ok here is the other problem...
dy/dx = (y^2+y)/(x^2+x)
i see that it is separable, but could it also be Bernoulli because of the y squared?
Yes, it is also a Bernouilli equation.
Fernando Revilla
Don't worry, we quickly make another book.
The equation is equivalent to:
$\displaystyle y'+\left(\dfrac{-1}{x^2+1}\right)y=\left(\dfrac{1}{x^2+1}\right)y^2$
That is, it has the form $\displaystyle y'+p(x)y=q(x)y^n$ (Bernouilli equation).
Fernando Revilla