Solve \begin {align*} y^{\prime } & =\frac {-y}{2x-ye^{y}}\\ 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} Let anstaz be\begin {align*} \xi & =g\left ( y\right ) \\ \eta & =0 \end {align*}
Substituting this into (1) gives\[ -\omega ^{2}\frac {dg}{dy}-\omega _{x}g=0 \] But \(\omega ^{2}=\frac {y^{2}}{\left ( 2x-ye^{y}\right ) ^{2}},\omega _{x}=\frac {d}{dx}\left ( \frac {-y}{2x-ye^{y}}\right ) =\frac {2y}{\left ( 2x-ye^{y}\right ) ^{2}}\). The above becomes\begin {align*} -\frac {y^{2}}{\left ( 2x-ye^{y}\right ) ^{2}}\frac {dg}{dy}-\frac {2y}{\left ( 2x-ye^{y}\right ) ^{2}}g & =0\\ -y^{2}\frac {dg}{dy}-2yg & =0\\ \frac {dg}{dy}+\frac {2}{y}g & =0 \end {align*}
This is linear ode. The solution is \[ g=\frac {c_{1}}{y^{2}}\] Hence\begin {align*} \xi & =\frac {1}{y^{2}}\\ \eta & =0 \end {align*}
But taking \(c_{1}=1\). The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{-\frac {1}{y^{2}}\left ( \frac {-y}{2x-ye^{y}}\right ) }\\ & =y\left ( 2x-ye^{y}\right ) \end {align*}
The next step is to determine the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} Since\(\ \eta =0\), then in this special case \(R=c_{1}=y\). To find \(S\) we use \(dS=\frac {dx}{\xi }\) or \(dS=y^{2}dx\). Hence \(S=c_{1}^{2}x+c_{2}=c_{1}^{2}x\) by taking \(c_{2}=0\). Therefore \(S=y^{2}x\) since \(c_{1}=y\).\begin {align} R & =y\tag {2}\\ S & =y^{2}x\nonumber \end {align}
Now that \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) are found, the ODE \(\frac {dS}{dR}=\Omega \left ( R\right ) \) is setup. The ODE comes out to be function of \(R\) only, so it is quadrature. This is the main idea of this method. By solving for \(R\) we go back to \(x,y\) and solve for \(y\left ( x\right ) \). How to find \(\frac {dS}{dR}\)? There is an equation to determine this given by\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*}
Everything on the RHS is known. \(S_{x}=y^{2},R_{x}=0,S_{y}=2yx,R_{y}=1\). Substituting into the above gives \begin {align*} \frac {dS}{dR} & =\frac {y^{2}+\omega \left ( x,y\right ) 2yx}{\omega \left ( x,y\right ) }\\ & =\frac {y^{2}+\left ( \frac {-y}{2x-ye^{y}}\right ) 2yx}{\left ( \frac {-y}{2x-ye^{y}}\right ) }\\ & =y^{2}e^{y} \end {align*}
Now we need to express the RHS in terms of \(R,S.\) From (2) we see that \(y=R\), hence the above becomes\[ \frac {dS}{dR}=R^{2}e^{R}\] This is quadrature. Solving gives \[ S=\left ( R^{2}-2R+2\right ) e^{R}+c_{1}\] Convecting back to \(x,y\,\) gives\[ y^{2}x=\left ( y^{2}-2y+2\right ) e^{y}+c_{1}\]