When specifying a function using the arrow, you must specify the function name. Your function names will be y1 and y2, not y1(x) and y2(x).
Also, your expression is somewhat better simplified with the  combine  command rather than the  simplify.

Download simplify_expression_new.mw

Download aabb_new.mw

I think this example reveals a serious bug in the  isolve  command that may appear in other examples. Let's consider a very simple example: find all integer solutions to the equation (x - y)*z = 1  . Obviously, the product of two integers is equal to 1 if and only if both factors are equal to 1 or both factors are equal to -1, that is, we obtain the set of two systems  {x - y = 1, z = 1}  or  {x - y = -1, z = -1} . But Maple only solves the first system and for some reason ignores the second one:

isolve((x-y)*z=1);

                      {x = _Z1+1, y = _Z1, z = 1}

Your problem has infinitely many solutions (see result of the  solve  command)). As an example at the end we find 1 solution:

Download Matrix_equation.mw

See help on the  Explore  command.

By default, Maple works in the complex domain. Use  LinearAlgebra:-DotProduct  command with the option conjugate=false to work in the real domain:

restart;
with(LinearAlgebra):
with(Typesetting):
Settings(typesetprime = true):
Settings(typesetdot = true):

v := <diff(rho(t), t)*cos(theta(t)) - rho(t)*diff(theta(t), t)*sin(theta(t)), diff(rho(t), t)*sin(theta(t)) + rho(t)*diff(theta(t), t)*cos(theta(t)), diff(z(t), t)>;

u_rho := <cos(theta(t)), sin(theta(t)), 0>;

DotProduct(v,u_rho, conjugate=false);
simplify(%)

I don't think your boundary value problem can be solved exactly, probably only numerically. To do this, I gave some numerical values ​​to all the parameters. Maple also does not accept a value at infinity as a boundary value. I replaced infinity with the number 50. You yourself can take as infinity some number that for your task is reasonable to take as infinity.

Download ode_exact_new.mw

Here are 2 solutions - numerical and symbolic. The  extrema command acknowledges the symbolic answer that was first received by the  identify  command:

restart;
with(geom3d):
point(A,0,0,4): point(B,6,-2,6): point(C,4,-8,4): v:=[1,-1,2]:
point(M,x,y,z):
line(d,[C,v]):
f:=distance(M,d);
v1:=<0,0,4>-<x,y,z>: v2:=<6,-2,6>-<x,y,z>:
p:=v1.v2 assuming real;
 Optimization:-Minimize(f, {x^2 + (y - 2)^2 + (z + 1)^2 = 29, p=0}); # Numerical solution
identify(evalf(%)); # Symbolic solution
extrema(f, {x^2 + (y - 2)^2 + (z + 1)^2 = 29, p=0}, {x,y,z}, s);
s;

The  P  procedure solves your problem:

restart;

P:=proc(Eq)
local Eq1;
Eq1:=(lhs-rhs)(Eq)=0;
select(t->has(t,diff),lhs(Eq1))=-sort(remove(t->has(t,diff),lhs(Eq1)));
end proc:

Eq:=diff(u(x, t), t, t) + 3 + 2*diff(u(x, t), t) + 4*t + x^2 + x^3/3 + diff(u(x, t), t, x, x) + diff(u(x, t), x, x, x, x) = x*t^2;

P(Eq);

@JAMET  You wrote "I wish to write the following in this order (for instance)
<2275,3648,4275>,<975,448,1273>,<275,252,373>,<33,56,65>,<15,8,17>,<3,4,5> Is it possible ?" 

You made a mistake - the first and second triplets are not Pythagorean. I fixed it. The procedure   FromLargestToSmallest  solves your problem:

restart;
FromLargestToSmallest:=proc(t::Vector)
local L, M1, M2, M3, A1, A2, A3, t1, k, r, R;
uses LinearAlgebra;
if t[1]^2+t[2]^2 <> t[3]^2 then error `The actual procedure argument is not a Pythagorean triple` fi;
M1, M2, M3 := <1,-2,2;2,-1,2;2,-2,3>, <1,2,2;2,1,2;2,2,3>, <-1,2,2;-2,1,2;-2,2,3>;
A1, A2, A3 := ([M1,M2,M3]^~(-1))[];
t1:=t; L[1]:=t1; k:=1;
while not Equal(t1,<3,4,5>) do
R:=[A1.t1,A2.t1,A3.t1];
for r in R do
if r[1]>0 and r[2]>0 and r[3]>0 then k:=k+1; L[k]:=r; t1:=r fi;
od;  od;
convert(L,list)[];
end proc:

Examples of use:

FromLargestToSmallest(<2275,3648,4275>);
FromLargestToSmallest(<2225, 3648, 4273>);

restart;
with(plots):
F:=(x,y)->x^3-4*x*y+2*y^3:
Curve:=plots:-implicitplot(F(x,y), x=-3..3, y=-3..3, color=red, thickness=2, gridrefine=3, scaling=constrained):
Y:= [fsolve(F(1,y))]; # The list of y-values
G:=unapply(implicitdiff(F(x,y), y, x), x, y);
L:=[seq(G(1,Y[i]),i=1..3)]; # The list of derivatives at the points [1,Y[i]]
T:=[seq(Y[i]+L[i]*(x-1), i=1..3)]; # The list of tangents
Tangents:=plot(T, x=-0.1..2, color=blue):
Points:=pointplot([seq([1,Y[i]], i=1..3)], symbol=solidcircle, color=blue, symbolsize=9):
Names:=textplot([[1,Y[1],A],[1,Y[2],B],[1,Y[3],C]], font=[times,18],align=above):
plots:-display(Curve,Tangents,Points,Names, size=[800,400]);

Here is a direct calculation of the distance between two lines, using no packages, but only Maple kernel commands. We find this distance as the length of a segment connecting 2 points on these lines, and the segment must be perpendicular to each of the lines. I think this approach is useful for understanding what exactly is the distance between 2 lines:

restart;
A := <0, 0, 0>:
B := <a, 0, 0>:
C := <a, a, 0>:
DD := <0, a, 0>:
S := <0, 0, h>:
v1:=C-A; v2:=DD-S; # Direction vectors of the first and second lines 
X1:=A+s*v1; # Arbitrary point on the first line
X2:=S+t*v2; # Arbitrary point on the second line
u:=X1-X2; # Vector connecting points X1 and X2
Sol:=solve({v1.u=0, v2.u=0}, {s,t}) assuming positive;
simplify(eval(sqrt(u.u), Sol)) assuming positive;

Will this be what you want:

restart;
S__x := sin(r__a):
S__y := cos(r__a):
S__z := sin(r__a):
lower_limit := 0: upper_limit := 10: step_size := .3: num_steps := ceil((upper_limit-lower_limit)/step_size): r_values := [seq(i*step_size+lower_limit, i = 0 .. num_steps)]:
                      
arrows := [seq(plots:-arrow([0, 0, r__a], [S__x(r__a), S__y(r__a), S__z(r__a)], color = blue), r__a in r_values)]:

plots:-display(arrows, axes = normal);

restart;
phi := proc(k,x,L)
  if (type(k,even)) then sqrt(2)*sin(Pi*k*x/L)/sqrt(L)
  else sqrt(2)*cos(Pi*k*x/L)/sqrt(L)
  end if;
end proc:
 	
A:=Int(phi(m,x,L)*_h^2/m2*diff(phi(n,x,L),x,x),x=-L/2..L/2);
value(A) assuming m::posint,n::posint,m=n;
value(A) assuming m::posint,n::posint,m<>n;

We place point  A  at the origin of coordinates, point  B  on the axis  Ox  to the right of  A , points C and  D  lie in the upper half-plane. We introduce 2 unknowns: s - the angle between the axis  Ox  and AD, t - the angle between the axis  Ox  and  BC. We use vectors for calculations. We find s  and  t  numerically by solving a system of 2 equations with 2 unknowns:

restart;
with(plottools): with(plots): 
local D:
Dist:=(X,Y)->sqrt((X[1]-Y[1])^2+(X[2]-Y[2])^2):
Angle:=(U,V)->arccos((U.V)/sqrt(U.U)/sqrt(V.V)):
A:=<0,0>: B:=<170,0>: C:=<170+170*cos(t),170*sin(t)>: D:=<180*cos(s),180*sin(s)>:
Sys:={Dist(C,D)=290,Dist(A,C)=Dist(B,D)};
Sol:=fsolve(Sys, {s=0..Pi,t=0..Pi});
C:=eval(C, Sol): D:=eval(D, Sol):
P:=curve(convert~([A,B,C,D,A],list), color=red, thickness=2):
T:=textplot([[convert(A,list)[],"A"],[convert(B,list)[],"B"],[convert(C,list)[],"C"],[convert(D,list)[],"D"]], font=[times,bold,18], align={left, above}):
display(P,T, scaling=constrained);
DA:=A-D: DC:=C-D: CD:=-DC: CB:=B-C:
eval(s,Sol); # Angle A
eval(Pi-t,Sol); # Angle B
Angle(DA,DC);  # Angle D
Angle(CD,CB);  # Angle C