= Solution
The reference <Brownian motion> must start at $x$, and both laws must use the same stopped-path rule. Under the reference <probability measure> let $R_s=x+W_s$, let $\tau$ be its exit time from $(a,b)$, and put $T=t\wedge\tau$ and $c=(\delta-1)/2$. The \b[<Girsanov density for a stopped Bessel process>], as a functional of the stopped reference path, is
$$
\boxed{D(R)=\exp\left\{c\int_0^T\frac{dW_s}{R_s}-\frac{c^2}{2}\int_0^T\frac{ds}{R_s^2}\right\}.}
$$
Use the <predictable process> $\theta_s=cR_s^{-1}\mathbf1_{\{s\leq\tau\}}$, with any harmless choice at the endpoint. On $[0,t]$, $\int\theta_s^2ds\leq c^2t/a^2$. The <Novikov condition> holds; under the density $D$, the <Girsanov theorem> makes $\widetilde W_s=W_s-\int_0^s\theta_u du$ a <Brownian motion>. Before exit, $dR_s=cR_s^{-1}ds+d\widetilde W_s$, which is the required <stochastic differential equation>. The drift is a <locally Lipschitz function> on $(0,\infty)$, so <pathwise uniqueness> and <uniqueness in law> hold up to this exit time.
For a version involving just the endpoint and an ordinary integral, the <Itô formula> for $\log R$ gives
$$
\int_0^T\frac{dW_s}{R_s}=\log\frac{R_T}{x}+\frac12\int_0^T\frac{ds}{R_s^2}.
$$
Thus
$$
\boxed{D(R)=\left(\frac{R_T}{x}\right)^{(\delta-1)/2}\exp\left\{-\frac{(\delta-1)(\delta-3)}8\int_0^T\frac{ds}{R_s^2}\right\}.}
$$
The stopped density is <measurable> from the stopped path, so it is also the <Radon-Nikodym derivative> of the stopped-path laws. The stopping times in this formula are evaluated on the reference path, rather than taken from a different realization of $X$. A zero-start reference <Brownian motion> would have a different deterministic initial value and could not dominate this law.
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