What is the solution for y^{''}+4y^'+4y=xe^{2x} ?

The solution for y^{''}+4y^'+4y=xe^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{2x}x)/(16)-1/32 e^{2x}

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y^{''}+4y^'+4y=xe^{2x}

y^{^{''}} + 4 y^{^{'}} + 4 y = x e^{2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = x e^{2 x}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

What is the solution for y^{''}+4y^'+4y=xe^{2x} ?
The solution for y^{''}+4y^'+4y=xe^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{2x}x)/(16)-1/32 e^{2x}