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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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