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

The solution for y^{''}+2y^'=x^2 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+x/4

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{- 3 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 100 y = csc (10 x)

y^{^{''}} + y^{^{'}} + y = x^{4} + 4 x^{3} + 12 x^{2}

(D^{2} + 5 D + 6) y = (D + 1) x

y^{^{''}} + 4 y^{^{'}} = 2 sec^{3} (2 t)

What is the solution for y^{''}+2y^'=x^2 ?
The solution for y^{''}+2y^'=x^2 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+x/4