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

The solution for y^{''}+4y=x-x^2 is y=c_{1}cos(2x)+c_{2}sin(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} cos (2 x) + c_{2} sin (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} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2}}{4} + \frac{x}{4} + \frac{1}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2}}{4} + \frac{x}{4} + \frac{1}{8}

y^{^{''}} - 6 y^{^{'}} + 5 y = e^{- 3 x}

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

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

\frac{d ^{2} y}{d t ^{2}} + 2 \cdot \frac{d y}{d t} + 2 y = 11 t^{3}

3 y^{^{''}} + 27 y^{^{'}} + 42 y = 25 sin (t)

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