1.12.16 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 \(\left ( X,Y\right ) \) are found similar to the above which gives

\begin{align*} X & =\frac {y}{x}\\ Y & =\ln y \end{align*}

What is left is to find \(\frac {dY}{dX}\). This is given by

\[ \frac {dY}{dX}=f\left ( X\right ) \]

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 \(\frac {dy}{dx}=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}=\frac {-3+X}{-1-X}\), hence

\begin{align*} \frac {dY}{dX} & =\frac {\frac {-3+X}{-1-X}}{-X^{2}+X\left ( \frac {-3+X}{-1-X}\right ) }\\ & =\frac {1}{X}\frac {X-3}{X^{2}+2X-3}\end{align*}

Which is a quadrature. In Lie method, for first order ode, we always obtain \(\frac {dY}{dX}=f\left ( X\right ) \). Integrating the above gives

\begin{align*} \int dY & =\int \frac {1}{X}\left ( \frac {X-3}{X^{2}+2X-3}\right ) dX\\ Y & =\ln \left ( X\right ) -\frac {1}{2}\ln \left ( X+3\right ) -\frac {1}{2}\ln \left ( X-1\right ) +c_{1}\end{align*}

Final step is to replace \(X,Y\) 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.