Short answer: Your question is great! I wish more people asked questions like you.
Long answer: Of course, anything can be improved. :P
Let's address your specific concerns:
How to title the question so that it can be relevant for someone apart from just me? Do I describe the problem as its given, along with unnecessary physical details? Do I describe the topic of the problem? Do I write "Help!"?
The topic should be the minimum description necessary to uniquely identify your problem. Or, if not uniquely identify, at least give the reader a good idea of what they're walking into if they open the link. The current title "Solution of 2nd order linear ODE with regular singular points, and complex exponents at singularity" is good, but a bit on the long side; unfortunately, I don't see how to make it shorter without removing valuable pieces of information.
Do not use filler text that could apply to any problem (e.g. "I need help to" or "Help!" or "Can't solve" or ...)
What are the right tags to use apart from solution-verification?
This is question-dependent. First, pick the most specific tag relating to the issue you're having. Just start typing keywords into the tag box and see if a tag exists already (don't create new tags). For instance, if you're trying to solve the heat-equation, there's a tag for that. Sometimes, there isn't a "super specific" tag you can use.
You should always have a "general topic" tag. These are tags that are typically titles of courses (like differential-equations or algebra-precalculus). If you can find any other ones that apply to your question, tack them on!
The only other suggestion I could make is that:
Your question is long. Erring on the side of long is good, though. To make your question shorter, I'd recommend cutting some of the step-by-step algebraic simplification. For instance:
Consider when $s = 0$:
\begin{align}
&\quad 5s^2 - 6s + 1 = 0 \\
&\equiv s = \frac{6 \pm \sqrt{36 - 20}}{10} \\
&\equiv s = \frac{6 \pm \sqrt{16}}{10} \\
&\equiv s = \frac{6 \pm 4}{10} \\
&\equiv s = 1 \vee s = \frac{1}{5}\\
\end{align}
Could become
Consider when $s = 0$:
\begin{align*}
&\quad 5s^2 - 6s + 1 = 0 \\
&\implies s = 1 \vee s = \frac{1}{5}\\
\end{align*}
Chances are, you didn't mess up on your quadratic formula. So, you can skip some simplification. Same thing goes for other simple algebra stuff.
But, all in all, your question is awesome. Keep up the good work, and please ask more good questions. :)