What is the solution for 2y^{''}+8y^'+8y=e^{-2x}+x ?

The solution for 2y^{''}+8y^'+8y=e^{-2x}+x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^2)/4+(x-1)/8

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2y^{''}+8y^'+8y=e^{-2x}+x

2 y^{^{''}} + 8 y^{^{'}} + 8 y = e^{- 2 x} + x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{2}}{4} + \frac{x - 1}{8}

Solution steps

2 y^{''} (x) + 8 y^{'} (x) + 8 y = e^{- 2 x} + x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{2}}{4} + \frac{x - 1}{8}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{2}}{4} + \frac{x - 1}{8}

What is the solution for 2y^{''}+8y^'+8y=e^{-2x}+x ?
The solution for 2y^{''}+8y^'+8y=e^{-2x}+x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^2)/4+(x-1)/8