Math 246, Spring 2011, Professor David Levermore

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1 Math 246, Spring 2011, Professor David Levermore 7. Exact Differential Forms and Integrating Factors Let us ask the following question. Given a first-order ordinary equation in the form 7.1 = fx, y, dx when do its solutions satisfy a relation of the form Hx, y = c where c is an arbitrary constant? Such an Hx, y is called an integral of 7.1. This question is easily answered if we assume that all functions involved are as differntiable as we need. Suppose that such an Hx, y exists, and that y = Y x is a solution of differential equation 7.1. Then Hx, Y x = H, Y, where is any point in the interval of definition of Y. By differentiating this equation with respect to x we find that x Hx, Y x + Y x y Hx, Y x = 0. Therefore, wherever y Hx, Y x 0 we see that Y x = xhx, Y x y Hx, Y x. For this to hold for every solution of 7.1, we must have or equivalently dx = xhx, y y Hx, y, 7.2 fx, y = xhx, y y Hx, y, wherever y Hx, y 0. The question then arises as to whether we can find an Hx, y such that 7.2 holds for any given fx, y? It turns out that this cannot always be done. In this section we explore how to seek such an Hx, y Exact Differential Forms. The starting point is to write equation 7.1 in a so-called differential form 7.3 Mx, y dx + Nx, y = 0, where Mx, y fx, y = Nx, y. There is not a unique way to do this. Just pick one that looks natural. If you are lucky then there will exist a function Hx, y such that 7.4 x Hx, y = Mx, y, y Hx, y = Nx, y. When this is the case the differential form 7.3 is said to be exact. 1

2 2 It turns out that there is test you can easily apply to find out if you are lucky. It derives from the fact that mixed partials commute namely, the fact that for any twice differentiable Hx, y one has y x Hx, y = x y Hx, y. This fact implies that if 7.4 holds for some twice differentiable Hx, y then Mx, y and Nx, y satisfy y Mx, y = y x Hx, y = x y Hx, y = x Nx, y. In other words, if the differential form 7.3 is exact then Mx, y and Nx, y satisfy 7.5 y Mx, y = x Nx, y. The remarkable fact is that the converse holds too. Namely, if the differential form 7.3 satisfies 7.5 for every x, y then it is exact i.e. there exists an Hx, y such that 7.4 holds. Moreover, the problem of finding Hx, y is reduced to evaluating two integrals. We illustrate this fact with examples. Example: Solve the initial-value problem dx + ex y + 2x 2y + e = 0, y0 = 0. x Solution: Express this equation in the differential form e x y + 2x dx + 2y + e x = 0. Because y e x y + 2x = e x = x 2y + e x = e x, this differential form satisfies 7.5 and is thereby exact. We can therefore find Hx, y such that 7.6 x Hx, y = e x y + 2x, y Hx, y = 2y + e x. You can now integrate either equation, and plug the result into the other equation to obtain a second equation to integrate. If we first integrate the first equation in 7.6 then we find that Hx, y = e x y + 2x dx = e x y + x 2 + hy. Here we are integrating with respect to x while treating y as a constant. The function hy is the constant of integration. We plug this expression for Hx, y into the second equation in 7.6 to obtain e x + h y = y Hx, y = 2y + e x. This reduces to h y = 2y. Notice that this equation for h y only depends on y. Taking hy = y 2, we see the general solution satisfies The initial condition y0 = 0 implies that Therefore Hx, y = e x y + x 2 + y 2 = c. c = e = 0. y 2 + e x y + x 2 = 0.

3 3 The quadratic formula then yields y = ex + e 2x 4x 2, 2 where the positive square root is taken so that solution satisfies the initial condition. This is a solution wherever e 2x > 4x 2. Alternative Solution: If we first integrate the second equation in 7.6 then we find that Hx, y = 2y + e x = y 2 + e x y + hx. Here we are integrating with respect to y while treating x as a constant. The function hx is the constant of integration. We plug this expression for Hx, y into the first equation in 7.6 to obtain e x y + h x = x Hx, y = e x y + 2x. This reduces to h x = 2x. Notice that this equation for h x only depends on x. Taking hx = x 2, we see the general solution satisfies Hx, y = e x y + x 2 + y 2 = c. Because this is the same relation for the general solution that we had found previously, the evaluation of c is done as before. The points to be made here are the following: It does not matter which equation in 7.4 that you integrate first. If you integrate with respect to x first then the constant of integration hy will depend on y and the equation for h y should only depend on y. If you integrate with respect to y first then the constant of integration hx will depend on x and the equation for h x should only depend on x. In either case, if your equation for h involves both x and y you have made a mistake! Sometimes the differential equation will be given to you alrea in differential form. In that case, use that form as the starting point. Example: Give an implicit general solution to the differential equation Solution: Because xy 2 + y + e x dx + x 2 y + x = 0. y xy 2 + y + e x = 2xy + 1 = x x 2 y + x = 2xy + 1. this differential form satisfies 7.5 and is thereby exact. You can therefore find Hx, y such that x Hx, y = xy 2 + y + e x, y Hx, y = x 2 y + x. By integrating the second equation you obtain Hx, y = x 2 y + x = 1 2 x2 y 2 + xy + hx. When you plug this expresion for Hx, y into the first equation you obtain xy 2 + y + h x = x Hx, y = xy 2 + y + e x,

4 4 which yields h x = e x. Notice that this only depends on x! Taking hx = e x, the general solution is Hx, y = 1 2 x2 y 2 + xy + e x = c. We will now derive formulas for Hx, y in terms of definite integrals. These formulas will encode the two steps given above. They thereby show that those steps can always be carried out. To see this, consider an exact differential form 7.7 Mx, y dx + Nx, y = 0, where y Mx, y = x Nx, y. Now seek Hx, y such that 7.8 x Hx, y = Mx, y, y Hx, y = Nx, y. By integrating the first equation with respect to x you obtain Hx, y = Mr, y dr + hy, where is an arbitrary point. When you plug this expresion for Hx, y into the second equation and use the fact that y Mr, y = r Nr, y, you obtain Nx, y = y Hx, y = = y Mr, y dr + h y r Nr, y dr + h y = Nx, y N, y + h y. This yields h y = N, y, which only depends on y because is a number. Let hy = y y I N, s ds, The general solution of 7.6 thereby satisfies Hx, y = Mr, y dr + where y I is an arbitrary value. y y I N, s ds = c. If you had integrated the second equation in 7.8 first then you would have found that general solution of 7.7 satisfies Hx, y = Mr, y I dr + y y I Nx, s ds = c. The above formlas give two expressions for the same function Hx, y. Rather than memorize these formulas, I strongly recommend that you simply learn the steps underlying them. Remark: Our recipe for separable equations can be viewed as a special case of our recipe for exact differential forms. Consider the separable first-order ordinary differential equation dx = fxgy. It can be put into the differential form fx dx 1 gy = 0.

5 This differential form is exact because 1 y fx = 0 = x gy You can therefore find Hx, y such that Indeed, you find that = 0. x Hx, y = fx, y Hx, y = 1 gy. Hx, y = Fx Gy, where F x = fx and G y = 1 gy. The general solution thereby satisfies This is precisely the recipe we derived earlier. Hx, y = Fx Gy = c Integrating Factors. Suppose you had considered the differential form Mx, y dx + Nx, y = 0, and found that is not exact. Just because you were unlucky the first time, do not give up! Rather, seek a nonzero function µx, y such that the differential form 7.9 µx, ymx, y dx + µx, ynx, y = 0 is exact! This means that µx, y must satisfy This means that µ satisfies y µx, ymx, y = x µx, ynx, y Mx, y y µ + [ y Mx, y]µ = Nx, y x µ + [ x Nx, y]µ. This is a first-order linear partial differential equation for µ. Finding its general solution is equivalent to finding the general solution of the original ordinary differential equation. Fortunately, you do not need this general solution. All you need is one nonzero solution. Such a µ is called an integrating factor for the differential form 7.9. A trick that sometimes yields a solution of 7.10 is to assume either that µ is only a function of x, or that µ is only a function of y. When µ is only a function of x then y µ = 0 and 7.10 reduces to the first-order linear ordinary differential equation dµ dx = ymx, y x Nx, y µ. Nx, y This equation will be consistent with our assumption that µ is only a function of x when the fraction on its right-hand side is independent of y. In that case you can integrate the equation to find the integrating factor µx = e Ax, where A x = ymx, y x Nx, y Nx, y Similarly, when µ is only a function of y then x µ = 0 and 7.10 reduces to the first-order linear ordinary differential equation dµ = xnx, y y Mx, y µ. Mx, y. 5

6 6 This equation will be consistent with our assumption that µ is only a function of y when the fraction on its right-hand side is independent of x. In that case you can integrate the equation to find the integrating factor µy = e By, where B y = xnx, y y Mx, y. Mx, y This will be the only method for finding integrating factors that we will use in this course. Remark: Rather than memorize the above formulas for µx and µy in terms of primitives, I strongly recommend that you simply follow the steps by which they were derived. Namely, you seek an integrating factor µ that satisfies You then expand the partial derivatives as y [Mx, y µ] = x [Nx, y µ]. Mx, y y µ + [ y Mx, y] µ = Nx, y x µ + [ x Nx, y] µ. If this equation reduces to an equation that only depends on x when you set y µ = 0 then there is an integrating factor µx. On the other hand, if this equation reduces to an equation that only depends on y when you set x µ = 0 then there is an integrating factor µy. We will illustrate this approach with the following examples. Example: Give an implicit general solution to the differential equation 2e x + y 3 dx + 3y 2 = 0. Solution: This differential form is not exact because y 2e x + y 3 = 3y 2 x 3y 2 = 0. You therefore seek an integrating factor µ such that Expanding the partial derivatives gives y [2e x + y 3 µ] = x [3y 2 µ]. 2e x + y 3 y µ + 3y 2 µ = 3y 2 x µ. Notice that if y µ = 0 then this equation reduces to µ = x µ, whereby µx = e x is an integrating factor. See how easy that was! Because e x is an integrating factor, you know that e x 2e x + y 3 dx + 3e x y 2 = 0 is exact. Of course, you should check that this is exact. If it is not then you made a mistake in finding µ! You can therefore find Hx, y such that x Hx, y = e x 2e x + y 3, y Hx, y = 3y 2 e x. By intgrating the second equation you see that Hx, y = y 3 e x +hx. When this expression for Hx, y is plugged into the first equation you obtain y 3 e x + h x = x Hx, y = 2e x + y 3 e x, which yields h x = 2e 2x. Upon taking hx = e 2x, the general solution satisfies Hx, y = y 3 e x + e 2x = c. In this case the general solution can be given explicitly as y = ce x e x 1 3,

7 7 where c is an arbitrary constant. Example: Give an implicit general solution to the differential equation 2xy dx + 2x 2 e y = 0. Solution: This differential form is not exact because y 2xy = 2x x 2x 2 e y = 4x. You therefore seek an integrating factor µ such that Expanding the partial derivatives gives y [2xyµ] = x [2x 2 e y µ]. 2xy y µ + 2xµ = 2x 2 e y x µ + 4xµ. Notice that if x µ = 0 then this equation reduces to y y µ = µ, whereby µy = y is an integrating factor. See how easy that was! Because y is an integrating factor, you know that You can therefore find Hx, y such that 2xy 2 dx + y2x 2 e y = 0 is exact. x Hx, y = 2xy 2, y Hx, y = 2x 2 y ye y. By intgrating the first equation you see that Hx, y = x 2 y 2 + hy. When this expression for Hx, y is plugged into the second equation you obtain 2x 2 y + h y = y Hx, y = 2x 2 y ye y, which yields h y = ye y. Upon taking hy = 1 ye y, the general solution satisfies Hx, y = x 2 y ye y = c. In this case you cannot solve for y explicitly. Remark: Integrating factors for the linear equations can be viewed as a special case of the foregoing method. Consider the linear first-order ordinary differential equation + axy = fx. dx It can be put into the differential form axy fx dx + = 0. This differential form is generally not exact because when ax 0 we have y axy fx = ax x 1 = 0. You therefore seek an integrating factor µ such that y [ axy fx µ ] = x µ. Expanding the partial derivatives gives axy fx y µ + axµ = x µ. Notice that if y µ = 0 then this equation reduces to x µ = axµ, whereby an integrating factor is µx = e Ax where A x = ax.

8 8 Because e Ax is an integrating factor, you know that You can therefore find Hx, y such that e Ax axy fx dx + e Ax = 0 is exact. x Hx, y = e Ax axy fx, y Hx, y = e Ax. By intgrating the second equation you see that Hx, y = e Ax y+hx. When this expression for Hx, y is plugged into the first equation you obtain e Ax axy + h x = x Hx, y = e Ax axy fx, which yields h x = e Ax fx. The general solution thereby satisfies Hx, y = e Ax y Bx = c, where A x = ax and B x = e Ax fx. This is equivalent to the recipe we derived previously.

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