Internally z is Float(16,-3); op(0,z) gives Float and op(z) gives 16,-3. lprint(z) gives .16e-1, which I assume is some sort of canonical form.

So the 0.016 is just for display. While it might be possible to recreate that display routine, probably sprinf gives you the ultimate in control for this.

@scottwr For inner nested loops, use print for the things you want to see. So in this case just use print(x*y) instead of x*y. 

@mmcdara That was my comment. In the paper they condense products of subscripted symbols into single subscripted symbols, where the new subscripts are the set union of the subscripts of the factors. After looking further at the paper, I think it is a bad idea to do this since one loses information.

@Carl Love True, but the standard way to abbreviate input for anything nontrivial would be with diff_table. For example

declare(u(x));
U:=diff_table(u(x));

then U[x,x] is used to abbreviate diff(u(x),x,x) and U[] for u(x).

@FZ Sorry, I don't understand what they are doing.

@FZ Please expand the image. As far as I can see, they use the first integral, which didn't work for your case unless you know the form for the integrating factor.

@FZ Your equation has a derivative on the rhs, unlike your textbook example, which can be done like this:

Download Hamiltonian3.mw

@FZ Sorry, I don't understand; r1 or r2 are not in the form of your first equation.

Hamiltonian2.mw

@Tom GIV Yes, that's correct.

@janhardo I managed to get various other solutions using various combinations of dchange and HINT=, but never sol11. Here's one attempt:

Download pdsolve.mw

@Andiguys You can manipulate the pieces with op, see: Q_12.mw

The empty otherwise just means there are no other possibilities.

@salim-barzani I simply followed the logic of the paper. I don't know about bilinear forms in this context; perhaps the nonlinear parts hare are bilinear forms? Here the linear part was set to zero, which implies the nonlinear part is also zero.

@FZ This is _a, not a; just a name Maple chose for the dummy integration variable.

@vv Thanks. (I'm still waiting for this to finish.) But I am also playing around with subsets of the eigenvalues (in disguised form), for which this will work, and be faster.

@acer Thanks. Much faster than @vv's.