yes.. bernoullis eqn...
dy/dx +Py = Q y^x , where P and Q are functions of x alone
You shouldn't regard delays in replying on this forum as unusual. I have just been doing other things for a while.
I'm not sure I agree with your solution of the Bernoulli equation. Can you show your steps, please?
multiply the whole eqn by 1/u^2
u^-2 . du/dx + 1/(ux) = 1/x^2
now, eqn becomes
-dz/dx + z/x = x^-2
dz/dx - z/x = -x^-2
P=-1/x , Q=-x^2
IF = 1/x
z . 1/x = inegral -x^-2 * 1/x dx + c