Properties of this ODE Given the ODE $$
    \{e^{\alpha\sin\theta}\Psi'(\theta)\}'+
     e^{\alpha\sin\theta}\beta\cos^{2}\theta \Psi(\theta) =
     \lambda e^{\alpha\sin\theta}\Psi(\theta).
$$
where $\theta \in [-\pi,\pi]$, $||\Psi||_{L^2} < \infty$.
I was wondering whether there is anything we can say about this problem: 
Does it have a (unique) solution?
Does it have analytic solutions (besides trivial ones)?
What can we say about the eigenvalues $\lambda$ and so on?
The problem is that I am not very familiar with ODEs (and Sturm-Liouville problems) so I thought that many people here might directly see what this ODE means? 
If anything is unclear or not precisely specified, please do not hesitate to ask that
 A: Normally, there are two conditions required for a Sturm-Liouville problem, such as
$$
   \Psi(-\pi)=\Psi(\pi),\;\;\; \Psi'(-\pi)=\Psi(\pi)\;\;(or\; \Psi'(-\pi)=-\Psi'(\pi)).
$$
Periodic problems are a little trickier than problems with separated conditions, but the problem here is a regular problem because the coefficient of the highest-order derivative does not vanish, and the other coefficients are regular on $[-\pi,\pi]$. Other periodic-type conditions are also possible, but the above are the most common.
If $\Psi_{1},\Psi_{2}$ are solutions with the same endpoint conditions, but with different parameters $\lambda_{1},\lambda_{2}$, then
$$
     \int_{-\pi}^{\pi}e^{\eta\sin\theta}\Psi_{1}(\theta)\Psi_{2}(\theta)\,d\theta = 0.
$$
The orthogonality condition follows from putting the differential equation into the symmetric form
$$
    \{e^{\eta\sin\theta}\Psi'(\theta)\}'+
     e^{\eta\sin\theta}\xi\cos^{2}\theta \Psi(\theta) =
     \lambda e^{\eta\sin\theta}\Psi(\theta).
$$
(A similar form holds if $\eta$ is not constant.) To see how the orthogonality condition arises, define the operator $L$ by
$$
           Lf = \{e^{\eta\sin\theta}f'(\theta)\}'.
$$
The corresponding adjoint equation is
$$
\begin{align}
  (Lf)g-f(Lg) & = \{e^{\eta\sin\theta}f'\}'g-f\{e^{\eta\sin\theta}g'\}' \\
     & = \{ e^{\eta\sin\theta}(f'g-fg')\}'.
\end{align}
$$
Therefore, if $\Psi_{j}$ are eigenfunctions with eigenvalues $\lambda_{j}$ which satisfy the same periodic-type conditions,
$$
\begin{align}
 (\lambda_{1}-\lambda_{2})\int_{-\pi}^{\pi}e^{\eta\sin\theta}\Psi_{1}\Psi_{2}d\theta
      &= \int_{-\pi}^{\pi} (L\Psi_{1})\Psi_{2}-\Psi_{1}(L\Psi_{2})\,d\theta \\
      &= e^{\eta\sin\theta}(f'g-fg')|_{-\pi}^{\pi} = 0.
\end{align}
$$
Assuming $\lambda_{1}\ne \lambda_{2}$, the stated orthogonality condition holds:
$$
            \int_{-\pi}^{\pi}e^{\eta\sin\theta}\Phi_{1}\Phi_{2}\,d\theta=0.
$$
Assuming $\eta, \xi$ are real, the possible eigenvalues $\lambda$ are real, and the eigenspace corresponding to eigenvalue $\lambda$ are either one- or two-dimensional.
Added: For your problem, there are unique classical solutions $f_{\lambda}$, $g_{\lambda}$ which satisfy
$$
      \begin{array}\\ f_{\lambda}(-\pi)=1 \\ f_{\lambda}'(-\pi)=0 \end{array}\;\;\;
      \begin{array}\\  g_{\lambda}(-\pi)=0 \\ g_{\lambda}'(-\pi)=1 \end{array}.
$$
The function $h=\cos\alpha f+\sin\alpha g$ is a solution of the equation, and will satisfy the conditions $h(-\pi)=h(\pi)$, $h'(-\pi)=h'(\pi)$ for some $\alpha \in [0,2\pi)$ iff the following determinant is $0$:
$$
       D(\lambda) = \left|\begin{array}\\ f_{\lambda}(\pi)-f_{\lambda}(-\pi) & g_{\lambda}(\pi)-g_{\lambda}(-\pi) \\
            f_{\lambda}'(\pi)-f_{\lambda}'(-\pi) & g_{\lambda}'(\pi)-g_{\lambda}'(-\pi)\end{array}\right|
$$
You can verify that $D(\lambda)=0$ iff there exists $\alpha\in[0,2\pi)$ such that
$$
            \left[\begin{array}\\ \cos\alpha \\ \sin\alpha\end{array}\right]
$$
is in the null space of the $2\times 2$ matrix whose determinant is $D$. Equivalently,
$h=\cos\alpha f_{\lambda}+\sin\alpha g_{\lambda}$ is periodic, along with $h'$. The order of the zero of $D$ should be the dimension of the eigenspace corresponding to the eigenvalue $\lambda$. The set of $\lambda$ for which you will have a solution at all for your equation should be a discrete set of real $\lambda$, and it is possible that $D'(\lambda)=0$, in which case the solution might not be unique because both $f_{\lambda}$ and $g_{\lambda}$ are already periodic, along with their first derivatives.
