$ (n^2-\lambda \cos^2\phi )\Phi(\phi)+(\sin\phi\cos\phi) \Phi '(\phi)-(\cos^2\phi) \Phi''(\phi)=0 $ I wish to solve the following linear 2nd order homogeneous ODE with trigonometric coefficients on the interval $[-\pi /2, \pi /2]$.
$$
(n^2-\lambda \cos^2\phi )\Phi(\phi)+(\sin\phi\cos\phi) \Phi '(\phi)-(\cos^2\phi) \Phi''(\phi)=0
$$
for $n\in \mathbb{Z}$ and $\lambda$ to be determined, however, I suspect there will be countably many eigenvalues $\lambda$ that allow for a solution.
This equation arises when solving the Laplacian differential equation $\nabla^2f=0$ where $f(r,\theta, \phi)$ is given in spherical coordinates (almost, in fact I started with a slight variation). I have used the separation of variables method, setting $f(r,\theta, \phi)=R(r)\Theta(\theta)\Phi(\phi)$ to isolate and solve the $R$ and $\Theta$ components, and this is what remains.
I learned the solution for deriving and solving the 2-dimensional problem in polar coordinates and am attempting spherical without looking at a standard solution. I can't get close to even guessing what the solutions might look like, and each ansatz I have tried leads nowhere, for example, $\Phi(\phi)=\sin^a(\phi)\cos^b(\phi)$.
I suspect that a series solution may help but I only have experience with infinite polynomial series. Hints would be much appreciated!
 A: Lets define $\varphi=\frac{\pi}{2}-\phi$. Then your equation becomes
$$-\sin^{2}\varphi\Phi^{\prime\prime}\left(\varphi\right)-\sin\varphi\cos\varphi\Phi^{\prime}\left(\varphi\right)+\left(n^2-\lambda\sin^{2}\varphi\right)\Phi\left(\varphi\right)=0$$
or
$$\Phi^{\prime\prime}\left(\varphi\right)+\cot\varphi\Phi^{\prime}\left(\varphi\right)+\left(\lambda-\frac{n^{2}}{\sin^{2}\varphi}\right)\Phi\left(\varphi\right)=0$$
This equation is the well-known equation for the Associated Legendre polynomials as you can see here, and the solutions in your variable are
$$\Phi\left(\varphi\right)=P^{n}_{\ell}\left(\cos\varphi\right)\rightarrow\Phi\left(\phi\right)=P^{n}_{\ell}\left(\sin\phi\right)$$
where $\lambda=\ell\left(\ell+1\right)$. If you want to solve it yourself you should define $x\equiv\cos\varphi=\sin\phi$ and get the usual equation for $x$, then as you said you'll need to use the power series method.
A: The usual trick when dealing with spherical coordinates is to change your dependent variables.  With standard spherical coordinates, the usual choice is $w = \cos \phi$;  but given your coordinate definitions, I'd encourage you to try $w = \sin \phi$ instead.  This lets you rewrite the equation in terms of $d\Phi/dw$ by noting
$$
\frac{d \Phi}{d \phi} = \frac{d w}{d \phi} \frac{d \Phi}{d w} = \cos \phi \frac{d \Phi}{d w} = \sqrt{1 - w^2} \frac{d \Phi}{d w}.
$$
The second derivative $d^2\Phi/d\phi^2$ can also be found by a similar application of the chain rule.  
Once you've done this (and assuming that your work is correct so far), you'll have an ODE in terms of $w$ that is more amenable to power-series techniques.  In particular, it will have singular points where $w = \pm 1$, and you'll probably want the solutions for $\Phi$ to be regular at these points.
