test:=module()
uses DT = DocumentTools;
export abcd,tt,vv;
abcd:=proc()
    tt:=2000;
    vv:=DT:-GetProperty('Slider0', 'value');
    DT:-SetProperty('Label0', 'caption', vv);
end proc;

end module;

Your module test, which I've copied above, contains a procedure abcd which sets the values of some of the exports of the module. However, abcd is merely defined; it is never executed. So, right above the end module put the line abcd().

LinearAlgebra:-GramSchmidt({< 22,11,5 >, < 13,6,3 >, < -5,-2,1 >});

The imaginary parts are due to round-off error. Note that they are quite small. Their exact value is zero. It is impossible to avoid these imaginary parts if you solve a cubic by the cubic formula and and then substitute floating-point values for the coefficients. A plot shows that the cubic polynomial has three real roots. So, just throw away the imaginary parts. You can apply the Re command to each solution.

You want to use an Array of Records. See ?record .

I suggest that you enter your graph the normal way, by using the command GraphTheory:-Graph. GRAPHLN is intended for internal use. If you entered it the normal way, then the error pointed out by Kitonum would not have happened.

restart:
Eckert:= 0.5:
sigma:= 15:

eq1:= diff(f(eta), eta)^2-f(eta)*diff(f(eta), eta$2) =
diff(f(eta), eta$3)+k_1*(2*diff(f(eta), eta)*diff(f(eta), eta$3)
+diff(f(eta), eta$2)^2-f(eta)*diff(f(eta), eta$4)):

N:= 10:
bcs1:= f(0) = 0, D(f)(0) = 1, D(f)(N) = 0, (D@@2)(f)(N) = 0:

eq2:= diff(theta(eta), eta$2)+sigma*f(eta)*diff(theta(eta), eta) +
sigma*Eckert*(diff(f(eta), eta$2)^2+k_1*diff(f(eta), eta) *
diff(f(eta), eta$2)^2-f(eta)*diff(f(eta), eta$2)*diff(f(eta), eta$3)) = 0:

bcs2 := theta(0) = 1, (D(theta))(0) = -2.876:

Sol:= proc(k1)
     if not k1::realcons then return 'procname'(args) end if;          
     dsolve(
          eval({eq||(1..2), bcs||(1..2)}, k_1= k1),
          numeric, method= bvp[middefer], maxmesh= 2^13, abserr= 1e-4
     )
end proc:

plots:-animate(
     plots:-odeplot,
     [Sol(k1), [[eta, f(eta)], [eta, theta(eta)]], eta= 0..2],
     k1= 1.5..2.5
);

Yes, you can know the order of the error. The order of the error is an inherent property of the method; it does not depend on whether the exact solution is known.

Using solve instead of fsolve will give both solutions in this case, although I can't say that this is generally true. Certainly fsolve always gives at most one solution to a transcendental equation.

n:= LinearAlgebra:-RowDimension(A):
P:= (LinearAlgebra:-IdentityMatrix(n)-A)^(-1).A;

The syntax requires

dsolve({bs} union sys1);

Note that you used {} on sys1 and not on bs. You need to take that into account in your call to dsolve.

But there is a deeper problem. The system cannot be solved for the last constant. You can impose the first five boundary conditions like this:

dsolve({bs[1..5]} union sys1);

This will give you a system with one constant. For now, I leave it to you or someone else to address why it cannot be solved for this constant using your conditions.

This will be easiest to do in the case where the curves in the original are all one color and you want to convert them to all one color. Let's suppose that you want to convert them all to black. Then do

macro(C= ColorTools):
subs(
     indets(P, specfunc(anything, COLOUR))[] =
     COLOUR(RGB, C:-RGB24ToRGB(C:-NameToRGB24("black"))[]),
     P
);

Note carefully that COLOUR is spelled with a U and that ColorTools is spelled without a u (a result of a Canadian identity crisis :-) ).

If you are dealing with a case of multiple colors, something can still be done---let me know.

Use

return fff, ggg;

or

return [fff, ggg];

Use the former if you are prepared to accept multiple outputs. Use the latter if you want to treat the multiple outputs as a single entity.

@jcrose 

First, convert the decimals in your Matrix to exact fractions with

polyGLS:= convert(polyGLS, rational):

Then

RRF:= LinearAlgebra:-ReducedRowEchelonForm(polyGLS);

This command completes in about 4 minutes on my computer, and uses only about 300M of memory. I get the bottom three rows of the matrix going to 0. This can be viewed with

RRF[5..7, ..];

I am using Maple 18. If you are using earlier Maple and cannot duplicate my results, please let me know.

You need to use one more level of unevaluation quotes:

Z2:= ' 'sin(-a*x+b)' ';
subs(sin= cos, Z2);

Another option is to use inert functions. The inert form of a function is obtained by prepending % to the name. For example,

Z2:= %sin(-a*x+b);
Z4:= subs(%sin= %cos, Z2);

When the expression is the way that you want it, use value to change the inert functions to their active forms:

value(Z4);

Change x < y to is(x < y). Or is that still too awkward?