230_pdfsam_math 54 differential equation solutions odd

230_pdfsam_math 54 differential equation solutions odd - y...

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Chapter 4 y 0 2 ( x )= 1 2 x 3 / 2 sin x + x 1 / 2 cos x. Thus, functions v 1 ( x )and v 2 ( x ) in a particular solution, y p ( x )= v 1 ( x ) y 1 ( x )+ v 2 ( x ) y 2 ( x ) , satisfy the system v 0 1 x 1 / 2 cos x + v 0 2 x 1 / 2 sin x =0 , v 0 1 ± 1 2 x 3 / 2 cos x x 1 / 2 sin x ² + v 0 2 ± 1 2 x 3 / 2 sin x + x 1 / 2 cos x ² = x 5 / 2 x 2 = x 1 / 2 . From the ±rst equation, we express v 0 1 = v 0 2 tan x and substitute this expression into the second equation. After some algebra, the result simpli±es to v 0 2 = x cos x v 0 1 = v 0 2 tan x = x sin x. Integrating, we get v 1 ( x )= Z x sin xdx = x cos x sin x + C 1 , v 2 ( x )= Z x cos xdx = x sin x +cos x + C 2 . With C 1 = C 2 =0 , y p ( x )=( x cos x sin x ) x 1 / 2 cos x +( x sin x +cos x ) x 1 / 2 sin x = x 1 / 2 .
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Unformatted text preview: y ( t ) = x 1 / 2 + x 1 / 2 ( c 1 cos x + c 2 sin x ) . EXERCISES 4.7: Qualitative Considerations for Variable-Coecient and Nonlinear Equations, page 208 1. Let Y ( t ) := y ( t ). Then, using the chain rule, we get dY dt = y ( t ) d ( t ) dt = y ( t ) , 226...
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