F := 2*y*sin(beta*x)+6*z*cos(beta*x)+24*sin(beta*x)*cos(beta*x):
W := coeffs(F, {sin, cos}(beta*x), 'V'):
[W], [V];
if member(sin(beta*x), [V], 'i') then op(i, [W]) else 0 fi;
if member(cos(beta*x), [V], 'i') then op(i, [W]) else 0 fi;
 

Replace relevant lines (from Eq3..Sols) with:
Eq3 := factor(expand(value(algsubs(subs(params, tr3), Eq2))))
Eq31 := collect(numer(lhs(Eq3)), exp(z))
sys1 := [coeffs(Eq31, exp(z)), c=c0];
vars := indets([sys1], name) minus {c0};
Sols := map(allvalues, [solve(sys1, vars)]); NSols := nops(Sols);

Alternatively you can try:
Eq3 := factor(expand(value(algsubs(subs(params, tr3), Eq2))))
vars := {c, mu, seq(a[i], i = -1 .. 1), seq(b[j], j = -1 .. 1)}
Sols := map(allvalues, [solve]({c=c0, identity(Eq3,z)},vars)); NSols := nops(Sols);

This groups by leading coefficient:

Q:=proc(E)
local X;
if type(E, `+`) then X := [op]('E') else X := ['E'] fi;
ListTools:-Categorize(proc(x, y) lcoeff(x) = lcoeff(y) end, X);
end;
Q(pn);
map(add, [Q(expand(pn))]);
(add@map)(add, [Q(expand(pn))]);
(add@map)(proc(E) ``(lcoeff(E))*normal(E/lcoeff(E)) end@add, [Q(expand(pn))]);
 

Have you tried looking at:

dsolve(ode, y(x), 'formal_solution', 'coeffs' = 'mhypergeom');

After You've assigned Leq add this line:
Leq := expand(Leq):

Q:=map(x->`if`(type(x,even),1,0), LinearAlgebra:-RandomMatrix(3,3));
plots[sparsematrixplot](Q, matrixview);
 

Normalizer:=radnormal;
d1 := LinearAlgebra:-Determinant(A):
frontend(collect,[d1, indets(d1, radext(Non(constant))),distributed,factor],[{Non}(radext(Non(constant))),{}]);
or
d1 := LinearAlgebra:-Determinant(A):
radnormal(d1);
frontend(collect,[%, indets(%, radext(Non(constant))),distributed,factor],[{Non}(radext(Non(constant))),{}]);

fsolve(SYS,VAR,fulldigits);
 

In Maple I would suggest care and caution when mixing trigs, exps with algebra.
You're solving this:

A:=(20 + 20*T + 2*T*(T + 1))*exp(-T) - 10*exp(-2*T) - 2*T - 10.0 = 0;
S:=(solve,((E,V)->V))(expand(A),exp(T));
plot([S],T=0..10,view=[DEFAULT,-1..5],axes=boxed,scaling=constrained,color=[grey,grey,red]);

solve(evalc(Im(S[1]))=0);

This works better:

eq := a*x^3 + b*x^2 + c*x + d;
s := x1 + x2 + x3 = -b/a;
sp := x1*x2 + x1*x3 + x2*x3 = c/a;
p := x1*x2*x3 = -d/a;
eliminate({eq, s, p, sp}, {x});
sol := solve(%[2], {x1, x2, x3}, explicit):
#easier way to solve:
#sol := solve(identity(eq-a*(x-x1)*(x-x2)*(x-x3), x), {x1, x2, x3}, explicit=true);
# evaluate without assignment:
# eval( radnormal([sol], rationalized), [a=1, b=-5, c=6, d=-1]);
a := 1;
b := -5;
c := 6;
d := -1;
radnormal([sol], rationalized);
 

As demonstrated here:

f := proc(a)
   option operator, arrow;
   proc(b)
      option operator, arrow;
   a/b;
   end;
end;

Try this:

with(PolyhedralSets):
V := [x, y, z];
p1 := x+z-1;
H1 := PolyhedralSet([p1 <= 0], V):
H2 := PolyhedralSet([p1 = 0], V):
H3 := PolyhedralSet([p1 >= 0], V):
C := PolyhedralSet([seq(n >= 0, n = V), seq(n <= 1, n = V)], V):
S := seq(h intersect C, h = [H1, H2, H3]);

ev:= edgeoptions = [color = "black", thickness=0], vertexoptions = [color = "red", symbolsize = 1]:
plots[display](
Plot(S[1], faceoptions = [color = pink, glossiness = 0.25, transparency=0.33], ev),
Plot(S[2], faceoptions = [color = aquamarine, glossiness = 0.25, transparency=0], ev),
Plot(S[3], faceoptions = [color = turquoise, glossiness = 0.25, transparency=0.33], ev),
view=[(-1..2)$3], scaling = constrained, lightmodel = light3, axes = boxed, insequence=false
);
 

.

y = interp(op(map2(map2, op, [1,2], [[-1, -4], [1, 0], [2, 0], [3, -4]])), x);
 

Converting into fractions we get:

Sum(binomial(196, j)*(1/2)^(321/10+j)*(Sum(binomial(109, l)*(-1)^(j+l)*(Sum((1/2)^((31/10)*h)*GAMMA(-41/10+h)*GAMMA(382/41+(10/41)*l+(10/41)*j+(31/41)*h)/(factorial(h)*(321/31+(10/31)*j+h)*GAMMA((10/41)*j+(10/41)*l+6321/410+(31/41)*h)), h = 0 .. infinity)), l = 0 .. 109)), j = 0 .. 196);

I do not really understand what you are saying so my question follows:
 

We can indentify any point P using: P = [x, y, c]; where c is the color and (x,y) is the coordinate in the plane.
We have a rule: For all ( P[1] = [x[1], y[1], c[1]], P[2] = [x[2], y[2], c[2]] ) (c[1]=c[2]) implies not (y[1]=y[2]).
An observation: for every set of points satisfying above rule, the cardinality of (subset of points) having only particular (y, c) coordinates is not greater than 1.

You've given us a set of 63 points, call it set S. Which are you trying to do:
1) choose a subset of S that satisifies the rule
2) partition S into subsets satisfying the rule

[{x = 4, x = 5, x = 6, x = 7, x = 2, x = 3, x = 1}, {y = 1, c = 1}], [{x = 4, x = 5, x = 6, x = 7, x = 2, x = 3, x = 1}, {c = 2, y = 2}], [{x = 4, x = 5, x = 7, x = 1}, {c = 3, y = 3}], [{x = 4, x = 5, x = 7, x = 1}, {y = 4, c = 3}], [{x = 6, x = 2, x = 3}, {c = 4, y = 3}], [{x = 6, x = 2, x = 3}, {y = 4, c = 4}], [{x = 5, x = 7, x = 3, x = 1}, {c = 5, y = 5}], [{x = 5, x = 7, x = 3, x = 1}, {y = 6, c = 5}], [{x = 4, x = 6, x = 2}, {c = 6, y = 6}], [{x = 4, x = 6, x = 2}, {c = 6, y = 5}], [{x = 7, x = 2, x = 1}, {c = 7, y = 9}], [{x = 7, x = 2, x = 1}, {c = 7, y = 7}], [{x = 7, x = 2, x = 1}, {c = 7, y = 8}], [{x = 5, x = 3}, {c = 8, y = 7}], [{x = 5, x = 3}, {c = 8, y = 8}], [{x = 5, x = 3}, {c = 8, y = 9}], [{x = 4, x = 6}, {c = 9, y = 8}], [{x = 4, x = 6}, {c = 9, y = 7}], [{x = 4, x = 6}, {c = 9, y = 9}];