What is the solution for y^{''}-x^2+y^'=0 ?

The solution for y^{''}-x^2+y^'=0 is y=c_{1}+c_{2}e^{-x}+(x^3)/3-x^2+2x

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y^{''}-x^2+y^'=0

y^{^{''}} - x^{2} + y^{^{'}} = 0

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

Solution steps

y^{''} (x) - x^{2} + y^{'} (x) = 0

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

\int \frac{1 - x ^{2}}{x} d x

f (x) = 7^{x}

\frac{2}{3} t^{3} \cdot e^{- 8 t}

\frac{d}{d x} (\frac{4}{x ^{2}} - \frac{2}{x})

3 y^{3} e^{x} d x + e^{- x} d y = 3 y^{3} d x

What is the solution for y^{''}-x^2+y^'=0 ?
The solution for y^{''}-x^2+y^'=0 is y=c_{1}+c_{2}e^{-x}+(x^3)/3-x^2+2x