Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-\lambda^2y(x),$$ defined in $(-\infty,0]$ or $[0,\infty)$ or $(-\infty,+\infty)$ and parameter $a>0$. Therefore, the SL coefficient functions $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$. It has a regular singularity $x=0$. We basically hope for homogeneous Dirichlet b.c.
It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to the Kummer's equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$. It has two (1st kind & 2nd kind) independent solutions. Let's then follow the standard argument.
Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained. Seen from this condition for $\alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum as suggested in the comments.
Question
Is it possible to know the limit circle/point classification of this ODE near $0,\pm\infty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.