solve the Bernoulli equation xy' - y = xy^2 
Solve the Bernoulli equation $xy' - y = xy^2$.

I started with diving both sides by $x$, and ended up with $y' - \frac{y}{x} = y^2$. Then, I divided both sides by $y^2$ and got $\frac{y'}{y^2} - \frac{1}{xy} = 1$. Can someone help me finish this problem?

Start over.  Divide through by $y^2$.  Express in terms of differentials (dy and dx).  The LHS is an exact differential.
