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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = (sec (x))^{3}

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

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

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

y^{^{''}} - 12 y^{^{'}} + 36 y = 18 e^{6 x}

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