Solve \begin {align*} y^{\prime } & =\frac {-y^{2}}{e^{x}-y}\\ y^{\prime } & =\omega \left ( x,y\right ) \end {align*}
The symmetry condition results in the PDE\[ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \] End of the problem shows how this is solved for \(\xi ,\eta \) which results in\begin {align*} \xi \left ( x,y\right ) & =1\\ \eta \left ( x,y\right ) & =y \end {align*}
The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{y-\left ( \frac {-y^{2}}{e^{x}-y}\right ) }\\ & =\frac {1-ye^{-x}}{y} \end {align*}
The next step is to determine what is called the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. So we are looking for \(S\left ( R\right ) \) function. This is done by using the standard characteristic equation by writing\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{1} & =\frac {dy}{y}=dS \tag {1} \end {align}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x}+\eta \frac {\partial }{\partial y}\right ) S\left ( x,y\right ) =1\). Which is a first order PDE. This is solved for \(S\), which gives (1) using the method of characteristic to solve first order PDE which is standard method. Starting with the first pair of ODE gives\[ \frac {dy}{dx}=y \] Integrating gives \(\ln \left \vert y\right \vert =x+c\) or \(y=ce^{x}\) where \(c\) is constant of integration. In this method \(R\) is always \(c\). Hence\[ R\left ( x,y\right ) =ye^{-x}\] \(S\left ( x,y\right ) \) is now found from the first equation in (1) and the last equation which gives\begin {align*} dS & =\frac {dx}{\xi }\\ dS & =\frac {dx}{1}\\ dS & =dx\\ S & =x \end {align*}
Hence\begin {align*} R & =ye^{-x}\\ S & =x \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. \(S_{x}=1,R_{x}=-ye^{-x},S_{y}=0,R_{y}=e^{-x}\). Substituting gives\begin {align*} \frac {dS}{dR} & =\frac {1}{-ye^{-x}+\frac {-y^{2}}{e^{x}-y}e^{-x}}\\ & =\frac {ye^{-x}-1}{ye^{-x}} \end {align*}
But \(R=ye^{-x}\), hence the above becomes\[ \frac {dS}{dR}=\frac {R-1}{R}\] This is just quadrature. Integrating gives \begin {align*} S & =\int \frac {R-1}{R}dR\\ & =R-\ln R+c_{1} \end {align*}
This solution is converted back to \(x,y\). Since \(S=x,R=ye^{-x}\), the above becomes\[ x=ye^{-x}-\ln \left ( ye^{-x}\right ) +c_{1}\] Which is the solution to the original ODE.
Finding Lie symmetries for this example
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 \begin {align*} \xi & =c_{1}x+c_{2}y+c_{3}\\ \eta & =c_{4}x+c_{5}y+c_{6} \end {align*}
Hence \(\xi _{x}=c_{1},\xi _{y}=c_{2},\eta _{x}=c_{4},\eta _{y}=c_{5}\) and (14) becomes\begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta & =0\\ c_{4}+\omega \left ( c_{5}-c_{1}\right ) -\omega ^{2}c_{2}-\omega _{x}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\omega _{y}\left ( c_{4}x+c_{5}y+c_{6}\right ) & =0 \end {align*}
But \(\omega =\frac {-y^{2}}{e^{x}-y},\omega _{x}=\frac {y^{2}e^{x}}{\left ( e^{x}-y\right ) ^{2}},\omega _{y}=\left ( -\frac {2y}{e^{x}-y}-\frac {y^{2}}{\left ( e^{x}-y\right ) ^{2}}\right ) \) and the above becomes\[ c_{4}+\frac {-y^{2}}{e^{x}-y}\left ( c_{5}-c_{1}\right ) -\left ( \frac {-y^{2}}{e^{x}-y}\right ) ^{2}c_{2}-\frac {y^{2}e^{x}}{\left ( e^{x}-y\right ) ^{2}}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\left ( -\frac {2y}{e^{x}-y}-\frac {y^{2}}{\left ( e^{x}-y\right ) ^{2}}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) =0 \] Need to do this again. I should get \(c_{3}=1,c_{5}=1\) and everything else zero.\begin {align*} \xi & =1\\ \eta & =y \end {align*}