I was hoping for some guidance on solving this PDE. I have taken the rho and fund the partial derivative wrt $t$ and $x$ and have plugged this into the general equation. I am unsure how to proceed from this point with the given initial conditions. Could someone provide me with a hint please?
My thought process is the partial derivative of rho wrt $t$ and $x$ at $t=0$ is 0 and hence I am left with $ke^{-x^2/2} = 0$ and hence $k = 0$. This would make my equation simplify to $$\frac{\partial\rho}{\partial t} + v \frac{\partial\rho}{\partial x} = 0$$.
Write $$\rho(x,t) = e^{\frac{-kx}{v}}w(x,t).$$ Use this form to solve the PDE problem $$\frac{\partial\rho}{\partial t} + v \frac{\partial\rho}{\partial x} + k\rho = 0 \quad\quad x\in \mathbb {R} ,t>0, $$ $$\rho(x,t=0) = e^\frac{-x^2}{2} \quad\quad x \in \mathbb {R.}$$
This Ansatz on $\rho$ leads to $$ \rho_t = e^{-kx/v} w_t \qquad\text{and}\qquad \rho_x = e^{-kx/v} w_x - \tfrac{k}{v} e^{-kx/v} w \, . $$ Injecting this into the PDE, the linear transport equation (a.k.a linear advection equation) $ w_t + v w_x =0 $ satisfied by $w = e^{kx/v} \rho$ is obtained. The solution to the initial-value problem $$ w(x,0) = e^{kx/v} e^{-x^2/2} = w_0(x) $$ can be obtained by using the method of characteristics (see e.g. this Wikipedia example): $$ w(x,t) = w_0(x-vt) = e^{k(x-vt)/v} e^{-(x-vt)^2/2} . $$ Finally, we know $\rho = e^{-kx/v} w$, i.e. $$ \rho(x,t) = e^{-kt} e^{-(x-vt)^2/2} . $$ Note that the method of characteristics can be applied directly to the initial PDE, without introducing $w$ (see e.g. this related post).