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384_pdfsam_math 54 differential equation solutions odd

# 384_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 So, v 1 ( x ) = g ( x ) W 1 ( x ) W [ x, x 2 , x 3 ]( x ) dx = x 4 x 4 2 x 3 dx = 1 4 x 2 + c 1 , v 2 ( x ) = g ( x ) W 2 ( x ) W [ x, x 2 , x 3 ]( x ) dx = x 4 ( 2 x 3 ) 2 x 3 dx = 1 3 x 3 + c 2 , v 3 ( x ) = g ( x ) W 3 ( x ) W [ x, x 2 , x 3 ]( x ) dx = x 4 ( x 2 ) 2 x 3 dx = 1 8 x 4 + c 3 , where c 1 , c 2 , and c 3 are constants of integration. Substitution back into (6.24) yields y p ( x ) = 1 4 x 2 + c 1 x + 1 3 x 3 + c 2 x 2 + 1 8 x 4 + c 3 x 3 = 1 24 x + c 1 x + c 2 x 2 + c 3 x 3 . Since { x, x 2 , x 3 } is a fundamental solution set for the homogeneous equation, taking c 1 , c 2 , and c 3 to be arbitrary constants, we obtain a general solution to the original nonhomogeneous equation. That is, y ( x ) = 1 24 x + c 1 x + c 2 x 2 + c 3 x 3 . 9. To find a particular solution to the nonhomogeneous equation, we will use the method of variation of parameters. We must first calculate the four determinants W [ e x , e x , e 2 x ]( x ), W 1 ( x ), W 2 ( x ), W 3 ( x ). Thus, we have W [ e x , e x , e 2 x ]( x ) = e x e x e 2 x e x e x 2 e 2 x e
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