3.1.12.21 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 ansatz 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*} X & =c_{1}\\ & =y-\ln x \end{align*}

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

\begin{align*} \frac {dy}{\eta } & =dY\\ dy & =dY\\ Y & =y \end{align*}

Where constant of integration is set to zero. What is left is to find \(\frac {dY}{dX}\). This is given by

\begin{equation} \frac {dY}{dX}=\frac {Y_{x}+Y_{y}\omega }{X_{x}+X_{y}\omega } \tag {2}\end{equation}

But

\begin{align*} X_{x} & =-\frac {1}{x}\\ X_{y} & =1\\ Y_{x} & =0\\ Y_{y} & =1 \end{align*}

Hence (2) becomes

\begin{align*} \frac {dY}{dX} & =\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 \(y=X+\ln x\). The above becomes

\begin{align*} \frac {dY}{dX} & =\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

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

Converting back to \(x,y\) gives

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