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

The solution for y^{''}+4y^'+4y=xe^x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^xx)/9-2/27 e^x

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{x} x}{9} - \frac{2}{27} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{x} x}{9} - \frac{2}{27} e^{x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{x} x}{9} - \frac{2}{27} e^{x}

y^{^{''}} - 7 y^{^{'}} + 10 y = 6 x + e^{2 x}

4 y^{^{''}} - 2 y^{^{'}} = 4 x, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 12 y = (4 x + 7) e^{- 2 x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{3 x}

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 6 e^{2 x} + 16 x^{2}

What is the solution for y^{''}+4y^'+4y=xe^x ?
The solution for y^{''}+4y^'+4y=xe^x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^xx)/9-2/27 e^x