solve the Bernoulli equation xy' - y = xy^2

Question

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 1

Let $g(x)=\frac{1}{y(x)}$ then $$g'(x)=\frac{-y'(x)}{(y(x))^2}$$ so we reduce the equation to $$-g'(x)-\frac{g(x)}{x} = 1$$

Can you finish from here?

Answer 2

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

Answer 3

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