What is the solution for y^{''}+4y^'+4y=8x^2 ?

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

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y^{''}+4y^'+4y=8x^2

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 x^{2}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 8 x^{2}

Solve linear ODE:

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

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

\frac{d}{d x} (x e^{x} y)

\frac{d ^{3}}{d x ^{3}} (cos (x) - sin (x))

\int \frac{6}{x ( x + 2 )} d x

x \to 3 + lim (\frac{∣ x - 2 ∣}{x - 2})

\frac{\partial}{\partial x} (56 x^{2} - 8 y^{2} - 16 x - 31 + 1 - 8 x)

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