Your examples just appear as blank spaces in your Question.

@Carl Love My results so far are that y' = 0 only at t = +infinity regardless of epsilon.

@adel-00 Okay, that makes sense to me now. I'll be thinking about it. Yes, I think that there's an easy way to do it.

Plotting epsilon against t makes no sense to me because they are both independent variables.

@J4James Do you mean 0 to 2 times the increment?

@J4James I don't know a way. There is no help page for that error message. Someone will probably chime in here soon with an analysis of the error. They might find a way around.

@J4James I am not getting that error with M17, but I do get it with M16. I don't know what it means. The error appears related to the call to dsolve and not to the looping structure. Here is the matrix that I end up with in M17:

Are you using _Z as a variable? If yes, then change it. If no, then this error message is a sign of a serious bug in solve.

Trying to restrict solve to real solutions is unreliable, IMO. If you want to try, then do

with(RealDomain);

which loads a different version of solve.

@John Fredsted The with command is needed to overload operators. But I don't see why you have any issue with with. It works regardless of these GUI issues.

@micahdavid I don't see how deleting a group of strictly print and lprint statements could make a difference with the error. There must be something else going on.

@brian bovril I don't know anything about that.

@brian bovril The labeling with non-unique data can be done like this:

restart:
S:= [3, 4, 5, 6, 9, 9, 28, 30, 35]:
SL:= ["A","B","C","D","E","F","G","H","I"]:
assign(Labels ~ (SL) =~ S); #Create remember table.
AllP:= combinat:-setpartition(SL,3):
lnp:= evalf(ln((`*`(S[]))^(1/3))):

Var:= (P::{list,set}({list,set}))->
     evalf(`+`(map(b-> abs(ln(`*`(Labels~(b)[]))-lnp), P)[]))
:

Min:= proc(S::{list,set}, P::procedure)
local M:= infinity, X:= (), x, v;
     for x in S do
          v:= P(x);
          if v < M then  M:= v;  X:= x  end if
     end do;
     X
end proc:

ans:= Min(AllP, Var);
      [["A", "E", "I"], ["B", "F", "G"], ["C", "D", "H"]]
subsindets(ans, string, Labels);
              [[3, 9, 35], [4, 9, 28], [5, 6, 30]]

@Carl Love This is probably faster:

not has(L, false)

@fluff The answer to all six of your questions is yes. I handle the n-door case in the next Answer.

@brian bovril Labeling the entries is trivial. Just create a remember table at the beginning and apply it at the end. Like this:

restart:
S:= {3, 4, 5, 6, 8, 9, 28, 30, 35}:
SL:= [A,B,C,D,E,F,G,H,I]:
assign(Labels ~ (S) =~ SL); #Create remember table.
AllP:= [seq(P, P= Iterator:-SetPartitions(S, [[3,3]], compile= false))]:
lnp:= evalf(ln((`*`(S[]))^(1/3))):

Var:= proc(P::({list,set}(set)))
local r:= evalf(`+`(map(b-> abs(ln(`*`(b[]))-lnp), P)[]));
end proc:

Min:= proc(S::{list,set}, P::procedure)
local M:= infinity, X:= (), x, v;
     for x in S do
          v:= P(x);
          if v < M then  M:= v;  X:= x  end if
     end do;
     X
end proc:

ans:= Min(AllP, Var);
              [{3, 9, 35}, {4, 8, 28}, {5, 6, 30}]
subsindets(ans, posint, Labels);
               [{I, A, F}, {B, E, G}, {C, D, H}]