1.2.7 Example 8 \(e^{2y}\left ( y^{\prime }\right ) ^{3}+\left ( e^{2x}+e^{3x}\right ) y^{\prime }-e^{3x}=0\)

\[ e^{2y}p^{3}+\left ( e^{2x}+e^{3x}\right ) p-e^{3x}=0 \]

Isolating \(y\)

\begin{align} e^{2y}p^{3} & =e^{3x}-\left ( e^{2x}+e^{3x}\right ) p\nonumber \\ e^{2y} & =\frac {e^{3x}-\left ( e^{2x}+e^{3x}\right ) p}{p^{3}}\nonumber \\ 2y & =\ln \left ( \frac {e^{3x}-\left ( e^{2x}+e^{3x}\right ) p}{p^{3}}\right ) \nonumber \\ y & =\frac {1}{2}\ln \left ( \frac {e^{3x}-\left ( e^{2x}+e^{3x}\right ) p}{p^{3}}\right ) =f\left ( x,p\right ) \tag {1}\end{align}

Using (2A)

\begin{align*} \frac {dx}{dp} & =\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}}\\ & =\frac {-1}{p\left ( p-1\right ) }\end{align*}

This ode is just quadrature. Integrating gives

\begin{equation} x=-\ln \left ( p-1\right ) +\ln \left ( p\right ) +c \tag {2}\end{equation}

Eliminating \(p\) between (1,2) gives

\begin{align*} y & =\frac {1}{2}\ln \left ( -\left ( e^{-x+c}-1\right ) ^{2}\left ( e^{3x}e^{-x+c}+e^{2x}\right ) \right ) \\ & =\frac {1}{2}\ln \left ( -\left ( c_{1}e^{-x}-1\right ) ^{2}\left ( c_{1}e^{3x}e^{-x}+e^{2x}\right ) \right ) \\ & =\frac {1}{2}\ln \left ( -\left ( c_{1}e^{-x}-1\right ) ^{2}\left ( c_{1}e^{2x}+e^{2x}\right ) \right ) \\ & =\frac {1}{2}\ln \left ( -\left ( c_{1}e^{-x}-1\right ) ^{2}e^{2x}\left ( 1+c_{1}\right ) \right ) \\ & =\frac {1}{2}\ln \left ( -\left ( 1+c_{1}\right ) \left ( c_{1}e^{-x}-1\right ) ^{2}\right ) +\frac {1}{2}\ln \left ( e^{2x}\right ) \\ & =\frac {1}{2}\ln \left ( -\left ( 1+c_{1}\right ) \left ( c_{1}e^{-x}-1\right ) ^{2}\right ) +x \end{align*}

This is good example where solving using parametric method is much easier than otherwise. Actually, sympy 1.13.3 and Mathematica V 14.2 were not able to solve this ode. They probably do not have this method implemented.