Relativity and differential equation I have a question regarding Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation ISBN 978-0-7167-0344-0. It is a book about Einstein's theory of gravitation.
In page 166 of chapter 6.2 about Hyperbolic Motion, the authors present a person feeling constant acceleration $g$ along the direction $x^1$. The authors get the following equations:

$$a^0 = \frac{du^0}{d \tau} = gu^1$$
$$a^1 = \frac{du^1}{d \tau} = gu^0$$

And the system is solved by the authors to get:

$$t=g^{-1} \sinh{g\tau}$$
$$x=g^{-1} \cosh{g\tau}$$

So obviously $u^0 = t$ and $u^1 = x$, but I am not sure about the change of variable between $t$ and $\tau$ to make the appearance of the Lorentz factor.
How do the authors solve the differential equation?
 A: solution:
$${\frac {d}{d\tau}}u_{{0}} \left( \tau \right) -gu_{{1}} \left( \tau
 \right) =0\tag 1
$$
$${\frac {d}{d\tau}}u_{{1}} \left( \tau \right) -gu_{{0}} \left( \tau
 \right) =0\tag 2
$$
and the constraint condition that 
$$dsq=\left( {\frac {d}{d\tau}}u_{{0}} \left( \tau \right)  \right) ^{2}-
 \left( {\frac {d}{d\tau}}u_{{1}} \left( \tau \right)  \right) ^{2}=
\epsilon\tag 3
$$
where $\epsilon=0$ or $1$ 
with $\frac{d}{d\tau}eq(1)$ and equation (2) we obtain:
$${\frac {d^{2}}{d{\tau}^{2}}}u_{{0}} \left( \tau \right) -{g}^{2}u_{{0}
} \left( \tau \right) =0
\tag 4$$
$\Rightarrow$
$$u_0(\tau)=1/2\,{\frac { \left( gA+B \right) {{\rm e}^{g\tau}}}{g}}+1/2\,{\frac {
 \left( -B+gA \right) {{\rm e}^{-g\tau}}}{g}}\tag 5
$$
where $A=u_0(0)$ and $B=D(u_0)(0)$ are arbitrary initial conditions 
with equation (3) and (2) we get for dsq
$$dsq(\tau)=\left( {\frac {d}{d\tau}}u_{{0}} \left( \tau \right)  \right) ^{2}-
 (g\,u_0(\tau))^2=
\epsilon
$$
thus:  for $dsq(0)=\epsilon$ we can obtain the initial condition
$B=B(A)$ and get:
$$B=\sqrt {{g}^{2}{A}^{2}+\epsilon}$$
with  $A=0$ and $\epsilon=1$ we get:
$$u_0(\tau)=1/2\,{\frac {{{\rm e}^{g\tau}}}{g}}-1/2\,{\frac {{{\rm e}^{-g\tau}}}{g
}}=\frac{1}{g}\sinh(g\tau)
$$
$$u_1(\tau)={\frac {1/2\,{{\rm e}^{g\tau}}+1/2\,{{\rm e}^{-g\tau}}}{g}}=\frac{1}{g}\cosh(g\tau)
$$
and for 
$A=1 \,,\epsilon=0$ we get
$$u_0(\tau)=e^{g\,\tau}$$
$$u_1(\tau)=e^{g\,\tau}$$
