3.4.10.16 Example \(y^{\prime }=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}\)

This is homogeneous ODE of Class A of form \(y^{\prime }=F\left ( \frac {y}{x}\right ) \), hence from the lookup table

\begin{align*} \xi & =x\\ \eta & =y \end{align*}

Canonical coordinates \(\left ( R,S\right ) \) are found similar to the above which gives

\begin{align*} R & =\frac {y}{x}\\ S & =\ln y \end{align*}

What is left is to find \(\frac {dS}{dR}\). This is given by

\[ \frac {dS}{dR}=G\left ( R\right ) \]

Which is the same as above

\[ \frac {dS}{dR}=\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}}\]

But in this problem, the only difference is that \(\frac {dy}{dx}=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}=\frac {-3+R}{-1-R}\), hence

\begin{align*} \frac {dS}{dR} & =\frac {\frac {-3+R}{-1-R}}{-R^{2}+R\left ( \frac {-3+R}{-1-R}\right ) }\\ & =\frac {1}{R}\frac {R-3}{R^{2}+2R-3}\end{align*}

Which is a quadrature. In Lie method, for first order ode, we always obtain \(\frac {dS}{dR}=G\left ( R\right ) \). Integrating the above gives

\begin{align*} \int dS & =\int \frac {1}{R}\left ( \frac {R-3}{R^{2}+2R-3}\right ) dR\\ S & =\ln \left ( R\right ) -\frac {1}{2}\ln \left ( R+3\right ) -\frac {1}{2}\ln \left ( R-1\right ) +c_{1}\end{align*}

Final step is to replace \(R,S\) back with \(x,y\) which gives

\[ \ln y=\ln \left ( \frac {y}{x}\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}+3\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}-1\right ) +c_{1}\]

This can be solved for \(y\) if an explicit solution is needed.