z = 1 2 x 3 4 y + 3 y dt

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1 Exact First Order Differential Equations This Lecture covers material in Section 2.6. A first order differential equations is exact if it can be written in the form M(x, ) + N(x, ) d dx = 0, where M = N x. Before showing how to solve these we need to review some multivariable calculus, especiall the two-variable chain rule. This will also help to motivate wh equations of this form are important in phsics. Let ψ(x, ) be a function of two variables. Then we can think of z = ψ(x, ) as a surface in three-dimensional space where z is the height above the x-plane. Now suppose the x and are functions of t (time) so that (x(t), (t)) gives a curve in the x-plane. Then z(t) = ψ(x(t), (t)) gives a curve in three-dimensional space. Suppose we desire to know the rate of change of z with respect to t. According to the twovariable chain rule the answer is dz dt = ψ x dx dt + ψ d dt. This formula is derived in Calculus III. Here I will give an intuitive motivation for wh it works. Suppose ψ(x, ) is just a plane. Then we have z = ψ(x, ) = Ax + B + C for some constants A, B and C. Here A is the slope of the plane with respect to the x direction, B is the slope of the plane with respect to the direction and C is the intercept with the z-axis. 1 ( )

2 2 z 3 P 4 6 x z = 1 2 x We want to compute the change in z as t changes from t 0 to t 0 + t. Let x = x(t 0 + t) x(t 0 ) and = (t 0 + t) (t 0 ) be the changes in x and, respectivel. For convenience let x 0 = x(t 0 ) and 0 = (t 0 ). Then the change in z is z = ψ(x 0 + x, 0 + ) ψ(x 0, 0 ) = A x + B. We divide both sides b t to obtain z t = A x t + B t. Now we can find the derivative of z with respect to t b taking limits as t 0. This gives dz dt = Adx dt + Bd dt. But notice that A = x ψ and B = ψ. Thus, we have dz dt = ψ dx x dt + ψ d dt which is ( ). This shows that the two-variable chain rule works for planes. In general if ψ(x, ) is reasonabl smooth it can be approximated near each point b a tangent plane. It can be shown that this gives the two-variable chain rule for an function of two variables that

3 is smooth enough that its graph has a tangent plane at each point in an open set containing the point of interest. Now, let z = ψ(x, ) be a surface. But suppose z is some quantit that is conserved, like energ. That is we now have ψ(x, ) = C. The slice of the surface through z = C is called level curve. Example. Let z = ψ(x, ) = x Then the level curve for z = 1 is a circle of radius 1 that floats one unit above the x-plane. Example. Let z = ψ(x, ) = 3x + 3. The level curve for z = 2 is the line 3x + 3 = 2, or = 3x + 5, that is it floating two units above the x-plane. For now suppose is a function of x (at least implicitl). As we change x we cause to change so that z = ψ(x, ) stas on the same level curve. Since z is not changing we have dz/dx = 0. The two-variable chain rule, using x for t, gives 0 = dz dx = dψ(x, (x)) dx = ψ dx x dx + ψ d dx. Therefore, ψ x + ψ = 0. If ψ x and ψ are known functions what we have is a differential equation in. We will be doing the inverse of this process. differential equation in the form 3 That is, given at M(x, ) + N(x, ) d dx = 0 we will solve it for (x) (or at least a relation between x and ) b finding a surface ψ(x, ) such that M = ψ x and N = ψ, and

4 4 then using an initial condition to find the desired level curve. In man applications M, N is given as a force field and then ψ is a potential energ function. If energ is conserved, the dnamics are restricted to a level curve of z = ψ(x, ). Enough talk, let s do some examples. Example 1. Solve (2x + ) + (x + 2) = 0, with (3) = 1. Solution. We want to find a function ψ(x, ) such that So, we integrate. ψ = ψ x dx = ψ x = 2x + & ψ = x x + dx = x 2 + x + C 1 (), where C 1 () an arbitrar function of. The idea is we are finding the class of all functions whose partial derivative with respect to x gives 2x +. But we also need for ψ = x + 2. So, we integrate. ψ = ψ d = x + 2 d = x C 2 (x), where C 2 (x) can be an function of x. We now have two classes of functions, each satisfing one of the two conditions. If we could find a function that is in both classes that would do the trick. The answer is obvious. Let ψ(x, ) = x 2 + x + 2. This function is in both classes and thus satisfies both the needed conditions. Now we consider the initial condition, (3) = 1, that is, x = 3 = = 1. Then ψ(3, 1) = = 13.

5 Thus, the level curve we want is 5 x 2 + x + 2 = 13. We will leave as a relation. Below are plots of the surface z = x 2 + x + 2 with the level 13 curve and a projection of this curve into the x-plane. Extra Credit. Prove that this curve x 2 + x + 2 = 13 is an ellipse and find its focal points. You can do this b reviewing how to rotate graphs with rotation matrices and the properties of ellipses. Then rotate the graph 45 o so that its major axis lies along the x-axis.

6 6 Example 2 (Not!). Solve (2x + 2) + (x + 2) = 0, with (3) = 1. We integrate. ψ = 2x + 2 dx = x 2 + 2x + C 1 (). ψ = x + 2 d = x C 2 (x). Now look closel. Since 2x x there is no function that meets both conditions. The method fails! What this means in phsical terms is that the force field 2x + 2, x + 2 does not arise from a potential function; in such a sstem energ is not conserved. Note: This example can be converted to a separable equation because it is homogeneous. We work through this in the Appendix at the end of these notes. What we need is a quick test to see if ψ exists for a given equation so that we don t waste a lot of time barking up the wrong tree. Theorem! Given two functions M(x, ) and N(x, ), there exists a function ψ(x, ) such that ψ x = M(x, ) & ψ = N(x, ), if and onl if M = N x, in an open rectangle containing the point of interest. Check this for the two examples above. This is Theorem in our textbook. Your textbook gives a proof, but another proof is covered in Calculus III that uses Green s Theorem. If ou are a Math major read both and compare them. However, one direction is eas: if ψ exists, then ψ x = ψ x = M = N x. This theorem is the motivation for the definition we gave at the beginning of an exact first order differential equation.

7 Example 3. Find the general solution to cos x + e x + (sin x + xe x ) = 0. Solution. Let M = cos x + e x and N = sin x + xe x. Then M = cos x + e x + xe x = N x. Thus, it is exact. We integrate. ψ = M dx = sin x + e x + C 1 () and ψ = N d = sin x + e x + C 2 (x). We let ψ(x, ) = sin x + e x. The general solution is then sin x + e x = C. Example 4. Solve 4x = 0, with (1) = 1. Solution. It is exact since (4x 3 ) = 0 and (4 3 ) x = 0. Then ψ = 4x 3 dx = x 4 + C 1 () and ψ = 4 3 d = 4 + C 2 (x). We let ψ = x Since (1,1) is our initial condition we see that our solution is x = 2. Below is a graph of this curve projected into the x-plane.

8 8 Note. The equation in Example 4 is separable and can be solved easil b that method. In fact all separable first order differential equations are also exact. See if ou can prove this. Example 5. Solve 4x 4 + 4x 3 = 0, with (1) = 1. Solution. We check for exactness. (4x 4 ) = 0 while (4x 3 ) x = 4 3. Thus, it is not exact. But wait! Notice that this example is exactl the same as Example 4, but that we have multiplied through b x. So, if we now multiple through b 1 x we get 4x = 0. Thus, the solution is same as in Example 4!

9 Integrating factors. 9 This last example motivates the following idea. Suppose we have a differential equation of the form M + N = 0 which is not exact. Can we find a function µ(x, ) such that µm + µn = 0 is exact? The answer is, not alwas, but sometimes ou can. When this works we call µ an integrating factor. Finding such as µ can be trick. Here we show three special cases where an integrating factor µ can be found. Each relies on an assumption about µ that can be tested for. Case 1. Suppose a suitable µ exists and that it is a function of x onl. Case 2. Suppose a suitable µ exists and that it is a function of onl. Case 3. Suppose a suitable µ exists and that it can be written as a function dependent onl on the product x. In all cases we need to find µ such that (µm) = (µn) x so that we have exactness. B the product rule this is equivalent to requiring µ M + µm = µ x N + µn x. ( )

10 10 Case 1. If Case 1 holds then µ = 0 and we can think of µ x as µ. Then ( ) becomes or µm = µ N + µn x µ µ = M N x. N If our assumption is correct then, since µ /µ depends onl on x, we know that (M N x )/N depends onl on x. Then the integrals below are well defined. 1 µ dµ = M N x dx N Thus, µ = e M Nx N dx. In fact, this gives us a test to determine when this method will work. If M N x N depends onl on x it follows that µ depends onl on x. Example 6. Find the general solution to 2 + x 3 + x = 0. Solution. Since ( 2 + x 3 ) = 2 and (x) x = are not equal, this equation is not exact. But 2 x depends onl on x. Thus, we let = 1 x µ = e 1 x dx = x. So, we multipl through b x to get x 2 + x 4 + x 2 = 0.

11 Let M = x 2 + x 4 and N = x 2. Then M = 2x = N x, so we have exactness. Now we find ψ as before. ψ = M dx = 1 2 x x5 + C 1 () ψ = N d = 1 2 x2 2 + C 2 (x) 11 Thus, we let ψ = 1 2 x x5, so the general solution is or if ou prefer Solving for gives 1 2 x x5 = C, 5x x 5 = C. C 2x 5 = ±. 5x 2 Case 2. This is so similar to Case 1 that we leave it to ou to develop the method and find the formula for µ(). Case 3. Recall equation ( ): µ M + µm = µ x N + µn x. Let v = x and remember we are assuming µ can be rewritten as a function of v. Thus, µ = µ(v) = dµ v dv = dµ dv x = xµ, and µ x = µ(v) x = dµ v dv = dµ dv = µ, where µ means the derivative with respect to v. Now ( ) becomes xµ M + µm = µ N + µn x,

12 12 which gives µ µ = N x M xm N. If the right hand side depends onl on v = x then the assumption we are making is valid, and thus µ = e Perhaps an example would help. Example. Solve with (1) = π. N x M xm N dv. 5x x cos x + x4 + cos x d dx = 0, Solution. Let M = 5x x cos x and N = x4 +cos x. Then M = sin x & N x = 4x3 sin x Thus, the given equation is not exact. We now search for an integration factor. Case 1. Case 2. M N x N = 4x3 x 4 +cos x N x M 4x3 M = 5x 4 +cos x x = = 4x 3 x 4 + cos x 4x 4 5x 4 + cos x No good! Rats!! Case 3. N x M xm N = Let v = x. Now, 4x3 5x 4 + cos x (x 4 + cos x) = 1 x! Eureka!!! µ(v) = e 1 v dv = e ln v +C = C v = C x ;

13 we will use µ = x On ward! We multipl the original equation b x to get 5x 4 + cos x + (x 5 + x cos x) = 0. Let M = 5x 4 + cos x and N = x 5 + x cos x. We double check that it is in fact exact. M = 5x 4 + cos x x sin x = N x. Now the hunt is on for ψ! ψ = M dx = x 5 + sin x + C 1 () 13 and ψ = N d = x 5 + sin x + C 2 (x). Thus, ψ = x 5 +sin x and the general solution is x 5 +sin x = C. Since (1) = π ou can check that C = π. Thus, the solution is x 5 + sin x = π. Appendix A. Example 2. Recall that Example 2 was not exact, but that it was noted that it is homogeneous. Here we will solve it. 2x 2 = = 2 2 ( ) x x ( 2 2v ) = 1 + 2v, x where v = /x. It follows that = v + xv. Now we have x dv dx Thus, = 2 2v 1 + 2v v = 2 2v 1 + 2v v1 + 2v 1 + 2v = 2 3v 2v v

14 v 2 + 3v + v dv = 2 We rewrite the left integrand as Now and 1 4v v 2 + 3v + 2 = x dx ( 4v + 3 2v 2 + 3v v 2 + 3v + 2 4v + 3 2v 2 + 3v + 2 dv = ln 2v2 + 3v C 1 2v 2 + 3v + 2 dv = dv ( 2v ) /8 Let u = 2v Then du = 2dv. The integral becomes du u 2 + /8 = 4 du 8u 2 / + 1. Let w = 2 2u. Then dw = 2 2du. Now the integral becomes, 2 dw w = 2 arctan(w) + C. Putting all this together gives ln x + C = 1 2 ln 2v2 + 3v arctan ) ( ) 4v + 3. Now we find C. We had (1) = 1 and since v = /x we get v = 1. Thus, ( ) C = ln + 1 ( ) arctan. Thus, our solution is given b the relation ln x = 1 2 ln(2v2 + 3v + 2) + 1 arctan ( ) 4v + 3

15 ln ( ) 1 1 ( ) arctan, 15 where v = /x. I tried to graph this using an implicitplot commend. I got no where. I kept getting an error message: Error, (in implicitplot) could not evaluate expression. So, even though I was able to solve the differential equation, the solution is so convoluted it was not of much use to me. So, I used a numerical method on a computer. > DEplot(diff((x),x)=(-2*x-2*(x))/(x+2*(x)),(x), x= ,[[(1)=1]],= ,arrows=line); But, notice something odd is happening just after x = 1.9. The graph of the solution curve suddenl shoots off straight. Look at the slope field. Notice the slopes are tending toward negative infinit just after x = 1.9. After that point the computer s solution curve is no longer valid. That computer will not tell ou this. You have to understand what is going on with the math in order to use solution properl.