z > 0 means Re(z)>0, Im(z)=0. So,

sys:=[
-V^2 + N + P=0,
-(V^2 + N - P)^2 - 4*A*V^2>0,
g*(A*V^2 + N*V^2 - N*P)<0,  #g<>0
V>0, V<1,
N>0, P>0]:
SolveTools:-SemiAlgebraic(sys, [N, P, V, A]);

N :: posint ==> empty set.

Here the fact that odetest gives 0 must be undestood as: the solution is valid in some interval, but not necessarily everywhere.
(I met several such situations).

Actually, directly
eval(ode, sol);

is obviously false if lhs is <0,  e.g. _C1=0, x=Pi^2.

A more correct solution is obtained with:

Sol:=dsolve(ode, implicit);

The only problem is that your system may have no solution for alpha >= 0.1

Note that alpha=0, upsilon=anything is always a solution, and depending on data, this could be the only one.
To avoid alpha=0 simply replace {alpha = 0.001 .. infinity, upsilon = 0.1 .. infinity}; sometimes the solution will satisfy alpha>0.1

f := x -> exp(x/a)/(a*(1 + exp(x/a))^2):
int(f(x)*exp(-I*w*x), x = -infinity .. infinity):
IntegrationTools:-Change(%, exp(x)=y) assuming a>0;

         Pi*csch(Pi*a*w)*a*w

As stated it's trivial. Probably you have some other conditions.

solve({x^2 + y^2 + (z-3)^2 = 25, z = 0},  explicit);

The exact solution is possible, but I am sure you know this.

restart;
pde:=diff(u(x,t),t)=diff(u(x,t),x,x);
ibc := u(x,0)=1, u(0,t)=0, u(1,t)=0;
dsolu := pdsolve({pde,ibc});
U:=rhs(dsolu);
V:=diff(U,x);
plot(eval(V,x=0), t=0..0.5);

StringTools:-FormatTime("%X");
StringTools:-FormatTime("%Y-%m-%d %X");

You have again here an ode which has a factor not containing y'.

factor(lhs(ode) );
      2*(y(x)*x + 1)*(diff(y(x), x)*x + y(x))

I don't know why you do this, because in my opinion dsolve should not be given such arguments (unless you want to test something).
Anyway, dsolve seems to be fooled by it and gives twice the same general solution (_C1  ---> - _C1).
The singular solution y = -1/x  is from the first factor, which has nothing to do with the second (linear ODE) factor, so it is "essential";  only by (intentional?) coincidence, -1/x  is obtained for _C1 = 0.

You have again here an ode which has a factor not containing y'.

factor(lhs(ode) );
      2*(y(x)*x + 1)*(diff(y(x), x)*x + y(x))

I don't know why you do this, because in my opinion dsolve should not be given such arguments (unless you want to test something).
Anyway, dsolve seems to be fooled by it and gives twice the same general solution (_C1  ---> - _C1).
The singular solution y = -1/x  is from the first factor, which has nothing to do with the second (linear ODE) factor, so it is "essential";  only by (intentional?) coincidence, -1/x  is obtained for _C1 = 0.

Maybe an unusual phenomenon closely follows StudentT(13/3); why exclude it?
Even a dimension can be noninteger; see Sierpinski's triangle whose (Hausdorff) dimension is log[2](3).

radnormal(arctanh(A)-arctanh(B));
                      0

Maple does "generic" simplifications.
See the simpler example:

ex:=x/(x+1)/(x+y):
eval(ex,x=0);    # generic only, not valid when y=0
                               0
eval(ex,y=0);
                  1/(x+1)

eval(%, x=0);
                               1

This is obvious and not related to Maple. If A+B*y' = 0  (or A dx+B dy = 0) is exact, then  A/B +y ' = 0 is no longer exact (in general): it has B as integrating factor.

p[1]:=(a+b)*x^2+(a+c)*x*y+d*z:
p[2]:=(a-b)*x+e*y*z+f*z^2:
q[1]:=x^2+2*x*y+3*z:
q[2]:=x+4*y*z:
coeffs(p[1]-q[1],[x,y,z]), coeffs(p[2]-q[2],[x,y,z]):
solve([%]);

           {a = 1, b = 0, c = 1, d = 3, e = 4, f = 0}

There are many examples of  sequences of functions f(n,x)
which are convergent (n-->oo) but diff(f(n,x),x) diverges.
A simple one:  f(n,x) = sin(n^2*x)/n.