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

The solution for y^{''}+4y^'+4y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x+7/8

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

What is the solution for y^{''}+4y^'+4y=x^2-2x ?
The solution for y^{''}+4y^'+4y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x+7/8