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}

x^{^{''}} + x = 2 e^{t}

y^{^{'''}} - 3 y^{^{''}} + 4 y = 12 e^{2 x} + 4 e^{3 x}

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

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

y^{^{''}} + y^{^{'}} - 12 y = 4, y (0) = 0, y^{^{'}} (0) = 0

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)