= Solution
The $x$ <Euler-Lagrange equation> has the first integral
$$
e^{2Ht}\dot x=v_0,
$$
where the constant follows from the initial data. Combining this with <proper-time normalization> gives the <Timelike geodesic in the expanding flat slicing of de Sitter spacetime>:
$$
-\dot t^2+e^{2Ht}\dot x^2=-1
\quad\Longrightarrow\quad
\dot t=\sqrt{1+v_0^2e^{-2Ht}}.
$$
Consequently
$$
\frac{d\tau}{dt}=\frac1{\sqrt{1+v_0^2e^{-2Ht}}}.
$$
Integrating from $t=0$, where $\tau=0$, gives
$$
\boxed{\tau(t)=\frac1{2H}\log\left[
\frac{\sqrt{1+v_0^2e^{-2Ht}}+1}
{\sqrt{1+v_0^2e^{-2Ht}}-1}
\frac{\sqrt{1+v_0^2}-1}
{\sqrt{1+v_0^2}+1}
\right].}
$$
As $t\to-\infty$, the first ratio tends to $1$, so
$$
\boxed{\tau\longrightarrow\tau_-
=-\frac1{2H}\log\frac{\sqrt{1+v_0^2}+1}
{\sqrt{1+v_0^2}-1}>-\infty.}
$$
Thus $t$ tends to $-\infty$ at the finite proper time $\tau_-$.
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