Why might this answer to a differential question be false? Is there a more rigorous proof or counterexample? I'm trepidatious to accept an answer here because there's a couple points the author hasn't cleared up
Why doesn't this differential technique work?
$y''/y$ certainly exists if the solution exists wherever $y \neq 0$, and it is integrable.
Secondly, $y''/y = t$ which is an invertible function. In the worst case scenario, suppose we define a new function, the inverse Airy function's derivative over a compact set $Ai'^{-1}(t).$
We can represent $t$ then as $Ai'^{-1}(Ai'(t)) = t$ over varying compact intervals corresponding to the appropriate branch of its inverse, so clearly an antiderivative does exist in terms of $y',$ though it is then still an open-question as to how we derive it.
What are the flaws in our reasonings here?
 A: In a nice answer to the post you are referring to it was demonstrated that $y''/y$ is not the first derivative of a function $f(y,y')\,.$ That's the reason we can't separate variables when we solve the ODE $y''=y\,g(x)\,.$
This is maybe not the end of the world. Apparently,
$$
\frac{y''}{y}-\frac{y'^2}{y^2}
$$
is the second derivative of the function $f(y)=\log y\,.$ More generally, for arbitrary twice differentialbe $f\,,$
$$
\frac{d}{dx}f(y)=f'(y)\,y'\,,\quad\frac{d^2}{dx^2}f(y)=f'(y)\,y''+f''(y)\,y'^2\,.
$$
When $f$ is invertible this means that the ODE
\begin{align}\tag{1}
f'(y)\,y'&=g(x)\,,\\[3mm]
\end{align}
has the solution
\begin{align}
\textstyle y=f^{-1}\Big(\int_0^x g(z)\,dz\Big)+\Big(y(0)-f^{-1}(0)\Big)\,.
\end{align}
For the ODE
\begin{align}
f'(y)\,y''+f''(y)\,y'^2&=g(x)\tag{2}
\end{align}
we get first
$$
f'(y)\,y'=\textstyle\int g(x)\,dx=:h(x)
$$
which is an ODE of the form (1) and has a known solution.
I believe the old Book Differntialgeleichungen by E. Kamke (not sure if it was ever translated) is a good reference for such "non ordinary" ODEs.
