@delvin Your 1st line implicitly defines U as a function of xi. So, the next line should be

eq:= 6*c^2*U + 9*c*U^2 + 3*U^3 - diff(U, xi$2);

@jediknight I get this:

restart:
GainQ:= 8*wx^2*(m-1)/(Pi^2*sqrt((m*wx^2-1)^2+Q^2*wx^2*(wx^2-1)^2*(m-1)^2));
params:= [m=6];
Qs:= [0.20, 0.30, 0.40, 0.70, 1.0, 6.0]:
a:= subs(params, GainQ);
da:= numer(diff(a, wx));
for q in Qs do
    x:= fsolve(subs(Q= q, da), wx= 0.01..infinity);
    Top:= subs(wx= x, Q= q, a);
    printf(`Q = %a, wx_max= %a, Top = %a\n`, q, x, Top)
od:  

Q = .20, wx_max= .4257659439, Top = 2.043988894
Q = .30, wx_max= .4507684910, Top = 1.415858396
Q = .40, wx_max= .4927877155, Top = 1.124695001
Q = .70, wx_max= .7476605392, Top = .8641647894
Q = 1.0, wx_max= .8878848311, Top = .8306147086
Q = 6.0, wx_max= .9972137076, Top = .8110221752

Are any of these unexpected? If so, why?

What is your intended purpose of this line of code?

v := t->vs(t); l/~nvs;

The first part, v:= t->vs(t), is semantically equivalent to v:= vs; thus it serves no practical purpose. The second part, l/~nvs, has no practical effect at all because it doesn't change the value of any variable.

@mmcdara I think that convert(..., unit_free) is the way to go. If you encounter some situation where that doesn't work (I don't have any reason to suspect that such a situation exists, but Maple often surprises), then I think a reasonable alternative is

eval(..., Units:-Unit= 1)

@tomleslie I did see the bug that you described earlier today when I tried the same thing, but now it's gone. Try doing restart. If that doesn't fix it, try using 1-D input. (I'm quite curious which of those two fixes it.) Anyway, using my code from above with the only change being semilogplot instead of loglogplot, I get

@acer Is explicitly combining addends necessary for the OP's Maple 2015? It doesn't seem necessary in Maple 2022 (if accepting the SI unit).

b:= 7*Unit(mm)+3*Unit(m);
convert(b, unit_free);  

@mmcdara Good Answer; voted up.

I guess that you didn't try searching "trace" in the help. The command to trace is simply trace(...procedure name(s)...). Thus IF1 can be reduced to

IF1:= (__w1, __w2, __x1, __x2, __x3)->
    evalf(
        Int(
            eval(MM, [w1= __w1, w2= __w2, x1= __x1, x2= __x2, x3= __x3])
           , [b2= 1..2, b3= 1..2]
           , method= _d01ajc
           , epsilon= 0.001
         )
    )
:

And do

trace(IF1):

before calling NLPSolve.

@The function I think that the breadth of my Reply has been misinterpreted. I was only objecting to this statement by @vv , which I thought was too simplistic:

• For a) it's enough to compute simplify(A . B), obtaining the unit matrix, without any assumption on p.

In his followup Reply to my objection, he changed that to

• A and B having polynomial entries, A.B = Id ==> A invertible, that's all. [Emphasis added.]

That modified statement I no longer think too simplistic. Both A and B having polynomial entries (i.e., no variables in denominators) is sufficient to guarantee that A is invertible for all p. (*1)  Regarding the determinant-based solution: If A has polynomial entries, and Determinant(A) is a nonzero constant, then A is invertible for all p. (*2)

Regarding the two proposed counterexamples, A1:= <<p>> and A2:= <p, 0; 0, 1/p>: The symbolic inverse of A1 is <<1/p>>, which is not a polynomial matrix. For A2, Determinant(A2) = 1, a nonzero constant, but A2 is not a polynomial matrix. So neither of these are counterexamples to (*1) or (*2). I had proposed A1 as a counterexample to vv's initial statement, which didn't mention polynomials.

@Rouben Rostamian  The OP's plaintext transcription contains all the information needed to Answer the Question, if it can be answered at all. Yes, the so-called ODE is not an ODE in the usual sense, but nonetheless dsolve will give a symbolic solution to it (actually 3 solutions). The Question is whether any or all of those solutions are correct. 

I've tried several things, and at this point I'm leaning towards "they're not correct", but that's just a guess.

@vv The problem is simple, but it's not that simple. Suppose A = <<p>>. Both Maple and I would claim that A is non-singular with A^(-1) = <<1/p>>. (There is a formal context to make this entirely valid that I can elaborate upon if you want.) But the problem at hand asks us for the isolated singularities (without putting it in those words). For A = <<p>>, p=0 is one of those isolated singularities. Maple's assumptions don't handle isolated singularities.

@JAMET Change this line:

seq(plot([[1/2*cos(Pi/6+i*1/5), 1/2*sin(Pi/6+i*1/5)],[0,0]],i=1..10),color=black)

to

seq(plot([[1/2*cos(Pi/6+i*1/5), 1/2*sin(Pi/6+i*1/5)],[0,0]], color= black),i=1..10)

The only difference is switching the position of i= 1..10 with color= black.

@nm Why not just redefine solve as a local procedure? Then the only change that you need to make is adding a half line of code. Example:

restart:
Program:= proc(expr, A)
local eq, solve:= eq-> :-solve(simplify(eq), _rest);
    eq:= expr;
    (solve(eq, _rest) assuming `if`(A::{list,set}, A[], A))
end proc
: 
Program(sqrt(exp(y))=tanh(x), [y::real, x>0], y);
                         2 ln(tanh(x))

Or you could make the local redefinition simply solve:= PDEtools:-Solve;

@nm The reason that assuming works differently from assume here is that the assume is in effect when sqrt(exp(y)) is entered, which causes that to immediately simplify to exp(y/2) under the y::real assumption. The code that causes this to happen is lines 83-85 of sqrt:-ModuleApply.

Without any assumptions in effect at the time that the sqrt is entered, it returns exp(y)^(1/2). This expression does not have an unevaluated sqrt. Thus, there's no way to go back to lines 83-85 after the assuming takes effect.

@Carl Love I wrote:

• You can't compare what you've done in 2D Input with anything possible in Maple V because Maple V doesn't have 2D Input.

Actually, I've thought of a valid way to make the code comparison in many or most cases: Any syntactically correct^[*1] 2D Input can be converted to 1D input, and indeed Maple does this conversion anyway (without showing you) for its own internal purposes before executing the code in the kernel. If you want to see this equivalent 1D input, highlight the 2D Input => right click for the context menu => 2-D Math => Convert To => 1-D Math Input. (This is the best way to debug syntactically correct 2D Input, IMO.) Then that 1D input can be copy-and-pasted to some other Maple version to make the comparison.

By converting to 1D input, you can discover many cases where identical sequences of keystrokes produce different executable results and hence different output in 2D- and 1D-Input. This case of the "extra dot" is one such; more details are below.

[*1] By syntactically correct I mean that Maple interprets the 2D Input as being something that the kernel can execute. It doesn't mean correct in the sense of doing what the user intended.
  

@C_R You wrote:

• I overlooked the extra dot. I will delete my reply. There is nothing to investigate.

Please don't delete Replies or other material which has been responded to. Here are 3 reasons why (there are probably many others):

1. It makes the thread confusing to read in the future.
2. It makes the responder's response seem like a non sequitur.
3. It  makes the responder feel that their effort was partially wasted, particularly the effort of writing a response that would remain understandable for years to come. 

Instead, you can add an addendum saying, for example, that you've been convinced otherwise about an earlier statement.

Regarding the "extra dot", I do think that there is something for Maplesoft to fix, or at least document clearly or have it trigger a warning. I don't think that "investigation" is required because it's a clear-cut case of different syntactically correct interpretations of identical character sequences of input in 2D and 1D.

It has always been the case (or at least back to my 1st Maple version---V r4) that the range operator .., when used infix, could be replaced by three or more dots (with no intervening spaces) with no difference in interpretation. So, 

0.2...2

is interpreted the same as (0.2) .. (2). But in 2D Input, it's interpreted as (0.2) .. (0.2), because the 3rd dot is parsed as a decimal point rather than as part of the range operator.

@jediknight You wrote:

• In conclusion, the argument "maximize" is not optimized in Maple V. Is this correct to say so?

No, that's not quite correct. It's not an argument but rather a command and a procedure. Arguments are the things in parentheses that are passed to procedures. Since the documented purpose of maximize is to optimize, it presumably does that in most cases. The command has a bug for the case at hand---a deviation from its documented behavior and purpose. It should never return a nonreal value.