3.5.3.12 Example 12

\begin{align*} \left ( y^{\prime }\right ) ^{2} & =e^{4x-2y}\left ( y^{\prime }-1\right ) \\ \ln \left ( y^{\prime }\right ) ^{2} & =\left ( 4x-2y\right ) +\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \left ( y^{\prime }\right ) ^{2}-\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ 2y & =4x-\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \\ & =xf+g \end{align*}

Where \(f=2,g=-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \).  Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives

\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}

Using values for \(f,g\) the above simplifies to

\begin{equation} p-2=\left ( \frac {2-p}{2p^{2}-2p}\right ) \frac {dp}{dx} \tag {2A}\end{equation}

The singular solution is when \(\frac {dp}{dx}=0\) which gives \(p=2\). From (1) this gives

\[ y=2x-\frac {1}{2}\ln 4 \]

The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2) becomes

\begin{align*} \frac {dp}{dx} & =\left ( p-2\right ) \left ( \frac {2p^{2}-2p}{2-p}\right ) \\ & =2p\left ( 1-p\right ) \end{align*}

is now separable. Solving for \(p\) gives

\[ p=\frac {1}{1+ce^{-2x}}\]

Substituting the above solutions of \(p\) in (1) gives

\begin{align*} y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( \frac {1}{1+ce^{-2x}}\right ) ^{2}}{\frac {1}{1+ce^{-2x}}-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {-e^{4x}}{c\left ( c+e^{2x}\right ) }\right ) \end{align*}