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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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