Solution of a Stochastic Differential Equation I am trying to prove that the solution of the SDE:
$$
Z_t = Y_t + \int_0^t Z_s dX_s
$$
is:
$$
Z_t = \mathcal{E}(X)_t \left(y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s\right)
$$
where $\mathcal{E}(X)_t := \exp(X_t - \frac{1}{2}\langle X\rangle_t)$ is the stochastic exponential, and $y_0$ is the initial value of Y, and the initial value of $X$ is 0.
So far I have shown, by Ito's formula applied to $Y_t \mathcal{E}(X)_t^{-1}$, that:
$$
Y_t = \mathcal{E}(X)_t\left( y_0 + \int_0^t \mathcal{E}(X)_s^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s - \int_0^t Y_s\mathcal{E}(X)_s^{-1}dX_s + \int_0^t Y_s \mathcal{E}(X)_s^{-1} d\langle X\rangle_s\right)
$$
which feels like it's going in the right direction as it's beginning to resemble the solution, but I don't know how to finish the proof.
Furthermore, I want to prove uniqueness of the solution after this, but I am not sure how I would do this?
 A: Writing
\begin{align}
E_t&:=\mathcal{E}(X)_t\,,\\
F_t&:=y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s
\end{align}
Applying Ito to $Z_t=E_tF_t$ gives
\begin{align}
dZ_t&=E_t\,dF_t+F_t\,dE_t+d\langle E,F\rangle_t\\[3mm]
&=dY_t-d\langle X,Y\rangle_t+F_t\,E_t\,dX_t+d\langle X,Y\rangle_t\\[3mm]
&=dY_t+Z_\,dX_t\,
\end{align}
as expected. Here I used
\begin{align}
dF_t&=E_t^{-1}\,dY_t-E_t^{-1}\,d\langle X,Y\rangle_t\,,\\[3mm]
dE_t&=E_t\,dX_t\,,\\[3mm]
d\langle E,F\rangle_t&=E_t E_t^{-1}\,d\langle X,Y\rangle_t=d\langle X,Y\rangle_t\,,\\[3mm]
d\langle X,\langle X,Y\rangle\rangle_t&=0\,.
\end{align}
Regarding uniqueness:
Let $Z$ and $Z'$ be solutions with $Z_0=Z'_0$. Then $\widetilde{Z}:=Z-Z'$ is a solution of
$$
d\widetilde{Z}_t=\widetilde{Z}_t\,dX_t
$$
with $\widetilde{Z}_0=0\,.$ Since this equation has a unique strong solution
of the form
$$
\widetilde{Z}_t=\widetilde{Z}_0\,e^{X_t-\frac{1}{2}\langle X\rangle_t}
$$
it follows that $\widetilde{Z}_t\equiv 0$.
