# Difference Between Feynman-Kac PDE and Kolmogorov Backward Equation

Suppose $X_u$ satisfies the SDE $$dX_u = a(u, X_u) du + b(u,X_u) dW_u\;.$$ Fix $T > 0$ and let $X_t = x$ for $0 \leq t \leq T$ be given. For any measurable function $h$, define $$g(t,x) = E^{t,x}(h(X_T))\;.$$ Then $g$ satisfies the Feynman-Kac equation $$g_t(t,x) + a(t, x) g_x(t,x) + \frac{1}{2}b^2(t,x)g_{xx}(t,x) = 0$$ with terminal condition $$g(T,x) = h(x) \;.$$ Fair enough, but now consider the Kolmogorov backward equation. Let $p(t,x; T, y)$ be the transition density of $X_u$; i.e., if $X_t = x$, then $X_T$ has density $p(t,x; T,y)$ in the $y$ variable. Then for each $T$ and $y$, $p$ satisfies the Kolmogorov backward equation $$p_t(t,x; T,y) + a(t,x)p_x(t,x;T,y) + \frac{1}{2}b^2(t,x)p_{xx}(t,x;, T,y) = 0$$ with terminal condition $$p(T,x; T,y) = u_T(x)$$ for some function $u_T(x)$

As far as I can tell, these are the same equations; the only difference between the solutions being the terminal conditions. Is this a correct understanding? Is there some greater insight as to why these aren't both called the same equation? I see that $g$ and $p$ are related by $$g(t,x) = \int_{-\infty}^\infty h(y) p(x,t;T,y)\,dy$$ but I'm not seeing anything insightful from this relationship, and am afraid I'm missing something. I've read the answer here but am still a bit unsure of the difference.