This is exercise 5.6 from SDE by Oksendal.
solve:
$dY_t=rdt+ \alpha Y_tdB_t$
Where $r,\alpha $ are real constants and $B(t)$ is a 1-dimensional Brownian motion.
The book has a hint for this problem,but I can't solve it yet.
Is there any other hint?
Thanks.
The hint given by the book is to multiply the equation by the integrating factor $$ F_t = \exp\left(-\alpha B_t + \alpha^2t/2\right) $$ So the new SDE will be $$ F_tdY_t = rF_tdt + \alpha Y_t dB_t $$ Continuing analogy with the integrating factor, can you check if there is an easy way to write the RHS as $dg(Y_t,t)$?