Separable diff. eqn: $(1+x^2)y' = x^2y^2, x > 0$ I have been given a step-by-step answer which I just cannot understand or follow.
$\begin{eqnarray}
&(1+x^2)y' &= x^2y^2 + y\cdot1 \\
\iff& \frac{1}{y^2} &= \frac{x^2}{1+x^2}
\end{eqnarray}$
From there on it's a matter of integrating and using a intitial value I was given, steps I understand.


*

*What is going on in the first step when a term $y\cdot1$ was added from nowhere?

*I have no clue how he got from step 1 to step 2.
What I do know:
Separable functions of the form $G(y(x))y' = h(x)$ can be rewritten $D(G(y(x))) = h(x)$ and I suspect something like that happens from step 1 to step 2; I just do not see it.

it is for $y\neq 0$ equivalent to $$\frac{dy}{y^2}=\frac{x^2}{1+x^2}dx$$
