If we have $X_t$ as a Brownian motion with variance $\sigma^{2}t$ and $X_0 = x$ how can I find the density, $p_{t}(x,y)$ of $Y_t = X_t + t\mu$ with $\mu \in \mathbb{R}$?
I know the expectation of $Y_t = x + t\mu$ and that for $X_t$ (with mean zero and variance $t$) $p_{t}(x,y) = \frac{1}{\sqrt{2\pi t}} e^{\frac{-(x-y)^2}{2t}}$
It's quite simple actually. Just observe that $t\mu$ is deterministic, so you're simply increasing the mean of your Gaussian by $t\mu$.