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))

x \to 2 + lim (2 x - [∣ x ∣])

x \to 0 lim (\frac{( x ^{2} + x )}{x})

\int \frac{ln ( x )}{2 x + 3} d x

x \to 2 lim (\frac{x ^{2} - 9}{x + 2})

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))