Can anyone help me with the following SDE?
Solve the following stochastic differential equation: $$dY_t=aY_tdt+(b(t)+cY_t)dB_t$$ with $Y_0=0$.
Hint: Try a solution of the form $Z_tH_t$ where $Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$ and $dH_t=F(t)dt+G(t)dB_t$ for some adapted process F and G which need to be determined.
Thanks gt6989b for your input!
Please correct me if I'm wrong anywhere.
Since, $$Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$$ Hence we get: $dZ_t = Z_t(cdB_t-\frac{1}{2}c^2)dt$
and we also have $dH_t=F(t)dt+G(t)dB_t$
Then we apply Ito's Lemma on $Z_tH_t$, which yields \begin{eqnarray*} d(Z_tH_t) &=& Z_tdH_t+H_tdZ_t+dH_tdZ_t \\ &=& Z_t(F(t)dt+G(t)dB_t)+Z_tH_t(cdB_t-\frac{1}{2}c^2dt)+Z_tcG(t)dt \\ &=& Z_t(F(t)-\frac{1}{2}c^2H_t+cG(t))dt+(Z_tG(t)+cZ_tH_t)dB_t \end{eqnarray*}
By letting $Y_t = Z_tH_t$, we compare between the expressions $dY_t$ and $d(Z_tH_t)$ in the $dt$ and $dB_t$ terms respectively. And we get the following:
\begin{eqnarray} Y_t &=& Z_tH_t \\ G(t) &=& \frac{Z_t}{b(t)} \\ F(t) &=& (a+\frac{1}{2}c^2)H_t - c\frac{Z_t}{b(t)} \end{eqnarray}
Note: $F(t) = P$ and $G(t) = Q$ for your P and Q respectively.
But now, I don't really understand how would this result help me in solving the original $dY_t$ SDE.