4.13.2.4 Example 4

\begin{equation} y^{\prime }=\frac {y}{x}\ln \left ( xy-1\right ) \tag {1}\end{equation}

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

\begin{align*} m & =\frac {f+xf_{x}}{f-yf_{y}}\\ & =\frac {\frac {y}{x}\ln \left ( xy-1\right ) +x\left ( \frac {-y\ln \left ( xy-1\right ) }{x^{2}}+\frac {y^{2}}{x\left ( xy-1\right ) }\right ) }{\frac {y}{x}\ln \left ( xy-1\right ) -y\left ( \frac {\ln \left ( xy-1\right ) }{x}+\frac {y}{xy-1}\right ) }\\ & =\frac {\frac {y^{2}}{xy-1}}{-\frac {y^{2}}{xy-1}}\\ & =-1 \end{align*}

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

\[ v^{\prime }=\frac {v\ln \left ( v\right ) }{x}\]

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