What is the solution for (x^2+y)+(x+e^y)y^'=0 ?

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

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(x^2+y)+(x+e^y)y^'=0

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{d y}{d x} = \frac{cos ( 10 x )}{e ^{10 y}}

x \to 0 lim (\frac{2 ^{x}}{3 ^{x}})

\frac{\partial}{\partial x} (ln (3 x + 5 y))

x \to 0 lim (\frac{x ^{2}}{x ^{2} + 1})

\int_{1}^{2} \frac{9}{x ^{3} + 5 x} d x

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