This is one of those questions whose answer can be straight forward but 1. I want to convince myself that my intuition is right or ... 2. convince myself that the answer is not trivial. Here it goes.
I have the following ODE's \begin{alignat}1 r^{2}\psi^{\prime\prime}+r\psi^{\prime}&-\psi(p^{2}+r^{2}q(r))&&=0 \tag{Equ 1.} \\ r^{2}\phi^{\prime\prime}+r\phi^{\prime}&-\phi p^{2}&&=0 \tag{Equ 2.} \end{alignat} A particular solution to Equ 2. requiring continuity at $r=0$ is given by $\phi=r^{p}$.
My question is the following: consider $\varphi(r)=\psi-\phi=\psi-r^{p}$. (i.e $\varphi$ is the difference of the solutions of Equ 1 and Equ 2 respectively.) Can I claim that $\varphi(r)$ is bounded on [0,1]?. Here I assume $q$ to be nice enough (Analytic is what I am using right now, and $p$ is a positive integer).
My guess it is that $\varphi$ it is indeed bounded. I guess this, based on the fact that I think $\psi$ is analytic (since $q$ is). Hence $\varphi$ will be continuous on a compact set.
Any ideas will be appreciated.