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

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

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(dy)/(dx)=(e^{x-y})/(1+e^x),y(1)=0

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

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

Solution steps

\frac{d y}{d 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))

\frac{d}{d x} (x^{\frac{1}{11}} (x - 1)^{2})

x \to \infty lim (6^{-})

y = lo g_{10} (e^{x})

\int \frac{1}{( x + 2 ) ( x + 5 )} d x

\frac{\partial}{\partial x} (\frac{x}{y} + sin (2 x + 3 y))

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