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*}
The canonical ode is\[ \frac {dS}{dR}=\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}}\] The above is the same ode in canonical coordinates for any ode of the form \(y^{\prime }=F\left ( \frac {y}{x}\right ) \). We just need to express \(y^{\prime }\) as function of \(R\). In this case the above becomes\begin {align*} \frac {dS}{dR} & =\frac {\frac {1+3R^{2}}{2R}}{-R^{2}+R\left ( \frac {1+3R^{2}}{2R}\right ) }\\ & =\frac {3R^{2}+1}{R^{3}+R} \end {align*}
Integrating gives\[ S=\ln \left ( R\left ( R^{2}+1\right ) \right ) +c_{1}\] Final step is to replace \(R,S\) back with \(x,y\) which gives\begin {align*} \ln y & =\ln \left ( \frac {y}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \right ) +c_{1}\\ y & =c_{2}\frac {y}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \\ 1 & =\frac {c_{2}}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \\ \frac {y^{2}}{x^{2}} & =c_{3}x-1\\ y^{2} & =c_{3}x^{3}-x^{2} \end {align*}
Hence\begin {align*} y & =\pm \sqrt {c_{3}x^{3}-x^{2}}\\ & =\pm x\sqrt {c_{3}x-1} \end {align*}
Finding \(\xi ,\eta \) from symmetry condition for the above ode This shows how to find \(\xi ,\eta \) directly also. The condition of symmetry is given above in equation (14) as \begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} Try Ansatz \begin {align*} \xi & =c_{0}+c_{1}x\\ \eta & =c_{2}+c_{3}y \end {align*}
And given\begin {align*} \omega & =\frac {1}{2}\frac {x^{2}+3y^{2}}{xy}\\ \omega ^{2} & =\frac {1}{4}\frac {\left ( x^{2}+3y^{2}\right ) ^{2}}{x^{2}y^{2}}\\ \omega _{x} & =\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\\ \omega _{y} & =\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}} \end {align*}
Hence (14) becomes\[ \eta _{x}+\frac {1}{2}\frac {x^{2}+3y^{2}}{xy}\eta _{y}-\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\xi -\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}}\eta =0 \] Therefore the above becomes\[ \frac {1}{2}\frac {x^{2}+3y^{2}}{xy}c_{3}-\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\left ( c_{0}+c_{1}x\right ) -\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}}\left ( c_{2}+c_{3}y\right ) =0 \] Using the computer the above simplifies to \[ \frac {x}{y}\left ( c_{3}-c_{1}\right ) +\frac {1}{2}c_{2}\frac {x}{y^{2}}-\frac {1}{y}\left ( \frac {1}{2}c_{0}\right ) -\frac {1}{x}\frac {3}{2}c_{2}+\frac {3}{2}c_{0}\frac {y}{x^{2}}=0 \] Hence \begin {align*} c_{3}-c_{1} & =0\\ \frac {1}{2}c_{2} & =0\\ -\frac {1}{2}c_{0} & =0\\ -\frac {3}{2}c_{2} & =0\\ \frac {3}{2}c_{0} & =0 \end {align*}
Solving gives \(c_{0}=0,c_{2}=0\) and \(c_{3}=c_{1}\). Hence the solution is \begin {align*} \xi & =c_{1}x\\ \eta & =c_{3}y \end {align*}
Let \(c_{1}=1\), therefore \(c_{3}=1\) and we obtain\begin {align*} \xi & =x\\ \eta & =y \end {align*}
Which is the result we used in solving the above problem. Notice that any scaler will also work. Hence\begin {align*} \xi & =5x\\ \eta & =5y \end {align*}
And\begin {align*} \xi & =10x\\ \eta & =10y \end {align*}
This will also give same solution.