What is the solution for (1+e^x)y*y^'=e^y ?

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

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(1+e^x)y*y^'=e^y

(1 + e^{x}) y \cdot y^{^{'}} = e^{y}

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

Solution steps

(1 + e^{x}) y \cdot y^{'} (x) = e^{y}

Solve separable ODE:

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

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

x y^{^{'}} - y = 0, y (0) = x

y^{^{'}} - y = 0, y (0) = 2

\frac{d y}{d x} y = arctan (3) x^{2}

\frac{d y}{d t} = 0.04 - \frac{y}{( \frac{4320}{7} )}

y^{^{'}} = \frac{2 x y - 2 x}{x ^{2} + 3}

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