Why doesn't this differential technique work? Suppose we try to solve the ODE $y' = y,$ we can rearrange this into $y'/y = 1,$ integrate and obtain $\ln(y) = x.$
However, let's try this on $y'' = xy$, then we rearrange this into $y''/y = x,$ but the solution is Ai$(x)$ and Bi$(x)$, the airy functions. Can this technique of rearranging then integrating be salvaged to explain this result or does it fail in some way? Is there a known way to integrate to derive a function $f$ in terms to $y'$ and $y$?
 A: The difference between the two examples is the following. The expression $y'/y$ is the derivative of $\ln y$. Thus, it is possible to "integrate" $y'/y$.
On the other hand $y''/y$ is not the derivative of any expression $f(y,y')$. This is quite easy to see. We set $p=y'$ and use the chain rule to obtain
$$
 \frac{d f(y,y')}{dx} = \frac{\partial f}{\partial y} \,y' + \frac{\partial f}{\partial p} y''\,.
$$
No, we need that $\partial f/\partial y = 0$ and $\partial f/\partial p=1/y$ in order to find the antiderivative $f(y,y')$. But if the function $f$ exists (and is sufficiently often differential), we have the Schwarz's Theorem
$$ \frac{\partial}{\partial y }\frac{\partial f}{\partial p} = \frac{\partial}{\partial p }\frac{\partial f}{ \partial y}.$$
In other words
$$ -\frac{1}{y^2}= \frac{\partial}{\partial y} \frac{1}{y} =  \frac{\partial}{\partial p} 0 = 0$$
which is a contradiction. Thus, no antiderivative of $y''/y$ exists.
A: You'd need to create antiderivative based on treating the differential as a fraction:   $$\frac{d^2 y}{dt^2} = yt \implies \int\frac{1}{y}d^2y = \int t dt^2$$
These techniques aren't defined, and so this technique of "integrating" to get a solution is unfortunately not something that has been formalized yet.
