xy' + y^2 = 1, y(1) = 0
xy' = 1 - y^2
dy / (1-y^2) = dx / x
Need to take into account the possiblity of having "divided away" solution. We check if = +/- 1 are solutions by inserting them into the equation. They aren't solutions.
Moving on. We get (1/2)ln|1+y| - (1/2)ln|1-y| = ln|x| + C
This is where I'm stuck. Since we have the initial condition y(1) = 0 we need to figure out C in this step before moving on, right? What I've done in previous exercises is to isolate y and take it from there, but I don't know how to do it in this case. I've tried doing the following:
ln| (1+y) / (1-y) | = ln|x|^2 + 2C
Multiply with e and we get
| (1+y) / (1-y) | = x^2 + e^(2C)
But, well, I can't get it to work out.