@Preben Alsholm 

Probably I was wrong, it's a very old version I seem to recall.
Anyway, the main reason for the alias is to not use a(t) at all.

@Markiyan Hirnyk 

When I say easy, I really mean easy, at least for an undergraduate level.
We are both professional mathematicians, as I understand.
Anyway, I am going to answer in the future only to polite questions, i.e. without "Roly-poly toy", "empty words" etc.

restart:
alias(a=a(t)):
a - a(t);   ###  it used to work!
   a - a(t)

@John Fredsted 

You are right here. The image of the connected set (0, oo) ^ 4 by a continuous function is connected, so the function  must be constant since the only possible values are I and -I. The continuity is easy but not automatic, because the principal branch of sqrt is not continuous in C.

Preben's solution is correct, the function w being C^1 in the convex open domain {(a,k) : a>0, k>0}.

OK, you win. It was my mistake to answer your "objections" following the pattern:

Assertion: p>0, q>0 implies p*q>0.

Objections:

1. Empty words.
2. Unbased words.
3. Prove it.
4. Prove it with Maple.
5. Refresh your logic.
6. Go to 1.

@Markiyan Hirnyk 

@Markiyan Hirnyk 

But t is >0; you had no objections about this. Now you come back? Infinite loop?
As I see you want to test my patience.

@Markiyan Hirnyk 

I have mentioned that this follows from ex^2=-1.

is( (- sqrt(p)* t )^2  < 0  )  assuming p>=0, t>0;

    false

is( (- sqrt(p)* t )^2  < 0  )  assuming p<0, t>0;

    true

@Markiyan Hirnyk 

argument( - sqrt(p)* t ) assuming p<0, t>0;

@Markiyan Hirnyk 

If you indicate which assertion is not obvious, I will try to turn my empty words into full ones.

@Markiyan Hirnyk 

What unbased words?

 p =    

p<0 implies argument(sqrt(p))  = Pi/2  ==>   ex = - (positive)*sqrt(p)  has the argument -Pi/2.

For omega=10^6, the graph of u(t) e.g. looks like this for t in [0.5,1]:

@adel-00 

It is nothing wrong with your system. It is linear and has a unique "nice" solution from a mathematical point of view.

It is the oscillatory nature of the solution. Have you read my simple example (previous answer)? The same situation happens here.
You can obtain a series solution but because Omega is so large, the Order of the series must also be large and then the roundoff errors will be huge. But even if an exact solution would be possible, it is not clear to me how are you going to use it -- unless you need it only in a small neighborhood (-eps,eps) or [0,eps).

@John Fredsted 

Instead of using "proc valued procs", why don't you use procs with 2 arguments an then curry (or rcurry) if needed.

h:=(x,y) -> 2*x+3*y;

curry(h,a);

curry(h,a)(b);

@Preben Alsholm 

But this way, unapply will be executed each time g is invoked, isn't it?