What is the solution for (D^2+4D+4)y=xe^{-x} ?

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

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(D^2+4D+4)y=xe^{-x}

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 9 t^{2}

y^{^{''}} - 3 y^{^{'}} - 40 y = (t + 1) e^{t}

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}

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

y^{^{''}} + 0.7 y + 4 = cos (3 t)

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