Posted by Raj on Tuesday, December 10, 2013 at 8:39pm.
xy' + 1 = y^2
xy' = y^2-1
y' = (y^2-1)/x
dy/(y^2-1) = dx/x
arctanh(y) = log(x)
or,
log (1-y)/(1+y) = 2log(x)+c
and you can massage that into exponentials and wind up with
y = 1-e^(2cx^2) / 1+e^(2cx^2)