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Unformatted text preview: Homework 1  Due on January 22 (Thursday)  Solution 1 1. Find the general solutions of the following 1st order ODE: (a) xy + 2 y = sin x First, find the integrating factor ( x ): x ( x ) y + 2 ( x ) y = ( x )sin x . [ x ( x )] = 2 ( x ) ( x ) + x ( x ) = 2 ( x ) x ( x ) = ( x ) ( x ) / ( x ) = 1 /x [ln  ( x )  ] = 1 /x ln  ( x )  = ln  x  ( x ) = x Thus we have x 2 y + 2 xy = [ x 2 y ( x )] = x sin x . Hence, integrating by parts, we get x 2 y ( x ) = Z x sin xdx = x ( cos x ) Z ( cos x ) dx = x cos x +sin x + c. So the general solution is y ( x ) = cos x x + sin x x 2 + c x 2 . (b) (1 + x 2 ) y + 4 xy = 1 1 + x 2 First, find the integrating factor ( x ): (1 + x 2 ) ( x ) y + 4 x ( x ) y = ( x ) 1 + x 2 . (1 + x 2 ) ( x ) / = 4 x ( x ) 2 x ( x ) + (1 + x 2 ) ( x ) = 4 x ( x ) (1 + x 2 ) ( x ) = 2 x ( x ) ( x ) / ( x ) = 2 x/ (1 + x 2 ) [ln  ( x )  ] = [ln  1 + x 2  ] ln  ( x )  = ln  1 + x 2  ( x ) = 1 + x 2 Thus we have (1 + x 2 ) 2 y + 4 x (1 + x 2 ) y = [(1 + x 2 ) 2 y ] = 1. Hence, by integrating, we get (1 + x 2 ) 2 y ( x ) = x + c . So the general solution is y ( x ) = x + c (1 + x 2 ) 2 . MATH2400  Introduction to Differential Equations  Spring 2004  JeongRock Yoon Homework 1  Due on January 22 (Thursday)  Solution 2 2. Solve the following initial value problems:2....
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 Spring '04
 Yoon
 Differential Equations, Equations

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