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Unformatted text preview: y (0) = ( c 1 e t + c 2 e 5 t + c 3 e 8 t ) t =0 = c 1 + c 2 + c 3 = 3 , y (0) = ( c 1 e t + 5 c 2 e 5 t + 8 c 3 e 8 t ) t =0 = c 1 + 5 c 2 + 8 c 3 = 6 , y (0) = ( c 1 e t + 25 c 2 e 5 t + 64 c 3 e 8 t ) t =0 = c 1 + 25 c 2 + 64 c 3 = 12 c 1 = 1 , c 2 = 3 , c 3 = 1 . Therefore, y ( t ) = e t 3 e 5 t + e 8 t is the solution to the given initial value problem. 35. Since the roots of the auxiliary equation, r 2 + 1 = 0, are r = i , the functions y 1 ( ) = cos and y 2 ( ) = sin are two linearly independent solutions to the corresponding homogeneous equation, and its general solution is given by y h ( ) = c 1 cos + c 2 sin . 254...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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