue A. It can be shown that the limit in ( 18) exists for every n x n square matrix A. + ··· 2! = [ 1 0] A2 01 + [ a Ob 0] + [ a /2! 0 2 Thus eA = [ e. At is obtained simply by multiplying each element of A by t . (22) In Problem 32 we ask you to conclude that (eArl= e-A. (23) In particular. The exponential matrix eA satisfies most of the exponential relations that are familiar in the case of scalar exponents. (17) gives e At =I+At+A . 0 0 (19) 0 (20) 0 obtained by exponentiating each diagonal element of D. ~ J. then Eq. 1. (17) yields (21) the n x n identity matrix.) . It therefore follows that eA = I+A+. then eA+B = eA e8 . If t is a scalar variable. 2! n! 2 t2 n tn (24) (Of course. In Problem 31 we ask you to show that a useful law of exponents holds for n x n matrices that commute: If AB = BA. For instance.+···+A .:.412 Chapter 5 Linear Systems of Differential Equations for each integer n . if 0 is the n x n zero matrix. The exponential of the n x n diagonal matrix D= is the n x n diagonal matrix 6 [.+ ···. • The n x n analog of the 2 x 2 result in Example 2 is established in the same way. then substitution of At for A in Eq. so the exponential of the diagonal2 x 2 matrix A is obtained simply by exponentiating each diagonal element of A. the matrix eA is nonsingular for every n x n matrix A (reminiscent of the fact that ez =I= 0 for all z). It follows from elementary linear algebra that the column vectors of eA are always linearly independent. Example 5 In Example 1 we found that the system x' = Ax with . is given by x(O) = xo (26) (27) and this solution is unique. Conversely. In particular. it follows that the matrix exponential eAt is a fundamental matrix for the linear system x' = Ax. if we already know a fundamental matrix <P(t) for the linear system x' = Ax. (4')) and eA·O = e0 = I (the identity matrix) yield (28) So we can find the matrix exponential eAt by solving the linear system x' = Ax. then the solution of the initial value problem x' =Ax. it is the fundamental matrix X(t) such that X(O) =I. dt (25) in analogy to the formula D 1 (ekt) = kekt from elementary calculus. Because the matrix eAr is nonsingular. THEOREM 2 Matrix Exponential Solutions If A is an n x n matrix. Thus the matrix-valued function X(t) =eAt satisfies the matrix differential equation X' = AX.414 Chapter 5 Linear Systems of Differential Equations Matrix Exponential Solutions It happens that term-by-term differentiation of the series in (24) is valid. Theorem 1 implies the following result. that is. Therefore. with the result d t2 (e At ) = A+ A2 t +A3 dt 2! + ··· t2 = A ( I+ At +A2 2! + ··· ) · . d _(eAr)= AeAt. Thus the solution of homogeneous linear systems reduces to the task of computing exponential matrices. then the facts that eAt = <P(t)C (by Eq. 2. Hence.5. 2.7 Matrix Exponentials and Linear Systems has fundamental matrix -21 [ c)(t) = -3:-21 Hence Eq. using Theorem 2. the solution of the initial value problem in (29) is given by 0 (19 + 243t [ (29 + 234t)e 21 + 351t2)e21 ] • 39e21 • . (28) gives with 415 • Example 6 Use an exponential matrix to solve the initial value problem x' = 2 3 4] [0 0 2 0 2 6 x. and so we do not yet have the three linearly independent solutions needed for a fundamental matrix. (29) Solution The coefficient matrix A in (29) evidently has characteristic equation (2. x(O)= un r. But we note that A is the same matrix whose matrix exponential 0 was calculated in Example 4.A) 3 = 0 and thus the triple eigenvalue A = 2. It is easy to see that the eigenvector equation Thus there has (to within a constant multiple) the single solution v = [ 1 0 0 is only a single eigenvector associated with the eigenvalue A = 2. Then one wouldusing Eqs. In this way. an explicit computation similar to that in Eq. .6-assemble the linearly independent solutions of the differential equation x' = Ax in (29).) Thus the matrix A. (31).21 is nilpotent: (A . (31) (This particular result is a special case of the Cay ley-Hamilton theorem of advanced linear algebra.21) 3 = 3 4] [ 00 0 OJ 0 0 [0 0 0 0 0 0 0 0 0 6 = 0.416 Chapter 5 Linear Systems of Differential Equations Remark: The same particular solution x(t) as in Example 6 could be found using the generalized eigenvector method of Section 5. For such a matrix. c 3 so that the particular solution x(t) = c 1x 1 (t) + c2 x2 (t) + c3 x 3 (t) satisfies the initial condition in (29).6. (30) A similar result holds for any 3 x 3 matrix A having a triple eigenvalue r. The final step would be to determine values of the coefficients c 1 . General Matrix Exponentials The relatively simple calculation of eAt carried out in Example 4 (and used in Example 6) was based on the observation that if A= 3 4] [2 0 0 2 0 2 6 3 . then A .r) 3 = 0. (30) will show that (A. according to which every matrix satisfies its own characteristic equation. One would start by finding the chain of generalized eigenvectors corresponding to the triple eigenvalue A = 2 of the matrix A.rl is nilpotent.rl) 3 = 0. in which case its characteristic equation reduces to (A . (27) in Section 5. and it follows that the exponential series here terminating because of Eq. c2 . At this point it should be apparent that-especially if the matrix exponential eAt is readily available (for instance. from a computer algebra system)-the method illustrated in Example 6 can well be more "computationally routine" than the generalized eigenvector method. we can rather easily calculate the matrix exponential e At for any square matrix having only a single eigenvalue. then the n x n matrix <l>(t) = [ x. A has n linearly independent generalized eigenvectors u 1.1 satisfies the initial condition X(O) = I. and r: eAt u. 2! (34) using (33) and the fact that eu1 = eM I.) Even if we do not yet know eAt explicitly.1)! J. As we saw in Section 5. We have therefore outlined a proof of the following theorem..l)ut +(A. THEOREM 3 Computation of eAt Let u 1 . .(t) x2(t) · · · Xn(t) J (35) is a fundamental matrix for the system x' = Ax.. .6. of A and has a rank r ~ 1 such that (A-UYu =0 but (33) (If r = 1. u 2 .... Un. (32) motivates a method of calculating eAt for any n x n matrix A whatsoever. . we can calculate x explicitly in terms of A. u. .'u (r. . then u is an ordinary eigenvector such that Au= 'Au. the specific fundamental matrix X(t) = <l>(t)ci>(0).. we can consider the function x(t) = which is a linear combination of the column vectors of eAt and is therefore a solution of the linear system x' = Ax with x(O) = u. Indeed. If the linearly independent solutions x 1(t). Un.5. Un be n linearly independent generalized eigenvectors of the n x n matrix A . let xi (t) be the solution of x' = Ax given by (34). For each i. Each generalized eigenvector u is associated with an eigenvalue A. Xn(t) ofx' =Ax are calculated using (34) with the linearly independent generalized eigenvectors u 1. x2(t). and rank r of the generalized eigenvector ui. substituting u = ui and the associated eigenvalue A. If the fundamental matrix <l>(t) is defined by (35). u2 .7 Matrix Exponentials and Linear Systems 417 The calculation in Eq.. and thus is the desired matrix exponential eAt.A. . Finally. u2.AI) 2 u+ ··· tr-1 2 + (A-uy.. so x(t) = e At [ u t +(A. . .. then (36) Example 7 Find e At if (37) . A. 1 ~ i ~ n. the matrix eA is nonsingular with (eArl =e-A. that AB = BA. 37. Apply Theorem 3 to calculate the matrix exponential eAt for each of the matrices in Problems 35 through 40. we know from Theorem 4 of Section 5.) 32. (1) has the form x(t) = Xc(t) + Xp(t). x' = [ 12 9 6 3 31. A= = I cosh t + A sinh t. A= [ ~ 33. Apply this fact to find a general solution of x' = Ax. so our task now is to find Xp(t). 34. (Suggestion: Group the terms in the product of the two series on the right-hand side to obtain the series on the left. and Xp (t) is a single particular solution of the original nonhomogeneous system in (1). Suppose that A=[~ eAt 6]. Conclude that 39. Prove that eA+B = eAe8 . A= [ -~ ~ l x. Preceding sections have dealt with Xc(t). 29. 35. ~] ! ~ ~] 0 1 and apply this fact to find a general solution of x' = Ax. Each of these may be generalized to nonhomogeneous linear systems. ! . A= 20 10 0 20 5 3 3 2 0 40. such as inflow of liquid to a cascade of brine tanks or an external force acting on a mass-and-spring system. Deduce from the result of Problem 31 that. Suppose that the n x n matrices A and B commute. for every square matrix A. u2 3] u n u 30] u n A=[~ n ~] 3 1 0 3 1 0 0 36. and verify that it is equivalent to the solution found by the eigenvalue method. x' = [ g 0 : 30. 4 2 0 0 4 4 2 0 IIIJ Nonhomogeneous Linear Systems In Section 2.420 28. A= 1 0 4 1 38. A= Show that A 2n = I and that A 2n+I = A if n is a positive integer. nonhomogeneous terms typically correspond to external influences. Suppose that x. (2) where • • Xc(t) = c 1x 1 (t) + c2 xz(t) + · · · + CnXn(t) is a general solution of the associated homogeneous system x' = Ax. Verify that it is equivalent to the general solution found by the eigenvalue method. Given the nonhomogeneous first-order linear system x'=Ax+f(t) (1) where A is ann x n constant matrix and the "nonhomogeneous term" f(t) is a given continuous vector-valued function.5 we exhibited two techniques for finding a single particular solution of a single nonhomogeneous nth-order linear differential equation-the method of undetermined coefficients and the method of variation of parameters. In a linear system modeling a physical situation. x' Chapter 5 Linear Systems of Differential Equations =[ 20 1~ ~ 0 30 g] 5 x. . Show that eAr = I cos 2t + ~A sin 2t.3 that a general solution of Eq. that is. -~- - •··• • Example 1 Find a particular solution of the nonhomogeneous system (3) Solution The nonhomogeneous term f = [ 3 2t linear trial particular solution of the form f [ is linear. (4) gives the particular solution X = [ Xj X2 of (3) described in scalar form by r Xt(t) = 4t + 17. Substitution of these coefficients in Eq.• •• --·-·m• • • .•.• · - •. a2.25. xz (t) = -6t . (5) We solve the first two equations in (5) for a 1 = 4 and a 2 = .-A ~ • •. . (3). With these values we can then solve the last two equations in (5) for b1 = 17 and b2 = ..--.8 Nonhomogeneous Linear Systems 421 Undetermined Coefficients First we suppose that the nonhomogeneous term f(t) in (1) is a linear combination (with constant vector coefficients) of products of polynomials. exponential functions. Moreover. the choice of this general form is essentially the same as in the case of a single equation (discussed in Section 2. 0.. (1). We will therefore confine the present discussion to illustrative examples..6.5).• "''"'"' •• ••· . we modify it only by using undetermined vector coefficients rather than undetermined scalars. Then the method of undetermined coefficients for systems is essentially the same as for a single linear differential equation. then attempt to determine the coefficients in Xp by substitution in Eq.. and sines and cosines.· -• ••--~-··-·-~··. . (4) Upon substitution of x = Xp in Eq.·----- . we get We equate the coefficients oft and the constant terms (in both x 1. • . We make an intelligent guess as to the general form of a particular solution Xp.• • • • ••.5.· . at.and x 2-components) and thereby obtain the equations 3at + 2a2 = 7at + 5a2 + 2 = 3bt + 2b2 + 3 = 7bt + 5b2 = 0. so it is reasonable to select a Xp (t) = at +b = :~ Jt + [ :~ J.25. 5 that the method of variation of parameters may be applied to a linear differential equation with variable coefficients and is not restricted to nonhomogeneous terms involving only polynomials. as t ---+ +oo.21 + be-2t exhibits duplication with the complementary function in (10).8 Nonhomogeneous Linear Systems 423 As illustrated in Fig. For a system.5). (9) In Example 1 of Section 5. but also by all lower (nonnegative integral) powers of t as well. In the case of duplicate expressions in the complementary function and the nonhomogeneous terms. and we would then have six scalar coefficients to determine.21 as our trial solution. • Variation of Parameters Recall from Section 2.5.4 we found the solution Xc(t) = CJ [ _ 1 ] -2t 3 e + c2 [ 2 ] e 5t 1 (10) of the associated homogeneous system. and all the resulting terms must be included in the trial solution.21 + ce. the usual first choice for a trial solution must be multiplied not only by the smallest integral power oft that will eliminate duplication. there is one difference between the method of undetermined coefficients for systems and for single equations (Rule 2 in Section 2.8. It is simpler to use the method of variation of parameters. We would therefore select Xp(t) = at 2 e.2. a uniform density of 2lbjgal-the same as the salt density in the inflow to tank 1. our next topic. The method of variation of parameters for systems enjoys the same flexibility and has a concise matrix formulation that is convenient for both practical and theoretical purposes.21 + bte. • 120 100 80 ~ x 3 (t) ~ 100 80 x 2(t) ~ 60 40 20 00 10 x 1(t) ~ 40 20 30 40 50 60 FIGURE 5. Example 3 Consider the nonhomogeneous system X1 = [ 4 2 3 -l J [ 15 te 4J X- -2t . and sinusoidal functions.2. exponentials. 5. The salt amount solution curves defined in (8). we see the salt in each of the three tanks approaching. .8. A preliminary trial solution Xp (t) = ate. (15) and (16) in (11) yields • ' (t)u(t) + •(t)u' (t) = P(t). indeed. . (17) reduces to •(t)u'(t) = f(t). (13). satisfy Eq.... so that (21) Upon substitution of (21) in (15). . (17) But (18) because each column vector of •(t) satisfies Eq. Therefore. Eq. Thus we seek a particular solution of the form Xp(t) = •(t)u(t). The derivative of xp(t) is (by the product rule) x~(t) = c)' (t)u(t) + •(t)u' (t). (15) We must determine u(t) so that Xp does.. Xn to rewrite the complementary function in ( 12) as (14) where c denotes the column vector whose entries are the coefficients c 1 . as stated in the following theorem. we finally obtain the desired particular solution. Our idea is to replace the vector "parameter" c with a variable vector u(t). x2. c2 . . (19) Thus it suffices to choose u(t) so that (20) that is. en. given that we have already found a general solution (11) (12) of the associated homogeneous system x' = P(t)x.424 Chapter 5 Linear Systems of Differential Equations We want to find a particular solution Xp of the nonhomogeneous linear system x = P(t)x + f(t).(t)u(t) + f(t). (13) We first use the fundamental matrix •(t) with column vectors x 1. . . ( 11 ). (16) Hence substitution of Eqs. we get the solution (25) of the nonhomogeneous initial value problem x' = P(t)x + f(t). for we need only a single particular solution. x(a) = Xa.8 Nonhomogeneous Linear Systems 425 THEOREM 1 Variation of Parameters If c)(t) is a fundamental matrix for the homogeneous system x' = P(t)x on some interval where P(t) and f(t) are continuous. then a particular solution of the nonhomogeneous system x' = P(t)x + f(t) is given by (22) This is the variation of parameters formula for first-order linear systems.1 = e . (26) Equations (22) and (25) hold for any fundamental matrix 4>(t) of the homogeneous system x' = P(t)x. In the constant-coefficient case P(t) = A we can use for 4>(t) the exponential matrix eA1. In solving initial value problems it often is convenient to choose the constant of integration so that Xp (a) = 0.5. If we add this particular solution and the complementary function in (14 ). (22) is immaterial. x(a) = Xa. x(a) =0 in (24) to the solution X c(t) = 4>(t)4>(a).At.1f(t) dt (23) of the nonhomogeneous system in (11). The choice of the constant of integration in Eq. substitution of 4>(t) = eAt in (22) yields the particular solution (27) . Then. because (eAt) .1xa of the associated homogeneous problem x' = P(t)x. and thus integrate from a tot: (24) If we add the particular solution of the nonhomogeneous problem x' = P(t)x + f(t). we get the general solution x(t) = 4>(t)c + 4>(t) J 4>(t). the particular fundamental matrix such that 4>(0) = I.that is. consider the simple equation -=e dy dx .Numerical Methods IIIIJ _N~~!:!_~cal A. y ). y 0 ) and attempts to thread its way through the slope field of a given differential equation y ' = f (x . Y2 ).x2 • But it is known that every antiderivative of f (x) = e-x2 is a nonelementary function-one that cannot be expressed as a finite combination of the familiar functions of elementary calculus. Any attempt to use the symbolic techniques of Chapter 1 to find a simple explicit formula for a solution of ( 1) is therefore doomed to failure. y!). YI). and now moves a tiny distance along the slope segment through this new starting point (xi.x2 (1) A solution of Eq.PP_~~~~_mation: Euler's Method It is the exception rather than the rule when a differential equation of the general form dy dx = f(x. • At (xi. As a possible alternative. (1) is finitely expressible in terms of elementary functions. y) can be solved exactly and explicitly by elementary methods like those discussed in Chapter 1. Yo) and moves a tiny distance along the slope segment though (x0 . (1) is simply an antiderivative of e. Hence no particular solution of Eq. 430 . • The plotter pen starts at the initial point (xo. This takes it to the point (xi . y 1) the pen changes direction. an old-fashioned computer plotter-one that uses an ink pen to draw curves mechanically-can be programmed to draw a solution curve that starts at the initial point (x0 . The procedure the plotter carries out can be described as follows. For example. This takes it to the next starting point (x2. y 0 ). Hence a horizontal change of h from X11 to Xn+I corresponds to a vertical change of m · h = h · f(xn. Yn+l = Yn + h · j(Xrz. Yn) from Yn to Yn+l. Y2). Then the resulting polygonal curve looks like a smooth. y3) .. Yn) ism = j(x11 . suppose that each "tiny distance" the pen travels along a slope segment-before the midcourse correction that sends it along a fresh new slope segment-is so very small that the naked eye cannot distinguish the individual line segments constituting the polygonal curve. y 2 ). 6. yi). However. Yn+1) are given in terms of the old coordinates by Xn+l = Xn FIGURE 6. Yn). x 3. y Solution curve FIGURE 6. Leonhard Euler-the great 18th-century mathematician for whom so many mathematical concepts.. we ordinarily do not sketch the corresponding polygonal approximation. and his idea was to do all this numerically rather than graphically. In order to approximate the solution of the initial value problem dy dx = f(x.. y(x3). Y2 · y3 . y0 ) and applying the formulas x. y 1). (2) we first choose a fixed (horizontal) step size h to use in making each step from one point to the next. . this is (in essence) how most of the solution curves shown in the figures of Chapter 1 were computer generated. y (x2 ). Xn. Indeed. . y3 ).2. · · · .1... .. the numerical result of applying Euler's method is the sequence of approximations Y1.. . on an approximate solution curve. However. . Instead.1. y). Given the initial value problem in (2). (xJ.. Yn) to (Xn+l. (x2. . x 2. of the exact (though unknown) solution y (x ) of the initial value problem.. The step from (Xn.. Yo) Y1 + h · f(x. The slope of the direction segment through (x11 . Y2) the pen changes direction again. . (x3. Then the step from (xn.1 Numerical Approximation: Euler's Method 431 • At (x2. methods. Y2) . y0 ) and after n steps have reached the point (xn. (x3..6. Figure 6. Yn) to the next point (xn+l. at the points x 1 . Yn).. Yn+l ) . y3). y(xo) =Yo. Euler's method with step size h consists of starting with the initial point (x 0 . continuously turning solution curve of the differential equation. Y2) to calculate successive points (x 1. Yo). This takes it to the next starting point (x3.. Yn). These results typically are presented in the form of a table of approximate values of the desired solution. formulas. Yn+J) is illustrated in Fig. Suppose we've started at the initial point (x 0 . In this figure we see a polygonal curve consisting of line segments that connect the successive points (xo.1. and results are named-did not have a computer plotter. Yn. = xo X3 = X2 +h +h Y1 = Yo Y2 = Y3 = X2 = XJ +h + h · f (xo. . and now moves a tiny distance along the slope segment through (x2 . The first few steps in approximating a solution curve. · · · to the true values y(xi). y (x n). . (x2.1 . Therefore the coordinates of the new point (xn+ 1 .1 illustrates the result of continuing in this fashion-by a sequence of discrete straight-line steps from one starting point to the next.2.1.. Yt) Y2 + h · j(x2. + h. This error. Thus the smaller the step size.7. Thus the tangent line in Fig. is called the local error in Euler's method.001. rather than merely an approximation to the actual value y(x 11 ). The percentage errors at the final point x = 1 range from 7.1.1 down to only 0. introduced at each step in the process.6 would be the total error in Yn+l if the starting point Yn in (6) were an exact value.1 Numerical Approximation: Euler's Method y 435 I (xn+l• Yn+ .1.6 is tangent to the "wrong" solution curve-the one through (x11 . h = 0. The local error indicated in Fig. 6. The local error in Euler's method. 7 illustrates this cumulative error in Euler's method. Figure 6. Yn) rather than the actual solution curve through the initial point (x0 .1. Yn) departs from the solution curve through (x11 . y ! C"m"l"i" '""' (xn . y(O) = 1.8 we observe that.1.8 shows the results obtained in approximating the exact solution y (x) = 2ex .6. the more slowly does the error grow with increasing distance from the starting point. For instance. so that x 11 is an integral multiple of 0.1. with h = 0. but the value Yn is shown only when n is a multiple of 100. We show computed values only at intervals of L\x = 0. 6. 6. The cumulative error in Euler's method. By scanning the columns in Fig. for each fixed step size h. it is the amount by which the polygonal stepwise path from (xo.) I I I I I I X FIGURE 6. Yo). Yo) departs from the actual solution curve through (x0 .08% with h = 0.1 of the initial value problem dy dx = x + y. But by scanning the rows of the table we see that for each fixed x. 6. .6. using the successively smaller step sizes h = 0. Yn) l Approximate values Xz X FIGURE 6. is the amount by which the tangent line at (x 11 . the computation required 1000 Euler steps.1. the error decreases as the step size h is reduced. and h = 0.005. the error Yactual .1. as illustrated in Fig.6.25% with h = 0.1.1. h = 0. The usual way of attempting to reduce the cumulative error in Euler's method is to decrease the step size h.Yapprox increases as x gets farther from the starting point xo = 0.02. But Yn itself suffers from the accumulated effects of all the local errors introduced at the previous steps. y 11 ).1.x . y0 ). The table in Fig.1.001. 6.001. 1 . .0192 3.5282 1.Ywith y with h =0. the computer itself will contribute roundoff error at each stage because only finitely many significant digits can be used in each calculation.7812 2. Once an appropriate computer program has been written. For example.6493 3. 6.4238 Actual y 1.5807 1.4872 2. whereas if h is too small.0442 2.1.2380 1.··y \fi'fh . because the cumulative effect of roundoff error in the latter case might exceed combined cumulative and roundoff error in the case h = 0.3275 2.9461 2. y) in the initial value problem in (2).4 0. and on the specific computer used.8. In addition to the local and cumulative errors discussed previously. do we not simply choose an exceedingly small step size (such as h = 10.3832 FIGURE 6.3 0.12).1 in Fig.2 0. 1.1 and about 1 min 50s with h = 0. 6.1000 1. An Euler's method computation with h = 0.7 0. the data in Fig.3205 2.5831 1.1102 1.3620 1.7966 2.1082 1.1.3977 1. But 50 steps are required to reach x = 1 with h = 0.0001 will introduce roundoff errors 1000 times as often as one with h = 0. The subject of error propagation in numerical algorithms is treated in numerical analysis courses and textbooks. with the expectation that very great accuracy will result? There are two reasons for not doing so.3993 1.1103 1.3997 1.001.0 1. the approximations inherent in Euler's method may not be sufficiently accurate. . <··· n··· <no!· = h.1098 1.5 0. the computer hardly cares how many steps it is asked to carry out.1.001 1. and 1000 steps with h = 0.:::::moos 1.4230 . then roundoff errors may accumulate to an unacceptable degree or the program may require too much time to be practical. Why.5836 1.12 it would require over 3000 years! The second reason is more subtle.2200 1. on the exact code in which the program is written.2416 1.4266 0. one step size is-in principle-just as convenient as another. 200 steps with h = 0.1 might actually produce more accurate results than those obtained with h = 0.2426 1.3917 1. Approximating the solution of dyjdx = x successively smaller step sizes.1.3261 2.0388 2. + y. then.5719 1.005.8159 3.2428 1.6422 3.8 requires only 10 steps.02.1 0.7974 2.436 Chapter 6 Numerical Methods X h ¥:: Oi'l .001.0082 3.0431 2.6511 3.9757 3.0227 2. after all. It depends on the nature of the function f (x .7210 1. But with h = 10.1974 2. The "best" choice of h is difficult to determine in practice as well as in theory. A computer is almost always used to implement Euler's method when more than 10 or 20 steps are required. The first is obvious: the time required for the computation. y(O) = 1 with The column of data for h = 0.6161 2.1875 >' wi~Ji ~-"'~ . Thus it required slightly over one second to approximate y(l) with h = 0.7933 2.0170 3.6 0. With a step size that is too large.0001.9 1. so Euler's method can be carried out with a hand-held calculator. Hence with certain differential equations. h = 0.8 0.8 were obtained using a hand-held calculator that carried out nine Euler steps per second.2998 2. 10) then generates the Xn. y (xo) = Yo (1) has a unique solution y(x) on the closed interval [a . The number e = y(l). and a careful study of this method yields insights into the workings of more accurate methods. b]. 6. 1.:::. because many of the latter are extensions or refinements of the Euler method. You should begin this project by implementing Euler's method with your own calculator or computer system.1 is not often used in practice.:.:. y(O) = 1. Theorem 1 tells what degree of accuracy we can expect when we use Euler's method. y ) . y(l) = 0. Y] = eu1er(O. f x. where y(x) is the solution of the initial value problem dyjdx = 4/(1 + x 2 ). and assume that y(x) has a continuous second derivative on [a. then to some of the problems for this section. The number JT = y(l).442 Chapter 6 Numerical Methods For instance.1.14159 as specific values of solutions of certain initial value problems. But Euler's method has the advantage of simplicity. ..8. we need some way to measure the accuracy of each. y(O) = 0. ln2.:. THEOREM 1 The Error in the Euler Method Suppose that the initial value problem dy d x = f (x. (This would follow from the assumption that f. 2. 2. How many subintervals are needed to obtain-twice in successionthe correct value of the target number rounded to three decimal places? 1.. mainly because more accurate methods are available. 3. 200. where y(x) is the solution of the initial value problem dyjdx = y.) Then there exists a constant C such that . b].69315. subintervals (doubling n each time). 3.71828.:::. and JT . Test your program by first applying it to the initial value problem in Example 1. b] with a= x 0 .and Yn-data shown in the first two columns of the table of Fig. apply Euler's method with n = 50. IIIJ A Closer Look at the Euler Method The Euler method as presented in Section 6. 1. The number ln 2 = y(2). where y(x) is the solution of the initial value problem dyjdx = 1jx . where c ~ y(x) ~ d for all x in [a. Also explain in each problem what the point is-why the indicated famous number is the expected numerical result. 0. 100. Famous Numbers Investigation The following problems describe the numbers e. In each case. and j y are all continuous for a ~ x ~ b and c ~ y ~ d. the MATLAB command [X.:::. To compare two different methods of numerical approximation. Birkhoff and G.. .1. Repeat with h/2.1.. 1.-C. the error is bounded by a [predetermined] constant C multiplied by the step size h. Because C tends to increase as the maximum value of Iy'' (x) I on [a . In practice. then (2) for each n = 1. it follows that C must depend in a fairly complicated way on y . The constant C deserves some comment. 6. 4th ed. Continue until the results obtained at one stage agree.2 A Closer Look at the Euler Method 443 the following is true: If the approximations y 1 . 1989). . 2.Yn in (2) denotes the [cumulative] error in Euler's method after n steps in the approximation. Then the approximate values obtained at this stage are considered likely to be accurate to the indicated number of significant digits.. we begin with a step size h = 0. 3. with step size hjlO we get 10 times the accuracy (that is. The table in Fig. 6. y(xk) at points of [a. b] are computed using Euler's method with step size h > 0. (New York: John Wiley. Yk to the actual values y(xt). y(x3 ). 1/10 the maximum error) as with step size h. similarly. Indeed..1 shows the approximate values of y(l) obtained with successively smaller values of h. k. exclusive of roundoff error (as though we were using a perfect machine that made no roundoff errors). y(x 2 ). Ordinary Differential Equations.14. y3. perhaps one of those listed in Figs. that (on a given closed interval) halving the step size cuts the maximum error in half.6. Example 1 Carry out this procedure with the initial value problem dy dx 2xy 1 +x 2 ' y(O) = 1 (3) to approximate accurately the value y(l) of the solution at x = 1. 2. The data suggest that the true value of y (l) is exactly 0. Consequently. but one can be found in Chapter 7 of G. Remark: The error Yactual Yapprox = y(Xn). It follows. • We will omit the proof of this theorem. . y 2 . Solution Using an Euler method program. hj4.13 and 6.5. at each stage halving the step size for the next application of Euler's method. Rota. we can-in principle-get any degree of accuracy we want by choosing h sufficiently small. and actual computation of a value of C such that the inequality in (2) holds is usually impractical.to an appropriate number of significant digits-with those obtained at the previous stage. . . Apply Euler's method to the initial value problem in (1) with a reasonable value of h. The theorem can be summarized by saying that the error in Euler's method is of order h.to reach x = 1. that is.. the exact solution of the initial value problem in (3) is y (x) = 1/(1 + x 2 ) . b] increases. . the following type of procedure is commonly employed. and so forth.04 requiring n = 25 steps. so the true value of y ( 1) is exactly 4.2.. for instance. • . 3. Figure 6.04 0. (4) x x+h X FIGURE 6. It uses the predicted slope k = f(xn .0003125 0.10 0. True and predicted values in Euler's method. y (x + h.2. · .50000 0.Yapprox 1/h.11 0.50007 0.50003 Actual y(l) 0.Appro~iinat~y(l) 0.000625 0.50000 0.50000 0.444 Chapter 6 Numerical Methods . suppose that after carrying out n steps with step size h we have computed the approximation Yn to the actual value y(xn) of the solution at Xn = x 0 + nh. Yn) of the graph of the solution at the left-hand endpoint of the interval [xn. y(x +h)) The final column of the table in Fig. 1 .50054 0.10 FIGURE 6.2.50000 IErrorl/h 0.005 0. the error bound in (2) appears to hold with a value of C slightly larger than 0.50220 0. y-value i} Error An Improvement in Euler's Method As Fig.11 0. 6.2.1. Yn) at x = Xn has already been calculated.50013 0. Thus Now that Un+l ~ y(xn+t) has been computed.01 0.1. Why not average these two slopes to obtain a more accurate estimate of the average slope of the solution curve over the entire subinterval [xn.in this computation.1 displays the ratio of the magnitude of the error to h. Given the initial value problem dy dx = f(x.50000 0. . Euler's method is rather unsymmetrical.50109 0.0025 0. y).50000 0. it is known as the improved Euler method. that is.3 shows the geometry behind this method. Xn+tl? This idea is the essence of the improved Euler method.2.. Table of values in Example 1.11 0. We now turn our attention to a way in which increased accuracy can easily be obtained. Xn + h] as if it were the actual slope of the solution over that entire interval.00125 0.50027 0.2 shows.11 0.11 0. / I I I I /Slope y'(x) I I I I I I y(xo) =Yo. Observe how the data in this column substantiate Theorem l . .50451 0. 6. the approximate slope k 1 = f(xn.2. we can take as a second estimate of the slope of the solution curve y = y(x) at x = Xn +I· Of course.50000 0..02 0.50000 0.2. We can use the Euler method to obtain a first estimate-which we now call Un+l rather than Yn+ 1-of the value of the solution at Xn+I = Xn +h.Predicted . IYactual .11 0. 3. y(x2 ). . xz. kz = f(xn+l• Un+I). Remark: The final formula in (5) takes the "Euler form" Yn+l = Yn +h ·k + kz if we write k = kt 2 for the approximate average slope on the interval [xn.2 A Closer Look at the Euler Method y 445 (xn . Un+l = Yn + h · k1. respectively. The improved Euler method: Average the slopes of the tangent lines at (Xn. ••• of the [exact] solution y = y(x) at the points x 1. Yn) I I I I I I I I I n+l X FIGURE 6. to the [true] values y(xJ).. Yn+I = Yn +h · !Ckt +kz) (5) to compute successive approximations Yl· yz. y3. x3. . then it is used to correct itself. y(x 3 ). First a predictor un+I of the next y-value is computed. Thus the improved Euler method with step ... the improved Euler method with step size h consists in applying the iterative formulas kt = f(xn.. y(xo) =Yo. ALGORITHM The Improved Euler Method dy dx Given the initial value problem - = f(x. Xn+Il· • The improved Euler method is one of a class of numerical techniques known as predictor-corrector methods.6.2. y). Yn). . Yn) and (Xn+l· Un+f). Answer: Under the assumption that the exact solution y = y(x) of the initial value problem in (4) has a continuous third derivative. to the values y(x 1). as compared with the single function evaluation required for an ordinary Euler step.11) + (0.2. and with step size h /10 we get 100 times the accuracy (that is. U2 = 1.2 + 1. .1 + 1. this means that the improved Euler method is more accurate than Euler's method itself.11. . YI = 1 + (0. 1/100 the maximum error) as with step size h.1 we calculate UJ = 1 + (0. y2. the error in the Euler approximation to y(l) is 7 . but the error in the improved Euler approximation is only 0.11 + (0. The table in Fig.11 + (0. (6) and (7).1 is used.05) · [(0 + 1) + (0. y) = x +yin Eqs. y2. But the factor h2 in (8) means that halving the step size results in 1/4 the maximum error.446 Chapter 6 Numerical Methods size h consists of using the predictor Un+l = Yn + h · j(Xn. When the same step size h = 0.05) · [(0. 6. the predictor-corrector formulas for the improved Euler method are Un+l = Yn + h · (Xn + Yn). and so forth.25%.24%. it can be proved-see Chapter 7 of Birkhoff and Rota-that the error in the improved Euler method is of order h 2 • This means that on a given bounded interval [a. Yn+l = Yn + h · ~ [Cxn + Yn) + (Xn +l + Un+l)] · With step size h = 0.1)] = 1. of the actual solution of the initial value problem in (4).24205.11) = 1..1.x . This advantage is offset by the fact that about twice as many computations are required. y(x3).1 + 1. We naturally wonder whether this doubled computational labor is worth the trouble. . • Example 2 Figure 6. Remark: Each improved Euler step requires two evaluations of the function f (x. b ]. y). each approximate value Yn satisfies the inequality (8) where the constant C does not depend on h.231)] = 1.. Because h 2 is much smaller than h if h itself is small. y(x2).1) · (0.1.231..1. With f(x. y (0) = 1 (9) with exact solution y(x) = 2ex.1) · (0+ 1) = 1.. Yn) (6) and the corrector (7) iteratively to compute successive approximations y 1. Y 2 = 1.4 compares the results obtained using the improved Euler method with those obtained previously using the "unimproved" Euler method.8 shows results of applying Euler's method to the initial value problem dy dx = x + y.1 + 1. .0 1. and P). y(x3). y(x2) . y(xn) and now want to compute Yn+! ~ y(xn+d· Then y(xn+!). ..y(xn) = 1 Xn 1 Xn+l y 1 (x) dx = l xn +h y 1 (x) dx Xn (2) by the fundamental theorem of calculus. Next. Yn to the actual values y(xJ). +2 1 (3) Hence we want to define Yn+ l so that Yn+l ~ Yn h[Y (Xn) + 2y +6 1 Xn h) + 2y +2 1 ( Xn h) + Y (Xn+ !)J. Y2. y3. 6. It is called the Runge-Kutta method.sin2.y). This figure suggests-although it does not prove-the existence of a threshold initial population such that • Beginning with an initial population above this threshold. the population oscillates (perhaps with period P?) around the (unharvested) stable limiting population y (t) = M. whereas • The population dies out if it begins with an initial population below this threshold.y(xn) ~ h[Y (Xn) + 4y 6 1 ( 1 ( Xn h) + Y (Xn+ !) J.3 The Runge-Kutta Method 2. suppose that we have computed the approximations YJ.. y). ... A numerical approximation program was used to plot the typical solution curves for the case k = M = h = P = 1 that are shown in Fig.5 -1.6.0 ~--~~~~~~~ 2 4 5 models a logistic population that is periodically harvested and restocked with period P and maximal harvesting/restocking rate h. .5 (2nt) -0. Solution curves = y(l. Do the observations indicated here appear to hold for your population? ofdyfdt FIGURE 6. We now discuss a method for approximating the solution y = y(x) of the initial value problem dy dx = f(x. after the German mathematicians who developed it.12. y(xo) =Yo (1) that is considerably more accurate than the improved Euler method and is more widely used in practice than any of the numerical methods discussed in Sections 6. 0.y). M. With the usual notation. Carl Runge (1856-1927) and Wilhelm Kutta (1867-1944).5 453 Periodic Harvesting and Restocking The differential equation dy = ky(M.12.1 and 6.0 .2.nt.2.>. Simpson's rule for numerical integration yields y(Xn+!).2. +2 1 (4) . h. .h sin dt p 1. Use an appropriate plotting utility to investigate your own logistic population with periodic harvesting and restocking (selecting typical values of the parameters k. Xn+ Il using the Euler method to predict the ordinate there. b] with a = x 0 is of order h 4 .. Xn+Il· The Runge-Kutta method is a fourth-order method-it can be proved that the cumulative error on a bounded interval [a.. but does not depend on the step size h. the result is the iterative formula (6) The use of this formula to compute the approximations y 1.454 Chapter 6 Numerical Methods we have split 4y' (xn +~h) into a sum of two terms because we intend to approximate the slope y' (Xn + ~h) at the midpoint Xn + ~ h of the interval [Xn . with the following estimates. (5d) • This is the Euler method slope at Xn+I. (Sa) • This is the Euler method slope at Xn. (6) takes the "Euler form" Yn+I = Yn +h ·k if we write (7) for the approximate average slope on the interval [xn . successively constitutes the Runge-Kutta method. (8) where the constant C depends on the function f(x. Note that Eq. The following example illustrates this high accuracy in comparison with the lower-order accuracy of our previous numerical methods. (Thus the iteration in (6) is sometimes called the fourth-order Runge-Kutta method because it is possible to develop Runge-Kutta methods of other orders. using the improved slope k3 at the midpoint to step to Xn+I· When these substitutions are made in (4 ). y3. b]. (5b) • This is an estimate of the slope at the midpoint of the interval [xn. we replace the [true] slope values y'(xn). y' (xn + ~h). andy' (xn+I). y' (xn + ~h). . On the right-hand side in (4).) That is. . y) and the interval [a. respectively. (5c) • This is an improved Euler value for the slope at the midpoint. Xn+ 1] in two different ways. Y2. In the first step we use the formulas in (5) and (6) to calculate kt = 0 + 1 = 1. + (1 + (0.2. so only two steps are required to go from x = 0 to x = 1. k2 = (0 + 0. . y(O) = 1 with the same step size h = 0. It is customary to measure the computational labor involved in solving dyjdx = f(x. We see that even with the larger step size.25). (1. .5) + (0. . + (0.3. 6. the Runge.5 YI = 1 + 6[1 + 2 · (1. Runge-Kutta and improved Euler results for the initial value problem dyjdx = x + y.0 1.6.1 and again in Example 2 of Section 6.03% 0.4366 FIGURE 6. Thus the Runge-Kutta method gave over four times the accuracy with only 40% of the labor. y(0)=1 (9) that we considered in Fig.x . Figure 6.25).05% 1.4347. 6.24% 1.7974 3. y) that are required.0 0. Figure 6.1.5) + 2 · (1.5)) = 1.8 of Section 6.3125.1.7949 3. In this comparison the Runge. 2. In Example 1. a larger step size than in any previous example. . y) numerically by counting the number of evaluations of the function f (x. whereas the improved Euler method required 20 such evaluations (two for each of 10 steps). The relative error in the improved Euler value at x = 1 is about 0.625)) = and then 0. y) = x + y (four at each step). To make a point we use h = 0.7969 3.2 shows the results obtained by applying the improved Euler and Runge-Kutta methods to the problem dyjdx = x + y.1.7969.1.2.0000 1.Kutta method is about 2000 times as accurate.5 1.00012%.0000 1. y(O) = 1.25) + (1 k4 = (0.00% 0.1 presents these results together with the results (from Fig.5).25) + (1 k3 = (0 + 0. (1)) = 1.3 The Runge-Kutta Method Example 1 455 We first apply the Runge-Kutta method to the illustrative initial value problem dy dx =x+y. but for the Runge-Kutta value it is 0.625.5. Runge-Kutta 0. • .4282 0.Kutta method required eight evaluations of f(x. Computer programs implementing the Runge-Kutta method are listed in the project material for this section.3125] ~ 1. but requires only twice as many function evaluations. Similarly. The exact solution of this problem is y (x) = 2ex . (1.3.4347 0. as the improved Euler method. the second step yields y 2 ~ 3.14% 0.3.0000 1.1.00% 0.625) + 2.5.24%. the Runge-Kutta method gives (for this problem) four to five times the accuracy (in terms of relative percentage errors) of the improved Euler method.4) of applying the improved Euler method with step size h = 0. £ 1 X X = x 2 + y 2 . together with the percentage error in the more accurate approximation. y' 18. .2.£ x . y(O) = -1.£ X . y 2 y' = 2x 5 .£ 2 = X + . y(l) = 2. Thus.. Use the Runge-Kutta method with a programmable calculator or computer to approximate the solution for 10 years. lm (with tn+1 = tn +h)..£ X .6v. 11.£ 2. 1 . ).. Because of linear air resistance.1. y' = . t2. y(O) = 1. shot straight upward from the ground with an initial velocity of 49 m js. Consider again the crossbow bolt of Example 2 in Section 1. 3.2. As in Problem 25 of Section 6. you bail out of a helicopter and immediately open your parachute.245. What percentage of the limiting velocity 20 ft/s has been attained after 1 second? After 2 seconds? 26.9. y (0) = 0.£ x . y(l) = 3. t3. first with step size h = 0. Use the Runge-Kutta method with a programmable calculator or computer to approximate the solution for 0 . 2.£ X . by beginning with the velocity data in Fig.y . VJ.2y. then with step size h = 0. 0 ~ x That is.04)v.£ 2 = In y. the linear acceleration a = dvjdt of a moving particle is given by a formula dvjdt = f(t. As in Problem 26 of Section 6.1. Suppose that the velocity v(t) is approximated using the Runge-Kutta method to solve numerically the initial value problem dv dt imate to six decimal places the values of the solution at five equally spaced points of the given interval. y(O) = 0. Make a table showing the approximate values and the actual value. the formulas in Eqs. 0 ~ x . y' 19. rounding off approximate P-values to four decimal places.. In these initial value problems.£X . (a) Use a calculator or computer implementation of the Runge. tn+I ].£ 2 = sinx +cosy.£ x ~ 2 = x 213 + y 213. v(O) = 49 dP = 0. Start with step size h = 1.3. 1 . y(2) = 3. . = f(t. primes denote derivatives with respect to x. once a table of approximate velocities has been calculated. Vm at the successive times t 1. The formula in (20) would give the correct increment (from Yn to Yn+I) if the acceleration an remained constant during the time interval [tn. y(O) = 1. 0 . Vn) ~ v'(tn) is the particle's approximate acceleration at time tn. 2 ~ x ~ 3 16. 0. 0 . We can do this by beginning with the initial position y(O) = Yo and calculating (20) 24. where the velocity v = dyjdt is the derivative of the function y = y(t) giving the position of the particle at timet. - S 1 - 25. y' 21.Kutta method to approximate v(t) for 0 ~ t ~ 10 using both .£ 1 = 4(y 1)2 .1 and then with h = 0. Eq.460 Chapter 6 Numerical Methods (with t in months). y' 12. first with step size h = 0.1125 .£ 3 A computer with a printer is required for Problems 17 through 24. y' = x 2 + y 2 .8. 0 . v(O) = 0 (with t in seconds and v in ft/s).£ 2 = X + ~. . y' 22. In each problem find the exact solution of the given initial value problem. and 0. y' 23. What percentage of the limiting population of 75 deer has been attained after 5 years? After 10 years? A programmable calculator or a computer will be useful for Problems 11 through 16.£ x . Then apply the RungeKutta method twice to approximate (to five decimal places) this solution on the given interval. its velocity function v = dyjdt satisfies the initial value problem dv dt = -(0. v(O) = vo. y' = y. y(2) =? 28.. y(2) = 3.£ t .JY. xy'=y 2 .0225 P . 0 . 6. for x an integral multiple of 0. y(2) =? 13. (20) provides a simple way to calculate a table of corresponding successive positions. yy' = 2x 3 .0003P 2 dt with exact solution v(t) = 294e. 2 .1. and then use successively smaller step sizes until successive approximate solution values at x = 2 agree rounded off to five decimal places. 0 .1~x~2 15.£ X . Now suppose that we also want to approximate the distance y(t) traveled by the particle. Throughout. xy' = 3x. 0. . .05. where an = f(tn. (19) 17. y(O) = 0. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h.2.2. starting with t0 = 0 and v0 .. so your downward velocity satisfies the initial value problem dv dt = 32. y(-2) = 0.8.£ 1 Use the Runge-Kutta method with a computer system to find the desired solution values in Problems 27 and 28. (5) and (6) are applied-with t and v in place of x and y-to calculate the successive approximate velocity values v 1... y (0) = 2. y' 20.2. y' =X+ 4y 2 .y(1)=1. y' .1.05. 8 (Example 3) and proceeding to follow the skydiver's position during her descent to the ground. rounding off approximate v-values to three decimal places.0. 0 .025 to approx- Velocity-Acceleration Problems In Problems 29 and 30. y(-1) = 1· -1 S 1 + y2 . v). y(O) = 1. v). primes denote derivatives with respect to x.2.£ 2 14. 29. v2 . This process is illustrated in the project for this section. y(O) = 1. Throughout. 27.£ I 2 2 = x . use the Runge-Kutta method with step sizes h = 0. first with step size h = 6 and then with h = 3. suppose the deer population P(t) in a small forest initially numbers 25 and satisfies the logistic equation (n = I. 4. the bolt's velocity and position functions during ascent and descent are given by the following formulas.245t? (c) If the exact solution were unavailable. explain how you could use the Runge-Kutta method to approximate closely the bolt's times of ascent and descent and the maximum height it attains. According to the results of Problems 17 and 18 in Section 1. Now consider again the crossbow bolt of Example 3 in Section 1. v(O) = 49.465 + (909.1/ 25 ) . As output it produces the resulting column vectors x andY of x. one need only change the initial line of the program. y). but because of air resistance proportional to the square of its velocity. Do these approximate position valueseach rounded to two decimal places-agree with the exact solution y(t) = 7350 (1 461 Beginning with this initial value problem. the initial value y. its velocity function v(t) satisfies the initial value problem dv .11 lists TI-85 and BASIC programs implementing the Runge-Kutta method to approximate the solution of the initial value problem dy dx = x + y. Ascent: v(t) - e.103827]t). 12 exhibits a MATLAB implementation of the Runge-Kutta method.091) In (cos(0.388) tan(0.3.OOll)vlvl.8.8. 30. dt y(t) = = (94.3 The Runge-Kutta Method n = 100 and n = 200 subintervals. You should begin this project by implementing the Runge-Kutta method with your own calculator or computer system. 108.9. Y] = rk(O. 6.3. Display the results at intervals of 1 second. Descent: v(t) = -(94. and the desired number n of subintervals.(909.091) In (cosh(O.= -(O.3. then to some of the problems for this section.103827]t)). repeat parts (a) through (c) of Problem 25 (except that you may need n = 200 subintervals to get four-place accuracy in part (a) and n = 400 subintervals for two-place accuracy in part (b)). 10) then generates the first and third columns of data shown in the table in Fig. Then the rk function takes as input the initial value x.465 . y) has been defined.2. y(t) = 108. 6.6119])).4. Do the two approximations-each rounded to four decimal places-agree both with each other and with the exact solution? (b) Now use the velocity data from part (a) to approximate y(t) for 0 ~ t ~ 10 using n = 200 subintervals.andy-values.8. To apply the Runge-Kutta method to a different equation dyjdx = f(x. .6119]). the MATLAB command [X. 1. It still is shot straight upward from the ground with an initial velocity of 49 mjs.478837.l03827[t. Suppose that the function f describing the differential equation y' = f (x. one need only change the value of N specified in the second line of the program. For instance. Test your program by applying it first to the initial value problem in Example 1. Figure 6. y(O) =1 considered in Example 1 of this section.388) tanh(0. Display the results at intervals of 1 second. To increase the number of steps (and thereby decrease the step size). the final value x1 of x.3 Application Runge-Kutta Implementation Figure 6.478837. 1. in which the function f is defined.[0.[0.103827[t.6. The comments provided in the final column should make these programs intelligible even if you have little familiarity with the BASIC and TI programming languages. let us write x = [~ J and f =[~ J. Yn ).. . Our goal is to apply the methods of Sections 6. Euler Methods for Systems For example.. With this assurance we can proceed to discuss the numerical approximation of this solution. We can then make the step from Xn to the next approximation Xn+I ~ x(tn+I) by any one of the methods of Sections 6. the iterative formula of Euler's method for systems is Xn+l = Xn + hf(t. ' of the exact solution of the system in (1). (4) . y' = g(t . and are vector-valued functions. x). where tn+I = tn + h for n ~ 0.. (3) Then the initial value problem in ( 1) is x' = f(t. Yn + h g (tn . x0 ).. tz. Xn. Yn ).. x(to) = xo (1) for a system of m first-order differential equations.3 to the initial value problem x' = f(t.3. y (to) = Yo	677.169	1
Numerical Analysis . In other words. and very often are found to fail in circumstances when a less obvious numerical process would have succeeded. While scientists have always been concerned to some extent with numerical computation. the modern discipline of numerical analysis is almost entirely a product of the period since 1950. do it accurately. during which there has been an explosion in the availability of electronic computers. especially on computers with floating point arithmetic. do it quickly.Numerical Analysis Numerical analysis is concerned with the accurate and efficient evaluation of mathematical expressions. Direct translations of expressions from mathematical theory are seldom optimal. and do it cheaply. to obtain accurate numerical approximations to quantities which may not have explicit mathematical expressions. to organize computations so that there is minimum accumulation of error in floating point arithmetic 2. so that they consume the least possible resources 3. There are three main issues: 1. It might be thought that numerical evaluation consists largely of translating textbook formulas into a computer programming language – a mere coding exercise – but this is very far from the case. to organize computations efficiently. differential equations. Approximation error occurs when the computed expression is not exactly equal to the theoretical quantity even in exact arithmetic. Indeed. Errors arise from three sources: (i) errors in the input data. Efficiency is usually measured by counting basic floating point . (ii) computation errors due to finite precision arithmetic. but can evaluate functions which are defined only indirectly. an indirect method is often preferred for numerical evaluation even when a direct expression exists. Computation error appears because of the difference between exact arithmetic and the finite-length arithmetic available on digital computers and hand calculators. since a slightly slower accurate program is invariably preferred to a faster one with unreliable accuracy. and (iii) approximation error. in fact it might be considered to be the special concern of the statistician.2 Numerical Analysis The focus on numerical results means that one is not limited to direct expressions. While efficiency and accuracy are both aims. it is accuracy which takes precedence. or an infinite series is evaluated only to a finite number of terms. or as solutions to equations. series. An integral is replaced with a sum. for example. for example through integrals. The first of these is not under the control of the calculation. An important example is the maximum likelihood estimator for a nonlinear statistical model (see Nonlinear Regression). many problems cannot be solved by simple . Another consideration is to minimize the use of computer memory and other space requirements. especially for large jobs. You must still understand the program's purpose and limitations to know whether it applies to your particular situation or not. More recent concerns which arise from modern computer architecture include parallel computing (designing algorithms so that they can be evaluated in parallel streams on fast computers with multiple processors) and local referencing (minimizing unnecessary paging of virtual memory). subtractions. It is also desirable to keep programs simple and understandable. which cannot be modified and must be taken somewhat on trust. thus making the programs easy to maintain and to modify. Users must often choose between using compact programs which can be tinkered with for their own use.Numerical Analysis 3 operations (flops). multiplications. An understanding of the numerical methods being used and an idea of when they will perform well or poorly is necessary even for users of standard statistical package programs (see Software. and using sophisticated high-performance software from public libraries. Most biostatisticians can benefit from familiarity with numerical analysis. Biostatistical). such as additions. and divisions. More importantly. . If you develop your own software. The condition ˜ number κ of f at x is defined. Conditioning The concept of conditioning refers to the intrinsic difficulty of a numerical problem. A problem is ill-conditioned if it is sensitive to perturbations in the data. which contains a lot of advice on routines to use. Pointers to available software are given at the end of the article. An introduction with some statistical orientation is by Thisted [13]. Suppose that x is the exact argument to a function f but unfortunately only an approximation x is available.4 Numerical Analysis application of a standard program. An elegant and elementary introduction to the fundamental ideas of numerical analysis is given by Stewart [11]. with respect to a given norm (\|\| · \|\|). [9]. This article discusses the basic ideas of accuracy and describes briefly key topics in numerical analysis which are treated at more depth in separate articles. and well-conditioned if it is not. A justifiably popular text on scientific computing is Press et al. Other good general texts on numerical analysis are by Atkinson [3] and Stoer & Bulirsch [12]. a knowledge of numerical analysis can help avoid numerical pitfalls that can occur easily in a number of problems. Conditioning is often quantified by a condition number which refers to the amplification of relative errors. f (x) = (x − 1)6 is an illconditioned function near x = 1. For example. For univariate. For general multivariate functions. the computation of regression coefficients from a multiple regression is ill-conditioned when the design matrix X displays collinearity. the condition number is essentially κ = \|xf (x)/f (x)\|. then errors in x are magnified in the evaluation ˜ of f (x). If κ = 10k . Conditioning also depends on the quantity of interest. differentiable functions. while the opposite is true if κ is small.Numerical Analysis by the relation \|\|x − x\|\| ˜ \|\|f (x) − f (x)\|\| ˜ ≈κ \|\|f (x)\|\| \|\|x\|\| 5 for x near x. and will do . A least squares regression may be ill-conditioned from the point of view of the regression coefficients but well-conditioned from the point of view of the fitted values. Stability A stable algorithm is one which evaluates a function to the accuracy allowed by the function's condition number. A stable algorithm therefore will evaluate a well-conditioned function accurately. the specific definition of condition number depends on the problem. If κ is large. while f (x) = x 1/2 is well-conditioned for any x > 0. then k is roughly the number of significant figures of accuracy we can expect to lose in the computation. For example. some of which are of great interest to manufacturers of hand calculators. s 2 is computed to be 0. A commonly taught formula for the variance is 1 s = n−1 2 n xi2 − nx 2 . which is 1191 × 101 in 4-digit arithmetic. For example. Similarly. 64 using four-digit decimal arithmetic. There are many other algorithms for computing the sample variance. The first formula is unstable. which evaluates to [(−1)2 + 02 + 12 ]/2 = 1 – the correct answer in this case. see Chan et al. . s 2 = (xi − x)2 /(n − 1). while the second formula is stable. Since the data are given to two significant figures. [4] for a discussion. In this case xi2 = 3844 + 3969 + 4096.6 Numerical Analysis as well as can be expected on an ill-conditioned problem. 63. Alternative algorithms are available: for example. Therefore. a 100% error compared with the true value of 1. nx 2 = 3(632 ) is 1191 × 101 to 4 digits. i=1 where n is the sample size and the xi are the observations. it might be thought that carrying four significant figures through the calculation will leave a more than adequate safety margin. consider the problem of computing the sample variance of the three numbers: 62. For example. Let f˜(x) be the approximation to f (x) which arises from an . because of the need to compute quantities which do not grow without bound. although rounding error occurred not in the subtraction but in the previous summation. Frequently. In this case. and the error is revealed when the difference is taken of the two large and nearly equal quantities. This is often called subtractive cancellation. of numerical analysis is to limit the growth in size of intermediate quantities in calculations. Even the small example above gives an example of this.Numerical Analysis 7 The error in the first formula above arises in the rounding errors of xi2 and nx 2 . not statistically useful quantities in their own right. There is often a close relationship between stability in numerical analysis and in statistics. a summation is generally stable if the summands are all of one sign. One concern. parameters which are statistically interpretable because they measure some invariant characteristic of a problem appear also in a stable algorithm. therefore. the partial sums cannot be greater in absolute value than the final sum. It is a general principle that one cannot add a large value to a floating point number and later subtract it without losing accuracy. as the xi − x are the wellknown residuals. while in the textbook formula xi2 and nx 2 are merely intermediate quantities. For example. Floating Point Arithmetic There are an infinite number of real numbers. A thorough treatment of rounding-error analyses can be found in Higham [6]. fraction f . but only a finite number can be represented on a computer. In general. where x is close ˜ ˜ to x. and the resulting arithmetic is called binary . The algorithm is called backwardly stable if f˜(x) can be shown to be equal to the exact evaluation of f at x. a (backwardly) stable algorithm will compute a wellconditioned function accurately.8 Numerical Analysis algorithm. In this way. 2. Although proving error bounds is an important part of modern numerical analysis. rounding-error analyses are less valued for their final bounds than for the insight they provide about a numerical algorithm. Therein lies the fundamental difference between exact and computer arithmetic. Numbers are represented on computers in floating point form.e. the specific bounds obtained are usually pessimistic and are seldom used in practice. i. and exponent e. alluded to above.597 × 10−3 is a base-10 floating point number with four figures of accuracy. Most computers use base 2. and will compute an ill-conditioned function as accurately as is allowed by its conditioning. f × β e in terms of a base β. an arithmetic operation that produces an exponent that is too small is said to have underflowed. more generally. we outline two applications of interest to biostatisticians. most numbers cannot be represented exactly on floating point arithmetic of a fixed word length. 9 Finite computer arithmetic produces three types of errors. Linear Equations and Matrix Computations The theory and practice of solving a linear system Ax = b for x. Even within the limits of the exponent. the whole subject of computations involving matrices. The resulting inaccuracy is called rounding error. numerical analysts have influenced statisticians to move away from the normal equations for the regression coefficients in favor of methods . Here.Numerical Analysis arithmetic. In least squares regression of a response vector y on a design matrix X. Similarly. the result is said to have overflowed. When an arithmetic operation produces a number with an exponent that is too large. It is a central concern of numerical analysis that rounding errors do not accumulate during a long computation (see Floating Point Arithmetic). and. is now very well developed (see Matrix Computations). Section 15. then the singular values entirely capture the idea of ill-conditioning and collinearity for the least squares problem. Conditioning for the least squares problem is determined by that of X. Numerical linear algebra is dealt with in more detail in the article on Matrix Computations. .4]. and there are those who propose its routine use in regression computations for that reason [9. say. where U and V are orthogonal and D is diagonal containing the singular values. where Q is an orthogonal matrix and R is upper triangular. while the normal equations are not. The singular value decomposition therefore gives statisticians the means to quantify collinearity.10 Numerical Analysis based on the decomposition X = QR. The condition number of X is usually defined to be the ratio of the largest to the smallest singular value. by dividing by the sample standard deviation of the column. If the columns of X are standardized. which can be analyzed through the singular value decomposition X = UDVT . This is because the QR approach is backwardly stable. Numerical Analysis Optimization and Nonlinear Equations 11 Optimization means to find that value of x which maximizes or minimizes a given function f (x). Interpolation and Approximation The purpose here is accurately to approximate complex functions with ones which are easy to evaluate. Details are given in the article on Optimization and Nonlinear Equations. A closely related problem is that of solving nonlinear equations. Many algorithms for optimizing f (x) are. derived from algorithms for solving ∂f/∂x = 0. rational function approximations to the standard normal distribution function and its inverse allow it to be computed rapidly within statistical programs. This is a central concern in statistics. where ∂f/∂x is the derivative vector of f with respect to x. in fact. maximum likelihood. posterior mode (see Bayesian Methods) and Mestimation (see Robustness) are defined in terms of optimizing an appropriate objective function. because statistical estimation principles such as least squares. Typical . Numerical optimization strategies come into play when the statistical model is nonlinear and analytic estimators of the parameters are not available. For example. Unfortunately. and polynomials (see. . Numerical Integration After matrix computations. Numerical analysts were also early users of the internet. rational functions. Chapter 5] and Polynomial Approximation). the picture is less clear in high dimensions. A large number of sophisticated and reliable methods are available for numerical integration in one dimension. for example.org. Press et al. [9.netlib. Statisticians have made a substantial contribution to highdimensional integration through the development of efficient Monte Carlo methods. A survey of integration methods is given in the article on Numerical Integration. numerical integration is one of the largest areas of numerical analysis. for statisticians wanting to evaluate mixture models or Bayesian marginal posteriors.12 Numerical Analysis methods include series expansions. A great many approximation formulas are given in Abramowitz & Stegun [1]. and a wide range of software is available online. Its URL is Available Software The final goal of numerical analysis is to make numerical methods generally available through high-quality portable software. Netlib is the most extensive collection of numerical programs. EISPACK. Commercial subroutine libraries include the NAG Library (Numerical Algorithms Group) and the IMSL Mathematics and Statistics Libraries. and the SLATEC library – an enormous library of FORTRAN programs.harvard. Other libraries of note include the QUADPACK library [8] for numerical integration. The journal ACM Transactions of Mathematical Software is a source of refereed software. also searchable by GAMS. The two libraries have now been combined and updated as LAPACK [2]. The Guide to Available Mathematical Software (GAMS) at Another popular commercial source is Numerical Recipes [9]. and other routines have also been incorporated into the interactive matrix programming language. Numerical Recipes supplies smaller. These are published.Numerical Analysis 13 Worthy of special mention are the LINPACK library [5] for linear algebra and the EISPACK library [10] for eigenvalue computations.gov provides a virtual database of documented and supported programs. MATLAB [7].edu/nr. understandable programs. LINPACK. which may be modified by users . documented and freely available. searchable by program and problem type. both from the Argonne National Laboratory.nist. accessible through and have gained wide acceptance by statisticians and other scientists.	677.169	1
PRODUCT DESCRIPTION The student will be able to: * understand the connections between algebra and geometry and uses the one- and two-dimensional coordinate systems to verify geometric conjectures. determine the coordinates of a point that is a given fractional distance less than one from one end of a line segment to the other in one- and two- dimensional coordinate systems, including finding the midpoint. process skills to understand the connections between algebra and geometry and uses the one- and two-dimensional coordinate systems to verify geometric conjectures. *derive and use the distance, slope, and midpoint formulas to verify geometric relationships, including congruence of segments [and parallelism or perpendicularity of pairs of lines	677.169	1
Introduction to Analytic Number textbook is designed to teach undergraduates the basic ideas and techniques of number theory, with special consideration to the principles of analytic number theory. The first five chapters treat elementary concepts such asMore... This introductory textbook is designed to teach undergraduates the basic ideas and techniques of number theory, with special consideration to the principles of analytic number theory. The first five chapters treat elementary concepts such as divisibility, congruence and arithmetical functions. The topics in the next chapters include Dirichlet's theorem on primes in progressions, Gauss sums, quadratic residues, Dirichlet series, and Euler products with applications to the Riemann zeta function and Dirichlet L-functions. Also included is an introduction to partitions. Among the strong points of the book are its clarity of exposition and a collection of exercises at the end of each chapter. The first ten chapters, with the exception of one section, are accessible to anyone with knowledge of elementary calculus; the last four chapters require some knowledge of complex function theory including complex integration and residue calculus. Tom Apostol is an Emeritus Professor of Mathematics at the California Institute of Technology	677.169	1
Solving Linear Equations 2 Day Review Compressed Zip File Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files. 4.99 MB \| 13 pages PRODUCT DESCRIPTION This is a Pre-Algebra Common Core Review on Solving Linear Equations. Students will practice solving equations by combining like terms, using the distributive property, with variables on both side, and literal equations. Students will also translate verbal problems, explain errors in student work and review important unit vocabulary	677.169	1
Check Your Delivery Options In which way is thinking dependent on the situational context? How is information transferred between different co-existing representations of the same situation? In this book, these questions are explored in the context of middle school students working with linear functions. Conceptual understanding and reference are discussed in a Peircian semiotic framework. A detailed, formal analysis based on situation calculus goes deeply into the questions of linkages between representations, the meaning of numbers, and the informational content of different mathematical representations. The study shows how our reasoning can be dependent on the present perceived situation, and how this may hinder us to "reason backwards" like understanding how a table we created from an observation relates back and represents that very observation. This has implications for modeling in particular and education and transfer in general.	677.169	1
McDougal Littell Math Course 1 - Student Textbook (2007) The focus of the early chapters in McDougal Littell Math Course 1 is on numbers, their operations, and their algebraic representations. Students will build their understanding of these concepts through a variety of models, such as base-ten pieces, rulers, and verbal models. They will also apply their skills to problem-solving situations and use estimation to check reasonableness. Topics from other math strands, such as measurement conversions, area, averages, and data displays, are introduced early in the course and then integrated and expanded upon throughout. Later chapters in McDougal Littell Math Course 1 include topics such as integers, functions, and probability. The number and variety of problems, ranging from basic to challenging, give students the practice they need to develop their math skills. Every lesson in McDougal Littell Math Course 1 has both skill practice and problem solving, including multi-step problems. These types of problems often appear on standardized tests and cover a wide variety of math topics. To help students prepare for standardized tests, McDougal Littell Math Course 1 provides instruction and practice on standardized test questions in a variety of formats – multiple choice, short response, extended response, and so on. NB: Neither Glossaries nor Index are included in the pdf version of the textbook!	677.169	1
Notes About the First Test and Software A Special Note about the First Test: The first test will include a "take home" portion in which you will write and use a computer "program" to compute and graph partial sums of Fourier series. The word "program" is misleading. Instead of a programming language like C or Pascal or Fortran, you should use a mathematics package such as Mathcad or Maple. See problem 5.13 (page 56) and problems 9.8 and 9.9 at the end of chapter 9 (pages 119 and 120). It would be a very good idea to start now to make sure that you can use one of these packages with reasonable proficiency. In particular, you will need to know how to handle functions, indices, summations, integration and graphing. Notes on Software: Maple, Mathcad, Mathematica, and Matlab are the math packages generally available. Of these, I would suggest using Maple, especially if you've already used it in previous classes. It is easy to use (compared to Matlab or Mathematica); it gives excellent results, and, best of all, we have an arrangement This is the end of the preview. Sign up to access the rest of the document. This note was uploaded on 11/07/2011 for the course MA 460 taught by Professor Staff during the Fall '11 term at University of Alabama in Huntsville.	677.169	1
Carl Friedrich Gauss referred to Mathematics as the Queen of the Sciences. Pure Mathematics is used throughout the world as an essential tool in many fields including Natural science, Engineering, Medicine, Finance and the social sciences and Applied Mathematics. Mathematics prepares you for careers as actuaries, computer scientists, mathematicians, statisticians, teachers and more. Advanced skills in mathematics are necessary in many science-related careers. A degree in mathematics is also a strong background for jobs in the computer Industries as well as many related fields For those with a substantial background in mathematics and passion for numbers, an unlimited number of career opportunities are available. Dr. Ujwala Deshmukh received. Department conducts its annual program 'Mithimatica' from the academic year 2012 – 13 . Under this, we organize various events such as Seminar Contest, Poster Competition and Quiz Competition. The aim of this program is to spread awareness among the students about the applications of Mathematics and to large extent we have achieved this. We conduct remedial lectures for Madhava Competition, a national level competition for the maths students. The aim of this, is to prepare students for various competitive exams such as entrance exams for IIT, TIFR, actuaries. This may facilitate them to appear for various government exams also such as UPSC/MPSC. Dr. Ms. Alka Kanetkar Workshop on syllabus of F.Y.B.Sc. (Revised for CBSGS) in Mathematics B.N. Banodkar college of Science , Thane July 11, 2011 Workshop on semester based , credit and Grading system ( Sem I and II ) Dr. Mrs. Ujwala N. Deshmukh Appointed as referee for journal of American Mathematical Society Recognized as Ph.D. guide (Graph Theory) in Mumbai University from August 2012. Appointed as a referee for Indian Journal of pure and Applied Mathematics. Co-ordinator of M.Sc. Mathematics Received Mr. Prabhat Dwivedi Teacher's Enrichment working on Multivariable calculus and its application, organized by school of Mathematics TIFR. Short term course on skill on soft skills Development organized by UGC Academic staff college, University of Mumbai. Mrs. Ujjwala D. Kurkute Organized one day event In Memory of Bhaskaracharya 900th Birth Anniversary Teacher's Enrichment working on Multivariable calculus and its application, organized by school of Mathematics TIFR. Attended Seven day workshop in Mumbai University Indian women for Mathematics . Attended Two day National Seminar in Patkar College Mr. Laxman S. Naik Organized one day event In Memory of Bhaskaracharya 900th Birth Anniversary Refresher Course In Mathematics ICT Matunga . Done Election Duty for Parliamentary Election Mr. Krishna Maurya Organized one day event In Memory of Bhaskaracharya 900th Birth Anniversary Done Election Duty for Parliamentary Election Three students of third year Bachelor's programme, Miss Gunjan Soni, Master Pradeep Mishra and Master Sumit Vishvakarma participated in an intercollegiate festival of K. J. Somaiya College, 'Rasayan 2015' and won first prize in two events namely 'Chem – O – Hunt' and 'Best Out Of Waste'. The first batch of T.Y.Bsc (Mathematics) passed out in 1981. Late Prof. M. L.Vaidya was the first Head of the Department. Dr. Alka V. Kanetkar is the current Head of the Mathematics Department. The Department consists of experienced, disciplined and knowledgeable staff members. In the Mathematics Department, teaching is done by traditional and innovative techniques in order to cope up with the changing trends in education. Mathematics Association was formed under the guidance of Dr. Alka V. Kanetkar and staff members of the Department. The association organizes various activities in mathematics for the students. Students participate actively in all the events. M.Sc. in Mathematics was started in the academic year 2012-13 to encourage the students for Higher education in Mathematics. Students from our department have pursued studies in higher Mathematics in India in reputed institutes like TIFR , IIT etc. as well as abroad	677.169	1
NCERT Solutions For Class 8 Class 8 NCERT Science and Maths solution can be found here. Class 8 Science and Maths NCERT solution have been aligned chapter wise and subject wise and hence the students will be at ease to use the answers at their disposal. NCERT book is the holy grail of the students studying in CBSE board and hence it is used for clearing the concept of each and every chapter. These solutions are available for the students to download in the PDF format and are readily used. These solutions will be helpful in learning the concepts and putting the understanding to test. Students having any doubt in their solutions can turn for a reliable approach through the PDFs that we are providing here. If, the students still has any doubt regarding the solution, then they can clear it by calling our mentor at 9243500430. Call us and our executive will connect the students with our subject matter expert.	677.169	1
280768 ISBN: 047028076X Edition: 10 Publication Date: 2008 Publisher: Wiley AUTHOR Karl E. Byleen, Raymond A. Barnett, Michael R. Ziegler, Dave Sobecki SUMMARY Featuring rich applications and integrated coverage of graphing utilities, this hands-on trigonometry text guides students step-by-step, from the right triangle to the unit-circle definitions of the trigonometric functions.Karl E. Byleen is the author of 'Analytic Trigonometry with Applications', published 2008 under ISBN 9780470280768 and ISBN 0470280	677.169	1
These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are:Strings and Languages, Sequences and Strings, Sequence of Elemen... Latest slides uploaded in discrete mathematics These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are:Combinatorics, Elementary Counting, Techniques, License Plates,... These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are:Discrete Mathematics, Consisting of Unconnected Parts, Continuou... These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are:Sequences and Summations, Image of Function, Subset of Set of In... These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are:Functions, Non Empty Set, Domain of Function, Codomain of Functi... These study notes are very easy to understand elementary discrete math and very helpful to built a concept about the foundation of computers.The key points discuss in these notes are: Inductive Proofs and Inductive, Mathematical Induction, Dominos...	677.169	1
Product Description: A knowledge of one or more high level symbolic mathematics programs is rapidly becoming a necessity for mathematics users from all fields of science. The main aim of this book is to provide a solid grounding in Maple, one of the best known of these programs. The authors have sought to combine efficiency and economy of exposition with a complete coverage of Maple. The book has twelve chapters of which eight are completely accessible to anyone who has completed the usual calculus and linear sequences as taught in American universities. These cover the great majority of Maple's capabilities. There are also three chapters on Maple programming. These can be read without prior programming experience, but a knowledge of a high level programming language (Basic, Fortran, C etc.) will help a lot. There is also a chapter on some relevant aspects of algebra. Although the book is complete in its coverage of Maple, there is no 'fat' in the book. Above all, the book is designed to enable the reader to extract value from Maple without wasting time and effort in the learning process. It is the fastest track to Maple expertise. REVIEWS for	677.169	1
Summary These authors understand what it takes to be successful in mathematics, the skills that students bring to this course, and the way that technology can be used to enhance learning without sacrificing math skills. As a result, they have created a textbook with an overall learning system involving preparation, practice, and review to help students get the most out of the time they put into studying. In sum, Sullivan and Sullivan'sTrigonometry:Enhanced with Graphing Utilitiesgives students a model for success in mathematics. Author Biography Mike Sullivan is a Professor of Mathematics at Chicago State University and received a Ph.D. in mathematics from Illinois Institute of Technology. Mike has taught at Chicago State for over 30 years and has authored or co-authored over fifty books. Mike has four children, all of whom are involved with mathematics or publishing: Kathleen, who teaches college mathematics; Mike III, who co-authors this series and teaches college mathematics; Dan, who is a Pearson Education sales representative; and Colleen, who teaches middle-school mathematics. When he's not writing, Mike enjoys gardening or spending time with his family, including nine grandchildren. Mike Sullivan III is a professor of mathematics at Joliet Junior College. He holds graduate degrees from DePaul University in both mathematics and economics. Mike is an author or co-author on more than 20 books, including a statistics book and a developmental mathematics series. Mike is the father of three children and an avid golfer who tries to spend as much of his limited free time as possible on the golf course.	677.169	1
Summary and Info This book is designed to offer applied mathematicians, physicists, chemists, engineers, geophysicists, and other scientists an elementary level explanation of integral equations of the first kind. It maintains a casual, conversational approach. The book emphasizes understanding, while deliberately avoiding special methods of highly limited application.	677.169	1
Secondary Mathematics Click on the title above to go to the textbook adoption website. It is updated with information on the secondary textbook adoption process during the 2015-2016 school year. Connected Mathematics Project (Grades 6-8) The Connected Mathematics Project (CMP) was funded by the National Science Foundation between 1991 and 1997 with the goal of developing a mathematics program for grades 6, 7, and 8. The result was Connected Mathematics, a complete mathematics program that helps students develop understanding of important concepts, skills, procedures, and ways of thinking and reasoning in number, geometry, measurement, algebra, probability, and statistics. CMP is published by Pearson Prentice Hall. Prentice Hall Mathematics: Algebra1, Geometry, Algebra2 (Grades 9-12) Prentice Hall Algebra 1, Geometry, Algebra 2 teaches for understanding by incorporating an interwoven strand of thinking and reasoning into problem solving. Students learn in different ways and at different paces. Unique, built-in resources differentiate instruction to support all levels of learners in becoming successful problem solvers. Differentiating instruction helps all students develop conceptual understanding, fosters mathematical reasoning, and refines problem-solving strategies. Math Pathways Video Resources Lessons That Count is an engaging new Channel 10 series of supplementary video lessons for Algebra 1, Algebra 2, College Algebra and Pre-Calculus. Imagined and created by our own Fort Collins High School math teacher Sara Slagle and her Math Honor Society students, these wonderful, short lessons can be used in class or at home. They are fun, based on the math standards, and teach important math concepts.	677.169	1
CHAPTER ONE Number Systems and Errors Introduction It simply provides an introduction of numerical analysis. Number Representation and Base of Numbers Here we consider methods for representing numbers on computers. 1. Normalized Floating-point Representation It describes how the numbers are stored in the computers. CHAPTER 1 NUMBER SYSTEMS AND ERRORS 1 . Human Error It causes when we use inaccurate measurement of data or inaccurate representation of mathematical constants. 2. Truncation Error It causes when we are forced to use mathematical techniques which give approximate, rather than exact answer. 3. Round-off Error This type of errors are associated with the limited number of digits numbers in the computers. This preview has intentionally blurred sections. Sign up to view the full version. CHAPTER 1 NUMBER SYSTEMS AND ERRORS Effect of Round-off Errors in Arithmetic Operation Here we analysing the different ways to understand the nature of rounding errors. 1. Rounding off Errors in Addition and Subtraction It describes how addition and subtraction of numbers are performed in a computer. 2. Rounding off Errors in Multiplication It describes how multiplication of numbers are performed in a computer. CHAPTER 1 NUMBER SYSTEMS AND ERRORS 3. Rounding off Errors in Division It describes how division of numbers are performed in a computer. 4. Rounding off Errors in Powers and roots It describes how the powers and roots of numbers are performed in a computer. This preview has intentionally blurred sections. Sign up to view the full version. CHAPTER TWO Solution of Nonlinear Equations Introduction Here we discuss the ways of representing the different types of nonlinear equation f(x) = 0 and how to find approximation of its real root . Simple Root's Numerical Methods Here we discuss how to find the approximation of the simple root (non-repeating) of the nonlinear equation f(x) = 0. CHAPTER 2 SOLUTI ON OF NONLI NEAR EQUATI ONS 1. Method of Bisection This is simple and slow convergence method (but convergence is guaranteed) and is based on the Intermediate Value Theorem. Its strategy is to bisect the interval from one endpoint of the interval to the other endpoint and then retain the half interval whose end still bracket the root. 2. False Position Method This is slow convergence method and may be thought of as an attempt to improve the convergence characteristic of bisection method. Its also known as the method of linear interpolation. This preview has intentionally blurred sections. Sign up to view the full version. CHAPTER 2 SOLUTI ON OF NONLI NEAR EQUATI ONS 3. Fixed-Point Method This is very general method for finding the root of nonlinear equation and provides us with a theoretical framework within which the convergence properties of subsequent methods can be evaluated. The basic idea of this method is convert the equation f(x) = 0 into an equivalent form x = g(x). 4. Newtons Method This is the end of the preview. Sign up to access the rest of the document. This note was uploaded on 11/12/2010 for the course MATH 267 taught by Professor Chandrasekhar during the Spring '10 term at Anna University.	677.169	1
Evaluating Expressions PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 1.23 MB \| 4 pages PRODUCT DESCRIPTION Evaluating Expressions. There are 3 levels of proficiency: Novice, Intermediate and Advanced. Each level has 8 problems for a total of 24 problems. You can use this resource to differentiate instruction. In my class, I assigned the both the Novice and Intermediate sections to everyone. I made the Advanced section "bonus problems." This resource can be used as independent practice, homework and/or extra credit	677.169	1
Matrix Calculator This calculator performs all matrix, vector operations. You can add, subtract, find length, find dot and cross product, check if vectors are dependant. For every operation, calculator will generate a detailed explanation	677.169	1
Resource Added! Type: Lesson Plan Description: 3 Appropriate Pedagogy: 2 Reviewer Comments: This is chapters 2-16 of the FHSST free mathematics text from South Africa. The topics in these chapters cover 10th grade mathematics, from rational numbers to probability. Chapter 2 reviews earlier mathematics, including the laws of arithmetic, numbers, and sets. Chapter 3, Rational Numbers, is quite satisfactory. Chapter 4, Exponentials, runs through all the laws of exponents in a perfunctory way. The important law of the zero power is given to explanation at all.Chapter 5, Estimating Surds, contains material generally left out of American texts. Indeed 'surds', meaning irrational roots, is a term not used in American texts. Chapter 6, Estimating Irrational Numbers, is a useful exercise. Chapter 7, Number Patterns, contains material of genuine mathematical interest and is exemplary. Chapter 8, Finance, brings interesting and useful topics generally ignored in American texts. Chapter 9, Factorising, gives lots of examples but consistently confuses expressions and equations. Chapter 10, Equations and Inequalities, is huge and strange. In an American Algebra II course, this chapter would fill a semester. Apparently two or more chapters were merged here. ABHISHEK BHAVSAR useless full fake nothing is there related to logarithm in this book nor basic nor laws Table of Contents Mathematics, Grades 10-12, BasicsChapter 1, Introduction to Book	677.169	1
Product Description: Kaufmann and Schwitters have built this text's reputation on clear and concise exposition, numerous examples, and plentiful problem sets. This traditional text consistently reinforces the following common thread: learn a skill; practice the skill to help solve equations; and then apply what you have learned to solve application problems. This simple, straightforward approach has helped many students grasp and apply fundamental problem solving skills necessary for future mathematics courses. Algebraic ideas are developed in a logical sequence, and in an easy-to-read manner, without excessive vocabulary and formalism. The open and uncluttered design helps keep students focused on the concepts while minimizing distractions. Problems and examples reference a broad range of topics, as well as career areas such as electronics, mechanics, and health, showing students that mathematics is part of everyday life. The text's resource package--anchored by Enhanced WebAssign, an online homework management tool--saves instructors time while also providing additional help and skill-building practice for students outside of class. REVIEWS for Algebra for College Students	677.169	1
9tyh Grade Honors/Gifted Geometry Information Sheet This class is a course for those students who are pursuing a college -prep diploma. Failure in this class could affect graduation. This class will require daily study and consistent effort in all areas. If you discover you're having problems with the material, please see me immediately to set up a time for before-school help. MATERIALS: loose-leaf notebook pencils a scientific calculator of your own grid paper ruler compass Be sure to have all materials in class every day. You will be assigned a book to keep at home and a book to use in the classroom. It is your responsibility to keep each book in good condition. CLASSROOM EXPECTATIONS: 1. Be seated and ready to work when the bell rings. If you are not seated, you will be counted as tardy. 2. Chewing gum, eating, and drinking in class are not permitted. 3. Keep daily assignments neat and orderly in your notebook. Place the date and assignment on each page. 4. Personal notes and work done for other classes during this class will be confiscated and not returned to the student. ABSENCES/MAKEUP WORK: MAKEUP WORK IS YOUR RESPONSIBILITY!!!!!!!!! If you have an excused absence on the day of any graded activity, you must make up the work within three days. Failure to do so will result in a zero. Class assignments will be made daily; therefore any missed assignment should be completed within one-two days or return to school. If you need assistance with the assignments, you may make a before-school appointment. TARDY/ATTENDANCE Please see the attendance portion of your agenda book (page 12) for these policies. TERM PERCENTAGES Tests 40% Assignments 34% Quizzes 10% Exam 16% (8% midterm and 8% final exam) NOTE: ANYTHING YOU HAND IN FOR CREDIT MUST BE DONE IN PENCIL. YOU WILL NOT This preview has intentionally blurred sections. Sign up to view the full version. BE GIVEN CREDIT FOR WORK DONE IN INK. ERASABLE INK IS NOT AN ACCEPTABLE SUBSTITUTE. Tests: Tests will be given at the end of each chapter each chapter with at least a three day notice. Assignments: Assignments will be made daily. These will need to be completed before the next class. I do not accept late work!!!!! Quizzes: You should expect a quiz almost every day over the previous day's This is the end of the preview. Sign up to access the rest of the document. This note was uploaded on 01/24/2012 for the course HUM 9999 taught by Professor Variousprofessorslisted during the Summer '06 term at Abilene Christian University.	677.169	1
0132700ctory Mathematics Best-selling author Nigel Cook's new fourth edition of Introductory Mathematics is a complete prep-math book. Clear and easy-to-understand, this book employs an integrated approach, reinforcing all math topics to show the value of math as a tool, enabling readers to retain all information. Beginning with fractions and decimal numbers and proceeding on to build a solid foundation in exponents, the metric system, algebra, trigonometry, logarithms, graphs, and binary math, this easy-to-understand book will give readers the math skills needed for their careers. A valuable reference book for those in the technical trades, including electricians, automotive mechanics, printing employees, landscapers, surveyors, machinists, carpenters, food service employees, manufacturers, welders, plumbers, masons, drafters	677.169	1
Open Mathematics: Track 1 Featuring: VideoAudio The Severn Valley Railway is one of Britain's best known steam railways. Over much of its length, there is only a single track, with passing points at various points along the line. As with any commercial rail operation, the timetable needs to meet passenger needs and health and safety requirements. The five video tracks in this album follow the work of railway employees as they monitor and develop the service to suit passenger demand and ensure safety systems are in place. The material forms part of the course MU120 Open MathematicsGraphs are a common way of presenting information. However, like any other type of representation, graphs rely on shared understandings of symbols and styles to convey meaning. Also, graphs are normally drawn specifically with the intention of presenting information in a particularly favourable or unfavourable light, to convince you of an argument or to influence your decisions. This free course, Exploring distance time graphs, will enable you to explain, construct, use and interpret distance-time graphs. Communication is as vital in mathematics as in any language. This free course, Language, notation and formulas, will help you to express yourself clearly when writing and speaking about mathematics. You will also learn how to answer questions in the manner that is expected by the examiner. This free course looks at Babylonian mathematics. You will learn how a series of discoveries has enabled historians to decipher stone tablets and study the various techniques the Babylonians used for problem-solving and teaching. The Babylonian problem-solving skills have been described as remarkable and scribes of the time received a training far in advance of anything available in medieval Christian Europe 3000 years later. This free course, Modelling heat transfer, is the fourth in the series of five courses on mathematical modelling. In this course you will be taken through the whole modelling process in detail, from creating a first simple model, through evaluating it, to the subsequent revision of the model by changing one of the assumptions. The problem that will be examined is one based on heat transfer. The course assumes you have studied the courses Modelling pollution in the Great Lakes, Analysing skid marks and Developing modelling skills. Number systems and the rules for combining numbers can be daunting. This free course, Number systems, will help you to understand the detail of rational and real numbers, complex numbers and integers. You will also be introduced to modular arithmetic and the concept of a relation between elements of a set.	677.169	1
Introduction to Integration (Integral Calculus) A great way to start learning Calculus through video lectures and quizzes course on Integral Calculus. It comprises of a total of 5 hours of videos and quizzes. This is perfect for secondary school students seeking a good primer on Integral Calculus. It is also great as a refresher for everyone else. However prior knowledge in Differential Calculus is a MUST before learning this topic. The course is arranged from the very basic introduction and progresses swiftly with increasing depth and complexity on the subject. It is recommended that the students do not skip any part of the lectures, or jump back and forth, because good understanding of the fundamental is important as you progress. Quizzes are included on 7 subtopics to strengthen your understanding and fluency on this topic. So it is advisable that you attempt all the questions. The course is delivered by an experienced teacher with five years of experience teaching students on a one to one basis. The instructor understands the difficulties that students normally face to become competent in mathematics. So words and examples were carefully chosen to ensure that everybody gets the most out of this series of lectures. This is a MUST course for all secondary school students. Have fun learning! Who is the target audience? Secondary and High Schools students taking Calculus 1 (Integral Calculus) Students preparing for Calculus 1 (Integral Calculus) tests / exams at 'O' Level or its equivalence Any adults looking for a refresher in Basic Calculus 1 Students Who Viewed This Course Also Viewed What Will I Learn? Understand the relationship between Integral and Differential Calculus Provides the method to answer the questions posed in the previous quiz. Answers to Quiz on Area Under a Curve (cont) 07:21 +– Volume of Revolution 7 Lectures 01:07:21 This video describes how to estimate Volumes of Revolution by calculating the volumes of thin cylinders constructed inside the revolved area under a curve. Estimating Volumes of Revolution 12:36 This video describes how to estimate Volumes of Revolution by calculating the volumes of thin cylinders constructed inside the revolved area under a curve. This video is a sequence of the previous video. Estimating Volumes of Revolution (cont) 09:52 This video describes how to accurately calculate Volume of Revolution using Definite Integral. Determining Volumes of Revolution using Definite Integral 04:49 This video describes how to accurately calculate Volume of Revolution using Definite Integral. This video is a sequence to the previous video. Determining Volumes of Revolution using Definite Integral (cont) 03:13 This tests the students ability to use Definite Integral to solve problems involving Volume of Revolution. Volume of Revolution 4 questions Provides the method to answer the questions posed in the previous quiz. Answers to Quiz on Volume of Revolution 16:28 Provides the method to answer the questions posed in the previous quiz. A Penang based Engineer cum Tutor, a father of four, with a passion to teach and help students excel in Math. He is also known as MrMaths by his students. He was graduated with Bachelors Degree in Electronics Engineering from the University of Sheffield in England. He has fourteen years of working experience in electronics industry with Intel Corp and Dell Corp and years of Mathematics teaching and home tutoring experience. In the past he has also taught Physics. Today he is a full time Math tutor Since January 2012, he started tutoring outside of Penang and outside of Malaysia using Video Conferencing on the internet. Teaching in both Bahasa Melayu (Malay) and in English. He is friendly and always treat his students as if they are his own children.	677.169	1
Please contact your nearest Dymocks store to confirm availability Email store This book is available in following stores Provides less mathematically minded students with a gentle introduction to basic mathematics and some more advanced topics. Covering algebra, trigonometry, calculus and statistics, it manages to combine clarity of presentation with liveliness of style and sympathy for studentsGÇÖ needs. It is straightforward, pragmatic and packed full of illustrative examples, exercises and self-test questions. The essentials of formal mathematics are lucidly explained, with terms such as GÇÿintegralGÇÖ or GÇÿdifferential equationGÇÖ fully clarified. Provides a gentle introduction to basic mathematics and some more advanced topics	677.169	1
1 PRELUDES 1.1 Nature and Mathematics: Physics as Natural Philosophy 1.2 Contemporary Physics: Classical and Modern 1.3 Standards for Measurement 1.4 Units of Convenience and Unit Conversions 1.5 The Meaning of the Word Dimension 1.6 The Various Meanings of the Equal Sign 1.7 Estimation and Order of Magnitude 1.8 The Distinction Between Precision and Accuracy 2 A MATHEMATICAL TOOLBOX 2.1 Scalar and Vector Quantities 2.2 Multiplication of a Vector by a Scalar 2.3 Parallel Transport of Vectors 2.4 Vector Addition by Geometric Methods:Tail-to-Tip Method 2.5 Determining Whether a Quantity Is a Vector* 2.6 Vector Difference by Geometric Methods 2.7 The Scalar Product of Two Vectors 2.8 The Cartesian Coordinate System and the Cartesian Unit Vectors 2.9 The Cartesian Representation of Any Vector 2.10 Multiplication of a Vector Expressed in Cartesian Form by a Scalar 2.11 Expressing Vector Addition and Subtraction in Cartesian Form 2.12 The Scalar Product of Two Vectors Expressed in Cartesian Form 2.13 Determining the Angle Between Two Vectors Expressed in Cartesian Form 2.14 Equality of Two Vectors 2.15 Vector Equations 2.16 The Vector Product of Two Vectors 2.17 The Vector Product of Two Vectors Expressed in Cartesian Form 2.18 Variation of a Vector 2.19 Some Aspects of Vector Calculus 3 KINEMATICSⅠ 3.1 Rectilinear Motion 3.2 Position and Changes in Position 3.3 Average Speed and Average Velocity 3.4 Instantaneous Speed and Instantaneous Velocity 3.5 Average Acceleration 3.6 Instantaneous Acceleration 3.7 Rectilinear Motion with a Constant Acceleration 3.8 Geometric Interpretaions* 4 KINEMATICSⅡ 4.1 The Position, Velocity, and Acceleration Vectors in Two Dimensions 4.2 Two-Dimensional Motion with a Constant Acceleration 4.3 Motion in Three Dimensions 4.4 Relative Velocity Addition and Accelerations 4.5 Uniform Circular Motion: A First Look 4.6 The Angular Velocity Vector 4.7 The Geometry and Coordinates for Describing Circular Motion 4.8 The Position Vector for Circular Motion 4.9 The Velocity and Angular Velocity in Circular Motion 4.10 Uniform Circular Motion Revisited 4.11 Nonuniform Circular Motion and the Angular Acceleration 4.12 Nonuniform Circular Motion with a Constant Angular Acceleration 5 NEWTON#S LAWS OF MOTION 5.1 Fundamental Particles* 5.2 The Fundamental Forces of Nature* 5.3 Newton#s First Law of Motion and a Qualitative Conception of Force 5.4 The Concept of Force and Its Measurement 5.5 Newton#s Second Law of Motion 5.6 Newton#s Third Law of Motion 5.7 Limitations to Applying Newton#s Laws of Motion 5.8 Inertial Reference Frames: Do They Really Exist?* 5.9 Second Law and Third Law Force Diagrams 5.10 Weight and the Normal Force of a Surface 5.11 Tensions in Ropes, Strings, and Cables 5.12 Static Friction 5.13 Kinetic Friction at Low Speeds 5.14 Kinetic Friction Proportional to the Particle Speed* 5.15 Fundamental Forces and Other Forces Revisited* 5.16 Noninertial Reference Frames* 6 THE GRAVITATIONAL FORCE AND THE GRAVITATIONAL FIELD 6.1 How Did Newton Deduce the Gravitational Force Law? 6.2 Newton#s Law of Universal Gravitation 6.3 Gravitational Force of a Uniform Spherical Shell on a Particle 6.4 Gravitational Force of a Uniform Sphere on a Particle 6.5 Measuring the Mass of the Earth 6.6 Artificial Satellites of the Earth 6.7 Kepler#s First Law of Planetary Motion and the Geometry of Ellipses 6.8 Spatial Average Position of a Planet in an Elliptical Orbit* 6.9 Kepler#s Second Law of Planetary Motion 6.10 Central Forces, Orbital Angular Momentum, and Kepler#s Second Law* 6.11 Newton#s Form for Kepler#s Third Law of Planetary Motion 6.12 Customized Units 6.13 The Gravitational Field 6.14 The Flux of a Vector* 6.15 Gauss#s Law for the Gravitational Field* 7 HOOKE#S FORCE LAW AND SIMPLE HARMONIC OSCILLATION 7.1 Hooke#s Force Law 7.2 Simple Harmonic Oscillation 7.3 A Vertically Oriented Spring 7.4 Connection Between Simple Harmonic Oscillation and Uniform Circular Motion 7.5 How to Determine Whether an Oscillatory Motion Is Simple Harmonic Oscillation 7.6 The Simple Pendulum 7.7 Through a Fictional Earth in 42 Minutes 7.8 Damped Oscillations* 7.9 Forced Oscillations and Resonance* 8 WORK,ENERGY,AND THE CWE THEOREM 8.1 Motivation for Introducing the Concepts of Work and Energy 8.2 The Work Done by Any Force 8.3 The Work Done by a Constant Force 8.4 The Work Done by the Total Force 8.5 Geometric Interpretation of the Work Done by a Force 8.6 Conservative, Nonconservative, and Zero-Work Forces 8.7 Examples of Conservative, Nonconservative,and Zero-Work Forces 8.8 The Concept of Potential Energy 8.9 The Gravitational Potential Energy of a System near the Surface of the Earth 8.10 The General Form for the Gravitational Potential Energy 8.11 The Relationship Between the Local Form for the Gravitational Potential Energy and the More Genral Form* 8.12 The Potential Energy Function Associated with Hooke#s Force Law 8.13 The CWE Theorem 8.14 The Escape Speed 8.15 Black Holes* 8.16 Limitations of the CWE Theorem: Two Paradoxical Examples* 8.17 The Simple Harmonic Oscillator Revisited 8.18 The Average and Instantaneous Power of a Force 8.19 The Power of the Total Force Acting on a System 8.20 Motion Under the Influence of Conservative Forces Only: Energy Diagrams* 9 IMPULSE, MOMENTUM,AND COLLISIONS 9.1 Momentum and Newton#s Second Law of Motion 9.2 Impulse-Momentum Theorem 9.3 The Rocket:A System with Variable Mass* 9.4 Conservation of Momentum 9.5 Collisions 9.6 Disintegrations and Explosions 9.7 The Centripetal Acceleration Revisited* 9.8 An Alternative Way to Look at Force Transmission* 9.9 The Center of Mass 9.10 Dynamics of a System of Particles 9.11 Kinetic Energy of a System of Particles 9.12 The Velocity of the Center of Mass for Collisions* 9.13 The Center of Mass Reference Frame* 10 SPIN AND ORBITAL MOTION 10.1 The Distinction Between Spin and Orbital Motion 10.2 The Orbital Angular Momentum of a Particle 10.3 The Circular Orbital Motion of a Single Particle 10.4 Noncircular Orbital Motion 10.5 Rigid Bodies and Symmetry Axes 10.6 Spin Angular Momentum of a Rigid Body 10.7 The Time Rate of Change of the Spin Angular Momentum 10.8 The Moment of Inertia of Various Rigid Bodies 10.9 The Kinetic Energy of a Spinning System 10.10 Spin Distorts the Shape of the Earth* 10.11 The Precession of a Rapidly Spinning Top* 10.12 The Precession of the Spinning Earth* 10.13 Simultaneous Spin and Orbital Motion 10.14 Synchronous Rotation and the Parallel Axis Theorem 10.15 Rolling Motion Without Slipping 10.16 Wheels* 10.17 Total Angular Momentum and Torque 10.18 Conservation of Angular Momentum 10.19 Conditions for Static Equilibrium 11 SOLIDS AND FLUIDS 11.1 States of Matter 11.2 Stress,Strain, and Young#s Modulus for Solids 11.3 Fluid Pressure 11.4 Static Fluids 11.5 Pascal#s Principle 11.6 Archimedes# Principle 11.7 The Center of Buoyancy* 11.8 Surface Tension* 11.9 Capillary Action* 11.10 Fluid Dynamics: Ideal Fluids 11.11 Equation of Flow Continuity 11.12 Bernoulli#s Principle for Incompressible Ideal Fluids 11.13 Nonideal Fluids* 11.14 Viscous Flow* 12 WAVES 12.1 What Is a Wave? 12.2 Longitudinal and Transverse Waves 12.3 Wavefunctions,Waveforms,and Oscillations 12.4 Waves Propagating in One, Two, and Three Dimensions 12.5 One-Dimensional Waves Moving at Constant Velocity 12.6 The Classical Wave Equation for One-Dimensional Waves* 12.7 Periodic Waves 12.8 Sinusoidal (Harmonic) Waves 12.9 Waves on a String 12.10 Reflection and Transmission of Waves 12.11 Energy Transport via Mechanical Waves 12.12 Wave Intensity 12.13 What is a Sound Wave?* 12.14 Sound Intensity and Sound Level* 12.15 The Acoustic Doppler Effect* 12.16 Shock Waves* 12.17 Diffraction of Waves 12.18 The Principle of Superposition 12.19 Standing Waves 12.20 Wave Groups and Beats* 12.21 Fourier Analysis and the Uncertainty Principles* 13 TEMPERATURE,HEAT TRANSFER,AND THE FIRST LAW OF THERMODYNAMICS 13.1 Simple Thermodynamic Systems 13.2 Temperature 13.3 Work,Heat Transfer,Temperature,and Thermal Equilibrium 13.4 The Zeroth Law of Thermodynamics 13.5 Thermometers and Temperature Scales 13.6 Temperature Conversions Between the Fahrenheit and Celsius Scales* 13.7 Thermal Effects in Solids and Liquids: Size 13.8 Thermal Effects in Ideal Gases 13.9 Calorimetry 13.10 Reservoirs 13.11 Mechanisms for Heat Transfer* 13.12 Thermodynamic Processes 13.13 Energy Conservation:The First Law of Thermodynamics and the CWE Theorem 13.14 The Connection Between the CWE Theorem General Statement of Energy Conservation 13.15 Work Done by a System on Its Surroundings 13.16 Work Done by a Gas Taken Around a Cycle 3.17 Applying the First Law of Thermodynamics:Changes of State 14 KINETIC THEORY 14.1 Background for the Kinetic Theory of Gases 14.2 The Ideal Gas Approximation 14.3 The Pressure of an Ideal Gas 14.4 The Meaning of the Absolute Temperature 14.5 The Internal Energy of a Monatomic Ideal Gas 14.6 The Molar Specific Heats of an Ideal Gas 14.7 Complications Arise for Diatomic and Polyatomic Gases 14.8 Degrees of Freedom and the Equipartition of Energy Theorem 14.9 Specific Heat of a Solid* 14.10 Some Failures of Classical Kinetic Theory 14.11 Quantum Mechanical Effects* 14.12 An Adiabatic Process for an Ideal Gas 15 THE SECOND LAW OF THERMODYNAMICS 15.1 Why Do Some Things Happen,While Other Do Not? 15.2 Heat Engines and the Second Law of Thermodynamics 15.3 The Carnot Heat Engine and Its Efficiency 15.4 Absolute Zero and the Third Law of Thermodynamics 15.5 Refrigerator Engines and the Second Law of Thermodynamics 15.6 The Carnot Refrigerator Engine 15.7 The Efficiency of Real Heat Engines and Refrigerator Engines 15.8 A New Concept:Entropy 15.9 Entropy and the Second Law of Thermodynamics 15.10 The Direction of Heat Transfer:A Consequence of the Second Law 15.11 A Statistical Interpretaiton of the Entropy* 15.12 Entropy Maximization and the Arrow of Time* 15.13 Extensive and Intensive State Variables* 16 ELECTRIC CHARGES, ELECTRICAL FORCES,AND THE ELECTRIC FIELD 16.1 The Discovery of Electrification 16.2 Polarization and Induction 16.3 Coulomb#s Force Law for Pointlike Charges:The Quantification of Charge 16.4 Charge Quantization 16.5 The Electric Field of Static Charges 16.6 The Electric Field of Pointlike Charge Distributions 16.7 A Way to Visualize the Electric Field:Electric Field Lines 16.8 A Common Molecular Charge Distribution: The Electric Dipole 16.9 The Electric Field of Continuous Distributions of Charge 16.20 Motion of a Charged Particle in a Uniform Electric Field:An Electrical Projectile 16.11 Gauss#s Law for Electric Fields* 16.12 Calculating the Magnitude of the Electric Field Using Gauss#s Law* 16.13 Conductors* 16.14 Other Electrical Materials* 17 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 17.1 Electrical Potential Energy and the Electric Potential 17.2 The Electric Potential of a Pointlike Charge 17.3 The Electric Potential of a Collection of Pointlike Charges 17.4 The Electric Potential of Continuous Charge Distributions of Finite Size 17.5 Equipotential Volumes and Surfaces 17.6 The Relationship Between the Electric Potential and the Electric Field 17.7 Acceleration of Charged Particles Under the Influence of Electrical Forces 17.8 A New Energy Unit: The Electron-Volt 17.9 An Electric Dipole in an External Electric Field Revisited 17.10 The Electric Potential and Electric Field of a Dipole* 17.11 The Potential Energy of a Distribution of Pointlike Charges 17.12 Lightning Rods 18 CIRCUIT ELEMENTS,INDEPENDENT VOLTAGE SOURCES,AND CAPACITORS 18.1 Terminology, Notation, and Conventions 18.2 Circuit Elements 18.3 An Independent Voltage Source:A Source of Emf 18.4 Connections of Circuit Elements 18.5 Independent Voltage Sources in Series and Parallel 18.6 Capacitors 18.7 Series and Parallel Combinations of Capacitors 18.8 Energy Stored in a Capacitor 18.9 Electrostatics in Insulating Material Media* 18.10 Capacitors and Dielectrics* 18.11 Dielectric Breakdown* 19 ELECTRIC CURRENT, RESISTANCE, AND DC CIRCUIT ANALYSIS 19.1 The Concept of Electric Current 19.2 Electric Current 19.3 The Piece de Resistance:Resistance and Ohm#s Law 19.4 Resistance Thermometers 19.5 Characteristic Curves 19.6 Series and Parallel Connections Revisited 19.7 Resistors in Series and in Parallel 19.8 Electric Power 19.9 Electrical Neworks and Circuits 19.10 Electronics 19.11 Kirchhoff#s Laws for Circuit Analysis 19.12 Electric Shock Hazards* 19.13 A Model for a Real Battery 19.14 Maximum Power Transfer Theorem 19.15 Basic Eletronic Instruments:Voltmeters, Ammeters,and Ohmmeters 19.16 An Introduction to Transients in Circuits: A Series RC Circuit* 20 MAGNETIC FORCES AND THE MAGNETIC FIELD 20.1 The Magnetic Field 20.2 Applications 20.3 Magnetic Forces on Currents 20.4 Work Done by Magnetic Forces 20.5 Torque on a Current Loop in a Magnetic Field 20.6 The Biot-Savart Law 20.7 Forces of Parallel Currents on Each Other and the Definition of the Ampere 20.8 Gauss#s Law for the Magnetic Field* 20.9 Magnetic Poles and Current Loops 20.10 Ampere#s Law* 20.11 The Displacement Current and the Ampere-Maxwell Law* 20.12 Magnetic Materials* 20.13 The Magnetic Field of the Earth* 21 FARADAY#S LAW OF ELECTROMAGNETIC INDUCTION 21.1 Faraday#s Law of Electromagnetic Induction 21.2 Lenz#s Law 21.3 An ac Generator 21.4 Summary of the Maxwell Equations of Electromagnetism 21.5 Electromagnetic Waves* 21.6 Self-Inductance* 21.7 Series and Parallel Combinations of Inductors* 21.8 A Series LR Circuit* 21.9 Energy Stored in a Magnetic Field* 21.10 A Parallel LC Circuit* 21.11 Mutual Inductance* 21.12 An Ideal Transformer* 22 SINUSOIDAL AC CIRCUIT ANALYSIS 22.1 Representations of a Complex Variable 22.2 Arithmetic Operations with Complex Variables 22.3 Complex Potential Differences and Currents: Phasors 22.4 The Potential Difference and Current Phasors for Resistors,Inductors,and Capacitors 22.5 Series and Parallel Combinations of Impedances 22.6 Complex Independent ac Voltage Sources 22.7 Power Absorbed by Circuit Elements in ac Circuits 22.8 A Filter Circuit 22.9 A Series RLC Circuit 23 GEOMETRIC OPTICS 23.1 The Domains of Optics 23.2 The Inverse Square Law for Light 23.3 The Law of Reflection 23.4 The Law of Refraction 23.5 Total Internal Reflection 23.6 Dispersion 23.7 Rainbows* 23.8 Objects and Images 23.9 The Cartesian Sign Convention 23.10 Image Formation by Spherical and Plane Mirrors 23.11 Ray Diagrams for Mirrors 23.12 Refraction at a Single Spherical Surface 23.13 Thin Lenses 23.14 Ray Diagrams for Thin 23.15 Optical Instruments 24 PHYSICAL OPTICS 24.1 Existence of Light Waves 24.2 Interference 24.3 Young#s Double Slit Experiment 24.4 Single Slit Diffraction 24.5 Diffraction by a Circular Aperture 24.6 Resolution 24.7 The Double Slit Revisited 24.8 Multiple Slits: The Diffraction Grating 24.9 Resolution and Angular Disperaion of a Diffraction Grating 24.10 The Index of Refraction and the Speed of Light 24.11 Thin-Film Interference* 24.12 Polarized Light* 24.13 Polarization by Absorption* 24.14 Malus#s Law* 24.15 Polarization by Reflection:Brewster#s Law* 24.16 Polarization by Double Refraction* 24.17 Polarization by Scattering* 24.18 Rayleigh and Mie Scattering* 24.19 Optical Activity* 25 THE SPECIAL THEORY OF RELATIVITY 25.1 Reference Frames 25.2 Classical Galilean Relativity 25.3 The Need for Change and the Postulates of the Special Theory 25.4 Time Dilation 25.5 Lengths Perpendicular to the Direction of Motion 25.6 Lengths Oriented Along the Direction of Motion: Length Contraction 25.7 The Lorentz Transformation Equations 25.8 The Relativity of Simultaneity 25.9 A Relativistic Centipede 25.10 A Relativistic Paradox and Its Resolution* 25.11 Relativistic Velocity Addition 25.12 Cosmic Jets and the Optical Illusion of Superluminal Speeds* 25.13 The Longitudinal Doppler Effect 25.14 The Transverse Doppler Effect* 25.15 A General Equation for the Relativistic Doppler Effect* 25.16 Relativistic Momentum 25.17 The CWE Theorem Revisited 25.18 Implications of the Equivalence Between Mass and Energy 25.19 Space-Time Diagrams* 25.20 Electromgnetic Implications of the Special Theory* 25.21 The General Theory of Relativity* 26 AN APERITIF: MODERN PHYSICS 26.1 The Discovery of the Electron 26.2 The Discovery of X-rays 26.3 The Discovery of Radioactivity 26.4 The Appearance of Planck#s Constant h 26.5 The Photoelectric Effect 26.6 The Quest for an Atomic Model: Plum Pudding 26.7 The Bohr Model of a Hydrogenic Atom 26.8 The Bohr Correspondence Principle 26.9 A Bohr Model of the Solar System?* 26.10 Problems with the Bohr Model 26.11 Radioactivity Revisited 26.12 Carbon Dating 26.13 Radiation Units,Dose,and Exposure* 26.14 The Momentum of a Photon 26.15 The de Broglie Hypothesis 27 AN INTRODUCTION TO QUANTUM MECHANICS 27.1 The Heisenberg Uncertainty Principles 27.2 Implications of the Position-Momentum Uncertainty Principle 27.3 Implications of the Energy-Time Uncertainty Principle 27.4 Observation and Measurement 27.5 Particle-Waves and the Wavefunction 27.6 Operators* 27.7 The Schrodinger Equation*	677.169	1
... Show More methods (with examples), and gives a set of exercises and complete solutions for each method. The idea is that by studying the principles and applying them to the exercises, the reader will gain problem-solving ability as well as general mathematical insight. Eventually, the reader should be able to produce results that have "the whole air of intuition." Organized according to specific techniques in separate chapters, techniques include induction and the pigeonhole principle, among others. Arranged in order of increasing difficulty, the book presents a wide variety of problem sets designed to illustrate significant mathematical ideas. Each chapter also includes a moderate amount of the "theory" behind each problem-solving principle it presents. An essential resource for every student of mathematics and every professional who needs to solve mathematical	677.169	1
Math 105 Week 2 Learning Goals 1 Overview This week, we will be learning about partial derivatives of functions oi' two variabies. Those are analogs of the derivative of a function of a single variable, and many aspects of these will be familiar. However, Math 105 Week 6 Learning Goals 1 Overview This week will be about the Fundamental Theorem of Calculus that connects differential calculus with integration. We use this result to evaluate definite integrals of functions f that cannot be evaiuate with Riema Math 105 Week 8 Learning Goals 1 Overview Over the past few weeks, we have covered numerous techniques for evaluating definite and indefinite integrals in various forms, but these techniques cannot be used on every integral. In these cases we can use numeri Math 105 Week 12 Learning Goals 1 Overview Last week we started on innite series. In particular we learned about two tests of convergence and/or divergence: namely the divergence test and the integral test. This week we will learn a few more such tests de Math 105 Week 9 Learning Goals 1 Overview This week we give a very brief introduction to random variables and probability theory. Most observable phenomena have at least some element of randomness associated to what we observe and how we observe it. Exper Math 105 Week 11 Learning Goals 1 Overview In rst-semester calculus, you learned what it meant to talk about the limit of a function. We begin this week by discussing what it means to talk about the limit of an innite list of numbers (which we call an inn Math 105 Week 1 Learning Goals 1 Overview In this introductory week, we will (iefine scalar functions of several variables, and discuss their geometric interprc-itatious. This is a subset of the material contained in sections 12.1 and 12.2 of the textbook. Chapter 7: Vectors and the Geometry of Space Section 7.6 Surfaces in Space Written by Richard Gill Associate Professor of Mathematics Tidewater Community College, Norfolk Campus, Norfolk, VA With Assistance from a VCCS LearningWare Grant In this lesson we Multiple choice (continued) 3. Consider a differential equation dy/dt = f (y). Shown in A-D is the phase line (state space) diagram (f (y) versus y). Which of the following is the correct pairing of these sketches with the sketch of a solution y(t) to the Multiple choice Each multiple choice question is worth 2 pts. No partial points will be given for work shown. Enter your answers using the bubbles on the front page. 1. According to the graphs in the figure below, which of the following is true? (a) f (x) A. Multiple choice questions Enter your choice for each multiple choice question in the box at the bottom of the page. There are two pages at the end of the exam that can be used for rough work. No partial marks will be given for this section. A.1 Let f ( Math 105 Week 13 Learning Goals 1 Overview We continue our study of power series, focusing on the relation between known functions and their power series representations. 2 Learning Objectives These should be considered a minimum, rather than a comprehens	677.169	1
Video Info Properties of Real Numbers \| Algebra 1: Module 5: Topic 2 In this program, students learn about the properties of real numbers, which can be real numbers, variables, or algebraic expressions and they define how the things we call numbers should behave	677.169	1
About this book This book, written at the level of a first course in calculus and linear algebra, offers a lucid and concise explanation of mathematical wavelets. Evolving from ten years of classroom use, its accessible presentation is designed for undergraduates in a variety of disciplines (computer science, engineering, mathematics, mathematical sciences) as well as for practising professionals in these areas. This unique text starts the first chapter with a description of the key features and applications of wavelets, focusing on Haar's wavelets but using only high school mathematics. The next two chapters introduce one-, two-, and three-dimensional wavelets, with only the occasional use of matrix algebra. The second part of this book provides the foundations of least squares approximation, the discrete Fourier transform, and Fourier series. The third part explains the Fourier transform and then demonstrates how to apply basic Fourier analysis to designing and analyzing mathematical wavelets. The book explains in a nice way the nature and computation of mathematical wavelets, which provide a framework and methods for the analysis and synthesis of signals, images, and other arrays of data. A useful text for engineers, financiers, scientists, and students looking for explanation of wavelets. a "Journal of Information and Optimization Sciences "Giving practice first and theory later, the author avoids discouraging readers whose main subject is not mathematics. The book is written in a very comprehensible and lively style. The text is essentially self-contained since many of the facts employed from analysis, linear algebra and functional analysis are stated and partially proved in the book."	677.169	1
Introduction to Analysis (Pearson New International Edition) Description For one- or two-semester junior or senior level courses in Advanced Calculus, Analysis I, or Real Analysis. This text prepares students for future courses that use analytic ideas, such as real and complex analysis, partial and ordinary differential equations, numerical analysis, fluid mechanics, and differential geometry. This book is designed to challenge advanced students while encouraging and helping weaker students. Offering readability, practicality and flexibility, Wade presents fundamental theorems and ideas from a practical viewpoint, showing students the motivation behind the mathematics and enabling them to construct their own proofs.	677.169	1
Alg 1 -- Operations with Rational Expressions -- Scavenger Hunt Compressed Zip File Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files. 0.64 MB PRODUCT DESCRIPTION I use relays to review a lot of different subjects in my Algebra 1 classes. This one reviews Chapter 11 of the Prentice Hall CA Algebra 1 textbook: Rational Expressions -- simplifying, adding, subtracting, multiplying, and dividing simplifying rational expressions, solving rational equations, and review standards from previous chapters. Common Core Standards: Rewrite rational expressions. A-APR 7. Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Previous California Standards for Algebra 1: 12.0 -- Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. 13.0 -- Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques	677.169	1
...operational research and general IT skills, this course will help you to find a suitable career. Course overview The course is structured to allow students... Aprende sobre: Problem Solving, Skills and Training, GCSE Mathematics... Ver más ...and measurement and management of portfolio risks, enabling you to seek employment in financial institutions, including regulator bodies. You will be trained in programming... Aprende sobre: IT risk, Risk Management, Financial Training... Ver más ...To introduce students to the concept of subspace in a concrete situation. To provide a foundation for the study of linear problems both within mathematics and in other subjects... Aprende sobre: Skills and Training, Basic IT training, Basic IT... Ver más ...for a range of career options as well as combining this with fundamental and applied physics. Along with problem-solving skills and computational training... Aprende sobre: GCSE Mathematics, GCSE Physics, Problem Solving... Ver más ...enabling you to specialise in either pure or applied mathematics, or a combination of these topics. Apart from engaging in study of essential mathematical theory and technique... Aprende sobre: Problem Solving, GCSE Mathematics, Financial Training... Ver más ...Many of our students, as part of their everyday work, are involved in data analysis, the interpretation of statistics, the optimal design and control of systems... Aprende sobre: Medical training... Ver más ...The programme is ideal if you are considering a career move into statistics, or if your work already involves aspects of data collection and exploration... Aprende sobre: Financial Modelling, Financial Training... Ver más ...If you have completed an industrial placement year , you will also study the module: Mathematics, Statistics and Operational Research Project... Aprende sobre: Communication Training, Problem Solving, GCSE Mathematics... Ver más ...the skills to apply that knowledge in the solution of problems. Become an effective communicator of mathematics who is able to employ ICT to enhance that communication... Aprende sobre: GCSE Mathematics, GCSE Mathematics... Ver más ...usually study both Pure and Applied Mathematics in the first year, with Statistics being a likely third component. Queen's offers several different degree... Aprende sobre: GCSE Mathematics, GCSE Physics, Quality Training... Ver más ...as well as financial mathematics, thus giving you the opportunity to increase your knowledge and abilities in these areas. Depending on your choices... Aprende sobre: Part Time, Financial Modelling, Financial Training... Ver más ...so you will need to spend 120 days in this capacity in order to achieve Qualified Teacher Status. These will take place in Year two and Year three... Aprende sobre: GCSE Mathematics, Learning Teaching... Ver más ...Mathematics has a key role to play in many aspects of modern business. This degree programme combines the development of mathematical concepts and advanced... Aprende sobre: Financial Training, Business School, GCSE Mathematics... Ver más ...Compulsory modules Further Mathematics: Part 1 of 3 Further Mathematics: Part 2 of 3 Further Mathematics: Part 3 of 3 Mathematics: Part 1 of 3 Mathematics... Aprende sobre: GCSE Mathematics, Part Time... Ver más	677.169	1
Differential Calculus Differential Calculus How would you like to follow in the footsteps of Euclid and Archimedes? Would you like to be able to determine precisely how fast Usain Bolt is accelerating exactly 2 seconds after the starting gun? Differential calculus deals with the study of the rates at which quantities change. It is one of the two principal areas of calculus (integration being the other). 3.Derivative applications The reason we study calculus, and the reason it was invented, is for its many uses in real-world problems. In particular, derivatives let us optimize functions and study their rates of change. Here we cover optimization, rates of change, L'Hopital's rule, the mean value theorem, and more!	677.169	1
Complex Numbers (Part-1) An introductory guide for mastering basic concepts of Complex Numbers quizzes in the course help the students to test their understanding. The videos can be completed in an hour. This course of Complex numbers (part 1) is sure to make understand your concepts easily and this will serve as a great foundation for other advanced courses in complex numbers which you may want to take up in the future. Good Luck! Who is the target audience? The lectures in this course will be helpful to K-11 and K-12 grade students, who want an introduction to the topic,"Complex numbers". Other than K-11 and K-12 grade students, this course will help the students of high school and who are studying Algebra-2 or those who are doing an advance course in Mathematics and want to review the topic complex numbers. This course is a bonus for people who have keen interest in the subject of Mathematics and also for the people who have great passion for the subject! I am a certified (post graduate trained) mathematics instructor (private tutor) with over 24 years of teaching experience to middle school, high school, senior secondary level and intermediate level covering various school boards including CBSE, ICSE, ISC, IGCSE, 2-year IB Diploma (International Baccalaureate) covering AL,SL and HL courses. I have so far made 200+ videos in various Math topics.	677.169	1
PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 0.18 MB \| 2 pages PRODUCT DESCRIPTION This worksheet is to be used with algebra tiles and reinforces the idea that an equation must stay equal when you manipulate/simplify terms or isolate the variables. It's designed to align with 7.EE.A.1, 7.EE.A.2, and 7.EE.A.3. One example of how to model an equation using algebra tiles is included and then there are seven other blank problems	677.169	1
Math 300: Mathematical Computing Math 300 Syllabus Welcome to Math 300 - Mathematical Computation. The goal of this course is to make you more sophisticated in your knowledge of computing in mathematics. Anyone can use a browser and a word processor, but mathematicians and teachers need an array of more specialized techniques to do and communicate mathematics in myriad formats. Mathematicians need unique powerful tools to analyze their problems, and use multiple platforms for those ends. To that end, we try to familiarize you with some of the most common aspects of operating systems, networking, typesetting, and applications that mathematicians use. Instructor: Kevin Cooper Office: Neill 322 Office Hours: 3:00-5:00 MWF: these will ordinarily be in room 120 Phone: 5-4771 Email: kcooper@math.wsu.edu Tests: There will be two tests worth a total of 200 points. In addition there will be two quizzes worth 50 points each. Assignments: There will be several assignments worth 300-400 points. These will typically involve solving a problem and writing about the solution, and then typesetting that writing in some way. Several of the assignments in this course include substantial writing components. You will be graded on writing as well as computational understanding. Thus, technical proficiency alone will not suffice to do well in the class. Text: This is it. There are some HTML text pages available at this site, as well as somewhat more complete notes in portable document format. There are other resources available on the Web to which we provide links. Academic Integrity: Because much of the work in this class is done electronically, some students find it too tempting to copy the work of others. While we encourage collaboration and helpfulness among students, ultimately students must demonstrate that they have learned something by turning in their own work. Assignments or exams that show clear evidence of plagiarism (copying) will receive scores of zero, or in egregious cases might lead to a failing grade in the class. This can apply regardless of whether the student in question was the one copying, or the one copied. Protect your own work. Topics: Working Remotely HTML, CSS, MathML LaTeχ - document formatting Matlab and Python programming Students with disabilities Reasonable accommodations are available for students with a documented disability. If you have a disability and need accommodations to fully participate in this class, please either visit or call the Access Center (Washington Building 217; 509-335-3417) to schedule an appointment with an Access Advisor. All accommodations MUST be approved through the Access Center. The second exam will take place on Friday, 16 December at 8:00 AM. It will be written as a 50 minute test, but you may have the whole two hours for it. As always, all paper notes will be permissible, but no electronic devices may be used. There is a sample exam available.	677.169	1
Translating Sentences Into Equations PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 2.3 MB \| 12 pages PRODUCT DESCRIPTION Enjoy this NO PREP mini unit that includes translating sentences into equations. It has notes and examples that your students can refer back to when they need help. This product includes: • A table of contents • Notes that show how to translate sentences into equations • Examples of how to translate sentences into equations	677.169	1
Precalculus PRECALCULUS, Fourth Edition focuses on teaching the essentials of what a student needs to fulfill their precalculus requirement and to fully prepare them to succeed in calculus. It provides students with an integrated review of algebra andMore... PRECALCULUS, Fourth Edition focuses on teaching the essentials of what a student needs to fulfill their precalculus requirement and to fully prepare them to succeed in calculus. It provides students with an integrated review of algebra and trigonometry while focusing on essential calculus concepts. Faires and DeFranza prepare students for calculus by providing a solid grounding in analysis and graphing, tools necessary to make a successful transition to calculus. This streamlined text provides all the mathematics that students need--it doesn't bog them down in review, or overwhelm them with too much, too soon. The authors are careful to keep this book, unlike many of the precalculus books on the market, at a length that can be covered in one term. J. Douglas Faires is a Emeritus Professor of Mathematics at Youngstown State University, where he received his undergraduate degree. His masters and doctoral degrees were awarded by the University of South Carolina. His mathematical interests include analysis, numerical analysis, mathematics history, and problems solving. Dr. Faires has won numerous awards, including the Outstanding College-University Teacher of Mathematics by the Ohio Section of MAA and five Distinguished Faculty awards from Youngstown State University, which also awarded him an Honorary Doctor of Science award in 2006. Faires served on the Council of Pi Mu Epsilon for nearly two decades, including a term as President, was the Co-Director of the American Mathematics Competitions AMC-10 and AMC-12 examinations for 8 years, and has been a long-term judge for the COMAP International Contest in Mathematical Modeling. He has authored or co-authored more than 20 books, including recent MAA publications to assist young students with mathematical problem solving	677.169	1
BUYER OFFERS: We are a retail store with set pricing and unfortunately we can't fulfil any requests to sell items for less than the listed price. Description: For a subject that is a challenge at all levels of education, this chart covers principles for basic algebra, intermediate algebra and college algebra courses. Topics covered include: set theory operations of real numbers algebraic terms steps for solving a first-degree equation with one variable steps for solving a first-degree inequality with one variable order of operations factoring special factoring hints rational expressions complex fractions synthetic division roots & radicals rational expressions in equations radical operations radical expressions in equations quadratic equations complex numbers	677.169	1
Modeling with Functions Unit one is about the descriptive, not the quantitative. This unit is about seeing relationships between variables in different ways, recognizing that there is a special kind of relationship we call a function, and understanding properties that are special to functions of all different types. The unit studies relationships between different variables and how to represent these in a variety of different ways - especially with graphs and tables (equations are not emphasized until Unit 2). We also introduce unit conversions repeatedly during the the unit so that students come to see converting units as foundational to understanding relationships between different quantities.	677.169	1
About this product Description Description This volume is the first to offer a comprehensive, research-based, multi-faceted look at issues in early algebra. In recent years, the National Council for Teachers of Mathematics has recommended that algebra become a strand flowing throughout the K-12 curriculum, and the 2003 RAND Mathematics Study Panel has recommended that algebra be the initial topical choice for focused and coordinated research and development [in K-12 mathematics]. This book provides a rationale for a stronger and more sustained approach to algebra in school, as well as concrete examples of how algebraic reasoning may be developed in the early grades. It is organized around three themes: * The Nature of Early Algebra * Students' Capacity for Algebraic Thinking * Issues of Implementation: Taking Early Algebra to the Classrooms. The contributors to this landmark volume have been at the forefront of an effort to integrate algebra into the existing early grades mathematics curriculum. They include scholars who have been developing the conceptual foundations for such changes as well as researchers and developers who have led empirical investigations in school settings. Algebra in the Early Grades aims to bridge the worlds of research, practice, design, and theory for educators, researchers, students, policy makers, and curriculum developers in mathematics education.	677.169	1
Calculus 1 vs. Physics Comparing the concepts and notations: position, velocity, acceleration K.K._TG Objective: To help Physics students find similarities and differences in the explanations of concepts already covered in Calculus Topic: Rate of Change I Inverse Functions and Logarithms Table 1 gives data from an experiment in which a bacteria culture started with 100 bacteria in a limited nutrient medium; the size of the bacteria population was recorded at hourly intervals. The number of bacteria N is a Sect 1.1: Four Ways to Represent a Function Functions arise whenever one quantity depends on another. For example, the human population of the world P depends on the time t. By definition, a function f is a rule that assigns to each element x in a set A e Exponential Functions The function f (x) = 2x is called an exponential function because the variable, x, is the exponent. It should not be confused with the power function g (x) = x2, in which the variable is the base. In general, an exponential function TNCC Department of Mathematics Expectations of Students enrolled in College Credit Math Courses In order for you to be successful in your mathematics courses, the faculty of the mathematics department has developed the following common expectations of all FUNCTIONS AND MODELS 1.1 Four Ways to Represent a Function In this section, we will learn about: The main types of functions that occur in calculus. EXAMPLE B The human population of the world P depends on the time t. The table gives estimates of the wor FUNCTIONS AND MODELS 1.3 New Functions from Old Functions TRANSLATIONS Lets first consider translations. If c is a positive number, then the graph of y = f(x) + c is just the graph of y = f(x) shifted upward a distance of c units. This is because each Attention! Test 4/Final Exam is comprehensive and covers Chapters 7,8,9,10,11 and 13. This RN practice test only covers Chapters 11 and 13. Be F' . sure to review all four of the practice tests and review materials for Units 1-4 to be prepared for Test 4/ MTH 164-001 Unit Test 3 Practice Test To receive full credit, show all steps in a clear, organized manner. Circle answers. Round answers to 2 decimal places where appropriate. All fractions and radicals must be in simplest form. The actual unit test will mum+m1osfis Q05w€()@j Unit Test 1 Practice Test To receive full credit, show all steps in a clear, organized manner. Circle answers. Round answers to 2 decimal places where appropriate. All fractions and radicals must be in simplest form. The actual unit t	677.169	1
Often calculus and mechanics are taught as separate subjects. It shouldn't be like that. Learning calculus without mechanics is incredibly boring. Learning mechanics without calculus is missing the point. This textbook integrates both subjects and highlights the profound connections between them. This is the deal. Give me 350 pages of your attention, and I'll teach you everything you need to know about functions, limits, derivatives, integrals, vectors, forces, and accelerations. This book is the only math book you'll need for the first semester of undergraduate studies in science. With concise, jargon-free lessons on topics in math and physics, each section covers one concept at the level required for a first-year university course. Anyone can pick up this book and become proficient in calculus and mechanics, regardless of their mathematical background. Based on course material used by the author at Yale University, this practical text addresses the widening gap found between the mathematics required for upper-level courses in the physical sciences and the knowledge of incoming students. This superb book offers students an excellent opportunity to strengthen their mathematical skills by solving various problems in differential calculus. By covering material in its simplest form, students can look forward to a smooth entry into any course in the physical sciences. A supplementary text for introductory courses in Calculus-Based Physics. Designed for students who plan to take or who are presently taking calculus-based physics courses. This book will develop necessary mathematical skills and help students gain the competence to use precalculus, calculus, vector algebra, vector calculus, and the statistical analysis of experimental data. Students taking intermediate physics, engineering, and other science courses will also find the book useful–and will be able to use the book as a mathematical resource for these intermediate level courses. The book emphasizes primarily the use of mathematical techniques and mathematical concepts in Physics and does not go into their rigorous developments. Vectors and tensors are among the most powerful problem-solving tools available, with applications ranging from mechanics and electromagnetics to general relativity. Understanding the nature and application of vectors and tensors is critically important to students of physics and engineering. Adopting the same approach used in his highly popular A Student's Guide to Maxwell's Equations, Fleisch explains vectors and tensors in plain language. Written for undergraduate and beginning graduate students, the book provides a thorough grounding in vectors and vector calculus before transitioning through contra and covariant components to tensors and their applications. Matrices and their algebra are reviewed on the book's supporting website, which also features interactive solutions to every problem in the text where students can work through a series of hints or choose to see the entire solution at once. Audio podcasts give students the opportunity to hear important concepts in the book explained by the author. Understanding Physics – Second edition is a comprehensive, yet compact, introductory physics textbook aimed at physics undergraduates and also at engineers and other scientists taking a general physics course. Written with today's students in mind, this text covers the core material required by an introductory course in a clear and refreshing way. A second colour is used throughout to enhance learning and understanding. Each topic is introduced from first principles so that the text is suitable for students without a prior background in physics. At the same time the book is designed to enable students to proceed easily to subsequent courses in physics and may be used to support such courses. Mathematical methods (in particular, calculus and vector analysis) are introduced within the text as the need arises and are presented in the context of the physical problems which they are used to analyse. Particular aims of the book are to demonstrate to students that the easiest, most concise and least ambiguous way to express and describe phenomena in physics is by using the language of mathematics and that, at this level, the total amount of mathematics required is neither large nor particularly demanding. 'Modern physics' topics (relativity and quantum mechanics) are introduced at an earlier stage than is usually found in introductory textbooks and are integrated with the more 'classical' material from which they have evolved. This book encourages students to develop an intuition for relativistic and quantum concepts at as early a stage as is practicable. The text takes a reflective approach towards the scientific method at all stages and, in keeping with the title of the text, emphasis is placed on understanding of, and insight into, the material presented.	677.169	1
Adventure Time: The Flip Side Mathematical Edition Product Description All Finn and Jake need to do is convince the Monkey Wizard to kidnap Painting Princess, shouldn't be too hard...right? Finn is suffering from Quest Deficiency and needs his pals to help him turn up a righteous quest that will fulfill all his adventuring needs! The back of a questing board might seem like the easiest answer, but our hero is about to find out that the easy way our an create a ton of trouble... It's a race ag This customizable Mathematical Practices Poster is designed on the poster and would interest those who like mathematical practices, mathematical practices poster, common core, mathematics, common core mathematics, middle school, and high school stuff. Origami Design Secrets: Mathematical Methods for an Ancient Art, Second Edition , New, Free Shipping The magnum opus of one of the world's leading origami artists, the second edition of Origami Design Secrets reveals the underlying concepts of origami and how to create original origami designs. Containing step-by-step instructions for 26 models, this book is not just an origami cookbook or list of instructions— it introduces the fundamental building blocks of origami, building up to ad What does your math course have to do with the latest TV shows or Hollywood movies? Plenty-if you're using the right text. Mathematical Ideas, Twelfth Edition brings the best of Hollywood into the classroom through descriptions of video clips from popular cinema and television. Well-known author John Hornsby's innovative approach is enhanced with great care in this revision, and refined to serve the needs of you and your instructor. Streamlined and updated, it offers a modernized design, new bubble pointers for Example annotations, and much more. It retains the consistent features, friendly writing style, clear examples, and exercise sets for which this text is known. One of the biggest issues college math instructors face is capturing and keeping student interest. Over the years, John Hornsby has refined a creative solution-- bringing the best of Hollywood into his mathematics classroom. Mathematical Ideas applies this same strategy of engaging students through video clips from popular cinema and television to the textbook. Alongside fresh data and tools, this Eleventh Edition uses up-to-the-minute images as well as old favorites of math being done in Hollywood. In addition, examples are clarified with additional annotations, chapter summaries are made more intuitive to aid review, and chapter tests now include specific section references, making it easier for students to refer back to topics that need more attention. With great care and effort, the... Less Description New Mathematical Mysteries: The Beauty and Magic of Numbers by Brand: Basic Books Product Description Why seemingly unrelated mathematical truths are connected in simple and beautiful equations continues to stump even mathematicians. This recreational math book takes the reader on a fantastic voyage into the world of natural numbers. From the earliest discoveries of the ancient Greeks to various fundamental characteristics of the natural number sequence, Clawson explains fascinating This customizable Maths Mathematical Equations Clock Version Two is designed on the clock and would interest those who like math, maths, mathematical, mathematics, arithmetic, mathematician, and teacher	677.169	1
Product Description: This new edition is written for students preparing for technical, engineering technology or scientific careers. It begins with thorough coverage of topics in precalculus, calculus and differential equations, with an emphasis on how they relate to technical applications. The student has the opportunity to solve problems, much as they will in their future career, through the text's extensive applications and integrated use of technology. The text and exercises are designed to help students learn mathematical concepts and skills. Once these skills have been learned, students are encouraged to use graphing calculators to solve difficult problems with greater ease. Hints, notes and cautions are found throughout the text provide problem-solving techniques. ALSO AVAILABLE Student Solutions Manual, ISBN: 0-8273-7417-8 INSTRUCTOR SUPPLEMENTS CALL CUSTOMER SUPPORT TO ORDER Instructor's Guide, ISBN: 0-8273-7416-X 21 Day Unconditional Guarantee REVIEWS for Technical Calculus with Analytic	677.169	1
tell me the syllabus of math cbse of class 9 Answers CBSE Class 9 Math Syllabus Number System And Number Sense Algebra Coordinate Geometry Lines Angles and Triangles Quadrilaterals and Area Of Parallelograms and Triangles Circles Area of Plane Figures and Solid Shapes Volume of solids Introduction to Trigonometry Introduction to Statistics and Probability	677.169	1
Core-Plus Mathematics Course 1 Core-Plus Mathematics, is a standards-based, four-year integrated series covering the same mathematics concepts students learn in the Algebra 1-Geometry-Algebra 2-Precalculus sequence. Concepts from algebra, geometry, probability, and statistics are integrated, and the mathematics is developed using context-centered investigations. Developed by the CORE-Plus Math Project at Western Michigan University with funding from the National Science Foundation (NSF), Core-Plus Mathematics is written for all students to be successful in mathematics.	677.169	1
A Transition to Abstract Mathematics, Second Edition: Learning Mathematical Thinking and Writing After mastering the art of the proof process, the reader may pursue two independent paths. The latter parts are purposefully designed to rest on the foundation of the first, and climb quickly into analysis or algebra. Maddox addresses fundamental principles in these two areas, so that readers can apply their mathematical thinking and writing skills to these new concepts. From this exposure, readers experience the beauty of the mathematical landscape and further develop their ability to work with abstract ideas. * Covers the full range of techniques used in proofs, including contrapositive, induction, and proof by contradiction * Explains identification of techniques and how they are applied in the specific problem * Illustrates how to read written proofs with many step by step examples * Includes 20% more exercises than the first edition that are integrated into the material instead of end of chapter * The Instructors Guide and Solutions Manual points out which exercises simply must be either assigned or at least discussed because they undergird later results	677.169	1
...In the third Year (5th & 6th semester), the student will drop one subject and study the remaining two subjects for Master Degree Programme... Learn about: Mathematical Statistics, Actuarial Mathematics Economics, Mathematics Series, Mathematical Statistics Statistics, Mathematics Algebra, Mathematics Series... ...the Colley method, and the Massey method, and how to adapt each ranking method to integrate momentum. What is the target audience? This course starts... Learn about: Colley method, Massey method, Math brackets... ... as a key life skill and will help form a basis for more advanced study. The course syllabus encourages students to increase their confidence, helping them develop a feel for numbers, patterns and relationships. The IGCSE Maths course places a strong emphasis on solving problems and interpreting results... Learn about: Applied Mathematical, Skills and Training, Confidence Training... ... will find in your own SAT test. If you have problems solving any of the questions then watch the video lesson to get an explanation. This course is ideal for any students who are sitting the SAT math exam... Learn about: Exam Prep, SAT Math Exam, Math Exam... ...... Learn about: Financial Training...	677.169	1
This book presents the fundamentals of probability theory, the mathematical science that deals with the laws of random phenomena. These laws play an extremely important role in physical and other fields of natural science, in technology and engineering, economics, linguistics and so forth. The material covered ranges over the following problems: the concept of probability, sequences of independent trials, Markov chains, random variables and distribution functions, numerical characteristics of random variables, the law of large numbers, characteristic functions, the classical limit theorem, the theory of infinitely divisible distribution laws, the theory of stochastic processes, and elements of the theory of queues. The theory of probability is presented as a mathematical discipline, however the examples given not only illustrate the general propositions of the theory but provide links with problems that occur in the natural sciences. This is a text for students of mathematical departments of colleges and universities. It will also be found of definite interest to specialists in a wide range of fields (physicists, engineers, economists, linguists and others) that the science of probability touches on.	677.169	1
Home Welcome to Socratica! Our channel is for everyone who wants to learn. We make high-quality educational videos for people of all ages. From science and math, to the arts and humanities - we're here to help. Our videos are clear, concise, and beautiful. Subscribe today, and join us at Socratica on a journey of lifelong learning. Abstract Algebra deals with groups, rings, fields, and modules. These are abstract structures which appear in many different branches of mathematics, including geometry, number theory, topology, and more. They even appear in scientific topics such as quantum mechanics. Our Python Tutorials will help you learn Python quickly and thoroughly. We start with "Hello World" and then move on to data structures (sets, lists, tuples and dictionaries). Next we'll cover classes, and give a variety of in-depth examples and applications.	677.169	1
Algebra 1 End of the Year Project Word Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 0.02 MB \| 3 pages PRODUCT DESCRIPTION This is a great project idea to connect topics in Algebra 1 with the real world. In the past, I have used this project as something to be presented after End of Course Exams. The students will take an Algebra 1 topic of their choice and relate to something in the real world. I have had students come back with projects on the following: Reading Music (functions with scales); Catering (conversions with recipes); Real Estate (multiple topics); Statistics of various activities or surveys (Graphs, data,and statistics; Agriculture (various topics) to name a few. Once the students get going on their project, they have fun with it and enjoy seeing what their peers have come up with. I have given this project as both an outside project to do on their own time and as an in class project where they have had a week to research, write their paper, and create a visual aid. The project helps bring the real world into the classroom and gives the students great research and presentation practice. The project does include a class grading rubric for the presentations where they grade each other and a teacher grading rubric for the paper and	677.169	1
60976195. Orion Tech TX, USA $6.79 FREE New: New 0060976195. Orion Tech TX, USA $6.79 FREE Used Very Good(1 Copy): Very Good Cozy Book Cellar MA, USA $6.882 Copies): Good Good condition. Wonder Book - Member ABAA/ILAB MD, USA $7.60 Good 5 1/2 x 8. In very good, clean condition. Explains the basic mathematical functions and provides guidance for understanding and solving them. Includes: estimation, quick-checking answers, converting metric measures, percentages, and how to master algebra using nine easy steps, and more. Adagio Books CO 1993 Copyright In Softcover Format, Math Magic: The Human Calculator Shows How to Master Everyday Math Problems In Seconds (1993 Copyright) LR21. Bookworksonline TN, USA $11.23 FREE Used Very Good(1 Copy): Very Good Book Owner's info half title page. SuzyQBooks UT, USA $11.30 FREE Used Very Good(1 Copy): Very Good Text in English (350 pages). Very good condition, tight binding, unmarked pages. Seeks to improve math education by introducing tricks for performing accurate mental calculations and fun alternate approaches to common math class difficulties. Students are motivated to learn by the chance to show off by quickly solving difficult-looking math problems. Included are ways to simplify problems, tricks for special instances, and explanations of common numerical systems like measurements, calendars, time zones, and prices. Bookshelf S-72. Keywords: Arithmetic; General; Mathematics; Non-Fiction; Reference; Study and teaching (Elementary) Louis Morin QC, CAN $13.28 FREE About the Book Scott Flansburg's heartfelt belief is that there are no "mathematical illiterates," just people who have not learned how to make math work for them. But millions of otherwise successful adults are afraid to balance their checkbooks and don't know how to figure interest on savings or credit. Millions of students dread their math classes and live in fear of the math section of the SAT. But, as Scott Flansburg demonstrates, anyone can put these phobias to rest and deal with essential everyday mathematical calculations with confidence. In "Math Magic, "Scott Flansburg shows the reader: How to master the basics; How addition, subtraction, multiplication, and division really work; How to simplify calculations through estimation; How to quick-check your answers; How measurements work and how to quickly convert metric measures to more familiar ones; How to figure tips and taxes and be sure you're getting the right change; How to figure percentages; and How to master algebra using the nine easy steps to algebra. "Math Magic" is for all of us who need and want to improve our understanding of math. Flansburg makes math what you may never have imagined it to be: easy and fun.	677.169	1
This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Math 4500/6500 Exam #1 This take-home exam covers the material from Chapter 1 through 4 of Cheney/Kincaid, from floating point numbers through Polynomial Interpolation. The exam is intended to be done with the aid of a computer program such as Mathematica Pick 5 of the following problems. 6500 students must include problem 4. All students must include a computer problem in their exam. You are encouraged to use the computer on as many problems as you like. It is ok to use Mathematica to do algebra for you, for instance. You are permitted to use You are not permitted to use Maple (or Mathematica or MATLAB) The internet (except for Mathematica help) Our book Other books Your notes Other peoples notes Your brain Other peoples brains... View Full Document This note was uploaded on 03/12/2012 for the course MATH 4500 taught by Professor Staff during the Spring '08 term at University of Georgia Athens.	677.169	1
reliminaries Tools for analysis Convergent sequences Continuous functions Differentiation Elementary functions as solutions of differential equations Integration: Two fundamental theorems Integration: Further topics Approximation by Taylor polynomials Sequences and series of functions The Euclidean space ${\mathbb R}^n$ Continuity, compactness, and connectedness Metric spaces Differentiating functions of several variables Local approximation of real-valued functions Approximating nonlinear mappings by linear mappings Images and inverses: The inverse function theorem The implicit function theorem and its applications Integrating functions of several variables Iterated integration and changes of variables Line and surface integrals Consequences of the field and positivity axioms Linear algebra Index	677.169	1
About A-level Further Mathematics Further mathematics, as the title implies, is more difficult and challenging. For those of you who are thinking about doing a degree in Mathematics, Engineering or Physics, then Further mathematics is certainly for you. Indeed, many of the top-ranked universities will not consider you unless you have Further mathematics as an entry qualification – and most probably with at least a B grade if not an A. In order to succeed in Further Mathematics you obviously have to both good at mathematics and willing to put in quite bit more hard work. Having said that, those who enjoy mathematics anyway tend to thrive with Further Mathematics. Given the above, we do not recommend that just anyone takes the Further Mathematics A-level – you need to be getting a comfortable A grade in the plain vanilla Mathematics A-level to be able to cope with and excel in the Further mathematics. At AGF Tutoring we prefer the Pearson Edexcel International A-level (IAL) which consists of six modules: FP1 & FP2 or FP3 then any four of the remaining available modules except C12 and C34 (as you will have done these in the Mathematics A-level) Full specifications of the IAL in Further Mathematics can be found here. How often can I sit the IALs? Theoretically you can take them as often as you wish though most students prepare themselves properly so that they succeed on their first attempt. Most students sit the C12 (a combined paper) and perhaps one more paper (M1, S1 or D1) in one sitting and then six months later they sit the remaining 3 modules. There are many though who do not succeed on their first attempt, or they do not attain the grade that they desired or deserved and therefore elect to re-sit the specific examination. A prospective test taker ought to be careful, however, not to sit the same examination too on too many occasions as the universities may take this into account when assessing the candidate's application. How many hours per week of tutoring would be ideal? Generally, we recommend that a serious student does 2 hours per week. The tutorials can be organized so that the tutee has a whole week to prepare the relevant homework for the next relevant tutorial. This system maximizes the progress and performance. Important links Coming soon Practice materials – Recommended textbooks There are not many books that one can confidently call classics and which are just as valid and useful today as they were 40 years ago. Of such books, Bostock and Chandler's suite of books for A-level Mathematics remain our all-time favourites. They are also error-free which is very rare with any book, especially mathematics books. Therefore, for the serious student who wants to read around the various topics a little more, derive excellent explanations and examples as well as numerous practice questions we highly recommend the following by Bostock & Chandler in addition to those essential books endorsed by Edexcel for the specific modules for the respective examinations:	677.169	1
Schaum's Outline of Basic Business Mathematics helps beginning Business students learn the practical application of mathematical concepts used in the business world including stock market applications, appreciation rates, and averaging inventory controls. This book differs fromSchaum's Outline of Business Mathematics in that it focuses exclusively on business (rather than business and finance) and uses basic math in its applications. The book will review course fundamentals in easy-to-understand language with illustrative examples. The outline supplements business mathematics texts and is best suited to two-yearcollege business courses. Business Mathematics + Mymathlab With Pearson Etext Access Card By Clendenen Brand new - ordered 2nd by mistake for my daughter!FREE SHIPPING WITHIN USA Your	677.169	1
books.google.com complete key to The teachers assistant, or, A system of practical arithmetic	677.169	1
Factoring Packet and Answer Key PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 0.11 MB \| 14 pages PRODUCT DESCRIPTION This homework packet provides a 7-10 day guide for teaching 8th graders the beginning of how to do basic factoring and expanding, how to factor quadratic equations, how to factor quadratic equations when the coefficient a is not equal to 1, and difference of squares. This packet includes 6 pages of homework assignments that range from basic calculations to more involved problems. An answer key is included for all	677.169	1
This site is sponsored by a Swedish non-profit organization called Mattecentrum. It's an online community where students can study U.S. high school math courses for free through theory, examples, and video lessons. This site offers a free course on aviation technology and air traffic management. Through tutorials and interactive activities, students learn scientific concepts involved in RADAR, sound, and meteorology as it applies to flight, while applying principles of algebra, geometry and calculus. : Age Range: 11-18 The Monterey Institute for Technology and Education has made a fantastic resource available for free for individual high school and college students to further their education. When you get to the site, the subjects can be selected from the menu at the left..: Age Range: 11 and up (readiness for Algebra or Calculus) GCF Learn English : Learning English and want more practice? This program can help you improve your ability to read and understand English.	677.169	1
A Geometric Approach Product Description: This is an undergraduate textbook suitable for linear algebra courses. This is the only textbook that develops the linear algebra hand-in-hand with the geometry of linear (or affine) spaces in such a way that the understanding of each reinforces the other. The text is divided into two parts: Part I is on linear algebra and affine geometry, finishing with a chapter on transformation groups; Part II is on quadratic forms and their geometry (Euclidean geometry), including a chapter on finite subgroups of 0 (2). Each of the 23 chapters concludes with a generous helping of exercises, and a selection of these have solutions at the end of the book. The chapters also contain many examples, both numerical worked examples (mostly in 2 and 3 dimensions), as well as examples which take some of the ideas further. Many of the chapters contain "complements" which develop more special topics, and which can be omitted on a first reading. The structure of the book is designed to allow as much flexibility as possible in designing a course, either by omitting whole chapters or by omitting the "complements" or specific examples. REVIEWS for	677.169	1
Principles of Mathematics/Introduction/This book needs your help Anyone who has a good working knowledge of the subjects covered by this book can help. We aim to keep the material self-contained and rigorous, but accessible. It might help to think of mathematics as an exploration. Rather than a single directed, linear journey, we want to get a good working knowledge of the mathematical terrain. Each section sets out on a journey not far from somewhere we have already been, and it builds up principle by principle a map of the new terrain being surveyed until that new terrain becomes wholly familiar.	677.169	1
This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 4. MATRICES 170 4. Matrices 4.1. Definitions. Definition 4.1.1 . A matrix is a rectangular array of numbers. A matrix with m rows and n columns is said to have dimension m n and may be represented as follows: A = a 11 a 12 a 1 n a 21 a 22 a 2 n . . . . . . . . . . . . a m 1 a m 2 a mn = [ a ij ] Definition 4.1.2 . Matrices A and B are equal , A = B , if A and B have the same dimensions and each entry of A is equal to the corresponding entry of B . Discussion Matrices have many applications in discrete mathematics. You have probably encountered them in a precalculus course. We present the basic definitions associated with matrices and matrix operations here as well as a few additional operations with which you might not be familiar. We often use capital letters to represent matrices and enclose the array of numbers with brackets or parenthesis; e.g., A = a b c d or A = a b c d . We do not use simply straight lines in place of brackets when writing matrices because the notation a b c d has a special meaning in linear algebra. A = [ a ij ] is a shorthand notation often used when one wishes to specify how the elements are to be represented, where the first subscript i denotes the row number and the subscript j denotes the column number of the entry a ij . Thus, if one writes a 34 , one is referring to the element in the 3rd row and 4th column. This notation, however, does not indicate the dimensions of the matrix. Using this notation, we can say that two m n matrices A = [ a ij ] and B = [ b ij ] are equal if and only if a ij = b ij for all i and j . Example 4.1.1 . The following matrix is a 1 3 matrix with a 11 = 2 , a 12 = 3 , and a 13 =- 2 . h 2 3- 2 i 4. MATRICES 171 Example 4.1.2 . The following matrix is a 2 3 matrix. - 2 2 5 4.2. Matrix Arithmetic. Let be a scalar, A = [ a ij ] and B = [ b ij ] be m n matrices, and C = [ c ij ] a n p matrix. (1) Addition: A + B = [ a ij + b ij ] (2) Subtraction: A- B = [ a ij- b ij ] (3) Scalar Multiplication: A = [ a ij ] (4) Matrix Multiplication: AC = " n X k =1 a ik c kj # Discussion Matrices may be added, subtracted, and multiplied, provided their dimensions... View Full Document This note was uploaded on 11/27/2011 for the course MCH 108 taught by Professor Penelopekirby during the Fall '08 term at FSU.	677.169	1
Browse related Subjects ... Read More use various graphing calculators to solve problems more quickly. Perhaps most important-this book effectively prepares readers for further courses in mathematics. Read Less Fine. Hardcover. Instructor Edition: Same as student edition with additional notes or answers. SKU: 9780840068545-2-0-1 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780840068521Customer Reviews Algebra and Trigonometry with Analytic Geometry by Earl William Swokowski An ordinary text This is a common secondary level text. Like most current texts of this type, it fails to give the reader any understanding of the breadth of this subject as developed over the last 150 years. There are a number of useful tricks of interest to engineers, computer graphics specialists, and other 3D modelers that this text does not even point to for additional reading. However, for its intended audience, it delivers adequately but still does not entice the imagination much	677.169	1
Logic concepts are more mainstream than you may realize. There's logic every place you look and in almost everything you do, from deciding which shirt to buy to asking your boss for a raise, and even to watching television, where themes of such shows as CSI and Numbers incorporate a variety of logistical studies. Logic For Dummies explains... more... Improve your score on the Analytical Reasoning portion of the LSAT If you're like most test-takers, you find the infamous Analytical Reasoning or "Logic Games" section of the LSAT to be the most elusive and troublesome. Now there's help! LSAT Logic Games For Dummies takes the puzzlement out of the Analytical Reasoning section of the exam and shows... more... An easy-to-understand primer on advanced calculus topics Calculus II is a prerequisite for many popular college majors, including pre-med, engineering, and physics. Calculus II For Dummies offers expert instruction, advice, and tips to help second semester calculus students get a handle on the subject and ace their exams. It covers intermediate... more... The fun and easy way® to understand the basic concepts and problems of pre-algebra Whether you're a student preparing to take algebra or a parent who needs a handy reference to help kids study, this easy-to-understand guide has the tools you need to get in gear. From exponents, square roots, and absolute value to fractions, decimals, and percents,... more... Basic Math and Pre-Algebra Workbook For Dummies, 2nd Edition helps take the guesswork out of solving math equations and will have you unraveling the mystery of FOIL in no time. Whether you need to brush up on the basics of addition, subtraction, multiplication, and division or you're ready to tackle algebraic expressions and equations, this... more... The fun and friendly guide to really understanding math U Can: Basic Math & Pre-Algebra For Dummies is the fun, friendly guide to making sense of math. It walks you through the "how" and "why" to help you master the crucial operations that underpin every math class you'll ever take. With no-nonsense lessons, step-by-step instructions, practical... more... Basic Math & Pre-Algebra For Dummies, 2nd Edition (9781119293637) was previously published as Basic Math & Pre-Algebra For Dummies, 2nd Edition (9781118791981). While this version features a new Dummies cover and design, the content is the same as the prior release and should not be considered a new or updated product. Tips for simplifying... more...	677.169	1
Could we change the names to be more descritive or explain what the current names mean? i'll likely have to give some explaination anyway. But, if they had titles like "8th grade algebra" it woud be less confusing on the surface.	677.169	1
This ebook is available for the following devices: iPad Windows Mac Sony Reader Cool-er Reader Nook Kobo Reader iRiver Story This book surveys the work of the Second International Congress on Mathematical Education, and presents it as a picture of developing trends in mathematical education. At the end of August 1972 around 1400 people from seventy-three countries gathered for the Second International Congress on Mathematical Education in Exeter, UK. This book surveys the work of this conference, and presents it as a picture of developing trends in mathematical education. A number of themes emerged from the Congress. For example, there was great concern with the relationship between mathematics and the way in which the formation of mathematical concepts in affected by the use of language or the means in which children form the concepts from which mathematics can be drawn.	677.169	1
Algebra & Functions: Simplifying an expression by collecting like terms. The Laws of indices Expanding an expression Factorising an expression The laws of indices for all rational exponents The use and manipulation of Surds Rationalising the denominator of a fraction when it's a surd. Quadratic Functions: Plotting the graphs of Quadratic Functions Solving Quadratic equations using factorisation Completing the square Solving quadratic equations by completing the square Solving quadratic equations by using the formula Sketching graphs of quadratic formulae Equations and equalities: Solving simultaneous linear equations by elimination Solving simultaneous linear equations by substitution Using substitution when one equation is linear and the other is quadratic Solving linear equations Solving quadratic equations Sketching Curves: Sketching the graphs of cubic functions Interpreting graphs of cubic functions Sketching the reciprocal functions Using the intersection points of graphs of functions to solve equations The effect of transformation f(x+a) and f(x-a) The effect of transformations f(ax) and af(x) Performing transformations on the sketches of curves Co-ordinate Geometry in the (x,y) plane: The equation of a straight line in the form y=mx+c or ax+by+c=0 The gradient of a straight line The equation of the straight line of the form y=y1=m(x-x1) The formula for finding the equation of a straight line The conditions for two straight lines to be parallel or perpendicular Sequences and Series: The nth term of a sequence Sequence generated by a recurrence relationship Arithmetic sequences/series Using notation Differentiation: The derivative of f(x) as the gradient of the tangent to the graph y=f(x) Finding the gradient formula of simple functions The gradient formula for a function where the powers of x are real numbers Expanding or simplifying functions to make them easier to differentiate Finding second order derivative Finding the rate of change of a function at a particular point Finding the equation of a tangent and normal to a curve at a point	677.169	1
5 6 7 8 9 Is there a difference between knowing an answer and understanding a solution in mathematics? Adapted from Kenneth Goodmans The Psycholinguistic Nature of the Reading Process 10 The Difference Between Teaching and Learning 11 The Difference Between Teaching and Learning or How to Teach a Dog French 37 Encapsulation Example 4th Grade Question How many different ways can you arrange 3 objects with 2 objects, 2 at a time? 38 Encapsulation Example 4th Grade Question A. How many different ways can you arrange 3 objects with 2 objects, 2 at a time? or B. How many ways can you make a gym outfit if you have red, blue, or white tee shirts and red or white shorts? (remember the movie Clueless?) (question from Dr. John Sleisky, Authenticity and Test Items In Large Scale Assessment SSMA Centennial Conference, Downers Grove, IL, Nov. 1, 2001) 39 7.8 % of high school students who take ALGEBRA I get a college degree Predictors of College Success 40 23.1% of high school students who take GEOMETRY get a college degree 7.8 % of high school students who take ALGEBRA I get a college degree 41 39.5% of high school students who take42 6243 7444 Answers in the Tool Box, U.S. Department of Education by Clifford Adelman 79.8% of high school students who take CALCULUS get a college degree 74 45 Predictors of College Success Answers in the Tool Box, a study by U.S. Department of Education researcher Clifford Adelman, examined more than 20 variables-- including high school courses, educational aspirations, race, socioeconomic status (SES), on-time versus late high school graduation, and parenthood prior to age 22--to determine what really influenced the college completion rates of over 10,000 students. 46 Of all the high school indicators of academic preparation, the one that is the strongest is taking rigorous and intense courses in high school. Predictors of College Success 47 Of all the high school indicators of academic preparation, the one that is the strongest is taking rigorous and intense courses in high school. Taking rigorous and intense high school courses has a greater impact on African-American and Latino students than on white students. 48 49 Socioeconomic status had some impact (but it was minimal after the first year of college), and race did not have a statistically significant impact at all.	677.169	1
Solving Trigonometric Equations (B-8) PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 1.06 MB \| 30 pages PRODUCT DESCRIPTION This is the eighth in a series of major trigonometry topics, adapted from lessons that I wrote for homeschoolers. It covers the basics of what it means to solve a trigonometric equation as well as the methods of square roots, squaring, factoring (common factor, difference of two squares, trinomials), multiple angles, and finding approximate solutions versus exact solutions. There are many comments with each example building to the next idea. Step-by-step examples and problem solutions. Three to 4 lessons depending on available time. Download the preview for suggestions and complete content for	677.169	1
Description This three-volume set addresses the interplay between topology, functions, geometry, and algebra. Bringing the beauty and fun of mathematics to the classroom, the authors offer serious mathematics in a lively, reader-friendly style. Included are exercises and many figures illustrating the main concepts. It is suitable for advanced high-school students, graduate students, and researchers. The three-volume set includes A Mathematical Gift I, II, and III. Create a review Contents Part I: Invitation to topology (Viewing figures globally): Introduction The Euler characteristic Vortices created by winds and the Euler characteristic Curvature of a surface and the Euler characteristic The story of dimension: Introduction Learning to appreciate dimension What is dimension? Three-dimensional figures Physics and dimension Part II: The legacy of trigonometric functions: Introduction Trigonometric functions and infinite series Elliptic functions Intersection of geometry and algebra: Introduction The Poncelet closure theorem The Poncelet theorem for circles The Poncelet theorem in the world of complex numbers Proof of the Poncelet theorem using plane geometry Conclusion Part III: The story of the birth of manifolds The prelude to the birth of manifolds The birth of manifolds The story of area and volume from everyday notions to mathematical concepts Transition from the notion of ""size"" to the concept of ""area"" Scissors-congruent polygons Scissors-congruent polyhedra.	677.169	1
Find a North Richland Hills Trigon ...Topics usually covered in Algebra I include number systems, properties of real numbers, order of operations (PEMDAS), prime numbers and factoring, exponents, simplifying expressions, solving equations and inequalities of one variable, properties of linear equations and inequalities, solving syste...	677.169	1
Single Variable Calculus Concepts and Stewart's CALCULUS, Fifth Edition has the mathematical precision, accuracy, clarity of exposition and outstanding examples and problem sets that have characterized the first four editions. In this Fifth Edition, Stewart retains the focus on problemMore... Stewart's CALCULUS, Fifth Edition has the mathematical precision, accuracy, clarity of exposition and outstanding examples and problem sets that have characterized the first four editions. In this Fifth Edition, Stewart retains the focus on problem solving and the pedagogical system that has worked so well for students in a wide variety of colleges and universities throughout the world. He has made refinements to the exposition and examples, to ensure that students have the best materials available. Further support for students and instructors is now available through a vast array of supplementary material. Functions And Models Four Ways to Represent a Function Mathematical Models: A Catalog of Essential Functions New Functions from Old Functions Graphing Calculators and Computers Review Principles of Problem Solving Limits The Tangent and Velocity Problems The Limit of a Function Calculating Limits Using the Limit Laws The Precise Definition of a Limit Continuity Review Problems Plus Derivatives Derivatives and Rates of Change Writing Project: Early Methods for Finding Tangents The Derivative as a Function Differentiation Formulas Applied Project: Building a Better Roller Coaster Derivatives of Trigonometric Functions The Chain Rule Applied Project: Where Should a Pilot Start Descent? Implicit Differentiation	677.169	1
Record details Publisher:[Place of publication not identified] :[publisher not identified],2010. Content descriptions Summary, etc.: Passing grades in two years of algebra courses are required for high school graduation. Algebra II Essentials For Dummies covers key ideas from typical second-year Algebra coursework to help students get up to speed. Free of ramp-up material, Algebra II Essentials For Dummies sticks to the point, with content focused on key topics only. It provides discrete explanations of critical concepts taught in a typical Algebra II course, from polynomials, conics, and systems of equations to rational, exponential, and logarithmic functions. This guide is also a perfect reference for parents who need to review critical algebra concepts as they help...	677.169	1
This page requires that JavaScript be enabled in your browser. Learn how » Using Real-World Data in Your Class Carlo Barbieri Learn how both Mathematica and Wolfram\|Alpha can investigate extensive data about the world. This Wolfram Technologies for STEM Education: Virtual Conference for Education talk demonstrates how using real-world data in the class can be used effectively with Wolfram technologies. This presentation employs techniques of hierarchical distance clustering and machine learning in the Wolfram Language to solve a famous and vexing problem of authorship attribution in medieval history. Walk through the development of deep learning applications for the processing and analysis of images in the Wolfram Language. Real-world examples provide practical insights on how to effectively leverage neural ... Discover how to solve PDEs over regions or find eigenvalues and eigenfunctions over regions. Use the latest Wolfram Language functionality to create better PDE models and gain a deeper understanding ... Jon McLoone of Wolfram Technical Communication and Strategy delivers a fast-paced overview of some of the 500+ new functions and thousands of improvements in the latest version of the Wolfram Language ... See how easy it is to use the Wolfram Language to solve real-world statistics and probability problems with quantity data, enhanced time series support, and over 150 distributions, including random matrices.	677.169	1
Further Mathematics A level Course introduction If you have a talent for mathematics and would like to get more out of the subject, then Further Maths is your opportunity to excel. As a student you will complete two different A Levels in mathematics over two years. In the first year you will complete A Level Maths in one of our fast-track groups. Then in the second year you will complete A Level Further Maths As a result, you will have scaled the heights of Mathematics at further education level and will be well prepared to pursue a maths related subject at a top university. Why choose Brock for Further Mathematics A level? At Brock we have a team of highly experienced and dedicated teachers who specialise in teaching A Level Maths and Further Maths. You will learn in dedicated Further Maths groups in both the first and second years, studying alongside students with similar abilities. The benefit of this is that you can challenge yourself to progress faster. Furthermore, your lessons will take place in high quality maths teaching rooms located within our new state-of-the-art STEM building. Teaching & learning In addition to traditional classroom learning, we give you extra opportunities for support and enrichment. Indeed, every year Brock students take part in the regional Senior Team Maths Challenge competition, pitting themselves against teams from other local schools and colleges. We also facilitate trips further afield, including visits to central London and elsewhere for stimulating talks by leading researchers. Minimum entry requirements Five GCSEs at grade A-C or grade 4 to include: • Maths grade A/A (7) • At least 4 A/A* in other GCSEs What can I do with this qualification? Many of our most successful students follow this course. This is largely because an A Level in Further Mathematics is a highly regarded qualification that can support you in applying to top universities. Importantly, it provides a springboard to studying Maths and other highly mathematical subjects like Physics and Chemistry at degree level. It will also help you with many other degree subjects such as those related to Engineering and Computer Science. Other courses of interest I chose to study at Brock because of the vast number of opportunities that were available to me. I was able to challenge myself by doing five AS subjects and four A Level subjects, and I had all the support I needed to obtain good grades	677.169	1
Description These 22 lessons provide initial instruction or intervention on how to solve various types of linear equations and inequalities, including finding solutions for absolute value equations and inequalities. Since the study of mathematics revolves around solving equations, the focus of the first six lessons is to teach students how to solve linear equations of one variable by inspection, in one step, in two-steps, and using multi-steps. A3.1 Identifying Properties of Equality A3.2 Solving Equations by Inspection A3.3 Solving One-Step Linear Equations A3.4 Solving Two-Step Linear Equations A3.5 Solving Multi-Step Linear Equations A3.6 Rewriting Formulas After learning to translate sentences into algebraic equations with one unknown, students work on solving all types of real-world problems in the last three lessons. A4.1 Translating Sentences into Algebraic Equations A4.2 Solving Consumer/Business Problems Using Equations of One Variable A4.3 Solving Geometry Problems Using Equations of One Variable A4.4 Solving Mixture and Rate Problems Using Equations of One Variable The next seven lessons introduce students to linear inequalities of one variable, a mathematical concept with many consumer applications. An algebraic inequality has an infinite number of solutions. A5.1 Solving Linear Inequalities by Inspection A5.2 Solving One-Step Linear Inequalities A5.3 Solving Two-Step Linear Inequalities A5.4 Solving Multi-Step Linear Inequalities A5.5 Solving Conjunction Inequalities A5.6 Solving Disjunction Inequalities A5.7 Solving Problems Using Inequalities of One Variable Absolute value is used to describe distance without regard to direction. This is what makes absolute value equations and inequalities ideal for describing situations in which a "target range" exists. In the five lessons of this module, students solve various types of absolute value equations and inequalities. Modeling situations with expressions, equations, and inequalities is important for computer programmers. A6.1 Solving Basic Absolute Value Equations A6.2 Solving Advanced Absolute Value Equations A6.3 Solving Inequalities Using "Absolute Value Is Less Than" A6.4 Solving Inequalities Using "Absolute Value is Greater Than" A6.5 Solving Problems Using Absolute Value Equations and Inequalities Student print materials and Teachers' Notes are available for download at Download the free Elevated Math app to view two complimentary lessons or buy the lessons individually	677.169	1
An Introduction to Number Theory (MIT Press) Author:Harold M. Stark ISBN 13:9780262690607 ISBN 10:262690608 Edition:1st MIT Press paperback e Publisher:The MIT Press Publication Date:1978-05-30 Format:Paperback Pages:360 List Price:$48.00 &nbsp &nbsp The majority of students who take courses in number theory are mathematics majors who will not become number theorists. Many of them will, however, teach mathematics at the high school or junior college level, and this book is intended for those students learning to teach, In addition to a careful presentation of the standard material usually taught in a first course in elementary number theory, this book includes a chapter on quadratic fields which the author has designed to make students think about some of the "obvious" concepts they have taken for granted earlier. The book also includes a large number of exercises, many of which are nonstandard.	677.169	1
Services Funko Puts the Pop! in Pop Culture Whether you are in high school or college, taking the course for your first time or tackling higher-level math, this guide is an essential resource for reviewing this fundamental area of mathematics. This item is Non-Returnable. Details ISBN: 9781423220381 Publisher: BarCharts, Inc. Imprint: QuickStudy Date: Feb	677.169	1
Math 54H: Linear Algebra and Differential Equations Introduction In this course, we will learn some of the most basic concepts in linear algebra and differential equations, two different but related subjects. Many physical phenomena are governed by differential equations. For example, the rotation of the solar planets around the sun can be accurately described by a set of time-dependent differential equations. Linear algebra provides the basic tools necessary to solve these differential equations (most commonly) on a computer. Enrollment is entirely done online. So if you are waitlisted, check online often to see if you can enroll. If you have questions about enrollment, please immediately contact Thomas Brown of the Math Department at thomasbrown@math.berkeley.edu. 1A and 1B or equivalent. It is important to note that calculus courses at most institutions either have no differential equations, or less than Berkeley's Math 1B. Transfer students who have taken such a course need to learn on their own that differential equations material from There will be no handouts to be distributed in the class. All material is posted on the class home page at Announcements will be made through the class website, bspace, or email. There are a total of 100 points you can earn toward your final grade in the course. There will be two midterm exams. The better of the two is worth 25 points and the worse 15 points. The final exam is worth 30 points. All exams will be graded by the GSI and myself. In addition to exams, there will be up to 13 homeworks and 12 quizzes. Only the best 10 homeworks and best 10 quizzes will be counted towards the final grade, with each homework and quiz worth 1.5 points. Homework is in general due on Friday during discussion, except the week of Nov. 25, when it is due on the following Monday (Dec. 2.) The homework is a written assignment to be done individually, although group discussion is allowed. Quizzes are on the days homework is due, to be given at the begining of discussion sections for 15 minutes. But there will be no quiz on Nov. 29 or Dec. 2. Note that this means there will be homeworks and quizzes in the midterm exam weeks as well. We will have a review and provide a sample test before each exam. Since the GSI is limited in his work hours, he will only grade ONE problem of his choice in each homework. In addition, he will make most quiz problems to be similar to the problems in homework that is due on the day of the quiz. Doing all the homework problems is worth 1 point; and doing the graded problem correctly is worth additional 0.5 point. Your final letter grade will be determined based on your own performance. Below is a break-down of grade range A+: raw score > 95. A- to A: raw score between 87 and 95. B- to B+: raw score between 75 and 87. C- to C+: raw score between 65 and 75. D: raw score between 60 and 65. F: raw score < 60. I hope everyone will excel in this course. We will give no credit for homework turned in after the due date. The exams will be cumulative, and there will be no make-up exams or quizzes. However, you can skip or cancel one of the midterms (but not both) if you notify the instructor within 48 hours after the test. In this case, your other midterm and final will be worth 30 points and 40 points, respectively. Grades of Incomplete will be granted only for dire medical or personal emergencies that cause you to miss the final, and only if your work up to that point has been satisfactory. The University policy on academic integrity can be found at Any forms of cheating on homework, quizzes or exams will be actively investigated and reported to ensure honesty and fairness in the class.	677.169	1
2 Starting With Confidence Please start the questions on the dice card straight away. Write your answers on paper. Please do not write on the card. The answers are on the back. Use the thermometer sheet to record your result. Please put your Starting With Confidence work on your table now. I will want to look at it There will be a test on this work next lesson 2 3 AS level Maths Three modules in the first year make up the AS course – C1 (Core 1) – C2 (Core 2) – M1 (Mechanics 1) Three modules in the second year make up the A2 course – C3 (Core 3) – C4 (Core 4) – M2 (Mechanics 2) or D1 (Decision 1) 3 5 Survival Kit You must bring this with you every lesson. Look after it – keep it in a ring binder or hard backed folder. Your Survival Kit helps you organise your revision. You provide paper, pens, pencils, rulers, calculators. Organise your work. Dividers are very useful. 5 6 Lesson structure Before the lesson – Watch the videos – Complete the Survival Kit – Hand in assignment (first lesson of the week) During the lesson – Test on assignment (first lesson of the week) – Activities – Start assignment (second lesson of the week) After the lesson – Complete the assignment – every question – Assignments available on the VLE 6 7 Assignments You are expected to complete every question Each assignment should take you about three hours You will need support Facebook (BHASVIC MATHS) email m.macve@bhasvic.ac.uk textbook online support Sunday at 6:00 p.m. Survival Kit one-to-ones available on request Subject Extensions Drop In With The Doubles 7 8 Absence If you know that you are going to miss a lesson for any reason, you should tell me in advance in order to be set work so that you do not fall behind. This work must be completed before the next lesson that you attend. If you are unwell and you miss a lesson you should contact me or another student to find out what work you missed. You should catch up on missed work. 8	677.169	1
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Trigonomic ratios, ti 89 titanium 6th root, why is it that in GCF the variable have to be in common and the lowest exponent and In LCM it they do not have to be in common and it is the highest exponent, free online hard year 7 algebra lesson, matlab code richardson second order	677.169	1
Course Description: This course is designed for participants who need a refresher course in pre-calculus, or for new teachers who need insights, additional resources or a different perspective into the content taught in Pre-Calculus. Pre-Calculus: A Graphing Approach, 4th Edition. Demana, Waites, Clemens, Foley. Participants, however, may use any other Pre-Calculus text. Course videos that contain 21 hours of instruction are provided for the student. Also, four Practice Problem Sets are provided to each student on the course homepage in Moodle, the learning management system for all TeacherStep® courses. Graphing Calculator: Participants are expected to use a graphing calculator. While participants may choose any graphing calculator, instruction and the suggested text will use the TI-83 or equivalent. Knowledge and competence for use of other graphing calculators will be the sole responsibility of participants. Course Requirements: This course is delivered online via Moodle, including four videos to view online as well. Moodle provides the practice problems, a short quiz for each of the course sections and a cumulative Final Exam. Course Topics: Course topics include but not limited to: Functions, polynomials, asymptotes, graphing functions, angles, trig, and laws of sines and cosines. Registration: When you are ready to Register, please add the course to your Cart and follow the steps to Checkout to complete your enrollment. When your online registration and payment are received, you will be sent your course materials and access to the Course Online Classroom.	677.169	1

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