@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?

@vv 

I asked for an example using a plot. Let a = e (Euler's number) and a(n) be a sequence of rational numbers converging to e. At what "closeness" of a(n) to e does it become apparent that for irrational a, i.e., a = e, the function y(x) is not periodic.

@Kitonum 

This is true, because v*v´ = (v^2/2)´. After integrating the ODE, Your argument becomes obvious.

@dharr 

...you discovered this by "looking closely," which, for me as a Maple novice, was impressively instructive. According to this, the conditions under which terms of algebraic structure can be simplified into trigonometric structure are unknown.

@janhardo 

...the implicit solution is obtained by multiplying the given ODE by 2 and then integrating. The initial value is irrelevant in this case. Only the general solution is obtained. The situation is different if the given ODE is converted into explicit form. Then, y(x) must be assumed to be nonzero to ensure continuity on the right-hand side. This also applies to the initial value. Only with special treatment can the abscissa x=0 be included as the boundary of the continuity region. Maple apparently checks for the possibility of b=0 in the background and then rejects the solution as an initial value problem.

... ic=b ? It should be not zero, because of continuity of the "right side". Try it.