Let $A_0=A>0$ and let $$dA_t = (rA_t - x)dt + \sigma dB_t,$$ where $B_t$ is standard Brownian motion and $r,x$ and $\sigma$ are positive constants. Let $T= \inf \{ t: A_t = 0 \}$ and $$G(A)=\Bbb{E}[\exp(-rT)].$$ How do I find the function $G$? I think \begin{equation} rG(A)=(rA-x)G'(A)+\frac{1}{2}\sigma^2G''(A). \tag{*}\label{eq:1} \end{equation} Is this correct? Is there a closed-form solution? Clearly, $G(0)=1$. It should also be true that $\lim_{A \to \infty}G(A)=0$.
Also, it should be true that $G'(A)<0$ at all $A>0$. This would mean that $G''$ does not change sign, and so $G$ must be strictly convex. Thus, $G'$ is strictly monotone in $A$ going from some unknown negative number $G'(0)$ to $0$ as $A$ goes to infinity.
One way to try and solve \eqref{eq:1} could be to change the independent variable to $G'$. If we denote $G'=\frac{dG}{dA}$ by $M$, we have $$\frac{dA}{dM} = \frac{1}{G''}=\frac{\frac{1}{2}\sigma^2}{rG-(rA-x)M}$$ and $$\frac{dG}{dM} = \frac{M}{G''}=\frac{\frac{1}{2}\sigma^2M}{rG-(rA-x)M},$$ which could be a solvable system of first-order equations.
Further, if we denote $rG-(rA-x)M$ by $S$, we can transform the above system in $(A(M),G(M))$ into a system in $(A(M),S(M))$ with $$\frac{dA}{dM} =\frac{\frac{1}{2}\sigma^2}{S}$$ and $$\frac{dS}{dM} =r\frac{dG}{dM}-r\frac{dA}{dM}M - (rA-x)=- (rA-x),$$ where we have used $\frac{dG}{dM}=\frac{\frac{1}{2}\sigma^2M}{S}=\frac{dA}{dM}M$. This system should be solvable by separation of variables.
In particular, we have $$ S=C\exp\left(-{\frac{\frac{1}{2}rA^2-xA}{\frac{1}{2}\sigma^2}}\right), $$ which can be verified by differentiation. The constant of integration $C$ satisfies $C=r+xM(0)$, where $M(0)=G'(0)$ is the slope of $G$ at $A=0$.
I think I should be able to integrate $S$ to recover $M$ and then integrate $M$ to recover $G$, but I'm stuck.
Any help with this problem will be very much appreciated. Thanks!