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

The solution for (D^2+2D+1)y=xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^3)/6

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(D^2+2D+1)y=xe^{-x}

(D^{2} + 2 D + 1) y = x e^{- x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6}

Solution steps

(D^{2} + 2 D + 1) y = x e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6}

\frac{d ^{2} y}{d x ^{2}} - y = 1

y^{^{''}} + y^{^{'}} + 1.25 y = 3 cos (x)

y^{^{''}} = 9 + (y^{^{'}}) 2

y^{^{''}} + 6 y^{^{'}} - 27 y = - 105 t + 54 t^{2}

y^{^{''}} - 6 y^{^{'}} + 5 y = x, y^{1} (0) = e^{x}

What is the solution for (D^2+2D+1)y=xe^{-x} ?
The solution for (D^2+2D+1)y=xe^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^3)/6