386_pdfsam_math 54 differential equation solutions odd

386_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 whose roots are r = 1, 1, and 3. Therefore, the functions y 1 ( t ) = e t , y 2 ( t ) = e t , and y 3 ( t ) = e 3 t form a fundamental solution set. Substituting back e t = x we find that y 1 ( x ) = e t = x , y 2 ( x ) = e t = ( e t ) 1 = x 1 , y 3 ( x ) = e 3 t = ( e t ) 3 = x 3 form a fundamental solution set for the homogeneous equation (6.25). Next, we apply the variation of parameters to find a particular solution to the original equation. A particular solution has the form y p ( x ) = v 1 ( x ) x + v 2 ( x ) x 1 + v 3 ( x ) x 3 . (6.26) To find functions v 1 ( x ), v 2 ( x ), and v 3 ( x ) we use formula (11) on page 340 of the text. We compute W [ x, x 1 , x 3 ]( x ) = x x 1 x 3 1 x 2 3 x 2 0 2 x 3 6 x = x x 2 3 x 2 2 x 3 6 x x 1 x 3 2 x 3 6 x = 16 , W 1 ( x ) = ( 1) 3 1 W [ x 1 , x 3 ]( x ) = x 1 x 3 x
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