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))

\frac{d}{d t} (5 e^{t})

f (x) = x^{4} - 5 x^{2} + 4, at x = 1

f (x) = x^{3} (x - 9)^{2}

\frac{d}{d x} (6 x^{5} - 15 x^{4} + 30 x^{2} - 30 x + 9)

x \to - 2 lim (x^{5})

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))