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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = \frac{2 e ^{x}}{e ^{x} + e ^{- x}}

y^{^{''}} - 4 y^{^{'}} + 3 y = 9 x^{2} - 36 x + 37

y^{^{''}} + 3 y^{^{'}} = 2 x^{4} + x^{2} e^{- 3 x} + sin (3 x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + t, y (0) = 0, y^{^{'}} (0) = 4

x^{^{''}} + 4 x^{^{'}} + 5 x = 13 sin (2 t)

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