= Solution
Write the <diffusion process> as $dX_t=b(X_t)dt+\sigma(X_t)dB_t$, with $a=\sigma\sigma^{\mathsf T}$. Its <diffusion generator> is
$$
L\varphi=\frac12\sum_{i,j}a^{ij}\partial_{ij}\varphi+\sum_i b^i\partial_i\varphi.
$$
The <Itô formula> says that $\varphi(X_t)-\varphi(X_0)-\int_0^tL\varphi(X_s)ds$ is a <local martingale>, and is a true <martingale> when the stopped terms are integrable. Taking <expectations> gives <Dynkin formula for a diffusion>. This is the bridge from infinitesimal stochastic dynamics to differential equations.
For the <transition semigroup> $P_tg(x)=\mathbb E_xg(X_t)$, the <Markov property> gives $P_{t+s}=P_tP_s$. On a smooth generator domain where differentiation is valid, differentiating the semigroup identity yields the <Kolmogorov backward equation> $\partial_tP_tg=L(P_tg)$, with initial value $g$. If a transition density $p(t,x,y)$ exists, integration against smooth compactly supported tests and integration by parts give the <Fokker-Planck equation>
$$
\partial_tp=L_y^*p,
\qquad
L^*p=\frac12\sum_{i,j}\partial_{ij}(a^{ij}p)-\sum_i\partial_i(b^ip).
$$
The backward equation acts on the starting point, while this forward equation describes the evolving probability density.
Here is a full representation proof with a source and potential. Suppose $X$ is nonexplosive, $V\ge0$ is bounded, $f,g$ are bounded, and $u$ is a bounded <classical solution> on a finite slab of
$$
u_t+Lu-Vu=-f,\qquad u(T,x)=g(x).
$$
For a process started from $x$ at time $t$, put $D_s=\exp(-\int_t^sV(X_r)dr)$. The <Itô product rule> gives
$$
d(D_su(s,X_s))=-D_sf(s,X_s)ds+D_s\nabla u(s,X_s)\sigma(X_s)dB_s.
$$
Therefore $D_su(s,X_s)+\int_t^sD_rf(r,X_r)dr$ is a <local martingale>. It is bounded in absolute value by $\|u\|_\infty+(T-t)\|f\|_\infty$, so it is a true <martingale>, without requiring a global derivative bound. Its endpoint <expectations> give the <parabolic Feynman-Kac formula with a source>
$$
\boxed{u(t,x)=\mathbb E_{t,x}\left[e^{-\int_t^TV(X_r)dr}g(X_T)
+\int_t^Te^{-\int_t^sV(X_r)dr}f(s,X_s)ds\right]}.
$$
This proves uniqueness in the bounded classical class. Setting $V=f=0$ recovers the backward heat or diffusion representation; the minus sign in the killing term produces the negative exponential.
For an elliptic problem, stop at the first exit $\tau$ from a bounded regular domain $D$. Suppose $\mathbb E_x\tau<\infty$, $u\in C^2(D)\cap C(\bar D)$ is bounded and solves $(L-V)u=-f$ with boundary data $g$. The same calculation, stopped at $t\wedge\tau$, gives a <martingale> dominated by $\|u\|_\infty+\|f\|_\infty\tau$. Continuity puts $X_\tau$ on the boundary. Taking the limit and using <dominated convergence> gives the <elliptic Feynman-Kac formula>
$$
\boxed{u(x)=\mathbb E_x\left[e^{-\int_0^\tau V(X_r)dr}g(X_\tau)
+\int_0^\tau e^{-\int_0^sV(X_r)dr}f(X_s)ds\right]}.
$$
For $V=f=0$, harmonic functions for $L$ give stopped <martingales> and are averages of their exit-boundary values. For <Brownian motion> $L=\tfrac12\Delta$, this is the probabilistic solution of the <Dirichlet problem>. Suitable smooth coefficients and uniform ellipticity give familiar settings where the required diffusion, exit-time integrability and <classical solutions> exist. The representation itself follows from the stated regularity and integrability conditions, rather than claiming classical solvability for arbitrary coefficients.
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