Hyperbolic harmonic oscillator The classical harmonic oscillator can be associated to the differential equation:
$$y''+\omega^2y=0$$
and solutions
$$y=A\cos(\omega t)+B\sin(\omega t)$$
or $$y=A\cos(\omega t+\delta)$$
The harmonic oscillator is associated to springs and/or small oscillation problems, or Hooke law force motion. What about the hyperbolic version of this equation? That is:
$$y''-k^2y=0$$
with 
$$y=A\sinh(kt)+B\cosh(kt)$$
What are the physical problems associated to this equation?
Remark: A "Wick-like" rotation in $\omega$ transforms the HO into the hyperbolic one at the level of the solutions. 
 A: Yes, there are many physical problems that do follow such equations. You even mentioned one of them in your question. 
A pendulum held against gravity near to its unstable equilibrium point locally follows such an equation
The differential equation that governs the motion of a pendulum is given as
$$ \dfrac{d^2\theta}{dt^2}+g \sin(\theta) = 0$$
But if you let $g \rightarrow -g$ (reverse the direction of gravity) then the equation would read as 
$$ \dfrac{d^2\theta}{dt^2}-g \sin(\theta) = 0$$
The transformation which I did transformed the equations in such a way that the pendulum is now near its unstable equilibrium point i.e. the initial position of the bob of the pendulum is more near the angle $\pi$ than to $0$. And now if we assume $\theta <<1$ then what we essentially have is (the nearer it is to $\pi$ the more accurate the solution)
$$ \dfrac{d^2\theta}{dt^2}=g\theta$$
Hence this physical situation has a local solution (around its unstable equilibrium point) as
$$\theta (t) =A\cosh(gt) +B\sinh(gt)$$
Another interesting way to come up with the same solution is to transform $\theta \rightarrow -\theta$ instead of $g \rightarrow -g$, this is an equivalent transformation and if we transform both $\theta \rightarrow -\theta$ and $g \rightarrow -g$ the differential equation remains unchanged. I'll leave this to you to figure out why this happens.
A: I had a professor give a really good homework problem that demonstrated the similarities of the two situations. The math is obviously important but the picture is what's helped me build an intuition for harmonic (and exponential) motion: 
Imagine a plank of wood resting on two rollers on either side of the center of mass so as to hold it up (located at $\pm a$, say). If the plank is moved slightly, the torques provided by the (currently unmoving) rollers must stay the same since the plank doesn't start rotating. Because the lever arms (distance from the CM to the rollers) changes, this means the normal force must be greater on the side with shorter arm. So if the plank moves to the right, the normal force at $+a$ has a greater magnitude than that at $-a$. 
1) If the rollers are now spinning inward, they generate a frictional force pushing the plank back to the center whose magnitude is proportional to that of the normal force from the roller. If the plank is at the center, these frictional forces are equal and nothing happens. If it moves slightly to the right, the roller at $+a$ has a greater normal force $\to$ it's frictional force to the left increases and the plank returns towards its original position. If you work out Newton's laws, you'll find the CM of the plank undergoes exactly harmonic motion. 
2) If the rollers are spinning the opposite direction (away from the center) and you move the plank to the right the exact opposite effect happens. Again, the normal force from the $+a$ roller is greater and so its frictional force to the right will be greater and the plank will continue accelerating (exponentially) to the right. As it accelerates to the right, the difference in frictional forces becomes greater so it seems reasonable to expect you would have an exponential motion of the plank. 
A: This equation would correspond to the physical situation where you had a force that was repulsive and somehow grew proportionally to the distance you are from the equilibrium point $y=0$. As such, the particle in question will be accelerated to ever increasing speeds away from $y=0$ and this motion is not bounded. Eventually relativistic effects would kick in, and you would have to modify your framework. I know of no real physical system that follows this equation. And in fact, the energy of this problem is unbounded just like the trajectory, so the source of this force field would have to be able to provide infinite energy to your particle. 
Also, such a system is not an oscillator as it does not oscillate. It just goes off to infinity. 
