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.

Limiting cases II: repeated roots of a differential equation

The solution to a polynomial-ish differential equation (the formal name being "linear homogenous time-invariant differential equation") with repeated roots is not completely unintuitive. While it's not immediately obvious where the solution to $(D-rI)^2y(t)=0$

$$y=(c_1+c_2t)e^{rt}$$
comes from, it is pretty clear in the case $r=0$, where $D^2y(t)=0$ is solved by

$$y=c_1+c_2t$$
... so it seems that the linear function comes from integrating twice, or more correctly, inverting the same differential operator twice.

Let's try to derive our desired equation $y=(c_1+c_2t)e^{rt}$ via a limit. It doesn't seem like this would arise in the limit of an equation like $y=c_1e^{r_1t}+c_2e^{r_2t}$, but once again -- this is an arbitrary-constant-problem. Much like how we switched to definite integrals (i.e. fixed the limits/boundary conditions of the integral) before taking the limit in Part 1, we must fix the initial conditions here too.

For those new to this series, here's the reason we switch to an initial conditions approach/co-ordinate system:

Most people have the right idea, that you need to take the solution for non-repeated roots, and take the limit as the roots approach each other. This is correct, but it's a mistake to take the limit of the general solution $c_1e^{r_1t}+c_2e^{r_2t}$, which is what most people try to do when they see this problem, and are then puzzled since it gives you a solution space of the wrong dimension.

This is wrong, because $c_1$ and $c_2$ are arbitrary mathematical labels, and have no reason to stay the same as the roots approach each other. You can, however, take the limit while representing the solution in terms of your initial conditions, because these can stay the same as you change the system.

You can think of this as a physical system where you change the damping and other parameters to create a repeated-roots system as the initial conditions remain the same -- this is a simple process, but if you instead try to ensure $c_1$ and $c_2$ remain the same, you'll run into infinities and undefined stuff.

This is exactly what happens here, there simply isn't a repeated-roots solution with the same $c_1$ and $c_2$ values, but you obviously do have a system/solution with the same initial conditions.

Taken from my answer on Math Stack Exchange.

We consider the differential equation

$$(D-I)(D-rI)y(t)=0$$
And tend $r\to1$. The solution to the equation in general is

$$y(t) = {c_1}{e^t} + {c_2}{e^{rt}}$$
 If we let $y(0) = a,\,\,y'(0) = b$, then it shouldn't be hard to show that the solution we're looking for is

$$y(t)=\frac{ra-b}{r-1}e^t-\frac{a-b}{r-1}e^{rt}$$
This is where we must tend $r\to1$. Doing so is simply algebraic manipulation and a bit of limits:

$$\begin{array}{c}y(t) = \frac{{\left( {ra - b} \right){e^t} - \left( {a - b} \right){e^{rt}}}}{{r - 1}} = \frac{{\left( {ra - b} \right) - \left( {a - b} \right){e^{(r - 1)t}}}}{{r - 1}}{e^t}\\ = \frac{{(r - 1)a + \left( {a - b} \right) - \left( {a - b} \right){e^{(r - 1)t}}}}{{r - 1}}{e^t}\\ = \left[ {a + \left( {a - b} \right)\frac{{1 - {e^{(r - 1)t}}}}{{r - 1}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right)\frac{{{e^{(r - 1)t}} - {e^{0t}}}}{{r - 1}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right){{\left. {\frac{d}{{dx}}\left[ {{e^{xt}}} \right]} \right|}_{x = 0}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right)t} \right]{e^t}\end{array}$$
Which indeed takes the form

$$y(t) = \left( {{c_1} + {c_2}t} \right){e^t}$$
With $c_1,\,\,c_2$ such that $y(0)=a,\,\,y'(0)=b$.

Here's a visualisation of the limit, with varying values of $r$:


And here's an interactive version with a slider for r.