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 first step is to verify that \(\bar {x}=\epsilon x,\bar {y}=\epsilon y\) leaves the ode invariant. \[ \frac {d\bar {y}}{d\bar {x}}=\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\frac {\epsilon y^{\prime }}{\epsilon }=y^{\prime }\] Hence the ode becomes\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =3-2\frac {\bar {y}}{\bar {x}}\\ y^{\prime } & =3-2\frac {\epsilon y}{\epsilon x}\\ & =3-2\frac {y}{x} \end {align*}
Verified. Now the ode is solved. The tangent curves are computed directly from the Lie group symmetry given above \begin {align*} \xi & =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}=x\\ \eta & =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}=y \end {align*}
The canonical coordinates \(\left ( R,S\right ) \) are now found. Using\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x} & =\frac {dy}{y}=dS \tag {1} \end {align}
The first pair gives\begin {align*} \frac {dy}{dx} & =\frac {y}{x}\\ \ln y & =\ln x+c_{1}\\ y & =cx \end {align*}
Hence \begin {align*} R & =c\\ & =\frac {y}{x} \end {align*}
Now we find \(S\) from the last pair of equations\begin {align*} \frac {dy}{y} & =dS\\ S & =\ln y \end {align*}
What is left is to find \(\frac {dS}{dR}\). This is given by\[ \frac {dS}{dR}=G\left ( R\right ) \] To find \(G\left ( R\right ) \), we use \(dS=S_{x}dx+S_{y}dy=\frac {1}{y}dy\) and \(dR=R_{x}dx+R_{y}dy=-\frac {y}{x^{2}}dx+\frac {1}{x}dy\). Hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {1}{y}dy}{-\frac {y}{x^{2}}dx+\frac {1}{x}dy}\\ & =\frac {\frac {dy}{dx}}{-\frac {y^{2}}{x^{2}}+\frac {y}{x}\frac {dy}{dx}}\\ & =\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}} \end {align*}
But \(\frac {dy}{dx}=3-2\frac {y}{x}=3-2R\), hence\begin {align*} \frac {dS}{dR} & =\frac {3-2R}{-R^{2}+R\left ( 3-2R\right ) }\\ & =\frac {3-2R}{3\left ( R-R^{2}\right ) } \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 {3-2R}{3\left ( R-R^{2}\right ) }dR\\ S & =\ln R-\frac {1}{3}\ln \left ( R-1\right ) +c_{1} \end {align*}
Final step is to replace \(R,S\) back with \(x,y\) which gives\begin {align*} \ln y & =\ln \frac {y}{x}-\frac {1}{3}\ln \left ( \frac {y}{x}-1\right ) +c_{1}\\ y & =c_{1}\frac {\frac {y}{x}}{\left ( \frac {y}{x}-1\right ) ^{\frac {1}{3}}}\\ \left ( \frac {y}{x}-1\right ) ^{\frac {1}{3}} & =c_{1}\frac {1}{x}\\ \frac {y}{x}-1 & =c_{2}\frac {1}{x^{3}}\\ y & =\left ( c_{2}\frac {1}{x^{3}}+1\right ) x \end {align*}