The Exact Form and General Integrating Factors

Transcription

1 7 The Exact Form an General Integrating Factors In the previous chapters, we ve seen how separable an linear ifferential equations can be solve using methos for converting them to forms that can be easily integrate In this chapter, we will evelop a more general approach to converting a ifferential equation to a form (the exact form ) that can be integrate through a relatively straightforwar proceure We will see just what it means for a ifferential equation to be in exact form an how to solve ifferential equations in this form Because it is not always obvious when a given equation is in exact form, a practical test for exactness will also be evelope Finally, we will generalize the notion of integrating factors to help us fin exact forms for a variety of ifferential equations The theory an methos we will evelop here are more general than those evelope earlier for separable an linear equations In fact, the proceures evelope here can be use to solve any separable or linear ifferential equation (though you ll probably prefer using the methos evelope earlier) More importantly, the methos evelope in this chapter can, in theory at least, be use to solve a great number of other first-orer ifferential equations As we will see though, practical issues will reuce the applicability of these methos to a somewhat smaller (but still significant) number of ifferential equations By the way, the theory, the computational proceures, an even the notation that we will evelop for equations in exact form are all very similar to that often evelope in the later part of many calculus courses for two-imensional conservative vector fiels If you ve seen that theory, look for the parallels between it an what follows 71 The Chain Rule The exact form for a ifferential equation comes from one of the chain rules for ifferentiating a composite function of two variables Because of this, it may be wise to briefly review these ifferentiation rules First, suppose φ is a ifferentiable function of a single variable y (so φ = φ(y) ), an that y, itself, is a ifferentiable function of another variable t (so y = y(t) ) Then the composite function φ(y(t)) is a ifferentiable function of t whose erivative is given by the (elementary) chain rule t φ(y(t))] = φ (y(t)) y (t) 129

2 130 The Exact Form an General Integrating Factors A less precise (but more suggestive) escription of this chain rule is! Example 71: Then an Let t sin( t 2) = φ y φ(y(t))] = t y t φ y φ(y(t))] = t y t y(t) = t 2 an φ(y) = sin(y) φ(y(t)) = sin ( t 2), = y sin(y)] t t 2 ] = cos(y) 2t = cos ( t 2) 2t (In practice, of course, you probably o not explicitly write out all the steps liste above) Now suppose φ is a ifferentiable function of two variables x an y (so φ = φ(x, y) ), while both x an y are ifferentiable functions of a single variable t (so x = x(t) an y = y(t) ) Then the composite function φ(x(t), y(t)) is a ifferentiable function of t, an its erivative can be compute using a chain rule typically encountere later in the stuy of calculus; namely, φ(x(t), y(t))] = t t + y t (71) In practice, it is usually easier to compute this erivative by simply replacing the x an y in the formula for φ(x, y) with the corresponing formulas x(t) an y(t), an then computing that formula of t to compute the above erivative Still, this chain rule (an other chain rules involving functions of several variables) can be quite useful in more avance applications Our particular interest is in the corresponing chain rule for computing φ(x, y(x))], which we can obtain from equation (71) by simply letting x = t Then t an equation (71) reuces to = 1, y = y(t) = y(x), φ(x, y(x))] = + y y t = y, (72) For brevity, we will henceforth refer to this formula as chain rule (72) (not very original, but better that constantly repeating the chain rule escribe in equation (72) ) Don t forget the ifference between / an / If φ = φ(x, y), then while φ = the erivative of φ(x, y) assuming x is the variable an y is a function of x = the erivative of φ(x, y) assuming x is the variable an y is a constant

3 The Exact Form, Define 131! Example 72: Then = an, by chain rule (72), Assume y is some function of x (ie, y = y(x) ) an y 2 + x 2 y ] = 2xy, φ(x, y(x))] = φ(x, y) = y 2 + x 2 y + y If, for example, y = sin(x), then the above becomes = y 2 + x 2 y ] = 2y + x 2, = 2xy + 2y + x 2] y φ(x, y(x))] = 2x sin(x) + 2 sin(x) + x 2] cos(x) On the other han, if y = y(x) is some unknown function, then, after replacing φ with its formula, we simply have y 2 + x 2 y ] = 2xy + 2y + x 2] y (73) In our use of chain rule (72), y will be an unknown function of x, an the right sie of equation (72) will correspon to one sie of whatever ifferential equation is being consiere 72 The Exact Form, Define Let R be some region in the XY plane We will say that a first-orer ifferential equation is in exact form (on R ) if an only if both of the following hol: 1 The ifferential equation is written in the form (74a) where M(x, y) an N(x, y) are known functions of x an y an 2 There is a ifferentiable function φ = φ(x, y) on R such that at every point in R = M(x, y) an = N(x, y) (74b) We will refer to the above φ(x, y) as a potential function for the ifferential equation 1 In practice, the region R is often either the entire XY plane or a significant portion of it There are a few technical issues regaring this region, but we can eal with these issues later In the meantime, little harm will be one by not explicitly stating the region Just keep in min that if we say an certain equation is in exact form, then it is in exact form on some region R, an that the graph of any solution y = y(x) erive will be restricte being a curve in that region 1 Referring to φ as a potential function comes from the theory of conservative vector fiels In fact, it is not common terminology in other ifferential equation texts Most other texts just refer to this function as φ

4 132 The Exact Form an General Integrating Factors! Example 73: Consier the ifferential equation 2xy + 2y + x 2] y (75) This equation is in the form with M(x, y) = 2xy an N(x, y) = 2y + x 2 Moreover, if we glance back at example 72, we immeiately see that, letting φ(x, y) = y 2 + x 2 y, we have an = = y 2 + x 2 y ] = 2xy = M(x, y) y 2 + x 2 y ] = 2y + x 2 = N(x, y) everywhere in the XY plane So equation (75) is in exact form (an we can take R to be the entire XY plane) In the next several sections, we will iscuss how to etermine when a ifferential equation is in exact form, how to convert one not in exact form into exact form, how to fin a potential function, an how to solve the ifferential equation using a potential function However, we will evelop the material backwars, starting with solving a ifferential equation given a known potential function, an working our way towars ealing with equations not in exact form This ens up being the natural (an least confusing) way to evelop the material But first, a few general comments regaring exact forms an potential functions: 1 Just being written as oes not guarantee that a ifferential equation is in exact form; there still might not be a φ(x, y) with = M(x, y) an = N(x, y) For example, we will iscover that there is no φ(x, y) satisfying = 3y + 3y3 an = xy2 x So is not in exact form 3y + 3y 3 + xy 2 x ] y

5 The Exact Form, Define A single ifferential equation will have several potential functions In particular, aing any constant to a potential function yiels another potential function After all, if 0 = M(x, y) an 0 = N(x, y) an φ 1 (x, y) = φ 0 (x, y) + c for some constant c, then, since any erivative of any constant is zero, 0 = 1 = M(x, y) an 0 = 1 = N(x, y) Moreover, as you will verify in exercises 72 an 73 (starting on page 154), any ifferential equation that can be written in one exact form will actually have several exact forms, all corresponing to ifferent (but relate) potential functions Because of the uniqueness results concerning solutions to first-orer ifferential equations, it oes not matter which of these potential functions is use to solve the equation Thus, in the following, we will only worry about fining a potential function for any given ifferential equation 3 Many authors refer to an equation in exact form as an exact equation We are avoiing this terminology because, unlike linear an separable ifferential equations, the exactness of an equation can be estroye by legitimate rearrangements of that equation For example, y + x 2 y = e 2x is a linear ifferential equation whether it is written as above or written as Consier, on the other han, y = e2x x 2 y 2xy + 2y + x 2] y As we saw in example 73, this ifferential equation is in exact form (ie, is an exact equation ) However, if we rewrite it as y = 2xy 2y + x 2, we have an equation that is not in exact form (ie, is not an exact equation )

6 134 The Exact Form an General Integrating Factors 73 Solving Equations in Exact Form Using a Known Potential Function Observe that, if φ = φ(x, y) is any sufficiently ifferentiable function of x an y, an = M(x, y) an = N(x, y), then the ifferential equation can be rewritten as + y Chain rule (72) then tells us that the left sie of this equation is just the erivative of φ(x, y(x)) So this last equation can be written more concisely as φ(x, y)] = 0 (with y = y(x) ) Not only is this concise, it is easily integrate: φ(x, y)] φ(x, y) = c Think about this last equation for a moment It escribes the relation between x an y = y(x) assuming y satisfies the ifferential equation In other wors, the equation φ(x, y) = c is an implicit solution to the ifferential equation for which it is a potential function All this is important enough to be restate as a theorem Theorem 71 (importance of a potential function) Let φ(x, y) be a potential function for a given first-orer ifferential equation ifferential equation can be written as Then that + y This, in turn, can be rewritten as which can be integrate, φ(x, y)] = 0 with y = y(x), φ(x, y)] = 0,

7 Solving Equations in Exact Form 135 to obtain the implicit solution where c is an arbitrary constant φ(x, y) = c The above theorem contains all the steps for fining an implicit solution to a ifferential equation that can be put in exact form, provie you have a corresponing potential function Of course, you have probably alreay notice that this theorem can be shortene to the following: Corollary 72 Let φ(x, y) be a potential function for a given first-orer ifferential equation Then φ(x, y) = c is an implicit solution to that ifferential equation In solving at least the first few ifferential equations in the exercises at the en of this chapter, you shoul use all the steps escribe in theorem 71 simply to reinforce your unerstaning of why φ(x, y) = c Then feel free to cut out the intermeiate steps (ie, use the corollary) An, of course, on t forget to see if the implicit solution can be solve for y in terms of x, yieling an explicit solution! Example 74: Consier the ifferential equation 2xy + 2y + x 2] y From example 73, we know this is in exact form an has corresponing potential function Since φ(x, y) = y 2 + x 2 y y 2 + x 2 y ] = 2xy an Our ifferential equation can be rewritten as y 2 + x 2 y ] + y 2 + x 2 y ] = 2y + x 2 ( y 2 + x 2 y ]) y By chain rule (72), this reuces to y 2 + x 2 y ] = 0 with y = y(x) Integrating this, y 2 + x 2 y ], yiels the implicit solution y 2 + x 2 y = c Rewriting this as y 2 + x 2 y c = 0 an then solving for y provies the explicit solution y = x2 ± x 4 + 4c 2

8 136 The Exact Form an General Integrating Factors Fining the Potential Function Let use now consier a more ifficult problem: Fining a potential function, φ(x, y), for a given ifferential equation that has been written in the form This means fining a function φ(x, y) satisfying both = M(x, y) an = N(x, y) A relatively straightforwar proceure for fining this φ will be outline in a moment But first, let us make a few observations regaring this pair of partial ifferential: 1 We are not assuming the given ifferential equation is in exact form, so there might not be a solution to this above pair of partial ifferential equations Our metho will fin φ if it exists (ie, if the equation is in exact form), an will lea to an obviously impossible equation otherwise 2 Because the above pair of equations are partial ifferential equations, we must treat x an y as inepenent variables Do not view y as a function of x in solving for φ 3 If you are acquainte with methos for recovering the potential for a conservative vector fiel, then you will recognize the following as one of those methos Now, here is the proceure: Basic Proceure for Fining a Potential Function Assume we have a ifferential equation in the form, For purposes of illustration, we will use the ifferential equation 2xy x ] y To fin a potential function φ(x, y) for this ifferential equation (if it exists), o the following: 1 Ientify the formulas for M(x, y) an N(x, y), an, using these formulas, write out the pair of partial ifferential equations to be solve, = M(x, y) an = N(x, y) (In oing so, we are naively assuming the given ifferential equation is in exact form) For our example, 2xy + 2 }{{} M(x,y) + } x 2 {{ + 4 } N(x,y) ] y So the pair of partial ifferential equations to be solve is = 2xy + 2 an = x 2 + 4

9 Solving Equations in Exact Form Integrate the first equation in the pair with respect to x, treating y as a constant = M(x, y) In computing the integral of M with respect to x, you get a constant of integration Keep in min that this constant is base on y being a constant, an may thus epen on the value of y Consequently, this constant of integration is actually some yet unknown function of y call it h(y) For our example, = 2xy + 2] φ(x, y) = x 2 y + 2x + h(y) (Observe that this step yiels a formula for φ(x, y) involving one yet unknown function of y only We now just have to etermine h(y) to know the formula for φ(x, y) ) 3 Replace φ in the other partial ifferential equation, = N(x, y), with the formula just erive for φ(x, y), an compute the partial erivative Keep in min that, because h(y) is a function of y only, h(y)] = h (y) Then algebraically solve the resulting equation for h (y) For our example: = N(x, y) x 2 y + 2x + h(y) ] = x x 2 + h (y) = x h (y) = 2 4 Look at the formula just obtaine for h (y) Because h (y) is a function of y only, its formula must involve only the variable y, no x may appear If the x s o not cancel out, then we have an impossible equation This means the naive assumption mae in the first step was wrong; the given ifferential equation was not in exact form, an there is no φ(x, y) satisfying the esire pair of partial ifferential equations In this case, Stop! Go no further in this proceure! On the other han, if the previous step yiels h (y) = a formula of y only (no x s ), then integrate both sies of this equation with respect to y to obtain h(y) (Because h(y) oes not epen on x, the constant of integration here will truly be a constant, not a function of the other variable)

10 138 The Exact Form an General Integrating Factors For our example, the last step yiele h (y) = 2 The right sie oes not contain x, so we can continue an integrate to obtain h(y) = h(y) y = 2 y = 2y + c 1 (We will later see that the constant c 1 is not that important) 5 Combine the formula just obtaine for h with the formula obtaine for φ in step 2 In our example, combining the results from step 2 an the last step above yiels φ(x, y) = x 2 y + 2x + h(y) where c 1 is an arbitrary constant = x 2 y + 2x + 2y + c 1 (Because of the arbitrary constant from the integration of h (y), the formula obtaine actually escribes all possible φ s satisfying the esire pair of partial ifferential equations If you look at our iscussion above, it shoul be clear that this formula will always be of the form φ(x, y) = φ 0 (x, y) + c 1 where φ 0 (x, y) is a particular formula an c 1 is an arbitrary constant But, as note earlier, we only nee to fin one potential function φ(x, y) So you can set c 1 equal to your favorite constant, 0, or keep it arbitrary an see what happens) If the above proceure oes yiel a formula φ(x, y), then it immeiately follows that the given equation is in exact form an φ(x, y) is a potential function for the given ifferential equation But on t forget that the goal is usually to solve the given ifferential equation, The function φ = φ(x, y) is not that solution It is a function such that the ifferential equation can be rewritten as φ(x, y)] = 0 Integrating this yiels the implicit solution φ(x, y) = c from which, if the implicit solution is not too complicate, we can obtain an explicit solution y = y(x)! Example 75: Consier solving 2xy x ] y

11 Solving Equations in Exact Form 139 As just illustrate, this equation is in exact form, an has a potential function φ(x, y) = x 2 y + 2x + 2y + c 1 where c 1 can be any constant So the ifferential equation can be rewritten as which integrates to ] x 2 y + 2x + 2y + c 1 = 0, x 2 y + 2x + 2y + c 1 = c 2 This is an implicit solution for the ifferential equation Solving this for y is easy, an (after letting c = c 2 c 1 ) gives us the explicit solution y = c 2x x Note that, in the last example, the constants arising from the integration of h (y) an φ / were combine at the en It is easy to see that this will always be possible Other Ways to Fin a Potential Function There are other ways to fin a function φ(x, y) satisfying both = M(x, y) an Two, in particular, are worth mentioning: = N(x, y) 1 The first is the obvious moification of the one alreay given in which the roles of / = M an / = N are interchange That is, instea of you integrating / = M with respect to x, an then plugging the result into / = N, integrate / = N with respect to y, an then plug the result into / = M The integration of / = N will yiel some formula involving x, y an an unknown function of x, g(x) Plugging this formula into / = M shoul yiel a orinary ifferential equation for g(x) which oes not contain y If the y s o not cancel out, the esire φ oes not exist Otherwise, g(x) can be obtaine by integration, an then combine with the formula just obtaine for φ(x, y) This is usually the preferre metho when N(x, y) y is much easier to compute than M(x, y) Inee, it usually is a goo iea to scan these two integrals an, if one looks much easier to compute, compute that one, an plug the result into the partial ifferential equation corresponing to the other integral Don t forget to check to see if equation resulting from that just involves the appropriate variable 2 The other metho is one often attempte by beginners who o not unerstan why it shoul not be use: First inepenently integrate both / = M an / = N Then stare at the two ifferent formulas obtaine for φ(x, y) (each involving a ifferent unknown function) an try to guess what single formula for φ(x, y) (without unknown functions) matches the results from the two integrations

12 140 The Exact Form an General Integrating Factors Fight any temptation to take this approach Yes, with a little luck an skill, you can get φ this way But it is usually more work, it oesn t easily warn you when φ(x, y) oes not exist, an is more likely to result in errors Why use a metho involving two two integrations an two unknown functions when you can use a metho involving just one one integration an one unknown function with a straightforwar way to etermine that one function? 74 Testing for Exactness Part I The proceure just iscusse for fining a potential function φ for (76) oes not tell us whether such a φ even exists until step 4, after a possibly tricky integration Fortunately, there is a simple test that that can often tell us when seeking that φ woul be futile This test is base on the fact that, for any sufficiently ifferentiable φ(x, y), 2 φ = 2 φ Now let R be any region in the XY plane on which φ is sufficiently ifferentiable an satisfies = M(x, y) an = N(x, y) Then, at every point in R, M = ] = 2 φ = 2 φ = ] = N So, for equation (76) to be in exact form over R, we must have M = N (77) at every point in R If this equation oes not hol, ifferential equation (76) is not in exact form over that region no corresponing potential function φ exists What we have not shown is that equality (77) necessarily implies that ifferential equation (76) is in exact form In fact, equality (77) oes imply that, given any point in R, the equation is in exact form over some subregion of R containing that point Unfortunately, showing that an escribing those regions takes more evelopment than is appropriate here For now, let us just say that, in practice, the equality (77) implies that ifferential equation (76) is probably in exact form over the given region, an it is worthwhile to seek a corresponing potential function φ via the metho outline earlier Let us summarize what has just been erive, accepting the term suitably ifferentiable as simply meaning that the necessary partial erivatives can be compute:

13 Testing for Exactness Part I 141 Theorem 73 (test for probable exactness) Let M(x, y) an N(x, y) be two suitably ifferentiable functions of two variables over a region R in the XY plane, an consier the ifferential equation 1 If 2 If M = N on R, then the above ifferential equation is not in exact form on R M = N on R, then the above ifferential equation might be in exact form on R It is worth seeking a corresponing potential function! Example 76: Here Consier the ifferential equation 3y + 3y 3 M = + xy 2 x ] y 3y + 3y 3 ] = 3 + 9y 2 an So N = xy 2 x ] = y 2 1 M = N telling us that the given ifferential equation is not in exact form over any region! Example 77: Here Consier the ifferential equation y x 2 + y 2 + x y x 2 + y 2 M(x, y) = y an N(x, y) = x 2 + y 2, x x 2 + y 2 are well efine an ifferentiable everywhere on the XY plane except (x, y) = (0, 0) So let s take R to be the entire XY plane with the origin remove Computing the partial erivatives, we get M = y ] x 2 + y 2 = 1(x2 + y 2 ) y(2y) (x 2 + y 2 ) 2 = y2 x 2 (x 2 + y 2 ) 2 an N = ] x x 2 + y 2 = 1(x2 + y 2 ) x(2x) (x 2 + y 2 ) 2 = y2 x 2 (x 2 + y 2 ) 2

14 142 The Exact Form an General Integrating Factors So these two partial erivatives are equal throughout R, an our test for probable exactness tells us that the given ifferential equation might be in exact form on R it is worthwhile to try to fin a corresponing potential function For many, the test escribe above in theorem 73 will suffice Those who wish a more complete test shoul jump to section 77 starting on page 150 (where we will also finish solving the ifferential equation in example77) 75 Exact Equations: A Summary To review: If you suspect that a given ifferential equation,, is in exact form, then you can quickly check for at least probable exactness by computing M / an N / an seeing if M = N If the two partial erivatives are equal, then follow the proceure for fining a potential function φ(x, y) outline on pages 136 to 138 If that proceure is successful an yiels a φ(x, y), then finish solving the given ifferential equation using the fact that the ifferential equation can be rewritten as φ(x, y)] = 0, the integration of which yiels the implicit solution φ(x, y) = 0 If this equation can be solve for y in terms of x, o so If the given ifferential equation is not in exact form, then there is a possibility that it can be put into an exact form using appropriate integrating factors We will iscuss these next By the way, on t forget that these equations may be solvable by other means For example, the equation use to illustrate the proceure for fining φ was a linear ifferential equation, an coul have been solve a bit more quickly using the methos from chapter 5

15 Converting Equations to Exact Form Converting Equations to Exact Form Basic Notions Obviously, the first step to converting a given first-orer ifferential equation to exact form is to get it into the form Then apply the test for (probable) exactness With luck, the test result will be positive More likely, it will not To see how we might further convert our equation to exact form, it may help to recall why we want the exact form It is so that the left sie of the ifferential equation can be ientifie as an orinary erivative of some formula of x an y(x), φ(x, y(x)) We ha a similar situation with linear equations Given a linear equation y + p(x)y = f (x), we foun that, after multiplying it by an integrating factor µ to get µ y + µpy = µf, we coul ientify the equation s left sie as a complete erivative of a formula of x an y(x), namely, µ(x)y(x)] The same iea can be applie to convert an equation not in exact form to one that is in exact form! Example 78: Consier the ifferential equation 3y + 3y 3 + xy 2 x ] y In example 76, we saw that this equation is not in exact form But look what happens after we multiply through by µ = x 2 y 2, ( x 2 y 2 3y + 3y 3 + xy 2 x ] ) y We get with 3x 2 y 1 + 3x 2 y }{{} M new (x,y) + x 3 x 3 y }{{ 2 } N new (x,y) ] y M new = 3x 2 y 1 + 3x 2 y ] = 3x 2 y 2 + 3x 2 = 3x 2 3x 2 y 2

16 144 The Exact Form an General Integrating Factors an So N new = x 3 x 3 y 2] = 3x 2 3x 2 y 2 M new = N new telling us that the equation is now (probably) in exact form (over any region where y never equals 0 ) We will refer to any nonzero function µ = µ(x, y) as an integrating factor for a first-orer ifferential equation M + N y if an only if multiplying that equation through by µ, µm + µn y, yiels a ifferential equation in exact form 2 This integrating factor may be a function of x or of y or of both x an y Notice that, because µm }{{} new M is exact, our test for exactness tells us that + µn }{{} y new N µm] = µn], Fining Integrating Factors General Approach Fining an integrating factor for an equation starts with the requirement just erive, µm] = µn] (78) Remember, M an N will be known formulas of x an y So equation (78) is a ifferential equation in which the unknown function is our integrating factor, µ Unfortunately, it is a rather nontrivial partial ifferential equation (unlike the partial ifferential equations in section 72), an a complete iscussion of how to solve it for µ is beyon the scope of any introuctory ifferential equations course Fortunately, there are some common cases where this partial ifferential equation reuces to an equation we can hanle The cases we will consier are where there is an integrating factor µ that is either as function of x only, or a function of y only, or a simple formula of x an y In all cases, the approach is basically the same: 2 You can show that the integrating factors foun for linear equations are just special cases of the integrating factors consiere here If there is any anger of confusion, we ll refer to the integrating factors now being iscusse as the more general integrating factors

17 Converting Equations to Exact Form Decie on which case you think is appropriate, an make the corresponing assumptions on µ 2 Expan equation (78) by computing out the erivatives as far as possible, taking into account the assumptions mae on µ 3 See if the resulting equation can be solve for a µ satisfying the assumptions mae If so, o so If not, start over using ifferent assumptions on µ (unless you ve run out of reasonable options) We ll illustrate the above approach in a moment Before that, however, a few more comments shoul be mae: 1 Only one integrating factor is neee So go ahea an assign convenient values to the arbitrary constants that arise in solving for µ (just as in fining an integrating factors for linear equations) 2 Once you have foun an integrating factor µ, remember why you wante it Use it to rewrite your ifferential equation in exact form, an then solve the ifferential equation as escribe earlier in this chapter 3 There are tests to etermine if there are integrating factors that are functions of just x or just y Moreover, there are formulas for µ that can be use if any of the tests are satisfie (see exercise 76 on page 156) DO NOT WASTE YOUR TIME TRYING TO USE THESE FORMULAS! They are har to memorize correctly an are worth learning only if you expect to compute many, many integrating factors over a relatively short time frame Chances are, you won t have that nee, just a nee to unerstan the basic concepts an to compute the occasional integrating factor Now, let s look at the common cases: First Case: µ Being a Function of x Only If you think the integrating factor µ coul be function of x only, then assume it, an see what happens when you compute out equation (78) Uner this assumption, µ = µ(x), µ = µ an µ = 0 Consequently, equation (78) shoul immeiately reuce to an orinary ifferential equation for µ Moreover, since µ is suppose to be just a function of x here, this ifferential equation for µ shoul not contain any y s Thus, if the y s o not cancel out, the assumption that µ coul be just a function of x is wrong go to the next case But if the y s o cancel out, then solve the ifferential equation just obtaine for a µ = µ(x)! Example 79: Consier the ifferential equation Here M 1 + y 3 + xy 2 y = 1 + y 3 ] = 3y 2 an N = xy 2 ] = y 2

18 146 The Exact Form an General Integrating Factors So M / = N / The equation is not exact, but, with luck, we can fin a function µ so that µ 1 + y 3] + µ xy 2] y is exact This integrating factor µ must satisfy ( ]) µ 1 + y 3 = ( ]) µ xy 2 Let s see if µ coul be a function of just x Assume µ = µ(x) Then µ = µ an µ = 0 Using this, we have µ ( ]) µ 1 + y 3 = ( ]) µ xy 2 ] 1 + y 3 + µ ] 1 + y 3 = µ ] xy 2 + µ ] xy y 3] + µ 0 + 3y 2] = µ ] xy 2 + µ y 2] 3y 2 µ = xy 2 µ + y2 µ The y s o cancel out, leaving us with 3µ = x µ + µ, which simplies to µ = 2µ x So there is an integrating factor µ that is a function of x alone Moreover, to fin such an integrating factor, we nee only fin one (nonzero) solution to the above simple, separable orinary ifferential equation Proceing to o so, we get 1 µ µ = 2 x ln µ = 2 ln x + c = ln x 2 + c µ = ±e ln x2 + c = Ax 2 Since only one nonzero integrating factor is neee, we can take A = 1, giving us µ(x) = x 2 as our integrating factor Unfortunately, there is always the possibility that the y s will not cancel out

19 Converting Equations to Exact Form 147! Example 710: It is easily verifie that 6xy + 5 x 2 + y ] y is not in exact form To be a corresponing integrating factor, µ must satisfy Assume µ is a function of x only Then an, thus, ( ) ( µ6xy] = µ5 x 2 + y ]) µ = µ an µ = 0, ( ) ( µ6xy] = µ5 x 2 + y ]) µ 6xy] + µ µ 6xy] = 5 x 2 + y ] + µ ( 5 x 2 + y ]) 0 6xy + µ6x] = µ 5 x ] + µ10x] µ = 10xµ 6xµ 4x 2 + 5y = 4xµ 4x 2 + 5y Here, the y s o not cancel out, as they shoul if our assumption that µ epene only on x were true Hence, that assumption was wrong This equation oes not have an integrating factor that is a function of x only We will have to try something else Secon Case: µ Being a Function of y Only This is just like the first case, but with the roles of x an y switche If you think the integrating factor µ coul be function of y only, then assume it, an see what happens when you compute out equation (78) Uner this assumption, µ = µ(y), µ = 0 an µ = µ y Again, equation (78) shoul immeiately reuce to an orinary ifferential equation for µ, only this time the ifferential equation for µ shoul not contain any x s If the x s o not cancel out, our assumption that µ coul be just a function of y is wrong an we can go no further with this hope But if the x s o cancel out, then solving the ifferential equation just obtaine for a µ = µ(y) yiels the esire integrating factor! Example 711: As just seen in example 710, 6xy + 5 x 2 + y ] y oes not have an integrating factor epening only on x So instea, let s try to fin one that epens on just y Assuming this, µ = µ(y), µ = 0 an µ = µ y

20 148 The Exact Form an General Integrating Factors Combining this with equation (78): ( ) ( µ6xy] = µ 5x 2 + 5y ]) µ 6xy] + µ ] µ 6xy = 5x 2 + 5y ] + µ 5x 2 + 5y ] µ y 6xy] + µ6x] = 0 5x 2 + 5y ] + µ10x] µ = 10xµ 6xµ 6xy = 2µ 3y The x s cancel out, as hope, an we have a simple ifferential equation for µ = µ(y) Solving it: 1 µ µ y y = 2 3y y ln µ = 2 3 ln y + c = ln y 2 / 3 + c µ = Ay 2 / 3 Taking A = 1 gives the integrating factor µ(y) = y 2 / 3 Thir Case: µ Being a Simple Function of Both Variables Of course, it is quite possible that no function of just x or just y will be an integrating factor for a given equation In this case, the best we can usually hope for is that there is a relatively simple function of x an y that will work as an integrating factor This means making a goo guess at µ(x, y), an verifying that it satisfies equation (78) One guess sometimes worth trying is µ(x, y) = x α y β where the exponents, α an β, are constants to be etermine To etermine the values of the constants so that the guess works (if such constants exist), just plug this formula for µ into equation (78) an see if it reuces to an equation that can be solve for α an β If so, fin those values an use µ(x, y) = x α y β (with the values just foun for α an β ) as your integrating factor If not, keep searching (or consier ealing with the ifferential equation using the graphical an numerical methos we ll iscuss in chapter 8)! Example 712: Consier the ifferential equation from example 76, 3y + 3y 3 + xy 2 x ] y

21 Converting Equations to Exact Form 149 Plugging µ(x, y) = x α y β into equation (78) for our ifferential equation yiels: (µm) = (µn) ( x α y β 3y + 3y 3]) = ( x α y β xy 2 x ]) ( 3x α y β+1 + 3x α y β+3) = ( x α+1 y β+2 x α+1 y β) 3(β + 1)x α y β + 3(β + 3)x α y β+2 = (α + 1)x α y β+2 (α + 1)x α y β Combining like terms then gives which, in turn, hols if an only if 3β + α + 4]x α y β + 3β α + 8]x α y β+2 = 0, 3β + α + 4 = 0 an 3β α + 8 = 0 This last pair of equations constitute a simple system of linear equations, 3β + α + 4 = 0 3β α + 8 = 0 which can be easily solve by any of a number of ways, yieling α = 2 an β = 2 Thus, the ifferential equation we starte with, 3y + 3y 3 + xy 2 x ] y, oes have an integrating factor of the form µ(x, y) = x α y β, an it is µ(x, y) = x 2 y 2 (just as was use in example 78 on page 143)

22 150 The Exact Form an General Integrating Factors 77 Testing for Exactness Part II Simple Connectivity an the Complete Test for Exactness A more complete test for exactness than given in theorem 73 can be escribe if we are more careful about escribing our situation So suppose we have an equation, an that, on some open region R of the XY plane, all of the following hol: 1 The functions M(x, y) an N(x, y), along with the erivatives M / an N /, are continuous everywhere in R 2 At each point (x, y) in R, M = N That region R will sai to be simply connecte if each an every simple close curve (ie, loop) in R encloses only points in R If any simple close curve in R encloses any point not in R, then we will say that R is not simply connecte If you think about it, you will realize that saying a region is simply connecte is just a precise way of saying that the region has no holes An if you think a little more about the situation, you will realize that, if our open region R has holes (ie, is not simply connecte), then it is probably because these are points where M(x, y) or N(x, y) or their partial erivatives fail to exist! Example 713: Again, consier the ifferential equation y x 2 + y + x y 2 x 2 + y 2 As note in example 77, M(x, y) = y x 2 + y 2 an N(x, y) = x x 2 + y 2 are well efine an ifferentiable an satisfy M = N everywhere in the region R consisting of the XY plane with the origin remove Removing this point (the origin) creates a hole in R This point is also a point not in R but which is enclose by any loop in R aroun the origin Now we can state the full test for exactness (Its proof will be briefly iscusse at the en of this section) For the intereste reaer

23 Testing for Exactness Part II 151 Theorem 74 (complete test for exactness) Let R be a simply-connecte open region in the XY plane, an let M(x, y) an N(x, y) be two continuous functions on R whose partial erivatives are also continuous on R Then is in exact form on R if an only if at every point in R M = N This theorem assures us that, if our region R is simply connecte, then we can (in theory at least) use the proceure outline on pages 136 to 138 to fin the corresponing potential function φ(x, y) on R, an from that, erive an implicit solution φ(x, y) = c to our ifferential equation Theorem 74 oes not efinitely say the ifferential equation is not in exact form if R is not simply connecte Whether the equation is or is not be in exact form over all of R is still uncertain What is certain, however, is the following immeiate consequence of theorem 74 Corollary 75 Assume M(x, y) an N(x, y) are two continuous functions on some open region R of the XY plane Assume further that, on R, the partial erivatives of M an N are continuous an satisfy M = N Then is in exact form on each simply-connecte open subregion of R Thus, even if our original region is not simply connecte, we can at least pick any open, simply connecte subregion R 1, an (in theory at least) use the proceure outline in section 73 to fin a corresponing potential function φ 1 (x, y) on R 1, an from that, erive the implicit solution φ 1 (x, y) = c to our ifferential equation, vali on subregion R 1 But then, why might there not be a potential function φ(x, y) vali on the entire region R? Let s go back to an example starte earlier to see! Example 714: Let us continue our consieration of the ifferential equation y x 2 + y + x y 2 x 2 + y 2 from example 77 As just note in the previous example, the region R consisting of all the XY plane except for the origin (0, 0) is not simply connecte But we can partition it into the left an right half-planes R + = {(x, y) : x > 0} an R = {(x, y) : x < 0}, which are simply connecte Theorem 74 assures us that our ifferential equation is in exact form over each of these half-planes, an, inee, you can easy show that all the corresponing

24 152 The Exact Form an General Integrating Factors potential functions on these regions for our ifferential equation are given by ( y ) φ + (x, y) = Arctan + c + on R + x an ( y ) φ (x, y) = Arctan x + c on R where c + an c are arbitrary constants But coul there be a potential function φ on all of R corresponing to our ifferential equation? If so, then φ woul also be a potential function over the left an right half-planes R + an R, an, as just note, there woul be constants c + an c such that ( y ) φ(x, y) = Arctan + c + for x > 0 x an ( y ) φ(x, y) = Arctan x + c for x < 0 Since φ must be continuous everywhere except at the origin, it must, in particular, be continuous at any point on the positive Y axis, (0, y) with y > 0 So, letting x 0 from the positive sie, we have ( y φ(0, y) = lim φ(x, y) = lim Arctan x 0 + x 0 + x ) + c + Using the substitution t = y / x an recalling the limiting values of the Arctangent function, we can rewrite the above as φ(0, y) = lim Arctan(t) + c + = π t c + Likewise, letting x 0 from the negative sie, we have Together, the above tells us that φ(0, y) = lim φ(x, y) x 0 = lim x 0 Arctan ( y x ) + c = lim t Arctan(t) + c = π 2 + c π 2 + c = φ(0, 1) = π 2 + c +, which, of course, means that c = π + c + Because of the arbitrariness of the constants ae to potential functions, we may, for simplicity, let c + = 0 Then ( ) y Arctan if x > 0 x ( ) φ(x, y) = y Arctan + π if x < 0 x π 2 if x = 0 an y > 0 But look at what must now happen at a point on the negative Y axis, say, at (0, 1) ( ) 1 lim φ(x, 1) = lim Arctan = lim x 0 + x 0 + x Arctan(t) = π t 2

25 Testing for Exactness Part II 153 an ( ) 1 lim φ(x, 1) = lim Arctan x 0 x 0 x + π = lim t + Arctan(t) + π = 3π 2 So lim φ(x, 1) = lim φ(x, 1) x 0 + x 0 Thus, there are points in R at which φ(x, y) is not continuous, contrary to the fact that a potential function on R must be continuous everywhere on R An thus, the answer to our question Coul there be a potential function φ on all of R corresponing to our ifferential equation? is No The above example illustrates that, while we can partition a non-simply connecte region into simply-connecte subregions an then fin all possible potential functions for our ifferential equation over each subregion, it may still be impossible to paste together these regions an potential functions to obtain a potential function that is well efine across all the bounaries between the partitions Is this truly a problem for us, whose main interest is in solving the ifferential equation? Not really We can still solve the given ifferential equation All we nee to o is to choose our simply-connecte partitions reasonably intelligently! Example 715: Let s solve the initial-value problem y x 2 + y + x y = 0 with y(1) = 3 2 x 2 + y 2 using the potential functions from the last example Since (x, y) = (1, 3) is in the right half plane, it makes sense to use φ + from the last example Letting c + = 0, this potential function is ( y ) φ + (x, y) = Arctan for x > 0 x So our ifferential equation has an implicit solution ( y ) Arctan = C + for x > 0 x Taking the tangent of both sies, letting A = tan(c + ), an then solving for y yiels the general solution y = Ax for x > 0 Combine with the initial conition, this is 3 = y(1) = A 1 So A = 3 an the solution to our initial-value problem is y = 3x for x > 0 (We will leave the issue of whether we truly nee to restrict x to being positive as an exercise for the intereste)

26 154 The Exact Form an General Integrating Factors Proving Theorem 74 This is one theorem we will not attempt to prove A goo proof woul require a review of integrals over curves in the plane an Green s theorem, both of which are subjects you may recall seeing near the en of your calculus course Moreover, if you replace the equation with F(x, y) = M(xy)i + N(x, y)j an replace the phrase in exact form with a conservative vector fiel, then theorem 74 becomes the theorem escribing when a two-imensional vector fiel is conservative The proofs of these two theorems are virtually the same, an since the theorem about conservative vector fiels is in any goo calculus text, this author will save space an writing by irecting the intereste stuent to reviewing the relevant chapters in his/her ol calculus book Aitional Exercises 71 For each choice of φ(x, y) given below, fin a ifferential equation for which the given φ is a potential function, an then solve the ifferential equation using the given potential function a φ(x, y) = 3xy b φ(x, y) = y 2 2x 3 y c φ(x, y) = x 2 y xy 3 φ(x, y) = x Arctan(y) 72 The following concern the ifferential equation y = 1 y y 2x (79) a Verify that the above ifferential equation can be rewritten as y 2 2x ] + 2xy y, an then verify that this is an exact form for equation (79) by showing that is a corresponing potential function φ(x, y) = xy 2 x 2 b Solve equation (79) using the above potential function c Note that we can also rewrite equation (79) as e xy2 x 2 y 2 2x ] + e xy2 x 2 2xy y Show that this is also an exact form by showing that is a corresponing potential function ψ(x, y) = e xy2 x 2

27 Aitional Exercises Assume φ(x, y) is a potential function corresponing to Show that ψ 1 (x, y) = e φ(x,y) an ψ 2 (x, y) = sin(φ(x, y)) are also potential functions for this ifferential equation, though corresponing to ifferent exact forms 74 Each of the following ifferential equations is in exact form Fin a corresponing potential function for each, an then fin a general solution to the ifferential equation using that potential function (even if it can be solve by simpler means) a 2xy + y 2 + 2xy + x 2] y b 2xy 3 + 4x 3 + 3x 2 y 2 y c 2 2x + 3y 2 y 1 + 3x 2 y 2 + 2x 3 y + 6y ] y e 4x 3 y + x 4 y 4] y f 1 + ln xy + x y y g 1 + e y + xe y y h e y + xe y + 1 ] y 75 For each of the following ifferential equations, i verify that the equation is not in exact form, ii iii fin an integrating factor, an solve the given ifferential equation (using the integrating factor just foun) a 2y 3 + 4x 3 y 3 3xy 2] y b y + y 4 3x ] y c 2x 1 y + 4x 2 y 3 ] y x tan(y)] y e 3y + 3y 2 + 2x + 4xy ] y

28 156 The Exact Form an General Integrating Factors f 2x(y + 1) y g 1 + y 4 + xy 3 y h 4xy + 3x 2 + 5y ] y for y > 0 i x 2 y 2 + 7x 3 y + x y ] y 76 The following problems concern the ifferential equation (710) Assume M an N are continuous an have continuous partial erivatives, an let P an Q be the functions given by P = M N N an Q = N M M a Show that, if P is a function of x only (so all the y s cancel out), then µ(x) = e P(x) is an integrating factor for equation (710) b Show that, if Q is a function of y only (so all the x s cancel out), then µ(y) = e Q(x) is an integrating factor for equation (710)