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

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

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

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

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

Solution steps

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

Solve exact ODE:

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

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

\int \frac{sin ( 2 x )}{12 + cos ^{2} ( x )} d x

\frac{2}{( s ^{3} - 7 s ^{2} + 2 s + 40 )}

arccos (∣ x + 1 ∣)

f (x) = 2 x^{2} - 5 x + 3, at a = 2

\frac{\partial}{\partial x} (ln (2 x^{2} + x y))

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