We aren't given points on the solution curve, so do indefinite integrals:
The RHS is easy: .
For the LHS use the method of partial fractions: , so:
for some constant A, B.
Any way you solve it, I get A = -1/2 and B = 1/2.
Thus the LHS from above is:
Equating the LHS and the RHS:
where C = C' - C'' is a new integration constant.
where stands as the new integration constant.
This is kind of ugly as it stands. But if we square both sides we get rid of the ugly absolute values. The trouble is that once we do this we might be adding extra solutions to the equation. We need to check these at the end.
Expanding and moving everything to the LHS:
This looks terrible, but is simply a quadratic in y. So use the quadratic formula:
After some simplification:
As it happens, both solutions check in the original differential equation.
So pick one, say . (The choice between the two is merely a matter of picking an "a" value.)
PS I just wanted to be kind of geeky and say that I really enjoyed solving this one. I haven't seen one quite like this before.