Question

Find the integral of 

(y2-1) dx-2dy = 0

Answer 1

$(y^{2}−1)dx−2dy=0$

$y^{2}−1−2y′=0$

$y^{2}−1=2y'$

$\frac{2}{y^{2}-1}dy=dx$

$\intop\frac{2}{y^{2}-1}dy=\intop dx$

$\frac{1}{2}.2 in(\frac{y-1}{y+1})=x+ C$

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

$\frac{y-1}{y+1}=Ae^{x}$

$y-1=A(y+1)e^{x}$

$y(1-Ae^{x})=Ae^{x}+1$

$y=\frac{1+Ae^{x}}{1-Ae^{x}}$