What is the solution for 0.25y^{''}+y^'+y=x^2-2x ?

The solution for 0.25y^{''}+y^'+y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-4x+3.5

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0.25y^{''}+y^'+y=x^2-2x

0.25 y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 4 x + 3.5

Solution steps

0.25 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 + 3.5

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 4 x + 3.5

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = x

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (2 x) + x

\frac{1}{2} x^{^{''}} + \frac{1}{2} x^{^{'}} + 6 x = 20 cos (3 t)

y^{^{''}} + 4 y^{^{'}} + 6 y = 4 e^{- 2 t} cos (t)

(d^{2} - 3 d + 2) y = x e^{x}

What is the solution for 0.25y^{''}+y^'+y=x^2-2x ?
The solution for 0.25y^{''}+y^'+y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-4x+3.5