I have mostly only seen application of Feynman-Kac where the RHS is 0, but I have recently stumpled upon a case where a deterministic function is included, and so the boundary problem is as stated: \begin{align*} \frac{\partial F}{\partial t}+\mu(t,x)\frac{\partial F}{\partial x}+\frac{1}{2}\sigma^2(t,x)\frac{\partial^ 2 F}{\partial x^2}-r(t,x)F+k(t,x)&=0\\ F(T,x)&=\Phi(x) \end{align*}
How would a solution for $F$ look like? (Note: I have answered my own question, but would like criticism of any mistake or errors I might have made)
My take on this was to seperate the problem into something known, so here goes. Proof: Firstly we split up the problem into two boundary problems with the first being $F$ dependant and the other $K$ dependant. Firstly, consider the function $$G(t,x)=E_{t,x}\left[\mathrm e^{-\int_t^Tr(u,x_u)du}\Phi(x)\right]$$ Hence we have \begin{align*} \frac{\partial G}{\partial t}+\mu(t,x)\frac{\partial G}{\partial x}+\frac{1}{2}\sigma^2(t,x)\frac{\partial^ 2 G}{\partial x^2}-r(t,x)G&=0\\ G(T,x)&=\Phi(x) \end{align*} Secondly, consider the function $$H(t,x)=\int_t^T E_{t,x}\left[\mathrm e^{-\int_t^s r(u,x_u)du}k(s,x_s)\right]ds$$ Which again must staisfy the PDE \begin{align*} \frac{\partial H}{\partial t}+\mu(t,x)\frac{\partial H}{\partial x}+\frac{1}{2}\sigma^2(t,x)\frac{\partial^ 2 H}{\partial x^2}-r(t,x)H&=k(t,x)\\ H(T,x)&=0 \end{align*} Lastly impose that $F(t,x)=G(t,x)+H(t,x)$ in which by linearity of derivatives the yields exatly the original boundary problem. Which would then yield: $$F(t,x)=E_{t,x}\left[\mathrm e^{-\int_t^Tr(u,x_u)du}\Phi(x)\right]+\int_t^TE_{t,x}\left[\mathrm e^{-\int_t^sr(u,x_u)du}k(t,s)\right]ds$$