Regularly, we can find the wave equation with a external force described as:
\begin{equation} \frac{\partial^2 u}{\partial t^2} - c^2 \Delta u = f(\vec x, t) \end{equation}
where $t$ is the time variable and $\vec x$ is some position in space.
I also saw in some papers the following notation:
\begin{cases} \frac{1}{k} \frac {\partial u}{\partial t} = - \nabla \cdot \vec{v} + f \\ \rho \frac{\partial \vec v}{\partial t} = - \nabla u \end{cases}
Here $\vec v$ is the velocity vector and $u$ is the pressure field.
Are those equivalent? (even considering $\rho$ and $k$ unitary for sake of simplicity). If they are, how this equivalence is found?
Of course. You can differentiate the first equation w.r.t $t$ and obtain
$$ \frac{1}{k}\frac{\partial^2 u}{\partial t^2} = -\nabla\cdot \frac{\partial \vec{v}}{\partial t} + \frac{\partial f}{\partial t} = \frac{1}{\rho} \nabla \cdot (\nabla u) + \frac{\partial f}{\partial t} $$
Then $\nabla \cdot (\nabla u) = \Delta u$, $\frac{k}{\rho} = c^2$ and $\frac{\partial f}{\partial t} := f(x,t)$