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:

\[(D-1)(D-2)y=0\]

Further Insight: 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 \(c\).

Further Insight: 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.