OH BOY who's ready to board the hype train for differential equations!?
*more cricket sounds*
Anyway... in case you're wondering, what we're gonna do today is we're gonna look into homogeneous linear differential equations with constant coefficients, that is, differential equations in the form
(, , , , are arbitrary real constants), and we're gonna talk about their general solutions! (yay)
By the way, if you don't know about homogeneous linear differential equations (really long name I know) and how to solve them, look it up here.
Now, there is this thing with a nice name called "characteristic equation" or "auxiliary equation":
which determines the nature of the general solution of .
In fact, we have this theorem stating that when all roots of equation are distinct, then a general solution for equation will be in the form
where , , , are arbitrary constants (may be real or complex, if there are complex roots, they are also rewritten to the trigonometric form using Euler's formula but that's beside the main point), , , , are the distinct roots of equation .
And if some of the roots are repeated, for example, say that one of the roots is repeated times , then the term in the general solution associated with that repeated root would be in the form
Now two questions might arise from our heads, why exactly must it be in the form ? And where did the part in come from?
There's one explanation that uses the principle of superposition (not the one involving waves!), but here I'm gonna present you an alternative approach to derive and , so hold on tight and wear your seatbelts!
Before we do anything crazy, let's look at something simpler, a second order homogeneous linear DE with constant coefficients:
We will now try to derive a general solution that satisfies .
Instead of "guessing" that the solution would look like the form , let's instead start at its characteristic equation shall we?
This is just a simple quadratic equation, let , be its two roots, then using Vieta's formula, we have
Back to , since , we have
Now, here's the interesting part, notice that if we let , then , the equation above becomes
Would you look at that! What we have now is a separable differential equation, which can be easily solved:
Integrating both sides:
is an arbitrary constant, so we'll replace it with :
Back substituting :
Brilliant! We now have a first order linear differential equation, to solve this, we multiply both sides by its integrating factor :
Case 1: Equation has two distinct roots, i.e.
Relabeling the constants, we have
which is exactly what we wanted, it is a general solution that satisfies , and is in the same form as
Let's now see what happens if the characteristic equation has two equal real roots.
Case 2: Equation has two equal real roots, i.e.
Relabeling the constants and rearranging, we have which is also what we wanted, it is a general solution and it is in the same form as .
Now what about third order homogeneous linear DEs with constant coefficients? Fear not! We're about to find out!
Again, we shall start from its characteristic equation:
Ah, a good-ol' cubic equation, let , , be its three roots, using Vieta's formula, we have
Case 1: Equation has 3 distinct roots
As , from , we have
Substitute , we get
Déjà vu? Hey! That's because we're back to the second order case again! In fact, if we let and (, , are constants, ), we have with , being the roots of its characteristic equation .
Hence we're dealing with a second order homogeneous linear DE with constant coefficients, which we know its general solution, and because ,
This is a first order linear differential equation, multiplying both sides by its integrating factor , we will get
Integrating both sides and multiplying both sides by , we have
Relabeling the constants:
Case 2: Equation has one repeated root, without loss of generality, let's assume that
In this case, all the steps will be identical to case 1 until up to , where integrating both sides of would instead give Multiplying both sides by and relabeling the constants, we have
Case 3: Equation has 3 equal real roots, i.e.
Similar to case 1, but when solving for in would instead give
Multiplying both sides by its integrating factor gives
Integrating both sides:
Multiplying both sides by and relabeling the constants, we have
We have shown that if the roots of a characteristic equation of a second or third order homogeneous linear DE with constant coefficients are all distinct, then its general solution will be in the form , a linear sequence of powers of , which makes sense, since comes up a lot in calculus, it's a special magic number with special properties. It also makes it reasonable to guess that solutions to might be in the form .
Things also get a bit more interesting when there are repeated roots. You might have noticed in and , when there are repeated roots, they cause the exponent of the terms to become zero, causing it to be equal to some constant . When constants are integrated, linear terms of appear, in higher orders, the terms are integrated several more times and become terms, then , , . This is one of the reasons integer powers of appear in when repeated roots are involved.
Of course, the derivations above can be extended to higher orders, an inductive way to prove for all orders might also exist, but this note is already too long and is left as an exercise to the reader :D (I'm so nice).
This might be the longest note I've ever written, so if you made it all the way here, then you are awesome. :)
Toodles, gotta go!