Example y′ = −y __

\begin{align*} y^{\prime } & =\frac {-y}{2x-ye^{y}}\\ y^{\prime } & =\omega \left ( x,y\right ) \end{align*}

\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}

\begin{align*} \xi & =g\left ( y\right ) \\ \eta & =0 \end{align*}

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*}

\begin{align*} \xi & =\frac {1}{y^{2}}\\ \eta & =0 \end{align*}

\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

3.4.10.9 Example \(y^{\prime }=\frac {-y}{2x-ye^{y}}\)