Let $W_t$ be Wiener process and $B_t$ a Brownian Bridge pinned at $0$ between $t$ and $T$. That is: $B_t = B_T = 0$. It is well known that $B_t$ has (among many) the following two equivalent (in terms of moments) solutions/representations:
Non-anticipatory version: $$ B_t = (T-t) \int^t_0 \frac{1}{T-s} dW_s $$
Anticipatory version (need to know $W_T$ at $t < T$): $$ B_t = W_t - \frac{t}{T}W_T $$
It's relatively simple to show that the non-anticipatory representation satisfies the SDE (see this answer here for example - mind that I interchange $W$ and $B$):
$$ dB_t = \left( \frac{-B_t}{T-t}\right) dt + dW_t $$
Question: is it possible to find a SDE satisfied by the anticipatory version? Note that this SDE shall necessarily be not Wiener-driven (see @Nate comment)