3.4.10.21 Example \(y^{\prime }=-\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}\)

\begin{align*} y^{\prime } & =-\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}\\ & =\omega \left ( x,y\right ) \end{align*}

Using anstaz’s it is found that

\begin{align*} \xi & =x\\ \eta & =1 \end{align*}

Hence

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

The first two give

\[ \frac {dy}{dx}=\frac {1}{x}\]

Hence

\[ y=\ln x+c_{1}\]

Therefore

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

And \(S\) is found from either \(\frac {dy}{\eta }=dS\,\) or \(\frac {dx}{\xi }=dS\). Since \(\eta =1\), it is simpler to use \(\frac {dy}{\eta }=dS\) instead.

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

Where constant of integration is set to zero. 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 {2}\end{equation}

But

\begin{align*} R_{x} & =-\frac {1}{x}\\ R_{y} & =1\\ S_{x} & =0\\ S_{y} & =1 \end{align*}

Hence (2) becomes

\begin{align*} \frac {dS}{dR} & =\frac {\omega }{-\frac {1}{x}+\omega }=\frac {1}{-\frac {1}{x\omega }+1}\\ & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}\right ) }}\end{align*}

But \(y=R+\ln x\). The above becomes

\begin{align*} \frac {dS}{dR} & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}xe^{-2\left ( R+\ln x\right ) }+\frac {1}{4}\sqrt {\left ( e^{-2\left ( R+\ln x\right ) }\right ) ^{2}x^{2}+4e^{-2R+\ln x}}\right ) }}\\ & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}\frac {xe^{-2R}}{x^{2}}+\frac {1}{4}\frac {1}{x}\sqrt {e^{-4R}+4e^{-2R}}\right ) }}\\ & =\frac {1}{1-\frac {1}{\left ( -\frac {1}{4}e^{-2R}+\frac {1}{4}\sqrt {e^{-4R}+4e^{-2R}}\right ) }}\end{align*}

Integrating gives

\[ S=\frac {\sqrt {\frac {1+4e^{2R}}{e^{4R}}}e^{2R}\operatorname {arctanh}\left ( \frac {1}{\sqrt {1+4e^{2R}}}\right ) }{\sqrt {1+4e^{2R}}}\]

Converting back to \(x,y\) gives

\[ y=\frac {\sqrt {\frac {1+4e^{2\left ( y-\ln x\right ) }}{e^{4\left ( y-\ln x\right ) }}}e^{2\left ( y-\ln x\right ) }\operatorname {arctanh}\left ( \frac {1}{\sqrt {1+4e^{2\left ( y-\ln x\right ) }}}\right ) }{\sqrt {1+4e^{2\left ( y-\ln x\right ) }}}\]