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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} + y^{^{'}} = sec (t)

y^{^{''}} - 4 y = \frac{e ^{2 x}}{x}

y^{^{'''}} + y^{^{'}} = tan (t)

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

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