Approach for Riccati's equation 
I understand the proof. I just don't understand the approach which doesn't seem intuitive in any sense to me but in the end works out serendipitously. My question is: can the approach be understood intuitively?
 A: How I understand it intuitively:
Dividing both sides by $y^2$ gives on the left hand side a term of the form $y'/y^2$, which one easily recognizes as $-\frac{d}{dx}(1/y)$ so that's why one would try the above substitution.
The problem is, that on the right hand side, the right hand most term becomes $r(x)/y^2$ which would still be quadratic after this substitution. Here is where the trick of adding the particular solution comes into play: the previously mentioned term disappears ( by the binomial theorem ), and only terms where $y$ appears with degree greater or equal to $-1$ or $0$ remain.
EDIT: Basically what I mean is, given:
$$y'=p(x)y^2+q(x)y+r(x)$$
and knowing that 
$$Y'=p(x)Y^2+q(x)Y+r(x)$$
how do we get rid of the $r(x)$ in the first equation, using the second?
You can either subract the second equation from the first and complete the square and so on, as I suggested in my second approach. Or you can have the good intuition and see that if instead of $y$ you plug into the first $y+a$ you get 
$$(y+a)'=p(x)(y+a)^2+q(x)(y+a)+r(x)=p(x)y^2+2p(x)ya+p(x)a^2+q(x)(y+a)+r(x)$$
where you recognize the apperence of the ODE in terms of $a$ and thus proceed by pluggin in the particular solution for $a$, so to get rid of the $r(x)$ term
