Solving a certain differential equation when assuming a surface of revolution is minimal

Question

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.

Answer 1

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 $$a\cosh()\frac{\cosh()}{a}=1+\sinh^2()$$

Answer 2

If you have ever experimented with two rings and some soap bubble solution you might recall that if the rings are pulled far enough apart the Catenoid surface snaps closed with both rings holding on to their "solution" , in parallel planes so to speak , this is also a solution to the problem , the Goldschmidt solution.