Let $q$ be a strictly positive integer and let $\beta \neq q$ be a real number. Consider a following Ordinary Differential Equation(ODE): \begin{equation} \frac{d^q r_t}{d t^q} + \frac{1}{t^\beta} r_t = 0 \end{equation} Using Mathematica I have checked that the fundamental solutions to that ODE reads: \begin{equation} r_t^{(\eta)} = t^\eta F_{0,q-1}\left[\begin{array}{c} \{\} \\ \left\{ 1+ \frac{\xi}{q-\beta} \right\}_{\xi=1}^\eta \cup \left\{ 1- \frac{\xi}{q-\beta} \right\}_{\xi=1}^{q-1-\eta} \end{array}; -\frac{t^{q-\beta}}{(q-\beta)^q}\right] \end{equation} where $\eta=0,\cdots,q-1$. Here $F_{p,q}$ is the hypergeometric function. The quantity on the bottom of the square brackets is understood as a concatenation of two sequences the first one of length $\eta$ and the second one of length $q-1-\eta$. In the special case $q=2$ the fundamental solutions read: \begin{equation} \left( \sqrt{t} J_{\frac{1}{-2+\beta}}\left[\frac{t^{1-\frac{\beta}{2}}}{\frac{\beta}{2}-1}\right] , \sqrt{t} J_{\frac{-1}{-2+\beta}}\left[\frac{t^{1-\frac{\beta}{2}}}{\frac{\beta}{2}-1}\right] \right) \end{equation} where $J_\beta$ is the Bessel function. The solutions match those given in Solution to a second order ordinary differential equation .
Is there a way to derive the solutions using some different method?