What is the solution for y^{''}-2y^'-3y=xe^{-x} ?

The solution for y^{''}-2y^'-3y=xe^{-x} is y=c_{1}e^{3x}+c_{2}e^{-x}-(e^{-x}x^2)/8-(e^{-x}x)/(16)

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

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{8} - \frac{e ^{- x} x}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{8} - \frac{e ^{- x} x}{16}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{e ^{- x} x ^{2}}{8} - \frac{e ^{- x} x}{16}

What is the solution for y^{''}-2y^'-3y=xe^{-x} ?
The solution for y^{''}-2y^'-3y=xe^{-x} is y=c_{1}e^{3x}+c_{2}e^{-x}-(e^{-x}x^2)/8-(e^{-x}x)/(16)