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I have $$\frac{dM_t}{M_t}=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t, \tag{1}$$ where $B_t$ and $W_t$ are two independent Brownian Motions, which was further presented as $$ M_t=\exp \left( -\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s -\frac{1}{2} \int_0^t \gamma_s^2 ds \right) \tag{2}$$
can anybody explain how $(1)$ was transitioned into $(2)$?
Apply Ito's lemma to $\ln M_t$, we obtain that \begin{align*} d\ln M_t &= \frac{1}{M_t} dM_t -\frac{1}{2} \frac{1}{M_t^2} d\langle M, M\rangle_t\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \frac{1}{M_t^2}\left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right) dt.\tag{Eq. 1} \end{align*} Here, since $dM_t=M_t\big[-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t\big]$, \begin{align*} d\langle M, M\rangle_t = \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt. \end{align*} From $(\textrm{Eq.} 1)$, \begin{align*} \ln M_t - \ln M_0 &= \int_0^t\left[-\frac{\mu}{\sigma} dW_s + \gamma_s dB_s -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_s^2\right) ds\right]\\ &=-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds. \end{align*} That is, \begin{align*} M_t = M_0 \exp\left(-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds\right). \end{align*}
