The differential equation $$R\dot{\theta}^2+\ddot{\theta}(R\theta-l)+g\cos\theta=0$$ is derived from considering a pendulum attached to the uppermost part of a disk. 
As in the picture above (but remember the string is attached to the intersection with the y-axis).
Anyways, I was approximating this by considering small oscillations, as usual. To end up with something like
$$\ddot{\theta}+\frac{R\dot{\theta}}{R\theta_0-l}+\frac{g\cos\theta_0}{R\theta_0-l}=0$$
But this has a complex solution, and a real valued log function, which doesn't seem like an oscillation at all or physically sound.
The solution manual to my book approximates the equation as $$\ddot{\epsilon}+\frac{g\sin\theta_0}{l-R\theta_0}\epsilon=\frac{g\cos\theta_0}{l-R\theta_0}$$ Where $\epsilon=\theta-\theta_0$. The equation doesn't make sense to me, especially the second term on the LHS.
So I was wondering if you guys could help me. Perhaps give me a hint or two... Or perhaps tell me what the author of the solutions manual did?
Thanks!
Let $\theta = \theta_0 + \epsilon$, plugging into your equation and keeping only terms up to order $\epsilon$ we get \begin{align} R \dot \theta^2 + \ddot \theta(R \theta - l) + g \cos \theta & = \ddot \epsilon (R \theta_0 - l) + g \cos (\theta_0 + \epsilon) \\ & = \ddot \epsilon(R \theta_0 - l) + g[\cos \theta_0 - \epsilon \sin \theta_0] = 0. \end{align} Rearranging gives the result from the manual.