Answer
\[\frac{y-1}{y+1}=\frac{C(x-2)^2}{x-1}\]
Work Step by Step
\[\frac{dy}{dx}=\frac{x(y^2-1)}{2(x-2)(x-1)}\]
Seperating the variables,
\[\frac{2}{y^2-1}dy=\frac{x}{(x-1)(x-2)}dx\]
\[\frac{2}{y^2-1}dy=\left[\frac{-1}{x-1}+\frac{2}{x-2}\right]dx\]
Integrating,
\[\int\frac{2}{y^2-1}dy=\int\left[\frac{-1}{x-1}+\frac{2}{x-2}\right]dx+\ln C\]
Where $\ln C$ is constant of integration
\[\frac{2}{2}\ln \left|\frac{y-1}{y+1}\right|=-\ln |x-1|+2\ln |x-2|+\ln C\]
\[\ln \left|\frac{y-1}{y+1}\right|=\ln \left|\frac{C(x-2)^2}{x-1}\right|\]
\[\frac{y-1}{y+1}=\frac{C(x-2)^2}{x-1}\]
Hence Solution of given differential equation is $\frac{y-1}{y+1}=\frac{C(x-2)^2}{x-1}$.
