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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{4} + \frac{1}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{4} + \frac{1}{8}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} - \frac{x}{4} + \frac{1}{8}

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

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e^{- 2 x}

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 4 x = 4 e^{- 2 t}

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

(D^{2} + D) y = sin (x)

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