1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,690 continue to offer high-quality educational resources for free. 5 00:00:10,690 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,270 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,270 --> 00:00:18,200 at ocw.mit.edu. 8 00:00:30,595 --> 00:00:32,350 HERBERT GROSS: Hi, today, we're going 9 00:00:32,350 --> 00:00:34,930 to emphasize the role of power series 10 00:00:34,930 --> 00:00:38,290 in the solution of linear differential equations. 11 00:00:38,290 --> 00:00:40,000 And let me just say at the outset 12 00:00:40,000 --> 00:00:43,870 that we have paid no attention to nonlinear differential 13 00:00:43,870 --> 00:00:45,740 equations in this course. 14 00:00:45,740 --> 00:00:49,420 And we won't pay any attention to the nonlinear equation 15 00:00:49,420 --> 00:00:52,750 simply because the nonlinear equation is very, very 16 00:00:52,750 --> 00:00:53,930 difficult to handle. 17 00:00:53,930 --> 00:00:57,190 We usually tackle it only by special cases. 18 00:00:57,190 --> 00:00:59,620 And, even of more importance, most 19 00:00:59,620 --> 00:01:01,630 of the problems that we have to tackle 20 00:01:01,630 --> 00:01:03,940 are at least reasonably well approximated 21 00:01:03,940 --> 00:01:06,380 by linear differential equations. 22 00:01:06,380 --> 00:01:09,130 Now what I'd like to do before we launch into power series 23 00:01:09,130 --> 00:01:11,380 today is to pull together everything 24 00:01:11,380 --> 00:01:14,930 that we've said so far about linear differential equations. 25 00:01:14,930 --> 00:01:17,680 And, rather than to drone on here saying that, 26 00:01:17,680 --> 00:01:20,470 I thought that I would write the summary on the board, 27 00:01:20,470 --> 00:01:22,960 and we could go through this fairly rapidly together. 28 00:01:22,960 --> 00:01:24,940 At any rate, the lecture for today 29 00:01:24,940 --> 00:01:29,470 is power series solutions, but the summary to date is this. 30 00:01:29,470 --> 00:01:33,670 Given the general linear differential equation L of y 31 00:01:33,670 --> 00:01:37,360 equals f of x with non-constant coefficients, 32 00:01:37,360 --> 00:01:41,290 this equation always has a general solution, 33 00:01:41,290 --> 00:01:45,700 provided p, q, and f are continuous. 34 00:01:45,700 --> 00:01:48,190 Not only does the solution exist, 35 00:01:48,190 --> 00:01:53,020 but the solution is always given by y sub h plus y sub p 36 00:01:53,020 --> 00:01:56,980 where y sub h is the general solution of L of y equals 0, 37 00:01:56,980 --> 00:01:59,480 and y sub p is any solution, in other words, 38 00:01:59,480 --> 00:02:03,400 a particular solution, of L of y equals f of x. 39 00:02:03,400 --> 00:02:06,220 And then you see the rest of our study 40 00:02:06,220 --> 00:02:10,990 has been how do you find y sub h and how do you find y sub p. 41 00:02:10,990 --> 00:02:14,530 The point is it's very easy to find y sub h 42 00:02:14,530 --> 00:02:17,140 if L sub y has constant coefficients. 43 00:02:17,140 --> 00:02:19,390 That was the first case that we tackled. 44 00:02:19,390 --> 00:02:22,720 If L of y has constant coefficient, 45 00:02:22,720 --> 00:02:28,510 then y sub h has one of these three forms, depending on what? 46 00:02:28,510 --> 00:02:30,670 This is the case that holds if the roots 47 00:02:30,670 --> 00:02:33,820 of the characteristic equation are real and unequal. 48 00:02:33,820 --> 00:02:38,090 This is the form that holds if the roots are real but equal. 49 00:02:38,090 --> 00:02:41,680 And this is the equation that holds if the reals are non-- 50 00:02:41,680 --> 00:02:44,200 if the roots are non-real where alpha 51 00:02:44,200 --> 00:02:48,790 is the real part of the root, and beta is the imaginary part 52 00:02:48,790 --> 00:02:51,370 of the root, OK? 53 00:02:51,370 --> 00:02:55,330 As far as y sub p was concerned, recall 54 00:02:55,330 --> 00:02:58,300 that the method of undetermined coefficients 55 00:02:58,300 --> 00:03:03,070 yields y sub p when L of y has constant coefficients 56 00:03:03,070 --> 00:03:09,400 and when f of x has the very special form e to the mx or x 57 00:03:09,400 --> 00:03:13,840 to the n or cosine mx or sine mx. 58 00:03:13,840 --> 00:03:17,320 We then turned our attention to variation of parameters, 59 00:03:17,320 --> 00:03:19,660 and the key point about variation of parameters 60 00:03:19,660 --> 00:03:23,210 was that, by use of variation of parameters, 61 00:03:23,210 --> 00:03:29,170 we could always obtain a y sub p when y sub h was known. 62 00:03:29,170 --> 00:03:35,140 By the way, in particular, notice 63 00:03:35,140 --> 00:03:39,520 that, if we are dealing with constant coefficients, then 64 00:03:39,520 --> 00:03:40,810 we do know y sub h. 65 00:03:40,810 --> 00:03:42,920 That's what we were just talking about earlier. 66 00:03:42,920 --> 00:03:45,580 So, in particular, the variation of parameters method 67 00:03:45,580 --> 00:03:49,630 will work whenever we have constant coefficients 68 00:03:49,630 --> 00:03:51,740 because we know y sub h in that case. 69 00:03:51,740 --> 00:03:53,230 But quite in general, you see, even 70 00:03:53,230 --> 00:03:55,460 if the coefficients aren't constant, 71 00:03:55,460 --> 00:03:57,310 variation of parameters works. 72 00:03:57,310 --> 00:04:01,420 And, in fact, once we know y sub h-- 73 00:04:01,420 --> 00:04:05,020 say y sub h is c1 u1 plus c2 u2, then 74 00:04:05,020 --> 00:04:11,200 y sub p is g1 u1 plus g2 u2 where g1 prime and g2 prime 75 00:04:11,200 --> 00:04:14,170 satisfy this pair of equations. 76 00:04:14,170 --> 00:04:17,950 The other important thing about variation of parameters 77 00:04:17,950 --> 00:04:20,529 was that it reduces the order of L of y 78 00:04:20,529 --> 00:04:25,930 equals 0 once one solution of the equation is known. 79 00:04:25,930 --> 00:04:29,280 And that, you see, completes our summary. 80 00:04:29,280 --> 00:04:34,430 What we want to tackle next is what's still lacking. 81 00:04:34,430 --> 00:04:39,280 And that is that the success of the method known as variation 82 00:04:39,280 --> 00:04:42,310 of parameters hinges on the fact that we 83 00:04:42,310 --> 00:04:47,360 have the general solution of the homogeneous equation. 84 00:04:47,360 --> 00:04:49,480 In other words, our next problem is 85 00:04:49,480 --> 00:04:52,270 to find the general solution of L of y 86 00:04:52,270 --> 00:04:56,420 equals 0 when the coefficients are not constant. 87 00:04:56,420 --> 00:04:59,740 In other words, if I can find this general solution of L of y 88 00:04:59,740 --> 00:05:03,610 equals 0 when coefficients are not constant, then, you see, 89 00:05:03,610 --> 00:05:06,430 I can use the method of variation of parameters, which 90 00:05:06,430 --> 00:05:09,940 always yields y sub p once y sub h is known, 91 00:05:09,940 --> 00:05:11,800 to find the particular solution. 92 00:05:11,800 --> 00:05:16,090 Then y sub p plus y sub h will be my general solution. 93 00:05:16,090 --> 00:05:18,550 At any rate, it's into this environment 94 00:05:18,550 --> 00:05:21,700 that we introduce the concept of power series. 95 00:05:21,700 --> 00:05:24,830 And the key theorem in using power series is this. 96 00:05:24,830 --> 00:05:28,480 Let's suppose that p of x and q of x are analytic. 97 00:05:28,480 --> 00:05:31,450 And, by the way, I've use that word for complex 98 00:05:31,450 --> 00:05:32,740 valued functions. 99 00:05:32,740 --> 00:05:35,140 Analytic meant that all of the derivatives existed. 100 00:05:35,140 --> 00:05:39,130 It meant that the function could be expanded or represented 101 00:05:39,130 --> 00:05:40,840 by a convergent power series. 102 00:05:40,840 --> 00:05:43,780 That's the meaning of analytic, even in the real case. 103 00:05:43,780 --> 00:05:46,120 When I say that p and q are analytic 104 00:05:46,120 --> 00:05:49,270 for absolute value of x minus x0 less 105 00:05:49,270 --> 00:05:51,760 than some value R, what I mean is 106 00:05:51,760 --> 00:05:55,360 that p and q can be represented by convergent power 107 00:05:55,360 --> 00:06:00,190 series for all x, which are within R of some fixed point 108 00:06:00,190 --> 00:06:00,940 x0. 109 00:06:00,940 --> 00:06:02,860 But, at any rate, all I'm saying is, 110 00:06:02,860 --> 00:06:05,800 if p and q are analytic in this interval, 111 00:06:05,800 --> 00:06:10,630 then every solution of this equation, every solution which 112 00:06:10,630 --> 00:06:15,430 exists, which is defined at x equals x0, is itself analytic, 113 00:06:15,430 --> 00:06:20,620 at least in the same interval that p and q were analytic. 114 00:06:20,620 --> 00:06:23,470 In other words, what this means is 115 00:06:23,470 --> 00:06:25,240 that, in the first part of our review, 116 00:06:25,240 --> 00:06:28,720 we said, lookit, as long as p, q, and f are continuous, 117 00:06:28,720 --> 00:06:30,950 there will always be a general solution. 118 00:06:30,950 --> 00:06:34,660 Now we're going one step further or maybe several steps further. 119 00:06:34,660 --> 00:06:38,650 We're saying, lookit, as long as p and q are not only 120 00:06:38,650 --> 00:06:41,530 just continuous, but they also happen to be analytic, 121 00:06:41,530 --> 00:06:45,820 we can not only guarantee that there'll be a general solution, 122 00:06:45,820 --> 00:06:48,820 but we can guarantee, more to the point, 123 00:06:48,820 --> 00:06:51,010 that whatever solution there is will 124 00:06:51,010 --> 00:06:54,730 be found in the form of a convergent power series. 125 00:06:54,730 --> 00:06:58,360 And that gives us the step that we need to solve problems 126 00:06:58,360 --> 00:06:59,860 by this technique. 127 00:06:59,860 --> 00:07:02,380 Now what I'm going to do is this. 128 00:07:02,380 --> 00:07:04,510 I am going to start with an example 129 00:07:04,510 --> 00:07:07,990 that we already knew how to solve before. 130 00:07:07,990 --> 00:07:10,240 First of all, notice that the equation y 131 00:07:10,240 --> 00:07:12,760 double prime plus y equals 0, first of all, 132 00:07:12,760 --> 00:07:15,260 has constant coefficients. 133 00:07:15,260 --> 00:07:17,578 And notice that the lead in to today's lesson 134 00:07:17,578 --> 00:07:19,870 was we we're going to tackle situations where we didn't 135 00:07:19,870 --> 00:07:21,400 have constant coefficients. 136 00:07:21,400 --> 00:07:23,750 This does have constant coefficients. 137 00:07:23,750 --> 00:07:30,260 And, even more to the point, solution is known. 138 00:07:30,260 --> 00:07:32,110 In fact, what is the solution here? 139 00:07:32,110 --> 00:07:35,830 It's y equals an arbitrary constant times sine x 140 00:07:35,830 --> 00:07:38,290 plus an arbitrary constant times cosine x. 141 00:07:38,290 --> 00:07:40,210 I picked one with constant coefficients 142 00:07:40,210 --> 00:07:42,190 where the solution was known simply 143 00:07:42,190 --> 00:07:44,860 so that we could illustrate what the method is. 144 00:07:44,860 --> 00:07:47,560 By the way, notice that, in this particular problem, 145 00:07:47,560 --> 00:07:51,310 p of x, which is the coefficient of y prime, is 0. 146 00:07:51,310 --> 00:07:53,920 q of x, the coefficient of y, is 1. 147 00:07:53,920 --> 00:07:56,560 And, certainly, 0 and 1 are analytic functions. 148 00:07:56,560 --> 00:07:58,730 In fact, their power series-- 149 00:07:58,730 --> 00:08:00,550 the power series for 0 is 0. 150 00:08:00,550 --> 00:08:02,500 And the power series for 1 is 1. 151 00:08:02,500 --> 00:08:05,260 So, certainly, our coefficients meet the requirement. 152 00:08:05,260 --> 00:08:08,080 And what we do is we start off saying, OK, 153 00:08:08,080 --> 00:08:11,350 if any solution exists, it must look like a power series. 154 00:08:11,350 --> 00:08:12,420 Let's go find it. 155 00:08:12,420 --> 00:08:14,320 Notice this is a glorified version 156 00:08:14,320 --> 00:08:15,700 of undetermined coefficients. 157 00:08:15,700 --> 00:08:19,180 We say, lookit, let's assume that our solution has the form 158 00:08:19,180 --> 00:08:21,370 of a power series, in other words, summation, 159 00:08:21,370 --> 00:08:25,690 n goes from 0 to infinity, a sub n x to the n where all I have 160 00:08:25,690 --> 00:08:28,810 to know now are the values of the a sub n's. 161 00:08:28,810 --> 00:08:32,140 And, if you don't like the sigma notation here, all we're saying 162 00:08:32,140 --> 00:08:34,539 is we're trying for a solution in the form y 163 00:08:34,539 --> 00:08:37,419 equals a0 plus a1 x plus a2 x squared, 164 00:08:37,419 --> 00:08:42,010 et cetera, where this series is a convergent power series. 165 00:08:42,010 --> 00:08:44,380 And why is convergent power series important? 166 00:08:44,380 --> 00:08:46,780 Well, the answer, recall from part one of our course 167 00:08:46,780 --> 00:08:50,470 that, in the interval of convergence, 168 00:08:50,470 --> 00:08:54,160 the power series behaves precisely like a polynomial. 169 00:08:54,160 --> 00:08:56,710 We can add power series term by term. 170 00:08:56,710 --> 00:08:58,720 We can subtract them term by term. 171 00:08:58,720 --> 00:09:01,030 We can multiply them the way we do polynomials. 172 00:09:01,030 --> 00:09:02,710 We can integrate term by term. 173 00:09:02,710 --> 00:09:04,570 We can differentiate term by term. 174 00:09:04,570 --> 00:09:08,070 We can rearrange the terms the way we wish, et cetera. 175 00:09:08,070 --> 00:09:09,820 In other words, it has all of the niceties 176 00:09:09,820 --> 00:09:12,040 of a finite polynomial expression. 177 00:09:12,040 --> 00:09:14,800 At any rate, what I then do is, because I'm 178 00:09:14,800 --> 00:09:16,810 assuming this is a convergent power series, 179 00:09:16,810 --> 00:09:19,650 I differentiate this thing term by term. 180 00:09:19,650 --> 00:09:22,510 And, if I differentiate this, notice what I'm doing. 181 00:09:22,510 --> 00:09:25,750 I differentiate each term here, but, because I can-- 182 00:09:25,750 --> 00:09:28,900 because I can differentiate term by term, 183 00:09:28,900 --> 00:09:31,150 a very nice generic way of doing this 184 00:09:31,150 --> 00:09:33,760 is notice that this is the n-th term. 185 00:09:33,760 --> 00:09:35,720 The derivative of this term would be what? 186 00:09:35,720 --> 00:09:40,822 n an x to the n minus 1, n an x to the n minus 1. 187 00:09:40,822 --> 00:09:43,060 Notice that, every time you differentiate, 188 00:09:43,060 --> 00:09:44,260 a term drops out. 189 00:09:44,260 --> 00:09:47,350 Like, when I differentiate this, the a0 term drops out. 190 00:09:47,350 --> 00:09:50,290 When I differentiate this, the a1 one term drops out. 191 00:09:50,290 --> 00:09:55,120 So what I do is I differentiate the general expression 192 00:09:55,120 --> 00:09:58,870 inside the sigma sign and then start the index of summation 193 00:09:58,870 --> 00:09:59,860 one up further. 194 00:09:59,860 --> 00:10:02,120 In other words, instead of going from 0 to infinity, 195 00:10:02,120 --> 00:10:03,850 I now go from 1 to infinity. 196 00:10:03,850 --> 00:10:06,340 In a similar way, y double prime, 197 00:10:06,340 --> 00:10:08,410 I differentiate this term by term. 198 00:10:08,410 --> 00:10:14,020 That's n times n minus 1 times a sub n times x to the n minus 2. 199 00:10:14,020 --> 00:10:16,910 And now my sum goes from 2 to infinity. 200 00:10:16,910 --> 00:10:18,790 That's what I've written over here. 201 00:10:18,790 --> 00:10:22,000 Now I'm trying to find out what the a's are equal to. 202 00:10:22,000 --> 00:10:23,960 I could actually call these y sub p's. 203 00:10:23,960 --> 00:10:26,590 See, I'm looking for a particular solution perhaps, 204 00:10:26,590 --> 00:10:29,473 not particular in the sense of a solution, but this is a trial. 205 00:10:29,473 --> 00:10:31,140 Maybe it should have been a T over here, 206 00:10:31,140 --> 00:10:34,760 y sub T. I'm looking for a trial solution 207 00:10:34,760 --> 00:10:36,590 over here, a trial solution. 208 00:10:36,590 --> 00:10:40,340 I take yT double prime, yT, plug it in here, 209 00:10:40,340 --> 00:10:41,630 and see what this means. 210 00:10:41,630 --> 00:10:46,100 And, simply replacing y double prime by this and y by this, 211 00:10:46,100 --> 00:10:48,500 I see that this is the equation that 212 00:10:48,500 --> 00:10:50,600 must be satisfied identically. 213 00:10:50,600 --> 00:10:53,600 Now the point is the only way a power series can be identically 214 00:10:53,600 --> 00:10:56,840 equal to 0 is if coefficient by coefficient 215 00:10:56,840 --> 00:10:58,520 the power series is 0. 216 00:10:58,520 --> 00:11:00,800 Now what has me bugged a little bit here 217 00:11:00,800 --> 00:11:03,110 is that the exponents don't match up. 218 00:11:03,110 --> 00:11:06,800 You see, notice that the general term here has an exponent n, 219 00:11:06,800 --> 00:11:10,460 and the general term here has an exponent n minus 2. 220 00:11:10,460 --> 00:11:12,770 A very nice trick that I can use here 221 00:11:12,770 --> 00:11:16,900 is I say, you know, why don't I jack up n by 2 222 00:11:16,900 --> 00:11:18,490 every place I see it? 223 00:11:18,490 --> 00:11:21,500 In other words, let me replace n by n plus 2. 224 00:11:21,500 --> 00:11:23,510 I'll jack it up by 2. 225 00:11:23,510 --> 00:11:25,440 And, to compensate for that, I'll 226 00:11:25,440 --> 00:11:27,530 lower the summation of index-- 227 00:11:27,530 --> 00:11:30,713 the index summation by 2 every place I see it. 228 00:11:30,713 --> 00:11:32,630 Now let me show you what that means so that we 229 00:11:32,630 --> 00:11:34,250 don't get too confused on that. 230 00:11:34,250 --> 00:11:37,130 Let's suppose I had written down here summation n goes 231 00:11:37,130 --> 00:11:39,320 from 0 to infinity a sub n. 232 00:11:39,320 --> 00:11:40,250 This means what? 233 00:11:40,250 --> 00:11:44,360 a0 plus a1 plus a2, et cetera. 234 00:11:44,360 --> 00:11:46,310 Suppose, for some reason, I would 235 00:11:46,310 --> 00:11:49,580 like to either lower or raise the index here. 236 00:11:49,580 --> 00:11:51,525 Let's suppose I want to lower the index. 237 00:11:51,525 --> 00:11:52,940 In other words, I want each-- 238 00:11:52,940 --> 00:11:57,130 I want the subscript here to be k less than what it was before. 239 00:11:57,130 --> 00:11:58,880 In other words, I'm going to replace a sub 240 00:11:58,880 --> 00:12:02,000 n by a sub n minus k. 241 00:12:02,000 --> 00:12:06,350 I claim that all I have to do is start my counting process 242 00:12:06,350 --> 00:12:09,560 at n equals k now, instead of n equals 0. 243 00:12:09,560 --> 00:12:11,540 In fact, look what this says. 244 00:12:11,540 --> 00:12:14,720 When n is k, n minus k is 0. 245 00:12:14,720 --> 00:12:17,010 So the first term here is a0. 246 00:12:17,010 --> 00:12:21,150 The next term would be k plus 1. k plus 1 minus k is 1. 247 00:12:21,150 --> 00:12:23,570 So the next term would be a1, et cetera. 248 00:12:23,570 --> 00:12:27,060 And this is just another way of saying this. 249 00:12:27,060 --> 00:12:29,780 So the idea is, if I want to get this 250 00:12:29,780 --> 00:12:32,600 to be an n, that means I have to raise everything 251 00:12:32,600 --> 00:12:35,330 by 2 inside the summation sign. 252 00:12:35,330 --> 00:12:39,330 So I will lower everything by 2 outside the summation sign. 253 00:12:39,330 --> 00:12:41,330 Now, to do that, notice what that's going to do. 254 00:12:41,330 --> 00:12:45,900 My sum here will now go from 0 to infinity. 255 00:12:45,900 --> 00:12:48,630 This will become n plus 2. 256 00:12:48,630 --> 00:12:50,110 This will become n plus 1. 257 00:12:50,110 --> 00:12:51,740 You see, I'm raising it by 2. 258 00:12:51,740 --> 00:12:54,900 Every place I'm seeing an n, I'm replacing it by n plus 2. 259 00:12:54,900 --> 00:12:58,970 This becomes a sub n plus 2, and this will become n. 260 00:12:58,970 --> 00:13:03,260 In summary, then this expression is the same as summation, 261 00:13:03,260 --> 00:13:07,700 n goes from 0 to infinity, n plus 2 times n plus 1 a sub n 262 00:13:07,700 --> 00:13:10,940 plus 2 x to the n plus summation, n goes from 0 263 00:13:10,940 --> 00:13:13,100 to infinity, an x to the n. 264 00:13:13,100 --> 00:13:14,870 And that must be identically zero. 265 00:13:14,870 --> 00:13:18,180 Now the point is that the exponents now line up. 266 00:13:18,180 --> 00:13:21,680 I'm starting the sum in both series at the same place. 267 00:13:21,680 --> 00:13:23,750 And, consequently, because these are 268 00:13:23,750 --> 00:13:25,970 assumed to be convergent power series, 269 00:13:25,970 --> 00:13:28,160 I can add them term by term. 270 00:13:28,160 --> 00:13:30,020 If I add them term by term, that means 271 00:13:30,020 --> 00:13:33,230 I come inside the summation sign here. 272 00:13:33,230 --> 00:13:35,210 x to the n is a common factor. 273 00:13:35,210 --> 00:13:36,780 I factor that out. 274 00:13:36,780 --> 00:13:38,000 And what's left is what? 275 00:13:38,000 --> 00:13:42,560 n plus 2 times n plus 1 times a sub n plus 2 276 00:13:42,560 --> 00:13:46,220 plus a sub n, that's this expression here. 277 00:13:46,220 --> 00:13:49,190 And this is what must be identically zero. 278 00:13:49,190 --> 00:13:52,420 Now the point is that the only way that a convergent power 279 00:13:52,420 --> 00:13:56,420 series can be identically zero is if each coefficient is 0. 280 00:13:56,420 --> 00:13:59,480 And that's how I now handle the undetermined coefficients. 281 00:13:59,480 --> 00:14:01,640 I come to the conclusion that, because this 282 00:14:01,640 --> 00:14:05,330 is to be identically zero, every one of these terms 283 00:14:05,330 --> 00:14:08,090 must be itself 0. 284 00:14:08,090 --> 00:14:12,260 Setting this equal to zero and solving for a sub n plus 2 285 00:14:12,260 --> 00:14:16,640 in terms of a sub n, I find that a sub n plus 2 286 00:14:16,640 --> 00:14:22,410 must be minus an over n plus 2 times n plus 1. 287 00:14:22,410 --> 00:14:27,260 And what this tells me is that I now know what a term looks 288 00:14:27,260 --> 00:14:31,100 like as soon as I know the one that comes two before it. 289 00:14:31,100 --> 00:14:35,900 You see, notice that this tells me how to find a sub n plus 2 290 00:14:35,900 --> 00:14:39,830 once I happen to know what a sub n is. 291 00:14:39,830 --> 00:14:42,500 Now, lookit, before I can get to two 292 00:14:42,500 --> 00:14:44,480 away from a particular subscript, 293 00:14:44,480 --> 00:14:47,180 I must be at least at the second subscript. 294 00:14:47,180 --> 00:14:50,360 In other words, it seems that this recipe will tell me 295 00:14:50,360 --> 00:14:53,750 how to find a sub 2 in terms of a sub 0, 296 00:14:53,750 --> 00:14:56,810 how to find a sub 4 in terms of a sub 2, 297 00:14:56,810 --> 00:14:59,660 how to find a sub 3 in terms of a sub 1, 298 00:14:59,660 --> 00:15:04,220 but it gives me no hold on what a sub 0 and a sub 1 must be. 299 00:15:04,220 --> 00:15:06,950 So I say, OK, what I will do is I 300 00:15:06,950 --> 00:15:09,800 will use this recipe, recognizing 301 00:15:09,800 --> 00:15:15,530 that I am free to pick a0 and a1 completely arbitrarily. 302 00:15:15,530 --> 00:15:17,180 Now look at what happens here. 303 00:15:17,180 --> 00:15:22,280 If I pick a0 and a1 arbitrarily, what does the recipe tell me? 304 00:15:22,280 --> 00:15:25,370 If I pick n to be 0, the recipe says what? 305 00:15:25,370 --> 00:15:30,340 a sub n plus 2, a sub 2, is minus a0 over n 306 00:15:30,340 --> 00:15:31,960 plus 2 times n plus 1. 307 00:15:31,960 --> 00:15:33,970 That's just 2 times 1. 308 00:15:33,970 --> 00:15:38,140 a sub 2 is just minus a0 over 2 factorial. 309 00:15:38,140 --> 00:15:40,690 In a similar way, the recipe tells me 310 00:15:40,690 --> 00:15:45,850 that a sub 4 is minus a sub 2 over 4 times 3. 311 00:15:45,850 --> 00:15:48,340 You see, in this case, I take n to be 2. 312 00:15:48,340 --> 00:15:50,530 So n plus 2 is 4. 313 00:15:50,530 --> 00:15:56,290 And, if I now say a sub n plus 2 is minus a sub n over n 314 00:15:56,290 --> 00:16:01,750 plus 2 times n plus 1, this just says that a4 is minus a sub 2 315 00:16:01,750 --> 00:16:03,370 over 4 times 3. 316 00:16:03,370 --> 00:16:06,050 I know what a sub 2 is from here. 317 00:16:06,050 --> 00:16:09,310 So I can now express a sub 2 in terms of a sub 0. 318 00:16:09,310 --> 00:16:12,130 In fact, it now turns out that a sub 4 319 00:16:12,130 --> 00:16:15,270 is a sub 0 over 4 factorially-- 320 00:16:15,270 --> 00:16:16,510 4 factorial. 321 00:16:16,510 --> 00:16:22,420 Similarly, a sub 6 is minus a sub 4 over 6 times 5. 322 00:16:22,420 --> 00:16:25,690 a sub 4 is a0 over 4 factorial. 323 00:16:25,690 --> 00:16:29,260 So, substituting that value of a4 in here, 324 00:16:29,260 --> 00:16:33,160 I get that a6 is minus a0 over 6 factorial. 325 00:16:33,160 --> 00:16:35,070 And, without beating this thing to death, 326 00:16:35,070 --> 00:16:40,030 notice that I can find every even subscript of a, a2, a4, 327 00:16:40,030 --> 00:16:45,280 a6, a8, et cetera, in terms of a0 times something. 328 00:16:45,280 --> 00:16:47,020 In fact, I think you can see what's 329 00:16:47,020 --> 00:16:48,730 starting to develop over here. 330 00:16:48,730 --> 00:16:55,210 In a similar way, picking n to be 1 so that n plus 2 is 3, 331 00:16:55,210 --> 00:16:59,830 the recipe a sub n plus 2 equals minus a sub n over n 332 00:16:59,830 --> 00:17:05,650 plus 2 times n plus 1 says that a3 is minus a1 over 3 times 2. 333 00:17:05,650 --> 00:17:08,440 That's minus a1 over 3 factorial. 334 00:17:08,440 --> 00:17:13,270 Similarly, a5 is minus a3 over 5 times 4. 335 00:17:13,270 --> 00:17:16,869 Noticing that a3 is minus a1 over 3 factorial, 336 00:17:16,869 --> 00:17:21,190 I see that a5 is a1 over 5 factorial. 337 00:17:21,190 --> 00:17:23,410 And, without going on further here, 338 00:17:23,410 --> 00:17:28,329 notice that a3, a5, a7, a9, a11, et cetera, 339 00:17:28,329 --> 00:17:31,263 will all be expressible in terms of a1. 340 00:17:31,263 --> 00:17:33,430 And, in fact, I think you see what's happening here, 341 00:17:33,430 --> 00:17:36,610 that the terms will alternate in sign, 342 00:17:36,610 --> 00:17:38,800 and the coefficients will be things like what? 343 00:17:38,800 --> 00:17:43,280 1, 1 over 3 factorial, 1 over 5 factorial, 1 over 7 factorial, 344 00:17:43,280 --> 00:17:43,780 et cetera. 345 00:17:43,780 --> 00:17:45,400 But I'll leave those details to you 346 00:17:45,400 --> 00:17:48,580 because I think that's the easy part to meditate upon. 347 00:17:48,580 --> 00:17:50,560 But now what we're saying is I now 348 00:17:50,560 --> 00:17:52,120 know what all these coefficients look 349 00:17:52,120 --> 00:17:55,140 like in terms of a0 and a1. 350 00:17:55,140 --> 00:17:56,390 Remember what I'm looking for. 351 00:17:56,390 --> 00:17:58,570 I'm looking for a power series solution 352 00:17:58,570 --> 00:18:02,830 that's y equals a 0 plus a1 x plus a2 x squared, et cetera. 353 00:18:02,830 --> 00:18:05,950 We're assuming that this is a convergent power series. 354 00:18:05,950 --> 00:18:09,070 That means I can group the terms any way that I want. 355 00:18:09,070 --> 00:18:11,740 In particular, since all of the even subscripts 356 00:18:11,740 --> 00:18:14,320 seem to be expressible all in terms of a0, 357 00:18:14,320 --> 00:18:16,990 and all the odd subscripts seem to be expressible in terms 358 00:18:16,990 --> 00:18:20,410 of a1, let me group the terms with even powers 359 00:18:20,410 --> 00:18:23,350 together and the terms with odd powers together. 360 00:18:23,350 --> 00:18:26,350 And now, remembering what a2 and a4 are-- remember, 361 00:18:26,350 --> 00:18:29,180 a2 was minus x squared-- 362 00:18:29,180 --> 00:18:32,320 was minus a0 over 2 factorial. 363 00:18:32,320 --> 00:18:35,740 a4 was a0 over 4 factorial, et cetera. 364 00:18:35,740 --> 00:18:38,620 I now factor out a0 from here. 365 00:18:38,620 --> 00:18:42,240 And, substituting in the values of a2, a4, a6, et 366 00:18:42,240 --> 00:18:44,290 cetera, that we found from before, 367 00:18:44,290 --> 00:18:48,340 this tells me that the coefficient that multiplies a0 368 00:18:48,340 --> 00:18:51,160 is 1 minus x squared over 2 factorial plus x 369 00:18:51,160 --> 00:18:53,830 to the fourth over 4 factorial, et cetera. 370 00:18:53,830 --> 00:18:59,320 And, similarly, since a3 is minus a1 over 3 factorial, 371 00:18:59,320 --> 00:19:03,040 and a5 was a1 over 5 factorial, et cetera, 372 00:19:03,040 --> 00:19:06,280 I can now express all of these in terms of a1. 373 00:19:06,280 --> 00:19:07,870 I factor out a1. 374 00:19:07,870 --> 00:19:11,480 And what's left this x minus x cubed over 3 factorial plus x 375 00:19:11,480 --> 00:19:13,870 to the fifth over 5 factorial, et cetera. 376 00:19:13,870 --> 00:19:18,490 And now it appears that my solution is this plus this. 377 00:19:18,490 --> 00:19:22,720 Keep in mind that a0 and a1 are arbitrary constants. 378 00:19:22,720 --> 00:19:24,760 By the ratio test, even if I didn't 379 00:19:24,760 --> 00:19:26,860 know what these series represented, 380 00:19:26,860 --> 00:19:31,930 I can show that these series converge absolutely 381 00:19:31,930 --> 00:19:35,560 for all finite values of x, but that's not too crucial here. 382 00:19:35,560 --> 00:19:39,130 The important point is that, even though it may be implicit, 383 00:19:39,130 --> 00:19:44,890 this does name some function of x, whatever it happens to be. 384 00:19:44,890 --> 00:19:48,880 This names some function of x, whatever it happens to be. 385 00:19:48,880 --> 00:19:54,850 I can find particular solutions by choosing a0 and a1 386 00:19:54,850 --> 00:19:58,120 to be certain fixed constants that I want them to be. 387 00:19:58,120 --> 00:20:00,640 These two power series are not constant multiples 388 00:20:00,640 --> 00:20:02,950 of one another, as you can see by a glance. 389 00:20:02,950 --> 00:20:05,950 Therefore, these two solutions are linearly independent. 390 00:20:05,950 --> 00:20:09,760 And, consequently, I now have a general solution here, 391 00:20:09,760 --> 00:20:12,040 even though this may be awkward for me. 392 00:20:12,040 --> 00:20:15,280 Well, the reason that I picked constant coefficients and y 393 00:20:15,280 --> 00:20:17,740 double prime plus y equals 0 was that I 394 00:20:17,740 --> 00:20:20,080 wanted our first experience to be pleasant, 395 00:20:20,080 --> 00:20:22,600 at least in the sense that we would get a feeling for what 396 00:20:22,600 --> 00:20:23,710 the answer meant. 397 00:20:23,710 --> 00:20:25,510 Keep in mind that, if I had never 398 00:20:25,510 --> 00:20:28,870 heard of sine x and cosine x, this solution here 399 00:20:28,870 --> 00:20:30,230 would make sense. 400 00:20:30,230 --> 00:20:32,860 But, if I happen to recall that the power series 401 00:20:32,860 --> 00:20:37,750 expansion for cosine x is precisely this, 402 00:20:37,750 --> 00:20:39,730 it just happens, you see, as an afterthought 403 00:20:39,730 --> 00:20:41,230 that this is cosine x. 404 00:20:41,230 --> 00:20:43,600 This also happens to be sine x. 405 00:20:43,600 --> 00:20:47,080 Notice now what this tells me is that the solution to I've found 406 00:20:47,080 --> 00:20:51,310 is a0 cosine x plus a1 sine x. 407 00:20:51,310 --> 00:20:53,620 And that just happens to agree with the result 408 00:20:53,620 --> 00:20:55,103 that I already knew. 409 00:20:55,103 --> 00:20:56,770 Now, rather than have you feel that I've 410 00:20:56,770 --> 00:20:58,540 wasted your time by giving you something 411 00:20:58,540 --> 00:21:01,690 that you've already known, let me go one step further 412 00:21:01,690 --> 00:21:03,910 and give you the same kind of a problem. 413 00:21:03,910 --> 00:21:06,250 Only now, I'm going to give you one 414 00:21:06,250 --> 00:21:08,620 that you couldn't solve before and, in fact, 415 00:21:08,620 --> 00:21:10,240 I don't think you could solve now 416 00:21:10,240 --> 00:21:11,830 if you don't use power series. 417 00:21:11,830 --> 00:21:15,130 And the example I have in mind is almost the same as the one 418 00:21:15,130 --> 00:21:16,060 we just did. 419 00:21:16,060 --> 00:21:18,760 Instead of y double prime plus y equals 0, 420 00:21:18,760 --> 00:21:22,870 I pick y double prime plus xy equals 0. 421 00:21:22,870 --> 00:21:25,060 And now you see I'm back-- 422 00:21:25,060 --> 00:21:28,870 not back, but now I have arrived at the case 423 00:21:28,870 --> 00:21:31,720 of non-constant coefficients. 424 00:21:31,720 --> 00:21:35,360 You see the coefficient of y is x, and that's not a constant. 425 00:21:35,360 --> 00:21:38,650 So you see now I'm at a bona fide problem where I couldn't 426 00:21:38,650 --> 00:21:41,770 find y sub h in the trivial way that I 427 00:21:41,770 --> 00:21:44,590 could if the coefficients all happened to be constant. 428 00:21:44,590 --> 00:21:46,570 Now the way I tackle this is precisely 429 00:21:46,570 --> 00:21:49,210 the same way as before. 430 00:21:49,210 --> 00:21:53,680 What I do is I again try for a solution in the form y 431 00:21:53,680 --> 00:21:56,940 equals summation, n goes from 0 to infinity, a sub n 432 00:21:56,940 --> 00:21:57,980 x to the n. 433 00:21:57,980 --> 00:22:00,400 And what I must do now is to determine 434 00:22:00,400 --> 00:22:02,200 what the a sub n's are. 435 00:22:02,200 --> 00:22:06,100 And, again, as before, I find y prime the same mechanical way 436 00:22:06,100 --> 00:22:06,910 as before. 437 00:22:06,910 --> 00:22:10,240 I differentiate term by term, meaning, generically, I 438 00:22:10,240 --> 00:22:12,140 differentiate this to get this. 439 00:22:12,140 --> 00:22:13,990 And, similarly, to get y double prime, 440 00:22:13,990 --> 00:22:17,680 I differentiate this term to get this one, OK? 441 00:22:17,680 --> 00:22:21,830 And now what I do, in terms of my so-called trial solution-- 442 00:22:21,830 --> 00:22:24,010 see, remember, these are trial solutions-- 443 00:22:24,010 --> 00:22:27,580 I'm going to try to feed these back into here 444 00:22:27,580 --> 00:22:30,780 and see if that will tell me what the coefficients a0, a1, 445 00:22:30,780 --> 00:22:33,040 a2, a3, et cetera have to look like. 446 00:22:33,040 --> 00:22:35,580 Now, if I do that, you see, I get what? 447 00:22:35,580 --> 00:22:39,130 y double prime is just this term here. 448 00:22:39,130 --> 00:22:41,560 And now I want x times y. 449 00:22:41,560 --> 00:22:46,270 That means I have to take this y and multiply it by x. 450 00:22:46,270 --> 00:22:49,570 Multiplying by x on the outside, since we're 451 00:22:49,570 --> 00:22:51,880 assuming that this is a convergent series, 452 00:22:51,880 --> 00:22:54,280 that means I can multiply term by term. 453 00:22:54,280 --> 00:22:56,410 Therefore, I can bring the x inside, 454 00:22:56,410 --> 00:23:00,100 and that makes this x to the n plus 1, rather than x to the n. 455 00:23:00,100 --> 00:23:02,530 And so I wind up with the fact that I now 456 00:23:02,530 --> 00:23:06,400 want to find the constants a0, a1, et cetera, 457 00:23:06,400 --> 00:23:07,922 from this particular relationship. 458 00:23:07,922 --> 00:23:09,130 You see what's happened here? 459 00:23:09,130 --> 00:23:10,840 This is just y double prime. 460 00:23:10,840 --> 00:23:13,300 This is y multiplied by x. 461 00:23:13,300 --> 00:23:15,130 That's why that n plus 1 is in here. 462 00:23:15,130 --> 00:23:17,300 And that must be identically zero. 463 00:23:17,300 --> 00:23:21,310 Now what I'm going to do is the same trick as before. 464 00:23:21,310 --> 00:23:24,190 What I'm going to do is I want to add these term by term. 465 00:23:24,190 --> 00:23:27,220 And I notice that their exponents are different. 466 00:23:27,220 --> 00:23:30,130 In fact, this exponent is three larger 467 00:23:30,130 --> 00:23:32,230 for each value of n than this one, 468 00:23:32,230 --> 00:23:36,470 right? n plus 1 minus the quantity n minus 2 is 3. 469 00:23:36,470 --> 00:23:40,060 So what I want to do is I want to lower this subscript-- 470 00:23:40,060 --> 00:23:41,960 this exponent by 3. 471 00:23:41,960 --> 00:23:44,460 So I'm going to inside the integral-- 472 00:23:44,460 --> 00:23:46,540 I'm sorry, inside the summation sign, 473 00:23:46,540 --> 00:23:50,390 I am going to replace n by n minus 3. 474 00:23:50,390 --> 00:23:53,860 And, in line with our previous note, to compensate for this, 475 00:23:53,860 --> 00:23:57,580 instead of my summation going from 0 to infinity here, 476 00:23:57,580 --> 00:24:00,268 it will now go from 3 to infinity. 477 00:24:00,268 --> 00:24:01,810 In other words, I am now simply going 478 00:24:01,810 --> 00:24:05,050 to replace this by the equivalent sum, 479 00:24:05,050 --> 00:24:12,310 n equals 3 to infinity, a sub n minus 3 x to the n minus 2, 480 00:24:12,310 --> 00:24:13,120 all right? 481 00:24:13,120 --> 00:24:18,370 And you see, if I do that, what I will end up with is simply 482 00:24:18,370 --> 00:24:19,960 this expression here. 483 00:24:19,960 --> 00:24:23,410 And now a new wrinkle comes up that didn't happen before, 484 00:24:23,410 --> 00:24:26,510 but I just wanted to make sure a few funny things happened here 485 00:24:26,510 --> 00:24:29,180 so I can tell you directly what's going on. 486 00:24:29,180 --> 00:24:31,090 And then we can drill with the remaining fine 487 00:24:31,090 --> 00:24:32,780 points in the exercises. 488 00:24:32,780 --> 00:24:34,750 But the key point here is, lookit, 489 00:24:34,750 --> 00:24:37,000 I've got the exponents lined up now. 490 00:24:37,000 --> 00:24:39,740 But look at the indices of summation. 491 00:24:39,740 --> 00:24:42,970 One begins at 2, and the other begins at 3. 492 00:24:42,970 --> 00:24:45,010 And it seems that they're out of phase. 493 00:24:45,010 --> 00:24:50,095 All I want you to observe is that, to make this start at 3, 494 00:24:50,095 --> 00:24:54,370 I could split off separately the term that corresponds to n 495 00:24:54,370 --> 00:24:55,102 equals 2. 496 00:24:55,102 --> 00:24:56,560 In other words, notice that, when n 497 00:24:56,560 --> 00:24:58,660 equals 2, what I have is what? 498 00:24:58,660 --> 00:25:03,460 2 times 1 times a sub 2 times x to the 0. 499 00:25:03,460 --> 00:25:08,980 In other words, all I have is 2 a sub 2, 2 a sub 2 500 00:25:08,980 --> 00:25:10,360 when n is equal to 2. 501 00:25:10,360 --> 00:25:12,700 So what I can do is I'll split that n 502 00:25:12,700 --> 00:25:14,280 equals 2 term off separately. 503 00:25:14,280 --> 00:25:16,480 That's 2 a sub 2. 504 00:25:16,480 --> 00:25:20,647 Then what's left is the sum as n goes from 3 to infinity. 505 00:25:20,647 --> 00:25:22,855 You see, splitting off these terms isn't hard at all. 506 00:25:22,855 --> 00:25:26,620 All you do is, if these don't match up, take the smaller 507 00:25:26,620 --> 00:25:30,310 index, and split off the number of terms necessary so 508 00:25:30,310 --> 00:25:33,190 that the resulting sum will begin at a higher 509 00:25:33,190 --> 00:25:34,900 index matching this one. 510 00:25:34,900 --> 00:25:36,670 Since they only differed by 1 here, 511 00:25:36,670 --> 00:25:39,320 I only have to split off the n equals 2 term. 512 00:25:39,320 --> 00:25:40,750 Now what I have is what? 513 00:25:40,750 --> 00:25:46,540 I have 2 a2 plus this summation, now going from 3 to infinity, 514 00:25:46,540 --> 00:25:50,270 plus this summation, which also goes from 3 to infinity. 515 00:25:50,270 --> 00:25:54,670 And now, since both the exponents and the indices 516 00:25:54,670 --> 00:25:58,510 match up, now I can add term by term for these convergent power 517 00:25:58,510 --> 00:25:59,260 series. 518 00:25:59,260 --> 00:26:02,320 I bring these inside one integral sign. 519 00:26:02,320 --> 00:26:06,460 See, the sigma of a sum is the sum of the sigmas. 520 00:26:06,460 --> 00:26:08,500 I add these up term by term. 521 00:26:08,500 --> 00:26:09,730 So that leaves me what? 522 00:26:09,730 --> 00:26:12,350 I have my 2 a2 still outside here. 523 00:26:12,350 --> 00:26:15,640 And then, inside the sigma notation, from 3 to infinity, 524 00:26:15,640 --> 00:26:16,480 I have what? 525 00:26:16,480 --> 00:26:22,300 n times n minus 1 times a sub n plus a sub n minus 3, 526 00:26:22,300 --> 00:26:25,810 that whole quantity, times x to the n minus 2. 527 00:26:25,810 --> 00:26:27,910 And that must be identically zero. 528 00:26:27,910 --> 00:26:30,310 Now the only way that this can be identically zero 529 00:26:30,310 --> 00:26:32,200 is if each coefficient is 0. 530 00:26:32,200 --> 00:26:34,900 By the way, notice that, since the first term that 531 00:26:34,900 --> 00:26:39,400 appears here corresponds to n equals 3, when n equals 3, 532 00:26:39,400 --> 00:26:41,140 the exponent here is 1. 533 00:26:41,140 --> 00:26:43,900 So notice that this is my only constant term. 534 00:26:43,900 --> 00:26:47,920 All the terms in here begin with at least a factor of x in them. 535 00:26:47,920 --> 00:26:50,830 Since my power series, to be identically zero, 536 00:26:50,830 --> 00:26:53,620 must have all of its coefficients identically zero, 537 00:26:53,620 --> 00:26:58,540 it means that not only must each of these be 0, but 2 a2, which 538 00:26:58,540 --> 00:27:01,390 is my constant term on the left-hand side, that must also 539 00:27:01,390 --> 00:27:02,590 be 0. 540 00:27:02,590 --> 00:27:05,300 So, in other words, to summarize this up, 541 00:27:05,300 --> 00:27:07,300 what we're saying so far is that, for this to be 542 00:27:07,300 --> 00:27:10,240 identically zero, a2 must be 0. 543 00:27:10,240 --> 00:27:12,760 And, once n is at least as big as 3, 544 00:27:12,760 --> 00:27:15,720 this condition here must be obeyed. 545 00:27:15,720 --> 00:27:16,560 What condition? 546 00:27:16,560 --> 00:27:20,310 That the expression in brackets must be 0 for each n 547 00:27:20,310 --> 00:27:22,320 once n is at least as big as 3. 548 00:27:22,320 --> 00:27:25,590 In other words, then, in summary, a2 is 0. 549 00:27:25,590 --> 00:27:30,360 And, for n at least as big as 3, a sub n is minus a sub n 550 00:27:30,360 --> 00:27:33,375 minus 3 over n times n minus 1. 551 00:27:33,375 --> 00:27:36,750 And keep in mind the way I got that quite trivially was 552 00:27:36,750 --> 00:27:38,520 I set this equal to 0. 553 00:27:38,520 --> 00:27:41,130 I transposed the a sub n minus 3, 554 00:27:41,130 --> 00:27:43,500 divided through by n times n minus 1, 555 00:27:43,500 --> 00:27:45,990 and solved for a sub n. 556 00:27:45,990 --> 00:27:48,900 What this tells me now is that, once I 557 00:27:48,900 --> 00:27:52,170 know a particular a, look at what this tells me. 558 00:27:52,170 --> 00:27:57,090 This says, to find a sub n, once I know what the a was three 559 00:27:57,090 --> 00:27:59,310 before this one, I'm home free. 560 00:27:59,310 --> 00:28:03,960 In other words, to find a sub 4, all I have to know is a sub 1. 561 00:28:03,960 --> 00:28:07,870 To find a sub 5, all I have to know is a sub 2. 562 00:28:07,870 --> 00:28:08,370 You see? 563 00:28:08,370 --> 00:28:14,010 To find a sub 6, all I have to know is a sub 3, et cetera. 564 00:28:14,010 --> 00:28:17,190 So the idea is I can now pick a0 and a1 565 00:28:17,190 --> 00:28:19,000 at random, just as before. 566 00:28:19,000 --> 00:28:22,380 And now, using this recipe with n equal to 3, 567 00:28:22,380 --> 00:28:27,120 I have that a sub 3 is minus a sub 0 over 3 times 2. 568 00:28:27,120 --> 00:28:32,040 a sub 6 is minus a sub 3 over 6 times 5. 569 00:28:32,040 --> 00:28:35,340 And, remembering what a3 is in terms of a0, 570 00:28:35,340 --> 00:28:41,490 this tells me that a6 is a0 over 5 times 6 times 3 times 2. 571 00:28:41,490 --> 00:28:44,610 And this is not a misprint that the four is missing here. 572 00:28:44,610 --> 00:28:47,190 This is not a factorial. 573 00:28:47,190 --> 00:28:49,320 You see, I have a way of generating 574 00:28:49,320 --> 00:28:50,790 what these coefficients look like, 575 00:28:50,790 --> 00:28:52,740 but now I've picked a problem where 576 00:28:52,740 --> 00:28:56,850 I may not remember or recognize what power series I'm getting. 577 00:28:56,850 --> 00:28:58,380 That's irrelevant again. 578 00:28:58,380 --> 00:28:59,910 All I'm saying is I can determine 579 00:28:59,910 --> 00:29:04,860 a3, a6, a9, et cetera, just knowing what a0 is. 580 00:29:04,860 --> 00:29:08,520 And, just knowing what a1 is, I can determine a4. 581 00:29:08,520 --> 00:29:11,130 Just by using this recipe with n equal to 4, 582 00:29:11,130 --> 00:29:14,146 a4 is minus a1 over 4 times 3. 583 00:29:14,146 --> 00:29:19,650 a7 is minus a4 over 7 times 6, using that same recipe. 584 00:29:19,650 --> 00:29:22,980 Putting in the value of a4 from here into here, 585 00:29:22,980 --> 00:29:28,110 I have that a7 is a1 over 7 times 6 times 4 times 3. 586 00:29:28,110 --> 00:29:29,610 And this keeps on going. 587 00:29:29,610 --> 00:29:32,280 I can now find a10 in terms of a7. 588 00:29:32,280 --> 00:29:34,560 That will give me a10 in terms of a1, 589 00:29:34,560 --> 00:29:37,030 et cetera, et cetera, et cetera. 590 00:29:37,030 --> 00:29:40,800 Finally, to find a5 or a8, notice 591 00:29:40,800 --> 00:29:45,930 that, using the same recipe, a5 is minus a2 over 5 times 4. 592 00:29:45,930 --> 00:29:49,680 But, since a2 is 0, a5 will be 0. 593 00:29:49,680 --> 00:29:52,470 And, similarly, because a5 is 0, a8 594 00:29:52,470 --> 00:29:57,150 will be 0 and so will a11 and a14, et cetera. 595 00:29:57,150 --> 00:30:01,650 Now the idea is the series that I was looking for, 596 00:30:01,650 --> 00:30:04,650 being a convergent series, can be rearranged. 597 00:30:04,650 --> 00:30:06,060 You see, the idea in this problem 598 00:30:06,060 --> 00:30:09,210 was that I would like to group the exponents according 599 00:30:09,210 --> 00:30:13,800 to whether they are divisible by 3, leave a remainder of 1 600 00:30:13,800 --> 00:30:16,680 when I divide by 3, or leave a remainder of 2 601 00:30:16,680 --> 00:30:18,190 when I divide by 3. 602 00:30:18,190 --> 00:30:20,610 So I group the terms this way-- see, 603 00:30:20,610 --> 00:30:24,540 a0 plus a sub 3 x cubed plus a sub 6 x to the sixth, 604 00:30:24,540 --> 00:30:28,560 et cetera, plus a1 x plus a4 x to the fourth, et cetera, 605 00:30:28,560 --> 00:30:31,830 plus a2 x squared plus a5 x fifth, et cetera-- 606 00:30:31,830 --> 00:30:34,230 the idea being that I know how to express 607 00:30:34,230 --> 00:30:37,860 a3, a6, a9 in terms of a0. 608 00:30:37,860 --> 00:30:43,050 I know how to express a4, a7, a10, et cetera, in terms of a1. 609 00:30:43,050 --> 00:30:47,160 And I know how to express a5, a8, et cetera, in terms of a2. 610 00:30:47,160 --> 00:30:50,520 In fact, in this simple case, which turned out nicely, 611 00:30:50,520 --> 00:30:52,990 they all turned out to be 0, these coefficients. 612 00:30:52,990 --> 00:30:54,210 So what's left is what? 613 00:30:54,210 --> 00:30:57,030 Replacing a4, a7, et cetera, by what 614 00:30:57,030 --> 00:31:00,270 they look like in terms of a1 and a3, a6, et 615 00:31:00,270 --> 00:31:02,760 cetera, by what they look like in terms of a0, 616 00:31:02,760 --> 00:31:07,380 I wind up with y equals a0 times 1 617 00:31:07,380 --> 00:31:11,190 minus x cubed over 3 times 2 plus x to the sixth 618 00:31:11,190 --> 00:31:15,450 over 6 times 5 times 3 times 2, et cetera, 619 00:31:15,450 --> 00:31:18,180 plus a1 times this thing here. 620 00:31:18,180 --> 00:31:20,670 Now the only difference between this problem 621 00:31:20,670 --> 00:31:23,070 and the previous one was that, in the previous one, 622 00:31:23,070 --> 00:31:26,250 I happened to recognize what well-known function had 623 00:31:26,250 --> 00:31:27,750 this as a power series. 624 00:31:27,750 --> 00:31:30,780 In this case, I don't know that, but this is no less 625 00:31:30,780 --> 00:31:34,800 a bona fide solution than we had in the previous case. 626 00:31:34,800 --> 00:31:36,690 This is a convergent power series. 627 00:31:36,690 --> 00:31:37,710 We give it a name-- 628 00:31:37,710 --> 00:31:39,270 h of x, k of x. 629 00:31:39,270 --> 00:31:42,990 If this were the solution to an important enough problem, 630 00:31:42,990 --> 00:31:45,672 we would say, lookit, instead of calling it h of x and k of x, 631 00:31:45,672 --> 00:31:47,630 let's give it a special name because it's going 632 00:31:47,630 --> 00:31:50,100 to come up over and over again. 633 00:31:50,100 --> 00:31:52,140 I'm not going to go into this. 634 00:31:52,140 --> 00:31:55,093 If I do go into it all, it'll be very lightly in the exercises. 635 00:31:55,093 --> 00:31:56,760 But things like that you may have heard, 636 00:31:56,760 --> 00:31:59,220 in terms of name dropping, Bessel functions 637 00:31:59,220 --> 00:32:01,500 and the like, Legendre polynomials, 638 00:32:01,500 --> 00:32:04,620 these were all special power series solutions 639 00:32:04,620 --> 00:32:07,260 to very special differential equations 640 00:32:07,260 --> 00:32:09,480 where the equation itself was so important 641 00:32:09,480 --> 00:32:11,730 an application that the power series that 642 00:32:11,730 --> 00:32:15,150 represented the equation was given a special name. 643 00:32:15,150 --> 00:32:17,750 So you see, in terms of power series solutions, 644 00:32:17,750 --> 00:32:21,800 one extremely powerful use of power series 645 00:32:21,800 --> 00:32:24,170 is that it's used to help us find 646 00:32:24,170 --> 00:32:29,000 solutions of the homogeneous equation L of y equals 0. 647 00:32:29,000 --> 00:32:32,630 And, once we have the general solution of the equation L of y 648 00:32:32,630 --> 00:32:36,500 equals 0, we can then use variation of parameters 649 00:32:36,500 --> 00:32:39,860 to find a solution of L of y equals f of x. 650 00:32:39,860 --> 00:32:41,600 Then we add these two solutions together 651 00:32:41,600 --> 00:32:43,100 to get the general solution. 652 00:32:43,100 --> 00:32:46,910 And that ends our theory on linear differential equations, 653 00:32:46,910 --> 00:32:50,690 except for one more topic called the Laplace transform 654 00:32:50,690 --> 00:32:53,660 that I would like to introduce for you to see. 655 00:32:53,660 --> 00:32:55,820 And so I will have one more lecture 656 00:32:55,820 --> 00:32:58,970 in the guise of linear differential equations in order 657 00:32:58,970 --> 00:33:01,430 that I can bring up the Laplace transform, 658 00:33:01,430 --> 00:33:03,470 but that won't be until next time. 659 00:33:03,470 --> 00:33:05,075 Until next time then, goodbye. 660 00:33:07,790 --> 00:33:10,190 Funding for the publication of this video 661 00:33:10,190 --> 00:33:15,080 was provided by the Gabriella and Paul Rosenbaum Foundation. 662 00:33:15,080 --> 00:33:19,220 Help OCW continue to provide free and open access to MIT 663 00:33:19,220 --> 00:33:24,655 courses by making a donation at ocw.mit.edu/donate.