= Solution
The <Itô formula> for $f(t,x)=e^{x+at}$ gives the \b[<stochastic differential equation>]
$$
\boxed{dZ_t=Z_t\,dB_t+(a+\tfrac12)Z_t\,dt,\qquad Z_0=1.}
$$
Write $N_t=\int_0^t Z_u\,dB_u$ for its <continuous local martingale> part. Its clock is the <quadratic variation>
$$
A_t=[Z]_t=[N]_t=\int_0^t Z_u^2du.
$$
Because $Z_t>0$, $A$ is strictly increasing. For $s<A_\infty$, its inverse $\tau_s$ satisfies $A_{\tau_s}=s$ and $d\tau_s/ds=Z_{\tau_s}^{-2}$. The <Dambis-Dubins-Schwarz theorem>, or the <Lévy characterization of Brownian motion> after the <optional time-change theorem>, makes $\beta_s=N_{\tau_s}$ a <Brownian motion> for the time-changed <filtration>. With $Y_s=Z_{\tau_s}$,
$$
Y_s=1+\beta_s+(a+\tfrac12)\int_0^sY_v^{-1}dv,
\qquad
\boxed{dY_s=d\beta_s+\frac{(2+2a)-1}{2Y_s}ds.}
$$
This is the <Bessel process> equation with \b[dimension $\boxed{2+2a}$], on the positive stochastic interval. This establishes the <Exponential Brownian-to-Bessel time change>.
The lifetime convention is important: the inverse is defined only for $s<A_\infty$. If $a<0$, the <strong law for Brownian motion> gives $B_t/t\to0$, so $Z_t\to0$ exponentially and $A_\infty<\infty$. Then $Y_s\to0$ as $s\uparrow A_\infty$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by <infinite occupation time of one-dimensional Brownian motion> on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\geq0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero <Bessel process> behavior is asserted.
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