Solving a certain differential equation when assuming a surface of revolution is minimal The problem is the following:
Consider the surface of revolution
$$
\textbf{q} (t, \mu) = (r(t)\cos(\mu),r(t)\sin(\mu),t)
$$
If $\textbf{q}$ is minimal, then $r(t) = a\cosh(t)+b\sinh(t)$ for $a,b$ constants.   
I'll skip the calculations. I've equated the mean curvature and $0$ and obtained the relation
$$
1+\dot{r}^2 = r\ddot{r}
$$
where each is understood to be a function of $t$. It's been a while since I've taken a class on differential equations, but since I "know" the solution, my plan was to check $r = \cosh(t)$ and $r = \sinh(t)$ are solutions to the above, and then conclude a linear combination of them is also a solution. However, $\cosh(t)$ worked fine, but I cannot really get $\sinh(t)$ to work the same. I get
$$
1+\dot{r}^2 = 1 + \cosh(t)^2 = 2 + \sinh(t)^2 \ne \sinh(t)^2
$$
is there perhaps an identity I'm not recalling/don't know? I also tried "guessing" $r(t) = a\cosh(t)+b\sinh(t)$, but that didn't work out too well either. Any suggestions?
Edit: The "solution" to check in the book was incorrect, which was kind of clear anyway since $\sinh(t)$ wasn't working. 

This ODE can be solved by separation of variables, To check that 
$$r(t):=a\cosh(\dfrac{t-t_o}{a})$$ for $t_0\in\mathbb{R}$, $a>0$, 
solve your ODE is easy.
Indeed
$$a\cosh()\frac{\cosh()}{a}=1+\sinh^2()$$
