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# Class Note for MATH 3321 at UH

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Date Created: 02/06/15

CHAPTERl Introduction to Differential Equations 11 Basic Terminology Most of the phenomena studied in the sciences and engineering involve processes that change with time For example7 it is well known that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t In mathematical terms this says that d i ky k a negative constant 1 dt where y yt is the amount of material present at time t If an object7 suspended by a spring7 is oscillating up and down7 then Newton7s Second Law of Motion F ma combined with Hooke7s Law the restoring force of a spring is proportional to the displacement of the object results in the equation dw k270 k quot 2 g y 7 7 a pos1t1ve constant where y yt denotes the position of the object at time t The basic equation governing the diffusion of heat in a uniform rod of nite length L is given W Bu i 7 3 at 812 where u uzt is the temperature of the rod at time t at position I on the rod Each of these equations is an example of what is known as a differential equation DIFFERENTIAL EQUATION A di ferential equation is an equation that contains an unknown function together with one or more of its derivatives Here are some additional examples of differential equations Example 1 2 fryiy M yiegjfv d2 d b 12 7 21 2y 413 8211 8211 c w w 0 Laplace7s equation dy 7 dgy d2y 71 d 7 4E73e i TYPE As suggested by these examples7 a differential equation can be classi ed into one of two general categories determined by the type of unknown function appearing in the equation If the unknown function depends on a single independent variable then the equation is an ordinary di er ential equation if the unknown function depends on more than one independent variable then the equation is a partial di erential equationi According to this classi cation the differential equations 1 and 2 are ordinary differential equations and 3 is a partial differential equation In Exam ple 1 equations a b and d are ordinary differential equations and equation c is a partial differential equation Differential equations both ordinary and partial are also classi ed according to the highest ordered derivative of the unknown function ORDER The order of a differential equation is the order of the highest derivative of the unknown function appearing in the equation Equation 1 is a rst order equation and equations 2 and 3 are second order equations In Example 1 equation a is a rst order equation b and c are second order equations and equation d is a third order equation In general the higher the order the more complicated the equation In Chapter 2 we will consider some rst order equations and in Chapter 3 we will study certain kinds of second order equationsi Higher order equations and systems of equations will be considered in Chapter 6 The obvious question that we want to consider is that of solving a given differential equation SOLUTION A solution of a di erential equation is a function de ned on some interval I in the case of an ordinary differential equation or on some domain D in two or higher dimensional space in the case of a partial differential equation with the property that the equation reduces to an identity when the function is substituted into the equation Example 2 Given the secondorder ordinary differential equation 12 y 7 21 y 2y 413 Example 1 show that a 12 213 is a solution b 2x2 31 is not a solution SOLUTION a The rst step is to calculate the rst two derivatives of y y 12 213 y 21 612 y 2 121 Next we substitute y and its derivatives into the differential equation 122 121 7 2121 612 212 213 77 413 Simplifying the lefthand side we get 212 1213 7 412 712a3 212 413 i 413 and 413 7 413 The equation is satis ed y 12 213 is a solution b The rst two derivatives of 2 are 2 212 31 2 4x 3 ZN 4 Substituting into the differential equation we have 124 7 2141 3 2212 31 7 413i Simplifying the lefthand side we get 412781276z4z26z 0413 The function 2 212 31 is not a solution of the differential equation I Example 3 Show that uz y cos I sinh y sin I cosh y is a solution of Laplace s equation 8211 8211 w wwi SOLUTION The rst step is to calculate the indicated partial derivatives 7 7 sin I sinh y cos I cosh y g 7 cos I sinh y 7 sin I cosh y 7 cos I cosh y sin I sinh y gig cos I sinh y sin I cosh yr Substituting into the differential equation we nd that 7 cos I sinh y 7 sin I cosh y cos I sinh y sin I cosh y 0 and the equation is satis ed uz y cos I sinh ysin z cosh y is a solution of Laplace s equation I Exercises 1 1 ll Classify the following differential equations With respect to type iiel7 ordinary or partial and order a y2 zyy sin 1 b y em tan 1 8211 8211 8211 c w27azay er 70 d2 3 cl zy2zl e y 75zy y e 71 f BuBz MailBy d2y dy 2 ds 721 g gigy Jrry i le l For each differential equation determine Whether or not the given functions are solutions 2 y 4y 0 sin 31 cos 21 2sin 21 d3 d TZ61 yI1sinIe 7 zz2coszexl z z 4 zy y 0 y1r1n1I7 MI 12 5l11y zy 7y 1 12 e 12 17 121l d3 d2 d 6 E 7 ST 6 0 61621 02637 cl Cg constants7 262 3631 4 8211 8211 2 2 3 2 7i 770 u1zylnI y u2zyz 731yl 812 8y2 8 ylliy27z yze 172 21sinhz172l Bu 7 k2 8211 i 7 739 k at 812 k u1zt 6 2t cos I 11217 6 2t sin 27ml Find the set of all solutions of each of the following differential equations 10 y 21 lnz 11 y 32 12 y 61 cos 21 dyi 13 i73l dz y 14 15 16 17 18 19 20 21 22 dy xy70i Determine values of 7 if possible7 so that the given differential equation has a solution of the 1 form y J i y 7 4y 0 y 2y 7 8y 0 y 7 6yl9y 0 y 7 4yH 5y 7 2y 0i y 72y 5y0i Determine values of 7 if possible7 so that the given differential equation has a solution of the form y I7 d2y dy 2 7 7 7 2 0 I dz2 zdz l y 12y zy 7 9y 0i 12y 7 31y 4y 0i 12 nParameter Family of Solutions General Solution Particular Solu tion Introduction You know from your experience in previous mathematics courses that the calculus of functions of several variables limits graphing differentiation integration and applications is more complicated than the calculus of functions of a single variable By extension therefore you would expect that the study of partial differential equations would be more complicated than the study of ordinary differential equations This is indeed the casel Since the intent of this material is to introduce some of the basic theory and methods for differential equations we shall con ne ourselves to ordinary differential equations from this point forward Hereafter the term di ereritial equation shall be interpreted to mean ordinary di erential equation Partial differential equations are studied in subsequent courses I We begin by considering the simple rstorder differential equation y 1 where f is some given function In this case we can nd y simply by integrating yfzdzFzC where F is an antiderivative of f and C is an arbitrary constant Not only did we nd a solution of the differential equation we found a whole family of solutions each member of which is determined by assigning a speci c value to the constant C In this context the arbitrary constant is called a parameter and the family of solutions is called a oneparameter family Remark ln calculus you learned that not only is each member of the family y FI C a solution of the differential equation but this family actually represents the set of all solutions of the equation that is there are no other solutions outside of this family I Example 1 The differential equation y 312 7 sin 21 has the oneparameter family of solutions 312 7 sin 21 dz IS 4 cos 21 C As noted above this family of solutions represents the set of all solutions of the equation I In a similar manner if we are given a second order equation of the form y fr then we can nd y by integrating twice with each integration step producing an arbitrary constant of integration Example 2 If y 61 4e27c then y 6x 4e dz 312 2e C1 and y 312 2621 C1 dz 13 e21 C11 C2 C1 C2 arbitrary constants The set of functions y13621C11C2 is a twoparameter family of solutions of the differential equation y 61 4621 Again from calculus we can conclude that this family actually represents the set of all solutions of the differential equation there are no other solutions I nPARAMETER FAMILY OF SOLUTIONS The examples given above are very special cases In general to nd a set of solutions of an n th order differential equation we would expect intuitively to integrate n times with each integration step producing an arbitrary constant of integration As a result we expect an n th order differential equation to have an npammeter family of solutions SOLVING A DIFFERENTIAL EQUATION To solve an n th order differential equation means to nd an n parameter family of solutions It is important to understand that the two nls here are the same For example to solve a fourthorder differential equation we need to nd a fourparameter family of solutions Example 3 Show that y Ce is a oneparameter family of solutions of y Icy k a given constantl Equation 1 in Section 11 SOL U T ON y Ce y kCem Substituting into the differential equation we get keel 7 k Gem keel we Thus y Ce is a oneparameter family of solutions You were shown in calculus that y Ce represents the set of all solutions of the equation I Example 4 Show that y C112 C21 213 is a twoparameter family of solutions of 12y 7 21y 2y 413 SOLUTION We calculate the rst two derivatives of y and then substitute into the differential equation y C112C2I2x3 y 2C11 C2 612 y 2C1 121 7 12201 12x 7 21 2011 02 612 2 0112 021 213 4131 Simplifying the lefthand side and rearranging the terms we get 01212 7 412 212 02 721 21 1213 7 1213 413 i 413 010 020 413 413 413 413 Thus for any two constants C1 C2 the function y C112 C21 213 is a solution of the differential equation The set of functions y C112 C21 213 is a twoparameter family of solutions of the equation In Chapter 3 we will see that this twoparameter family represents the set of all solutions of the equation I GENERAL SOLUTIONSINGULAR SOLUTIONS For most of the equations that we will study in this course an nparameter family of solutions of a given nth order equation will represent the set of all solutions of the equation In such cases the term general solution is often used in place of nparameter family of solutions Because it less cumbersome we will use the term general solution77 rather than nparameter family of solutions77 recognizing that there is possible imprecision in the use of the term in some cases an nparameter family of solutions may not be the set of all solutions 1 Solutions of an nth order differential equation which are not included in an n parameter family of solutions are called singular solutionsi Example 5 Consider the differential equation y 41y 1 y I2 C2 1 is a oneparameter family of solutions verify this In Section 212 you will learn how to solve this equation Also it is easy to see that the constant function y E 1 is a solution of the equation y E 1 implies y E 0i and 0 411 7 1 2 0 the equation is satis ed This solution is not included in the general solution because there is no value that you can assign to C that will produce the solution y E 1 y E 1 is a singular solutioni Additional examples of differential equations having singular solutions are given in the Exercises 13 PARTICULAR SOLUTION If speci c values are assigned to the arbitrary constants in the general solution of a differential equation then the resulting solution is called a particular solution of the equation Example 6 a y C65 is the general solution of the rst order differential equation y 5y see Example 3 y 20065 is a particular solution of the equation b y C112 C21 213 is the general solution of the second order differential equation 12y 7 21y 2y 413i Setting C1 2 and C2 3 we get the particular solution y 212 31 213 I The Differential Equation of an nParameter Family If we are given an nparameter family of curves then we can regard the family as the general solution of an nthorder differential equation and attempt to nd the equation The equation that we search for called the di erential equation of the family should be free of the parameters arbitrary constants and its order should equal the number of parameters The strategy for nding the differential equation of a given nparameter family is to differentiate the equation n times This produces a system of n 1 equations which can be used to eliminate the parameters Example 7 Given the oneparameter family y CI2 3 Find the differential equation of the family SOLUTION Since we have a oneparameter family we are looking for a rst order equationi Dif ferentiating the given equation we obtain y 2C1 We can solve this equation for C to get C Substituting this into the given equation gives 1 y 12 3 which simpli es to my 7 2y 6 0 This is the differential equation of the given oneparameter family I Example 8 Find the differential equation of the twoparameter family C yi1C2i z SOL UTION We are looking for a second order differential equation Differentiating twice we obtain the equations 2C1 if and H 7 y 12 13 Solving each of these equations for C1 we get C1 712M and C1 z y Therefore zgy 712M which simpli es to my Zy This is the differential equation of the given familyi Example 9 Find the differential equation of the twoparameter family y C1 cos 21 C2 sin 21 SOLUTION We differentiate twice y C1 cos 21 C2 sin 21 y 72C1 sin 21 2C2 cos 21 y 74C1 cos 21 7 4C2 sin 21 Multiplying the rst equation by 4 and adding it to the third equation7 we get y 4y 0 This is the differential equation of the given family I Remark These examples illustrate that there is no 77general method77 for nding the differential equation of a given nparameter family of functions You can only follow the strategy and try to nd some way to eliminate the parameters from the system of equations I 10 13 Initial Conditions InitialValue Problems As we noted in the preceding section we can obtain a particular solution of an nth order differential equation simply by assigning speci c values to the n constants in the general solution However in typical applications of differential equations you will be asked to nd a solution of a given equation that satis es certain preassigned conditions Example 1 Find a solution of 2 y 31 7 21 that passes through the point l 3 SOLUTION In this case we can nd the general solution by integrating y31272z dzzg712Ci The general solution is y 13 7 12 C To nd a solution that passes through the point 2 6 we set I 2 and y 6 in the general solution and solve for C 623722C874C whichimplies 02 3 Thus y z 7 12 2 is a solution of the differential equation that satis es the given condition In fact it is the only solution that satis es the condition since the general solution represented all solutions of the equation and the constant C was uniquely determined I Example 2 Find a solution of 12y 7 21y 2y 413 which passes through the point 1 4 with slope 2 SOLUTION As shown in Example 4 in the preceding section the general solution of the differential equation is y C112 C21 213 Setting 1 l and y 4 in the general solution yields the equation C1 C2 2 4 which implies C1 C2 2 The second condition slope 2 at z l is a condition on y We want y l 2 We calculate y y 2C11 C2 612 and then set I l and y 2 This yields the equation 201 C2 6 2 which implies 2C1 02 4 Now we solve the two equations simultaneously 11 C1 C2 2 2C1 C2 74 We get C1 76 C2 8 A solution of the differential equation satisfying the two conditions is 7612 81 213 It will follow from our work in Chapter 3 that this is the only solution of the differential equation that satis es the given conditions I INITIAL CONDITIONS Conditions such as those imposed on the solutions in Examples 1 and 2 are called initial conditions This term originated with applications where processes are usually observed over time starting with some initial state at time t 0 Example 3 The position yt of a weight suspended on a spring and oscillating up and down is governed by the differential equation y 9y 0 a Show that the general solution of the differential equation is yt C1 sin 3t C2 cos 3t b Find a solution that satis es the initial conditions y0 l y 0 72 SOLUTION a y C1 sin 3t C2 cos 3t y 3C1 cos 3t 7 SC sin 3t y 79C1 sin 3t 7 9C2 cos 3t Substituting into the differential equation we get y 9y 79C1 sin 3t 7 9C2 cos 3t 9 C1 sin 3t C2 cos 3t 0 Thus yt C1 sin 3t C2 cos St is the general solution b Applying the initial conditions we obtain the pair of equations y0 l C1 sin 0 C2 cos 0 C2 which implies C2 l yO 72 3C1 cos 0 7 SC sin 0 which implies C1 7 who A solution which satis es the initial conditions is yt 7 sin 3t cos St I Any nth order differential equation with independent variable I and unknown function y can be written in the form F y39vy wym lhymd 7 0 1 12 by moving all the nonzero terms to the lefthand side Since we are talking about an nth order equation gm must appear explicitly in the expression F Each of the other arguments may or may not appear explicitly For example the thirdorder differential equation 12y 7 21y few written in the form of equation 1 is 12y 7 21y 7 few 0 2y and Fzyy y ym z 7 21y 7 erTy Note that y does not appear explicitly in the equation However it is there implicitly For example y y y y nth ORDER INITIALVALUE PROBLEM An nth order initialvalue problem consists of an n th order differential equation FIyy7y 7y WW 0 together with n initial conditions of the form MC 1607 40 k1 y 6 k2 WAVE 767171 where c and kg 161 16771 are given numbers It is important to understand that to be an nth order initialvalue problem there must be n conditions same n of exactly the form indicated in the de nition For example the problem 0 Find a solution of the differential equation y 9y 0 satisfying the conditions y0 0 y7r 0 is not an initialvalueproblem the two conditions are not of the form in the de nition Similarly the problem 0 Find a solution of the differential equation y 7 3y 7 y 0 satisfying the conditions y0 l y 0 2 is not an initialvalue problem a third order equation requires three conditions yc k0 y c k1 y c k2 EXISTENCE AND UNIQUENESS The fundamental questions in any course on differential equations are 1 Does a given initialvalue problem have a solution That is do solutions to the problem exist 2 If a solution does exist is it unique That is is there exactly one solution to the problem or is there more than one solution 13 The initialvalue problems in Examples 17 2 and 3 each had a unique solution values for the arbitrary constants in the general solution were uniquely determined Example 4 The function y 12 is a solution of the differential equation y Z and y0 0 Thus the initialvalue problem y 2W y0 0 has a solution However7 y E 0 also satis es the differential equation and y0 0i Thus7 the initialvalue problem does not have a unique solution In fact7 for any positive number a the function 07 I g a yaw 17a7 zgta is a solution of the initialvalue problemi I Y Example 5 The oneparameter family of functions y CI is the general solution of There is no solution that satis es y0 l the initialvalue problem 9 y 7 W 1 does not have a solution I The questions of existence and uniqueness of solutions Will be addressed in the speci c cases of interest to us A general treatment of existence and uniqueness of solutions of initialvalue problems is beyond the scope of this course 14 Exercises 13 1 a b b b b C b b Show that each member of the oneparameter family of functions y 065x is a solution of the differential equation y 7 5y 0 Find a solution of the initial value problem y 7 5y 07 y0 2 Show that each member of the twoparameter family of functions y 01621 02671 is a solution of the differential equation y 7 y 7 2y 0 Find a solution of the initial value problem y 7 y 7 2y 0 y0 2 y 0 1 Show that each member of the oneparameter family of functions 7 1 y 7 Ce l is a solution of the differential equation y y y2l Find a solution of the initial value problem y y y2 yl 71 Show that each member of the threeparameter family of functions y 0212 C11 C0 is a solution of the differential equation y 0 Find a solution of the initial value problem y 0 yl l7 y l 4 y l 2 Find a solution of the initial value problem y 0 y2 y 2 y 2 0 Show that each member of the twoparameter family of functions y C1 sin 31 C2 cos 31 is a solution of the differential equation y 9y 0 Find a solution of the initial value problem y 9y 0 y7r2 y 7r2 1 Show that each member of the twoparameter family of functions y C112 C212 ln 1 is a solution of the differential equation 12y 7 3Iy 4y 0 Find a solution of the initial value problem 12y 7 3Iy 4y 0 yl 07 y l 1 Is there a member of the twoparameter family Which satis es the initial condition y0 y 0 0 15 Exercises 13 1 a b b b b C b b Show that each member of the oneparameter family of functions y 065x is a solution of the differential equation y 7 5y 0 Find a solution of the initial value problem y 7 5y 07 y0 2 Show that each member of the twoparameter family of functions y 01621 02671 is a solution of the differential equation y 7 y 7 2y 0 Find a solution of the initial value problem y 7 y 7 2y 0 y0 2 y 0 1 Show that each member of the oneparameter family of functions 7 1 y 7 Ce l is a solution of the differential equation y y y2l Find a solution of the initial value problem y y y2 yl 71 Show that each member of the threeparameter family of functions y 0212 C11 C0 is a solution of the differential equation y 0 Find a solution of the initial value problem y 0 yl l7 y l 4 y l 2 Find a solution of the initial value problem y 0 y2 y 2 y 2 0 Show that each member of the twoparameter family of functions y C1 sin 31 C2 cos 31 is a solution of the differential equation y 9y 0 Find a solution of the initial value problem y 9y 0 y7r2 y 7r2 1 Show that each member of the twoparameter family of functions y C112 C212 ln 1 is a solution of the differential equation 12y 7 3Iy 4y 0 Find a solution of the initial value problem 12y 7 3Iy 4y 0 yl 07 y l 1 Is there a member of the twoparameter family Which satis es the initial condition y0 y 0 0 15 17 18 19 20 21 22 23 24 25 y C1 C21e2 y C11 C21 y C11 C21 1i y C1 cos 31 C2 sin 31 y C1 sin 31 C2 y C162 cos 31 C2 y C1 C21 C31 y C11 C212 C313i Verify that the function ya1 is a solution of the initialvalue problem y Z y0 01 17

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