Laplace Transform for Linear ODEs with Variable Coefficients If you take the Laplace Transform of the time-domain ODE $x^2 y'' + xy' - 9y = 0$ and do some algebra, you get the new frequency-domain ODE $s^2 Y'' + 3sY' - 8Y = 0$.  If you then apply the same technique to this ODE, apparently dragging it one step further away from the time-domain, you end up with $r^2 '' + r' - 9 = 0$.
Surely it cannot be a coincidence that this is the original ODE up to new choices of variable names, right?  Why does this cyclic pattern occur, despite apparently moving further away from the time-domain into the "hyperfrequency-domain"?  What is the right way to think about the domain movement?
Also, how should we conceive of the impact this cyclic process would have on a non-homogeneous term, if the original ODE had one?  I can't really find such a term that is amenable to taking the Laplace Transform twice.

Suppose $\beta = \{e_0, e_1, e_1\}$ is a list of formal vector-like objects, where the symbol $e_0$ stands for $y = Y = \mathfrak Y$; the symbol $e_1$ stands for $xy' = sY' = r\mathfrak Y'$, and the symbol $e_2$ stands for $x^2y'' = s^2Y'' = r^2\mathfrak Y''$. Then the Laplace transform $\mathcal L$ induces an operator on $\operatorname{span} \beta$, given by $\mathcal L(e_0) = e_0$, $\mathcal L(e_1) = -e_0-e_1$, and $\mathcal L(e_2) = 2e_0 + 4e_1 + e_2$. The matrix of $\mathcal L$, with respect to the basis $\beta$, is therefore
$$ [\mathcal L]_\beta = \begin{pmatrix} 1 & -1 & 2 \\ 0 & -1 & 4\\ 0 & 0 & 1 \end{pmatrix}$$
and a quick calculation yields the surprising fact that $[\mathcal L]_\beta^2 = I$. This demonstrates that the Laplace transform is involutory over equidimensional equations, the type of equation which you are considering here (I suspect you may be able to extend what I have described here to higher order terms via induction, which may be illuminating). As for how to interpret the double-Laplace transform (or the hyperfrequency domain), or how to apply it to nonhomogeneous equations, I do not know.
I do know that there is a connection between the Laplace transform and the Fourier transform, which under certain conditions is also self-inverse. You may be able to find more answers by exploring that connection.
