At each step except the first, you take tau12, tau21 from the previous step.

Numeric only (of course).

ode:=((D@@2)(y))(r)+(D(y))(r)/r-sin(y(r))*cos(y(r))/r^2+sin(y(r))^2/r-sin(y(r))-sin(2*y(r)) = 0:
ds:=dsolve([ode, y(2)=1, D(y)(2)=1/2], numeric):
plots:-odeplot(ds, r=1e-6..100, color=red);

Nothing unusual. The integrator is very complex, so, equivalent functions could be integrated using different methods.

But it's easy to have the same unevaluated result using:
int(simplify(integrand2), x);  ##  :-)

In the absence of a better solution (the generated LaTeX code is not easy to manipulate), I'd replace in the worksheet e.g.

sys_set := convert(sys_vec, set);

with

sys_set := convert(sys_vec, set):
for eq in sys_set do print(eq) od; # display elements

You don't need Maple for this: z / |z|^2 = z is holomorphic, so the answer is 0.

The limit of a function F : A --> B, at a point u is defined when u is an accumulation point (= limit point = cluster point) of A.
In our case the standard choises for A, B are B = R and A = R \ {0} \ {1/(n*Pi) : n in Z\{0}}.
0 is an accumulation point for A, so the limit problem makes sense. It exists and equals 1 (using the standard limit : sin(t)/t, for t --> 0).

[A more techical note: the topology in A is by default the subspace topology induced by the natural topology in R; your neighbourhood refers to this topology]. 

Your ode is

ode := diff(y(x), x) = 15*exp(4*y(x)) - 5*y(x)

There is no explicit solution for this ODE; this is why the Qualitative theory of ODEs exists.
You can study the solutions taking y0 = y(0) as a parameter. You will see that the solution always blows up in a finite interval.

Actually in our case it's easy to see that f(y)>0 (=  rhs(ode)) and

int(1/f(y), y=y0 .. y(x)) = x - 0

so, the existence domain for x>0 is bounded, because LHS is obviously bounded.

P.S. It is a good idea to post the problem, not just in the worksheet; it is short, and the worksheet should be only optional. You will increase the chances to get answers.

Your function has an essential singularity at z=0. Maple (and probably other CAS too) cannot compute the series for such singularities. (The series has the form Sum(a(n)*z^n, n = -infinity .. infinity)).

But you can use Maple to obtain it by multiplying two series (try it first by hand).

This occurs frequently when the symbolic solution is expressed via mathematical functions with branches.

The numerical solution is continuous (in general), and it may coorespond to different branches (i.e. it "jumps" from one branch to another).

``||i   are valid names, so the assignments work.

But note that they are global variables. So, if you have

local a, b__c, t;

then the assignments will produce new global variables, e.g. :-a  will be 1, but a remains free (unassigned).

value(IntegrationTools:-Change(z, s=x-t, t)) assuming x>0;

But do you really need it?

You simply forgot to load the package. Start with:

with(Student[Calculus1]):

Selecting the homogeneous solutions is not complicated:

with(PDETools):
PDE := 2*((x1 - x2)^2 + (y1 - y2)^2)*(u1*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), x1) + v1*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), y1) + u2*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), x2) + v2*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), y2)) - ((u1 - v1)^2 + (u2 - v2)^2)*((x1 - x2)*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), u1) + (y1 - y2)*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), v1) - (x1 - x2)*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), u2) - (y1 - y2)*diff(f(x1, y1, x2, y2, u1, v1, u2, v2), v2)) = 0:
n:=2:
P:=rhs(PolynomialSolutions(PDE, degree = n)[]):
C:=[indets(P,suffixed(_C, integer))[]]:
T:=[coeffs(collect(P,C,'distributed'),C)]:
select(u -> degree(u)=n, T); 

[v1^2 + 2*v1*v2 + v2^2, (v1 + v2)*u1 + (v1 + v2)*u2, u1^2 + 2*u1*u2 + u2^2, u1*y2 + u2*y1 - v1*x2 - v2*x1, u1*y1 + u2*y2 - v1*x1 - v2*x2]

But it seems that you want to eliminate also other terms. This will need extra work (probably using Groebner bases).

There are essentially two methods.

1. Use DirectSearch.
GlobalOptima(f(x,y), {x=3..4, y=0..1}, maximize);
The approx max is (almost) 0.

2. Use calculus.
Solve the system [diff(f(x,y),x), diff(f(x,y),y)].
Maple finds a unique solution in the domain: x=4, y=0, ( f(4,0)=0 ).
It remains to compute the max on the boundary: f(x,0), f(x,1), f(3,y), f(4,y)
(it's easy).

Maple has not a package for solving general functional equations.
A few years ago I saw at a conference a group of hugarian mathematicians developping such a package, but I cannot find the references (they should be somewhere on my computer).
Anyway, the equations you mention do not seem to be difficult. The general solution for the first one (for x>0) is:

F(x) = piecewise( x<1, a(x), x^(-r)*a(1/x) )

where a(x) is an arbitrary function for 0 < x <= 1.