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# Ordinary Differential Equation MATH 2280

Weber State University

GPA 3.53

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This 14 page Class Notes was uploaded by Rosa Farrell on Wednesday October 28, 2015. The Class Notes belongs to MATH 2280 at Weber State University taught by Michael Wills in Fall. Since its upload, it has received 35 views. For similar materials see /class/230802/math-2280-weber-state-university in Math at Weber State University.

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Date Created: 10/28/15

FIRST ORDER DIFFERENTIAL EQUATIONS MIKE WILLS CONTENTS 1 Introduction 1 2 Alternative Texts 2 3i Calculus 3 Notation and Facts 2 4i Picardls Existence and Uniqueness Theorem 4 5i Separable Differential Equations 6 6i Autonomous Equations 7 7i Qualitative Analysis Isoclines 8 8 Linear Homogeneous Equations 9 9 The Superposition Principle and Inhomogeneous Equations 9 10 Method of Undetermined Coef cients 10 11 Integrating Factor 11 12 Variation of Parameters 12 13 Other First Order Equations 13 14 Exact Equations l4 1 INTRODUCTION As your correspondent was writing up lecture notes and selecting exercises for chapter 2 he realized that the presentation given in the text was not to his liking The purpose of this handout is to give a summary of the key ideas in what he hopes is a somewhat more orderly manneri Before we proceed we should make something clear for most math classes below the 3000 level at this university the textbooks are selected by the department As a professor your correspondent can choose any text that he pleases provided the said text covers the material that is needed for the course however it is usually best to go with the department recommended texti As such our textbook was selected passivelyi Having gone through the rst two chapters of the text your correspondent can safely say that he is not impressed The book uses a lot to say a little and the order of topics in chapter 2 is in this writer7s view confusing More importantly it is dif cult to identify where the book placed several of the key algorithmsi This is a problem in a introductory analytically oriented differential equation course such as ours where cookbook recipes are de rigeuri Hence the purpose of the handout is to present the core material from chapter 2 in an orderly logically consistent manner The examples in the text are reasonable so the presentation below will have relatively few examples Also the goal is to present the material in 1 2 MIKE WILLS an elementary manner As such mathematical rigor is given relatively short shrift Your instructor hopes that you nd the exposition helpful orderly and elegant Since not every one has taken vector calculus some concepts from vector calculus which we will need from time to time are also discussed 2 ALTERNATIVE TEXTS Here are some alternative texts that the readers may nd usefuli The last time your professor taught DEs it was out of BlanchardDevaneyHall BDH The isbn is 0495012653 Your professor likes the book a lot more than BoyceDiprima and now wishes held used this book despite its hefty price tag and even though supplemental material would be needed It should be pointed out that Amazon reviews for both our book and BDH are similar Another book that was considered is by Zill whose isbn is 0534418783 Some people swear by the book but it too is expensive and has decidedly mixed reviews s a graduate student teacher at UCSB your correspondent used Differential Equations and Linear Algebra by Farlow Hall McDill and West The isbn is 0131860615 Unfortunately it s not suitable for our course which is a pure diei coursei Also it seems to be especially unpopular with reviewersi For the reader with a background in linear algebra this would be a good reference For a more advanced perspective Hirsch and Smale is a very good text The isbn is 0123495504 The text assumes that the reader has had linear algebra and calculus 3 and perhaps even some real analysis The text by Perko isbn 0387974431 is also pretty good but it is arguably more difficult than Hirsch and Smalei Ultimately the best supplemental text for most readers would probably be Schaumls outline whose isbn is 0071456872 Notice that it is cheap Regardless of which differential equations texts one uses one would be well advised to have one7s calculus book onhand for the entire course 3 CALCULUS 3 NOTATION AND FACTS Differential equations involve to a certain extent the calculus of functions of more than one variable For the most part this is not a big deal since the ne details of vector calculus are not required for our course This section summarizes the key ideas that we will need Notation 31 The set R will for this handout and for our course be a rectangle in the plane Speci cally 31 RtyElR2la tgbc y dabgtltcdi To avoid silly irritations we will assume that a lt t lt b and that c lt y lt d The interior of R is 32 R0 ty E R2 l a lt tlt bclt y lt d ab gtlt cdi We will also assume that the point t0y0 is in Roi We will often have cause to deal with realvalued functions of more than one independent variable Speci cally functions whose domain is the rectangle R and whose codomain is i FIRST ORDER DIFFERENTIAL EQUATIONS 3 Example 32 ft y t2y3 lt11 m y 3t 7 ye iii f t y 0086 7 y M f t y i In the rst three examples R could be any rectangle that we please In the last example R could be any rectangle that does not intersect the line y it in the ty plane Remark 33 The book sometimes takes t be the rst independent variable and sometimes I Where feasible we will stick with t As in single variable calculus the idea of a limit is central in multi variable calculus De nition 34 Let f R A R be a given function Let t1 yl E R Then 33 ft7y L lim tgty t1y1 if given 6 gt 0 there exists 6 gt 0 such that lfty 7 Ll lt 6 whenever 34 xt7t12y7y12 lt6 and t y E R The de nition of a limit is a bit tricky to work with directly Fortunately we will largely deal with classes of functions in which the limit takes a back seat De nition 35 Let f R A R be a given function Let t1y1 E R Then f is continuous at t1y1 if 35 10057 9 7517 11 lim tgty t1y1 In many situations we will assume that the function under inspection f is continuous on all of R in which case we just say that f is continuous De nition 36 Suppose that f R A R is given At a point ty in the interior of R the partial derivative of f with respect to t is fth yft y 6f 36 l 4 7 33 h at f whenever the limit exists Similarly the partial derivative of f with respect to y is t h 7 t 8 3 7 i wyjyf j whenever the limit exists For the most part in this course we can compute partial derivatives using the rules we learned from calculus 1 For example if we compute the partial derivative with respect to t we simply pretend that y is a constant and take the derivative with respect to t Example 37 If fty t2y3 then ft Ztyg and fy 3t2y2 In singlevariable calculus one of the most important theorems is the chain rule There is a similar result in multi variable calculus which is typically presented as several distinct results We state here a version that is used in out text 4 MIKE WILLS Theorem 38 Chain Rule Suppose t and y Suppose that the partial derivatives of f R A R exist and that 45 and 1 are di erehtiable with respect to 739 Then f is di erehtiable with respect to 739 and df 38 i lt d7mhw where in this case the dot deiivative is with respect to 739 In particular ift 739 we df 39 7 lt gt w h If f is continuous we can talk about its partial integrals One can de ne partial integrals in terms of Riemann sums muc as we did in calculus However for our purposes all we need to do is integrate f with respect to one of the variables pretending that the other variable is constant There is a slight twist which we illustrate in the following examp e Example 39 Let fty t3y2 y We compute the partial integrals f fat and f fdyA 74y2 aw wweimww 3 10 4 tsys y2 fdy t3y2 ydy 7 3 t7 where 45 is an arbitrary function of y and 1 is an arbitrary function of t The point is that when partially integrating functions with more than one independent variable the arbitrary constant so beloved from calculus becomes an arbitrary function of the variables that are not involved in the integration 4 PICARD S EXISTENCE AND UNIQUENESS THEOREM The general rst order ordinary differential equation ODE is an equation of the form 41 Ft y yquot 0 Notice that F can be thought of as a function of three variables The primary objective is to nd a function y that solves this equation ldeally 45 will be differentiable and typically we hope that 45 is continuous as well Sometimes this is possible and sometimes it is not Equation 41 is a bit too general for our purposes The generic equation that we will work with is 42 y fty where we will typically assume that f is continuous on some rectangle R a b X 67 dl The corresponding initial value problem is y39 ft y We 907 where the point t0y0 is in the interior of R 43 FIRST ORDER DIFFERENTIAL EQUATIONS 5 A primary reason to study differential equations is because they model real world phenomena The only justi cation for such models is that they are expected to workl This means that the models are soluble in some sense and make accurate predictions In our context this translates to requiring at the very least that the initial value problem 43 have a unique solution There is a general theorem which we now state that gives suf cient conditions to guarantee this happy ending Theorem 41 Picard7s Existence and Uniqueness Theorem Consider the initial value problem ff and fy are continuous2 in R then there exists some interval I to 7 ht0h C ab and a function y such that 45 solves uniquely on the interval I Proof The proof of the result is sketched in the extra credit assignment from section 28 The basic idea is to de ne 44 9175 90 SHOWS and inductively 45 yn1t W fsynsds Then one shows that the sequence of functions converges near to to a solution of 43 Finally one shows that the difference between any two solutions of 43 is zero thus guaranteeing that there is really only one solution There are several potential pitfalls which have to be dealt with when proving the result The full details require a good deal of real analysis Some of the pitfalls are addressed in the text D Remark 42 To guarantee existence to a solution of the initial value problem 43 Peano showed that one only needs f to be continuous To guarantee uniqueness one needs a condition called Lipschitz continuity which is de ned in the exercises Remark 43 Picard7s theorem can be extended to rst order systems of ODEs Since any ODE can be converted to a system of ODEs as illustrated in the following example it then follows that we have a general existence and uniqueness theorem for ODEs Similar results for partial differential equations PDES require signi cant strengthening of the hypotheses and thus it is reasonable to state that there is no general existence theorem for PDEs Example 44 Convert the second order ode 46 v y 0 into a rst order system Solution 45 Let v y39 Then 47 v y 7y 1This is a paraphrase of something John Von Neumann a demigod to mathematicians said 2It is of passing interest that the continuity of f y does not guarantee the continuity of f 6 MIKE WILLS Hence a rst order system corresponding to equation 46 is given by y v 48 v iyi Having established a reasonable existence theorem we turn now to the question of actually nding the solutions One way to doing that is to explicitly compute the sequence of functions yn discussed in the proof of Picard s theorem and then nd the limit In practice this is usually Cll 39lCUlt to do analytically although it is a valid numerical technique For more systematic analytic approaches we need to classify and analyze rst order ODEsI 5i SEPARABLE DIFFERENTIAL EQUATIONS The simplest class of rst order ODEs to solve analytically are those that are separable De nition 51 A rst order ODE is separable if it can be written in the form 51 y ftgy We will assume for the purposes of this discussion that f and g are continuous From calculus we are already familiar with the special case when gy l for any y In that situation equation 51 reduces to 52 y39 fa which has general solution 53 y ftdt C where C is an arbitrary constant The hypotheses of Picardls theorem are satis ed and so the initial value problem 9 f t 54 9050 yo has a unique solution given by as y I fsds yo The reader will check that this is indeed a solution The only issue in this special case is that it may be difficult or impossible to nd an antiderivative for f in which case the computation of f f will be at best highly nontriviali We will informally call an equation of the form y39 directly integrable We turn now to the general separable case The equilibrium solutions of 51 are the roots of gy 0 If gy 0 has no roots there are no equilibriai If we assume that exists and is continuous then by Picardls theorem the graphs in the direction eld for the ODE of the nonconstant solutions do not intersect the graphs of the constant solutions In this situation the following algorithm works and can be justi ed using the product ruleI FIRST ORDER DIFFERENTIAL EQUATIONS 7 Since gy is continuous My 2 is continuous away from the zeros of g Since and My are continuous there exist functions Ft and such that F f and h We compute y ftyy dy 7 7 ftdt 56 WW WW hwy ftdt HmFma where C is an arbitrary constant Given an initial condition yt0 yo the initial value problem 9 f tyy 5 7 9750 yo has by Picard s theorem a unique solution provided 2 is continuous In this situation if yo is a zero of g the solution is yt yer If yo is not a zero of y then H is invertible near yo C Hy0 7 Ft0 and 58 W H1Flttgt Hltyogt 7 Fm This algorithm called separation of variables always works in principal but there are a few pit a s I It is easy to forget to check for the constant solutionsi Further it may be impossible to nd the roots ofg explicitlyi 2 It may be dif cult or impossible to compute H andor F explicitlyi 3 The last equation in 56 is usually valid only for the nonconstant solutions Hence the general solution to y39 is the last equation in 56 together with the equilibriai 4 Even if one has computed H and F explicitly nding H 1 may not be possible 5 Separation of variables may work even if gy is not continuous as long as g is continuous However one will likely lose uniqueness of solutions A canonical example is given in example 3 on page 71 in the text 6 AUTONOMOUS EQUATIONS A special case of a separable rst order ODE is an autonomous rst order ODEI De nition 61 A rst order autonomous ODE is an ODE that can be written in the form 61 y 9y Hence l and using the notation of the previous section Ft ti When 2 is continuous the solution to the initial value problem 9 99 6 2 we yo 8 MIKE WILLS is yt yo ifgy0 0 lfgy0 y 0 the solution is 63gt m Hm Hltyogt e to where the notation from the previous section is in use 7 QUALITATIVE ANALYSIS lSOCLINES ln principal separable rst order ODEs are about the easiest class of DEs to solve analytically However because of the difficulties inherent in integration the practice is often rather different Even when analytic methods work it is often helpful to draw pretty pictures to help get a feel for how solutions behave The simplest qualitative tool is the direction eld Drawing direction elds by hand can be tiresome However there are some techniques which cut down on the tedium Consider the rst order ODE 71 y fty De nition 71 Let c E R The set LC ty l fty c is called a level set or level curve of In the context of qualitative analysis of ODEs LC is also known as an isocline Typically LC will be a smooth curve or union of curves in the ty plane The key point is that all the slope lines in the direction eld that lie on LG will have the same slope namely 0 Hence sketching a direction eld reduces to sketching level curves and then on each level curve LC draw a series of slope lines each with slope c This is much easier than working out the slope at a bunch of points in a grid The only downside is that it adds an extra layer of information which may confuse matters Example 72 If ft y y7t2 then the isoclines are parabolas given by y ct2 Each parabola is centered on the yaxis and has vertex 0 0 Hence the slope lines on a given parabola have slope 0 Example 73 If ft y is a constant function with fty a then there is only one level set and it is the entire plane All the slope lines have the same slope namely a Example 74 If fty gy the ODE is autonomous the isoclines are hori zontal lines provided 9 is not constant Example 75 If ft y the ODE is directly integrable the isoclines are vertical lines provided f is not constant Example 76 If fty ty then the t and y axes are the isoclines for c 0 The other isoclines are hyperboli of the form y where c y 0 We sketch these examples at the end of the handout Notice in particular that sketching direction elds for autonomous equations is relatively easy FIRST ORDER DIFFERENTIAL EQUATIONS 9 8 LINEAR HOMOGENEOUS EQUATIONS De nition 81 A rst order ODE that can be written in the form 81 9 alttgty W is said to be linear If bt 0 then the ODE is in addition homogeneous If bt 0 the ODE is 39 or We will assume that a and b are continuous functions of t In this situation the initial value problem 9 aty W We yo has as the reader will check a unique solution by Picardls theorem The goal over the next few sections is to nd methods of nding solutions We start with the homogeneous equation Let us write Ly y39aty We would like to solve Ly 0 Suppose that yl and yg both solve Ly 0 Suppose further that c and d are real numbers Then LCyi dy2 Cyi dy2 atcy1 dy2 Cyi dy catyl daty2 8 3 Cy39i 04391 dy391 atyi 51491 dLy2 LCyi dy2 0 Hence Cyi dyg also solves Ly 0 This result is called the superposition principle We have shown in the jargon that L is a linear operator Notice that y39 aty 0 is separable and that y 0 is the only constant solution Using separation of variables the reader will carry out the details one can show that any solution including the constant solution can be written in the form 84 y Ce Am Ceifamdt where C is an arbitrary constant and A is an antiderivative of a This is the general solution of the general linear homogeneous rst order ODE We will denote this general solution by yh For reasons which elude this writer the book does not discuss the superposition principle until the chapter on second order equations Although one does not need the superposition principle for rst order ODEs your correspondent thinks that it brightens the discussion of linear equations Example 82 If at a is constant then yh 06quotquot and C yh0 82 9 THE SUPERPOSITION PRINCIPLE AND INHOMOGENEOUS EQUATIONS We now consider the ODE 91 9 alttgty blttgt where bt 0 In the notation of the previous section we are now studying the ODE 92 Ly my 10 MIKE WILLS Theorem 91 Suppose yp is a particular solution of 91 Then y yh yp is the general solution of 91 Proof We are given that Lyp bt and we know that Lyh 0 Hence 9 3 Lyh yp Lyh Lyp 0 W Thus yh yp solves 91 Conversely if Ly1 Mt that is yl is any particular solution then 94 MM 7 up 1191 MW W i W 0 and so yl 7 yp solves the homogeneous equation Ly 0 Consequently there exists a solution yk of the homogeneous equation such that yl yk ypl D The proof shows rst that yh yp is a solution of 91 and then shows that any solution of 91 can be written in the form yk yp for some yk a speci c solution to Ly 0 This is exactly what we needed to show This result is sometimes referred to as the extended superposition principle or extended linearity principle Since computing yh is straightforward we now search for ways of nding partic ular solutions to 9 1 10 METHOD OF UNDETERMINED COEFFICIENTS Example 101 If y ay b where a f 0 and b are constants then yp Q is a particular solution Hence the general solution is a b 101 y 06 quot a a Notice that C y0 7 g This was essentially a lucky guessi If a is constant and b is a nice function lucky guesses like this can often work The fancy name for the lucky guesses the method of undetermined coef cients Here is the basic idea Consider the equation 102 y39 ay btl Here a is constant and b is continuous We guess that yp looks similar to bt perhaps differing by some constants which we can determine by plugging our guess into the ODE The process is best illustrated with examples Example 102 Suppose y39 y e2 l Here bt e2 l We guess that yp mew Now y39p 2me2 l Plugging yp into the left hand side of the ODE we get 103 y yp 2me2 mew Smeml Hence yp solves the ODE provided 3m 1 that is m Hence the general solution is y Cequot Example 103 Suppose that y39 7 2y t2 The right hand side is a polynomial of degree 2 We therefore guess that yp mt2 nt k where mn and k are to be determined Now y39p th n Plugging yp into the left hand side of the ODE we get 104 y 72y 2mtn72mt27nt7k72mt2 2m7nt7kl Z7 Z7 FIRST ORDER DIFFERENTIAL EQUATIONS 11 Hence if yp is to solve the ODE we nd by equating coefficients that 72m l 2min 0 and 7k 0 Hence m 7 n 71 and k 0 Thus yp igit solves the ODE The general solution is 2 105 y 062 7 t3 7 ti Here is a summary of what will often work Throughout we assume that a is constant i If bt is a polynomial of degree n Take yp to be a polynomial of degree n with coefficients to be determined ii If bt lei try yp me where m is to be determined This will not work if a iki If bt k1 sinlt k2 coslt try yp m1 sinlt 7212 coslt where m1 and 7212 are to be determine iv If bt lei t try yp mlek mgt 7213 where 77117712 and 7213 are to be determined The interested reader will no doubt nd further generalizations The method breaks down if a is not constant or if one7s guess for yp turns out to be a solution to the homogeneous equationi There are various ways of dealing with this but the bottom line is that the method is simply not that systematic This is probably why the text barely gives the method a mention until chapter 3 in the context of second order equations However this method has the advantage of being fairly simple for sufficiently nice 12 The more systematic methods are not as simp e 11 INTEGRATING FACTOR Once again we look at the ODE 111 9 alttgty blttgt where a and b are continuous Let 6 be a nonzero continuous function Multi plying the ODE by M we obtain 112 W May 177 where we have suppressed the arguments of the functions a b and it since they are starting to irritate usi Suppose that we can choose p in such a way that 113 y W May By the product rule this currently hypothetical equation becomes 114 WWW W39Hwy Then 5 y May and so 116 Ma away from y 0 We now have a linear homogenous ODE in Mi If A is an antiderivative of a then 0 6A0 solves our ODE In other words we have found our candidate Mi Having done so we now note that 11 MW W May Mb 12 MIKE WILLS Integrating both sides with respect to t we nd that My fcb and so y i MIL If we want to nd the general solution as the text does then we will take all antiderivatives of M12 in the last equation If however we think that the superposi tion principle is pretty neat we will take the constant of integration to be zero to get a particular solution Either way will work Equation 33 on page 36 of he text gives the general solution The function u is called an integrating factori Here is the algorithm called the integrating factor method using the ap proach in this handout 1 Compute Mt efamdt where we take the constant of integration to be zero 2 Then Mtypt f Mtbtdt where the constant of integration is again zero 3 Hence ypt i f Mtbtdti 4 The general solution to y39 aty bt is l 67 mm L L L 118 in f tctbtdt Ma mwWWW The key advantage to the integrating factor method is that it is systematic and always works in princip er There are three problems rst the integrating factor method involves computing two integrals neither of which may be easy to do However in that situation there s no analytic remedy in any case no method will give an explicit solution with no integral signsi Second if the integrating factor method works and gives a nice formula then with hindsight the method of undetermined coef cients would probably have worked as well at least if a is constant Third the integrating factor method does not generalize to higher order equationsi Conversely the method of undetermined coef cients and the method of variation of parameters discussed below both generalize quite well 12 VARIATION OF PARAMETERS The idea of this method is rather cleveri As usual we consider the equation 12 1 y39 aty bti The general solution to the homogeneous equation is yh Ce famdti To get a particular solution we replace C with a function vt and try to gure out what 1 must be that is yp v67 f war In this context C is a parameter which depends on the initial condition and by replacing C with v we are allowing the parameter to vary with time It is not at all obvious that this trick is going to work ti 122 Since we are searching for a particular solution we take the constants of in tegration to be zero We have just shown that when 1 fbtef Mahdi then yp v67 f 001 is a particular solution of y39 aty bti The general solution is 123 y o v Juana It is the fth line in the above calculation where the miracle takes place We get a nice calculation which allows us to compute v explicitly albeit in terms of integrals which as usual may be impossible to computer When employing variation of parameters for a speci c ODE we usually repeat the above process rather than memorize the formula for v at the end Some of the exercises in 21 deal with this method Variation of parameters is probably the hardest of the three methods to im plement but it is systematic unlike undetermined coef cients and generalizable unlike integrating factori As such variation of parameters is arguably the best method for linear equations The text discusses variation of parameters for second order equations in chapter 3 13 OTHER FIRST ORDER EQUATIONS Most rst order ODEs are neither separable nor lineari Hence none of the above methods are applicable to the general situation There are a variety of techniques which allow certain classes of ODEs to be converted to linear or separable ODEsi Some of these are discussed in the exercises in section 24 Most of these techniques are highly specialized do not generalize and are quite frankly ad hour One class of equations that we consider in detail in this class is the collection of exact equations However for most rst order ODE S the canonical way to arrive at solutions is via numerical techniques Eulerls method is a good start to this method Your correspondent is reasonably happy with the presentation of Eulerls method in the text so our last section in this handout will be on exact equationsi 14 MIKE WILLS 14 EXACT EQUATIONS De nition 141 Let R be a rectangle in R2 and suppose that M N R A R2 are continuous Suppose that there exists a function 1 R A R2 such that i th y c implicitly de nes y as a function oft for some arbitrary constant c a 11 371 M iii N Then the rst order ODE 141 M Ny39 0 is said to be exact Let 739 t so we can think of y and hence 1 as functions of 739 as well Let us differentiate the equation 1 c with respect to 739 By the chain rule in the case where t 739 mm c gum 0 142 3970 6t ny MNy390 where the dot derivative is with respect to 739 Thus the equation 1 c implicitly de nes a solution to the exact equation If My and Ny are both continuous then by Picard s theorem the solution will be unique for a given initial condition It is not immediately clear how to check whether such a 1 exists Luckily there is a theorem 261 in the text which will help We restate it here for completeness but the text does a good job of stating and proving the result Theorem 142 If M N Mg and N are all continuous on R then M Ny39 0 is exact and only if My NE In this situation 143 it Mdt Ndy Thus a straightforward way to check if our candidate ODE is exact if simply to compute Mat and dey If one can make these two partial integrals equal the ODE is exact and we have found an implicit solution The proof given in the text can also be mimicked for speci c situations

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