What is the solution for y^'=((x^2y-y))/(y+1) ?

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

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y^'=((x^2y-y))/(y+1)

y^{^{'}} = \frac{( x ^{2} y - y )}{y + 1}

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

Solution steps

y^{'} (x) = \frac{( x ^{2} y - y )}{y + 1}

Solve separable ODE:

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

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