Solve \begin {align*} y^{\prime } & =3\sqrt {yx}\\ y^{\prime } & =\omega \left ( x,y\right ) \end {align*}
The symmetry condition results in the pde\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {1} \end {equation} Trying polynomial anstaz \begin {align*} \xi & =a_{0}+a_{1}x\\ \eta & =b_{0}+b_{1}y \end {align*}
And substituting these into (1) and simplifying gives\[ \left ( -9a_{1}+3b_{1}\right ) yx-3xb_{0}-3ya_{0}=0 \] Setting all coefficients to zero gives\begin {align*} -9a_{1}+3b_{1} & =0\\ b_{0} & =0\\ a_{0} & =0 \end {align*}
Hence \(a_{1}=\frac {1}{3}b_{1}\). Letting \(b_{1}=1\) then \(a_{1}=\frac {1}{3}\) and the infinitesimals are \begin {align*} \xi & =\frac {1}{3}x\\ \eta & =y \end {align*}
The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{y-\frac {1}{3}x\left ( 3\sqrt {yx}\right ) }\\ & =-\frac {y+x\sqrt {xy}}{x^{3}y-y^{2}} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} The first pair of equations gives\[ \frac {dy}{dx}=\frac {\eta }{\xi }=\frac {3y}{x}\] Solving gives\[ y=c_{1}x^{3}\] Hence \begin {equation} R=c_{1}=\frac {y}{x^{3}} \tag {2} \end {equation} And \(S\) is found from \[ dS=\frac {dx}{\xi }=3\frac {dx}{x}\] Integrating gives\begin {align*} S & =3\ln x+c_{1}\\ & =3\ln x \end {align*}
By choosing \(c_{1}=0\). Now that \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) are found, the ODE \(\frac {dS}{dR}=F\left ( R\right ) \) is determined. This is determined from\begin {align*} \frac {dS}{dR} & =\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}\\ & =\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}} \end {align*}
But \(S_{x}=\frac {3}{x},R_{x}=-3\frac {y}{x^{4}},S_{y}=0,R_{y}=\frac {1}{x^{3}}\). Substituting these into the above gives\begin {align*} \frac {dS}{dR} & =\frac {\frac {3}{x}}{-3\frac {y}{x^{4}}+\omega \left ( x,y\right ) \frac {1}{x^{3}}}\\ & =\frac {3x^{3}}{-3y+x\omega \left ( x,y\right ) } \end {align*}
But \(\omega \left ( x,y\right ) =3\sqrt {yx}\). The above becomes\begin {align} \frac {dS}{dR} & =\frac {3x^{3}}{-3y+3x\sqrt {yx}}\nonumber \\ & =\frac {x^{3}}{x\sqrt {yx}-y}\nonumber \\ & =\frac {-1}{\sqrt {\frac {y}{x^{3}}}-\frac {y}{x^{3}}} \tag {3} \end {align}
But \(R=\frac {y}{x^{3}}\) and the above becomes\[ \frac {dS}{dR}=\frac {-1}{R-\sqrt {R}}\] Which is a quadrature. Solving gives\begin {align*} \int dS & =\int \frac {-1}{R-\sqrt {R}}dR\\ S & =-2\ln \left ( \sqrt {R}-1\right ) +c_{1} \end {align*}
Converting back to \(x,y\) gives\begin {align*} 3\ln x & =-2\ln \left ( \sqrt {\frac {y}{x^{3}}}-1\right ) +c_{1}\\ \ln x^{3}+\ln \left ( \sqrt {\frac {y}{x^{3}}}-1\right ) ^{2} & =c_{1}\\ \ln \left ( x^{3}\left ( \sqrt {\frac {y}{x^{3}}}-1\right ) ^{2}\right ) & =c_{1}\\ x^{3}\left ( \sqrt {\frac {y}{x^{3}}}-1\right ) ^{2} & =c_{2} \end {align*}
Or\begin {align*} y_{1}\left ( x\right ) & =2x\left ( x^{2}+x\sqrt {xc_{1}}\right ) -x^{3}+c_{1}\\ y_{2}\left ( x\right ) & =-2x\left ( -x^{2}+x\sqrt {xc_{1}}\right ) -x^{3}+c_{1} \end {align*}