Example 4

\begin{matrix} (1) & y^{'} = \frac{y}{x} \ln (x y - 1) \end{matrix}

We start by checking if it is isobaric or not. Using

\begin{aligned} m & = \frac{f + x f_{x}}{f - y f_{y}} \\ = \frac{\frac{y}{x} \ln (x y - 1) + x (\frac{- y \ln (x y - 1)}{x^{2}} + \frac{y^{2}}{x (x y - 1)})}{\frac{y}{x} \ln (x y - 1) - y (\frac{\ln (x y - 1)}{x} + \frac{y}{x y - 1})} \\ = \frac{\frac{y^{2}}{x y - 1}}{- \frac{y^{2}}{x y - 1}} \\ = - 1 \end{aligned}

Hence the substitution $y = \frac{v}{x}$ will make the ode separable. Substituting this in (1) results in

v^{'} = \frac{v \ln (v)}{x}

Which is separable. This is solved for $v$ , and then $y$ is found from $y = \frac{v}{x}$ .