Thread: General Solution of y' + y^2 = 1 ?

Please help me find general solution of y' + y^2 = 1 and y(0)=2

y' = 1(y^-2)

I try to use Bernoulli's but it did not work because there is no y term

y' + p(x)y = q(x)y^n
v' + (1-n)p(x)v = (1-n)q(x)

use separation of variables:
dy/1-y^2=dx