What is the solution for y^{''}+4y^'+4y=8e^{-2}x ?

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

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y^{''}+4y^'+4y=8e^{-2}x

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 e^{- 2} x

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 8 e^{- 2} x

Solve linear ODE:

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

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

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