SORRY, THERE WERE SOME MINOR ERRORS IN THE LAST POST. THIS IS THE CORRECTED VERSION.
I tried to verify your answer. This is what I got:
dy/dx = - 4 c exp(2ix) / [ 1 - c exp(2ix) ]^2
y^4 + 1 = 2 [1 + 6 c^2 exp(4ix) + c^4 exp(8ix)] / [1 - c exp(2ix) ]^4
I tried it the following way (So far am not done with the solution)
The equation after separation of variables becomes
dy/ [y^4 + 1] = dx
y^4 + 1 = (y^2 + i)(y^2 - i)
= [y + (1 + i)/sqrt(2)] [y - (1 + i)/sqrt(2)] [y + (1 - i)/sqrt(2)] [y - (1 - i)/sqrt(2)]
So resolve 1/(y^4 + 1) into partial fractions and then integrate each linear factor separately. The final answer would be of the form x = ...
Finally, to answer this question (Q. 31) on the GRE, we don't need to solve this differential equation. Observe that the dy/dx = m = slope of the solution at various values of y.
For example at y = 0, dy/dx = m = 1 => 45 degrees
This rules out B, E.
At y = 1, -1, dy/dx = m = 2. So the slope of the curve should be symmetric on either side of the origin. This rules out D.
Finally at y = +infinity, -infinity, the slope should be tan 90 => 90 degrees. This rules out C.
The only answer left is A.