What is the solution for y^{''}+4y^'+4y=x^2e^{-2x} ?

The solution for y^{''}+4y^'+4y=x^2e^{-2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^4)/(12)

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''''}} - 2 y^{^{''}} + y = 11 t + 8

y^{^{''}} + \frac{4}{490} y^{^{'}} + 20 y = \frac{2}{245} cos (6 t)

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

What is the solution for y^{''}+4y^'+4y=x^2e^{-2x} ?
The solution for y^{''}+4y^'+4y=x^2e^{-2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^4)/(12)