Suppose a bead mass $m$ that is free to slide in on a wire that is rotating about the $z$ axis through origin $O$ with angular velocity $\omega$ and the wire is made into a parabolic shape such that $z=r^2/2a$ where $z$ is the vertically upward from $O$ and $r$ is the horizontal distance from $O$.
I have got the second order differential equation related to $r$ such that $(a^2+r^2) \ddot r +r\dot r^2 = (a^2 \omega^2-ga)r$.
I have now got to form the linearised equation about $r=0$, which is given as $a^2 \ddot r=(a^2 \omega^2-ga)r$ but I have no idea how to get to this linearised equation. My teacher told me it's something to do with computing $r(t)=p+\epsilon(t)$ where $p$ is the equilibrium position, which is $r=0$ in our case but I do not know how to do so.
My idea: Do I plug $r(t)=p+\epsilon(t)$ into the differential equation and collect order $\epsilon$ terms?