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 --> 00:00:18,220 at ocw.mit.edu. 8 00:00:33,950 --> 00:00:35,010 HERBERT GROSS: Hi. 9 00:00:35,010 --> 00:00:37,420 Last time we were discussing how to find 10 00:00:37,420 --> 00:00:40,660 the general solution of the homogeneous equation 11 00:00:40,660 --> 00:00:42,370 with constant coefficients. 12 00:00:42,370 --> 00:00:44,620 Our lesson today is going to be concerned 13 00:00:44,620 --> 00:00:46,840 with finding a particular solution 14 00:00:46,840 --> 00:00:49,360 of a linear differential equation 15 00:00:49,360 --> 00:00:51,040 with constant coefficients. 16 00:00:51,040 --> 00:00:53,950 But in which case, the right hand side of the equation 17 00:00:53,950 --> 00:00:55,870 is not identically 0. 18 00:00:55,870 --> 00:00:59,860 I should point out that there is a more general technique called 19 00:00:59,860 --> 00:01:02,320 the method of variation of parameters 20 00:01:02,320 --> 00:01:04,540 that we will talk about next time. 21 00:01:04,540 --> 00:01:06,850 And that's the method which is done in the book. 22 00:01:06,850 --> 00:01:09,880 You see that it talks about this problem in more detail. 23 00:01:09,880 --> 00:01:12,430 But it does turn out that for many types of problems, 24 00:01:12,430 --> 00:01:15,910 a very special case occurs, a case which is not mentioned 25 00:01:15,910 --> 00:01:18,730 in our textbook and which I thought is important enough 26 00:01:18,730 --> 00:01:21,190 that I'd like to present to you before we 27 00:01:21,190 --> 00:01:24,190 get to the stickier approach. 28 00:01:24,190 --> 00:01:27,430 At any rate, today's lesson is called 29 00:01:27,430 --> 00:01:29,510 undetermined coefficients. 30 00:01:29,510 --> 00:01:31,780 And it's a technique that's used for finding 31 00:01:31,780 --> 00:01:35,050 a particular solution of the equation y 32 00:01:35,050 --> 00:01:38,410 double prime plus 2ay prime plus b-y equals f of x. 33 00:01:38,410 --> 00:01:40,600 In other words, constant coefficients. 34 00:01:40,600 --> 00:01:42,850 But the right hand side is not necessarily 0. 35 00:01:42,850 --> 00:01:44,690 In fact, if the right hand side were 0, 36 00:01:44,690 --> 00:01:47,800 we'd be back to the lecture of last time. 37 00:01:47,800 --> 00:01:51,610 But we're going to solve this in the very special cases 38 00:01:51,610 --> 00:01:57,520 that f of x is either e to the m-x where m is a constant 39 00:01:57,520 --> 00:02:02,920 or f of x is either sine m-x or cosine m-x where m is, again, 40 00:02:02,920 --> 00:02:03,895 a constant. 41 00:02:03,895 --> 00:02:09,165 Or finally, in the case where f of x is x to the n, where n 42 00:02:09,165 --> 00:02:10,610 is a whole number. 43 00:02:10,610 --> 00:02:12,400 And the reason that we focus our attention 44 00:02:12,400 --> 00:02:13,900 on these three cases-- 45 00:02:13,900 --> 00:02:16,030 and we'll prove this in more detail 46 00:02:16,030 --> 00:02:18,790 during the course of the unit but not in the lecture-- 47 00:02:18,790 --> 00:02:21,740 is that this is the only family, so to speak, 48 00:02:21,740 --> 00:02:25,750 of functions whereby we can tell what 49 00:02:25,750 --> 00:02:27,880 you have to differentiate in order 50 00:02:27,880 --> 00:02:29,480 to get the given function. 51 00:02:29,480 --> 00:02:32,960 For example, we know that to wind up with e to the m-x, 52 00:02:32,960 --> 00:02:34,900 we essentially have to differentiate 53 00:02:34,900 --> 00:02:37,000 a constant times e to the m-x. 54 00:02:37,000 --> 00:02:39,940 To wind up with sine m-x or cosine m-x, 55 00:02:39,940 --> 00:02:42,640 we have to start with nothing worse than sine m-x 56 00:02:42,640 --> 00:02:44,020 or cosine m-x. 57 00:02:44,020 --> 00:02:46,630 And similarly, to wind up with a power of x we have 58 00:02:46,630 --> 00:02:48,790 to start with a power of x. 59 00:02:48,790 --> 00:02:51,460 That's where these constant coefficients are so important. 60 00:02:51,460 --> 00:02:54,970 You see, as long as the coefficients are constant that 61 00:02:54,970 --> 00:03:00,190 means I can't have any functions of x beefing up anything 62 00:03:00,190 --> 00:03:02,740 in my y, y prime, and y double prime. 63 00:03:02,740 --> 00:03:05,170 Consequently if the coefficients are constant 64 00:03:05,170 --> 00:03:08,320 and f of x is of one of these three forms, 65 00:03:08,320 --> 00:03:10,660 I have a very easy way of looking 66 00:03:10,660 --> 00:03:14,050 for trial solutions for a particular solution 67 00:03:14,050 --> 00:03:15,430 of the equation. 68 00:03:15,430 --> 00:03:17,530 Let me show you what that means in particular. 69 00:03:17,530 --> 00:03:20,530 In case one, if f of x is e to the m-x, 70 00:03:20,530 --> 00:03:23,110 a reasonable trial is y sub p. 71 00:03:23,110 --> 00:03:26,140 The particular solution is A-e to the m-x. 72 00:03:26,140 --> 00:03:29,380 And we then try to find what A is 73 00:03:29,380 --> 00:03:32,740 by plugging this expression into the equation 74 00:03:32,740 --> 00:03:35,900 and seeing what value of A balances the equation. 75 00:03:35,900 --> 00:03:37,330 The reason I underlined reasonable 76 00:03:37,330 --> 00:03:39,970 here is as we shall see later in the lecture, 77 00:03:39,970 --> 00:03:42,820 there are some peculiarities which may exist. 78 00:03:42,820 --> 00:03:45,850 But we're not going to worry about those just yet. 79 00:03:45,850 --> 00:03:50,290 If f of x is sine m-x, then a reasonable trial solution 80 00:03:50,290 --> 00:03:55,680 is A sine m-x plus B cosine m-x where A and B, again, 81 00:03:55,680 --> 00:03:57,460 are undetermined coefficients. 82 00:03:57,460 --> 00:03:58,630 They're constants. 83 00:03:58,630 --> 00:04:00,250 And we're going to try to figure out 84 00:04:00,250 --> 00:04:02,730 what they must be by feeding this trial 85 00:04:02,730 --> 00:04:04,540 solution into the equation. 86 00:04:04,540 --> 00:04:06,640 And I'll emphasize this more when we come to it 87 00:04:06,640 --> 00:04:08,440 in the context of an example. 88 00:04:08,440 --> 00:04:10,540 But notice that even though the right hand side 89 00:04:10,540 --> 00:04:13,570 side is sine m-x, the trial solution is not just 90 00:04:13,570 --> 00:04:18,040 A sine m-x, it's A sine m-x plus B cosine m-x. 91 00:04:18,040 --> 00:04:20,140 Because remember in taking derivatives, 92 00:04:20,140 --> 00:04:23,290 if I differentiate the cosine I can get the sine. 93 00:04:23,290 --> 00:04:27,910 If I differentiate the sine twice I get back to the sine. 94 00:04:27,910 --> 00:04:30,370 So I want both of these terms in here. 95 00:04:30,370 --> 00:04:33,500 And the final case if f of x is x to the n, 96 00:04:33,500 --> 00:04:37,840 a reasonable trial is y sub p equals some constant times x 97 00:04:37,840 --> 00:04:41,800 to the n plus some constant times x to the n minus 1 plus, 98 00:04:41,800 --> 00:04:44,950 et cetera, some constant A1 times x plus, 99 00:04:44,950 --> 00:04:46,140 et cetera, some constant. 100 00:04:46,140 --> 00:04:47,800 In other words, if the right hand side 101 00:04:47,800 --> 00:04:52,660 is a simple power of x where the exponent is a whole number, 102 00:04:52,660 --> 00:04:55,270 we simply try as a trial solution 103 00:04:55,270 --> 00:04:58,320 the most general polynomial of that degree. 104 00:04:58,320 --> 00:05:01,780 And I think the best way to show this now is by example. 105 00:05:01,780 --> 00:05:03,580 We will teach by example. 106 00:05:03,580 --> 00:05:05,530 And the rest of this lecture shall be devoted 107 00:05:05,530 --> 00:05:08,290 to a sequence of such examples. 108 00:05:08,290 --> 00:05:13,300 Example number one, y double prime minus 4y prime plus 3y 109 00:05:13,300 --> 00:05:16,007 equals e to the 5x. 110 00:05:16,007 --> 00:05:17,590 I look at this and I say, well, here I 111 00:05:17,590 --> 00:05:19,150 have constant coefficients. 112 00:05:19,150 --> 00:05:21,370 It appears that the only thing I can differentiate 113 00:05:21,370 --> 00:05:23,290 that will give me an e to the 5x back 114 00:05:23,290 --> 00:05:28,700 again is essentially some constant times e to the 5x. 115 00:05:28,700 --> 00:05:33,010 So for my trial solution, I let y sub p be A-e the 5x. 116 00:05:33,010 --> 00:05:37,620 Consequently y-p prime is 5 Ae to the 5x. 117 00:05:37,620 --> 00:05:41,750 And y-p double prime is 25 A-e to the 5x. 118 00:05:41,750 --> 00:05:46,520 If I now plug y-p double prime, y-p prime, and y-p 119 00:05:46,520 --> 00:05:49,790 into the original equation, I see that each of the 5x 120 00:05:49,790 --> 00:05:51,200 is a common factor. 121 00:05:51,200 --> 00:05:52,360 And I'm left with what? 122 00:05:52,360 --> 00:05:55,620 25A minus 4 times 5A. 123 00:05:55,620 --> 00:05:58,370 In other words, minus 20 A plus 3A. 124 00:05:58,370 --> 00:05:59,690 And what must that be? 125 00:05:59,690 --> 00:06:02,990 That must be identically e to the x. 126 00:06:02,990 --> 00:06:06,410 Let me write identically in here to emphasize the fact 127 00:06:06,410 --> 00:06:08,540 that whatever value of A I pick here, 128 00:06:08,540 --> 00:06:12,320 for this to be a solution the left-hand side of the equation 129 00:06:12,320 --> 00:06:14,240 and the right hand side of the equation 130 00:06:14,240 --> 00:06:16,940 must agree for every value of x. 131 00:06:16,940 --> 00:06:18,920 In particular, I notice that this 132 00:06:18,920 --> 00:06:22,820 says that 8A times e to the 5x must 133 00:06:22,820 --> 00:06:25,610 be identical to e to the 5x. 134 00:06:25,610 --> 00:06:27,710 Consequently, we've got an x equals 0. 135 00:06:27,710 --> 00:06:30,960 This says that 8A is equal to 1. 136 00:06:30,960 --> 00:06:33,950 Therefore, the only candidate for a solution 137 00:06:33,950 --> 00:06:36,380 here is that A be 1/8. 138 00:06:36,380 --> 00:06:41,060 And taking A to be 1/8, I find that my candidate for a trial 139 00:06:41,060 --> 00:06:44,120 solution is 1/8 e to the 5x. 140 00:06:44,120 --> 00:06:47,960 And a trivial check verifies that the only candidate 141 00:06:47,960 --> 00:06:49,350 is indeed a solution. 142 00:06:49,350 --> 00:06:51,500 In other words, a particular solution 143 00:06:51,500 --> 00:06:56,090 of the equation y double prime minus 4y prime plus 3y equals 144 00:06:56,090 --> 00:07:01,250 e to the 5x is y equals 1/8 e to the 5x. 145 00:07:01,250 --> 00:07:06,020 By the way, lest we lose track of what the general theory is, 146 00:07:06,020 --> 00:07:08,330 remember that to find the general solution 147 00:07:08,330 --> 00:07:12,050 of the equation, all we need is a particular solution 148 00:07:12,050 --> 00:07:16,440 plus the general solution of the reduced homogeneous equation. 149 00:07:16,440 --> 00:07:19,040 Recall that with the right hand side here 0, 150 00:07:19,040 --> 00:07:22,280 we already know that y double prime minus 4y prime plus 151 00:07:22,280 --> 00:07:26,400 3y equals 0 has as its general solution, 152 00:07:26,400 --> 00:07:30,800 y equals C1 e to the x plus C2 e to the 3x. 153 00:07:30,800 --> 00:07:34,400 So in other words then, the general solution 154 00:07:34,400 --> 00:07:37,130 of this equation is obtained merely 155 00:07:37,130 --> 00:07:39,620 by tacking on the general solution 156 00:07:39,620 --> 00:07:42,903 of the homogeneous equation to this one particular solution 157 00:07:42,903 --> 00:07:43,820 that we've found here. 158 00:07:43,820 --> 00:07:45,770 In other words, the general solution 159 00:07:45,770 --> 00:07:47,690 of the equation given in this example 160 00:07:47,690 --> 00:07:52,630 is y equals 1/8 e to the 5x plus C1 e to the x plus C2 e 161 00:07:52,630 --> 00:07:54,680 to the 3x. 162 00:07:54,680 --> 00:07:57,290 That takes care of our first illustration. 163 00:07:57,290 --> 00:08:01,130 For our next illustration, we will keep the same linear part 164 00:08:01,130 --> 00:08:01,880 of the equation. 165 00:08:01,880 --> 00:08:05,810 We'll again take y double prime minus 4y prime plus 3y. 166 00:08:05,810 --> 00:08:08,110 But now we'll let f of x be sine x. 167 00:08:08,110 --> 00:08:10,880 In other words, we want to find a particular solution 168 00:08:10,880 --> 00:08:13,940 of this particular differential equation. 169 00:08:13,940 --> 00:08:16,430 And again, constant coefficients, the right hand 170 00:08:16,430 --> 00:08:18,200 side is of the desired form. 171 00:08:18,200 --> 00:08:22,460 I simply try for a solution in the form y sub p 172 00:08:22,460 --> 00:08:25,310 is some constant times sine x, say A sine 173 00:08:25,310 --> 00:08:28,670 x, plus a constant B times cosine x. 174 00:08:28,670 --> 00:08:33,530 If I do this, notice that y-p prime is A cosine x minus B 175 00:08:33,530 --> 00:08:34,370 sin x. 176 00:08:34,370 --> 00:08:39,289 And y-p double prime is minus A sine x minus B cosine x. 177 00:08:39,289 --> 00:08:42,090 By the way, just in pausing here for a moment, 178 00:08:42,090 --> 00:08:44,750 notice that if I had tried for a trial solution 179 00:08:44,750 --> 00:08:49,410 either in the form A sine x alone or B cosine x alone, 180 00:08:49,410 --> 00:08:51,632 noticed that I would have got-- 181 00:08:51,632 --> 00:08:53,090 look what would have happened here. 182 00:08:53,090 --> 00:08:56,840 I would have got sine x's and cosine x terms on both-- 183 00:08:56,840 --> 00:08:59,660 A would have been a coefficient of a sine x term 184 00:08:59,660 --> 00:09:02,030 and a cosine x term. 185 00:09:02,030 --> 00:09:03,740 That means that A would have had to have 186 00:09:03,740 --> 00:09:05,420 been compared with two coefficients 187 00:09:05,420 --> 00:09:06,380 on the right-hand side. 188 00:09:06,380 --> 00:09:09,440 Namely the coefficient of sine x on the right hand is 1. 189 00:09:09,440 --> 00:09:12,500 The cosine x is 0 because it doesn't appear. 190 00:09:12,500 --> 00:09:15,073 And we could very easily have wound up with a contradiction. 191 00:09:15,073 --> 00:09:16,490 We would have found out that there 192 00:09:16,490 --> 00:09:19,220 was no value of A that could have satisfied 193 00:09:19,220 --> 00:09:21,200 two things simultaneously. 194 00:09:21,200 --> 00:09:24,260 But that will be emphasized more in the exercises. 195 00:09:24,260 --> 00:09:25,850 For now, just take my word for it 196 00:09:25,850 --> 00:09:28,020 that this is the trial that we make. 197 00:09:28,020 --> 00:09:30,200 Now if we make this trial, if we now 198 00:09:30,200 --> 00:09:33,800 replace y double prime by minus A sine x minus B 199 00:09:33,800 --> 00:09:39,020 cosine x and y prime by A cosine x minus B sin x and y 200 00:09:39,020 --> 00:09:45,080 by A sine x plus B cosine x, this equation then reads what? 201 00:09:45,080 --> 00:09:49,640 Minus A sine x minus B cosine x plus plus-- 202 00:09:49,640 --> 00:09:53,390 you see, minus this term-- plus 4B sine x minus 4A 203 00:09:53,390 --> 00:09:58,520 cosine x plus 3A sine x plus 3B cosine x. 204 00:09:58,520 --> 00:10:02,870 And that must equal 1 sine x plus 0 cosine x. 205 00:10:02,870 --> 00:10:04,790 You see, what I'm going to do is I'm 206 00:10:04,790 --> 00:10:08,390 going to equate the coefficient of sine x on the left hand 207 00:10:08,390 --> 00:10:11,180 side of the equation with the coefficient of sine 208 00:10:11,180 --> 00:10:13,470 x on the right hand side of the equation. 209 00:10:13,470 --> 00:10:15,980 I'm going to equate the coefficient of cosine x 210 00:10:15,980 --> 00:10:17,870 on the left hand side of the equation 211 00:10:17,870 --> 00:10:20,120 with the coefficient of cosine x on the right hand 212 00:10:20,120 --> 00:10:21,260 side of the equation. 213 00:10:21,260 --> 00:10:23,930 The fact that these coefficients are 1 and 0 is irrelevant. 214 00:10:23,930 --> 00:10:26,595 The theory remains the same. 215 00:10:26,595 --> 00:10:28,970 The reason that I can do this, if you want justification, 216 00:10:28,970 --> 00:10:32,630 in other words how do I know I can compare like terms. 217 00:10:32,630 --> 00:10:35,890 The answer is, well, for example, let x be 0. 218 00:10:35,890 --> 00:10:39,610 You see if x is 0, all of the sine terms on both sides 219 00:10:39,610 --> 00:10:41,410 of the equation drop out. 220 00:10:41,410 --> 00:10:44,000 On the other hand, cosine 0 is 1. 221 00:10:44,000 --> 00:10:47,080 So if I let x be 0, notice that the left hand 222 00:10:47,080 --> 00:10:53,920 side says that minus B minus 4A plus 3B must be zero. 223 00:10:53,920 --> 00:10:57,490 If I then assume, this being an identity, 224 00:10:57,490 --> 00:11:00,520 that this must also be true when x is pi over 2, 225 00:11:00,520 --> 00:11:04,480 if I let x be pi over 2 all the cosine terms drop out. 226 00:11:04,480 --> 00:11:06,700 Because the cosine of pi over 2 is 0. 227 00:11:06,700 --> 00:11:08,870 The sine of pi over 2 is 1. 228 00:11:08,870 --> 00:11:14,860 So now the equation would read minus A plus 4B plus 3A sine x. 229 00:11:14,860 --> 00:11:18,640 Well, sine x isn't in here because we let x be pi over 2. 230 00:11:18,640 --> 00:11:22,230 So it'd be minus A plus for 4B plus 3A times 1. 231 00:11:22,230 --> 00:11:26,070 This minus A plus 4B plus 3A has to equal 1. 232 00:11:26,070 --> 00:11:28,095 In other words, 1 sine pi over 2. 233 00:11:28,095 --> 00:11:29,470 In other words, regardless of how 234 00:11:29,470 --> 00:11:31,595 you want to look at this thing, the important point 235 00:11:31,595 --> 00:11:33,970 is is that looking at the left-hand side, 236 00:11:33,970 --> 00:11:35,980 comparing it with the identity that must 237 00:11:35,980 --> 00:11:37,480 equal on the right-hand side. 238 00:11:37,480 --> 00:11:38,730 We see that what? 239 00:11:38,730 --> 00:11:41,410 2a plus 4b, which is the coefficient 240 00:11:41,410 --> 00:11:43,780 of sine x on the left-hand side of the equation, 241 00:11:43,780 --> 00:11:45,820 must equal 1, which is the coefficient 242 00:11:45,820 --> 00:11:48,160 of sine x on the right-hand side of the equation. 243 00:11:48,160 --> 00:11:52,533 And similarly, minus 4a plus 2b, which 244 00:11:52,533 --> 00:11:54,700 is the coefficient of cosine x on the left-hand side 245 00:11:54,700 --> 00:11:57,640 of the equation, must equal 0, because that's 246 00:11:57,640 --> 00:11:59,950 the coefficient of cosine x on the right-hand side 247 00:11:59,950 --> 00:12:01,130 of the equation. 248 00:12:01,130 --> 00:12:03,760 In other words, if there are values of a and b 249 00:12:03,760 --> 00:12:05,710 that give us a solution, it must come 250 00:12:05,710 --> 00:12:08,620 from this pair of simultaneous linear equations. 251 00:12:08,620 --> 00:12:11,770 And the solution of this is simply a equals 1/10 252 00:12:11,770 --> 00:12:13,360 and b equals 1/5. 253 00:12:13,360 --> 00:12:16,180 And a trivial check shows, that if we 254 00:12:16,180 --> 00:12:20,560 let a be 1/10 and b equal 1/5, that we 255 00:12:20,560 --> 00:12:24,290 do get a particular solution to this differential equation. 256 00:12:24,290 --> 00:12:27,970 In other words, a particular solution of this equation 257 00:12:27,970 --> 00:12:33,940 is y sub p is 1/10 sine x plus 1/5 cosine x. 258 00:12:33,940 --> 00:12:35,650 By the way, the equation that we're 259 00:12:35,650 --> 00:12:38,800 dealing with has the same homogeneous equation, 260 00:12:38,800 --> 00:12:43,360 namely, y double prime minus 4y prime plus 3y equals 0, 261 00:12:43,360 --> 00:12:45,400 as we had in example one. 262 00:12:45,400 --> 00:12:47,860 Consequently, the general solution now is what? 263 00:12:47,860 --> 00:12:54,640 It's 1/10 sine x plus 1/5 cosine x plus c1e to the x plus c2e 264 00:12:54,640 --> 00:12:56,140 to the 3x. 265 00:12:56,140 --> 00:13:00,190 Finally, to get an example in which the right-hand side is 266 00:13:00,190 --> 00:13:02,710 a polynomial and x, we take, again, 267 00:13:02,710 --> 00:13:04,670 the same left-hand side as before-- 268 00:13:04,670 --> 00:13:09,310 y double prime minus 4y prime plus 3y equals x squared. 269 00:13:09,310 --> 00:13:12,700 We try for our trial solution in the form ax 270 00:13:12,700 --> 00:13:15,940 squared plus bx plus c. 271 00:13:15,940 --> 00:13:17,500 See, you might ask over here, how 272 00:13:17,500 --> 00:13:20,110 do you know that there aren't higher powers of x to bring in? 273 00:13:20,110 --> 00:13:22,732 I mean, why shouldn't there have been an x cubed term in here? 274 00:13:22,732 --> 00:13:24,940 Because when you differentiate, the x cubed term gets 275 00:13:24,940 --> 00:13:25,840 knocked down. 276 00:13:25,840 --> 00:13:28,400 Why couldn't it collect with terms over here? 277 00:13:28,400 --> 00:13:31,260 Notice, that if we had an x cubed term, 278 00:13:31,260 --> 00:13:33,460 that coefficient would have to be 0. 279 00:13:33,460 --> 00:13:36,190 Because if you replaced y by x cubed, 280 00:13:36,190 --> 00:13:37,930 the only place that an x cubed term 281 00:13:37,930 --> 00:13:41,260 would appear on the left-hand side is in the 3y term. 282 00:13:41,260 --> 00:13:45,100 You see, if y equals x cubed, y prime 283 00:13:45,100 --> 00:13:48,730 has an x squared as the highest power, y double prime and x. 284 00:13:48,730 --> 00:13:52,210 The only place that x cubed would appear is here. 285 00:13:52,210 --> 00:13:54,940 And since it doesn't appear at all on the right-hand side, 286 00:13:54,940 --> 00:13:57,040 its coefficient would have to be 0. 287 00:13:57,040 --> 00:13:58,540 So the trick is what? 288 00:13:58,540 --> 00:14:01,330 Don't worry about powers higher than this, in general. 289 00:14:01,330 --> 00:14:04,840 In general, start with this and just work down. 290 00:14:04,840 --> 00:14:08,763 A good rule of thumb is, when in doubt, put the extra terms in. 291 00:14:08,763 --> 00:14:10,180 Because the worst that will happen 292 00:14:10,180 --> 00:14:12,360 is that you will have done some work for nothing. 293 00:14:12,360 --> 00:14:15,550 In other words, if I put in an undetermined coefficient 294 00:14:15,550 --> 00:14:17,800 of a term which isn't supposed to be in there, 295 00:14:17,800 --> 00:14:20,400 then that coefficient will turn out to be 0. 296 00:14:20,400 --> 00:14:22,700 And that will tell me that that term isn't in there. 297 00:14:22,700 --> 00:14:24,240 On the other hand, if I leave out 298 00:14:24,240 --> 00:14:26,710 a term that should be in there, then I usually 299 00:14:26,710 --> 00:14:28,570 wind up with a contradiction, as we 300 00:14:28,570 --> 00:14:30,580 shall see later in our lesson. 301 00:14:30,580 --> 00:14:34,000 But the idea is, we try for a solution in the form ax 302 00:14:34,000 --> 00:14:37,390 squared plus bx plus c, in which case, 303 00:14:37,390 --> 00:14:40,030 yp prime would be 2ax plus b. 304 00:14:40,030 --> 00:14:43,060 yp double prime would simply be 2a. 305 00:14:43,060 --> 00:14:49,570 And consequently, if we now replace y double prime by 2a, 306 00:14:49,570 --> 00:14:54,730 y prime by 2ax plus b, and y by ax squared plus bx plus c 307 00:14:54,730 --> 00:14:57,280 in this equation, we wind up with 308 00:14:57,280 --> 00:14:59,890 the undetermined coefficients a, b, and c. 309 00:14:59,890 --> 00:15:04,720 Having to satisfy the identity, 2a minus 4 times the quantity, 310 00:15:04,720 --> 00:15:07,820 2ax plus b, plus 3 times the quantity, 311 00:15:07,820 --> 00:15:11,470 ax squared plus bx plus c, has to be identically 312 00:15:11,470 --> 00:15:16,960 equal to x squared, which means 1x squared plus 0x plus 0. 313 00:15:16,960 --> 00:15:19,180 Now, the only way that two quadratics 314 00:15:19,180 --> 00:15:22,450 can be identically equal is if they're equal coefficient 315 00:15:22,450 --> 00:15:23,620 by coefficient. 316 00:15:23,620 --> 00:15:25,840 Consequently, the coefficient of x 317 00:15:25,840 --> 00:15:28,780 squared on the left-hand side of the equation has to be 1. 318 00:15:28,780 --> 00:15:33,150 The coefficient of x on the left-hand side of the equation 319 00:15:33,150 --> 00:15:35,370 has to be 0, because that term is missing 320 00:15:35,370 --> 00:15:37,500 on the right-hand side, which is the same as saying 321 00:15:37,500 --> 00:15:38,880 it's multiplied by 0. 322 00:15:38,880 --> 00:15:42,240 And the constant term must also be 0. 323 00:15:42,240 --> 00:15:45,780 And see why you had to tack in the lower-order terms here? 324 00:15:45,780 --> 00:15:48,630 Because they do get multiplied and combined, 325 00:15:48,630 --> 00:15:51,150 and we have to utilize these. 326 00:15:51,150 --> 00:15:56,730 In other words, we do have a, b, and c appearing as coefficients 327 00:15:56,730 --> 00:15:59,820 not just of an x squared term, but on the x term, 328 00:15:59,820 --> 00:16:00,990 and the constant term. 329 00:16:00,990 --> 00:16:03,840 But again, we'll do more of that in the exercises. 330 00:16:03,840 --> 00:16:06,840 For the time being, simply observe 331 00:16:06,840 --> 00:16:09,930 that the only term that involves the a x squared 332 00:16:09,930 --> 00:16:12,780 on the left-hand side is 3a. 333 00:16:12,780 --> 00:16:16,110 The terms involving an x have coefficients what? 334 00:16:16,110 --> 00:16:18,450 3b minus 8a. 335 00:16:18,450 --> 00:16:20,370 And the constant term seems to have 336 00:16:20,370 --> 00:16:25,320 the form 2a minus 4b plus 3c. 337 00:16:25,320 --> 00:16:28,080 Consequently, 3a must be 1. 338 00:16:28,080 --> 00:16:30,330 3b minus 8a is 0. 339 00:16:30,330 --> 00:16:33,360 2a minus 4b plus 3c is 0. 340 00:16:33,360 --> 00:16:37,080 From this equation, we see immediately that a is 1/3. 341 00:16:37,080 --> 00:16:40,710 Knowing that a is 1/3, we put that into the second equation. 342 00:16:40,710 --> 00:16:44,400 Immediately, can solve for b, which turns out to be 8/9. 343 00:16:44,400 --> 00:16:47,310 Knowing that a is 1/3 and b is 8/9, 344 00:16:47,310 --> 00:16:50,640 we can now come into the third equation and solve for c, 345 00:16:50,640 --> 00:16:53,850 and c turns out to be 26/27. 346 00:16:53,850 --> 00:16:55,890 In other words, a particular solution 347 00:16:55,890 --> 00:17:00,030 to our differential equation in example three 348 00:17:00,030 --> 00:17:05,560 is 1/3x squared plus 8/9x plus 26/27. 349 00:17:05,560 --> 00:17:07,530 So the general solution of that equation 350 00:17:07,530 --> 00:17:14,339 is 1/3x squared plus 8/9x plus 26/27 plus c1e 351 00:17:14,339 --> 00:17:18,390 to the x plus c2e to the 3x, because this is still 352 00:17:18,390 --> 00:17:20,700 the general solution of the reduced equation. 353 00:17:20,700 --> 00:17:23,760 See, same reduced equation, homogeneous equation, 354 00:17:23,760 --> 00:17:25,859 in all three examples. 355 00:17:25,859 --> 00:17:28,740 Another example, an example now that says, 356 00:17:28,740 --> 00:17:32,730 what happens if the right-hand side isn't simply 357 00:17:32,730 --> 00:17:37,800 a term of the form described in cases 1, 2, and 3, but rather, 358 00:17:37,800 --> 00:17:40,080 is a combination of such terms? 359 00:17:40,080 --> 00:17:42,600 Suppose, for example, that I want to solve the differential 360 00:17:42,600 --> 00:17:47,100 equation, y double prime minus 4y prime plus 3y equals 361 00:17:47,100 --> 00:17:49,680 e to the 5x plus sine x. 362 00:17:49,680 --> 00:17:51,730 Maybe this looks familiar to you. 363 00:17:51,730 --> 00:17:55,050 In other words, if I let l of y denote y double prime minus 364 00:17:55,050 --> 00:17:59,930 4y prime plus 3y, notice that example number one was solving 365 00:17:59,930 --> 00:18:02,490 l of y equals e to the 5x. 366 00:18:02,490 --> 00:18:07,890 And example number two was solving l of y equals sine x. 367 00:18:07,890 --> 00:18:11,280 In other words, we seem to have solved this problem 368 00:18:11,280 --> 00:18:14,610 in the special case, where either one of these two terms, 369 00:18:14,610 --> 00:18:16,440 but not the other, appeared. 370 00:18:16,440 --> 00:18:18,420 And the interesting thing is, that once we 371 00:18:18,420 --> 00:18:20,640 can solve these problems separately, 372 00:18:20,640 --> 00:18:24,900 it turns out that the linearity of our differential equation 373 00:18:24,900 --> 00:18:27,870 allows us to use something which is called the superposition 374 00:18:27,870 --> 00:18:28,770 principle. 375 00:18:28,770 --> 00:18:33,360 In other words, that it allows us to, in many cases, 376 00:18:33,360 --> 00:18:36,330 solve more complicated right-hand sides 377 00:18:36,330 --> 00:18:40,080 simply by reducing it to a sum of simpler right-hand sides. 378 00:18:40,080 --> 00:18:42,420 In particular, what I'm saying in this particular case, 379 00:18:42,420 --> 00:18:46,350 was that in example one we saw that l of y 380 00:18:46,350 --> 00:18:49,950 equals z to the 5x had as a particular solution 381 00:18:49,950 --> 00:18:52,140 1/8e to the 5x. 382 00:18:52,140 --> 00:18:55,080 In example two, we saw that l of y 383 00:18:55,080 --> 00:18:58,230 equals sine x had as a particular solution 384 00:18:58,230 --> 00:19:01,840 sine x/10 plus cosine x/5. 385 00:19:01,840 --> 00:19:05,010 Now, the point is, by equals added to equals, 386 00:19:05,010 --> 00:19:09,300 this plus this is e to the 5x plus sine x. 387 00:19:09,300 --> 00:19:14,580 But by linearity, l of 1/8e to the 5x plus l 388 00:19:14,580 --> 00:19:18,660 of sine x/10 plus cosine x/5 is l 389 00:19:18,660 --> 00:19:25,000 of 1/8e to the 5x plus sine x/10 plus cosine x/5. 390 00:19:25,000 --> 00:19:29,100 In other words, l of u plus l of v by linearity 391 00:19:29,100 --> 00:19:31,830 is l of u plus v. 392 00:19:31,830 --> 00:19:33,390 In other words, then by linearity, 393 00:19:33,390 --> 00:19:36,330 adding these two results yields this. 394 00:19:36,330 --> 00:19:39,540 And that tells us that a particular solution 395 00:19:39,540 --> 00:19:42,410 to our equation in example four is simply 396 00:19:42,410 --> 00:19:49,400 y sub p is 1/8e to the 5x plus sine x/10 plus cosine x/5. 397 00:19:49,400 --> 00:19:50,220 See? 398 00:19:50,220 --> 00:19:53,790 More generally, if we want to solve a differential equation, 399 00:19:53,790 --> 00:19:59,670 l of y equals f of x plus g of x, then by linearity 400 00:19:59,670 --> 00:20:03,960 the solution y will simply be u plus v-- a particular solution 401 00:20:03,960 --> 00:20:07,890 will be u plus v, where u is any particular solution 402 00:20:07,890 --> 00:20:10,870 of the equation l of y equals f of x, and v 403 00:20:10,870 --> 00:20:13,340 is any particular solution of the equation l 404 00:20:13,340 --> 00:20:15,540 of y equals g of x. 405 00:20:15,540 --> 00:20:18,180 In other words, notice here that from this we 406 00:20:18,180 --> 00:20:22,620 see that l of u plus l of v is f of x plus g of x. 407 00:20:22,620 --> 00:20:27,750 By linearity, l of u plus l of v is l of u plus v. Therefore, 408 00:20:27,750 --> 00:20:31,440 l of u plus v is f of x plus g of x, which 409 00:20:31,440 --> 00:20:33,710 is exactly what we mean by saying that this 410 00:20:33,710 --> 00:20:35,840 is a solution of this equation. 411 00:20:35,840 --> 00:20:38,810 That's why there's no loss of generality 412 00:20:38,810 --> 00:20:43,340 if we elect to always think of the right-hand side 413 00:20:43,340 --> 00:20:45,110 as having just one term. 414 00:20:45,110 --> 00:20:46,970 Because if it's a sum of terms, we 415 00:20:46,970 --> 00:20:49,280 can solve the problem separately, 416 00:20:49,280 --> 00:20:52,670 one for each term on the right-hand side, and then 417 00:20:52,670 --> 00:20:56,090 just add the answers up. 418 00:20:56,090 --> 00:20:57,950 Now, let me come to an example that 419 00:20:57,950 --> 00:21:00,340 illustrates just a little bit of a pitfall, 420 00:21:00,340 --> 00:21:04,370 and I think that it's one that's worth falling into. 421 00:21:04,370 --> 00:21:07,130 Let's now take the problem y double prime 422 00:21:07,130 --> 00:21:11,090 minus 4y prime plus 3y equals e to the x. 423 00:21:11,090 --> 00:21:12,920 That doesn't look appreciably different 424 00:21:12,920 --> 00:21:14,850 than example number one. 425 00:21:14,850 --> 00:21:16,920 Let's try the same kind of a solution. 426 00:21:16,920 --> 00:21:19,190 In fact, it looks even easier than example one. 427 00:21:19,190 --> 00:21:21,410 Because in an example one, the right-hand side 428 00:21:21,410 --> 00:21:22,580 was e to the 5x. 429 00:21:22,580 --> 00:21:24,698 This is just simply e to the x. 430 00:21:24,698 --> 00:21:26,240 This one looks even easier to handle. 431 00:21:26,240 --> 00:21:29,630 Namely, notice that since the derivative of e to the x 432 00:21:29,630 --> 00:21:33,110 is still e to the x, that if we try for a solution in the form 433 00:21:33,110 --> 00:21:38,030 y sub p equals a to the x, then yp prime and yp double prime 434 00:21:38,030 --> 00:21:40,190 are also just ae to the x. 435 00:21:40,190 --> 00:21:44,570 Therefore, let's replace y, y prime, and y double prime by ae 436 00:21:44,570 --> 00:21:45,530 to the x. 437 00:21:45,530 --> 00:21:47,570 We can then factor l and e to the x. 438 00:21:47,570 --> 00:21:48,920 And what's left over here? 439 00:21:48,920 --> 00:21:54,200 We have an a minus 4a plus 3a, and on the right-hand side, 440 00:21:54,200 --> 00:21:55,580 e to the x. 441 00:21:55,580 --> 00:21:57,630 Now, look at this bracketed expression-- 442 00:21:57,630 --> 00:22:00,290 a very nasty thing has just happened. 443 00:22:00,290 --> 00:22:05,600 Notice that a minus 4a plus 3a, for any constant a-- 444 00:22:05,600 --> 00:22:08,480 in fact, for that matter, for any variable a, as long as a 445 00:22:08,480 --> 00:22:10,580 stays the same in all three expressions-- 446 00:22:10,580 --> 00:22:13,610 the bracketed expression is identically 0. 447 00:22:13,610 --> 00:22:17,900 And consequently, this says that our left-hand side must be 0, 448 00:22:17,900 --> 00:22:20,720 our right-hand side must be identically e to the x. 449 00:22:20,720 --> 00:22:22,100 This is impossible. 450 00:22:22,100 --> 00:22:26,420 First of all, e to the x cannot be identically 0. 451 00:22:26,420 --> 00:22:29,180 e to the x, in fact, is never equal to 0 452 00:22:29,180 --> 00:22:31,040 for any real value of x. 453 00:22:31,040 --> 00:22:34,190 And secondly, it appears that our undetermined coefficient, 454 00:22:34,190 --> 00:22:38,000 a, has disappeared completely from the equation. 455 00:22:38,000 --> 00:22:42,080 What went wrong here that didn't go wrong an example one? 456 00:22:42,080 --> 00:22:45,140 You may have recalled when we first started today's lecture, 457 00:22:45,140 --> 00:22:47,840 I said that if f of x was e to the mx, 458 00:22:47,840 --> 00:22:52,850 a reasonable trial solution was y equals ae to the mx, 459 00:22:52,850 --> 00:22:55,400 and emphasized that "reasonable" might not 460 00:22:55,400 --> 00:22:58,010 be good enough in certain tough cases. 461 00:22:58,010 --> 00:23:01,790 What happened here was the very interesting fact that e 462 00:23:01,790 --> 00:23:04,310 to the x, in other words, the right-hand side 463 00:23:04,310 --> 00:23:07,790 of this equation, happens to be a solution 464 00:23:07,790 --> 00:23:10,220 to the homogeneous equation. 465 00:23:10,220 --> 00:23:13,820 Remember, that when we took the equation, y double prime minus 466 00:23:13,820 --> 00:23:16,910 4y prime plus 3y equals 0, we saw 467 00:23:16,910 --> 00:23:20,345 that two particular solutions were e to the x, 468 00:23:20,345 --> 00:23:23,120 see, e to the x, and e to the 3x. 469 00:23:23,120 --> 00:23:27,680 Consequently, when we substituted in e to the x, 470 00:23:27,680 --> 00:23:29,648 ae to the x on the left-hand side 471 00:23:29,648 --> 00:23:31,190 here to see what was going to happen, 472 00:23:31,190 --> 00:23:34,000 we should have known that that term was going to drop out. 473 00:23:34,000 --> 00:23:34,680 Why? 474 00:23:34,680 --> 00:23:37,160 Because l of e to the x is 0. 475 00:23:37,160 --> 00:23:42,290 Consequently, by linearity, l of ae to the x is also 0. 476 00:23:42,290 --> 00:23:46,970 What were we really doing when we replaced y sub p by ae 477 00:23:46,970 --> 00:23:47,870 to the x? 478 00:23:47,870 --> 00:23:52,200 We were computing l of ae to the x. 479 00:23:52,200 --> 00:23:54,620 That's what we did in here when we computed this. 480 00:23:54,620 --> 00:23:57,110 We were computing l of ae to the x. 481 00:23:57,110 --> 00:24:00,240 And we should have known that that was going to be 0. 482 00:24:00,240 --> 00:24:02,430 We must be very, very careful, in other words-- 483 00:24:02,430 --> 00:24:03,740 this is the key point-- 484 00:24:03,740 --> 00:24:08,030 when we're given y double prime plus 2ay prime plus by 485 00:24:08,030 --> 00:24:10,490 is some function of x, where that function of x 486 00:24:10,490 --> 00:24:15,140 is not identically 0, that when we look for the trial solution, 487 00:24:15,140 --> 00:24:18,710 we must make sure that f of x is not 488 00:24:18,710 --> 00:24:23,070 a particular solution of the homogeneous equation. 489 00:24:23,070 --> 00:24:27,170 In other words, given this, the safest thing to do 490 00:24:27,170 --> 00:24:30,530 is to first solve the homogeneous equation. 491 00:24:30,530 --> 00:24:33,740 So you solve this one first. 492 00:24:33,740 --> 00:24:40,190 Then proceed as usual if l of f of x is not 0. 493 00:24:40,190 --> 00:24:42,800 In other words, if the right-hand side is not 494 00:24:42,800 --> 00:24:45,620 a particular solution of the homogeneous equation, 495 00:24:45,620 --> 00:24:46,970 proceed as usual. 496 00:24:46,970 --> 00:24:50,120 That means the way we did in the examples one through four. 497 00:24:50,120 --> 00:24:53,930 But if l of f of x is 0, certainly 498 00:24:53,930 --> 00:24:57,590 replacing the trial solution as being some constant times 499 00:24:57,590 --> 00:25:00,750 f of x, that's not going to give us anything. 500 00:25:00,750 --> 00:25:02,300 They a will drop out there. 501 00:25:02,300 --> 00:25:03,710 What we're saying is what? 502 00:25:03,710 --> 00:25:06,020 Given whatever trial solution you would've 503 00:25:06,020 --> 00:25:09,590 tried, if you didn't know that this was 0, 504 00:25:09,590 --> 00:25:13,430 replace that by x times that trial solution. 505 00:25:13,430 --> 00:25:15,930 And that, we'll explain in the course of the exercises. 506 00:25:15,930 --> 00:25:17,840 But the mechanics are similar to what 507 00:25:17,840 --> 00:25:20,210 happened in the lecture of last time 508 00:25:20,210 --> 00:25:22,280 when we got a repeated root, when 509 00:25:22,280 --> 00:25:24,680 we said that one of the roots was e to the rx 510 00:25:24,680 --> 00:25:26,990 and the other one was xe to the rx. 511 00:25:26,990 --> 00:25:28,850 In other words, quite mechanically, 512 00:25:28,850 --> 00:25:31,970 whenever the right-hand side of the equation 513 00:25:31,970 --> 00:25:34,610 satisfies the homogeneous equation, 514 00:25:34,610 --> 00:25:36,320 replace the trial solution that you 515 00:25:36,320 --> 00:25:40,220 would have used by the trial solution multiplied by x. 516 00:25:40,220 --> 00:25:43,010 In regards to example five, where we first 517 00:25:43,010 --> 00:25:46,310 ran into this mess right over here, what we're saying is, 518 00:25:46,310 --> 00:25:49,640 if we had solved the homogeneous equation first, 519 00:25:49,640 --> 00:25:51,770 we would have noticed right away that e 520 00:25:51,770 --> 00:25:55,220 to the x was a solution of the reduced equation, in which case 521 00:25:55,220 --> 00:25:58,860 we would not have tried ae to the x as a trial solution, 522 00:25:58,860 --> 00:26:03,020 but rather axe to the x. 523 00:26:03,020 --> 00:26:04,880 I leave the details out, because they're 524 00:26:04,880 --> 00:26:06,890 very easy for you to verify. 525 00:26:06,890 --> 00:26:09,570 We will have homework problems similar to this. 526 00:26:09,570 --> 00:26:12,530 And if you wish, you can verify this for yourself now. 527 00:26:12,530 --> 00:26:15,410 Otherwise, we'll do problems like this in the exercises. 528 00:26:15,410 --> 00:26:17,150 But to actually check, that if you 529 00:26:17,150 --> 00:26:21,320 try y sub p to be axe to the x, that we actually do 530 00:26:21,320 --> 00:26:23,510 find a value for a here. 531 00:26:23,510 --> 00:26:26,270 By the way, do not say, do I try axe 532 00:26:26,270 --> 00:26:29,570 to the x plus something times e to the x itself, 533 00:26:29,570 --> 00:26:32,360 because after all, I can wind up with an e to the x term 534 00:26:32,360 --> 00:26:34,340 by starting with an e to the x term? 535 00:26:34,340 --> 00:26:37,370 The answer is that as soon as you tack on a term, 536 00:26:37,370 --> 00:26:40,865 like be to the x, that when you take l of that, 537 00:26:40,865 --> 00:26:45,650 that whole term is going to drop out, because e to the x 538 00:26:45,650 --> 00:26:48,600 was a solution of the homogeneous equation. 539 00:26:48,600 --> 00:26:53,480 So this is the only term that you try in the solution. 540 00:26:53,480 --> 00:26:55,940 Now, the last example-- 541 00:26:55,940 --> 00:26:58,490 and this is one that we will not solve now, 542 00:26:58,490 --> 00:27:02,300 not because we don't have the time, but because we can't. 543 00:27:02,300 --> 00:27:04,830 And the reason that we can't is simply this. 544 00:27:04,830 --> 00:27:08,030 I asked you to find a particular solution of the equation, y 545 00:27:08,030 --> 00:27:11,170 double prime plus y equals secant x. 546 00:27:11,170 --> 00:27:15,850 Notice, the homogeneous equation, y double prime plus 547 00:27:15,850 --> 00:27:18,910 y equals 0, is very easy for us to solve. 548 00:27:18,910 --> 00:27:23,860 Namely, it's y equals c1 sine x plus c2 cosine x-- 549 00:27:23,860 --> 00:27:25,750 no trouble with the homogeneous case. 550 00:27:25,750 --> 00:27:28,480 The problem is that the right-hand side here 551 00:27:28,480 --> 00:27:31,420 is not of one of the three types that we talked about. 552 00:27:31,420 --> 00:27:34,930 Try to find all the functions whose derivatives 553 00:27:34,930 --> 00:27:37,060 can lead to secant x. 554 00:27:37,060 --> 00:27:39,370 It's an impossible task. 555 00:27:39,370 --> 00:27:42,070 There are infinitely many functions 556 00:27:42,070 --> 00:27:46,240 that one can find combinations of whose derivative yields 557 00:27:46,240 --> 00:27:47,590 secant x. 558 00:27:47,590 --> 00:27:49,570 In other words, this particular method 559 00:27:49,570 --> 00:27:51,940 that I did for you today, which is not in the text 560 00:27:51,940 --> 00:27:54,610 but which I felt was important, needs 561 00:27:54,610 --> 00:27:56,870 to have two things going for it. 562 00:27:56,870 --> 00:28:00,220 One is that on the left-hand side of the linear equation, 563 00:28:00,220 --> 00:28:02,590 there must be constant coefficients. 564 00:28:02,590 --> 00:28:05,740 And the other is, even if the coefficients are constant, 565 00:28:05,740 --> 00:28:09,070 the right-hand side must belong to one of the three families 566 00:28:09,070 --> 00:28:16,270 e to the mx, cosine mx or sine mx, or x to the mth power. 567 00:28:16,270 --> 00:28:18,250 Now, a more general method, which 568 00:28:18,250 --> 00:28:21,430 is called the method of variation of parameters, 569 00:28:21,430 --> 00:28:26,690 allows the left-hand side to have variable coefficients. 570 00:28:26,690 --> 00:28:30,740 It allows the right-hand side to be an arbitrary function of x. 571 00:28:30,740 --> 00:28:33,650 And the price that we pay for this nice general formula 572 00:28:33,650 --> 00:28:36,020 is the fact that it's a messy thing, 573 00:28:36,020 --> 00:28:39,260 both from a practical point of view to apply 574 00:28:39,260 --> 00:28:42,110 and from a theoretical point of view to justify. 575 00:28:42,110 --> 00:28:45,440 But that is what we'll be talking about next time. 576 00:28:45,440 --> 00:28:47,120 And that will be the lecture that 577 00:28:47,120 --> 00:28:50,650 corresponds to the material in the book, you see in the text. 578 00:28:50,650 --> 00:28:53,030 And until next time, what we want to do 579 00:28:53,030 --> 00:28:55,580 is just drill on undetermined coefficients 580 00:28:55,580 --> 00:28:59,150 in the particular case where they happen to be applicable. 581 00:28:59,150 --> 00:29:01,230 At any rate, then, until next time. 582 00:29:01,230 --> 00:29:01,730 Goodbye. 583 00:29:05,680 --> 00:29:08,080 Funding for the publication of this video 584 00:29:08,080 --> 00:29:12,970 was provided by the Gabriella and Paul Rosenbaum Foundation. 585 00:29:12,970 --> 00:29:17,110 Help OCW continue to provide free and open access to MIT 586 00:29:17,110 --> 00:29:22,396 courses by making a donation at ocw.mit.edu/donate.