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

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

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{3}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{3}{8}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{2} + \frac{3}{8}

\int (8 x^{5} - 5 x^{4} - 4 x^{3} - 6 x^{2} - 2 x - 3) d x

\frac{30000}{1.8 t + 9}

\int \frac{1}{4 - 3 x} d x

(2.5) (- 1.7)

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