Lecture 16 18.01 Fall 2006 Lecture 16: Diﬀerential Equations and Separation of Variables Ordinary Diﬀerential Equations (ODEs) dy = f (x) dx � Solution: y = f (x)dx. We consider these types of equations as solved. Example 1. � � � � d dy Example 2. +x y =0 or + xy = 0 dx dx � � d ( + x is known in quantum mechanics as the annihilation operator.) dx Besides integration, we have only one method of solving this so far, namely, substitution. Solving dy for gives: dx dy = −xy dx The key step is to separate variables. dy = −xdx y Note that all y-dependence is on the left and all x-dependence is on the right. Next, take the antiderivative of both sides: � � dy = − xdx y x2 ln |y| = − + c (only need one constant c) 2 c −x2 /2 |y| = e e (exponentiate) y = ae−x 2 /2 (a = ±ec ) Despite the fact that ec �= 0, a = 0 is possible along with all a �= 0, depending on the initial 2 2 conditions. For instance, if y(0) = 1, then y = e−x /2 . If y(0) = a, then y = ae−x /2 (See Fig. 1). 1 Lecture 16 18.01 Fall 2006 1 Y 0.8 0.6 0.4 0.2 0 −6 −4 −2 0 2 4 6 X Figure 1: Graph of y = e− x2 2 . In general: dy dx dy g(y) h(y)dy = f (x)g(y) = f (x)dx which we can write as = f (x)dx where h(y) = 1 . g(y) Now, we get an implicit formula for y: � H(y) = F (x) + c (H(y) = � h(y)dy; F (x) = f (x)dx) where H � = h, F � = f , and y = H −1 (F (x) + c) (H −1 is the inverse function.) In the previous example: f (x) = x; g(y) y; = −x2 ; 2 1 1 h(y) = = , g(y) y F (x) = H(y) = ln |y| �y� dy =2 . dx x Find a graph such that the slope of the tangent line is twice the slope of the ray from (0, 0) to (x, y) seen in Fig. 2. Example 3 (Geometric Example). 2 Lecture 16 18.01 Fall 2006 (x,y) Figure 2: The slope of the tangent line (red) is twice the slope of the ray from the origin to the point (x, y). dy = y ln |y| = |y| = 2dx (separate variables) x 2 ln |x| + c (antiderivative) ec x2 (exponentiate; remember, e2 ln |x| = x2 ) Thus, y = ax2 Again, a < 0, a > 0 and a = 0 are all acceptable. Possible solutions include, for example, y y y y y y = x2 (a = 1) = 2x2 (a = 2) = −x2 (a = −1) = 0x2 = 0 (a = 0) = −2y 2 (a = −2) = 100x2 (a = 100) Example 4. Find the curves that are perpendicular to the parabolas in Example 3. We know that their slopes, dy −x −1 = = dx slope of parabola 2y Separate variables: ydy = −x dx 2 3 Lecture 16 18.01 Fall 2006 Take the antiderivative: y2 x2 x2 y2 =− +c =⇒ + =c 2 4 4 2 which is an equation for a√family of ellipses. For these ellipses, the ratio of the x-semi-major axis to the y-semi-minor axis is 2 (see Fig. 3). Figure 3: The ellipses are perpendicular to the parabolas. Separation of variables leads to implicit formulas for y, but in this case you can solve for y. � � � x2 y =± 2 c− 4 4