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# 427 Class Note for MATH 251 at PSU

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Date Created: 02/06/15

Differential Equations LECTURE 8 Exact Equations The nal category of rst order differential equations we will consider are the so called exact equations These nonlinear equations have the form d 81 Mz7yN7vy0 dm where y is a function of z and 8M 7 8N 8y 7 8m 7 where these two derivatives are partial derivatives discussed more below We ll begin this lecture by brie y discussing how to compute the partial derivatives of a mul tivariable function Then we ll see a motivating example of how to solve an exact equation which will in turn lead to a discussion of why what we did in the example worked Finally we ll see some examples done with the solution method in mind 1 Multivariable Differentiation Suppose we have a function m y depending on two variables What does it mean to differ entiate this function Recall that if we have a function of a single variable 9a we can interpret the derivative as the rate of change of the output of 9a as x increases We d like to have a similar notion of differentiation for our multivariable function m The situation is of course complicated We no longer just have to worry about what happens as x increases we have to worry about how to measure the rate of change of fm y as y increases or even as x and y vary in different ways Geometrically we can think of the graph of fzy as specifying some surface in three dimensional space so talking about the slope at a point is vague in this way So what do we do It turns out and this is a topic that is explored more in a multivariable calculus class that if we can account for how m y varies as only z or only y vary we don t have to worry about mixtures of the two Let s consider how we want to express the derivative of fz y at a point mo yo Without loss of any generality we can consider what happens only as x varies since the y situation will be analogous and we ve already punted on considering mixing the two to a class that is more suited to discussing that Fixing y yo reduces our function of two variables fzy to a function of a single variable 9a m yo So we de ne the partial derivative of f with respect to x at 0343 denoted by gm yo fz9007y07 to be g m0 with 9a fmy0 Geometrically this is the slope of the curve on the graph of 1 corresponding to the line y yo at z me What does this mean computationally Our function g is de ned by treating y as the constant yo So its derivative will be equivalent to the derivative of 1 taken with respect to z while treating y as a constant The following examples should clarify the above discussion Differential Equations Lecture 8 Exact Equations EXAMPLE 81 Let ag y 952g yz Then 8f a 7 2mg 8f 7 2 gig 7 m 2y D EXAMPLE 82 Let ag y ysinz 8f 7 a 7 y cosz 8f 7 i a 7 s1nz D We will also need to be able to recognize the multivariable chain rule The relevant version of this says that if we have a function ltIgtmyz depending on some variable x and a function y depending on x then dlt1gt78lt1gt 9de M a a 87m 11 Elly 2 Exact Equations Before we talk in more detail about how to solve an exact equation7 let s work one example to get a feel for what an exact equation is and why the solution method works We ll rework this later after a more general discussion to reveal all the details EXAMPLE 83 Consider 2my79m2 2ym2l dig 0 dm The rst step in solving an exact equation is to nd a certain function May We ll see how to nd this function later and in fact this computation is where most of the work lies7 but for this example it will turn out that the desired function is May y2 x2 1 7 3x3 Notice that if we compute the partial derivatives of I we obtain 11z7y 2mg 7 9m2 Pgmy 2y m2 1 Looking back at the differential equation7 we can see that it can be rewritten as dy 11 7 Then7 thinking back to the chain rule as expressed in Equation 827 we see that the differential equation is just n4 O dlt 7 0 dm This tells us that our function I must be equal to some constant7 since its ordinary derivative is zero7 and thus the general solution is 342 z2lyi3z3c for some constant c which is just the constant of integration If we had an initial condition7 at this point we could solve for c and get the particular solution to the initial value problem D Differential Equations Lecture 8 Exact Equations Let s think about what we saw in the previous example An exact equation has the form dy M N i 0 any 90724 with Myz y Nmzy The key to nding the solution to an exact equation is to construct a function Qm y such that the differential equation turns into dQ 7 0 dm by using the multivariable chain rule 82 as we did above Thus we require that Q satisfy 907 y MW 24 Qym y Nz REMARK One of the standard facts of multivariable calculus is that mixed partial derivatives commute This is why we want My Nm My me and Nm Qym and these should be identical if such a function Q can exist lt s imperative to check that a function is actually exact by computing Mm and Ny and seeing that they coincide before proceeding with the solution process or there s no way it can work Once we have this function Q then we know O and hence Way 07 yielding an implicit general solution to the differential equation So we can see that the real work involves computing Qm How can we do it Let s revisit Example 83 this time lling in all the details We ll also add an initial condition into the mix EXAMPLE 84 Solve the initial value problem 1 2my79m22ym21d70 y02 Let s begin as we always should be checking that this differential equation is actually exact Comparing the equation to Equation 81 we have Mzy 2mg 7 9m2 and Nmy 2yz2 1 Then My 2x and Nm 2m hence the equation is exact Now how do we nd Qm y We have that Qm M and Qy N Thus we could compute Q in one of two ways QmyMdm or ampNay We ll need to be slightly careful here as we ll see but rst let s just note that it doesn t usually matter which of M or N you choose to integrate to get Q There are some examples in which one is clearly easier but it s a judgement call integrate whichever seems easier to you In this case they re equally easy so let s use the rst one Qzy 2xy 7 9m2 dm xzy 7 3m3 Notice the presence of the function Hg in the integral This is the equivalent of the constant of integration that we obtain when we integrate a single variable function since if we differentiate Q with respect to m any function that just depends on y will vanish So instead of tacking on a constant to the end we need to have a function of y Differential Equations Lecture 8 Exact Equations If we had integrated N with respect to y to get Q we would have needed an equivalent function of m something like Mm to play the role of the constant of integration It s very important that this not be forgotten Now all we need to do is to nd Mg and we ll have our Q How do we do this We know that if we differentiate Q with respect to m h will vanish which is utterly unhelpful However if we differentiate Q with respect to y we re in good shape since 71 will hang around and we know that Q N So we ll compute Qy and any terms in N that aren t in Q must be h y Analogously if we had started by integrating N with respect to y to nd Q we would want to differentiate Q with respect to z and compare with M Ok so we can see that Q 2 h y and N 2 2y 1 So since these are equal we must have h y 2y 1 and so hyh ydyy2y REMARK As we ve tended to do we re going to be a little careless with constants In general we ll drop the constant of integration from our computation of h since it will end up combining with the constant c that we get as part of the solution process Thus we have Way 9021 73953 y2 y 22 2 1y 7 3963 which is precisely the Q that we used in Example 83 From here on it s exactly the same as it was there we observe that the differential equation is just dQ 7 dm 7 and thus Qm y y2 m2 Dye 3x3 c for some constant c To compute c we ll use our initial condition y0 2 0 22 2 c a c 6 and so we have a particular solution of y2 x2 Dy 73mg 6 This is a quadratic equation in y so as we ve seen we can complete the square or use the quadratic formula to get an explicit solution which we want to do when we can y2m2lyi3m36 2 2 2 2 1 1 y2z21y63z3f 212412m32225 y 2 4 7m21i m412m32x225 m y 2 Now we use the initial condition to gure which whether we want the or the 7 in that i Since y0 2 we have Aim 71i5 2y0f 2 Thus we see that we ll want the solution and our particular solution is 7x2 1 24 12z3 2x2 25 2 73 Differential Equations Lecture 8 Exact Equations The following examples should be less long since we won t need to repeat the rationales behind each of the steps EXAMPLE 85 Solve the initial value problem 2my22 237x2yy y7l 1 First we need to put the equation into the form of Equation 81 mez 2 7 23 7 mzyy 0 Now we have Mzy 2zy2 2 and Nzy 723 7 y incorporating the minus into N is very important otherwise the derivatives won t work out Thus My 4mg Nm and the equation is exact The next step is to compute May In this example it s equally easy to integrate either M or N so let s use N ltIgtzy Ndy 2z2y 7 6dy xzyz 7 6y To nd Mm we compute Pm 2zy2 h m and notice that for this to be equal to M h m 2 Hence hz 2x and we have an implicit solution of mzyz 7 634 2m c Now we use the initial condition y71 1 17672cc77 So our implicit solution is m2y276y2m70 Again we can solve for the explicit solution by completing the square or using the quadratic formula 6 i 36 74x22x 7 WE 7 T 7 3iv972x377z2 2 and using the initial condition we see we want the 3 solution so the explicit particular solution is 37V97 2x377z2 M95 EXAMPLE 86 Solve the VP 2759 2 m72t747lnt 1y 0 y20 and nd the solution s interval of validity This is already in the form of Equation 81 so let s start by checking if its exact Mty 3 7 2t and Nt y lnt2 1 7 4 so My Nt Thus the equation is exact Now let s compute I In this case it will be easiest to integrate M ltIgtMdt 2 59 72tdtylnt217t2hy lt1gt t21 ylnt21hylnt2174N Differential Equations Lecture 8 Exact Equations so we conclude h y 7y and thus My 74y So our implicit solution is then ylnt217t2 74y 0 and we use our initial condition to compute c 74 Thus the particular solution is yM D7 77amp and this is very easy to solve explicitly Doing so we obtain 252 7 4 y im ni4 Now let s nd the interval of validity We don t have to worry about the logarithm since 252 1 gt 0 for all 25 Thus we only need to avoid division by zero so we need to avoid the following points 1m ni40 im n4 t2e471 tiv7 So there are three possible intervals of validity 700 7 e4 7 1 7 e4 71 e4 7 10md 4 7 1 The middle one contains 25 2 so our interval of validity is 7 e4 7 1 V54 7 1 D EXAMPLE 87 Solve 3y353my 7 1 23453 3my253my y 0 y1 2 We have Mm y 334353 7 1 and Nz y 23453 Bmyzegmy so My gyzesmy Qxysesmy Nm Thus the equation is exact We ll integrate M since it s a bit easier I Mdz 3y3531y 71 yzegmy 7 z My and by 2y53my 3my253my hy Comparing by to N we see that they are already identical so we must have h y 0 and hence My 0 since we re ignoring constants in h So we have yzegmy 7 m c and using the initial condition gives 0 456 7 1 Thus our implicit particular solution is yzesmy 7 m 456 71 and we re done because we won t be able to solve this explicitly D

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