Ma 530. Special Methods for First Order Equations. Separation of Variables. Consider the equation. M x,y N x,y y 0

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1 Ma 530 Consider the equation Special Methods for First Order Equations Mx, Nx, 0 1 This equation is first order and first degree. The functions Mx, and Nx, are given. Often we write this as Mx, Nx,d 0 2 Separation of Variables Equation (2) takes a simple form in the special case when Mx, Ax and Nx, B. Ax Bd 0 That is the variables separate. If we Integrate Ax Bd c. Example Which leads to x 2 d 0 x 2 d c. x c. Now consider the I.V.P. Integrating from x 0, 0 x, D.E. Ax Bd 0 I.C. x 0 0 x x0 Ax 0 Bd 0. 1

2 Example D.E. cos x 2 d 0 I.C. 0 x cosx 2 d 0 sin x x or sinx sin 3 0 sin x sinx Example Solve xd 0 This equation is not separable as is. Divide b x d x 0 ln x ln c or ln x c x k x k k x x 0. Example This example is a video slide show. Slide Example You will need Real Plaer to view this. To get it click on Real Plaer. First Order linear differential equations Clearl not all equations are as simple as the equation Ax Bd 0. Consider the equation ax d bx cx. Assuming ax 0 we divide b ax Px Qx 1 or d Px Qx 0. We want to solve 1. Consider first the homogeneous problem which is separable. ln Px c c Px e Px 0. d Px 0 Hence is the homogeneous solution. e c Px e Px ke 2

3 Non-homogeneous case: d Px Qx. Px We shall use variation of parameters. Note that an constant times e is also a solution of the homogeneous equation 1. Px To solve the nonhomogeneous equation we shall tr a function times e i.e. Now the D.E. v Px e v Px e Px vxe ve Px Px. ve Px Px Pxve Px Q v Qe Px v Qe Px c. Therefore the solution is ve Px ce Px Qe Px Px e homogeneous solution particular solution Example: x 1 x2 P 1 Q x 2. x1 Consider x 1 0 d x 0 orln x 1 c k/x 1. 1 Using the formula for the homogeneous solution, we have Px ke x1 ke. ke lnx1 k x 1 We now solve the nonhomogeneous equation. Since ve Px. The D.E. v x 1 v x 1 2 v x 1 3

4 Thus and therefore. v x 1 v x 1 v x 2 2 x 1 2 v x 2 x 1 x 3 x 2 v x4 4 x3 3 c c x 1 x4 4 x3 3 x 1 Remark: The variation of parameters method works because the assumption ve P leads to v Qe P. Since v e P d e P Qe P e P P Qe P. Therefore if we multipl the original equation b e P we get an integrable form right awa. Example x 1 x2 P 1 x 1 or as before. Summar: e P (Again) e x1 e lnx1 x 1 x 1 x 2 x 1 d x 1 x2 x 1 x 1 x4 4 x3 3 c To solve P Q multipl both sides b the integrating factor I e P. Then the L.H.S. becomes d e P e P Q and the solution is found b integrating both sides. This is called the Method of the Integrating Factor. We can use the above to solve the I.V.P. D.E. Px Qx I.C. x 0 0 4

5 Use the integrating factor and integrate both sides from x 0 to x. x I e x 0 Ptdt Example Here are two video slide show examples. Slide Example 1 Slide Example 2 You will need Real Plaer to view this. To get it click on Real Plaer. Example The equation d Px Qxn n anrealnumber is known as Bernoulli s equation. We shall suppose that n 0 or 1, since we alread know how to solve the equation for these two cases. Multipling b n ields Let z n1 Then z n 1 n n Px n1 Qx z Pxz Qx. 1 n This is a linear differential equation for z which can be solved. For example, consider the equation Let z 5 z 5 4 x d x4 1 x x 1 z 5 z x 1 z 5 x z 5 n 4 Thus the integrating factor is e 5 x e 5lnx x 5 so we have d x5 z 5x 5 5

6 Since z 5 x 5 z 5 x6 6 c z 5 x 6 cx5 5 5 x 6 cx5. Exact Differential Equations Definition: The differential expression Mx, Nx,d is called exact a function fx, that is differentiable in some region R of the x,-plane, i.e. exist and are continuous in R andsuchthat f x M f N x, R. Remark: Since dfx, f f d M Nd is exact dfx, M Nd. x Definition: The differential equation Mx, Nx,d 0 1 is called an exact differential equation if the left hand side is an exact differential. Remark: When the differential equation 1 is exact f x, f dfx, M Nd 0 2. Using this we ma solve the differential equation. For if x is the solution, then 2 ma be integrated with respect to x to ield fx, c 3. Conversel if 3 defines as a differential function of x, then this x is a solution of the differential equation. For 3 b the chain rule. Example: xd 0 Consider fx, x Since f x and f x df 0 f x f M N d 0. Here M and N x dfx, xd d 6

7 Therefore df f x f d xd 0 determines the solution. fx, x c c x d Check c d c. xd x c c x 2 x 2 x 2 x 0. Thus if we know that a certain differential equation is exact we can solve it. Question: When is a differential equation exact? The answer is given b following theorem. Theorem If Mx, and Nx, are continuous functions and have continuous partial derivatives in some region R of the x,-plane, then the expression Mx, Nx,d is an exact differential throughout R. M N x Remark: If f x M and f N, then f x M N x f x. Example: xd exact. Here M and N x so that M 0 N x and we see that this equation is Example: e x cos e x sin d We rewrite the equation as e x cos e x sin d 0. Thus M e x cos and N e x sin and therefore M e x sin and N x e x sin. Therefore this equation is exact. fx, such that f x M f N, i.e., f x ex cos fx, e x cos g e x cos g. g? We must have f N ex sin. Now f ex sin g e x sin g 0 g const c Therefore fx, e x cos c 7

8 solution is fx, k, i.e. e x cos c k k. Example Here is a video slide show example. Slide Example You will need Real Plaer to view this. To get it click on Real Plaer. Integrating factors Recall: Mx, Nx,d 0 is exact M N x. When the equation is exact, fx, such that f x M and f N and df M Nd 0. fx, c gives the solution to the equation. Clearl not ever differential equation is exact. Question: Can we make M Nd 0 exact when it is not? We want to find a function ux, such that when we multipl the differential equation b ux,, then um und 0 is exact. ux, is called an integrating factor. Example 2 xd 0 M N 2 x M 1 N x 1 Thus the equation is not exact. We multipl b a function u so that is exact. u u 2 xd 0 u u 2 x x u u u x 2 x u. This last equation is harder to solve in general than the original. However, we do not need the general solution. We need an u which when multiplied times the equation makes it exact. If we assume u x 0, then u u, i.e.u is onl a function of and our partial differential equation for u becomes the ordinar differential equation du d 2u 0 which has the solution u 1. Multipling the original equation b this u ields x 2 d 0 Since now M 1 and N 1 x 2 M 1 2 N x this new equation is exact. f x 1 8

9 f x g. Hence g 1 g c Therefore f x 2 g x 2 1 fx, x c and the solution is x k. Example xd 0 N x M M 1 and N x 1 Clearl this equation is not exact. Multipl b u and get u uxd 0. Then M u u and N x u u x x. u u u u x x Setting u 0 ields u 1 x 2 This new equation is exact. x 2 1 x d 0 f x x 2 and f x 2 f x hx f x x 2 h x x 2 solution is h 0 h c f x c x k 9