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

The solution for (dy)/(dx)=(e^y+e^{-y})/(e^x-e^{-x)} is y=ln(tan(c_{1}+(-ln(e^x+1)+ln(e^x-1))/2))

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(dy)/(dx)=(e^y+e^{-y})/(e^x-e^{-x)}

\frac{d y}{d x} = \frac{e ^{y} + e ^{- y}}{e ^{x} - e ^{- x}}

y = ln (tan (c_{1} + \frac{- ln ( e ^{x} + 1 ) + ln ( e ^{x} - 1 )}{2}))

Solution steps

\frac{d y}{d x} = \frac{e ^{y} + e ^{- y}}{e ^{x} - e ^{- x}}

Solve separable ODE:

y = ln (tan (c_{1} + \frac{- ln ( e ^{x} + 1 ) + ln ( e ^{x} - 1 )}{2}))

y = ln (tan (c_{1} + \frac{- ln ( e ^{x} + 1 ) + ln ( e ^{x} - 1 )}{2}))

\int (e^{2 x} + sin (2 x)) d x

\frac{d}{d x} (sin (x) + \frac{8}{3} cot (x))

\frac{d}{d x} (\frac{1}{2} x + 1)

\frac{d}{d x} (\frac{9 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\frac{d}{d x} (\frac{x ^{5} - 5 x}{5})

What is the solution for (dy)/(dx)=(e^y+e^{-y})/(e^x-e^{-x)} ?
The solution for (dy)/(dx)=(e^y+e^{-y})/(e^x-e^{-x)} is y=ln(tan(c_{1}+(-ln(e^x+1)+ln(e^x-1))/2))