@Kitonum How does Maple check the accuracy of satisfying the equation - is it done exactly symbolically or numerically with high precision?

@dharr 

It works. Now I can edit my ancient MC14 files in Maple much more conveniently.

@acer 

... and how is the last plot animated for monotonically increasing t?

@dharr  @ Kitonum

Using the last plot command in the attached file, I would like to plot a parameter-dependent set of points (curve (x;y)). I'm asking for some advice.test.mw

Download test.mw

@dharr 

I need to think carefully about your advice.

@Kitonum 

Thank you, it works. The command structure is logical. I learned something new again :-).

It's a pleasure to be able to read and understand your two (theoretically correct) solutions in Maple. Now I have a lot of practice to do to progress in using the software. :-)

Here is the source where I found this problem:
Monatshefte für Mathematik, Volume 64, Issue 4, Page 305, https://gdz.sub.uni-goettingen.de/id/PPN362162050_0064

https://doi.org/10.1007/BF01498605
For methodological reasons, I had introduced the goat and the wolf instead of the dots in the source.

@dharr 

Both speeds are constant and equal from the start. The wolf always runs toward the goat.

@janhardo 

...yes. A similar problem at school level is when a goat escapes along a straight line. In this case, the solution is almost obvious. However, an escape along a circle is much more difficult to model ;-) .

@WD0HHU 

There is only one goat and there is no wolf hunter anywhere.

@janhardo 

...it is.

@vv 

...there are "only" pure proofs of existence using sophisticated methods. I don't know of any concrete examples.

@janhardo 

As a contribution to a peaceful conclusion to the discussion, here is a simple proof:

It must be proven that the function y(x) = cos (a*x) + cos x is not periodic for irrational a.

Suppose/assume that y(x) = cos (a*x) + cos x is periodic with period T for irrational a. Then
y(x) = cos (a*x) + cos x = cos (a*(x + T)) + cos (x + T)
In particular, for x = 0 (this relation must hold for any real x):
cos 0 + cos 0 = cos (a*T) + cos T = 2
This means that
1.) cos (a*T) = 1 with a*T = 2*k*pi,
2.) cos T = 1 with T = 2*m*pi,
where k and m are integers. But then
a = (a*T)/T = k/m,
so a is the quotient of two integers and thus a rational number, contradicting the premise of the problem. Therefore, the assumption is false, y(x) is not periodic.
(There is no third truth value. ;-) )

@janhardo 

...has nothing to do with the purpose of my question.
Rather, what's interesting about this simple problem from the theory of real functions is that a non-periodic continuous function can be approximated by a sequence of periodic functions (which ultimately don't form a Fourier series and whose terms always have only two summands). And I would have liked to see this represented graphically.
I was already interested in this problem decades ago. But unfortunately, computer science wasn't that advanced yet.
Many thanks for all the contributions on the topic :-) .

@janhardo @sand15 902 

... for example, easy to show by contradiction that y(x) = cos(a*x) + cos(x) is not periodic for irrational a.
As an approach:
Assume y(x) is periodic. Then evaluate for x=0 and quickly find that, contrary to the assumption, a is rational. What does bijectivity have to do with this?