Download CBrayMath216-4-1

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SPEAKER 1: So we just saw that there's a strong similarity between the structure of solutions to an nth order
linear differential equation and the structure of the solutions to a matrix equation. In both cases, the solutions
form vector spaces. So in both cases, we're talking about linear algebra problems, at least from a certain point
of view. What we're going to find now is that there's another relationship between these kinds of things.
Here we go. Suppose we have an nth order linear differential equation. Note that this is not homogeneous. And
suppose that I have a solution. We're going to choose one and stick with it. I'm going to call this y sub p, p for
particular. This is a particular solution. Pick one, stick with it.
Now suppose, furthermore, that you have a fundamental set of solutions to this associated homogeneous
linear differential equation. By associated, I mean note that the left hand side is exactly the same. The big
difference is that the right hand side is now 0. Of course, it's only for homogeneous linear differential equations
that we know that the solutions form a vector space.
So with these together, we can actually write down all of the solutions to this non-homogeneous linear
differential equation. All of these solutions take the form of whatever that particular solution was that you
picked plus the complete collection of all homogeneous solutions. So keep in mind we had a fundamental set
of solutions, and all of the homogeneous solutions are linear combinations of that fundamental set.
So you see the similarity, I'm sure, between this theorem and a theorem that we saw concerning matrix
equations. And that is the complete collection of all solutions is pick any particular solution, and add to it the
complete collection of homogeneous solutions. So let's see a proof of this result.
We'll start by assuming that we have some function of this form. Let's plug it in and see what L turns out to be
for this function. By linearity we have this. By assumption, L of y1 is 0, because we are assuming that y1 is a
homogeneous solution, so that's 0. Likewise, these terms are all 0. Those all go away.
By assumption, y sub p of our particular solution to that equation, that means that that's equal to that. And all
together, we find that L of y is g of x. And so therefore, everything of this form is a solution.
Now we need to go the other way. Let's suppose that we have a solution, and let's show that it has the desired
form. So assuming that this is a solution, keep in mind that what that means is that L of y is equal to g of x.
Don't forget that by assumption, yp is a solution to that non-homogeneous equation, so L of yp is also g of x.
And therefore, by linearity, y minus yp is a homogeneous solution.
Now, this is a very useful observation. In particular, if y minus yp is a homogeneous solution, recall that all
homogeneous solutions are linear combinations of the fundamental set. And so therefore, we can write that.
And as required, solving for y, we have that it's the particular solution plus a homogeneous solution as
desired.
So this theorem is very similar to this theorem that we saw back in chapter 1. The theorem that we saw back in
chapter 1 is a little bit anti-climactic because, well, we already had a method for solving systems of equations,
even non-homogeneous ones. It was just a row reduction. And so the particular and homogeneous solution
theorem didn't seem to buy us much.
What we're going to find in this context, though, is that finding these homogeneous solutions is going to turn
out to be relatively easy. And there's a whole discussion, and it's not trivial, but it's something that we can do.
We're subsequently going to find that finding a single solution to the non-homogeneous equation is also
something that we can do. But finding one is hard enough. So this is barely all we can do in terms of particular
solutions. This is something that we can do for homogeneous solutions. And so we have, therefore, a way to
find all of the solutions to these differential equations. So we're going to rely on this theorem.
We don't have other alternatives. The method that we use to find this particular solution is only going to yield
one solution usually. So we need this connection relating homogeneous solutions in order to find all of the
solutions. This is a big theorem.
And let's see an example. Let's look at this differential equation here. Now at a glance, it's not hard to confirm
that that's a solution. Take that, plug it in. The equation works.
So we can call this our particular solution. We know from a previous example that these are the homogeneous
solutions. Namely, those are the solutions to this associated homogeneous equation. And keep in mind when I
say associated, I mean that the left-hand sides are the same. In other words, we're talking about the same L
operator.
So knowing as we do a particular solution, knowing as we do all the homogeneous solutions, we know that the
complete set of solutions is the sum of them, particular plus homogeneous.
So I think it's worth every student's time to take a few minutes and go back, look at the statement of the similar
theorem in chapter 1. Look at the proof of that similar theorem. Note that both the statement and the proof are
very similar. Very strong similarity between the kinds of equations that we're talking about here versus the kind
of equations we were talking about back in chapter 1. Very strong connections.