### Understanding polynomial-ish differential equations

This is a rather simple idea, perhaps not one you really had too many problems understanding to begin with. Given you know that $e^{\lambda x}$ solves first-order polynomial differential equations, it's not too much of a stretch to imagine it solves higher-order polynomial differential equations too. But let's talk about this anyway.

So suppose you have differential equation like:

$$y''-3y'+2=0$$
A more interesting way of writing this would be:

$$(D-1)(D-2)y=0$$
The fact that you can do such a factoring is a consequence of the fact that polynomials in $D$ form a commutative ring. The idea behind rings and fields and other such objects is to look for a bunch of properties that a familiar set -- like the integers or the real numbers -- satisfies, then drilling those properties down to the basic axioms that imply them, to generalise them to objects other than the integers or real numbers. Differentiation operators are a great example of such a ring.

Now, your first instinct may to look at the factorisation and claim that $(D-1)y=0$ or $(D-2)y=0$. But this isn't right -- you assumed, here, incorrectly, that $(D-1)^{-1}$ and $(D-2)^{-1}$ existed (and that when applied on 0, they give you 0). This is not right, though -- we know there are in fact multiple functions that give 0 when you take $(D-1)$ of them. Which functions, specifically? The functions that are in the null space of $D-1$, i.e. the functions which satisfy:

$$(D-1)f=0$$
And 0 isn't the only such function. Ok, I've been giving you silly tautologies for about three lines now, but the point I'm making is that when you take the inverse operator of $(D-1)$ of both sides, what you really get is:

$$(D-2)y=(D-1)^{-1}0=ce^{x}$$
For arbitrary $c$.

The way to think about this kind of a $c$ is that you don't really have an equal to relation, i.e. an equation, you have an equivalence relation -- the "=" sign there is really abuse of notation. And you're saying that $(D-2)y$ belongs in an equivalence class where all elements are of the form $ce^{x}$ (and your quotient group's "representative element" can be $e^x$. The same applies, for example, for _____ in calculus -- fill in the blank. Well, fill it in.

Anyway, what you now have is a first-order differential equation (or really differential equivalence) in $y$.

$$(D-2)y=c_1e^x$$
But it isn't homogenous. I don't really know how to motivate a solution for a non-homogenous differential equation, really -- all I can say is that because the right-hand-side is an exponential, we just know that we can get some hints as to what $(D-2)^{-1}(ce^x)$ is by applying $(D-2)(ce^x)$ -- and if the right-hand-side isn't an exponential, then you can make it a sum or integral of exponentials, which is what Laplace and Fourier transforms are all about.

In any case, performing $(D-2)$ on $c_1e^x$ gives us $(c_1-2)e^x$, which immediately gives us an example solution, or a particular solution, $(c_1+2)e^x$ -- and all other solutions can be formed by adding linear combinations of the elements of the null space, i.e. solutions to the homogenous equation $(D-2)y=0$. These elements we know to take the form $c_2e^{2x}$.

$$y=(D-2)^{-1}c_1e^x=(c_1+2)e^x+c_2e^{2x}$$
Or transforming arbitrary constants,

$$y=c_1e^x+c_2e^{2x}$$

Use this method to find a general form for the solution to $(D-\alpha_1)(D-\alpha_2)...(D-\alpha_n)y=0$. Formalise our method with induction, and prove this general form with induction.