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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 15 = x^{2}

y^{^{''}} + y = sec (x) \cdot tan (x)

y^{^{''}} - y = e^{2 t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 6 sin (x))

y^{^{''}} + 49 y = 13 sec^{2} (7 t)

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