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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = 12 x^{3} e^{x}

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 t}

y^{^{'''}} - 6 y^{^{''}} - 6 = cos (x)

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 3 + 3 x

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

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