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})

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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})