1.3.16 Example 16 \(y=\ln \left ( \cos p\right ) +p\tan p\) (d’Alembert)

\[ \ln \left ( \cos y^{\prime }\right ) +y^{\prime }\tan y^{\prime }=y \]

Solving for \(y\) gives

\begin{align} y & =\ln \left ( \cos p\right ) +p\tan p\tag {1}\\ y & =xf+g\nonumber \\ & =g\tag {1A}\end{align}

Where \(f=0\) and \(g\left ( p\right ) =\ln \left ( \cos p\right ) +p\tan p\). Important note: This ode has \(f=0\) which is strictly speaking is not of the form \(y=xf\left ( p\right ) +g\left ( p\right ) \). But Maple says this is d’Alembert. This is why it is included. I should make special case d’Almbert classification to handle this special case.

Taking derivative of (1A) w.r.t. \(x\) gives

\begin{align} p & =\frac {dg}{dp}\frac {dp}{dx}\nonumber \\ p & =\left ( -\frac {\sin p}{\cos p}+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =\left ( -\tan p+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =p\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\nonumber \\ 1 & =\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\end{align}

The singular solution is found by setting \(\frac {dp}{dx}=0\) which does not result in solution.

The general solution is found by finding \(p\) from (2). Since (2) is not linear in \(p\), then inversion is needed. Writing (1) as

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

Integrating gives

\begin{align*} x & =\tan p+c\\ p & =\arctan \left ( x-c\right ) \end{align*}

Substituting the above in (1) gives the solution

\begin{align*} y & =\ln \left ( \cos p\right ) +p\tan p\\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( \arctan \left ( x-c\right ) \right ) \tan \left ( \arctan \left ( x-c\right ) \right ) \\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( x-c\right ) \arctan \left ( x-c\right ) \end{align*}

This ode also have solution \(y=0\).