What is the solution for y^{''}+4y^'+4y=x^2+3 ?

The solution for y^{''}+4y^'+4y=x^2+3 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x/2+9/8

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{9}{8}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = x^{2} + 3

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{9}{8}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{9}{8}

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

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 3 e^{x}

4 y^{^{''}} - 3 y^{^{'}} = 4 x - 6

y^{^{''}} - 9 y^{^{'}} + 8 y = x, y (0) = e^{x}

What is the solution for y^{''}+4y^'+4y=x^2+3 ?
The solution for y^{''}+4y^'+4y=x^2+3 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x/2+9/8