# How to derive the solutions of a linear constant-coefficient differential equation

One thing I've hated about differential equations is how I need to guess the form of the solution.
e.g. it's easy to show that solutions to a linear constant-coefficient differential equation such as

$$y''' + 2 y'' + 3 y' + 4 y = 0$$

have the form of (some linear combination of) exponentials:
You can just plug in $y = e^{a x + b}$ and show that the equation is satisfiable.

But I feel there is something wrong if I must find the solutions through guess-and-check.
Yet what I was never told, and what I seem to never be able to find from looking online, is how to derive this fact rigorously, when I don't already have the intuition necessary to guess the form of the solution?

Sitting down and working through some math, I've come up with some nonsense that works quite beautifully:

1. Place the homogeneous equation into the following matrix form: $$\vec{y}\,' = A \vec{y}$$ where, for example, we have $\vec{y} = \begin{bmatrix} y & y' & \ldots \end{bmatrix}^T$

2. Drop the arrows and pretend everything is a scalar: $$y' = Ay$$

3. Separate the, uh, variables: $$\frac{dy}{dx} = A y$$ $$\frac{dy}{y} = A\,dx$$

4. Integrate: $$\int \frac{dy}{y} = \int A\,dx$$

$$\ln y = A x + b$$

$$y = e^{A x + b}$$

1. Profit!

That was fun while it lasted... but how do I derive it properly?

• I'm not sure this is any less rigorous than the same proof for a first-order equation, where the permissibility of separating variables is usually left unjustified...well, OK, here you definitely can't actually divide by $y$. – Kevin Carlson Nov 3 '14 at 5:51
• @KevinCarlson: It's way less rigorous. $dy/y$ is "vector division", for one thing... edit: yeah, you noticed it too. – Mehrdad Nov 3 '14 at 5:52

So, most of the math you've seen up to now is very simple, by mathematical standards. The sorts of problems that come up in algebra, trig, and calculus courses can often be solved just by starting from an equation, applying a set of transformations to both sides of the equation, and arriving at an answer.

Most of math doesn't work that way at all. I mean, consider Euclid's proof that there are infinitely many prime numbers, for instance. Would you ask something like

"Why did I have to guess that looking at $p_1 p_2 \cdots p_n + 1$ and its factorization was the right thing to do? Isn't there some way to just derive this automatically?"

No, of course not. There's nothing nonrigorous about Euclid's proof, and there's nothing nonrigorous about observing that the $k$-th derivative of $e^{ax+b}$ is $a^k e^{ax+b}$, and that taking that together with the fundamental theorem of algebra will give us all the solutions we seek.

• -1 because (1) Euclid's proof does not require guess and check, (2) I never said that the guess-and-check method is nonrigorous. – Mehrdad Nov 3 '14 at 7:24
• Sure it does. It required him to "guess" that the correct number to look at was $p_1 p_2 \cdots p_n + 1$, in precisely the same sense that this requires you to "guess" that the solution is of the form $e^{ax +b}$. – Daniel McLaury Nov 3 '14 at 7:25
• No it does not. There is no guesswork involved there. It simply assumes that the claim is true and then proceeds by showing that he can construct another prime, which implies the claim is wrong -- all of which is very logical and requires zero guesswork. If you don't see the striking contrast between the two methods then I'm afraid you can't answer my question. – Mehrdad Nov 3 '14 at 7:26
• Perhaps you don't really understand the usual solution here. Do you see why the solution to any linear differential equation $c_n y^{(n)} + c_{n-1} y^{(n-1)} + \cdots + c_1 y + c_0 = 0$ is always a linear combination of terms of the form $e^{ax + b}$? – Daniel McLaury Nov 3 '14 at 7:32
• In Euclid's proof, there is guesswork involved in the following sense. He has a list of $n$ prime numbers, and wants to create a number which is divisible by none of them. So he has to "guess" some function of $n$ numbers which will not be divisible by any of them, and then check that his function actually works. Of course this is not how we usually think about this, but it's no different from "guessing" an ansatz to a differential equation and then proving that it works. – Daniel McLaury Nov 3 '14 at 7:36