Example:

ex:= 7*diff(eta1(xi1), xi1)+8*diff(eta2(xi2), xi2)+9*eta1(xi1)^2;

 coeff(ex, diff(eta1(xi1), xi1));

                        7

From this we learn the interesting fact that linestyle and thickness aren't independent: The thickness affects not only the width but also the length of the segments (including the blank segments) for the segmented linestyles. So, when the thickness is large, the segmented linestyles look ridiculous.

plots:-display([
     seq(seq(
          plot(exp(-x^2)*sin(Pi*x), x= -2..2,
               thickness= T, linestyle= L,
               caption= sprintf("Linestyle = %d, thickness = %d", L, T)
          ), T= 0..70), L= 0..7)
     ], insequence= true
);

You shouldn't use I as a variable because it is reserved as the complex unit. So, I changed that to i. 

You can use the solve command to solve an equation, like this:

restart:
eqn:= U = i*R:
Soln:= R = solve(eqn, R);

You make Rthe second argument to solve because you want to solve eqn for R. The way that I wrote the solve command will only return the right side of the solved equation. That's why I put R = in front of solve.

Rather than a loop, use select, like this:

tneighbors:= proc(G::Graph)
description "Returns the set of vertices of degree two of a graph.";
uses GT= GraphTheory;
     {select(v-> GT:-Degree(G,v) = 2, GT:-Vertices(G))[]}
end proc:

Vertices returns a list of vertices rather than a set. The result of the select is thus a list. To convert a list L to a set, do {L[]}.

It is slightly more difficult to return a set or a list from a loop, although it is often necessary. Acer provided an example of how to handle that, returning a "table" from the loop and then converting it to a set. I would avoid the suggestion of Axel because it is a bad habit to start, and it is extremely inefficient when the set size is large.

The line that defines currentvertex needs to end with a semicolon.

Also, you should include the line

uses GraphTheory;

as the second line of your procedure.

The following code shows that there are no solutions where all the variables are distinct integers between 3 and 14. The key idea in the code is that if a valid solution exists then the set formed by evaluation of all the variables will be the set {3, 4, ..., 14}.

restart;
n:= 0:
Domain:= {$3..14}:
Basic:= {F,H,J,K,L}:
Nonbasic:= {A,B,C,M,E,G,N}:
Eqs:=
     {A = 33 - K - L, B = 1 + F - J, C = -15 - F + J + K + L, M = 15 + H - K,
     E = 16 - F - H + J + K, G = 34 - H - J - L, N = 18 - J - K}
:
for s in combinat:-permute(Domain, 5) do
     if {s[]} union eval(Nonbasic, eval(Eqs, Basic =~ s)) = Domain then
          n:= n+1;
          Sols[n]:= s
     end if
end do:
n;

                           0

You need to change cos x to cos(x). All function calls need parentheses is Maple.

Maple's own BVP solver works fine on this problem.

epsilon:= L[1]:
Sol:= dsolve({eq1,eq2,bcs1,bcs2}, numeric, method= bvp[midrich]):
plots:-odeplot(Sol, [[eta,f(eta)], [eta,theta(eta)]]);

NoNegatives:= (sol::set, Vars::list(name))-> ormap(v-> eval(v, sol) < 0, Vars):
remove(NoNegatives, remove(has, [sol], I), [a,b]);

You can do two things. The first is to load the plots package:

with(plots);

The second is to refer to the package without loading it by changing implicitplot3d to plots:-implicitplot3d.

@micahdavid Maple refuses to do your integral, but that has nothing to do with you having used indexed variables. Indeed, the integral is clearly improper and easily shown to be divergent. Note that when x[2] = 2, the inside integral is 0, and you're dividing by it.

You're approaching it the wrong way. Here's how to use a seq declaration for your edges. The following procedure takes a graph and any number of edges and returns true if there's more than one edge and all the edges have a common vertex.

proc55:= proc(G::Graph, E::seq(set))
     if not {E} subset GraphTheory:-Edges(G) then
          error "Found an edge not belonging to the graph."
     end if;
     evalb(nops({E}) > 1 and `intersect`(E) <> {})
end proc:

G:= GraphTheory:-CompleteGraph(5):
proc55(G, {1,2}, {2,3}, {5,2});
                              true
proc55(G, {1,2}, {2,3}, {3,4});
                             false
proc55(G, {1,2}, {2,3}, {3,6});
Error, (in proc55) Found an edge not belonging to the graph.

v1:=<1,2,3,4,5,6,7,8>:  v2:= <9,10,11,12,13,14,15,16>:
cols:= 10:
M:= Matrix(8,cols):
M[.., 1]:= v1:  M[.., -1]:= v2:

The reason that the second run produces different results is that when you re-execute an assume command (without an intervening restart), weird things happen. There is some combination of assumptions that will cause your integral to be done. I am trying to figure out what that combination is so that you can enter it directly. The situation is further complicated by your use of both assume and assuming on the same variables.

I think that the assumptions that various variables are constant are not needed.

Student:-Calculus1:-Roots(sin(x)^2 = exp(-x)*cos(x), x= -5..15, numeric);