The first partial differential equation, using the shorthand $\mathcal{L}u(t,x) = \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t)$, is: \begin{equation} \frac{\partial u}{\partial t}(x,t) + \mathcal{L}u(x,t) -Vu(x,t) + f(x,t) = 0 \tag{1}\end{equation}
and the other is \begin{equation} \frac{\partial v}{\partial t}(x,t) + \mathcal{L}v(x,t) -(V + W)* v(x,t) + f(x,t) = 0 \tag{2} \end{equation}
both with terminal condition: $v(T,x) = u(T,x) = g(x)$ and with constants $V$ and $W$.
So the only difference is that the second equation has the term $-Wv(x,t)$ included. I want to find out how I can relate the two solutions, so that if I have a solution of $(1)$ I can multiply it by something and get a solution of $(2)$.
I am wondering if this is possible. So far, I have written, using Feynman-Kac, solutions to both equations:
\begin{equation} u(x,t) = E^Q\left[ \int_t^T e^{-V (r-t)}f(X_r,r)dr + e^{-V(T-t)}g(X_T) \Bigg| X_t=x \right] \end{equation}
\begin{equation} v(x,t) = E^Q\left[ \int_t^T e^{-(V +G) (r-t)}f(X_r,r)dr + e^{-(V+G)(T-t)}g(X_T) \Bigg| X_t=x \right] \end{equation}
I can't seem to think of a way to connect the two together, but according to a book I'm reading there should be a way to do this (but there's no other details). Any help would be greatly appreciated. Thanks!
Suppose $u$ and $v$ solve $$ \left[\partial_{t}+\mathcal{L}-c\right]u+f=0 \qquad \text{and} \qquad \left[\partial_{t}+\mathcal{L}-\left(c+d\right)\right]v+e^{dt}f=0. $$ Moreover, suppose $u(T,\cdot)=g$ and $v(T,\cdot)=e^{dT}g$. If we assume enough regularity on $u$ and $v$ along with $\mu$, $\sigma$, and $g$, the Feynman-Kac formula tells us that $$ u(t,x)=\mathbb{E}\left[\int_{t}^{T}e^{-c(r-t)}f(r,X_{r})dr+e^{-c\left(T-t\right)}g(X_{T})\mid X_{t}=x\right] $$ and \begin{align*} v(t,x) & =\mathbb{E}\left[\int_{t}^{T}e^{-(c+d)(r-t)}e^{dr}f(r,X_{r})dr+e^{-(c+d)(T-t)}e^{dT}g(X_{T})\right]\\ & =\mathbb{E}\left[\int_{t}^{T}e^{-c(r-t)}f(r,X_{r})dr+e^{-c(T-t)}g(X_{T})\right]e^{dt}\\ & =e^{dt}u(t,x) \end{align*} where $X$ satisfies an Ito SDE with infinitesimal generator $\mathcal{L}$
Are you sure this is not what you/the text meant?
Note: a simpler argument to go from one PDE to the other is by a change of variables.