solve for y: $xy'=y^2-2y$ $$xy'=y^2-2y$$ $$y(1) = 1, x &gt;= 0$$

I get $C = 0$ which gives:

$$\left| \frac{y}{y-2}\right|=x^2$$ When solving for $y$ it gives $$\frac{2x^2}{1+x^2}$$, however the correct answer should be $$\frac{2}{1+x^2}$$.

Is $$\left| \frac{y}{y-2}\right|=x^2$$ correct? And if it is, how to continue from there?
(2018-11-25) "Amusing" revenge downvote, four years later.


Instead, try $$\left| \frac{y-2}{y}\right|=x^2.$$
(2018-11-25) "Amusing" revenge downvote, four years later.


Instead, try $$\left| \frac{y-2}{y}\right|=x^2.$$