I was trying to solve and get the general solution of the differential equation $y''-2y'+2y=x e^{x}\cos{2x}+e^x \sin{2x}+xe^x \sin{x}$ and I think that in order to get the particular solution it can be solved by the undetermined coefficients but it is very long, is there any other method or trick?

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    $\begingroup$ I believe undetermined coefficients is the easiest way $\endgroup$ – John Doe Apr 28 '19 at 2:11

What I should have done first is to let $y=z\,e^x$ to make the equation $$y''-2y'+2y=x e^{x}\cos{(2x)}+e^x \sin{(2x)}+xe^x \sin{(x)}$$ to become $$z''+z=x \cos{(2x)}+ \sin{(2x)}+x \sin{(x)}$$ which is more pleasant and easier to work using undetermined coefficients.

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  • $\begingroup$ can you explain why this change work like that with the equation? $\endgroup$ – Jack Talion Apr 28 '19 at 18:40

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