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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{1}{8} (y^{\frac{1}{2}}) (cos (y^{\frac{1}{2}}))^{2}

\frac{x ^{2} d y - y ^{2} d x}{( x - y ) ^{2}} = 0

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

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

x^{^{'}} = - 5 x, x (0) = 1

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