As I noted earlier, your equation is linear, so your focus on nonlinear equations is out of place. If you refer to the discussion of linear first order equations in your differential equations textbook, you will find a general solution obtained through the method of integrating factors. The transient and steady-state parts of the solution may be extracted from there.

What you should be asking is: What do the transient and steady-state solutions look like? I have attached a PDF in which I have analyzed a slightly more general differential equation than the one that you have asked about. In the following worksheet I use the results obtained in that PDF to calculate and plot the transient and steady-state graphs for your equation. Note that this has nothing to do with nonlinear equations.

Download worksheet: steady-state-of-an-ode-2.mw

Download documentation: steady-state-of-an-ode.pdf

Your equation has two solutions, and therefore sol produced by maple consists of two parts:

sol := {solution1}, {solution2}

In your code, you are attempting to convert sol to Matlab() and that backfires.  You need to pass the individual solution1 and solution2 to Matlab():

Matlab(sol[1][]);
Matlab(sol[2][]);

The purpose of the empty brackets is to remove the curly braces in sol.

The "triangle" that you have depicted is not a spherical triangle since the blue edge is not an arc of a great circle, and therefore the Pythagorean Theorem does not apply.

Aside: You have not shown your calculations, so I calculated everything from scratch.  My calculation shows that the length of the green curve is 

which evaluates to 0.9228460424.

restart;
f := x -> 2*2^x-2;
g := x -> -1/2*x^2+3/2+5;
fsolve(f(x)=g(x), x=-10..10, maxsols=2);
                -4.094614685, 1.787210719

This problem can be solved by hand with the method that I described in my yesterday's answer to another question.  The solution turns out to be

which you may verify by direct substitution.  I don't have the time this evening but I should be able to post the details that lead to that solution tomorrow.

=============

OK, it's tomorrow now, and here is the calculation that led to the result above.

Download poisson-problem-on-disk.mw

This problem may be solved through Fourier series. Maple can help with some of the calculations but much of the infrastructure needs to be calculated by hand.  The document attached at the bottom of this reply explains the general strategy and then produces the specific solution of the problem that you have posed.

Taking only 8 terms in the Fourier series produces a symbolically exact solution to the PDE which satisfies the boundary condition approximately with a residual error of O(10⁻⁷), which is quite good.  Here is the graph of the solution.

Documentation: poisson-problem-with-robin-boundary-conditions.pdf

A boundary value problem does not necessarily have a solution. Yours appears to be in that category. If, however, you have some flexibility in choosing the parameter values, then solutions do exist.

Here, in addition to your parameters t3, R, and M, I have introduced two new parameters—the length of the interval, L, and the value of G(0), d.  For instance, setting L=1 and d=0.9 results in a solution as shown below.  Another possible pair is L=0.6, d=1.

restart;
eq1 := diff(F(x), x$4) +2*R*(F(x)*diff(F(x), x$3) + G(x)*diff(F(x),x)) + 2*R*t3*(2*diff(F(x), x$2)*diff(F(x), x$3) + diff(F(x), x)*diff(F(x), x$4) + 3*diff(G(x), x)*diff(G(x), x$2)) = 0;
eq2 := diff(G(x),x$2) - 2*R*(diff(F(x),x)*G(x)- F(x)*diff(G(x), x)) - 2*R*t3*(diff(F(x),x$2)*diff(G(x),x) - F(x)*diff(G(x), x$2) ) = 0;
ic := F(0)=0,  F(L)=0, G(0)=d, G(L)=0, D(F)(0)=0, D(F)(L)=M;
params := t3=2, R = 0.5, M =0.5, L=1.0, d=1.0;   # original (bad choices)
params := t3=2, R = 0.5, M =0.5, L=0.6, d=1.0;   # good choices
params := t3=2, R = 0.5, M =0.5, L=1.0, d=0.9;   # another set of good choices
sys := eval({eq1,eq2,ic}, {params}):
dsol := dsolve(sys, numeric, method=bvp[middefer]);
plots:-odeplot(dsol, [[x,F(x)],[x,G(x)]], color=[red,blue]);

Maple produces the general solution with a little help.

restart;
kernelopts(version);
   Maple 2021.1, X86 64 LINUX, May 19 2021, Build ID 1539851

ode:=diff(y(x),x) = 2*(2*y(x)-x)/(x+y(x));
dsolve(ode, implicit);      # solve
subs(_C1 = ln(C), %):       # replace _C1 by ln(C)
(exp@lhs = exp@rhs)(%):     # exponentiate both sides
simplify(%);

Then you can determine C by applying the initial condition.

Since you are composing procedures, it would make life easier if each f[i] received a procedure and returned a procedure.  Here is how to do it. 
 

restart;
kernelopts(assertlevel=2):

f[1] := proc(x::procedure)::procedure;
	piecewise(0 <= t and t <= 1/2, (1/2)*x(2*t), 1/2 < t and t <= 1, 1/2*x(2*t-1));
	unapply(%, t);
end proc:

f[2] := proc(x::procedure)::procedure;
	t*x(t);
	unapply(%, t);
end proc:

# Illustration
L := [1,2,1]:
map(i->f[i], L):
`@`(%[]):
h := %(t->t^2);   # change t->t^2 to anything else, as needed

plot(h(t), t=0..1);

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restart;
with(Physics):
alias(x=x(t),y=y(t)):
phi := [ x, phi__1(x,y) ]:
J := < diff(phi[1], x), diff(phi[1], y);
       diff(phi[2], x), diff(phi[2], y) >;

This problem is hardly different from the one that you posted a few hours ago. Here is the same answer with minor modifications.

Download mw.mw

vv has already provided a solution. Here is an alternative approach.

>  restart; >  f:=x->1/(x^2-2*x); >  p1 := plot(f(x), x=-3..5, discont, view=-3..3, scaling=constrained, color=blue); f is symmetric about the line >  is( f(x+1) = f(-x+1) ); Therefore, the center of the circle lies on the line . The circle touches the graph at , and consequently, the y coordinate of its center is , where R is the radius. We conclude that the circle's equation is >  C := (x-1)^2 + (y-R+1)^2 = R^2; We calculate the points of intersection of the circle and the graph of f: >  xy := solve({C, y=f(x)}, [x,y]); Let us isolate the argument of the RootOf: >  select(has, xy, RootOf): eval(x, %[]): P := op(1, %); The quartic polynomial P has 4 roots in general, but when the circle is tangent to the graph of f,  those 4 roots collapse into 2 roots.  That happens when the discriminant of the quartic is zero. >  discrim(P, _Z) = 0; solve(%); We conclude that the radius is 3/2 and therefore the y coordinate of its center is 1/2. Here is its graph >  plots:-display(p1, plottools:-circle([1,1/2], 3/2, color=red));

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In your worksheet you have
motioneq  = (3.2)

That "=" should be ":=".

I know nothing about MapleSim, so I cannot help you in that respect.

Unless you have a very ancient version of maple, this should do what you want:

int~(<sin(x), cos(x)>, x=0..2);

I can think of a dozen or two different answers to your extremely vague question and it's likely that they will be unsatisfactory as long as you refuse to show your code. Anyway, here is one possibility:

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