Solve the SDE $dN=a N \log(\frac MN)dt+\sigma NdB_t,$ where $M,a \in \mathbb{R}$.
Apply Ito's formula to $ f = \log(N)$ gives
\begin{align*} df &= \frac{\partial f}{\partial N} dN +\frac{1}{2}\frac{\partial ^2 f}{\partial N^2} d\langle N\rangle_t\\ &= \left(\frac{\partial f}{\partial N}aN\log\left(\frac MN\right) +\frac{\sigma^2}{2}\frac{\partial ^2 f}{\partial N^2} \right)dt + \frac{\partial f}{\partial N}\sigma dB_t\\ &=a \log\left(\frac {M}{N}\right)dt+\sigma dB_t - \frac{\sigma^2}{2}dt\\ &=\left(alog(M) - af -\frac{\sigma^2}{2}\right)dt - \sigma dB_t \end{align*} which looks like an Ornstein-Uhlenbeck SDE. How can I proceed from here?
I am not sure, what I write can help. but I'll try it. rewrite as $$\frac{dN}{N}=a(\log M -\log N)dt+\sigma.dB_t\\ d(\log N)=a(\log M -\log N)dt+\sigma.dB_t$$this equation is very similar to mean-revert SDE's $dx_t=(\mu-x_t)dt+\sigma.dB_t$