Let $\mathbb{H}^2$ be the upper sheet of the hyperboloid defined by $x^2+y^2-z^2=-1$ in three-dimensional Minkowski space $(\mathbb{R}^3, g_M)$, where $g_M = \text{diag}(1,1,-1)$. In other words, consider the hyperboloid model of the hyperbolic plane.
I am having trouble showing that the unit speed geodesic at point $p \in \mathbb{H}^2$ in direction $v \in T_p\mathbb{H}^2$ such that $g(v,v) = 1$ is given by $\gamma(t) = \cosh(t)p + \sinh(t)v$.
I tried the following:
Parametrize $\mathbb{H}^2$ by \begin{equation} (r, \theta) \subseteq (0, \infty) \times (0, 2\pi) \mapsto (\sinh(r)\cos(\theta), \sinh(r)\sin(\theta), \cosh(r)) \end{equation}
Pulling back the Minkowski metric to our local coordinates I obtain
$$ g = dr^2 + \sinh^2(r)d\theta^2$$
Plugin this into the geodesic equation \begin{align} r'' - \theta'^2\sinh(r)\cosh(r) &= 0 \\ \theta''\sinh^2(r) + 2r'\theta'\sinh(r)\cosh(r) &= 0 \end{align}
Because the metric in independent of $\theta$ we find that $\theta'\sinh^2(r)$ is conserved. Plugging this into the first geodesic equation we get
$$r'' - k\frac{\cosh(r)}{\sinh^3(r)} = 0.$$
At this point I am stuck, as there doesn't seem to be a simple solution using these coordinates.
Thank you for any help!