I have a question about the geometric brownian motion ${\rm d}S = \mu {\rm d}t + \sigma dW$. I want to calculate $v(x)$ which is the expected time $\tau$ at which the particle first exits the strip $[a,b]$ with $a < b$ i.e. $$ v(x) = \mathbb E[\tau \,|\, X_0 = x] $$ I believe we can use Ito's lemma to obtain the differential equation $$ v'(x) \mu + \frac{\sigma^2}{2}v''(x) = 1 $$ and solve it, but I am unable to get to this equation. Can somebody please explain this?
Also if we were considering the same problem for standard Browian motion $~{\rm d}S = \mu {\rm d}t + \sigma {\rm d}W~$, we would get the same differential equation which implies the expected time in both cases is the same. I need help with understanding why this holds true.