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

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

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

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

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

Solution steps

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

\frac{d y}{d x} = - 10 x^{4} y

x \to 0 + lim (cos^{x} (\frac{π}{2} - x))

\int 5 cos (2 x) d x

y^{^{''}} + 4 y^{^{'}} + 4 = 0, y^{^{'}} (0) = 1

a re a f (x) = cos (x), - \frac{π}{2}, \frac{π}{2}

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