Hi Jim, today I found in the four roots U1p, -U1p, U1n, -U1n, only two of them are correct answers to your original equation. Except that obviously there are two zero roots solved the original equation, in the roots set only two of them are correct because the other two solved:
Expr4-2*coeff(Expr4,G12)*G12=0
This equation has a root which also results in Expr4=0 only when
u1^2=-1/a
where it is assummed G12!=0. Of course this is a coincidence.
Due to Maple does not simplify product of two radicals, and furthermore, sum of two radicals, the root cannot be verified. But that is the key to this question:
>simplify(subs(u1 = U1n, Expr4), size);
>PPTT1 := collect(expand(G12^2*2*c*(sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)+(1/2)*(-G33+G22)*(G22-G11)*c+a*G12^2)*2*a*(-sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)-(1/2)*(-G33+G11)*(G22-G11)*c+a*G12^2)), c);
>PPTT2 := collect(expand((c*((G11-G22)*sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)+a*G12^2*(G11+G22-2*G33)))^2), c);
>-PPTT1+PPTT2;
0
Note a=a(x1,x2) and c=c(x1,x2). The last equation obviously implies
-sqrt(PPTT1)+sqrt(PPTT2)=0
To guarantee the above holds, PPTT2>0 and G12>0 must be satisfied. Due to a(x1,x2), c(x1,x2) and the other four parameters G11, G12, G22 and G33 does not have a relation to guarantee PPTT2>0, it cannot be concluded that U1n is the corect root. Therefore the conclusion is which two roots in the four roots set are correct answers to the original equation is dependent on both the parameters G11, G12, G22, G33 and a(x1,x2), c(x1,x2).
-slimriver-

Hi Jim, today I found in the four roots U1p, -U1p, U1n, -U1n, only two of them are correct answers to your original equation. Except that obviously there are two zero roots solved the original equation, in the roots set only two of them are correct because the other two solved:
Expr4-2*coeff(Expr4,G12)*G12=0
This equation has a root which also results in Expr4=0 only when
u1^2=-1/a
where it is assummed G12!=0. Of course this is a coincidence.
Due to Maple does not simplify product of two radicals, and furthermore, sum of two radicals, the root cannot be verified. But that is the key to this question:
>simplify(subs(u1 = U1n, Expr4), size);
>PPTT1 := collect(expand(G12^2*2*c*(sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)+(1/2)*(-G33+G22)*(G22-G11)*c+a*G12^2)*2*a*(-sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)-(1/2)*(-G33+G11)*(G22-G11)*c+a*G12^2)), c);
>PPTT2 := collect(expand((c*((G11-G22)*sqrt(a*(-(-G33+G11)*(-G33+G22)*c+a*G12^2)*G12^2)+a*G12^2*(G11+G22-2*G33)))^2), c);
>-PPTT1+PPTT2;
0
Note a=a(x1,x2) and c=c(x1,x2). The last equation obviously implies
-sqrt(PPTT1)+sqrt(PPTT2)=0
To guarantee the above holds, PPTT2>0 and G12>0 must be satisfied. Due to a(x1,x2), c(x1,x2) and the other four parameters G11, G12, G22 and G33 does not have a relation to guarantee PPTT2>0, it cannot be concluded that U1n is the corect root. Therefore the conclusion is which two roots in the four roots set are correct answers to the original equation is dependent on both the parameters G11, G12, G22, G33 and a(x1,x2), c(x1,x2).
-slimriver-

For power series, convergent radius
r=limit(abs(a_n)/abs(a_n+1),n=infinity).
For general series, you need to prove the limit
for partial sum exist.

For power series, convergent radius
r=limit(abs(a_n)/abs(a_n+1),n=infinity).
For general series, you need to prove the limit
for partial sum exist.

It is indeed so awkward to write the soltuion in the "Arctan(y,x)" form, but not in arctan(y/x) form. The worse thing is Maple does not cancel the common factor 1+exp(2*t)*_C1^2 in y and x, to make it looks more "understandable".
Anyway, this explains why this equation fails Maple to solve it with an initial value x(0)=x0.