Differential equation in non-uniform circular motion

I have a question which states

An astronaut is conducting an experiment on a spaceship under conditions of zero gravity. A bead is threaded on a circular wire, and set in motion with angular velocity $\omega _0$ about the centre. If the coefficient of friction between the bead and the wire is $\mu$, show that the angular velocity $\omega$ at time $t$ satisfies the differential equation $\dot{\omega} = -\mu \omega ^ 2$ . Solve this equation, and hence find an expression for $\theta$, the angle turned after time $t$. Show that, according to this model, the bead will never come to a complete stop.

I have shown that the differential equation $\dot{\omega} = - \mu \omega ^2$ is satisfied. However, I am struggling to solve the differential equation. Am i correct in thinking it is a second order non-linear ordinary differential equation? If so how do I solve this. My textbook doesn't require you to know how to solve non-linear differential equations, so is there some special way to go about this?

This can be integrated directly. You have $$\frac{d\omega}{dt}=-\mu\omega^2,$$ and you can separate the variables: $$\frac{d\omega}{\omega^2}=-\mu dt.$$ This last one you can integrate on both sides and get $\omega$ as a function of $t$.
I have solved the equation by doing it in two steps. First, rather than thinking of it as $\ddot{\theta} = -\mu \dot{\theta} ^2$, treat it as $\frac{d\omega}{dt} = -\mu \omega ^2$, which can be solved by separating the variables: \begin{align} \int \frac{1}{\omega ^2} \ d\omega &= \int -\mu \ dt \\ -\frac{1}{\omega} &= c - \mu t \end{align} $c$ can be found by inputting the intial conditions, $t=0 , \ \omega = \omega _0$, to give $c = -\frac{1}{\omega _0}$ $$\omega = -\frac{\omega _0}{1 + \mu \omega _0 t}$$ From this, $\theta$ can be found by doing: \begin{align} \frac{d\theta}{dt} &= -\frac{\omega _0}{1 + \mu \omega _0 t} \\ \theta &= \int -\frac{\omega _0}{1 + \mu \omega _0 t} \ dt \\ \theta &= \frac{1}{\mu} \ln|1 + \mu \omega _0t|\end{align}