Prove identity in law for stochastic process driven by Brownian Motion Let $B = (B_t)_{t\geq 0}$ be a standard brownian motion started at $0$. Consider the two following stochastic equations:
\begin{equation}
\begin{split}
dX_t &=& (13 + 2X_t)\,dt + (6 + X_t)\,dB_t \\
dY_t &=& 2Y_t\,dt + Y_t\,dB_t
\end{split}
\end{equation}
with $X_0 = Y_0 = 1$. 
I am trying to show that the following identity in law holds:
$$
X_t - Y_t\left(1 + 6\int_0^t\frac{1}{Y_s}\,dB_s \right) \stackrel{law}{=} 7\int_0^tY_s\,ds 
$$
Combining $X$ and $Y$, it can be shown through Ito's formula that the solution for $X$ is given by
$$
X_t = Y_t\left(1 + 7\int_0^t\frac{1}{Y_s}\,ds + 6\int_0^t\frac{1}{Y_s}\,dB_s \right)
$$
But I am having trouble operating with the first equation to work out the law identity, and the solution for $X$ is not really helping me. Any hint is most welcome, as I am a bit lost now. I suspect the solution might not be trivial.
 A: Note that the solution of the SDE, $\mathrm dY_t = 2Y_t\mathrm dt + \mathrm dB_t\ (\mbox{for }Y_0 = 1)$, is a collection of log-normal random variables,
$$
Y_t = e^{\frac{3}{2}t + B_t}~(\mbox{for }t\geqslant 0)\,.
$$
Therefore, for $0\leqslant s\leqslant t$,
$$
\left\{\frac{Y_t}{Y_s}\right\}_{0\leqslant s\leqslant t}\quad \stackrel{\text{law}}{=}\quad \left\{Y_{s}\right\}_{0\leqslant s\leqslant t}\,.\tag{1}
$$
proof:

For $n\in\mathbb{N}$ and $0\leqslant t_1<t_2<\ldots<t_n\leqslant t$, the time-reversal properties of Brownian motion justify the following:


$
\,\,\,P\left(\frac{Y_t}{Y_{t_1}}\leqslant y_1,\ \ldots,\ \frac{Y_t}{Y_{t_n}}\leqslant y_n\right)\\
\begin{eqnarray*}
&=& P\left(\frac{3}{2}(t-t_1) + B_t - B_{t_1}\leqslant \log y_1,\ \ldots,\ \frac{3}{2}(t-t_n) + B_t - B_{t_n}\leqslant \log y_n \right) \\
&=& P\left( B_t - B_{t_1}\leqslant \log y_1 - \frac{3}{2}(t-t_1),\ \ldots,\ B_t - B_{t_n}\leqslant \log y_n - \frac{3}{2}(t-t_n) \right) \\
&=& P\left( -(B_{t-r_1}-B_t)\leqslant \log y_1 - \frac{3}{2}r_1,\ \ldots,\ -(B_{t-r_n}-B_t)\leqslant \log y_n - \frac{3}{2}r_n \right) \\
&=& P\left( B_{t-r_1}-B_t\leqslant \log y_1 - \frac{3}{2}r_1, \ldots, B_{t-r_n}-B_t\leqslant \log y_n - \frac{3}{2}r_n \right) \\
&=& P\left( B_{r_1}\leqslant \log y_1 - \frac{3}{2}r_1,\ \ldots,\ B_{r_n}\leqslant \log y_n - \frac{3}{2}r_n \right) \\
&=& P\left( \frac{3}{2}r_1 + B_{r_1}\leqslant \log y_1,\ \ldots,\ \frac{3}{2}r_n + B_{r_n}\leqslant \log y_n \right) \\
&=& P\left(Y_{r_1}\leqslant y_1,\ \ldots,\ Y_{r_n}\leqslant y_n\right),
\end{eqnarray*}
$


where we defined $r_i:=t-t_i\ .$



The equality in law, $(1)$, implies
$$
\int_0^t \frac{Y_t}{Y_s} \, ds \stackrel{\text{law}}{=} \int_0^t Y_s \, ds\,,
$$
which, from your application of Ito's lemma and as deduced by @saz, is the relationship to be proved. 
