@Carl Love Thanks for this.  I had seen this trick but had completely forgotten it.  I just uploaded a modified worksheet which includes a section that computes the travel time following your suggestion.

@Yee Voon The letter "I" represents the imaginary unit, that is, sqrt(-1), in Maple.  Use some other letter in your calculations.

@lucaud Your question refers to "inverted pendulum" without saying what it is that you want to do with it.  Because of that we have at least three different proposed ideas:

• My model of stabilizing an inverted pendulum through vertical oscillations of its pivot;
• Thomas Richard's suggestion of stabilizing an inverted pendulum attached to a horizontally moving cart;
• The photo that you posted in reply to Thomas Richard's suggestion which shows something totally different.

To get a useful response you should formulate your question much more clearly.

@MrMarc 

As I said earlier, your system is guaranteed to have the solution {x=0, y=0, z=0}.  If you want a nonzero solution, then you will have to impose the extra condition -100*b*p-p*r+100*a+100*p-r = 0 on the coefficients.  If we solve that condition for r, we get r = (100*(-b*p+a+p))/(p+1).

Plugging in the numbers that you have given for a, b, and p we get r = 13.28571429 which is close to your value of r = 13.29.  That's why it appears to you that you are getting a well-defined solution but that's only an illusion.

You should know that if r is anything other than what I have shown, then the only legitimate solution is {x=0, y=0, z=0}.  If r is exactly what I have shown, then x can be anything, while y = p*x.  For example, with your choice of p = 2.5 and x = 215054, we get y =  p*x = 537635 as you have noted.  But you could taken any other x, let's say x = 1234, in which case you will have y = 1234*p = 3085, and that would have been just as good a solution.

@Earl The change of variable from alpha to t in vv's calculations is not essential.  It's possible to do the calculation directly with alpha.
 

Download projective-alt.mw

@vv Thanks for the pointer to showstat(log) which clearly shows the use of the optional bracketed arguments in Maple's log[b](a) procedure.

What you have shown works for me.  Try re-executing your worksheet.

Aside: Where you have said PLOT(...) I would have said display(...).

@vv I looked up several help pages pertaining to Maple's procs but was unable to find a reference to the f[u](x) construct.  It must be hidden in an obscure place.

Carl, what I am asking is why is it that g[12] has any meaning at all?  Where is such a construct defined/documented?

@vv Your construction puzzles me.  Let's look at a simpler case:

g := x -> x^2;

Then g(5) returns 25.  I also see that g[12](5) returns 25 but I don't understand why.  What is the meaning of g[12]?  Would you please explain?

@Volker Lehner Short answer: To enter new material inside a procedure, type Shift-Enter.

For anything other than toy programs, I must echo acer's suggestion of using 1D input plus the "Worksheet" instead of the "Document" mode.  You will find instructions https://userpages.umbc.edu/~rostamia/math481/config/maple.html

As far as I know, you cannot convert your existing worksheet.  Configure Maple as instructed, then open a fresh new worksheet.

I cannot understand your equations.  Nor can Maple.  You can help yourself by writing a more readable code,  How many differential equations do you have?  How many unknowns?  I suggest something like:

de1 := diff(f(eta), eta, eta, eta) + ...
de2 := ...
de3 := ...
bc := f(0)=1, D(f)(0)=...

and then

dsol := dsolve([de1, de2, de3, ..., bc], numeric);

@Volker Lehner In your worksheet some of the entries are German text and some others are Maple code.  Therefore I assume that you know how to enter text and code as needed.

You entered my code as text in your document.  But you should have entered it as code. Here I have corrected that and have illustrated how to run the procedure.  My comments are in red.

Berechnungen-1-fixed.mw

@Preben Alsholm That's very nice analysis.  The "magic" happens between ineq1 and inqe2.  The calculation up to and including ineq1 is consistent with y(x)≥ 0. The strict inequality, y(x) > 0, enters in the separation of variables calculation that takes us from iineq1 to ineq2. Quite insightful.

@JohnS I just posted a response to oceanxinlie in which I explained some of the details of the calculations, including how Lcrit was calculated.  The title of the post is "Details".