Why is the exponential function substituted when deriving the characteristic equation? I'm currently learning about second order differential equations and how to solve them if the coefficients are constant.
I understand how the derivation of the characteristic equation works (e.g. I understand that when plugging in $e^{rx}$, we get a polynomial which can be solved to find solutions). What I don't understand, however, is why we even decide to plug $e^{rx}$ in in the first place. I have heard people say that this is just an "educated guess"; but why is it an educated guess?
 A: More than an educated guess, it is a consequence of applying the Fourier /Laplace transform to the linear ode.
Since you say you did not study yet about integral transforms (but you will face with that subject soon, most probably) then let me introduce a physical insight, leaving apart a rigorous mathematical assessment.
Probably you studied some physics (at least mechanics).
So a simple mass-spring system is governed by the the 2nd order diff. eq. $$m {\ddot x} + kx =0$$
and it is common experience that ,upon being perturbed, it oscillates with a sinusoidal motion.
It was later acquired in physics that if the perturbing force is sinusoidal, so will be the motion, and that the "answer" to a linear combination of sinusoidal forces was the same linear combination of the single sinusoidal answers.
This has been later rigorously systematized into the mathematical scheme of the Fourier series, then Fourier Transform and Laplace Transform, reaching to the conclusion that the family of functions
$e^{i\omega t}= \cos(\omega t)+i \sin(\omega t)$ or  $e^{st}$ for different values of $\omega , \; s$
represents a "very natural" basis (i.e. "simple", "generally and effectively applicable") for all the natural (causal) systems governed by a linear differential model.
