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.

First, if you differentiate $G(y(x))$ with respect to $x$ you get $\cfrac {dy}{dx} \cdot \cfrac {dG}{dy}$
So if you have an equation $G(y)=G(y(x))=F(x)$ and differentiate it, you get$$y'\frac {dG}{dy}=\frac {dF}{dx}$$
Going into reverse, if you have $y'g(y)=f(x)$ then you can put $G(y)=F(x)+c$ where $G(y)=\int g(y)dy$ and $F(x)=\int f(x) dx$
I'm not sure that this is quite what you have stated as what you know, because you don't seem to have differentiated your function $G$ with respect to $y$ - just multiplied by $y'$

Now your first target is therefore to put the expression into the form $y'g(y)=f(x)$. The equation is separable if you have that factor $y'$.
The first problem is that the $y\cdot 1$ term simply looks, as others have said, like a typo. So let's ignore that.
It is a simple manipulation to separate the terms - divide through by $y^2$ and $1+x^2$ to obtain $$y'\cdot \frac 1{y^2} = \frac {x^2}{1+x^2}$$Note that in the question the $y'$ which needs to be there has been dropped.
Once this is in place each side can be separately integrated, as you have noted.
