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
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.