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Let me offer another viewpoint. By a Feynman-Kac formula, a solution to $$ \partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;, ~~ m(t,0) = m(t,1) = 0\;, $$ admits the following stochastic representation $$ m(t, y) = \mathbb{E}_y \{ 1_{\{\tilde T \ge t\}} m_0(Z_t) \} $$ where $\mathbb{E}_y$ is an expected value over the process which satisfies $$ Z_t = y - b t + \sigma W_t $$ and $\tilde T$ is the first exit time of $Z$ from $(0,1)$. Alternatively, one can write $m(t,y)$ as $$ m(t, y) = \mathbb{E}_y \{ m_0(Z_t) \mid \tilde T \ge t \} \mathbb{P}_y(\tilde T \ge t) $$ Unfortunately, due to the truncation of $X$ by $T$, there does not seem to be a relation between this $m(t,y)$ and the transition density of $Y$.