I have to solve the following PDE:
$$ \begin{align} \frac{\partial F}{\partial t} + x^2\frac{\partial^2 F}{\partial x^2} + F + e^x &= 0 \\ F(T,x,y) &= x \end{align} $$
by using the Feynman-Kac formula. To be honest, in my studies I've never seen a version of the FK formula for solving this kind of problems, in which we have both a term "$F$" and a term "$e^x$", but looking on Wikipedia I found this formula: the solution of
$$ {\frac {\partial u}{\partial t}}(x,t)+\mu (x,t){\frac {\partial u}{\partial x}}(x,t)+{\tfrac {1}{2}}\sigma ^{2}(x,t){\frac {\partial ^{2}u}{\partial x^{2}}}(x,t)-V(x,t)u(x,t)+f(x,t)=0 $$
with initial condition $u(x,T)=\psi (x)$ is
$$ u(x,t)=E^{Q}\left[\int _{t}^{T}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr+e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }\psi (X_{T}){\Bigg |}X_{t}=x\right] $$
so I tried it using $V=-1$, $u = F$, $f = e^x$ and
$$ x_s = x_te^{-(T-t)+\sqrt{2}(W_s-W_t)} $$
(where $W$ is a Brownian motion). However, I don't get a correct result. Can anyone help me? Am I using the wrong formula?