3.4.10.20 Example \(y^{\prime }=\frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\)

\begin{align*} y^{\prime } & =\frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\\ & =\omega \left ( x,y\right ) \end{align*}

Using anstaz’s it is found that

\begin{align*} \xi & =x-y\\ \eta & =y-x \end{align*}

Hence

\begin{align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x-y} & =\frac {dy}{y-x}=dS \tag {1}\end{align}

The first two give

\[ \frac {dy}{dx}=\frac {\eta }{\xi }=\frac {y-x}{x-y}=-1 \]

Hence

\begin{equation} y=-x+c_{1} \tag {2}\end{equation}

Therefore

\begin{align*} R & =c_{1}\\ & =y+x \end{align*}

To find \(S\), since both \(\xi ,\eta \) depend on both \(x,y\), then \(\frac {dy}{\eta }=dS\,\) or \(\frac {dx}{\xi }=dS\) can be used. Lets try both to show same answer results.

\begin{align*} \frac {dy}{\eta } & =dS\\ dS & =\frac {dy}{y-x}\end{align*}

But from (2), \(x=c_{1}-y\). The above becomes

\begin{align*} dS & =\frac {dy}{y-\left ( c_{1}-y\right ) }\\ & =\frac {dy}{2y-c_{1}}\end{align*}

Hence

\[ S=\frac {1}{2}\ln \left ( 2y-c_{1}\right ) \]

But \(c_{1}=y+x\). So the above becomes

\begin{align} S & =\frac {1}{2}\ln \left ( 2y-\left ( y+x\right ) \right ) \nonumber \\ & =\frac {1}{2}\ln \left ( y-x\right ) \tag {3}\end{align}

Let us now try the other ode

\begin{align*} \frac {dx}{\xi } & =dS\\ dS & =\frac {dx}{x-y}\end{align*}

But from (2) \(y=-x+c_{1}\). The above becomes

\begin{align*} dS & =\frac {dx}{x-\left ( -x+c_{1}\right ) }\\ & =\frac {dx}{2x-c_{1}}\end{align*}

Therefore

\[ S=\frac {1}{2}\ln \left ( 2x-c_{1}\right ) \]

But \(c_{1}=y+x\). Therefore

\begin{align} S & =\frac {1}{2}\ln \left ( 2x-\left ( y+x\right ) \right ) \nonumber \\ & =\frac {1}{2}\ln \left ( x-y\right ) \tag {4}\end{align}

The constant of integration is set to zero when finding \(S\). What is left is to find \(\frac {dS}{dR}\). This is given by

\begin{equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\omega }{R_{x}+R_{y}\omega } \tag {5}\end{equation}

But, and using (4) for \(S\) we have

\begin{align*} R_{x} & =1\\ R_{y} & =1\\ S_{x} & =\frac {-1}{y-x}\\ S_{y} & =\frac {1}{y-x}\end{align*}

Hence (2) becomes

\begin{align*} \frac {dS}{dR} & =\frac {\frac {-1}{y-x}+\frac {1}{y-x}\omega }{1+\omega }\\ & =\frac {-\frac {\omega -1}{x-y}}{1+\omega }\\ & =\frac {1-\omega }{\left ( 1+\omega \right ) \left ( x-y\right ) }\\ & =\frac {1-\left ( \frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\right ) }{\left ( 1+\left ( \frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\right ) \right ) \left ( x-y\right ) }\\ & =-x-y\\ & =-\left ( x+y\right ) \\ & =-R \end{align*}

Hence

\begin{align*} \frac {dS}{dR} & =-R\\ S & =-\frac {R^{2}}{2}\end{align*}

Converting back to \(x,y\) gives

\[ \ln \left ( y-x\right ) =-\frac {\left ( y+x\right ) ^{2}}{2}\]