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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\int e^{- \frac{x}{10}} d x

\frac{d}{d x} (20 - 2 x)

\frac{d}{d x} (\frac{x ^{4}}{8} ln (x) - \frac{x ^{4}}{32})

y = \frac{8}{x + 1}

x \to 0 lim (x \cdot e^{\frac{1}{x}})

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