General form of solution to Stochastic Linear SDEs

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As a background, I am studying Stochastic Calculus to improve my Financial Mathematics understanding. One book I'm reading is Introduction to Stochastic Calculus with Applications 3rd edition by Klebaner. I am currently stuck with understanding his approach on page 132, where he says to look for a solution of the form

$X(t) = U(t)V(t)$

to general linear SDEs

$dX(t)=(\alpha(t)+\beta(t)X(t))dt+(\gamma(t)+\delta(t)X(t))dB(t)$.

where

$dU(t)=\beta(t)U(t)dt+\delta(t)U(t)dB(t)$

and

$dV(t)=a(t)dt+b(t)dB(t)$

His example, for a Brownian bridge, starts with the SDE:

$dX(t)=\frac{b-X(t)}{T-t}dt +dB(t)$, for $0\le{t}<T$, $X(0)=a$

and matches terms to the general linear SDE with $\alpha(t)=\frac{b}{T-t}$, $\beta(t)=-\frac{1}{T-t}$, $\gamma(t)=1$, and $\delta(t)=0$.

This I follow, but then he says to identify $U(t)$ and $V(t)$ using

$X(t)=U(t)\left(X(0)+\int_0^t\frac{\alpha(s)-\beta(s)\gamma(s)}{U(s)}ds+\int_o^t\frac{\gamma(s)}{U(s)}dB(s)\right)$

however that equation is just in terms of $X(t)$ and $U(t)$. I can't seem to understand his approach to this general solution as I'm not sure how the above maps to $V(t)$. I am hoping for clarification on how to use this method, either for the Brownian bridge example, or for a general Linear SDE. For example, how do I derive the $a(t)$ and $b(t)$? Thanks for the suggestions.

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Before using (5.31) you need to solve for $U_t$. For that you can use (5.28) or (5.26).

For the Brownian bridge example:

$$dU_t=-\frac{1}{T-t}U_t dt$$

This differential equation $u' + \frac{1}{T-t}u=0$, $u(0)=1$ solves to $u=\frac{T-t}{T}$.

Then you can plug in this value in (5.31).

By "identifying $U_t$ and $V_t$" I think the author really means solve for $U_t$ and deduce $V_t$ from it using (5.31), $V_t$ being the second part of the right term.