You really should get into the habit of using symbols, then replace these with numbers at the end. If you do that, it will easier to see why it can't give explicit solution.  In addition, your code will be easier to modify and maintain.

Your ODE is linear first oder. But the integrating factor produces integral which Maple can't integrate.

restart;
ode1:=-diff(q(t), t) - A*q(t)*(B - C*(d*cos(f*t) + e*sin(f*t))^2) +h=0;
dsolve(ode1);

The integral of exp(.....) above is the problem. Try numerical dsolve instead.

To get to the essense of it, keep throwing away uneeded things, and you'll get to something that looks like 

int(exp((t -cos(t)- sin(t)+ sin(t) + t)), t)

which has no closed form solution. It is the integral of things like exp(cos(t)) which makes Maple produces the result it did, with `Int` in it, as there is no antiderivative for these. 

sys_ode := diff(F0(zeta), zeta, zeta)-b^2*F0(zeta)+G0(zeta)^2 = 0, diff(G0(zeta), zeta, zeta)-b^2*G0(zeta) = 0, 2*F0(zeta)+diff(H0(zeta), zeta) = 0;

ics := F0(0) = 0, G0(0) = 1, H0(0) = 0, F0(infinity) = 0, G0(infinity) = 0;

sol:= dsolve([sys_ode,ics])

simplify(sol) assuming positive

you can eliminate  the remove has calls by doing

map(x->ListTools:-SearchAll(x,L1),L2)

do not know if this will speed it up much or not.

Another apprach to the fine answers given already is to use: (can't write Latex in the form, so I put screen shot)

Using Maple

g:=x->cos(x);
limit(1/abs(g(x)),x=0)+limit(1/abs(g(x)),x=-Pi)+limit(1/abs(g(x)),x=Pi)

    3

 But the lecturer's solution was non-imaginary.

restart;
ode := x^2*diff(z(x), x, x) + (1 + gamma + beta)*x*diff(z(x), x) + gamma*beta*z(x) - cos(ln(x)):
sol:=(dsolve({ode, z(1) = 1, D(z)(1) = -1}) assuming (gamma <> beta));

simplify(evalc(sol)) assuming x>0

You get same out using

remove(`=`,op~(0,indets(x)),symbol)

{f, g, h}

It works since op(0,f(x))  gives f  while op(0,f) gives symbol

there might be a better way to do ofcourse, but this is how I see it so far. But there should be a more robust way to do this using type(....,'function')

May be something like

restart;
x := -h(U) + f( f( g( f(U+W*X)*V) + g( f(W)*g(V) ) ) ):
foo:=x-> if type(x,'function') then return(op(0,x)) fi;
convert(foo~(convert(indets(x),list)),set)

{f, g, h}

In Section 14.5 - External Calling ..., page 489

I do not use Maple on Linux. But you have one ". You need two of these, one of each side. 

Here is the example from book. Try and see if this fixes it.

expr:=latex('int(1/(x^2+2),x=2..3)')
\int_{2}^{3}\! \left( {x}^{2}+2 \right) ^{-1}\,{\rm d}x

compiles to

And you can do things like this

expr:="int(1/(x^2+2),x=2..3)";
result:=cat(latex(parse(expr),output=string)," = ",latex(eval(parse(expr)),output=string));

Which gives latex which compiles to

It is not a bug, Isn't this known and due to when evalation and binding names to value happen? When you type f:=(x,y)->eq2 notice that Maple did not replace eq2 with its value y = 10 - 5*x at the time the function is defined.

When you called the function next, maple did not replace the x,y since eq2 was still name. Next, it evaluates eq2 and returns it value which is y = 10 - 5*x but by then it is too late. 

I think this is why unapply was invented.

but to do what you want, you could always write

          f:=(x0,y0)->eval(eq2,[x=x0,y=y0])

And now f(3,1) returns 1 = -5

ofcourse Maple can open a CDF file. Any software can open a CDF file, after all, it is just plain text file and not a binary file. Same for Mathematica notebooks. They are all plain text files.

Here is start of one such CDF file

(* Content-type: application/vnd.wolfram.cdf.text *)

(*** Wolfram CDF File ***)
(* http://www.wolfram.com/cdf *)

(* CreatedBy='Mathematica 12.1' *)

Now, the question is what you meant to ask is, can Maple run a CDF?  The answer is No. Only WRI software can run CDF files. You need either Mathematica itself or the Wolfram player installed to run a CDF file.

This is becuase it needs the Wolfram kernel software to run it. Just like one needs Maple kernel to Maple worksheet. No difference.

I am going to assume you meant fixed length string and u(0,t)=0=u(0,1) is a typo and you meant  u(0,t)=0=u(1,t)

restart;
pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2);
bc  := u(0,t)=0,u(L,t)=0;
ic  := u(x,0)=f(x),D[2](u)(x,0)=g(x);
sol:=pdsolve([pde, ic, bc],u(x,t)) assuming L>0;

To do animation, we need to put specific values. For example

L:=1;
c:=2;
g:=0;
f:=(8*x*(L-x)^2)/L^3;
pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2);
bc  := u(0,t)=0,u(L,t)=0;
ic  := u(x,0)=f,D[2](u)(x,0)=g;
sol:=pdsolve([pde, ic, bc],u(x,t));
sol:=subs(infinity=15,sol); #should be good enough

Another example

restart;
c:=2;
pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2);
bc  := u(-Pi,t)=0,u(Pi,t)=0;
ic  := u(x,0)=0,D[2](u)(x,0)=sin(x)^2;
sol:=pdsolve([pde, ic, bc],u(x,t));
sol:=subs(infinity=20,sol);

If you search the internet, there are hundreds of such examples out there. 

restart;
pde  := diff(u(x,t),t$2)= diff(u(x,t),x$2);
f    := x->piecewise(-1/2<x and x<1/2,5*cos(Pi*x),true,0);
ic   := u(x,0)=x, D[2](u)(x,0)=f(x);
sol  := pdsolve([pde,ic],u(x,t));
plots:-animate(plot, [rhs(sol),x=-6..6],t=0..10);

update

Thanks to the hint by Carl below, I can now save animation to gif file. Made a new one by making small change to the initial conditions given above just to make the wave motion a little bit more interesting looking. I am surprised how fast Maple saved the animation gif file to disk. This is good.

restart;
pde :=  diff(u(x,t),t$2)= diff(u(x,t),x$2);
f   :=  x->piecewise(-1/2<x and x<1/2,10*cos(Pi*x),true,0);
ic  :=  u(x,0)=f(x), D[2](u)(x,0)=0;
sol := pdsolve([pde,ic],u(x,t));

plots:-animate(plot, [rhs(sol),x=-10..10],t=0..10,frames=100);

restart;
PDE := diff(u(x, t), t) - VectorCalculus:-Laplacian(u(x, t), [x]) - u(x, t) + x - 2*sin(2*x)*cos(x) = 0;
IBC := D[1](u)(Pi/2, t) = 1, u(0, t) = 0, u(x, 0) = x;
pdsolve(eval(PDE), {IBC}, type = numeric);

Works OK on Maple 2020. 

Or you can try analytical solution

restart;
PDE := diff(u(x, t), t) - VectorCalculus:-Laplacian(u(x, t), [x]) - u(x, t) + x - 2*sin(2*x)*cos(x) = 0;
IBC := D[1](u)(Pi/2, t) = 1, u(0, t) = 0, u(x, 0) = x;
pdsolve([PDE,IBC]);

Both work. No error. Attached worksheet.

Download maple_sheet.mw

Removed as not needed.

L1:=[1,2,5,6,9];
andmap( x->x>0, L1 );

         true

L2:=[0,-2,5,6,9]:
andmap( x->x>0, L2 );

       false