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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}})

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))