The answer is order 2. Here is why:
First of all, the equation of a parabola with any horizontal
axis of symmetry y = k is
(y-k)² = 4p(x-h)
That has three arbitrary constants h, k and p.
Therefore a differential equation of all parabolas
having ANY horizontal axis of symmetry (not necessarily the
x-axis, would be of order 3, the number of arbitrary constants.
If we restrict the family of parabolas to those with axis of
symmetry y = 0, (where k = 0), the equations would be
y² = 4p(x-h)
y² = 4px - 4ph
or
y² = c1x + c2, where c1 = 4p and c2 = 4ph
That has 2 arbitrary constants, so the order would be 2.
That's the answer you were looking for. But I'll continue:
To find that differential equation, we differentiate
twice to get rid of the arbitrary constants:
y² = c1x + c2
2y*y' = c1
y*y' = c where c = c1/2
y*y" + y'*y' = 0
y*y" + (y')² = 0
That would be the 2nd order differential equation whose solution
is all parabolas with their axis of symmetry being the x-axis.
Edwin
