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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} - 9 y = \frac{9 x}{e ^{3 x}}

y^{^{''}} + y = 2 x sin (x)

y^{^{''}} - y^{^{'}} - 2 y = e^{3 x}

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