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?

Answer
xy^'_ y=xy^2
divide by  x then
y^'_ y/x= y^2
Alsultani d.e. denotes that;
y^'+P(x)y/n= Q(x)y^k, where k=1_n ,
k=2=1_n so n=_1 then multiply by ( _ 1)
We get
_ y^'+y/x = _ y^2
So P(x)= 1/x then I.F.=e^int.1/xdx= x then
_xy^'+y=_xy^2 I.e.
x/y= int.(_x)dx+c so
x/y=_x^2/2+c
y/x=1/(c_x^2/2) then
y= x/(c_x^2/2) so
y=2x/(c1_x^2)  where c1=2c
