What is the solution for y^{''}-4y=xe^{-x} ?

The solution for y^{''}-4y=xe^{-x} is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^{-x}x)/3+2/9 e^{-x}

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y^{''}-4y=xe^{-x}

y^{^{''}} - 4 y = x e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{- x} x}{3} + \frac{2}{9} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{- x} x}{3} + \frac{2}{9} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{- x} x}{3} + \frac{2}{9} e^{- x}

4 \frac{d ^{2} y}{d x ^{2}} - y = x e^{\frac{x}{2}}

y^{^{''}} + 9 y = 15

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

y^{^{''}} + y = 4 sin (3 x)

y^{^{''}} - 5 y^{^{'}} = 25 (H (t - 2) - H (t - 4))

What is the solution for y^{''}-4y=xe^{-x} ?
The solution for y^{''}-4y=xe^{-x} is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^{-x}x)/3+2/9 e^{-x}