MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � � Lecture 16 18.01Fall 2006 Lecture 16: Differential Equations and Separation of Variables Ordinary Differential Equations (ODEs) Example 1. dy = f(x)dx Solution: y = f(x)dx. We consider these types of equations as solved. Example 2. d + x y = 0 or dy + 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 for dy gives: dx dy = −xydx 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 y = − xdx 2xln |y| = − 2+ c (only need one constant c) |y| = e c e−x 2/2 (exponentiate) 2y = ae−x /2 (a = ±e c) cDespite the fact that e =� 0, a = 0 is possible along with all a =� 0, depending on the initial conditions. For instance, if y(0) = 1, then y = e−x 2/2 . If y(0) = a, then y = ae−x 2/2 (See Fig. 1). 1� � Lecture 16 18.01Fall 2006 −6 −4 −2 0 2 4 600.20.40.60.81XY2 Figure 1: Graph of y = e− x 2 . In general: dy = f(x)g(y)dx dy = f(x)dx which we can write as g(y) 1 h(y)dy = f (x)dx where h(y) = . 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: 2 f(x) = x; F (x) = −2 x; 1 1 g(y) = y; h(y) = g(y)= y, H(y) = ln |y| 2Lecture 16 18.01Fall 2006 dy � y � Example 3 (Geometric Example). = 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. (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 = 2dx (separate variables) y x ln |y| |y| = = 2 ln |x| + c (antiderivative) e c x 2 (exponentiate; remember, e 2 ln |x| = x 2 ) Thus, y = ax 2 Again, a < 0, a > 0 and a = 0 are all acceptable. Possible solutions include, for example, y = x 2 (a = 1) y = 2x 2 (a = 2) y = −x 2 (a = −1) y = 0x 2 = 0 (a = 0) y = −2y 2 (a = −2) y = 100x 2 (a = 100) 3� � � Lecture 16 18.01Fall 2006 Example 4. Find the curves that are perpendicular to the parabolas in Example 3. We know that their slopes, dy −1 −x = = dx slope of parabola 2y Separate variables: ydy = −xdx2 Take the antiderivative: 2 2 2 2y x x y2= − 4+ c = ⇒ 4+2= c 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 Exam Review Exam 2 will be harder than exam 1 — be warned! Here’s a list of topics that exam 2 will cover: 1. Linear and/or quadratic approximations 2. Sketches of y = f(x) 3. Maximum/minimum problems. 4. Related rates. 5. Antiderivatives. Separation of variables. 6. Mean value theorem. More detailed notes on all of these topics are provided in the Exam 2 review sheet.
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