What is the solution for y^'=(e^y)/((x^2-1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=(e^y)/((x^2-1))

y^{^{'}} = \frac{e ^{y}}{( x ^{2} - 1 )}

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

Solution steps

y^{'} (x) = \frac{e ^{y}}{( x ^{2} - 1 )}

Solve separable ODE:

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

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

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