Solve the SDE $dX_t=dB_t+\frac{c-X_t}{T-t}dt$ How to solve the following SDE characterizing Brownian motion with fixed end point $c$ at time $T$?
$$\mathrm dX_t=\mathrm dB_t+\frac{c-X_t}{T-t}\,\mathrm dt$$
I do not believe a strong solution exists, by ito lemma it would seem contradictory. Although, intuitively the solution should be something like
$$X_t=B_t+\frac{t(c-B_T)}{T}$$
Tips for solving the equation are welcome!
 A: Your SDE is a special case for Hull-White SDE, which can be solved
analytically. Recall that $X$ is a Hull-White process if $X$ satisfies
a SDE of the form: 
$$
dX_{t}=\beta_{t}(\alpha_{t}-X_{t})dt+\sigma_{t}dB_{t},
$$
where $\alpha,\beta,\sigma$ are deterministic processes (and are
usually identified as functions with domain $[0,\infty)$ and codomain
$\mathbb{R}$). For your case, $\alpha_{t}=c$, $\beta_{t}=\frac{1}{T-t}$,
$\sigma_{t}=1$.
The trick to solve such SDE is to consider a suitable integrating
factor $\mu$ (a deterministic process)
Apply Ito's lemma, then 
\begin{eqnarray*}
 &  & d(X_{t}\mu_{t})\\
 & = & \mu_{t}'X_{t}dt+\mu_{t}dX_{t}.
\end{eqnarray*}
Choose $\mu_{t}$ such that $\mu_{t}'X_{t}+\mu_{t}\beta_{t}(-X_{t})=0$.
That is, $\mu_{t}'=\beta_{t}\mu_{t}$. Note that, there is more than
one choice for $\mu$ but any choice is OK. Then we observe that 
$$
d(\mu_{t}X_{t})=\gamma_{1}(t)dt+\gamma_{2}(t)dB_{t}
$$
for some deterministic functions $\gamma_{1}$ and $\gamma_{2}$.
The above SDE can be integrated directly.
