Solve the differential equation $y'-xy^2 = 2xy$ I get it to the form $\left | \dfrac{y}{y+2} \right |=e^{x^2}e^{2C}$ but I'm not sure how to get rid of the absolute value and then solve for y. I've heard the absolute value can be ignored in differential equations. Is this true?

We have
$$y' - x{y^2} = 2xy \Rightarrow y' - 2xy = x{y^2} \Rightarrow y'.{y^{ - 2}} - 2x{y^{ - 1}} = x,\,\,\left( {y \ne 0} \right).$$
Put $z=y^{-1}$. Then
$$\begin{gathered}
  z' + 2xz =  - x \Rightarrow \frac{d}{{dx}}\left( {z{e^{{x^2}}}} \right) =  - x{e^{{x^2}}} \\
\Rightarrow z = {e^{ - {x^2}}}\int {\left( { - x{e^{{x^2}}}} \right)dx}  =  - \frac{1}{2}{e^{ - {x^2}}}\int {{e^{{x^2}}}d\left( {{x^2}} \right)}  \hfill \\
   =  - \frac{1}{2}{e^{ - {x^2}}}\left( {{e^{{x^2}}} + C} \right) =  - \frac{1}
{2} - \frac{C}{2}{e^{ - {x^2}}}. \hfill \\ 
\end{gathered} .$$
Finally, 
$$y = {z^{ - 1}} = \frac{1}{z} = \frac{1}{{ - \frac{1}{2} - \frac{C}{2}{e^{ - {x^2}}}}} =  - \frac{2}{{1 + C{e^{ - {x^2}}}}}.$$
Note: $y=0$ is also a solution of the equation.
