Consider the SDE $$X_t = -\int_0^t \frac{X_s}{1-s}ds + W_t$$ where $0\leq t < 1$, and $W_t$ is standard Brownian.
One can easily verify that this is a solution $$X_t = (1-t)\int_0^t \frac{1}{1-s}dW_s$$ We wish to show that $\lim\limits_{t\uparrow 1} X_t = 0$ a.s. I can't seem to get this strong result, but I am able to show that $X_t \xrightarrow{L^2} 0$, which implies that there exists a subsequence that converges to $0$ a.s. $$E(X_t^2) = (1-t)^2\int_0^t \frac{1}{(1-s)^2} ds = t(1-t)$$ by Ito's Isometry.
Also, note that $X_t$ is not a martingale. It is possible that the question is wrong, but it does seem reasonable. Any ideas?