What is the solution for 1/4 y^{''}+y^'+y=x^2-4x ?

The solution for 1/4 y^{''}+y^'+y=x^2-4x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-6x+11/2

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1/4 y^{''}+y^'+y=x^2-4x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 4 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 6 x + \frac{11}{2}

Solution steps

\frac{1}{4} y^{''} (x) + y^{'} (x) + y = x^{2} - 4 x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 6 x + \frac{11}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 6 x + \frac{11}{2}

y \cdot y^{^{'}} + 36 x = 0

y = x, y = \frac{x}{2}

\frac{d}{d x} (a e^{- 2 x})

\frac{d y}{d x}, 4 x^{2} y^{7} - 2 x = x^{5} + 4 y^{3}

\frac{d}{d x} (4 x (x^{2} - 2)^{2})

What is the solution for 1/4 y^{''}+y^'+y=x^2-4x ?
The solution for 1/4 y^{''}+y^'+y=x^2-4x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-6x+11/2