Loading ...
Global Do...
News & Politics
5
0
Try Now
Log In
Pricing
by Steven Holzner, PhD Differential Equations FOR DUMmIES ‰ Differential Equations FOR DUMmIES ‰ by Steven Holzner, PhD Differential Equations FOR DUMmIES ‰ Differential Equations For Dummies® Published by Wiley Publishing, Inc. 111 River St. Hoboken, NJ 07030-5774 www.wiley.com Copyright © 2008 by Wiley Publishing, Inc., Indianapolis, Indiana Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as per- mitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax 978-646-8600. Requests to the Publisher for permission should be addressed to the Legal Department, Wiley Publishing, Inc., 10475 Crosspoint Blvd., Indianapolis, IN 46256, 317-572-3447, fax 317-572-4355, or online at http://www.wiley.com/go/permissions. Trademarks: Wiley, the Wiley Publishing logo, For Dummies, the Dummies Man logo, A Reference for the Rest of Us!, The Dummies Way, Dummies Daily, The Fun and Easy Way, Dummies.com and related trade dress are trademarks or registered trademarks of John Wiley & Sons, Inc. and/or its affiliates in the United States and other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. Wiley Publishing, Inc., is not associated with any product or vendor mentioned in this book. LIMIT OF LIABILITY/DISCLAIMER OF WARRANTY: THE PUBLISHER AND THE AUTHOR MAKE NO REP- RESENTATIONS OR WARRANTIES WITH RESPECT TO THE ACCURACY OR COMPLETENESS OF THE CON- TENTS OF THIS WORK AND SPECIFICALLY DISCLAIM ALL WARRANTIES, INCLUDING WITHOUT LIMITATION WARRANTIES OF FITNESS FOR A PARTICULAR PURPOSE. NO WARRANTY MAY BE CRE- ATED OR EXTENDED BY SALES OR PROMOTIONAL MATERIALS. THE ADVICE AND STRATEGIES CON- TAINED HEREIN MAY NOT BE SUITABLE FOR EVERY SITUATION. THIS WORK IS SOLD WITH THE UNDERSTANDING THAT THE PUBLISHER IS NOT ENGAGED IN RENDERING LEGAL, ACCOUNTING, OR OTHER PROFESSIONAL SERVICES. IF PROFESSIONAL ASSISTANCE IS REQUIRED, THE SERVICES OF A COMPETENT PROFESSIONAL PERSON SHOULD BE SOUGHT. NEITHER THE PUBLISHER NOR THE AUTHOR SHALL BE LIABLE FOR DAMAGES ARISING HEREFROM. THE FACT THAT AN ORGANIZATION OR WEBSITE IS REFERRED TO IN THIS WORK AS A CITATION AND/OR A POTENTIAL SOURCE OF FUR- THER INFORMATION DOES NOT MEAN THAT THE AUTHOR OR THE PUBLISHER ENDORSES THE INFOR- MATION THE ORGANIZATION OR WEBSITE MAY PROVIDE OR RECOMMENDATIONS IT MAY MAKE. FURTHER, READERS SHOULD BE AWARE THAT INTERNET WEBSITES LISTED IN THIS WORK MAY HAVE CHANGED OR DISAPPEARED BETWEEN WHEN THIS WORK WAS WRITTEN AND WHEN IT IS READ. For general information on our other products and services, please contact our Customer Care Department within the U.S. at 800-762-2974, outside the U.S. at 317-572-3993, or fax 317-572-4002. For technical support, please visit www.wiley.com/techsupport. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic books. Library of Congress Control Number: 2008925781 ISBN: 978-0-470-17814-0 Manufactured in the United States of America 10 9 8 7 6 5 4 3 2 1 About the Author Steven Holzner is an award-winning author of science, math, and technical books. He got his training in differential equations at MIT and at Cornell University, where he got his PhD. He has been on the faculty at both MIT and Cornell University, and has written such bestsellers as Physics For Dummies and Physics Workbook For Dummies. Dedication To Nancy, always and forever. Author’s Acknowledgments The book you hold in your hands is the work of many people. I’d especially like to thank Tracy Boggier, Georgette Beatty, Jessica Smith, technical reviewer Jamie Song, PhD, and the folks in Composition Services who put the book together so beautifully. Publisher’s Acknowledgments We’re proud of this book; please send us your comments through our Dummies online registration form located at www.dummies.com/register/. Some of the people who helped bring this book to market include the following: Acquisitions, Editorial, and Media Development Project Editor: Georgette Beatty Acquisitions Editor: Tracy Boggier Copy Editor: Jessica Smith Editorial Program Coordinator: Erin Calligan Mooney Technical Editor: Jamie Song, PhD Editorial Manager: Michelle Hacker Editorial Assistants: Joe Niesen, Leeann Harney Cartoons: Rich Tennant (www.the5thwave.com) Composition Services Project Coordinator: Erin Smith Layout and Graphics: Carrie A. Cesavice, Stephanie D. Jumper Proofreaders: Caitie Kelly, Linda D. Morris Indexer: Broccoli Information Management Publishing and Editorial for Consumer Dummies Diane Graves Steele, Vice President and Publisher, Consumer Dummies Joyce Pepple, Acquisitions Director, Consumer Dummies Kristin A. Cocks, Product Development Director, Consumer Dummies Michael Spring, Vice President and Publisher, Travel Kelly Regan, Editorial Director, Travel Publishing for Technology Dummies Andy Cummings, Vice President and Publisher, Dummies Technology/General User Composition Services Gerry Fahey, Vice President of Production Services Debbie Stailey, Director of Composition Services Contents at a Glance Introduction .................................................................1 Part I: Focusing on First Order Differential Equations......5 Chapter 1: Welcome to the World of Differential Equations .........................................7 Chapter 2: Looking at Linear First Order Differential Equations................................23 Chapter 3: Sorting Out Separable First Order Differential Equations........................41 Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method........................................................................................................63 Part II: Surveying Second and Higher Order Differential Equations .................................................89 Chapter 5: Examining Second Order Linear Homogeneous Differential Equations....................................................................................................91 Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations..................................................................................................123 Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations ......................................................................................................................151 Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations..................................................................................................173 Part III: The Power Stuff: Advanced Techniques ..........189 Chapter 9: Getting Serious with Power Series and Ordinary Points........................191 Chapter 10: Powering through Singular Points ..........................................................213 Chapter 11: Working with Laplace Transforms ..........................................................239 Chapter 12: Tackling Systems of First Order Linear Differential Equations ...........265 Chapter 13: Discovering Three Fail-Proof Numerical Methods ................................293 Part IV: The Part of Tens ...........................................315 Chapter 14: Ten Super-Helpful Online Differential Equation Tutorials....................317 Chapter 15: Ten Really Cool Online Differential Equation Solving Tools ................321 Index .......................................................................325 Table of Contents Introduction..................................................................1 About This Book...............................................................................................1 Conventions Used in This Book .....................................................................1 What You’re Not to Read.................................................................................2 Foolish Assumptions .......................................................................................2 How This Book Is Organized...........................................................................2 Part I: Focusing on First Order Differential Equations.......................3 Part II: Surveying Second and Higher Order Differential Equations.........................................................................3 Part III: The Power Stuff: Advanced Techniques ................................3 Part IV: The Part of Tens........................................................................3 Icons Used in This Book..................................................................................4 Where to Go from Here....................................................................................4 Part I: Focusing on First Order Differential Equations ......5 Chapter 1: Welcome to the World of Differential Equations . . . . . . . . .7 The Essence of Differential Equations...........................................................8 Derivatives: The Foundation of Differential Equations .............................11 Derivatives that are constants............................................................11 Derivatives that are powers................................................................12 Derivatives involving trigonometry ...................................................12 Derivatives involving multiple functions ..........................................12 Seeing the Big Picture with Direction Fields...............................................13 Plotting a direction field ......................................................................13 Connecting slopes into an integral curve .........................................14 Recognizing the equilibrium value.....................................................16 Classifying Differential Equations ................................................................17 Classifying equations by order ...........................................................17 Classifying ordinary versus partial equations..................................17 Classifying linear versus nonlinear equations..................................18 Solving First Order Differential Equations ..................................................19 Tackling Second Order and Higher Order Differential Equations............20 Having Fun with Advanced Techniques ......................................................21 First Things First: The Basics of Solving Linear First Order Differential Equations ................................................................................24 Applying initial conditions from the start.........................................24 Stepping up to solving differential equations involving functions.........................................................25 Adding a couple of constants to the mix...........................................26 Solving Linear First Order Differential Equations with Integrating Factors ............................................................................26 Solving for an integrating factor.........................................................27 Using an integrating factor to solve a differential equation ...........28 Moving on up: Using integrating factors in differential equations with functions .................................................................29 Trying a special shortcut ....................................................................30 Solving an advanced example.............................................................32 Determining Whether a Solution for a Linear First Order Equation Exists ...........................................................................................35 Spelling out the existence and uniqueness theorem for linear differential equations ......................................................35 Finding the general solution ...............................................................36 Checking out some existence and uniqueness examples ...............37 Figuring Out Whether a Solution for a Nonlinear Differential Equation Exists.......................................................................38 The existence and uniqueness theorem for nonlinear differential equations......................................................39 A couple of nonlinear existence and uniqueness examples ...........39 Chapter 3: Sorting Out Separable First Order Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 Beginning with the Basics of Separable Differential Equations ...............42 Starting easy: Linear separable equations ........................................43 Introducing implicit solutions ............................................................43 Finding explicit solutions from implicit solutions ...........................45 Tough to crack: When you can’t find an explicit solution ..............48 A neat trick: Turning nonlinear separable equations into linear separable equations ..............................................................49 Trying Out Some Real World Separable Equations....................................52 Getting in control with a sample flow problem ................................52 Striking it rich with a sample monetary problem ............................55 Break It Up! Using Partial Fractions in Separable Equations....................59 Equations and Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .63 Exploring the Basics of Exact Differential Equations ................................63 Defining exact differential equations .................................................64 Working out a typical exact differential equation ............................65 Determining Whether a Differential Equation Is Exact..............................66 Checking out a useful theorem...........................................................66 Applying the theorem ..........................................................................67 Conquering Nonexact Differential Equations with Integrating Factors ............................................................................70 Finding an integrating factor...............................................................71 Using an integrating factor to get an exact equation.......................73 The finishing touch: Solving the exact equation ..............................74 Getting Numerical with Euler’s Method ......................................................75 Understanding the method .................................................................76 Checking the method’s accuracy on a computer.............................77 Delving into Difference Equations................................................................83 Some handy terminology ....................................................................84 Iterative solutions ................................................................................84 Equilibrium solutions ..........................................................................85 Part II: Surveying Second and Higher Order Differential Equations..................................................89 Chapter 5: Examining Second Order Linear Homogeneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .91 The Basics of Second Order Differential Equations...................................91 Linear equations...................................................................................92 Homogeneous equations.....................................................................93 Second Order Linear Homogeneous Equations with Constant Coefficients ........................................................................94 Elementary solutions ...........................................................................94 Initial conditions...................................................................................95 Checking Out Characteristic Equations ......................................................96 Real and distinct roots.........................................................................97 Complex roots.....................................................................................100 Identical real roots .............................................................................106 Getting a Second Solution by Reduction of Order ...................................109 Seeing how reduction of order works..............................................110 Trying out an example .......................................................................111 Linear independence .........................................................................115 The Wronskian....................................................................................117 Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .123 The General Solution of Second Order Linear Nonhomogeneous Equations ..................................................................124 Understanding an important theorem.............................................124 Putting the theorem to work.............................................................125 Finding Particular Solutions with the Method of Undetermined Coefficients......................................................................127 When g(x) is in the form of erx ..........................................................127 When g(x) is a polynomial of order n ..............................................128 When g(x) is a combination of sines and cosines ..........................131 When g(x) is a product of two different forms ...............................133 Breaking Down Equations with the Variation of Parameters Method ....135 Nailing down the basics of the method...........................................136 Solving a typical example..................................................................137 Applying the method to any linear equation..................................138 What a pair! The variation of parameters method meets the Wronskian......................................................................142 Bouncing Around with Springs ’n’ Things ................................................143 A mass without friction .....................................................................144 A mass with drag force ......................................................................148 Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .151 The Write Stuff: The Notation of Higher Order Differential Equations ..............................................................................152 Introducing the Basics of Higher Order Linear Homogeneous Equations.........................................................................153 The format, solutions, and initial conditions .................................153 A couple of cool theorems ................................................................155 Tackling Different Types of Higher Order Linear Homogeneous Equations.........................................................................156 Real and distinct roots.......................................................................156 Real and imaginary roots ..................................................................161 Complex roots.....................................................................................164 Duplicate roots ...................................................................................166 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .173 Mastering the Method of Undetermined Coefficients for Higher Order Equations.....................................................................174 When g(x) is in the form erx ...............................................................176 When g(x) is a polynomial of order n ..............................................179 When g(x) is a combination of sines and cosines ..........................182 Solving Higher Order Equations with Variation of Parameters..............185 The basics of the method..................................................................185 Working through an example............................................................186 Part III: The Power Stuff: Advanced Techniques...........189 Chapter 9: Getting Serious with Power Series and Ordinary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .191 Perusing the Basics of Power Series..........................................................191 Determining Whether a Power Series Converges with the Ratio Test ...................................................................................192 The fundamentals of the ratio test...................................................192 Plugging in some numbers ................................................................193 Shifting the Series Index..............................................................................195 Taking a Look at the Taylor Series .............................................................195 Solving Second Order Differential Equations with Power Series ...........196 When you already know the solution ..............................................198 When you don’t know the solution beforehand.............................204 A famous problem: Airy’s equation..................................................207 Chapter 10: Powering through Singular Points . . . . . . . . . . . . . . . . . .213 Pointing Out the Basics of Singular Points ...............................................213 Finding singular points ......................................................................214 The behavior of singular points .......................................................214 Regular versus irregular singular points.........................................215 Exploring Exciting Euler Equations ...........................................................219 Real and distinct roots.......................................................................220 Real and equal roots ..........................................................................222 Complex roots.....................................................................................223 Putting it all together with a theorem..............................................224 Figuring Series Solutions Near Regular Singular Points..........................225 Identifying the general solution........................................................225 The basics of solving equations near singular points ...................227 A numerical example of solving an equation near singular points........................................................................230 Taking a closer look at indicial equations.......................................235 Breaking Down a Typical Laplace Transform...........................................239 Deciding Whether a Laplace Transform Converges ................................240 Calculating Basic Laplace Transforms ......................................................241 The transform of 1..............................................................................242 The transform of eat............................................................................242 The transform of sin at ......................................................................242 Consulting a handy table for some relief ........................................244 Solving Differential Equations with Laplace Transforms........................245 A few theorems to send you on your way.......................................246 Solving a second order homogeneous equation ............................247 Solving a second order nonhomogeneous equation .....................251 Solving a higher order equation .......................................................255 Factoring Laplace Transforms and Convolution Integrals .....................258 Factoring a Laplace transform into fractions .................................258 Checking out convolution integrals .................................................259 Surveying Step Functions............................................................................261 Defining the step function .................................................................261 Figuring the Laplace transform of the step function .....................262 Chapter 12: Tackling Systems of First Order Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .265 Introducing the Basics of Matrices ............................................................266 Setting up a matrix .............................................................................266 Working through the algebra ............................................................267 Examining matrices............................................................................268 Mastering Matrix Operations......................................................................269 Equality................................................................................................269 Addition ...............................................................................................270 Subtraction..........................................................................................270 Multiplication of a matrix and a number.........................................270 Multiplication of two matrices..........................................................270 Multiplication of a matrix and a vector ...........................................271 Identity.................................................................................................272 The inverse of a matrix......................................................................272 Having Fun with Eigenvectors ’n’ Things..................................................278 Linear independence .........................................................................278 Eigenvalues and eigenvectors ..........................................................281 Solving Systems of First-Order Linear Homogeneous Differential Equations ..............................................................................283 Understanding the basics..................................................................284 Making your way through an example ............................................285 Solving Systems of First Order Linear Nonhomogeneous Equations .....288 Assuming the correct form of the particular solution...................289 Crunching the numbers.....................................................................290 Winding up your work .......................................................................292 Number Crunching with Euler’s Method ..................................................294 The fundamentals of the method .....................................................294 Using code to see the method in action..........................................295 Moving On Up with the Improved Euler’s Method ..................................299 Understanding the improvements ...................................................300 Coming up with new code.................................................................300 Plugging a steep slope into the new code.......................................304 Adding Even More Precision with the Runge-Kutta Method ..................308 The method’s recurrence relation....................................................308 Working with the method in code ....................................................309 Part IV: The Part of Tens............................................315 Chapter 14: Ten Super-Helpful Online Differential Equation Tutorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .317 AnalyzeMath.com’s Introduction to Differential Equations ...................317 Harvey Mudd College Mathematics Online Tutorial ...............................318 John Appleby’s Introduction to Differential Equations...........................318 Kardi Teknomo’s Page .................................................................................318 Martin J. Osborne’s Differential Equation Tutorial..................................318 Midnight Tutor’s Video Tutorial.................................................................319 The Ohio State University Physics Department’s Introduction to Differential Equations...................................................319 Paul’s Online Math Notes ............................................................................319 S.O.S. Math ....................................................................................................319 University of Surrey Tutorial ......................................................................320 Chapter 15: Ten Really Cool Online Differential Equation Solving Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .321 AnalyzeMath.com’s Runge-Kutta Method Applet ....................................321 Coolmath.com’s Graphing Calculator .......................................................321 Direction Field Plotter .................................................................................322 An Equation Solver from QuickMath Automatic Math Solutions...........322 First Order Differential Equation Solver....................................................322 GCalc Online Graphing Calculator .............................................................322 JavaView Ode Solver....................................................................................323 Math @ CowPi’s System Solver...................................................................323 A Matrix Inverter from QuickMath Automatic Math Solutions ..............323 Visual Differential Equation Solving Applet ..............................................323 Index........................................................................325 Introduction For too many people who study differential equations, their only exposure to this amazingly rich and rewarding field of mathematics is through a textbook that lands with an 800-page whump on their desk. And what follows is a weary struggle as the reader tries to scale the impenetrable fortress of the massive tome. Has no one ever thought to write a book on differential equations from the reader’s point of view? Yes indeed — that’s where this book comes in. About This Book Differential Equations For Dummies is all about differential equations from your point of view. I’ve watched many people struggle with differential equa- tions the standard way, and most of them share one common feeling: Confusion as to what they did to deserve such torture. This book is different; rather than being written from the professor’s point of view, it has been written from the reader’s point of view. This book was designed to be crammed full of the good stuff, and only the good stuff. No extra filler has been added; and that means the issues aren’t clouded. In this book, you discover ways that professors and instructors make solving prob- lems simple. You can leaf through this book as you like. In other words, it isn’t important that you read it from beginning to end. Like other For Dummies books, this one has been designed to let you skip around as much as possible — this is your book, and now differential equations are your oyster. Conventions Used in This Book Some books have a dozen confusing conventions that you need to know before you can even start reading. Not this one. Here are the few simple con- ventions that I include to help you navigate this book: Boldfaced text highlights important theorems, matrices (arrays of num- bers), keywords in bulleted lists, and actions to take in numbered steps. Monofont points out Web addresses. When this book was printed, some Web addresses may have needed to break across two lines of text. If that happens, rest assured that I haven’t put in any extra characters (such as hyphens) to indicate the break. So when using one of these Web addresses, type in exactly what you see in this book, pretending as though the line break doesn’t exist. What You’re Not to Read Throughout this book, I share bits of information that may be interesting to you but not crucial to your understanding of an aspect of differential equa- tions. You’ll see this information either placed in a sidebar (a shaded gray box) or marked with a Technical Stuff icon. I won’t be offended if you skip any of this text — really! oolish Assumptions This book assumes that you have no experience solving differential equations. Maybe you’re a college student freshly enrolled in a class on differential equa- tions, and you need a little extra help wrapping your brain around them. Or perhaps you’re a student studying physics, chemistry, biology, economics, or engineering, and you quickly need to get a handle on differential equations to better understand your subject area. Any study of differential equations takes as its starting point a knowledge of calculus. So I wrote this book with the assumption in mind that you know how to take basic derivatives and how to integrate. If you’re totally at sea with these tasks, pick up a copy of Calculus For Dummies by Mark Ryan (Wiley) before you pick up this book. How This Book Is Organized The world of differential equations is, well, big. And to handle it, I break that world down into different parts. Here are the various parts you see in this book. Differential Equations I start this book with first order differential equations — that is, differential equations that involve derivatives to the first power. You see how to work with linear first order differential equations (linear means that the derivatives aren’t squared, cubed, or anything like that). You also discover how to work with separable first order differential equations, which can be separated so that only terms in y appear on one side, and only terms in x (and constants) appear on the other. And, finally, in this part, you figure out how to handle exact differential equations. With this type of equation you try to find a func- tion whose partial derivatives correspond to the terms in a differential equa- tion (which makes solving the equation much easier). Part II: Surveying Second and Higher Order Differential Equations In this part, I take things to a whole new level as I show you how to deal with second order and higher order differential equations. I divide equations into two main types: linear homogeneous equations and linear nonhomogeneous equations. You also find out that a whole new array of dazzling techniques can be used here, such as the method of undetermined coefficients and the method of variation of parameters. Part III: The Power Stuff: Advanced Techniques Some differential equations are tougher than others, and in this part, I bring out the big guns. You see heavy-duty techniques like Laplace transforms and series solutions, and you start working with systems of differential equations. You also figure out how to use numerical methods to solve differential equa- tions. These are methods of last resort, but they rarely fail. Part IV: The Part of Tens You see the Part of Tens in all For Dummies books. This part is made up of fast-paced lists of ten items each; in this book, you find ten online differential equation tutorials and ten top online tools for solving differential equations. cons Used in This Book You can find several icons in the margins of this book, and here’s what they mean: This icon marks something to remember, such as a law of differential equa- tions or a particularly juicy equation. The text next to this icon is technical, insider stuff. You don’t have to read it if you don’t want to, but if you want to become a differential equations pro (and who doesn’t?), take a look. This icon alerts you to helpful hints in solving differential equations. If you’re looking for shortcuts, search for this icon. When you see this icon, watch out! It indicates something particularly tough to keep an eye out for. Where to Go from Here You’re ready to jump into Chapter 1. However, you don’t have to start there if you don’t want to; you can jump in anywhere you like — this book was writ- ten to allow you to do just that. But if you want to get the full story on differ- ential equations from the beginning, jump into Chapter 1 first — that’s where all the action starts. Focusing on First Order Differential Equations In this part . . . In this part, I welcome you to the world of differential equations and start you off easy with linear first order differential equations. With first order equations, you have first order derivatives that are raised to the first power, not squared or raised to any higher power. I also show you how to work with separable first order differen- tial equations, which are those equations that can be sep- arated so that terms in y appear on one side and terms in x (and constants) appear on the other. Finally, I introduce exact differential equations and Euler’s method. Welcome to the World of Differential Equations This Chapter Breaking into the basics of differential equations Getting the scoop on derivatives Checking out direction fields Putting differential equations into different categories Distinguishing among different orders of differential equations Surveying some advanced methods It’s a tense moment in the physics lab. The international team of high- powered physicists has attached a weight to a spring, and the weight is bouncing up and down. “What’s happening?” the physicists cry. “We have to understand this in terms of math! We need a formula to describe the motion of the weight!” You, the renowned Differential Equations Expert, enter the conversation calmly. “No problem,” you say. “I can derive a formula for you that will describe the motion you’re seeing. But it’s going to cost you.” The physicists look worried. “How much?” they ask, checking their grants and funding sources. You tell them. “Okay, anything,” they cry. “Just give us a formula.” You take out your clipboard and start writing. “What’s that?” one of the physicists asks, pointing at your calculations. math at lightning speed. “I’ve got it,” you announce. “Your formula is y = 10 sin (5t), where y is the weight’s vertical position, and t is time, measured in seconds.” “Wow,” the physicists cry, “all that just from solving a differential equation?” “Yep,” you say, “now pay up.” Well, you’re probably not a renowned differential equations expert — not yet, at least! But with the help of this book, you very well may become one. In this chapter, I give you the basics to get started with differential equations, such as derivatives, direction fields, and equation classifications. he Essence of Differential Equations In essence, differential equations involve derivatives, which specify how a quantity changes; by solving the differential equation, you get a formula for the quantity itself that doesn’t involve derivatives. Because derivatives are essential to differential equations, I take the time in the next section to get you up to speed on them. (If you’re already an expert on derivatives, feel free to skip the next section.) In this section, however, I take a look at a qualitative example, just to get things started in an easily digestible way. Say that you’re a long-time shopper at your local grocery store, and you’ve noticed prices have been increasing with time. Here’s the table you’ve been writing down, tracking the price of a jar of peanut butter: Month Price 1 $2.40 2 $2.50 3 $2.60 4 $2.70 5 $2.80 6 $2.90 peanut butter going to be a year from now? You know that the slope of a line is ∆y/∆x (that is, the change in y divided by the change in x). Here, you use the symbols ∆p for the change in price and ∆t for the change in time. So the slope of the line in Figure 1-1 is ∆p/∆t. Because the price of peanut butter is going up 10 cents every month, you know that the slope of the line in Figure 1-1 is: t p ∆ ∆ = 10¢/month The slope of a line is a constant, indicating its rate of change. The derivative of a quantity also gives its rate of change at any one point, so you can think of the derivative as the slope at a particular point. Because the rate of change of a line is constant, you can write: dt dp t p ∆ ∆ = = 10¢/month In this case, dp/dt is the derivative of the price of peanut butter with respect to time. (When you see the d symbol, you know it’s a derivative.) And so you get this differential equation: dt dp = 10¢/month 1 2 3 4 Time Price5 6 2.40 2.50 2.60 2.70 2.80 2.90 igure 1-1: e price of peanut butter by month. tion, and you can solve for price as a function of time like this: p = 10t + c In this equation, p is price (measured in cents), t is time (measured in months), and c is an arbitrary constant that you use to match the initial conditions of the problem. (You need a constant, c, because when you take the derivative of 10t + c, you just get 10, so you can’t tell whether there’s a constant that should be added to 10t — matching the initial conditions will tell you.) The missing link is the value of c, so just plug in the numbers you have for price and time to solve for it. For example, the cost of peanut butter in month 1 is $2.40, so you can solve for c by plugging in 1 for t and $2.40 for p (240 cents), giving you: 240 = 10 + c By solving this equation, you calculate that c = 230, so the solution to your differential equation is: p = 10t + 230 And that’s your solution — that’s the price of peanut butter by month. You started with a differential equation, which gave the rate of change in the price of peanut butter, and then you solved that differential equation to get the price as a function of time, p = 10t + 230. Want to see the solution to your differential equation in action? Go for it! Find out what the price of peanut butter is going to be in month 12. Now that you have your equation, it’s easy enough to figure out: p = 10t + 230 10(12) + 230 = 350 As you can see, in month 12, peanut butter is going to cost a steep $3.50, which you were able to figure out because you knew the rate at which the price was increasing. This is how any typical differential equation may work: You have a differential equation for the rate at which some quantity changes (in this case, price), and then you solve the differential equation to get another equation, which in this case related price to time. Note that when you substitute the solution (p = 10t + 230) into the differential equation, dp/dt indeed gives you 10 cents per month, as it should. Derivatives: The Foundation of Differential Equations As I mention in the previous section, a derivative simply specifies the rate at which a quantity changes. In math terms, the derivative of a function f(x), which is depicted as df(x)/dx, or more commonly in this book, as f'(x), indi- cates how f(x) is changing at any value of x. The function f(x) has to be con- tinuous at a particular point for the derivative to exist at that point. Take a closer look at this concept. The amount f(x) changes in a small distance along the x axis ∆x is: f(x + ∆x) – f(x) The rate at which f(x) changes over the change ∆x is: x f x x f x ∆ ∆ + - ^ ^ h h So far so good. Now to get the derivative dy/dx, where y = f(x), you must let ∆x get very small, approaching zero. You can do that with a limiting expres- sion, which you can evaluate as ∆x goes to zero. In this case, the limiting expression is: ∆ lim dx dy x f x x f x ∆ ∆ x 0 = + - " ^ ^ h h In other words, the derivative of f(x) is the amount f(x) changes in ∆x, divided by ∆x, as ∆x goes to zero. I take a look at some common derivatives in the following sections; you’ll see these derivatives throughout this book. Derivatives that are constants The first type of derivative you’ll encounter is when f(x) equals a constant, c. If f(x) = c, then f(x + ∆x) = c also, and f(x + ∆x) – f(x) = 0 (because all these amounts are actually the same), so df(x)/dx = 0. Therefore: f x c dx df x 0 = = ^ ^ h h So f(x + ∆x) – f(x) = c ∆x and (f(x + ∆x) – f(x))/∆x = c. Therefore: f x cx dx df x c = = ^ ^ h h Derivatives that are powers Another type of derivative that pops up is one that includes raising x to the power n. Derivatives with powers work like this: f x x dx df x n x n n 1 = = - ^ ^ h h Raising e to a certain power is always popular when working with differential equations (e is the natural logarithm base, e = 2.7128 . . ., and a is a constant): f x e dx df x a e ax ax = = ^ ^ h h And there’s also the inverse of ea, which is the natural log, which works like this: ln f x x dx df x x 1 = = ^ ^ ^ h h h Derivatives involving trigonometry Now for some trigonometry, starting with the derivative of sin(x): sin cos f x x dx df x x = = ^ ^ ^ ^ h h h h And here’s the derivative of cos(x): cos sin f x x dx df x x = = - ^ ^ ^ ^ h h h h Derivatives involving multiple functions The derivative of the sum (or difference) of two functions is equal to the sum (or difference) of the derivatives of the functions (that’s easy enough to remember!): f x a x b x dx df x dx d a x dx d b x ! ! = = ^ ^ ^ ^ ^ ^ h h h h h h tive of the first. For example: f x a x b x dx df x a x dx d b x b x dx d a x = = + ^ ^ ^ ^ ^ ^ ^ ^ h h h h h h h h How about the derivative of the quotient of two functions? That derivative is equal to the function in the denominator times the derivative of the function in the numerator, minus the function in the numerator times the derivative of the function in the denominator, all divided by the square of the function in the denominator: f x b x a x dx df x b x b x dx d a x a x dx d b x 2 = = - ^ ^ ^ ^ ^ ^ ^ ^ ^ h h h h h h h h h eeing the Big Picture with Direction Fields It’s all too easy to get caught in the math details of a differential equation, thereby losing any idea of the bigger picture. One useful tool for getting an overview of differential equations is a direction field, which I discuss in more detail in Chapter 2. Direction fields are great for getting a handle on differen- tial equations of the following form: , dx dy f x y = _ i The previous equation gives the slope of the equation y = f(x) at any point x. A direction field can help you visualize such an equation without actually having to solve for the solution. That field is a two-dimensional graph consisting of many, sometimes hundreds, of short line segments, showing the slope — that is, the value of the derivative — at multiple points. In the following sections, I walk you through the process of plotting and understanding direction fields. Plotting a direction field Here’s an example to give you an idea of what a direction field looks like. A body falling through air experiences this force: F = mg – γ v γ is the drag coefficient (which adds the effect of air friction and is measured in newtons sec/meter), and v is the speed of the object as it plummets through the air. If you’re familiar with physics, consider Newton’s second law. It says that F = ma, where F is the net force acting on an object, m is its mass, and a is its acceleration. But the object’s acceleration is also dv/dt, the derivative of the object’s speed with respect to time (that is, the rate of change of the object’s speed). Putting all this together gives you: F ma m dt dv mg v = = = - c Now you’re back in differential equation territory, with this differential equa- tion for speed as a function of time: dt dv g m v = - c Now you can get specific by plugging in some numbers. The acceleration due to gravity, g, is 9.8 meters/sec2 near the Earth’s surface, and let’s say that the drag coefficient is 1.0 newtons sec/meter and the object has a mass of 4.0 kilo- grams. Here’s what you’d get: . dt dv v 9 8 4 = - To get a handle on this equation without attempting to solve it, you can plot it as a direction field. To do so you create a two-dimensional plot and add dozens of short line segments that give the slope at those locations (you can do this by hand or with software). The direction field for this equation appears in Figure 1-2. As you can see in the figure, there are dozens of short lines in the graph, each of which give the slope of the solution at that point. The vertical axis is v, and the horizontal axis is t. Because the slope of the solution function at any one point doesn’t depend on t, the slopes along any horizontal line are the same. Connecting slopes into an integral curve You can get a visual handle on what’s happening with the solutions to a dif- ferential equation by looking at its direction field. How? All those slanted line segments give you the solutions of the differential equations — all you have to do is draw lines connecting the slopes. One such solution appears in Figure 1-3. A solution like the one in the figure is called an integral curve of the differential equation. 20 1 2 3 4 5 6 7 8 9 10 25 30 35 40 v t 45 50 igure 1-3: olution in direction field. 20 1 2 3 4 5 6 7 8 9 10 25 30 35 40 v t 45 igure 1-2: direction field. As you can see from Figure 1-3, there are many solutions to the equation that you’re trying to solve. As it happens, the actual solution to that differential equation is: v = 39.2 + ce–t/4 In the previous solution, c is an arbitrary constant that can take any value. That means there are an infinite number of solutions to the differential equation. But you don’t have to know that solution to determine what the solutions behave like. You can tell just by looking at the direction field that all solutions tend toward a particular value, called the equilibrium value. For instance, you can see from the direction field graph in Figure 1-3 that the equilibrium value is 39.2. You also can see that equilibrium value in Figure 1-4. 20 1 2 3 4 5 6 7 8 9 10 25 30 35 40 v t 45 50 igure 1-4: An quilibrium value in a direction field. Classifying Differential Equations Tons of differential equations exist in Math and Science Land, and the way you tackle them differs by type. As a result, there are several classifications that you can put differential equations into. I explain them in the following sections. Classifying equations by order The most common classification of differential equations is based on order. The order of a differential equation simply is the order of its highest deriva- tive. For example, check out the following, which is a first order differential equation: dx dy x 5 = Here’s an example of a second order differential equation: dx d y dx dy x 19 4 2 2 + = + And so on, up to order n: . . . dx d y dx d y dx d y dx dy x 9 16 14 12 19 4 0 n n n n 1 1 2 2 - + + + - + = - - As you might imagine, first order differential equations are usually the most easily managed, followed by second order equations, and so on. I discuss first order, second order, and higher order differential equations in a bit more detail later in this chapter. Classifying ordinary versus partial equations You can also classify differential equations as ordinary or partial. This classifi- cation depends on whether you have only ordinary derivatives involved or only partial derivatives. of an ordinary differential equation, relating the charge Q(t) in a circuit to the electromotive force E(t) (that is, the voltage source connected to the circuit): L dt d Q R dt dQ C Q E t 1 2 2 + + = ^ h Here, Q is the charge, L is the inductance of the circuit, C is the capacitance of the circuit, and E(t) is the electromotive force (voltage) applied to the cir- cuit. This is an ordinary differential equation because only ordinary deriva- tives appear. On the other hand, partial derivatives are taken with respect to only one vari- able, although the function depends on two or more. Here’s an example of a partial differential equation (note the squiggly d’s): , , x u x t t u x t α 2 2 2 2 2 2 2 = _ _ i i In this heat conduction equation, α is a physical constant of the system that you’re trying to track the heat flow of, and u(x, t) is the actual heat. Note that u(x, t) depends on both x and t and that both derivatives are partial derivatives — that is, the derivatives are taken with respect to one or the other of x or t, but not both. In this book, I focus on ordinary differential equations, because partial differ- ential equations are usually the subject of more advanced texts. Never fear though: I promise to get you your fair share of partial differential equations. Classifying linear versus nonlinear equations Another way that you can classify differential equations is as linear or non- linear. You call a differential equation linear if it exclusively involves linear terms (that is, terms to the power 1) of y, y', y", and beyond to y(n). For exam- ple, this equation is a linear differential equation: L dt d Q R dt dQ C Q E t 1 2 2 + + = ^ h clear: LQ" RQ C Q E t 1 + + = l ^ h On the other hand, nonlinear differential equations involve nonlinear terms in any of y, y', y", up to y(n). The following equation, which describes the angle of a pendulum, is a nonlinear differential equation that involves the term sin θ (not just θ): sin dt d L g θ θ 0 2 2 + = Handling nonlinear differential equations is generally more difficult than han- dling linear equations. After all, it’s often tough enough to solve linear differ- ential equations without messing things up by adding higher powers and other nonlinear terms. For that reason, you’ll often see scientists cheat when it comes to nonlinear equations. Usually they make an approximation that reduces the nonlinear equation to a linear one. For example, when it comes to pendulums, you can say that for small angles, sin θ ≈ θ. This means that the following equation is the standard form of the pendulum equation that you’ll find in physics textbooks: dt d L g θ θ 0 2 2 + = As you can see, this equation is a linear differential equation, and as such, it’s much more manageable. Yes, it’s a cheat to use only small angles so that sin θ ≈ θ, but unless you cheat like that, you’ll sometimes be reduced to using numerical calculations on a computer to solve nonlinear differential equa- tions; obviously these calculations work, but it’s much less satisfying than cracking the equation yourself (if you’re a math geek like me). olving First Order Differential Equations Chapters 2, 3, and 4 take a look at differential equations of the form f'(x) = f(x, y); these equations are known as first order differential equations because the derivative involved is of first order (for more on these types of equations, see the earlier section “Classifying equations by order.” tions in Chapters 2, 3, and 4. The following are some examples of what you can look forward to: As you know, first order differential equations look like this: f'(x) = f(x, y). In the upcoming chapters, I show you how to deal with the case where f(x, y) is linear in x — for example, f'(x) = 5x — and then nonlinear in x, as in f'(x) = 5x2. You find out how to work with separable equations, where you can factor out all the terms having to do with y on one side of the equation and all the terms having to do with x on the other. I also help you solve first order differential equations in cool ways, such as by finding integrating factors to make more difficult problems simple. Direction fields, which I discuss earlier in this chapter, work only for equa- tions of the type f'(x) = f(x, y) — that is, where only the first derivative is involved — because the first derivative of f(x) gives you the slope of f(x) at any point (and, of course, connecting the slope line segments is what direc- tion fields are all about). ackling Second Order and Higher Order Differential Equations As noted in the earlier section “Classifying equations by order,” second order differential equations involve only the second derivative, d2y/dx2, also known as y". In many physics situations, second order differential equations are where the action is. For example, you can handle physics situations such as masses on springs or the electrical oscillations of inductor-capacitor circuits with a differential equation like this: y" – ay = 0 In Part II, I show you how to tackle second order differential equations with a large arsenal of tools, such as the Wronskian matrix determinant, which will tell you if there are solutions to a second (or higher) order differential equation. Other tools I introduce you to include the method of undetermined coefficients and the method of variation of parameters. equations, which I also cover in Part II. With these high-end equations, you find terms like dny/dxn, where n > 2. The derivative dny/dxn is also written as y(n). Using the standard syntax, deriv- atives are written as y', y", y''', yiv, yv, and so on. In general, the nth derivative of y is written as y(n). Higher order differential equations can be tough; many of them don’t have solutions at all. But don’t worry, because to help you solve them I bring to bear the wisdom of more than 300 years of mathematicians. Having Fun with Advanced Techniques You discover dozens of tools in Part III of this book; all of these tools have been developed and proved powerful over the years. Laplace Transforms, Euler’s method, integrating factors, numerical methods — they’re all in this book. These tools are what this book is all about — applying the knowledge of hun- dreds of years of solving differential equations. As you may know, differential equations can be broken down by type, and there’s always a set of tools devel- oped that allows you to work with whatever type of equation you come up with. In this book, you’ll find a great many powerful tools that are just waiting to solve all of your differential equations — from the simplest to the seemingly impossible! Looking at Linear First Order Differential Equations This Chapter Beginning with the basics of solving linear first order differential equations Using integrating factors Determining whether solutions exist for linear and nonlinear equations As you find out in Chapter 1, a first order differential equation simply has a derivative of the first order. Here’s what a typical first order differen- tial equation looks like, where f(t, y) is a function of the variables t and y (of course, you can use any variables here, such as x and y or u and v, not just t and y): , dt dy f t y = _ i In this chapter, you work with linear first order differential equations — that is, differential equations where the highest power of y is 1 (you can find out the difference between linear and nonlinear equations in Chapter 1). For example: dt dy 5 = dt dt y 1 = + dt dt y 3 1 = + I provide some general information on nonlinear differential equations at the end of the chapter for comparison. irst Things First: The Basics of Solving inear First Order Differential Equations In the following sections, I take a look at how to handle linear first order dif- ferential equations in general. Get ready to find out about initial conditions, solving equations that involve functions, and constants. Applying initial conditions from the start When you’re given a differential equation of the form dy/dt = f(t, y), your goal is to find a function, y(t), that solves it. You may start by integrating the equa- tion to come up with a solution that includes a constant, and then you apply an initial condition to customize the solution. Applying the initial condition allows you to select one solution among the infinite number that result from the integration. Sounds cool, doesn’t it? Take a look at this simple linear first order differential equation: dt dy a = As you can see, a is just a regular old number, meaning that this is a simple example to start with and to introduce the idea of initial conditions. How can you solve it? First of all, you may have noticed that another way of writing this equation is: dy = a dt This equation looks promising. Why? Well, because now you can integrate like this: dy a dt x t y y 0 0 = # # Performing the integration gives you the following equation: y – y0 = at – at0 You can combine y0 – at0 into a new constant, c, by adding y0 to the right side of the equation, which gives you: y = at + c So, for example, if a = 3 in the differential equation, here’s the equation you would have: dt dy 3 = The solution for this equation is y = 3t + c. Note that c, the result of integrating, can be any value, which leads to an infi- nite set of solutions: y = 3t + 5, y = 3t + 6, y = 3t + 589,303,202. How do you track down the value of c that works for you? Well, it all depends on your initial con- ditions; for example, you may specify that the value of y at t = 0 be 15. Setting this initial condition allows you to state the whole problem — differential equation and initial condition — as follows: dt dy 3 = y(0) = 15 Substituting the initial condition, y(0) = 15, into the solution y = 3t + c gives you the following equation: y(t) = 3t + 15 Stepping up to solving differential equations involving functions Of course, dy/dt = 3 (the example from the previous section) isn’t the most exciting differential equation. However, it does show you how to solve a dif- ferential equation using integration and how to apply an initial condition. The next step is to solve linear differential equations that involve functions of t rather than just a simple number. This type of differential equation still contains only dy/dt and terms of t, making it easy to integrate. Here’s the basic form: dt dy g t = ^ h where g(t) is some function of t. Here’s an example of this type of differential equation: dt dy t t t 3 3 2 = - + dy = t3 dt – 3t2 dt + t dt Then you can integrate to get this equation: y t t t c 4 2 4 3 2 = - + + Adding a couple of constants to the mix The next step up from equations such as dy/dx = a or dy/dt = g(t) are equa- tions of the following form, which involve y, dy/dt, and the constants a and b: dt dy ay b = - How do you handle this equation and find a solution? Using some handy alge- bra, you can rewrite the equation like this: / / y b a dy dt a - = _ i Integrating both sides gives you the following equation: ln | y – (b/a) | = at + c where c is an arbitrary constant. Now get y out of the natural logarithm, which gives you: y = (b/a) + deat where d = ec. And that’s it! You’re done. Good job! olving Linear First Order Differential quations with Integrating Factors Sometimes integrating linear first order differential equations isn’t as easy as it is in the examples earlier in this chapter. But it turns out that you can often convert general equations into something that’s easy to integrate if you find an integrating factor, which is a function, µ(t). The idea here is to multiply the differential equation by an integrating factor so that the resulting equation can easily be integrated and solved. In the following sections, I provide tips and tricks for solving for an integrating factor and plugging it back into different types of linear first order equations. In general, first order differential equations don’t lend themselves to easy integration, which is where integrating factors come in. How does the method of integrating factors work? To understand, say, for example, that you have this linear differential equation: dt dy y 2 4 + = First, you multiply the previous equation by µ(t), which is a stand-in for the undetermined integrating factor, giving you: t dt dy t y t µ µ µ 2 4 + = ^ ^ ^ h h h Now you have to choose µ(t) so that you can recognize the left side of this equation as the derivative of some expression. This way it can easily be integrated. Here’s the key: The left side of the previous equation looks very much like differentiating the product µ(t)y. So try to choose µ(t) so that the left side of the equation is indeed the derivative of µ(t)y. Doing so makes the integration easy. The derivative of µ(t)y by t is: dt d t y t dt dy dt d t y µ µ µ = + ^ ^ ^ h h h 8 B Comparing the previous two equations term by term gives you: dt d t t µ µ 2 = ^ ^ h h Hey, not bad. Now you’re making progress! This is a differential equation you can solve. Rearranging the equation so that all occurrences of µ(t) are on the same side gives you: / t d t dt µ µ 2 = ^ ^ h h Now the equation can be rearranged to look like this: t d t dt µ µ 2 = ^ ^ h h Fine work. Integration gives you: ln |µ(t)| = 2t + b where b is an arbitrary constant of integration. µ(t) = ce2t where c is an arbitrary constant. So that’s it — you’ve solved for the integrating factor! It’s µ(t) = ce2t. Using an integrating factor to solve a differential equation After you solve for an integrating factor, you can plug that factor into the original linear differential equation as multiplied by µ(t). For instance, take your original equation from the previous section: t dt dy t y t µ µ µ 2 4 + = ^ ^ ^ h h h and plug in the integrating factor to get this equation: ce dt dy ce y ce 2 4 t t t 2 2 2 + = Note that c drops out of this equation when you divide by c, so you get the following equation (because you’re just looking for an arbitrary integrating factor, you could also set c = 1): e dt dy e y e 2 4 t t t 2 2 2 + = When you use an integrating factor, you attempt to find a function µ(t) that, when multiplied on both sides of a differential equation, makes the left side into the derivative of a product. Figuring out the product allows you to solve the differential equation. In the previous example, you can now recognize the left side as the derivative of e2t y. (If you can’t recognize the left side as a derivative of some product, in general, it’s time to go on to other methods of solving the differential equation). In other words, the differential equation has been conquered, because now you have it in this form: dt d e y e 4 t t 2 2 = _ i You can integrate both sides of the equation to get this: e2ty = 2e2t + c y = 2 + ce–2t You’ve got yourself a solution. Beautiful. The use of an integrating factor isn’t always going to help you; sometimes, when you use an integrating factor in a linear differential equation, the left side isn’t going to be recognizable as the derivative of a product of functions. In that case, where integrating factors don’t seem to help, you have to turn to other methods. One of those methods is to determine whether the differen- tial equation is separable, which I discuss in Chapter 3. Moving on up: Using integrating factors in differential equations with functions Now you’re going to take integrating factors to a new level. Check out this linear equation, where g(t) is a function of t: dt dy ay g t + = ^ h This one’s a little more tricky. However, using the same integrating factor from the previous two sections, eat (remember that the c dropped out), works here as well. After you multiply both sides by eat, you get this equation: e dt dy a e y e g t at at at + = ^ h Now you can recast this equation in the following form: dt d e y e g t at at = _ ^ i h To integrate the function g, I use s as the variable of integration. Integration gives you this equation: e y e g s ds c at as = + # ^ h You can solve for y here, which gives you the following equation: y e e g s ds ce at as at = + - - # ^ h And that’s it! You’ve got your answer! Of course, solving this equation depends on whether you can calculate the integral in the previous equation. If you can do it, you’ve solved the differential equation. Otherwise, you may have to leave the solution in the integral form. In this section, I give you a shortcut for solving some particular differential equations. Ready? Here’s the tip: In general, the integrating factor for an equation in this form: dt dy ay g t + = ^ h is this: exp t a dt µ = # ^ h In this equation, exp(x) means ex. As an example, try solving the following differential equation with the shortcut: dt dy y t 2 1 4 + = + Assume that the initial condition is y = 8, when t = 0 This equation is an example of the general equation solved in the previous section. In this case, g(t) = 4 + t, and a = 1⁄2. Using a, you find that the integrating factor is et/2, so multiply both sides of equation by that factor: e dt dy e y e te 2 4 / / / / t t t t 2 2 2 2 + = + Now you can combine the two terms on the left to give you this equation: dt d e y e te 4 / / / t t t 2 2 2 = + _ i All you have to do now is integrate this result. The term on the left and the first term on the right are no problem. The last term on the right is another story. You can use integration by parts to integrate this term. Integration by parts works like this: f x g x dx f b g b f a g a f x g x dx a b a b = - - # # l l ^ ^ ^ ^ ^ ^ ^ ^ h h h h h h h h et/2 y = 8et/2 + 2 t et/2 – 4et/2 + c where c is an arbitrary constant, set by the initial conditions. Dividing by eat gives you this equation: y = 4 + 2t + ce–at By applying the initial condition, y(0) = 8, you get y(0) = 8 8 = 4 + c Or c = 4. So the general solution of the differential equation is: y = 4 + 2t + 4e–t/2 In Chapter 1, I explain that direction fields are great tools for visualizing dif- ferential equations. You can see a direction field for the previously noted general solution in Figure 2-1. 0 0 1 2 3 4 5 6 7 8 9 10 y x 25 20 15 10 5 igure 2-1: The direction eld of the general solution. Solving an advanced example I think you’re ready for another, somewhat more advanced, example. Try solving this differential equation to show that you can have different integrat- ing factors: t dt dy y t 2 4 2 + = where y(1) = 4. To solve, first you have to find an integrating factor for the equation. To get it into the form: dt dy ay g t + = ^ h you have to divide both sides by t, which gives you this equation: dt dy t y t 2 4 + = 0 0 1 2 3 4 5 6 7 8 9 10 y x 25 20 15 10 5 igure 2-2: e graph of e general solution. exp exp t a dt t dt µ 2 = = # # ^ h Performing the integral gives you this equation: exp t t dt e t µ 2 ln t 2 2 = = = # ^ h So the integrating factor here is t2, which is a new one. Multiplying both sides of the equation by the integrating factor, µ(t) = t2, gives you: t dt dy ty t 2 4 2 3 + = Because the left side is a readily apparent derivative, you can also write it in this form: dt d yt t4 2 3 = _ i Now simply integrate both sides to get: yt2 = t4 + c Finally you get: y t t c 2 2 = + where c is an arbitrary constant of integration. Now you can plug in the initial condition y(1) = 4, which allows you to see that c = 3. And that helps you come to this solution: y t t 3 2 2 = + And there you have it. You can see a direction field for the many general solu- tions to this differential equation in Figure 2-3. You can see this function graphed in Figure 2-4. 2 –2 –1 0 1 2 y x 8 7 6 5 4 3 igure 2-4: e graph of a more advanced solution. –5 –2 –1 0 1 2 y x 4 3 2 1 0 –1 –2 –3 –4 igure 2-3: The direction field of a more advanced solution. Determining Whether a Solution for a inear First Order Equation Exists I show you how to deal with different kinds of linear first order differential equations earlier in this chapter, but the fact remains that not all linear differ- ential equations actually do have a solution. Luckily, a theorem exists that tells you when a given linear differential equa- tion with an initial condition has a solution. That theorem is called the exis- tence and uniqueness theorem. This theorem is worth knowing. After all, if a differential equation doesn’t have a solution, what use is it to search for a solution? In other words, this theorem represents another way to tackle linear first order differential equations. Spelling out the existence and uniqueness theorem for linear differential equations In this section, I explain what the existence and uniqueness theorem for linear differential equations says. Before I continue, however, note that a continuous function is a function for which small changes in the input result in small changes in the output (for example, f(x) = 1/x is not continuous at x = 0). Without further ado, here’s the existence and uniqueness theorem: If there is an interval I that contains the point to, and if the functions p(x) and g(x) are continuous on that interval, and if you have this differential equation: dx dy p x y g x + = ^ ^ h h then there exists a unique function, y(x), that is the solution to that differen- tial equation for each x in interval I that also satisfies this initial condition: y(to) = yo where yo is an arbitrary initial value. In other words, this theorem says that a solution exists and that the solution is unique. Thinking about the theorem in the previous section begs the question: What is the general solution to the following linear differential equation? dx dy p x y g x + = ^ ^ h h Note that this differential equation has a function p(x) and g(x), which pro- vides a more complex situation. So you can’t use the simple form I explain in the earlier section “Adding a couple of constants to the mix,” where a and b are constants like this: dx dy ay b = - The solution here is: y = (b/a) + ceat Now you face a more complex situation, with functions p(x) and g(x). A gen- eral solution to the general equation does exist, and here it is: y t s g s ds c µ µ = + # ^ ^ ^ h h h where the integrating factor is the following: exp t p t dt µ = # ^ ^ h h The integrals in these equations may not be possible to perform, of course. But together, the equations represent the general solution. Note that for linear differential equations, the solution, if there is one, is com- pletely specified, up to a constant of integration, as in the solution you get in the earlier section “Solving an advanced example”: y t t c 2 2 = + where c is a constant of integration. You can’t necessarily say the same thing about nonlinear differential equations — they may have solutions of completely different forms, not just differing in the value of a constant. Because the solution to a linear differen- tial equation has one form, differing only by the value of a constant, those solutions are referred to as general solutions. This term isn’t used when dis- cussing nonlinear differential equations, which may have multiple solutions of completely different forms. I discuss nonlinear first order differential equa- tions later in this chapter. and uniqueness examples In this section, I include a few examples to help you understand the existence and uniqueness theorem for linear differential equations. Example 1 Apply the existence and uniqueness theorem to the following equation to show that there exists a unique solution: dx dy y x 5 4 = - _ i Just kidding! This equation isn’t linear because the term (y – 5) is in the denom- inator of the right side. And, of course, because the equation isn’t linear, the existence and uniqueness theorem doesn’t apply. Did you catch that? Example 2 Try this differential equation (which I promise is linear!). Does a unique solu- tion exist? dx dy y x 2 4 2 + = where y(1) = 2. The equation is already in the correct form: dx dy p x y g x + = ^ ^ h h where p(x) = 2 and g(x) = 4x2. Note that p(x) and g(x) are continuous everywhere, so there’s a general solu- tion that’s valid on the interval, –∞ < x < ∞. In particular, the initial condition is y(1) = 2, which is definitely inside the interval that p(x) and g(x) are continuous (everything is inside that interval). So, yes, there exists a solution to the initial value problem. Example 3 Now take a look at this equation, which is similar to the example in the previ- ous section, and determine whether a unique solution exists: x dx dy y x 2 4 2 + = where y(1) = 2. dx dy p x y g x + = ^ ^ h h Here’s what the equation should look like: dx dy x y x 2 4 + = In other words: p x x 2 = ^ h and g(x) = 4x Note that p(x) and g(x) aren’t continuous everywhere. In particular, p(x) is discontinuous at x = 0, which makes the interval in which p(x) and g(x) are continuous on the interval 0 > x and 0 < x. Because the initial condition here is y(1) = 2, the point of interest is x = 1, which is inside the interval where p(x) and g(x) are continuous. Therefore, by the existence and uniqueness theorem, the initial value problem indeed has a unique solution. Cool, huh? iguring Out Whether a Solution for a Nonlinear Differential Equation Exists In the previous sections of this chapter, I cover linear first order differential equations in detail. But you may be wondering: Is there such a thing as a non- linear differential equation? You bet there is! A nonlinear differential equation simply includes nonlinear terms in y, y', y", and so on. Nonlinear equations are pretty tough, so I don’t delve into them a lot in this book. But I do want to discuss one important theorem related to solving these equations. You see, the existence and uniqueness theorem (which you use for linear equations, and which I cover earlier in this chapter) is analogous to another theorem that’s used for nonlinear equations. I explain this theorem and show some examples in the following sections. nonlinear differential equations Here’s the existence and uniqueness of solutions for nonlinear equations: Say that you have a rectangle R that contains the point (to, yo) and that the functions f and df/dy are continuous in that rectangle. Then, in an interval to – h < t < to + h contained in R, there’s a unique solution to the initial value problem: , , dt dy f t y y t y 0 0 = = _ _ i i Note that this theorem discusses the continuity of both f and df/dy instead of the continuity of both p(x) and g(x). Like the first theorem in this chapter, this theorem guarantees the existence of a unique solution if its conditions are met. Here’s another note: If the differential equation in question actually is linear, the theorem reduces to the first theorem in this chapter. In that case, f(t, y) = –p(t)y + g(t) and df/dy = –p(t). So demanding that f and df/dy be continuous is the same as saying that p(t) and g(t) be continuous. Here’s a side note that many differential equations books won’t tell you: The first theorem in this chapter guarantees a unique solution, but it’s actually a little tighter than it needs to be in order to guarantee just a solution (which isn’t necessarily unique). In fact, you can show that there’s a solution — but not that it’s unique — to the nonlinear differential equation merely by proving that f is continuous. A couple of nonlinear existence and uniqueness examples In the following sections, I provide two examples that put the nonlinear exis- tence and uniqueness theorem into action. Example 1 Determine what the two theorems in this chapter have to say about the fol- lowing differential equation as far as its solutions go: dx dy y x x 2 4 5 9 6 2 = - + + _ i where y(0) = –1. That means you need the nonlinear theorem. Note that for this theorem: , f x y y x x 2 4 5 9 6 2 = - + + _ _ i i and dy df y x x 2 4 5 9 6 2 2 = - - + + _ i These two functions, f and df/dy, are continuous, except at y = 4. So you can draw a rectangle around the initial condition point, (0, –1) in which both f and df/dy are continuous. And the existence and uniqueness theorem for nonlinear equations guarantees that this differential equation has a solu- tion in that rectangle. Example 2 Now determine what the existence and uniqueness theorems say about this differential equation: dx dy y / 1 5 = where y(1) = 0. Clearly, this equation isn’t linear, so the first theorem is no good. Instead you have to try the second theorem. Here, f is: f(x, y) = y1/5 and df/dy is: dy df y 5 /4 5 = - Now you know that f(x, y) is continuous at the initial condition point given by: y(1) = 0 But df/dy isn’t continuous at this point. The upshot is that neither the first theorem nor the second theorem have anything to say about this initial value problem. On the other hand, a solution to this differential equation is still guaranteed because f(x, y) is continuous. However, it doesn’t guarantee the uniqueness of that solution. Sorting Out Separable First Order Differential Equations This Chapter Figuring out the fundamentals of separable differential equations Applying separable differential equations to real life Advancing with partial fractions Some rocket scientists call you, the Consulting Differential Equation Expert, into their headquarters. “We’ve got a problem,” they explain. “Our rockets are wobbling because we can’t solve their differential equation. All the rockets we launch wobble and then crash!” They show you to a blackboard with the following differential equation: dx dy y x 2 2 2 = - “It’s not linear,” the scientists cry. “There’s a y2 in there!” “I can see that,” you say. “Fortunately, it is separable.” “Separable? What does that mean?” they ask. “Separable means that you can recast the equation like this, where x is on one side and y is on the other,” you say while showing them the following equation on your clipboard: (2 – y2) dy = x2 dx “You can integrate the equation with respect to y on one side, and x on the other,” you say. “We never thought of that. That was too easy.” to the first order.) I explain the basics of separable equations here, such as determining the difference between linear and nonlinear separable equations and figuring out different types of solutions, such as implicit and explicit. I also introduce you to a fancy method for solving separable equations involving partial fractions. Finally, I show you a couple of real world applications for separable equations. When you’re an expert at these equations, you too can solve problems for rocket scientists. Beginning with the Basics of Separable Differential Equations Separable differential equations, unlike general linear equations in Chapter 2, let you separate variables so only variables of one kind appear on one side, and only variables of another kind appear on the other. Say, for example, that you have a differential equation of the following form, in which M and N are functions: , , M x y N x y dx dy 0 + = _ _ i i And furthermore, imagine that you could reduce this equation to the follow- ing form, where the function M depends only on x and the function N depends only on y: M x N y x dy 0 + = ^ _ h i This equation is a separable equation; in other words, you can separate the parts so that only x appears on one side, and only y appears on the other. You write the previous equation like this: M(x) dx + N(y) dy = 0 Or in other words: M(x) dx = –N(y) dy If you can separate a differential equation, all that’s left to do at that point is to integrate each side (assuming that’s possible). Note that the general form of a separable differential equation looks like this: M x N y dx dy 0 + = ^ _ h i x y dx dy 0 2 + = And if you’re still not convinced, check out this one, which is also separable but not linear: x y dx dy 1 0 9 3 + - = _ i In the following sections, I ease you into linear separable equations before tackling nonlinear separable equations. I also show you a trick for turning nonlinear equations into linear equations. (It’s so cool that it’ll impress all your friends!) Starting easy: Linear separable equations To get yourself started with linear separable equations, say that you have this differential equation: dx dy x 0 2 - = This equation qualifies as linear. This also is an easily separated differential equation. All you have to do is put it into this form: dy = x2 dx And now you should be able to see the idea behind solving separable differ- ential equations immediately. You just have to integrate, which gives you this equation: y x c 3 3 = + where c is an arbitrary constant. There’s your solution! How easy was that? Introducing implicit solutions Not all separable equation solutions are going to be as easy as the one in the previous section. Sometimes finding a solution in the y = f(x) format isn’t ter- ribly easy to get. Mathematicians refer to a solution that isn’t in the form y = f(x) as an implicit solution. Coming up with such a solution is often the best you can do, because solving a separable differential equation involves tion; I show you how to find an explicit solution from an implicit solution in the next section.) Try this differential equation to see what I mean: dx dy y x 2 2 2 = - How about it? One of the first things that should occur to you is that this isn’t a linear differential equation, so the techniques in the first part of this chap- ter won’t help. However, you’ll probably notice that you can write this equa- tion as: (2 – y2) dy = x2 dx As you can see, this is a separable differential equation because you can put y on one side and x on the other. You can also write the differential equation like this: x y dx dy 2 0 2 2 - + - = _ i You can cast this particular equation in terms of a derivative of x, and then you integrate with respect to x to solve it. After integration, you wind up with the following: / x dx d x 3 2 3 - = - _ i Note that: / y dx dy dx d y y 2 2 3 2 3 - = - _ _ i i because of the chain rule, which says that: dx df dy df dx dy = So now you can write the original equation like this: x y dx dy dx d x y y 2 3 2 3 0 2 2 3 3 - + - = - + - = _ e i o If the derivative of the term on the right is 0, it must be a constant this way: x y y c 3 2 3 3 3 - + - = e o –x3 + 6y – y3 = c To see how the solutions look graphically, check out the direction field for this differential equation in Figure 3-1. (I introduce direction fields in Chapter 1.) Finding explicit solutions from implicit solutions The implicit solution in the previous section, with terms in y and y3, isn’t ter- ribly easy to cram into the y = f(x) format. In this section, you discover that you can find an explicit solution to a separable equation by using a quadratic equation, which is the general solution to polynomials of order two. Try another, somewhat more tractable problem. Solve this differential equation: dx dy y x x 2 1 9 6 4 2 = - + + _ i where y(0) = –1. –4 –4 4 3 2 1 0 –1 –2 –3 y x 4 3 2 1 0 –1 –2 –3 igure 3-1: The direction field of a nonlinear separable equation. dx dy x x 9 6 4 2 = + + there would be no problem. After all, you would just integrate. But you’ve probably noticed that pesky 2(y – 1) term in the denominator on the right side. Fortunately, you may also realize that this is a separable differential equation because you can put y on one side and x on the other. Simply write the equation like this: 2(y – 1) dy = (9x2 + 6x + 4) dx Now you integrate to get this equation: y2 – 2y = 3x3 + 3x2 +4x + c Using the initial condition, y(0) = –1, substitute x = 0 and y = –1 to get the following: 1 + 2 = c Now you can see that c = 3 and that the implicit solution to the separable equation is: y2 – 2y = 3x3 + 3x2 +4x + 3 If you want to find the explicit solution to this and similar separable equa- tions, simply solve for y with the quadratic equation because the highest power of y is 2. Solving for y using the quadratic formula gives you: y x x x 1 3 3 4 4 3 2 ! = + + + You have two solutions here: one where the addition sign is used and one where the subtraction sign is used. To match the initial condition that y(0) = –1, however, only one solution will work. Which one? The one using the sub- traction sign: y x x x 1 3 3 4 4 3 2 = - + + + In this case, the solution with the subtraction is valid as long as the expres- sion under the square root is positive — in other words, as long as x > –1. You can see the direction field for the general solutions to this differential equation in Figure 3-2. As I note in Chapter 1, connecting the slanting lines in a direction field gives you a graph of the solution. You can see a graph of this particular function in Figure 3-3. –5 –2 3 2 1 0 –1 y x 2 1 0 –1 –2 –3 –4 igure 3-3: A graph of e solution of a separable equation with initial onditions. –5 –2 3 2 1 0 –1 y x 1 0 –1 –2 –3 –4 igure 3-2: The direction field of a separable equation with initial onditions. explicit solution Most of the time, you can find an explicit solution from an implicit solution. But every once in a while, getting an explicit solution is pretty tough to do. Here’s an example: sin dx dy y y x 1 2 2 = + _ i where y(0) = 1. As you get down to work (bringing to bear all your differential equation skills!), the first thing that may strike you is that this equation isn’t linear. But, you’ll also likely note that it’s separable. So simply separate the equation into y on the left and x on the right, which gives you this equation: sin y y dy x dx 1 2 2 + = _ i This equation subsequently becomes sin y dy y dy x dx 2 + = Now you can integrate to get this: ln|y| + y2 = –cos x + c Next, take a look at the initial condition: y(0) = 1. Plugging that condition into your solution gives you this equation: 0 + 1 = –1 + c or c = 2 So your solution to the initial separable equation is: ln|y| + y2 = –cos x + 2 This is an implicit solution, not an explicit solution, which would be in terms of y = f(x). In fact, as you can see from the form of this implicit solution, get- ting an explicit solution would be no easy task. tion field for this differential equation, which indicates what the integral curves look like, in Figure 3-4. A neat trick: Turning nonlinear separable equations into linear separable equations In this section, I introduce you to a neat trick that helps with some differen- tial equations. With it, you can make a linear equation out of a seemingly non- linear one. All you have to do to use this trick is to substitute the following equation, in which v is a variable: y = xv In some cases, the result is a separable equation. As an example, try solving this differential equation: dx dy xy y x 2 3 4 4 = + –3 –4 –3 –2 –1 0 1 2 3 4 –2 –1 0 1 y x 2 3 igure 3-4: The direction field of a separable equation th a hard- to-find explicit solution. dx dy x y y x 2 3 3 = + What do you do now? Keep reading to find out. Knowing when to substitute You can use the trick of setting y = xv when you have a differential equation that’s of the following form: , dx dy f x y = _ i when f(x, y) = f(tx, ty), where t is a constant. You can see that substitution is possible, because substituting tx and ty into this differential equation gives you the following result: dx dy txt y t y t x 2 3 3 4 4 4 4 = + which breaks down to: dx dy xy y x 2 3 4 4 = + Substituting y = xv into this differential equation gives you: v x dx dv x xv xv x 2 3 4 4 + = + ^ ^ h h This equation now can be simplified to look like this: x dx dv v v 1 3 4 = + You now have a separable equation! Separating and integrating Continuing with the example from the previous section, you can now sepa- rate the terms, which gives you: x dx v v dv 1 4 3 = + ln ln x v c 4 1 = + + ^ _ h i where c is a constant of integration. Bearing in mind that, where k is a constant: ln(x) + ln(k) = ln(kx) and that: n ln(x) = ln(xn) you get: v4 + 1 = (kx)4 where: c = –ln(k) Where does all this get you? You’re ready to substitute with the following: v x y = This substitution gives you: x y kx 1 4 4 + = d ^ n h So: y4 + x4 = mx8 where m = k4. And solving for y gives you the following: y = (mx8 – x4)1/4 And there’s your solution. Nice work! rying Out Some Real World eparable Equations In the following sections, I take a look at some real world examples featuring separable equations. Getting in control with a sample flow problem To understand the relevance of differential equations in the real world, here’s a sample problem to ponder: Say that you have a 10-liter pitcher of water, and that you’re mixing juice concentrate into the pitcher at the same time that you’re pouring juice out. If the concentrate going into the pitcher has 1⁄4 kg of sugar per liter, the rate at which the concentrate is going into the pitcher, which I’ll call rin, is 1⁄100 liter per second, and the juice in the pitcher starts off with 4 kg of sugar, find the amount of sugar in the juice, Q, as a function of time, t. Because this problem involves a rate — dQ/dt, which is the change in the amount of sugar in the pitcher — it’s a differential equation, not just a simple algebraic equation. I walk you through the steps of solving the equation in the following sections. Determining the basic numbers When you start trying to work out this problem, remember that the change in the amount of sugar in the pitcher, dQ/dt, has to be the rate of sugar flow in minus the rate of sugar flow out, or something like this: dt dQ rate of sugar flow in rate of sugar flow out = - _ _ i i Now you ask: What’s the rate of sugar flow in? That’s easy; it’s just the con- centration of sugar in the juice concentrate multiplied by the rate at which the juice concentrate is flowing into the pitcher, which I’ll call rin. So, your equation looks something like this: r rate of sugar flow in 4 kg/sec in = _ i Now what about the flow of sugar out? The rate of sugar flow out is related to the rate at which juice leaves the pitcher. So if you assume that the amount of juice in the pitcher is constant, then rin = rout = r. That, in turn, means the rate of the pitcher (10 liters), or Q/10. Here’s what your equation would look like: Q r rate of sugar flow out 10 kg/sec = _ i So that means: dt dQ r Qr 4 10 rate of sugar flow in rate of sugar flow out = - = - _ _ i i where the initial condition is: Q0 = 4 kg Solving the equation The equation at the end of the previous section is separable, and separating the variables, each on their own side, gives you this equation: dt dQ Qr r 10 4 + = Now that, you might say, is a linear differential in Q. And you’d be right. So you know that the equation is both linear and separable. You can handle this differential equation using the methods in Chapter 2. For instance, to solve, you find an integrating factor, multiply both sides by the integrating factor, and then see if you can figure out what product the left side is the derivative of and integrate it. Whew! It sounds rough, but note that the equation is of the following form: dt dy ay b + = The solution to this kind of differential equation is already found in Chapter 2; you use an integrating factor of eat. The solution to this kind of equation is: y = (b/a) + ce–at So you can see that the solution to the juice flow problem is: Q(t) = 2.5 + ce–rt/10 Because r = 1⁄100 liter per second, the equation becomes: Q(t) = 2.5 + ce–t/1000 Q0 = 4 kg you know that: Q = 2.5 + 1.5 e–t/1000 Note the solution as t → ∞ is 2.5 kg of sugar, and that’s what you’d expect. Why? Because the concentrate has 1⁄4 kg of sugar per liter, and 10 liters of water are in the pitcher. So 10/4 = 2.5 kg. The direction field for different values of Q0 appears in Figure 3-5. Notice that all the solutions tend toward the final Q of 2.5 kg of sugar, as you’d expect. You can see a graph of this solution in Figure 3-6. 0 0 300 600 900 1200 1500 1800 2100 2400 2700 3000 Q t 20 15 10 5 igure 3-5: The direction field of a flow problem solution. Striking it rich with a sample monetary problem You may not have realized that differential equations can be used to solve money problems. Well they can! And here’s a problem to prove it: Say that you’re deciding whether to deposit your money in the bank. You can calcu- late how your money grows, dQ/dt, given the interest rate of the bank and the amount of money, Q, that you have in the bank. As you can see, this is a job for differential equations. Figuring out the general solution Suppose your bank compounds interest continuously. The rate at which your savings, Q, grows, is: dt dQ rQ = where r is the interest rate that your bank pays. 0 0 300 600 900 1200 1500 1800 2100 2400 2700 3000 Q t 4 3 2 1 igure 3-6: e graph of e solution of a flow problem. equation for the rate at which your money grows, not the actual amount of money. Say that you have Q0 money at t = 0: Q(0) = Q0 How much money would you have at a certain time in the future? That’s easy enough to figure out. Separate the variables, each on their own side, like this: Q dQ r dt = Then integrate: ln|Q| = rt Finally, exponentiate both sides, which gives you the following equation: Q = cert To match the initial condition: Q(0) = Q0 the solution becomes: Q = Q0e rt So, in other words, your money would grow exponentially. Not bad. Compounding interest at set intervals Now I want you to examine the result from the previous section a little, deriv- ing it another way so that it makes more sense. If your bank compounded interest once a year, not continuously, after t years, you’d have this much money: Q = Q0(1 + r) t That’s because if your interest was 5 percent, after the first year, you would have 1.05Q0; at the end of the second year, 1.05 2Q0, and so on. Q = Q0(1 + r) 2t No, you wouldn’t. Why? Because that would pay you r percent interest twice a year. For example, if r = 8 percent, the previous equation would pay you 8 percent of your total savings twice a year. Instead, banks divide the interest rate they pay you by the number of times they compound per year, like this: Q Q r 1 2 t 0 2 = + c m In other words, if the bank compounds twice a year, and the annual interest rate is 8 percent, six months into the year it pays you 4 percent, and at the end of the year it pays another 4 percent. In general, if your bank compounds interest m times a year, after t years, you’d have: Q Q m r 1 mt 0 = + c m If you take the limit as m → ∞ — that is, as your bank starts to compound continuously — you get this equation: lim Q Q m r 1 m mt 0 = + "3 c m But that’s just the expansion for ert. So, as the bank compounds continuously, you get: lim Q Q m r Q e 1 m mt r 0 0 = + = "3 t c m And this result confirms the answer you got from solving the differential equation in the previous section. So if you had $25 invested, and you left it alone at 6 percent for 60 years, you’d have: Q = Q0e rt = 25e0.06(60) or: Q = Q0e rt = 25e0.06(60) = $914.96 Hmm, not such a magnificent fortune. How about if you add a set amount every year to the equation in the previous section? That would be better, wouldn’t it? Say that you add $5,000 a year. In that case, remember that the set amount would change the differential equa- tion for your savings, which was this: dt dQ rQ = The equation would change to this, where k is the amount you contribute regularly: dt dQ rQ k = + If you deposit regularly, k > 0; if you withdraw regularly, k < 0. Ideally, you should add or subtract k from your account continuously over the year to make your solution exact, but here you can just assume that you add or sub- tract k once a year. Putting this new equation into standard separable form gives you this: dt dQ rQ k - = This equation is of the following form: dt dy ay b + = The solution to this kind of equation is: y = (b/a) + ce–at In this case, that solution means: Q = cert – k/r What’s going on here? It looks like you have the solution for leaving money in the bank without adding anything minus the amount you’ve added. Can that be right? The answer is in c, the constant of integration. Here, the initial con- dition is: Q(0) = Q0 which means that: Q(0) = cer0 – k/r = c – k/r = Q0 c = Q0 + k/r So your solution turns out to be: Q = cert – k/r = (Q0 + k/r)e rt – k/r Working this out gives you: / / Q Q k r e k r Q e r k e 1 rt rt rt 0 0 = + - = + - _ _ i i That looks a little better! Now the first term is the amount that you’d earn if you just left Q0 in the account, and the second term is the amount resulting from depositing or withdrawing k dollars regularly. For example, say you started off with $25, but then you added $5,000 every year for 60 years. At the end of 60 years at 6 percent, you’d have: . , Q Q e r k e e e 1 25 0 06 5 000 1 . ( ) . ( ) rt rt 0 0 06 60 0 06 60 = + - = + - _ _ i i After calculating this out, you’d get: . , $ . $ , , $ , , Q e e 25 0 06 5 000 1 914 96 2 966 519 2 967 434 . ( ) . ( ) 0 06 60 0 06 60 = + - = + = _ i Quite a tidy sum. Break It Up! Using Partial Fractions n Separable Equations When a term in a separable differential equation looks a little difficult to inte- grate, you can use the method of partial fractions to separate it. This method is used to reduce the degree of the denominator of a rational expression. For example, using the method of partial fractions, you can express: x x 2 8 6 2+ - as the following equation: x x x x 2 8 6 2 1 4 1 2+ - = - - +
