Preliminary notation: $\wedge$ - logical "and", $J$ - bessel function of I kind, $Y$-bessel function of II kind
Problem: a) Solve: $\frac{d^{2}y}{dx^{2}} - [\frac{p(p+1)}{x^{2}} + c]y = 0$ $(\star)$ $(c, p \in \mathbb{R}$, $x \in \mathbb{R} \setminus \{0\}$) and distinguish bounded solutions for this equation in the neighbourhood of $x=0$.
My partial solution(a.k.a. what I did by far):
Let: $A:=c$ $\wedge$ $B:=p(p+1)$. Then: $(\star)$ is equivalent to: $$ x^{2}\frac{d^{2}y}{dx^{2}}-(Ax^{2}+B)y=0 $$ This equation is recognised as Bessesl differential equation. We proceed as follows:
Consideration of given algebraic form of coefficients of dependent variable $y$ implies existence of the solution in the form of general power series in the surrounding of $x=0$. Let's apply Frobenius method. Let: $y=y(x)=x^{\rho}\sum\limits_{n=0}^{+\infty}a_{n}x^{n}$ ($a_{0} \in \mathbb{R} \setminus \{ 0 \}$ $\wedge$ $\mid x \mid < +\infty$ $\wedge$ $\rho \in \mathbb{R}_{+}$). Then:
$y^{'}(x) = \sum\limits_{n=0}^{+\infty}(n+\rho)a_{n}x^{n+\rho-1} = x^{\rho} \sum\limits_{n=0}^{+\infty}(n+\rho)a_{n}x^{n-1}$ $\implies$ $y^{''}(x) = \sum\limits_{n=0}^{+\infty}(n+\rho)(n+\rho-1)a_{n}x^{n+\rho-2}$ $\implies$ $x^{2}y^{''}(x) = x^{\rho}\sum\limits_{n=0}^{+\infty}(n+\rho)(n+\rho-1)a_{n}x^{n}$ $\implies$ $x^{2}\frac{d^{2}y}{dx^{2}}-(Ax^{2}+B)y = 0$ $\iff$ $x^{\rho}\sum\limits_{n=0}^{+\infty}(n+\rho)(n+\rho-1)a_{n}x^{n} - (Ax^{2}+B)x^{\rho}\sum\limits_{n=0}^{+\infty}a_{n}x^{n} = 0$ $\implies$ $\sum\limits_{n=0}^{+\infty}(n+\rho)(n+\rho-1)a_{n}x^{n} - A\sum\limits_{n=0}^{+\infty}a_{n}x^{n+2} - B\sum\limits_{n=0}^{+\infty}a_{n}x^{n} = 0$ $\implies$ $\sum\limits_{n=0}^{+\infty}(n+\rho)(n+\rho-1)a_{n}x^{n} - \sum\limits_{n=2}^{+\infty}Aa_{n}x^{n} - \sum\limits_{n=0}^{+\infty}Ba_{n}x^{n} = 0$ $\implies$
$[\rho(\rho - 1)-B]a_{0} + [(1 + \rho)\rho - B]a_{1}x + \sum\limits_{n=2}^{+\infty}\Big\{ [(n+\rho)(n+\rho-1)-B]a_{n} - Aa_{n-2}\Big\}x^{n} = 0$ $\implies$ $\implies$
$\begin{cases}
\rho(\rho-1)-B=0 \\
[(1+\rho)\rho-B]a_{1}=0 \\
[(n+\rho)(n+\rho-1)-B]a_{n}-A a_{n-2} = 0
\end{cases}$ $\implies$
$\begin{cases}
B=\rho(\rho-1) \\
[(\rho+\rho^{2})-(\rho^{2}-\rho)]a_{1}=0 \\
a_{n} = \frac{A}{[(n+\rho)(n+\rho-1)-B]}a_{n-2}
\end{cases}$ $\implies$
$\begin{cases}
B=\rho(\rho-1) \\
2\rho a_{1} = 0 \\
a_{n} = \frac{A}{[(n+\rho)(n+\rho-1)-B]}a_{n-2}
\end{cases}$ $\implies$
$\begin{cases}
B=\rho(\rho-1) \\
a_{1} = 0 \\
a_{n} = \frac{A}{[(n+\rho)(n+\rho-1)-B]}a_{n-2}
\end{cases}$ $\implies$
$\begin{cases}
B=\rho(1-\rho) \\
\forall n \in \mathbb{N}_{0}: a_{2n+1}=0 \\
\forall n \in \mathbb{N}: a_{2n} = \frac{A}{[(2n+\rho)(2n+\rho-1)-B]}a_{2n-2}
\end{cases}$ ;
$\rho(\rho-1)-B=0$ $\iff$ $\rho^{2}-\rho-B = 0$. $\Delta = 1+4B$. $\rho = \frac{1\pm \sqrt{1+4B}}{2}$ $\implies$ $\rho = \frac{1+\sqrt{1+4B}}{2}$
$\implies$ $\forall n \in \mathbb{N}$: $a_{2n} = \frac{A}{[(n+\frac{1+\sqrt{1+4B}}{2})(n+\frac{1+\sqrt{1+4B}}{2}-1)-B]}a_{2n-2}$ $\implies$ $\forall n \in \mathbb{N}$: $a_{2n} = \frac{A^{n}}{\Big{[} \prod\limits_{k=0}^{n}(2k+\frac{1+\sqrt{1+4B}}{2})(2k+\frac{1+\sqrt{1+4B}}{2}-1)-B\Big{]}}a_{0}$
$\implies$ $y_{1}(x) = a_{0}\sum\limits_{n=0}^{+\infty}\frac{A^{n}x^{2n}}{ {\Big{[} \prod\limits_{k=0}^{n}(2k+\frac{1+\sqrt{1+4B}}{2})(2k+\frac{1+\sqrt{1+4B}}{2}-1)-B\Big{]}}} = \sqrt{x}J_{\frac{1}{2}\sqrt{1+4B}}(-i\sqrt{A}x)$
$\rho_{1} = \frac{1+\sqrt{1+4B}}{2}$ $\wedge$ $\rho_{2} = \frac{1-\sqrt{1+4B}}{2}$ $\implies$ $\rho_{1} - \rho_{2} = \sqrt{1+4B}$ $\implies$ $y_{2}(x) = b_{0}\sum\limits_{n=0}^{+\infty}\frac{(-1)^{n}A^{n}x^{2k-\sqrt{1+4B}}}{{\Big{[} \prod\limits_{k=0}^{n}(2k+\frac{1+\sqrt{1+4B}}{2})(2k+\frac{1+\sqrt{1+4B}}{2}-1)-B\Big{]}}} = \sqrt{x}Y_{\frac{1}{2}\sqrt{1+4B}}(-i\sqrt{A}x)$ \
$\implies$ $y(x) = C_{1}y_{1}(x) + C_{2}y_{2}(x)=C_{1}\sqrt{x}J_{\frac{1}{2}\sqrt{1+4B}}(-i\sqrt{A}x) + C_{2}\sqrt{x}Y_{\frac{\sqrt{1+4B}}{2}}(-i\sqrt{A}x)$ [$C_{1},C_{2} \in \mathbb{R}$ $\wedge$ $A:=c$ $\wedge$ $B:=p(p-1)$: $c,p \in \mathbb{R}$]
I checked my solution with mathematical package Maple and it turned out that I obtained correct result.
Note: I know that the solution can be expressed in more concise way by using Bessel function of III kind(a.k. Hankel function)
What is my problem: I do not know how to distinguish bounded solutions in the neighbourhood of $x=0$. I got stuck here completely.
My request Since I did a significant work, I do very request for help in the determination of the aforementioned bounded solution of the considered equation in the neighbourhood of x=0. Help very appreciated!