Solving SDE $dY = Adt + BYdW, Y(0) = Y_0$? I am trying to solve the following SDE:
$dY = Adt + BYdW$, where $Y(0)=Y_0$.
Thus far, I have done the following.
First, from Ito's Lemma, we have:
$d(ln(Y))=\frac{1}{Y}dY-\frac{1}{2Y^2}dY^2$.
Comparing to the original SDE, we see that:
$\frac{1}{Y}dY=d(ln(Y))+\frac{1}{2Y^2}dY^2$
so,
$d(ln(Y))+\frac{1}{2Y^2}(B^2Y^2)dt=\frac{1}{Y}Adt+B dW$.
I am not sure what to do from here. Any help would be greatly appreciated!
 A: Inspired by the geometric Brownian motion, set $Z_t=F(t,W_t,Y_t)$ with $F(t,w,y)=e^{-Bw+Ct}y$.
Then the Ito formula for products or generally multivariate functions gives
\begin{align}
dZ_t&=F_t\,dt+F_wdW_t+F_ydY_t+\tfrac12F_{ww}dt+F_{wy}d\langle W,Y\rangle_t
\end{align}
This allows to find a non-surprising value for $C$ that gives a formula for $Y_t$ that is not recursive, but contains a non-reducible integral with an integrand in $W_t$ and $t$.
A: Define $dZ_t=BZ_tdW_t,\,Z_0=1$. Then $d(Z_t^{-1})=-B(Z_t^{-1})dW_t+B^2(Z_t^{-1})dt$ which (use $\ln$ and Ito) implies $Z_t^{-1}=e^{\frac{B^2t}{2}-BW_t}$. Now consider $d(Y_tZ_t^{-1})$. We get by product rule and Ito heuristics
$$\begin{aligned}d(Y_tZ_t^{-1})&=Z_t^{-1}dY_t+Y_td(Z_t^{-1})+d(Y_t)d(Z_t)=\\
&=AZ_t^{-1}dt+B(Y_tZ_{t}^{-1})dW_t-B(Y_tZ_t^{-1})dW_t+B^2(Y_tZ_t^{-1})dt-B^2(Y_tZ_t^{-1})dt=\\
&=AZ_t^{-1}dt\end{aligned}$$
and therefore
$$Y_tZ_t^{-1}=Y_0+A\int_{[0,t]}e^{\frac{B^2s}{2}-BW_s}ds\implies Y_t=Y_0e^{-\frac{B^2t}{2}+BW_t}+A\int_{[0,t]}e^{-\frac{B^2}{2}(t-s)+B(W_t-W_s)}ds$$
