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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y = e^{2 t} - 2 t cos (3 t)

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

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

y^{^{''}} + 9 y = 2 csc^{2} (3 x)

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

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