= Solution
For the stated outward-accelerating hyperbola, take positive signed <proper acceleration> and an initially stationary spaceship. Then $\eta=a\tau$, so $\dot p=e^{-a\tau}$ and $\dot q=e^{a\tau}$. The initial values are $p(0)=-1/a$, $q(0)=1/a$. <Integration> gives
$$
\boxed{p(\tau)=-\frac1a e^{-a\tau},\qquad q(\tau)=\frac1a e^{a\tau}.}
$$
Taking their half-sum and half-difference yields the <hyperbolic motion>
$$
\boxed{t(\tau)=\frac1a\sinh(a\tau),\qquad x(\tau)=\frac1a\cosh(a\tau).}
$$
The printed position data alone do not specify the initial velocity. If the initial <rapidity> is an arbitrary $\eta_0$, the equally valid positive-acceleration solution is
$$
t(\tau)=\frac{\sinh(\eta_0+a\tau)-\sinh\eta_0}{a},\qquad
x(\tau)=\frac1a+\frac{\cosh(\eta_0+a\tau)-\cosh\eta_0}{a}.
$$
Thus the requested special expression implicitly selects $\eta_0=0$.
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