Math 293 prelim 2 solutions

# Math 293 prelim 2 solutions - Math 293 Prelim#2 Solutions 1...

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Math 293 Prelim #2 Solutions Spring 2000 1 . The equation x dy dx + y 2 = x 2 y 2 is separable since it can be rewritten as dy dx = xy 2 - y 2 x = y 2 ( x - x - 1 ) , from which we get y - 2 dy = ( x - x - 1 ) dx . Integrating both sides gives - y - 1 = x 2 2 - ln x + C , hence y = ( ln x - x 2 2 + C ) - 1 . Plugging in the initial condition y( 1 ) = 1 gives C = 3 2 , so y = ( ln x - x 2 2 + 3 2 ) - 1 . 2 . (a) An integrating factor for the equation y 0 + 4 y = e - x is e 4 dx = e 4 x . Multiplying the equation by this gives (e 4 x y) 0 = e 3 x . Integrating both sides gives e 4 x y = e 3 x 3 + C . Solving for y , we get y = e - x 3 + Ce - 4 x as the general solution. (b) Think of y 0 + 4 y = e - x as a second-order equation in which the coefficient of y 00 happens to be zero. To solve the associated homogeneous equation we look at the ‘quadratic’ equation 0 r 2 + r + 4 = 0, with the root r = - 4. This gives the general solution y = Ce - 4 x for the homogeneous equation. Next we find a particular solution of the nonhomogeneous equation by the method of undetermined coefficients, trying for a solution of the form y = Ae - x . Computing y 0 = - Ae - x and plugging into the nonhomogeneous equation, we find that A must be 1 3 , so we have the particular solution y = e - x 3 . Hence the general solution of the nonhomogeneous equation is y = e - x 3 + Ce - 4 x , the same answer as in part (a).

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