What is the solution for y^'=(e^{x+y})/(1+e^x),y(1)=0 ?

The solution for y^'=(e^{x+y})/(1+e^x),y(1)=0 is y=-ln(-ln(1+e^x)+1+ln(1+e))

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y^'=(e^{x+y})/(1+e^x),y(1)=0

y^{^{'}} = \frac{e ^{x + y}}{1 + e ^{x}}, y (1) = 0

y = - ln (- ln (1 + e^{x}) + 1 + ln (1 + e))

Solution steps

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

Solve separable ODE:

y = - ln (- ln (1 + e^{x}) + 1 + ln (1 + e))

y = - ln (- ln (1 + e^{x}) + 1 + ln (1 + e))

\int 2^{2 - 2 x} d x

\int_{0}^{1} (e^{x} - x) d x

\frac{1}{x - 1}, 24

x = y (2 - y), x = - y

x \to 343 lim (x^{\frac{2}{3}})

What is the solution for y^'=(e^{x+y})/(1+e^x),y(1)=0 ?
The solution for y^'=(e^{x+y})/(1+e^x),y(1)=0 is y=-ln(-ln(1+e^x)+1+ln(1+e))