Link between Fokker-Planck equation and Feynman-Kac formula

Question

According to the Feynman-Kac formula, we know the solution of the partial differential equation: $${\frac {\partial u}{\partial t}}(x,t)+\mu (x,t){\frac {\partial u}{\partial x}}(x,t)+{\tfrac {1}{2}}\sigma ^{2}(x,t){\frac {\partial ^{2}u}{\partial x^{2}}}(x,t)-V(x,t)u(x,t)+f(x,t)=0,$$ is given by: $$u(x,t)=E^{Q}\left[\int _{t}^{T}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr+e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }\psi (X_{T}){\Bigg |}X_{t}=x\right]$$, under the probability measure $Q$ such that $X$ is an Itô process driven by the equation $$dX=\mu (X,t)\,dt+\sigma (X,t)\,dW^{Q}.$$ In another hand, consider the Fokker-Planck equation: $${\displaystyle {\frac {\partial }{\partial t}}p(x,t)=-{\frac {\partial }{\partial x}}\left[\mu (x,t)p(x,t)\right]+{\frac {\partial ^{2}}{\partial x^{2}}}\left[D(x,t)p(x,t)\right],}$$ the solution $p(x,t)$ is the probability density function of the random variable, where the SDE of the random variable is given below: $${\displaystyle dX_{t}=\mu (X_{t},t)\,dt+\sigma (X_{t},t)\,dW_{t}}$$, where ${\displaystyle D(X_{t},t)=\sigma ^{2}(X_{t},t)/2},$ we can then rewrite the Fokker-Planck equation as the following form: $$\frac{\partial}{\partial t}p(x,t) +\left[ \mu(x,t)-\frac{\partial}{\partial x}D(x,t) \right]\frac{\partial}{\partial x}p(x,t) + \left[\frac{\partial}{\partial x} \mu(x,t) -\frac{1}{2} \frac{\partial^2}{\partial x^2}D(x,t) \right]p(x,t) -\frac{1}{2}D(x,t)\frac{\partial^2}{\partial x^2}p(x,t)$$, and then solve the partial differential equation above by using the Feynman-Kac formula, which is: $$u(x,t)=E^{Q}\left[e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }\psi (X_{T}){\Bigg |}X_{t}=x\right]$$ under the probability measure $Q$ such that $X$ is an Itô process driven by the equation $${\displaystyle dX_{t}=\left[\mu (X_{t},t)-{\frac {1}{2}}{\frac {\partial }{\partial X_{t}}}D(X_{t},t)\right]\,dt+{\sqrt {2D(X_{t},t)}}\circ dW_{t}.}$$ Is this idea right?