Typo

[Luyu CEN]

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.9

If we follow from equation(7) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.7
(9) should be
\begin{equation}
\sin^2(\phi)
\Phi'' +\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
instead of
\begin{equation}
\sin^2(\phi)
\Phi'' +2\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
There will be a coefficient 2 if we substitute $\Phi(\phi)=L(\cos(\phi))$ into the equation and use the chain rule to get its derivatives but I think not now.

[Victor Ivrii]

Thanks. There are also more serious omissions. Corrected.