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

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

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

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

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 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}{d x} ((2 x^{2} - 6)^{- 10})

x \to 0 lim (\frac{3}{x ( x + 3 ) ^{2}})

\int_{1}^{e} x ln (x) d x

\frac{d ^{2} y}{d x ^{2}} - y = 0

x \to 0 lim (3 (x)^{\frac{x}{2}})

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