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Unformatted text preview: Exercise Set 1 Math 4027 Due: February 1, 2005 1. Find the solution of the initial value problems: (a) y + 2 y = x, y (0) = 1, I Solution. Multiply by e 2 x to get ( e 2 x y ) = xe 2 x and then integrate to get e 2 x y = x 2 e 2 x 1 4 e 2 x + C. Solve for y and substitute y (0) = 1 to get C = 5 / 4. Hence y = 1 4 ( 2 x 1 + 5 e 2 x ) . J (b) y y = sin x, y (0) = 2. I Solution. Multiply by e x to get ( e x y ) = e x sin x and then integrate to get e x y = 1 2 e x (sin x + cos x ) + C. Solve for y and substitute y (0) = 2 to get C = 5 / 2. Hence y = 5 2 e x 1 2 (sin x + cos x ) . J 2. Find the general solution of (a) y + 2 y = x 2 , I Solution. Multiply by e 2 x to get ( e 2 x y ) = x 2 e 2 x and then integrate to get e 2 x y = x 2 2 e 2 x x 2 e 2 x + 1 4 e 2 x + C. Solve for y to get y = 1 4 ( 2 x 2 2 x + 1 ) + Ce 2 x . J (b) y y = sin2 x . I Solution. Multiply by e x to get ( e x y ) = e x sin2 x and then integrate to get e x y = 1 5 e x (sin2 x + 2cos2 x ) + C....
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This note was uploaded on 03/20/2011 for the course MATH 4027 taught by Professor Adkins during the Spring '06 term at LSU.
 Spring '06
 Adkins
 Math, Differential Equations, Equations

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