Consider the recurrence
$$g(h) = g(h-1) - \frac{1}{4}g(h-1)^2,$$
for $h \geq 0$ and $g(0)=1$. This recurrence occurs in many applications (For example fast minimum cut algorithms, the Galton-Watson process, etc). It is known that $g(h) = \Theta(1/h)$. In fact, the more accurace bounds of
$$ \frac{3}{h+3} \leq g(h) \leq \frac{4}{h+4},$$
are known. I recently saw a very "slick" idea of "guessing" the solution to this recurrence where you rewrite the recurrence as
$$ \frac{g(h)-g(h-1)}{1} = -\frac{1}{4}g(h-1)^2.$$
Now we approximate $g'(h) \approx (g(h)-g(h-1))/1$, and $g(h-1)\approx g(h)$ to get the differential equation
$$g'(h) \approx -\frac{1}{4}g(h)^2 \implies g(h) \approx \frac{4}{h}.$$
This is extremely close to the correct solution, so I'm wondering if there is a reference to this technique and if it is used to guess the correct solution to other (non-chaotic) quadratic recurrences?