3.4.10.7 Example $y^{\prime }=\frac{-y^2}{e^x-y}$

Solve

$$\begin{array}{rl}{y}^{\prime }& =\frac{-y^2}{e^x-y}\\ y^{\prime }& =\omega (x,y)\end{array}$$

The symmetry condition results in the PDE

$${\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta =0$$

End of the problem shows how this is solved for $\xi ,\eta$ which results in

$$\begin{array}{rl}\xi (x,y)& =1\\ \eta (x,y)& =y\end{array}$$

The integrating factor is therefore

$$\begin{array}{rl}\mu (x,y)& =\frac{1}{\eta -\xi \omega }\\ & =\frac{1}{y-\left(\frac{-y^2}{e^x-y}\right)}\\ & =\frac{1-y{e}^{-x}}{y}\end{array}$$

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

$$\begin{array}{rl}\frac{dx}{\xi }& =\frac{dy}{\eta }=dS\\ \text{(1)}& \frac{dx}{1}& =\frac{dy}{y}=dS\end{array}$$

The above comes from the requirements that $(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y})S(x,y)=1$. 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 $\ln \left|y\right|=x+c$ or $y=c{e}^x$ where $c$ is constant of integration. In this method $R$ is always $c$. Hence

$$R(x,y)=y{e}^{-x}$$

$S(x,y)$ is now found from the first equation in (1) and the last equation which gives

$$\begin{array}{rl}dS& =\frac{dx}{\xi }\\ dS& =\frac{dx}{1}\\ dS& =dx\\ S& =x\end{array}$$

Hence

$$\begin{array}{rl}R& =y{e}^{-x}\\ S& =x\end{array}$$

Now that $R(x,y),S(x,y)$ are found, the ODE $\frac{dS}{dR}=\mathrm{\Omega }\left(R\right)$ 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

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{\frac{dS}{dx}+\omega (x,y)\frac{dS}{dy}}{\frac{dR}{dx}+\omega (x,y)\frac{dR}{dy}}\\ & =\frac{S_x+\omega (x,y)S_y}{R_x+\omega (x,y)R_y}\end{array}$$

Everything on the RHS is known. $S_x=1,R_x=-y{e}^{-x},S_y=0,R_y=e^{-x}$. Substituting gives

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{1}{-y{e}^{-x}+\frac{-y^2}{e^x-y}{e}^{-x}}\\ & =\frac{y{e}^{-x}-1}{y{e}^{-x}}\end{array}$$

But $R=y{e}^{-x}$, hence the above becomes

$$\frac{dS}{dR}=\frac{R-1}{R}$$

This is just quadrature. Integrating gives

$$\begin{array}{rl}S& =\int \frac{R-1}{R}dR\\ & =R-\ln R+c_1\end{array}$$

This solution is converted back to $x,y$. Since $S=x,R=y{e}^{-x}$, the above becomes

$$x=y{e}^{-x}-\ln \left(y{e}^{-x}\right)+c_1$$

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

$$\begin{array}{}\text{(14)}& {\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta =0\end{array}$$

Try

$$\begin{array}{rl}\xi & =c_{1}x+c_{2}y+c_3\\ \eta & =c_{4}x+c_{5}y+c_6\end{array}$$

Hence ${\xi }_x=c_1,{\xi }_y=c_2,{\eta }_x=c_4,{\eta }_y=c_5$  and (14) becomes

$$\begin{array}{rl}{\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta & =0\\ c_4+\omega (c_5-c_1)-{\omega }^2{c}_2-{\omega }_x(c_{1}x+c_{2}y+c_3)-{\omega }_y(c_{4}x+c_{5}y+c_6)& =0\end{array}$$

But $\omega =\frac{-y^2}{e^x-y},{\omega }_x=\frac{y^2{e}^x}{{(e^x-y)}^2},{\omega }_y=(-\frac{2y}{e^x-y}-\frac{y^2}{{(e^x-y)}^2})$ and the above becomes

$$c_4+\frac{-y^2}{e^x-y}(c_5-c_1)-{\left(\frac{-y^2}{e^x-y}\right)}^2{c}_2-\frac{y^2{e}^x}{{(e^x-y)}^2}(c_{1}x+c_{2}y+c_3)-(-\frac{2y}{e^x-y}-\frac{y^2}{{(e^x-y)}^2})(c_{4}x+c_{5}y+c_6)=0$$

Need to do this again. I should get $c_3=1,c_5=1$ and everything else zero.

$$\begin{array}{rl}\xi & =1\\ \eta & =y\end{array}$$