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

The solution for y^{''}-y^'-2y=x^2 is y=c_{1}e^{2x}+c_{2}e^{-x}-(x^2)/2+x/2-3/4

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{x ^{2}}{2} + \frac{x}{2} - \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{x ^{2}}{2} + \frac{x}{2} - \frac{3}{4}

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{x ^{2}}{2} + \frac{x}{2} - \frac{3}{4}

\frac{d}{d x} ((3 x - 1)^{3})

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\frac{\partial}{\partial x} (2 cos (x + 2 y))

x^{2} sin^{4} (x) + x cos^{- 2} (x)

What is the solution for y^{''}-y^'-2y=x^2 ?
The solution for y^{''}-y^'-2y=x^2 is y=c_{1}e^{2x}+c_{2}e^{-x}-(x^2)/2+x/2-3/4