WEBVTT 00:00:05.000 --> 00:00:11.000 The task for today is to find particular solutions. 00:00:09.000 --> 00:00:15.000 So, let me remind you where we've gotten to. 00:00:12.000 --> 00:00:18.000 We're talking about the second-order equation with 00:00:17.000 --> 00:00:23.000 constant coefficients, which you can think of as 00:00:21.000 --> 00:00:27.000 modeling springs, or simple electrical circuits. 00:00:25.000 --> 00:00:31.000 But, what's different now is that the right-hand side is an 00:00:29.000 --> 00:00:35.000 input which is not zero. So, we are considering, 00:00:35.000 --> 00:00:41.000 I'm going to use x as your book does, keeping to a neutral 00:00:41.000 --> 00:00:47.000 letter. But, again, in the 00:00:43.000 --> 00:00:49.000 applications, and in many of the applications 00:00:48.000 --> 00:00:54.000 at any rate, it wants to be t. But, I make it x. 00:00:54.000 --> 00:01:00.000 So, the independent variable is x, and the problem is, 00:00:59.000 --> 00:01:05.000 remember, that to find a particular solution, 00:01:04.000 --> 00:01:10.000 and the reason why we want to do that is then the general 00:01:10.000 --> 00:01:16.000 solution will be of the form y equals that particular solution 00:01:17.000 --> 00:01:23.000 plus the complementary solution, the general solution to the 00:01:23.000 --> 00:01:29.000 reduced equation, which we can write this way. 00:01:30.000 --> 00:01:36.000 So, all the work depends upon finding out what that yp is, 00:01:34.000 --> 00:01:40.000 and that's what we're going to talk about today, 00:01:37.000 --> 00:01:43.000 or rather, talk about for two weeks. 00:01:40.000 --> 00:01:46.000 But, the point is, not all functions that you 00:01:43.000 --> 00:01:49.000 could write on the right-hand side are equally interesting. 00:01:47.000 --> 00:01:53.000 There's one kind, which is far more interesting, 00:01:50.000 --> 00:01:56.000 but more important in the applications than all the 00:01:54.000 --> 00:02:00.000 others. And, that's the one out of 00:01:56.000 --> 00:02:02.000 which, in fact, as you will see later on this 00:02:00.000 --> 00:02:06.000 week and into next week, an arbitrary function can be 00:02:03.000 --> 00:02:09.000 built out of these simple functions. 00:02:08.000 --> 00:02:14.000 So, the important function is on the right-hand side to be 00:02:13.000 --> 00:02:19.000 able to solve it when it's a simple exponential. 00:02:17.000 --> 00:02:23.000 But, if you allow me to make it a complex exponential, 00:02:22.000 --> 00:02:28.000 so, here are the important right-hand sides where we want, 00:02:27.000 --> 00:02:33.000 we want to be able to do it when it's of the form, 00:02:31.000 --> 00:02:37.000 e to the ax. In general, that will be, 00:02:36.000 --> 00:02:42.000 in most applications, a is not a growing exponential, 00:02:40.000 --> 00:02:46.000 but a decaying exponential. So, typically, 00:02:43.000 --> 00:02:49.000 a is negative. But, it doesn't have to be. 00:02:46.000 --> 00:02:52.000 I'll put it in parentheses, though, often. 00:02:48.000 --> 00:02:54.000 That's not any assumption that I'm going to make today. 00:02:52.000 --> 00:02:58.000 It's just culture. But, we want to be able to do 00:02:56.000 --> 00:03:02.000 it for sine omega x and cosine omega x. 00:02:59.000 --> 00:03:05.000 In other words, 00:03:03.000 --> 00:03:09.000 when the right-hand side is a pure oscillation, 00:03:06.000 --> 00:03:12.000 that's another important type of input both for electrical 00:03:11.000 --> 00:03:17.000 circuits, think alternating current, or the spring systems. 00:03:15.000 --> 00:03:21.000 That's a pure vibration is you're imposing pure vibration 00:03:19.000 --> 00:03:25.000 on the spring-mass-dashpot system, and you want to see how 00:03:24.000 --> 00:03:30.000 it responds to that. Or, you could put them 00:03:27.000 --> 00:03:33.000 together, and make these decaying oscillations. 00:03:32.000 --> 00:03:38.000 So, we could also have something like e to the ax times 00:03:36.000 --> 00:03:42.000 sine omega x, or times cosine omega x. 00:03:41.000 --> 00:03:47.000 Now, the point is, 00:03:45.000 --> 00:03:51.000 all of these together are really just special cases of one 00:03:50.000 --> 00:03:56.000 general thing, exponential, 00:03:52.000 --> 00:03:58.000 if you allow the exponent not to be a real number, 00:03:56.000 --> 00:04:02.000 but to be a complex number. So, they're all special cases 00:04:02.000 --> 00:04:08.000 of e to the, I'll write it alpha x, 00:04:07.000 --> 00:04:13.000 well, why don't we write it (a plus i omega) x, right? 00:04:14.000 --> 00:04:20.000 If omega is zero, 00:04:16.000 --> 00:04:22.000 then I've got this case. If a is zero, 00:04:20.000 --> 00:04:26.000 I got this case separating into its real and imaginary parts. 00:04:26.000 --> 00:04:32.000 And, if neither is zero, I have this case. 00:04:32.000 --> 00:04:38.000 But, I don't want to keep writing (a plus i omega) all the 00:04:36.000 --> 00:04:42.000 time. So, I'm going to write that 00:04:38.000 --> 00:04:44.000 simply as e to the alpha x. 00:04:41.000 --> 00:04:47.000 And, you understand that alpha is a complex number now. 00:04:45.000 --> 00:04:51.000 It doesn't look like a real number. 00:04:48.000 --> 00:04:54.000 Okay, so the complex number. So, the equation we are solving 00:04:52.000 --> 00:04:58.000 is which one? This pretty purple equation. 00:04:55.000 --> 00:05:01.000 And, we are trying to find a particular solution of it. 00:04:59.000 --> 00:05:05.000 And, the special functions we are going to use are these, 00:05:03.000 --> 00:05:09.000 well, this one in particular, e to the alpha x. 00:05:08.000 --> 00:05:14.000 That's going to be our input. 00:05:13.000 --> 00:05:19.000 Now, it turns out this is amazingly easy to do because 00:05:19.000 --> 00:05:25.000 it's an exponential because I write it in exponential form. 00:05:26.000 --> 00:05:32.000 The idea is simply to use a rule which, in fact, 00:05:32.000 --> 00:05:38.000 you know already, the rule of substitution. 00:05:38.000 --> 00:05:44.000 So, I'm going to write the equation in the form, 00:05:43.000 --> 00:05:49.000 so, there it is. It's y double prime plus A y 00:05:48.000 --> 00:05:54.000 prime plus B y equals f of x. 00:05:55.000 --> 00:06:01.000 But, I'm going to think of the left-hand side as the polynomial 00:06:02.000 --> 00:06:08.000 operator, AD plus B. A and B are constants, 00:06:07.000 --> 00:06:13.000 applied to y equals f of x. 00:06:10.000 --> 00:06:16.000 That's the way I write the thing. 00:06:13.000 --> 00:06:19.000 And, this part, I'm going to think of in the 00:06:16.000 --> 00:06:22.000 form. This is p of D, 00:06:18.000 --> 00:06:24.000 a polynomial in D. In fact, it's a simple 00:06:21.000 --> 00:06:27.000 quadratic polynomial. But, most of what I'm going to 00:06:25.000 --> 00:06:31.000 say today would apply equally well if we were a higher order 00:06:29.000 --> 00:06:35.000 polynomial, a polynomial of higher degree. 00:06:34.000 --> 00:06:40.000 And, just to reinforce the idea, I've given you one problem 00:06:38.000 --> 00:06:44.000 in your problem set when p is a polynomial of higher degree. 00:06:42.000 --> 00:06:48.000 I should say, the notes are written for 00:06:44.000 --> 00:06:50.000 general polynomials, not just for quadratic ones. 00:06:48.000 --> 00:06:54.000 I'm simplifying it by leaving it, today, I'll do what's in the 00:06:52.000 --> 00:06:58.000 notes, but I'll do it in the quadratic case to save a little 00:06:56.000 --> 00:07:02.000 time, and because that's the one you will be most concerned with 00:07:00.000 --> 00:07:06.000 in the problems. All right, so p of D y equals f 00:07:05.000 --> 00:07:11.000 of x. And now, there are just a 00:07:09.000 --> 00:07:15.000 couple of basic formulas that we're going to use all the time. 00:07:15.000 --> 00:07:21.000 The first is that if you apply p of D to a complex 00:07:19.000 --> 00:07:25.000 exponential, or a real one, it doesn't matter, 00:07:23.000 --> 00:07:29.000 the answer is you get just what you started with, 00:07:27.000 --> 00:07:33.000 with D substituted by alpha. So, it's p of alpha. 00:07:32.000 --> 00:07:38.000 In other words, put an alpha wherever you saw a 00:07:36.000 --> 00:07:42.000 D in the polynomial. And, what is this? 00:07:40.000 --> 00:07:46.000 Well, this is now just an ordinary complex number, 00:07:43.000 --> 00:07:49.000 and multiply that by what you started with, 00:07:46.000 --> 00:07:52.000 e to the alpha x. 00:07:48.000 --> 00:07:54.000 So, that's a basic formula. It's called in the notes the 00:07:52.000 --> 00:07:58.000 substitution rule because the heart of it is, 00:07:55.000 --> 00:08:01.000 you substitute for the D, you substitute alpha. 00:08:00.000 --> 00:08:06.000 Now, this hardly requires proof. 00:08:02.000 --> 00:08:08.000 But, let's prove it just so you see, to reinforce things and 00:08:07.000 --> 00:08:13.000 make things go a little more slowly to make sure you are on 00:08:12.000 --> 00:08:18.000 board all the time. How would I prove that? 00:08:15.000 --> 00:08:21.000 Well, just calculate it out, what in fact is (D squared plus 00:08:20.000 --> 00:08:26.000 AD plus B) times e to the alpha x. 00:08:25.000 --> 00:08:31.000 Well, it's D squared e to the alpha x by linearity plus AD 00:08:30.000 --> 00:08:36.000 times e to the alpha x plus B times e to the alpha x. 00:08:40.000 --> 00:08:46.000 Well, what are these? What's the derivative of e to 00:08:43.000 --> 00:08:49.000 the alpha x? It's just alpha times e to the 00:08:46.000 --> 00:08:52.000 alpha x. What's a second derivative? 00:08:50.000 --> 00:08:56.000 Well, if you remember from the exam, you can do tenth 00:08:53.000 --> 00:08:59.000 derivatives now. So, the second derivative is 00:08:56.000 --> 00:09:02.000 easy. It's alpha squared times e to 00:08:58.000 --> 00:09:04.000 the alpha x. 00:09:00.000 --> 00:09:06.000 In other words, this law, what I'm saying 00:09:03.000 --> 00:09:09.000 really is that this law is obviously, quote unquote, 00:09:06.000 --> 00:09:12.000 "true." Okay, I'm not even going to put 00:09:10.000 --> 00:09:16.000 it in quotes. It's obviously true for the 00:09:12.000 --> 00:09:18.000 operator, D, and the operator D squared. 00:09:16.000 --> 00:09:22.000 In other words, D of e to the alpha x equals 00:09:19.000 --> 00:09:25.000 alpha times e to the alpha x. 00:09:22.000 --> 00:09:28.000 D squared times e to the alpha 00:09:25.000 --> 00:09:31.000 x equals alpha squared times e to the alpha x. 00:09:30.000 --> 00:09:36.000 And, therefore, 00:09:34.000 --> 00:09:40.000 it's true for linear combinations of these as well by 00:09:37.000 --> 00:09:43.000 linearity. So, therefore, 00:09:39.000 --> 00:09:45.000 also true for p of D. And, in fact, 00:09:42.000 --> 00:09:48.000 so if you calculate it out, what is it? 00:09:45.000 --> 00:09:51.000 This is alpha squared e to the alpha x plus alpha e to the 00:09:50.000 --> 00:09:56.000 alpha x times the 00:09:54.000 --> 00:10:00.000 coefficient plus b times e to the alpha x. 00:10:00.000 --> 00:10:06.000 So, it's in fact exactly this. It's e to the alpha x times 00:10:04.000 --> 00:10:10.000 (alpha squared plus A alpha plus B). 00:10:10.000 --> 00:10:16.000 Now, how are we going to use 00:10:13.000 --> 00:10:19.000 this? Well, the idea is very simple. 00:10:16.000 --> 00:10:22.000 Remember, we're trying to solve this, I should have some 00:10:21.000 --> 00:10:27.000 consistent notation for these equations. 00:10:25.000 --> 00:10:31.000 Purple, I think, will be the right thing here. 00:10:28.000 --> 00:10:34.000 You are solving purple equations. 00:10:33.000 --> 00:10:39.000 The formulas which will solve them will be orange formulas, 00:10:38.000 --> 00:10:44.000 and we will see what we need as we go along. 00:10:42.000 --> 00:10:48.000 So, I would like to just formulate it, 00:10:45.000 --> 00:10:51.000 this solution, the particular solution now. 00:10:49.000 --> 00:10:55.000 I'm going to call it a theorem. It's really too simple to be a 00:10:54.000 --> 00:11:00.000 theorem. On the other hand, 00:10:57.000 --> 00:11:03.000 it's too important not to be a theorem. 00:11:02.000 --> 00:11:08.000 So, let's call it, as I called it in the notes, 00:11:06.000 --> 00:11:12.000 the exponential input theorem, which says it all. 00:11:11.000 --> 00:11:17.000 Theorem says it's important. Exponential input means it's 00:11:17.000 --> 00:11:23.000 taking f of x to be an exponential. 00:11:21.000 --> 00:11:27.000 It's an exponential input, and the theorem tells you what 00:11:27.000 --> 00:11:33.000 the response is. So, for that equation, 00:11:33.000 --> 00:11:39.000 I'm not going to recopy the equation for the purple 00:11:40.000 --> 00:11:46.000 equation, adequately indicated this way. 00:11:46.000 --> 00:11:52.000 There. Now try to take notes. 00:11:51.000 --> 00:11:57.000 For the purple equation, a solution is e to the alpha x. 00:11:59.000 --> 00:12:05.000 Somewhere I neglected to say 00:12:05.000 --> 00:12:11.000 that f of x, all right, so for the purple equals e to 00:12:09.000 --> 00:12:15.000 the alpha x, how about that? 00:12:14.000 --> 00:12:20.000 That equation, y double prime plus a y prime 00:12:17.000 --> 00:12:23.000 plus b y equals e to the alpha x. 00:12:22.000 --> 00:12:28.000 So, here's the exponential 00:12:25.000 --> 00:12:31.000 input. The solution is e to the alpha 00:12:28.000 --> 00:12:34.000 x divided by p of alpha. 00:12:34.000 --> 00:12:40.000 Now, that's a very useful formula. 00:12:36.000 --> 00:12:42.000 In fact, Haynes Miller, who also teaches this course, 00:12:40.000 --> 00:12:46.000 in his notes calls of the most important theorem in the course. 00:12:45.000 --> 00:12:51.000 Well, I don't have to totally agree with him, 00:12:49.000 --> 00:12:55.000 but it's certainly important. It's probably the most 00:12:53.000 --> 00:12:59.000 important theorem for these two weeks, anyway. 00:12:57.000 --> 00:13:03.000 But, you will have others as well. 00:12:59.000 --> 00:13:05.000 Okay, so that's a theorem. The theorem is going green. 00:13:03.000 --> 00:13:09.000 You can tell what they are by their color code. 00:13:09.000 --> 00:13:15.000 Well, in other words, what I've done is simply write 00:13:12.000 --> 00:13:18.000 down the solution for you, write down the particular 00:13:16.000 --> 00:13:22.000 solution. But let's verify it in general. 00:13:19.000 --> 00:13:25.000 So, the proof would be what? Well, I have to substitute it 00:13:24.000 --> 00:13:30.000 into the equation. So, the equation is p of D 00:13:27.000 --> 00:13:33.000 applied to y is equal to alpha x. 00:13:31.000 --> 00:13:37.000 And, I want to know, when I substitute that 00:13:34.000 --> 00:13:40.000 expression in, is it the case that when I plug 00:13:37.000 --> 00:13:43.000 it in, that the right-hand side, I calculate it out, 00:13:41.000 --> 00:13:47.000 apply p of D to it. Is it the case that I get e to 00:13:46.000 --> 00:13:52.000 the alpha x on the right? 00:13:49.000 --> 00:13:55.000 Well, all you have to do is do it. 00:13:51.000 --> 00:13:57.000 What is p of D applied to e to the alpha x divided by p of 00:13:56.000 --> 00:14:02.000 alpha? 00:14:00.000 --> 00:14:06.000 Well, p of D applied to e to the alpha x is p of alpha times 00:14:03.000 --> 00:14:09.000 e to the alpha x. 00:14:07.000 --> 00:14:13.000 That's the substitution rule. What about this guy? 00:14:10.000 --> 00:14:16.000 This guy is a constant, so it just gets dragged along 00:14:13.000 --> 00:14:19.000 because this operator is linear. If this applied to that is 00:14:17.000 --> 00:14:23.000 this, then if I apply it to one half that, I get one half the 00:14:21.000 --> 00:14:27.000 answer, and so on. So, the p of alpha 00:14:24.000 --> 00:14:30.000 is a constant and just gets dragged along. 00:14:27.000 --> 00:14:33.000 And now, they cancel each other, and the answer is, 00:14:30.000 --> 00:14:36.000 indeed, e to the alpha x. 00:14:34.000 --> 00:14:40.000 That's not much of a proof. I hope that to at least half 00:14:39.000 --> 00:14:45.000 this class, you're wondering, yes, but what if Peter had not 00:14:45.000 --> 00:14:51.000 caught the wolf? I mean, what if? 00:14:48.000 --> 00:14:54.000 What if? 00:15:01.000 --> 00:15:07.000 I'm looking stern. Okay, we will take care of it 00:15:04.000 --> 00:15:10.000 in the simplest possible way. We will assume that p of alpha 00:15:10.000 --> 00:15:16.000 is not zero. The case p of alpha is zero 00:15:13.000 --> 00:15:19.000 is, in fact, an extremely important 00:15:17.000 --> 00:15:23.000 case, one that makes the world go 'round, one that contributes 00:15:22.000 --> 00:15:28.000 to all sorts of catastrophes, and they occur first here in 00:15:27.000 --> 00:15:33.000 the solution of differential equations, and that's what 00:15:32.000 --> 00:15:38.000 controls all the catastrophes. But, there's a good side to it, 00:15:37.000 --> 00:15:43.000 too. It also makes a lot of good 00:15:39.000 --> 00:15:45.000 things happen. So, there are no moral 00:15:41.000 --> 00:15:47.000 judgments in mathematics. For the time being, 00:15:44.000 --> 00:15:50.000 let's assume p of alpha is not zero. 00:15:46.000 --> 00:15:52.000 And, that proof is okay because the p of alpha, 00:15:48.000 --> 00:15:54.000 being in the denominator, it's okay to be in the 00:15:51.000 --> 00:15:57.000 denominator if you're not zero. Okay, let's work in a simple 00:15:55.000 --> 00:16:01.000 example. Well, I'm picking the most 00:15:56.000 --> 00:16:02.000 complicated example I can think of. 00:16:00.000 --> 00:16:06.000 Simple examples, I'll leave for your practice 00:16:04.000 --> 00:16:10.000 and for the recitations, can start off with simple 00:16:08.000 --> 00:16:14.000 examples if you are confused by this. 00:16:12.000 --> 00:16:18.000 But, let's solve an equation, find a particular solution. 00:16:17.000 --> 00:16:23.000 So, y double prime plus minus y prime plus 2y is equal to 10 e 00:16:23.000 --> 00:16:29.000 to the minus x sine x. 00:16:29.000 --> 00:16:35.000 Gulp. Okay, so, the input is this 00:16:33.000 --> 00:16:39.000 function, 10 e to the minus x, 00:16:36.000 --> 00:16:42.000 it's a decaying oscillation. You're seeing those already on 00:16:40.000 --> 00:16:46.000 the computer screen if you started your homework, 00:16:44.000 --> 00:16:50.000 if you've done problem one on your homework. 00:16:47.000 --> 00:16:53.000 It's a decaying exponential, and I want to find a particular 00:16:52.000 --> 00:16:58.000 solution. Well, let's find a particular 00:16:55.000 --> 00:17:01.000 and the general solution. Find the general solution. 00:17:00.000 --> 00:17:06.000 Well, the main part of the work is finding the particular 00:17:05.000 --> 00:17:11.000 solution, but let's quickly, the general solution, 00:17:10.000 --> 00:17:16.000 let's find first the complementary part of it, 00:17:14.000 --> 00:17:20.000 in other words, the solution to the homogeneous 00:17:19.000 --> 00:17:25.000 equation. That's D squared minus D plus 00:17:22.000 --> 00:17:28.000 two. No, let's not. 00:17:26.000 --> 00:17:32.000 I don't want to solve messy quadratics. 00:17:31.000 --> 00:17:37.000 Okay, we're going to find a particular solution. 00:17:34.000 --> 00:17:40.000 I thought it was going to come out easy, and then I realized it 00:17:39.000 --> 00:17:45.000 wasn't because I picked the wrong signs. 00:17:43.000 --> 00:17:49.000 Okay, so if you don't like, just change the problem. 00:17:47.000 --> 00:17:53.000 I can do that, but you cannot. 00:17:49.000 --> 00:17:55.000 Don't forget that. So, we want a particular 00:17:53.000 --> 00:17:59.000 solution in our equation. It is this equals that. 00:17:57.000 --> 00:18:03.000 Now, let's complexify it to make this part of a complex 00:18:01.000 --> 00:18:07.000 exponential. So, the complex exponential 00:18:06.000 --> 00:18:12.000 that's relevant is ten times e to the (minus one plus i), 00:18:12.000 --> 00:18:18.000 you see that? x. 00:18:15.000 --> 00:18:21.000 What is this? This is the imaginary part of 00:18:19.000 --> 00:18:25.000 this complex exponential. So, this is imaginary part of 00:18:24.000 --> 00:18:30.000 that guy, e to the negative x times e to the i x, 00:18:29.000 --> 00:18:35.000 and the imaginary part of e to 00:18:33.000 --> 00:18:39.000 the i x. 00:18:39.000 --> 00:18:45.000 The ten, of course, just comes along for the ride. 00:18:42.000 --> 00:18:48.000 Okay, well, now, since this is a complex 00:18:44.000 --> 00:18:50.000 equation, I shouldn't call this y anymore by my notation. 00:18:48.000 --> 00:18:54.000 I like to call it y tilde to indicate that the solution we 00:18:52.000 --> 00:18:58.000 get to this is not going to be the original solution to the 00:18:56.000 --> 00:19:02.000 original problem, but you will have to take the 00:18:59.000 --> 00:19:05.000 imaginary part of it to get it. So, we are looking, 00:19:03.000 --> 00:19:09.000 now, for the complex solution to this complexified equation. 00:19:08.000 --> 00:19:14.000 Okay, what is it? Well, the complex particular 00:19:11.000 --> 00:19:17.000 solution I can write down immediately. 00:19:14.000 --> 00:19:20.000 It is ten, that, of course, just gets dragged 00:19:18.000 --> 00:19:24.000 along by linearity, times e to the (minus one plus 00:19:22.000 --> 00:19:28.000 i) times x. And, it's over this polynomial 00:19:27.000 --> 00:19:33.000 evaluated at this alpha. So, just write it down with, 00:19:31.000 --> 00:19:37.000 have faith. So, what do I get? 00:19:33.000 --> 00:19:39.000 The alpha is minus one plus i. 00:19:36.000 --> 00:19:42.000 I. square that, 00:19:37.000 --> 00:19:43.000 because I'm substituting this alpha into that polynomial. 00:19:41.000 --> 00:19:47.000 The reason I'm doing that is because the formula tells me to 00:19:45.000 --> 00:19:51.000 do it. That's going to be that 00:19:47.000 --> 00:19:53.000 solution. Okay, so it's minus one plus i, 00:19:50.000 --> 00:19:56.000 the quantity squared, minus minus one plus i plus two. 00:19:54.000 --> 00:20:00.000 All I've done is substitute 00:19:58.000 --> 00:20:04.000 minus one plus i for D in that polynomial, 00:20:01.000 --> 00:20:07.000 the quadratic polynomial. And now, all I want is the 00:20:07.000 --> 00:20:13.000 imaginary part of this. The imaginary part of this will 00:20:12.000 --> 00:20:18.000 be the solution to the original problem because this was the 00:20:18.000 --> 00:20:24.000 right hand side with the imaginary part of the 00:20:22.000 --> 00:20:28.000 complexified right hand side. Okay, now, let's make it look a 00:20:28.000 --> 00:20:34.000 little better, yp tilde. 00:20:32.000 --> 00:20:38.000 Clearly, what we have to do something nice to the 00:20:34.000 --> 00:20:40.000 denominator. So, I'll copy the numerator. 00:20:37.000 --> 00:20:43.000 That's e to the minus (one plus i) x 00:20:40.000 --> 00:20:46.000 and how about the denominator? Well, again, 00:20:43.000 --> 00:20:49.000 don't expand things out because it's already this long. 00:20:46.000 --> 00:20:52.000 And, what's the point of making it this long? 00:20:48.000 --> 00:20:54.000 You want to make it as long, right? 00:20:50.000 --> 00:20:56.000 Okay, then there is room here for one real number, 00:20:53.000 --> 00:20:59.000 and another real number times i, there's no more room. 00:20:57.000 --> 00:21:03.000 Okay, what's the real number? Okay, we're looking for the 00:21:02.000 --> 00:21:08.000 real part of this expression. So, just put it in and keep it 00:21:07.000 --> 00:21:13.000 mentally. So, minus one squared: 00:21:09.000 --> 00:21:15.000 that's one, plus i squared, that's minus one. 00:21:13.000 --> 00:21:19.000 One minus one is zero. I can forget about that term. 00:21:18.000 --> 00:21:24.000 The term gives me plus one for the real part, 00:21:21.000 --> 00:21:27.000 plus two. The answer is that the real 00:21:24.000 --> 00:21:30.000 part is three. How about the imaginary part? 00:21:28.000 --> 00:21:34.000 Well, from here, there's negative 2i, 00:21:31.000 --> 00:21:37.000 negative 2i. I'm expanding that out by the 00:21:38.000 --> 00:21:44.000 binomial theorem, or whatever you like to call 00:21:44.000 --> 00:21:50.000 that, minus 2i minus i makes minus 3i. 00:21:51.000 --> 00:21:57.000 Is that right? Minus 2i, minus i, 00:21:56.000 --> 00:22:02.000 minus 3i. So, it is ten thirds, 00:22:00.000 --> 00:22:06.000 and now in the denominator I have one minus i. 00:22:08.000 --> 00:22:14.000 I'll put that in the numerator, make it one plus i, 00:22:12.000 --> 00:22:18.000 but I have to divide by the product of one minus i and its 00:22:17.000 --> 00:22:23.000 complex conjugate. In other words, 00:22:20.000 --> 00:22:26.000 I'm multiplying both top and bottom by one plus i. 00:22:24.000 --> 00:22:30.000 And so, that makes here one squared plus one squared is two. 00:22:30.000 --> 00:22:36.000 And now, what's left is e to the negative x times cosine x 00:22:35.000 --> 00:22:41.000 plus i sine x. 00:22:40.000 --> 00:22:46.000 Now, of that, what we want is just the 00:22:43.000 --> 00:22:49.000 imaginary part. Well, let's see. 00:22:46.000 --> 00:22:52.000 Two goes into ten makes five, so that's five thirds. 00:22:52.000 --> 00:22:58.000 So, we're practically at our solution. 00:22:55.000 --> 00:23:01.000 The solution, then, finally, 00:22:58.000 --> 00:23:04.000 is going to be yp is the imaginary part of yp tilde. 00:23:05.000 --> 00:23:11.000 And, what's that? Well, what's the coefficient 00:23:08.000 --> 00:23:14.000 out front, first of all? It's five thirds, 00:23:12.000 --> 00:23:18.000 so let's pull out to five thirds before we forget it. 00:23:16.000 --> 00:23:22.000 And, we'll pull out the e to the negative x before we forget 00:23:21.000 --> 00:23:27.000 that. And then, the rest is simply a 00:23:24.000 --> 00:23:30.000 question of seeing what's left. Well, it's one. 00:23:28.000 --> 00:23:34.000 I want the imaginary part. So, the imaginary part is going 00:23:33.000 --> 00:23:39.000 to be one times cosine x, and then the other 00:23:37.000 --> 00:23:43.000 imaginary part comes from these two pieces, which is one times 00:23:42.000 --> 00:23:48.000 sine x. And, that should be the 00:23:47.000 --> 00:23:53.000 particular solution. Notice that most of the work is 00:23:52.000 --> 00:23:58.000 not getting this thing. It's turning it into something 00:23:56.000 --> 00:24:02.000 human that you can take the real and imaginary parts of. 00:24:02.000 --> 00:24:08.000 If we don't like this form, you can put it in the other 00:24:07.000 --> 00:24:13.000 form, which many engineers would do almost automatically, 00:24:12.000 --> 00:24:18.000 make it five thirds, e to the negative x, 00:24:16.000 --> 00:24:22.000 and what will that be? Well, you can use the general 00:24:21.000 --> 00:24:27.000 formula if you want. Remember, cosine, 00:24:24.000 --> 00:24:30.000 the two coefficients are one and one, so it's one and one. 00:24:30.000 --> 00:24:36.000 So, this is the square root of two. 00:24:33.000 --> 00:24:39.000 So, it is times, this part makes the square root 00:24:37.000 --> 00:24:43.000 of two times cosine of x minus pi over the angle. 00:24:44.000 --> 00:24:50.000 This is a phi. So, that's pi over four, 00:24:47.000 --> 00:24:53.000 minus pi over four. 00:24:58.000 --> 00:25:04.000 Okay, all right, now let's address the case 00:25:02.000 --> 00:25:08.000 which is going to occupy a lot of the rest of today, 00:25:07.000 --> 00:25:13.000 and in a certain sense, all of next time. 00:25:11.000 --> 00:25:17.000 What happens when p of alpha is zero? 00:25:16.000 --> 00:25:22.000 Well, in order to be able to handle this decently, 00:25:21.000 --> 00:25:27.000 it's necessary to have one more formula, which is very slightly 00:25:28.000 --> 00:25:34.000 more complicated than the substitution rule. 00:25:34.000 --> 00:25:40.000 But, it's the same kind of rule. 00:25:36.000 --> 00:25:42.000 I'm going to call this, or it is called the 00:25:38.000 --> 00:25:44.000 exponential. So, I'm going to first prove a 00:25:41.000 --> 00:25:47.000 formula, which is the analog of that, and then I will prove a 00:25:45.000 --> 00:25:51.000 green formula, which is what to do here if p 00:25:48.000 --> 00:25:54.000 of alpha turns out to be zero. 00:25:51.000 --> 00:25:57.000 But, in order to be able to prove that, we're going to be 00:25:54.000 --> 00:26:00.000 the analog of the orange formula. 00:25:56.000 --> 00:26:02.000 And, the analogue of the orange formula, that tells you how to 00:26:00.000 --> 00:26:06.000 apply p of D to a simple exponential. 00:26:05.000 --> 00:26:11.000 I need a formula which applies p of D to that simple 00:26:11.000 --> 00:26:17.000 exponential times another function. 00:26:14.000 --> 00:26:20.000 Now, I found I got into trouble by continuing to call that 00:26:20.000 --> 00:26:26.000 alpha. So, I'm now going to change the 00:26:24.000 --> 00:26:30.000 name of alpha to change alpha's name to a. 00:26:30.000 --> 00:26:36.000 But, it's still complex. I don't mean it's guaranteed to 00:26:34.000 --> 00:26:40.000 be complex. I mean it's allowed to be 00:26:37.000 --> 00:26:43.000 complex. So, a is now allowed to be a 00:26:40.000 --> 00:26:46.000 complex number. I'm thinking of it, 00:26:43.000 --> 00:26:49.000 in general, as a complex number, okay? 00:26:46.000 --> 00:26:52.000 I hope this doesn't upset you too much, but you know, 00:26:51.000 --> 00:26:57.000 you change x to t's, and y's to x's. 00:26:54.000 --> 00:27:00.000 This is no worse. All right, what we going to do? 00:26:58.000 --> 00:27:04.000 Well, I'm going to use this exponential shift rule, 00:27:02.000 --> 00:27:08.000 I'll call it, exponential shift rule or 00:27:06.000 --> 00:27:12.000 formula or law. That's the substitution rule 00:27:11.000 --> 00:27:17.000 for me. So, this is going to be 00:27:13.000 --> 00:27:19.000 exponential shift law. And, to apply, 00:27:17.000 --> 00:27:23.000 it tells you how to apply the polynomial to not D, 00:27:21.000 --> 00:27:27.000 not just the exponential, but the exponential times some 00:27:25.000 --> 00:27:31.000 function of x. What's that? 00:27:28.000 --> 00:27:34.000 And now, the rule is very simple. 00:27:32.000 --> 00:27:38.000 See, you can understand the difficulty. 00:27:34.000 --> 00:27:40.000 If you try to start differentiating, 00:27:37.000 --> 00:27:43.000 you're going to have to calculate second derivatives of 00:27:40.000 --> 00:27:46.000 the stuff, and God forbid, higher order equations. 00:27:44.000 --> 00:27:50.000 You would have to calculate fourth derivatives, 00:27:47.000 --> 00:27:53.000 fifth derivatives. You barely even want to 00:27:50.000 --> 00:27:56.000 calculate the first derivative. That's okay. 00:27:53.000 --> 00:27:59.000 But, second derivative, do I have to? 00:27:55.000 --> 00:28:01.000 No, not if you know the exponential shift rule, 00:27:58.000 --> 00:28:04.000 which says you can get rid of the e to the ax, 00:28:01.000 --> 00:28:07.000 make it pass to the left of the operator where it's not in any 00:28:06.000 --> 00:28:12.000 position to do any harm any longer, or upset the 00:28:09.000 --> 00:28:15.000 differentiation. And, all you have to do is, 00:28:14.000 --> 00:28:20.000 when it passes over that operator, it changes D to D plus 00:28:18.000 --> 00:28:24.000 a. So, the answer is, 00:28:21.000 --> 00:28:27.000 e to the ax. There, it's passed over. 00:28:24.000 --> 00:28:30.000 But, when it did so, it changed D to D plus a. 00:28:28.000 --> 00:28:34.000 And, what about the u? Well, the u just stayed there. 00:28:32.000 --> 00:28:38.000 Nothing happened to it. Okay, there's our orange 00:28:36.000 --> 00:28:42.000 formula. I guess we better put the thing 00:28:40.000 --> 00:28:46.000 around the whole business. Should I prove that, 00:28:43.000 --> 00:28:49.000 or the proof is quite easy. So, let's do it just again to 00:28:48.000 --> 00:28:54.000 give you a chance to try to see, now, if somebody gives you a 00:28:53.000 --> 00:28:59.000 formula like that, you first stare at it. 00:28:56.000 --> 00:29:02.000 You might try a couple of special cases, 00:28:59.000 --> 00:29:05.000 try it on a function and see if it works, but already, 00:29:03.000 --> 00:29:09.000 you probably don't want to do that. 00:29:08.000 --> 00:29:14.000 I mean, even if you took a function like x here, 00:29:10.000 --> 00:29:16.000 you'd have to do a certain amount of differentiating, 00:29:14.000 --> 00:29:20.000 and some quadratic thing here. You'd calculate and calculate 00:29:17.000 --> 00:29:23.000 away for a little while, and then if you did it 00:29:20.000 --> 00:29:26.000 correctly, the two would in fact turn out to be equal. 00:29:23.000 --> 00:29:29.000 But, you would not necessarily feel any the wiser. 00:29:26.000 --> 00:29:32.000 A better procedure in trying to understand something like this 00:29:30.000 --> 00:29:36.000 is say, well, let's keep the u general. 00:29:34.000 --> 00:29:40.000 Suppose we make D simple. For example, 00:29:36.000 --> 00:29:42.000 well, if D is a constant, of course there's nothing to 00:29:39.000 --> 00:29:45.000 happen because if this is just a constant, both sides of these 00:29:43.000 --> 00:29:49.000 are the same. This doesn't make any sense if 00:29:46.000 --> 00:29:52.000 p doesn't really have a D in it. Well, what's the simplest 00:29:49.000 --> 00:29:55.000 polynomial which would have a D in it? 00:29:52.000 --> 00:29:58.000 Well, D itself. So, let's take a special case. 00:29:54.000 --> 00:30:00.000 p of D equals D, and check the formula in that 00:29:58.000 --> 00:30:04.000 case; see if it works. So, the formula is asking us, 00:30:03.000 --> 00:30:09.000 what is D, that's the p of D, of e to the ax times u? 00:30:08.000 --> 00:30:14.000 I'm not going to put in the 00:30:12.000 --> 00:30:18.000 variable here because it's just a waste of chalk. 00:30:16.000 --> 00:30:22.000 Well, what is that? I know how to calculate that. 00:30:21.000 --> 00:30:27.000 I'll use the product rule. So, it's the derivative. 00:30:25.000 --> 00:30:31.000 I'll tell you what; let's do the other order first. 00:30:30.000 --> 00:30:36.000 So, it's e to the ax times the derivative of u plus the 00:30:35.000 --> 00:30:41.000 derivative of e to the ax, which is a times e to the ax 00:30:40.000 --> 00:30:46.000 times u. 00:30:45.000 --> 00:30:51.000 Do you follow that? This is the product rule. 00:30:48.000 --> 00:30:54.000 It's e to the ax times the derivative of u plus the 00:30:51.000 --> 00:30:57.000 derivative of e to the ax, which is this thing times u, 00:30:55.000 --> 00:31:01.000 the other factor. 00:30:58.000 --> 00:31:04.000 Now, is that right? I want to make it look like 00:31:01.000 --> 00:31:07.000 that. Well, to make it look like that 00:31:04.000 --> 00:31:10.000 I should first factor e to the ax out. 00:31:07.000 --> 00:31:13.000 And now, what's left? Well, if I factor e to the ax 00:31:11.000 --> 00:31:17.000 out, what's left is Du plus au, which is exactly (D plus a) 00:31:15.000 --> 00:31:21.000 operating on u, D u plus a u. 00:31:19.000 --> 00:31:25.000 Well, hey, that's just what the formula said it should be. 00:31:23.000 --> 00:31:29.000 If you make e to the x pass over D, it changes D to 00:31:27.000 --> 00:31:33.000 D plus a. Okay, now here's the main thing 00:31:32.000 --> 00:31:38.000 I want to show you. All right, now, 00:31:34.000 --> 00:31:40.000 well let's try, if this is true, 00:31:37.000 --> 00:31:43.000 also works out for D squared, then the formula is 00:31:41.000 --> 00:31:47.000 clearly true by linearity because an arbitrary p of D 00:31:45.000 --> 00:31:51.000 is just a linear combination with constant 00:31:49.000 --> 00:31:55.000 coefficients of D, D squared, and that constant 00:31:53.000 --> 00:31:59.000 thing, which we agreed there was nothing to prove about. 00:31:57.000 --> 00:32:03.000 Now, hack, you're a hack if you take D squared and start 00:32:01.000 --> 00:32:07.000 calculating the second derivative of this. 00:32:06.000 --> 00:32:12.000 Okay, it's question about hacks. 00:32:08.000 --> 00:32:14.000 I mean, it's just, you haven't learned the right 00:32:12.000 --> 00:32:18.000 thing to do. Okay, that will work, 00:32:14.000 --> 00:32:20.000 but it's not what you want to do. 00:32:17.000 --> 00:32:23.000 Instead, you bootstrap your way up. 00:32:20.000 --> 00:32:26.000 I have already a formula telling me how to handle this. 00:32:24.000 --> 00:32:30.000 And, you can be anything. Look at this not as D squared 00:32:28.000 --> 00:32:34.000 all by itself. Calculate, instead, 00:32:32.000 --> 00:32:38.000 D squared e to the ax times u. 00:32:36.000 --> 00:32:42.000 Think of that as D, the derivative of the 00:32:39.000 --> 00:32:45.000 derivative of e to the a x u. 00:32:42.000 --> 00:32:48.000 In other words, we will do it one step at a 00:32:45.000 --> 00:32:51.000 time. But you see now immediately the 00:32:48.000 --> 00:32:54.000 advantage of this. What's D of e to the a x u? 00:32:51.000 --> 00:32:57.000 Well, I just calculated that. 00:32:55.000 --> 00:33:01.000 Now, don't go back to the beginning. 00:32:57.000 --> 00:33:03.000 Don't go back to here. Use the formula. 00:33:02.000 --> 00:33:08.000 After all, you worked to calculate it, 00:33:04.000 --> 00:33:10.000 or I did. So, it's D of, 00:33:06.000 --> 00:33:12.000 and what's this inside? It's e to the ax times (D plus 00:33:10.000 --> 00:33:16.000 a) times u. Well, that looks like a mess, 00:33:15.000 --> 00:33:21.000 but it isn't because I'm taking D of e to the ax times 00:33:19.000 --> 00:33:25.000 something. And I already know how to take 00:33:23.000 --> 00:33:29.000 D of e to the ax times something. 00:33:25.000 --> 00:33:31.000 It doesn't matter what that something is. 00:33:30.000 --> 00:33:36.000 Here, the something was u. Here, the something is D plus 00:33:35.000 --> 00:33:41.000 (a times u) operating on u. But, the principle is the same, 00:33:40.000 --> 00:33:46.000 and the answer is what? Well, to take D of e to the ax 00:33:46.000 --> 00:33:52.000 times something, you pass the e to the ax over 00:33:50.000 --> 00:33:56.000 the D. That changes D to D plus a. 00:33:53.000 --> 00:33:59.000 And, you apply that to the 00:33:57.000 --> 00:34:03.000 other guy, which is (D plus a) applied to u. 00:34:03.000 --> 00:34:09.000 What's the answer? e to the a x times (D plus a) 00:34:06.000 --> 00:34:12.000 squared u. 00:34:09.000 --> 00:34:15.000 It's just what you would have gotten if you had taken e to the 00:34:14.000 --> 00:34:20.000 e to the x, pass it over, and then changed D to D plus a. 00:34:18.000 --> 00:34:24.000 Now, another advantage to doing 00:34:22.000 --> 00:34:28.000 it this way is you can see that this argument is going to 00:34:26.000 --> 00:34:32.000 generalize to D cubed. In other words, 00:34:31.000 --> 00:34:37.000 you would continue on in the same way by the process of 00:34:36.000 --> 00:34:42.000 mathematical, one word, mathematical, 00:34:39.000 --> 00:34:45.000 begins with an I, induction. 00:34:41.000 --> 00:34:47.000 By induction, you would prove the same 00:34:44.000 --> 00:34:50.000 formula for the nth derivative. If you don't know what 00:34:49.000 --> 00:34:55.000 mathematical induction is, shame on you. 00:34:53.000 --> 00:34:59.000 But it's okay. A lot of you will be able to go 00:34:57.000 --> 00:35:03.000 through life without ever having to learn what it is. 00:35:03.000 --> 00:35:09.000 And, the rest of you will be computer scientists. 00:35:08.000 --> 00:35:14.000 Okay, so that's the idea of this rule. 00:35:13.000 --> 00:35:19.000 Now, we can use it to calculate something. 00:35:18.000 --> 00:35:24.000 Let's see, I'm going to need green for this, 00:35:23.000 --> 00:35:29.000 I guess, for our better formula. 00:35:27.000 --> 00:35:33.000 The formula, now, that tells you what to do 00:35:32.000 --> 00:35:38.000 if p of alpha is zero. 00:35:39.000 --> 00:35:45.000 So, we're trying to solve the equation, D squared plus A D, 00:35:45.000 --> 00:35:51.000 we are trying to find a 00:35:49.000 --> 00:35:55.000 particular solution, e to the ax, 00:35:54.000 --> 00:36:00.000 let's say. Remember, a is complex. 00:35:58.000 --> 00:36:04.000 a could be complex. It doesn't have to be real. 00:36:05.000 --> 00:36:11.000 But, the problem is that p of alpha is zero. 00:36:09.000 --> 00:36:15.000 How do I get a particular solution? 00:36:12.000 --> 00:36:18.000 Well, I will write it down for you. 00:36:14.000 --> 00:36:20.000 So, this is part of that exponential input theorem. 00:36:18.000 --> 00:36:24.000 I think that's the way it is in the notes. 00:36:21.000 --> 00:36:27.000 I gave all the cases together, but I thought pedagogically 00:36:25.000 --> 00:36:31.000 it's probably a little better to do the simplest case first, 00:36:30.000 --> 00:36:36.000 and then build up on the complexity. 00:36:34.000 --> 00:36:40.000 So, what's yp? The answer is yp is e to the a, 00:36:38.000 --> 00:36:44.000 except now you have to multiply 00:36:43.000 --> 00:36:49.000 it by x out front. Where have you done something 00:36:48.000 --> 00:36:54.000 like that before? Yes, don't tell me. 00:36:51.000 --> 00:36:57.000 I know you know. But, what should go in the 00:36:56.000 --> 00:37:02.000 denominator? Clearly not p of alpha. 00:36:59.000 --> 00:37:05.000 What goes in the denominator is 00:37:04.000 --> 00:37:10.000 the derivative. Okay, but what if p prime of 00:37:09.000 --> 00:37:15.000 alpha is zero? Couldn't that happen? 00:37:14.000 --> 00:37:20.000 Yes, it could happen. So, we better make cases. 00:37:18.000 --> 00:37:24.000 This case is, the case where this is okay 00:37:22.000 --> 00:37:28.000 corresponds to the case where we're going to assume that alpha 00:37:27.000 --> 00:37:33.000 is a simple root, is a simple root of the 00:37:31.000 --> 00:37:37.000 polynomial, p. I don't know what to call the 00:37:35.000 --> 00:37:41.000 variable, p of D is okay. A simple zero, 00:37:40.000 --> 00:37:46.000 in other words, it's not double. 00:37:43.000 --> 00:37:49.000 Well, suppose is double. One of the consequences you 00:37:48.000 --> 00:37:54.000 will see just in a second, if it's a simple zero, 00:37:52.000 --> 00:37:58.000 that means this derivative is not going to be zero. 00:37:57.000 --> 00:38:03.000 That's automatic. Yeah, well, suppose it's not as 00:38:01.000 --> 00:38:07.000 simple. Well, suppose is a double root. 00:38:05.000 --> 00:38:11.000 How did a-- How did that get changed to, argh! 00:38:11.000 --> 00:38:17.000 [LAUGHTER] That's not an alpha. Oh, well, yes it is, 00:38:16.000 --> 00:38:22.000 obviously. Change! 00:38:18.000 --> 00:38:24.000 All of you, I want you to change. 00:38:21.000 --> 00:38:27.000 They should have something like in a search key where 00:38:27.000 --> 00:38:33.000 occurrences of alpha have been changed to a with a stroke of a, 00:38:34.000 --> 00:38:40.000 just your thumb. They don't have that for the 00:38:39.000 --> 00:38:45.000 blackboard, unfortunately. Well, too bad, 00:38:42.000 --> 00:38:48.000 for the future. Correctable blackboards. 00:38:46.000 --> 00:38:52.000 Well, what if a is double root? It can't be more than a double 00:38:51.000 --> 00:38:57.000 root because you've only got a quadratic polynomial. 00:38:55.000 --> 00:39:01.000 Quadratic polynomials only have two roots. 00:39:00.000 --> 00:39:06.000 So, the worst that can happen is that both of them are a. 00:39:04.000 --> 00:39:10.000 All right, in that case the formula should be yp is equal 00:39:09.000 --> 00:39:15.000 to, you are now going to need x squared up there times e 00:39:14.000 --> 00:39:20.000 to the ax, and in the denominator what you 00:39:18.000 --> 00:39:24.000 are going to need is the second derivative of, 00:39:22.000 --> 00:39:28.000 evaluated at a. Now, you can guess the way this 00:39:25.000 --> 00:39:31.000 going to go on. For higher degree things, 00:39:29.000 --> 00:39:35.000 if you've got a triple root, you will need here x cubed, 00:39:37.000 --> 00:39:43.000 except you're going to need a factorial there, 00:39:41.000 --> 00:39:47.000 too. So, don't worry about it. 00:39:45.000 --> 00:39:51.000 It's in the notes, but I'm not going to give you 00:39:49.000 --> 00:39:55.000 that for higher roots. I don't even know if I will 00:39:53.000 --> 00:39:59.000 give it to you for double root. Yes, I already did, 00:39:57.000 --> 00:40:03.000 so it's too late. It's too late. 00:40:01.000 --> 00:40:07.000 Okay, so we will make this two formulas according to whether a 00:40:06.000 --> 00:40:12.000 is a single or a double root. Okay, let's prove one of these, 00:40:11.000 --> 00:40:17.000 and all of that will be good enough for my conscience. 00:40:16.000 --> 00:40:22.000 Let's prove the first one. Mostly, it's an exercise in 00:40:21.000 --> 00:40:27.000 using the first exponential shift rule. 00:40:24.000 --> 00:40:30.000 Okay, this will be a first example actually seeing a work 00:40:29.000 --> 00:40:35.000 in practice as opposed to proving it. 00:40:34.000 --> 00:40:40.000 Okay, so what does that thing look like? 00:40:37.000 --> 00:40:43.000 So, what does the polynomial look like, which has a as a 00:40:42.000 --> 00:40:48.000 simple root? So, we're going to try to prove 00:40:46.000 --> 00:40:52.000 the simple root case. So, I'm just going to calculate 00:40:50.000 --> 00:40:56.000 what those guys actually look like. 00:40:53.000 --> 00:40:59.000 What does p of D look like if a is a simple root? 00:41:00.000 --> 00:41:06.000 Well, if it's a simple root, that means it has a factor. 00:41:04.000 --> 00:41:10.000 When it factors, it factors into the product of 00:41:08.000 --> 00:41:14.000 (D minus a) times something which isn't, D minus some other 00:41:13.000 --> 00:41:19.000 root. And, the point is that b is not 00:41:16.000 --> 00:41:22.000 equal to A. The roots are really distinct. 00:41:20.000 --> 00:41:26.000 Okay, what's, then, p prime, 00:41:23.000 --> 00:41:29.000 I'm going to have to calculate p prime of a. 00:41:27.000 --> 00:41:33.000 What is that? Well, let's calculate p prime 00:41:32.000 --> 00:41:38.000 of D first. It is, well, 00:41:34.000 --> 00:41:40.000 by the ordinary product rule, it's the derivative of this 00:41:38.000 --> 00:41:44.000 times, which is one times (D minus a) plus, 00:41:41.000 --> 00:41:47.000 that's one thing plus the same thing on the other side, 00:41:45.000 --> 00:41:51.000 the derivative of this, which is one times (D minus b). 00:41:49.000 --> 00:41:55.000 So, that's p prime. And therefore, 00:41:51.000 --> 00:41:57.000 what's p prime of a? It's nothing but, 00:41:55.000 --> 00:42:01.000 this part is zero, and that's a minus b. 00:41:59.000 --> 00:42:05.000 Of course, this is not zero because it's a simple root. 00:42:02.000 --> 00:42:08.000 And, that's the proof for you if you want, that if the root is 00:42:06.000 --> 00:42:12.000 simple, that p prime of a is guaranteed not to 00:42:10.000 --> 00:42:16.000 be zero. And, you can see, 00:42:12.000 --> 00:42:18.000 it's going to be zero exactly when b equals a, 00:42:15.000 --> 00:42:21.000 and that root occurs twice. But, I'm assuming that didn't 00:42:19.000 --> 00:42:25.000 happen. Okay, then all the rest we have 00:42:22.000 --> 00:42:28.000 to do is calculate, do the calculation. 00:42:24.000 --> 00:42:30.000 So, what I want to prove now is that with this p of D, 00:42:28.000 --> 00:42:34.000 what I'm trying to calculate that p of D times that guy, 00:42:32.000 --> 00:42:38.000 x e to the a x, except I'm going to 00:42:36.000 --> 00:42:42.000 write it as e to the a x times x, guess why, 00:42:39.000 --> 00:42:45.000 divided by p prime of a. 00:42:44.000 --> 00:42:50.000 This is my proposed particular solution. 00:42:47.000 --> 00:42:53.000 So, what I have to do is calculate it, 00:42:50.000 --> 00:42:56.000 and see that it turns out to be, what do I hope it turns out 00:42:54.000 --> 00:43:00.000 to be? What the right hand side of the 00:42:57.000 --> 00:43:03.000 equation, the input? The input is e to the ax. 00:43:01.000 --> 00:43:07.000 If this is true, 00:43:03.000 --> 00:43:09.000 then yp, a particular solution, indeed, nothing will be a 00:43:07.000 --> 00:43:13.000 particular solution. Of course, there could be 00:43:11.000 --> 00:43:17.000 others, but in this game, I only have to find one 00:43:14.000 --> 00:43:20.000 particular solution, and that certainly by far is 00:43:18.000 --> 00:43:24.000 the simple as one you could possibly find. 00:43:21.000 --> 00:43:27.000 So, I have to calculate this. And now, you see why I did the 00:43:25.000 --> 00:43:31.000 exponential shift rule because this is begging to be 00:43:29.000 --> 00:43:35.000 differentiated by something simpler than hack. 00:43:34.000 --> 00:43:40.000 Okay, you can also see why I violated the natural order of 00:43:38.000 --> 00:43:44.000 things and put the e to the ax on the left in order 00:43:43.000 --> 00:43:49.000 that it pass over more easily. So, the answer on the left-hand 00:43:48.000 --> 00:43:54.000 side is e to the ax times p of (D plus a). 00:43:51.000 --> 00:43:57.000 Now, what is (p of D) plus a? Write it in this form. 00:43:56.000 --> 00:44:02.000 It's going to be a minus b. 00:44:00.000 --> 00:44:06.000 So, p of (D plus a) is, change D to D plus a. 00:44:04.000 --> 00:44:10.000 So, the first factor is going 00:44:07.000 --> 00:44:13.000 to be D plus a minus b. 00:44:10.000 --> 00:44:16.000 And, what's the second factor? Change D to D plus a. 00:44:15.000 --> 00:44:21.000 It turns into D. 00:44:16.000 --> 00:44:22.000 All this is the result of taking that p of D, 00:44:21.000 --> 00:44:27.000 and changing D to D plus a. And now, this is to be applied 00:44:25.000 --> 00:44:31.000 to what? Well, e to the a x 00:44:28.000 --> 00:44:34.000 is already passed over. So, what's left is x. 00:44:33.000 --> 00:44:39.000 And, that's to be divided by the constant, 00:44:36.000 --> 00:44:42.000 p prime of a. Now, what does this all come 00:44:41.000 --> 00:44:47.000 out to be? e to the ax, 00:44:44.000 --> 00:44:50.000 what's D applied to x? One, right? 00:44:48.000 --> 00:44:54.000 And now, what's this thing applied to the constant one? 00:44:53.000 --> 00:44:59.000 Well, the D kills it, so it has no effect. 00:44:57.000 --> 00:45:03.000 It makes it zero. The rest just multiplies it by 00:45:02.000 --> 00:45:08.000 a minus b. So, the answer to the top is (a 00:45:07.000 --> 00:45:13.000 minus b) times one. And, the answer to the bottom 00:45:12.000 --> 00:45:18.000 is p prime of a, which I showed you by just 00:45:17.000 --> 00:45:23.000 explicit calculation is a minus b. 00:45:21.000 --> 00:45:27.000 And so the answer is, e to the a x comes 00:45:25.000 --> 00:45:31.000 out right. Now, the other one, 00:45:29.000 --> 00:45:35.000 the other formula comes out the same way. 00:45:31.000 --> 00:45:37.000 I'll leave that as an exercise. Also, I don't dare do it 00:45:34.000 --> 00:45:40.000 because it's much too close to the problem I asked you to do 00:45:38.000 --> 00:45:44.000 for homework. So, let's by way of conclusion, 00:45:41.000 --> 00:45:47.000 I'll do one more simple example, okay? 00:45:43.000 --> 00:45:49.000 And then, you can feel you understand something. 00:45:46.000 --> 00:45:52.000 I'm sort of bothered that I haven't done any examples of 00:45:49.000 --> 00:45:55.000 this more complicated case. So, I'll pick an easy version 00:45:53.000 --> 00:45:59.000 instead of the one that you have in your notes, 00:45:56.000 --> 00:46:02.000 which is the one you have for homework, which is even easier. 00:46:01.000 --> 00:46:07.000 So, this one's epsilon less easy. 00:46:04.000 --> 00:46:10.000 Y double prime minus 3 y prime plus 2 y equals e to the x. 00:46:10.000 --> 00:46:16.000 Okay, notice that one is a 00:46:14.000 --> 00:46:20.000 simple root. The one I'm talking about is 00:46:18.000 --> 00:46:24.000 the a here, which is one. One is a simple root of the 00:46:23.000 --> 00:46:29.000 polynomial D squared minus 3D plus two, 00:46:28.000 --> 00:46:34.000 isn't it? It's a zero. 00:46:32.000 --> 00:46:38.000 Put D equal one and you get one minus three plus 00:46:35.000 --> 00:46:41.000 two equals zero. It's a simple root because 00:46:38.000 --> 00:46:44.000 anybody can see that one is not a double root because you know 00:46:41.000 --> 00:46:47.000 from critical damping, if one were a double root, 00:46:44.000 --> 00:46:50.000 you know just what the polynomial would look like, 00:46:47.000 --> 00:46:53.000 and it wouldn't look like that at all. 00:46:49.000 --> 00:46:55.000 It would not look like D squared minus 3D plus two. 00:46:52.000 --> 00:46:58.000 It would look differently. 00:46:54.000 --> 00:47:00.000 Therefore, that proves that one is a simple root. 00:46:57.000 --> 00:47:03.000 Okay, what's the particular solution, therefore? 00:47:01.000 --> 00:47:07.000 The particular solution is x times e to the x divided by the 00:47:06.000 --> 00:47:12.000 derivative, the derivative evaluated at the point, 00:47:11.000 --> 00:47:17.000 so, what's p prime of D? It is 2D minus three. 00:47:15.000 --> 00:47:21.000 If I evaluate it at the point, one, it is negative one. 00:47:20.000 --> 00:47:26.000 So, if this is to be divided by negative one, 00:47:25.000 --> 00:47:31.000 in other words, it's minus x e to the X. 00:47:30.000 --> 00:47:36.000 And, if you don't believe it, you could plug it in and check 00:47:35.000 --> 00:47:41.000 it out. Okay, I'm letting you out one 00:47:39.000 --> 00:47:45.000 minute early. Remember that. 00:47:41.000 --> 00:47:47.000 I'm trying to pay off the accumulated debt.