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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + k y = 0

y^{^{'}} + 3 x e^{y} = 0

n = 1 \sum \infty \frac{3 ^{n}}{n 2 ^{n}}

\frac{\partial}{\partial y} (arctan (\frac{y}{x}))

y^{^{''}} + 18 y^{^{'}} + 97 y = 0

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