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Unformatted text preview: Exercise Set 5 Math 4027 Due: May 5, 2005 1. Solve the following differential equations: (a) y = x 2 /y . I Solution. The equation is separable, so rewrite it in the form yy = x 2 or in differential form y dy = x 2 dx and integrate to get an implicit equation y 2 2 = x 3 3 + C. A solution is any function ( x ) implicitly defined by this equation on some interval I , so long as ( x ) 6 = 0 for x I . J (b) y + y 2 sin x = 0 I Solution. Separate the variables to get y- 2 y =- sin x or in differential form y- 2 dy =- sin xdx and integrate to get an implicit equation- 1 y = cos x + C. This can be solved explicitly for y on any interval I for which cos x + C 6 = 0 for all x I to get y =- 1 cos x + C . There is also one constant solution, namely y = 0, which is lost in the process of separating the variables. J 2. Solve the initial value problem y = 3 x 2 3 y 2- 4 , y (1) = 0 , and determine the interval in which the solution is valid. ( Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.) I Solution. Separate the variables to get (3 y 2- 4) y = 3 x 2 or in differential form (3 y 2- 4) dy = 3 x 2 dx and integrate to get an implicit equation y 3- 4 y = x 3 + C, where C is an integration constant. The initial value y (1) = 0 means that y = 0 when x = 1, so this implies that C =- 1 and we get that the initial value problem is defined...
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