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 00:00:28,493 --> 00:00:29,250 HERBERT GROSS: Hi. 9 00:00:29,250 --> 00:00:32,549 Today we're going to begin our study of higher order 10 00:00:32,549 --> 00:00:34,300 differential equations. 11 00:00:34,300 --> 00:00:35,790 In particular, we're going to talk 12 00:00:35,790 --> 00:00:39,300 about a special equation called the linear differential 13 00:00:39,300 --> 00:00:42,010 equation, which came up, by the way, 14 00:00:42,010 --> 00:00:45,410 you'll notice, in first order equations as a special type. 15 00:00:45,410 --> 00:00:47,190 But we're going to generalize that. 16 00:00:47,190 --> 00:00:50,100 And the only liberty I'm going to take during the lecture 17 00:00:50,100 --> 00:00:53,790 is to restrict our study to the case of differential equations 18 00:00:53,790 --> 00:00:57,420 of order two so that we won't have unwieldy expressions 19 00:00:57,420 --> 00:00:58,560 all over the board. 20 00:00:58,560 --> 00:01:01,050 The idea is that what happens in the case of n 21 00:01:01,050 --> 00:01:04,920 equals 2 happens for all orders n, 22 00:01:04,920 --> 00:01:07,410 except for a little modification in the algebra. 23 00:01:07,410 --> 00:01:10,140 But we'll talk about that more in the exercises. 24 00:01:10,140 --> 00:01:12,270 At any rate, for the time being we simply 25 00:01:12,270 --> 00:01:15,660 call today's lecture Linear Differential Equations. 26 00:01:15,660 --> 00:01:17,310 And now I must define for you what 27 00:01:17,310 --> 00:01:19,800 I mean by a linear differential equation. 28 00:01:19,800 --> 00:01:22,080 And I'll motivate this more as we go along. 29 00:01:22,080 --> 00:01:25,070 For the time being, a linear differential equation 30 00:01:25,070 --> 00:01:28,200 is simply one which has this very special form. 31 00:01:28,200 --> 00:01:32,820 What you have is a y double prime term appearing, then 32 00:01:32,820 --> 00:01:36,900 some function of x alone multiplying y prime, 33 00:01:36,900 --> 00:01:41,340 plus some function of x alone multiplying y, and on the right 34 00:01:41,340 --> 00:01:43,120 some function of x alone. 35 00:01:43,120 --> 00:01:46,750 And an analogous result would hold for higher order. 36 00:01:46,750 --> 00:01:50,370 In other words, you have the various derivatives 37 00:01:50,370 --> 00:01:53,460 of y appearing, y by itself. 38 00:01:53,460 --> 00:01:56,700 The coefficients are always functions of x alone. 39 00:01:56,700 --> 00:02:01,140 And no power is-- or no derivative or y 40 00:02:01,140 --> 00:02:03,060 is raised to any power. 41 00:02:03,060 --> 00:02:05,340 In particular, notice here-- it's a very small point-- 42 00:02:05,340 --> 00:02:07,320 but notice here that I assume that the leading 43 00:02:07,320 --> 00:02:09,037 coefficient here is 1. 44 00:02:09,037 --> 00:02:10,620 For example, somebody might have said, 45 00:02:10,620 --> 00:02:12,078 couldn't you have had some function 46 00:02:12,078 --> 00:02:14,250 of x times y double prime? 47 00:02:14,250 --> 00:02:16,320 The answer is yes, I could have. 48 00:02:16,320 --> 00:02:18,360 But I'm assuming that if I had a function, 49 00:02:18,360 --> 00:02:21,510 say, r of x multiplying y double prime, 50 00:02:21,510 --> 00:02:23,790 I could have multiplied-- divided 51 00:02:23,790 --> 00:02:26,790 both sides of this equation through by that coefficient 52 00:02:26,790 --> 00:02:29,830 and wound up with an equation of this particular type. 53 00:02:29,830 --> 00:02:32,100 In other words, without loss of generality-- and this 54 00:02:32,100 --> 00:02:33,630 is very important, though. 55 00:02:33,630 --> 00:02:35,790 Many of my theorems are going to be messed up 56 00:02:35,790 --> 00:02:39,338 a little bit if the coefficients of y double prime isn't 1. 57 00:02:39,338 --> 00:02:41,880 In other words, the algebra will become a little bit tougher. 58 00:02:41,880 --> 00:02:45,030 I'll speak about that later in the lecture if I remember to. 59 00:02:45,030 --> 00:02:47,070 But the idea is, for the time being 60 00:02:47,070 --> 00:02:49,770 this is all we mean by a linear differential equation. 61 00:02:49,770 --> 00:02:52,620 And perhaps the best way to emphasize what we really 62 00:02:52,620 --> 00:02:54,540 mean by linear differential equation 63 00:02:54,540 --> 00:02:57,810 is to show what we mean by a nonlinear differential 64 00:02:57,810 --> 00:03:01,650 equation-- a differential equation which is not linear. 65 00:03:01,650 --> 00:03:04,890 For example, y double prime plus y prime 66 00:03:04,890 --> 00:03:10,080 squared plus y equals sine x would not be a linear equation, 67 00:03:10,080 --> 00:03:14,370 because the y prime term is being raised to a power other 68 00:03:14,370 --> 00:03:15,470 than the first power. 69 00:03:15,470 --> 00:03:17,110 See, it's squared. 70 00:03:17,110 --> 00:03:21,330 y double prime plus y times y prime equals e to the x 71 00:03:21,330 --> 00:03:25,230 is not a linear equation, because the multiplier 72 00:03:25,230 --> 00:03:29,460 of y prime is y as opposed to just a function of x, which is 73 00:03:29,460 --> 00:03:31,590 what the definition calls for. 74 00:03:31,590 --> 00:03:36,270 y double prime plus x y prime plus y squared equals x cubed 75 00:03:36,270 --> 00:03:37,650 is not linear. 76 00:03:37,650 --> 00:03:39,330 And the reason that it's not linear 77 00:03:39,330 --> 00:03:42,930 is because y appears to a power higher than the first. 78 00:03:42,930 --> 00:03:45,030 Maybe I should circle the things that 79 00:03:45,030 --> 00:03:47,340 prevent these things from being linear so that you 80 00:03:47,340 --> 00:03:49,710 can see what's happening here. 81 00:03:49,710 --> 00:03:54,300 y double prime plus e to the x y prime plus x cubed y equals tan 82 00:03:54,300 --> 00:03:57,900 y is not linear, because even though the left-hand side is 83 00:03:57,900 --> 00:04:01,290 fine, notice that the function on the right hand side is 84 00:04:01,290 --> 00:04:02,910 a function of y-- 85 00:04:02,910 --> 00:04:07,560 depends on y rather than just x alone. 86 00:04:07,560 --> 00:04:12,720 And to finish this off, an example of a linear equation 87 00:04:12,720 --> 00:04:13,530 would be what? 88 00:04:13,530 --> 00:04:15,030 Well, almost this one-- 89 00:04:15,030 --> 00:04:18,660 y double prime plus e to the x y prime plus x 90 00:04:18,660 --> 00:04:21,120 cubed y equals tangent x. 91 00:04:21,120 --> 00:04:25,230 You see a y double prime term, y prime to the first power being 92 00:04:25,230 --> 00:04:27,570 multiplied by a function of x alone, 93 00:04:27,570 --> 00:04:30,690 y to the first power being multiplied by a function of x 94 00:04:30,690 --> 00:04:34,200 alone, and the right-hand side being a function 95 00:04:34,200 --> 00:04:37,080 of x explicitly by itself. 96 00:04:37,080 --> 00:04:39,690 And notice, by the way, don't get caught 97 00:04:39,690 --> 00:04:41,640 in a psychological hangup here. 98 00:04:41,640 --> 00:04:45,700 When I say linear, it's modifying the equation, 99 00:04:45,700 --> 00:04:47,350 not what the coefficients look like. 100 00:04:47,350 --> 00:04:49,300 In other words, certainly we don't think of e 101 00:04:49,300 --> 00:04:51,240 to the x as a linear function. 102 00:04:51,240 --> 00:04:53,820 We don't think of x cubed as being linear. 103 00:04:53,820 --> 00:04:56,550 We don't think of tangent x as being linear. 104 00:04:56,550 --> 00:05:00,090 The coefficients do not have to be linear functions of x. 105 00:05:00,090 --> 00:05:02,370 They have to be functions of x alone. 106 00:05:02,370 --> 00:05:05,550 It's the equation that's called linear. 107 00:05:05,550 --> 00:05:07,350 And maybe the best way to explain 108 00:05:07,350 --> 00:05:10,090 that is in the following way. 109 00:05:10,090 --> 00:05:13,620 Let's look at y double prime plus p of xy prime plus q 110 00:05:13,620 --> 00:05:14,820 of xy. 111 00:05:14,820 --> 00:05:17,760 My claim is I can think of this as a function machine 112 00:05:17,760 --> 00:05:21,300 where the input is y, and the machine is told, look it-- 113 00:05:21,300 --> 00:05:24,570 whatever y comes in, differentiate it twice, 114 00:05:24,570 --> 00:05:27,690 add on p of x times the first derivative plus 115 00:05:27,690 --> 00:05:31,250 q of x times the function, and that will be the output. 116 00:05:31,250 --> 00:05:33,590 In other words, I could think of a machine 117 00:05:33,590 --> 00:05:36,110 where the input of the machine is y-- 118 00:05:36,110 --> 00:05:39,770 and let me call the machine the L machine to emphasize the word 119 00:05:39,770 --> 00:05:40,730 Linear-- 120 00:05:40,730 --> 00:05:42,080 see, what L does is this. 121 00:05:42,080 --> 00:05:46,400 The domain of L are functions which are twice differentiable. 122 00:05:46,400 --> 00:05:48,830 After all, for this to make sense 123 00:05:48,830 --> 00:05:51,990 I have to be able to differentiate y at least twice. 124 00:05:51,990 --> 00:05:56,810 So the input of the L machine is a twice differentiable function 125 00:05:56,810 --> 00:06:00,785 of x, and the output is some function of x. 126 00:06:00,785 --> 00:06:03,800 You see y prime, y double prime are functions of x, p 127 00:06:03,800 --> 00:06:06,320 and q are functions of x, y is a function of x. 128 00:06:06,320 --> 00:06:10,280 All these things combine, then, to give you a function of x. 129 00:06:10,280 --> 00:06:12,810 Let me show how this machine works for example. 130 00:06:12,810 --> 00:06:15,520 If the L machine is y double prime plus 131 00:06:15,520 --> 00:06:18,652 e to the x y prime plus x cubed y-- 132 00:06:18,652 --> 00:06:20,930 [? see, ?] L of y equals this-- 133 00:06:20,930 --> 00:06:23,940 if I feed in sine x to the L machine, 134 00:06:23,940 --> 00:06:25,550 what does the L machine do? 135 00:06:25,550 --> 00:06:28,790 It takes sine x, it differentiates it twice, 136 00:06:28,790 --> 00:06:32,090 adds on e to the x times the first derivative, 137 00:06:32,090 --> 00:06:36,590 and x cubed times the function itself, which is sine x. 138 00:06:36,590 --> 00:06:39,890 Going through this operation, which is trivial, L of sine x 139 00:06:39,890 --> 00:06:44,390 would be x cubed minus 1 sine x plus e to the x cosine x. 140 00:06:44,390 --> 00:06:46,010 By the way, to help you understand 141 00:06:46,010 --> 00:06:48,740 what we mean by a solution, notice that what we're saying 142 00:06:48,740 --> 00:06:53,690 is that if we were to refer back to this particular equation, 143 00:06:53,690 --> 00:06:57,980 y equals sine x would not be a solution of this equation, 144 00:06:57,980 --> 00:07:01,220 because when I feed sine x into the L machine, 145 00:07:01,220 --> 00:07:04,420 I do not get tangent x, which is what a solution would mean. 146 00:07:04,420 --> 00:07:08,570 In other words, in terms of our new notation, this says L of y 147 00:07:08,570 --> 00:07:10,850 equals tan x. 148 00:07:10,850 --> 00:07:12,950 So anything which is a solution means 149 00:07:12,950 --> 00:07:16,040 if I shove in y to the machine, the output would 150 00:07:16,040 --> 00:07:17,660 have to be tan x. 151 00:07:17,660 --> 00:07:19,820 To look at this from a different perspective, 152 00:07:19,820 --> 00:07:24,980 if the right-hand side of my equation had been this-- 153 00:07:24,980 --> 00:07:27,620 in other words, if tan x had been replaced 154 00:07:27,620 --> 00:07:32,090 by x cubed minus 1 sine x plus e to the x cosine x, 155 00:07:32,090 --> 00:07:36,020 then y equals sine x would have been a solution 156 00:07:36,020 --> 00:07:37,730 of this particular equation. 157 00:07:37,730 --> 00:07:39,650 But enough about that. 158 00:07:39,650 --> 00:07:42,380 And we'll emphasize that more in the exercises. 159 00:07:42,380 --> 00:07:46,310 The key point as to why the word "linear" is used 160 00:07:46,310 --> 00:07:51,530 is that linear modifies our L machine, not the coefficients 161 00:07:51,530 --> 00:07:53,360 of the differential equation. 162 00:07:53,360 --> 00:07:55,100 See, what did "linear" mean when we 163 00:07:55,100 --> 00:07:58,670 were dealing with ordinary linear change of variables 164 00:07:58,670 --> 00:08:01,220 back in block four of the course, when we talked 165 00:08:01,220 --> 00:08:04,130 about u equals ax plus by and talked 166 00:08:04,130 --> 00:08:06,000 about linear approximations? 167 00:08:06,000 --> 00:08:07,970 "Linear" meant, if the mapping was 168 00:08:07,970 --> 00:08:11,810 linear L of a constant times the input would 169 00:08:11,810 --> 00:08:14,940 be the constant times L of the input. 170 00:08:14,940 --> 00:08:17,430 In other words, you could factor the constant out. 171 00:08:17,430 --> 00:08:24,080 Notice, by the way, if I feed cu into my L machine, 172 00:08:24,080 --> 00:08:25,670 what will the output be? 173 00:08:25,670 --> 00:08:26,960 See, be very careful here. 174 00:08:26,960 --> 00:08:28,850 Don't be blinded by the y. 175 00:08:28,850 --> 00:08:32,490 y stands for the placeholder, the input. 176 00:08:32,490 --> 00:08:35,299 In other words, L of anything is the second derivative 177 00:08:35,299 --> 00:08:38,990 of that anything plus p times the first derivative plus 178 00:08:38,990 --> 00:08:42,630 q times that input, the anything that I put in here. 179 00:08:42,630 --> 00:08:46,430 So notice that L of cu would be the quantity cu 180 00:08:46,430 --> 00:08:52,760 double prime plus p of x times cu prime plus q of x times cu. 181 00:08:52,760 --> 00:08:58,160 In other words, L of cu is equal to this expression here. 182 00:08:58,160 --> 00:09:02,570 Notice that differentiating a constant times a function of x 183 00:09:02,570 --> 00:09:04,430 means that we skip over the constants 184 00:09:04,430 --> 00:09:06,530 and just differentiate the function of x. 185 00:09:06,530 --> 00:09:08,030 In other words, differentiating this 186 00:09:08,030 --> 00:09:11,690 twice would just be the constant times u double prime. 187 00:09:11,690 --> 00:09:13,670 I can take the constants out here, 188 00:09:13,670 --> 00:09:15,140 I can take the constants out here. 189 00:09:15,140 --> 00:09:17,060 In other words, c factors out. 190 00:09:17,060 --> 00:09:18,950 That's where linearity comes in. 191 00:09:18,950 --> 00:09:20,300 This is linear in c. 192 00:09:20,300 --> 00:09:23,030 I factor the c out, what's left is what? 193 00:09:23,030 --> 00:09:26,930 u double prime plus p u prime plus qu. 194 00:09:26,930 --> 00:09:28,730 That's what we're calling L of u. 195 00:09:28,730 --> 00:09:32,430 So L of c times u is c times L of u, 196 00:09:32,430 --> 00:09:33,770 which is a linear property. 197 00:09:33,770 --> 00:09:36,470 And by the way, again let me emphasize 198 00:09:36,470 --> 00:09:40,400 why linearity definition was given the way it was. 199 00:09:40,400 --> 00:09:43,010 For example-- I just said here that it's not 200 00:09:43,010 --> 00:09:45,200 true of nonlinear-- but for example, 201 00:09:45,200 --> 00:09:47,900 suppose I had an equation that had 202 00:09:47,900 --> 00:09:50,633 y prime multiplied by a function of y 203 00:09:50,633 --> 00:09:51,800 rather than a function of x. 204 00:09:51,800 --> 00:09:56,270 In other words, let's look at L of y equals y times y prime. 205 00:09:56,270 --> 00:09:59,600 If I now replace y by c times u, this 206 00:09:59,600 --> 00:10:02,780 becomes what? y is replaced by c times u, 207 00:10:02,780 --> 00:10:06,620 y prime is replaced by the derivative of cu. 208 00:10:06,620 --> 00:10:09,750 Well, what is the derivative of cu if c is a constant? 209 00:10:09,750 --> 00:10:11,990 It's c times u prime. 210 00:10:11,990 --> 00:10:13,490 In other words, the right-hand side 211 00:10:13,490 --> 00:10:16,940 here is just c squared u times u prime. 212 00:10:16,940 --> 00:10:19,670 u times u prime is L of u. 213 00:10:19,670 --> 00:10:24,800 In other words, in this example L of cu would not be c L of u. 214 00:10:24,800 --> 00:10:30,470 It would be c squared L of u, which 215 00:10:30,470 --> 00:10:33,650 is not the linear property that we're talking about. 216 00:10:33,650 --> 00:10:36,320 Finally, the other property of linearity 217 00:10:36,320 --> 00:10:40,970 is that if I replace my input by the sum of two 218 00:10:40,970 --> 00:10:45,590 differentiable functions, u1 and u2, then L of u1 plus u2 219 00:10:45,590 --> 00:10:48,770 turns out to be L of u1 plus L of u2. 220 00:10:48,770 --> 00:10:51,740 And the proof, again, is just by looking at the definition. 221 00:10:51,740 --> 00:10:55,970 If I feed u1 plus u2 into my L machine, I have what? 222 00:10:55,970 --> 00:10:59,510 u1 want plus u2 double prime plus p of x times u1 223 00:10:59,510 --> 00:11:03,770 plus u2 prime plus q of x times u1 plus u2. 224 00:11:03,770 --> 00:11:06,120 The derivative of the sum is the sum of the derivatives, 225 00:11:06,120 --> 00:11:09,170 so I can split this up into u1 double prime plus u2 double 226 00:11:09,170 --> 00:11:09,980 prime. 227 00:11:09,980 --> 00:11:14,720 Similarly, this is p u1 prime p u2 prime. 228 00:11:14,720 --> 00:11:17,660 This is q u1, q u2. 229 00:11:17,660 --> 00:11:20,090 I break down the u1 terms together, 230 00:11:20,090 --> 00:11:21,710 the u2 terms together. 231 00:11:21,710 --> 00:11:25,310 This by definition is L of u1. 232 00:11:25,310 --> 00:11:28,620 This, by definition of L, is L of u2. 233 00:11:28,620 --> 00:11:33,560 In other words, L of u1 plus u2 is L of u1 plus L of u2. 234 00:11:33,560 --> 00:11:35,780 By the way, conditions one and two 235 00:11:35,780 --> 00:11:39,830 can be stated equivalently by the one statement L 236 00:11:39,830 --> 00:11:46,610 of c1u1 plus c2u2 is c1 L of u1 plus c2 L of u2. 237 00:11:46,610 --> 00:11:49,160 The proof will be left as an exercise. 238 00:11:49,160 --> 00:11:51,710 It's a fairly trivial observation. 239 00:11:51,710 --> 00:11:54,050 And I think, as you may remember, 240 00:11:54,050 --> 00:11:57,680 that we did something at least similar to this in block four 241 00:11:57,680 --> 00:11:59,960 when we talked about linear mappings 242 00:11:59,960 --> 00:12:04,120 when we were mapping the xy plane into the uv plane, OK? 243 00:12:04,120 --> 00:12:08,150 But at any rate, let's see what all this means. 244 00:12:08,150 --> 00:12:10,880 Why are these properties so important? 245 00:12:10,880 --> 00:12:14,903 And so let's call our subtopic Properties of Linear Equations. 246 00:12:14,903 --> 00:12:16,820 And the first thing that I'd like to point out 247 00:12:16,820 --> 00:12:20,660 is that if the right-hand side of our linear differential 248 00:12:20,660 --> 00:12:23,420 equation is 0, the interesting fact 249 00:12:23,420 --> 00:12:27,950 is that any linear combination of solutions of that equation 250 00:12:27,950 --> 00:12:29,220 is again a solution. 251 00:12:29,220 --> 00:12:34,130 In other words, if I find some function u1 of x that satisfies 252 00:12:34,130 --> 00:12:37,730 the equation L of y equals 0-- in other words L of u1 is 0-- 253 00:12:37,730 --> 00:12:40,040 and I also know that L of u2 is 0, 254 00:12:40,040 --> 00:12:46,100 then the amazing thing is that L of c1u1 plus c2u2 is also 0. 255 00:12:46,100 --> 00:12:50,600 In other words, any linear combination of u1 and u2, 256 00:12:50,600 --> 00:12:54,500 where u1 and u2 are solutions of L of y equals 0, 257 00:12:54,500 --> 00:12:57,860 will also be a solution of L of y equals 0. 258 00:12:57,860 --> 00:12:59,990 And the proof, again, is trivial. 259 00:12:59,990 --> 00:13:03,440 Namely, by the properties of linearity, 260 00:13:03,440 --> 00:13:06,950 L of c1u1 plus c2u2 is what? 261 00:13:06,950 --> 00:13:11,630 It's c1 L of u1 plus c2 L of u2. 262 00:13:11,630 --> 00:13:13,430 L of u1 is 0. 263 00:13:13,430 --> 00:13:16,040 L of u2 is 0-- that was given, you see. 264 00:13:16,040 --> 00:13:18,510 Consequently, this is 0. 265 00:13:18,510 --> 00:13:21,530 See, a constant times 0 plus a constant times 0 is 0. 266 00:13:21,530 --> 00:13:24,590 And that's exactly what we wanted to show over here. 267 00:13:24,590 --> 00:13:29,090 A second important factor is that if I have a solution of L 268 00:13:29,090 --> 00:13:32,010 of y equals 0-- say, L of u is 0-- 269 00:13:32,010 --> 00:13:36,410 and I also have a solution v of the equation L of y equals f 270 00:13:36,410 --> 00:13:40,370 of x-- in other words, if L of u is 0 and L of v is f of x-- 271 00:13:40,370 --> 00:13:43,880 the amazing thing is, that if I add these two solutions 272 00:13:43,880 --> 00:13:46,760 together, the resulting function will 273 00:13:46,760 --> 00:13:51,650 be a solution of the equation L of y equals f of x. 274 00:13:51,650 --> 00:13:55,580 In other words, if L of u is 0 and L of v is f of x, 275 00:13:55,580 --> 00:14:00,020 then L of u plus v is also equal to f of x. 276 00:14:00,020 --> 00:14:02,120 What this means is that, whenever 277 00:14:02,120 --> 00:14:05,660 I add onto a solution of this equation, any solution 278 00:14:05,660 --> 00:14:08,660 of this equation I again get back 279 00:14:08,660 --> 00:14:10,880 a solution of this equation. 280 00:14:10,880 --> 00:14:15,350 The proof, again, by definition of linearity is trivial. 281 00:14:15,350 --> 00:14:21,200 Namely, by linearity L of u plus v is L of u plus L of v. 282 00:14:21,200 --> 00:14:23,750 But we're given that L of u is 0. 283 00:14:23,750 --> 00:14:26,930 We're given that L of v is f of x. 284 00:14:26,930 --> 00:14:30,110 And consequently, 0 plus f of x is f of x. 285 00:14:30,110 --> 00:14:33,260 So L of u plus v is f of x, as asserted. 286 00:14:33,260 --> 00:14:35,510 By the way, let me point out here-- 287 00:14:35,510 --> 00:14:39,080 do not overlook the power of linearity. 288 00:14:39,080 --> 00:14:42,440 There is a very big danger that what you might say here 289 00:14:42,440 --> 00:14:46,550 is, wasn't this result true just by adding equals to equals? 290 00:14:46,550 --> 00:14:49,340 In other words, couldn't I just add these two results? 291 00:14:49,340 --> 00:14:51,330 And the answer is yes, you can. 292 00:14:51,330 --> 00:14:53,600 But when you add equals to equals here, 293 00:14:53,600 --> 00:14:55,670 notice that what you get is what? 294 00:14:55,670 --> 00:15:01,550 L of u plus L of v is equal to 0 plus f of x. 295 00:15:01,550 --> 00:15:04,940 In other words, what you can prove by equals added to equals 296 00:15:04,940 --> 00:15:08,480 is that L of u plus L of v is f of x. 297 00:15:08,480 --> 00:15:12,860 It was linearity that allowed us to say that L of u plus L of v 298 00:15:12,860 --> 00:15:16,310 was the same as L of u plus v. And to show 299 00:15:16,310 --> 00:15:20,090 you that in terms of a simple analogy, 300 00:15:20,090 --> 00:15:21,950 let's suppose I take the function 301 00:15:21,950 --> 00:15:26,930 f to be defined by f of x is x minus 2 times x minus 3. 302 00:15:26,930 --> 00:15:32,210 Well, trivially, when x is 2 or x is 3, f of x is 0, right? 303 00:15:32,210 --> 00:15:35,560 In other words, f of 2 is 0, f of 3 is 0. 304 00:15:35,560 --> 00:15:38,130 Now, by equals added to equals, you 305 00:15:38,130 --> 00:15:43,140 can certainly say that f of 2 plus f of 3 is zero. 306 00:15:43,140 --> 00:15:47,280 But you can't say that f of the quantity 2 plus 3 is 0. 307 00:15:47,280 --> 00:15:50,040 In fact, 2 plus 3 is 5. 308 00:15:50,040 --> 00:15:52,500 If I put 5 in here, I get what? 309 00:15:52,500 --> 00:15:57,060 f of 5 is 5 minus 2 times 5 minus 3. 310 00:15:57,060 --> 00:15:59,580 That's 3 times 2, or 6. 311 00:15:59,580 --> 00:16:03,840 In other words, f of 2 plus f of 3 is 0, but f of 2 312 00:16:03,840 --> 00:16:06,420 plus 3 is not 0, it's 6. 313 00:16:06,420 --> 00:16:09,090 You see, equals added to equals gives you this result. 314 00:16:09,090 --> 00:16:11,250 But it's linearity that you need to be 315 00:16:11,250 --> 00:16:14,130 able to get from here to here. 316 00:16:14,130 --> 00:16:17,130 This was not a linear function, you see. 317 00:16:17,130 --> 00:16:20,610 Well, let me give you an example of how to use all this stuff. 318 00:16:20,610 --> 00:16:24,470 Let's find all solutions of the differential equation 319 00:16:24,470 --> 00:16:29,520 y double prime minus 4 y prime plus 3y equals 0. 320 00:16:29,520 --> 00:16:32,490 Special case where the right-hand side is 0. 321 00:16:32,490 --> 00:16:35,640 My coefficients of y and y prime, notice, 322 00:16:35,640 --> 00:16:36,960 are still functions of x. 323 00:16:36,960 --> 00:16:40,110 They happen to be constant, but the requirement 324 00:16:40,110 --> 00:16:42,570 that p of x and q of x be constants 325 00:16:42,570 --> 00:16:44,640 is certainly not outlawed. 326 00:16:44,640 --> 00:16:47,100 In other words, one special case of a function of x 327 00:16:47,100 --> 00:16:49,830 is the function of x, which is identically a constant. 328 00:16:49,830 --> 00:16:53,010 So this qualifies as a linear equation. 329 00:16:53,010 --> 00:16:55,470 Let me show you, then, how I tried to find 330 00:16:55,470 --> 00:16:57,390 all solutions of this equation. 331 00:16:57,390 --> 00:17:00,760 As we've mentioned before both in part one of our course 332 00:17:00,760 --> 00:17:04,260 and as a motivation for defining e to the ix, 333 00:17:04,260 --> 00:17:07,470 a trial solution of this equation 334 00:17:07,470 --> 00:17:10,170 involves using e to the rx. 335 00:17:10,170 --> 00:17:12,839 Because we differentiate e to the rx, you get e to the rx 336 00:17:12,839 --> 00:17:13,740 back again. 337 00:17:13,740 --> 00:17:17,400 If I differentiate e to the rx, I get r e to the rx. 338 00:17:17,400 --> 00:17:20,400 Differentiate again, I get r squared e to the rx. 339 00:17:20,400 --> 00:17:24,420 I plug that into here, factor out the e to the rx, 340 00:17:24,420 --> 00:17:28,170 and I wind up with e to the rx times r squared 341 00:17:28,170 --> 00:17:31,800 minus 4r plus 3 must equal 0. 342 00:17:31,800 --> 00:17:35,070 Since e to the rx is not 0, it must 343 00:17:35,070 --> 00:17:38,160 be that r squared minus 4r plus 3 is 0. 344 00:17:38,160 --> 00:17:41,610 And that says that either r is 1 or r is 3. 345 00:17:41,610 --> 00:17:46,470 Remembering what r is, it means that my trial solutions should 346 00:17:46,470 --> 00:17:49,770 be correct when r is 1 or r is 3. 347 00:17:49,770 --> 00:17:52,620 Leaving the details for you to verify, 348 00:17:52,620 --> 00:17:55,980 it does turn out that L of e to the x is 0-- 349 00:17:55,980 --> 00:17:57,360 see, r is 1-- 350 00:17:57,360 --> 00:17:58,890 L of e to the 3x-- 351 00:17:58,890 --> 00:18:00,580 r is 3-- is 0. 352 00:18:00,580 --> 00:18:04,920 In other words, if I were to replace y by e to the x 353 00:18:04,920 --> 00:18:08,550 or by e to the 3x, this equation is satisfied. 354 00:18:08,550 --> 00:18:10,710 And in terms of our definition of L, 355 00:18:10,710 --> 00:18:13,050 this is the abbreviation for writing this. 356 00:18:13,050 --> 00:18:18,420 Well, by our first property, the fact that this satisfies L of y 357 00:18:18,420 --> 00:18:23,010 equals 0 and this satisfies L of y equals 0, we had what? 358 00:18:23,010 --> 00:18:26,970 Any linear combination of these two must satisfy L of y 359 00:18:26,970 --> 00:18:27,900 equals 0. 360 00:18:27,900 --> 00:18:30,930 In other words, I now know in one fell swoop 361 00:18:30,930 --> 00:18:34,890 that every function of the form c1 e to the x plus c2 362 00:18:34,890 --> 00:18:39,250 e to the 3x must also be a solution of this equation. 363 00:18:39,250 --> 00:18:39,960 OK? 364 00:18:39,960 --> 00:18:42,960 By the way, that's another motivation-- 365 00:18:42,960 --> 00:18:45,130 and why we'll be doing this in block eight-- 366 00:18:45,130 --> 00:18:48,490 for going further into the study of vector spaces. 367 00:18:48,490 --> 00:18:50,610 Notice, in a sense what you're saying is, 368 00:18:50,610 --> 00:18:53,310 you have found a whole family of solutions 369 00:18:53,310 --> 00:18:55,065 which are linear combinations of e 370 00:18:55,065 --> 00:18:58,260 to the x and e to the 3x that somehow or other 371 00:18:58,260 --> 00:19:01,560 e to the 3x and e to the x behave like i 372 00:19:01,560 --> 00:19:03,210 and j did in the plane. 373 00:19:03,210 --> 00:19:06,740 Namely, every solution of this type can be written as what? 374 00:19:06,740 --> 00:19:10,890 A constant times e to the x plus a constant times e to the 3x. 375 00:19:10,890 --> 00:19:12,870 This is like a two-dimensional vector space. 376 00:19:12,870 --> 00:19:16,020 But I'm not going to pursue that any further right now. 377 00:19:16,020 --> 00:19:18,960 I just wanted to give you a preview of coming attractions. 378 00:19:18,960 --> 00:19:21,810 But at any rate, to summarize what we've done so far 379 00:19:21,810 --> 00:19:25,110 is that we have now found that one family of solutions 380 00:19:25,110 --> 00:19:28,320 of that equation is y equals c1 e to the x 381 00:19:28,320 --> 00:19:30,060 plus c to e to the 3x. 382 00:19:30,060 --> 00:19:33,750 I emphasize "one family," because so far, these 383 00:19:33,750 --> 00:19:37,320 are the solutions I found by assuming that the solution had 384 00:19:37,320 --> 00:19:40,800 to have the form y is e to the rx. 385 00:19:40,800 --> 00:19:42,630 I don't know as yet whether there 386 00:19:42,630 --> 00:19:44,040 are other types of solutions. 387 00:19:44,040 --> 00:19:45,990 I'll worry about that in a little while. 388 00:19:45,990 --> 00:19:49,230 At any rate, the next most natural question to ask here 389 00:19:49,230 --> 00:19:50,120 is this-- 390 00:19:50,120 --> 00:19:50,730 look it. 391 00:19:50,730 --> 00:19:53,010 I have two arbitrary constants. 392 00:19:53,010 --> 00:19:55,500 And because I have two arbitrary constants, 393 00:19:55,500 --> 00:19:58,830 it seems to me that, just like in the first order case, 394 00:19:58,830 --> 00:20:01,200 not only should I be able to find a solution that 395 00:20:01,200 --> 00:20:03,210 passes through a particular point, 396 00:20:03,210 --> 00:20:05,730 but I have another degree of freedom to play around with. 397 00:20:05,730 --> 00:20:09,090 Maybe I can require, not only if the curve passes 398 00:20:09,090 --> 00:20:12,780 through a given point, but that it have a particular slope when 399 00:20:12,780 --> 00:20:14,195 it passes through that point. 400 00:20:14,195 --> 00:20:15,570 In other words, with one constant 401 00:20:15,570 --> 00:20:17,280 I could make it pass through a point. 402 00:20:17,280 --> 00:20:20,640 With two, maybe I could make it pass through a point having 403 00:20:20,640 --> 00:20:21,720 a given slope. 404 00:20:21,720 --> 00:20:27,270 So the question now is, can I determine c1 and c2, 405 00:20:27,270 --> 00:20:30,540 such that for a given point x0, y0 406 00:20:30,540 --> 00:20:33,630 I can find a curve, a member of this family, 407 00:20:33,630 --> 00:20:37,350 which passes through the point x0, y0, 408 00:20:37,350 --> 00:20:39,630 and has slope equal to z0. 409 00:20:39,630 --> 00:20:43,330 Why I use z0 instead of m here will become clear, 410 00:20:43,330 --> 00:20:44,790 I hope, in a few moments. 411 00:20:44,790 --> 00:20:47,370 But the point is, to see if I can satisfy 412 00:20:47,370 --> 00:20:49,230 this system of equations, let's see 413 00:20:49,230 --> 00:20:54,120 what happens if I replace x and y by x0 and y0. 414 00:20:54,120 --> 00:20:59,250 One of my equations that must be satisfied by c1 and c2 is this. 415 00:20:59,250 --> 00:21:01,410 The derivative of this, which is a slope, 416 00:21:01,410 --> 00:21:05,940 is c1 e to the x plus 3 c2 e to the 3x. 417 00:21:05,940 --> 00:21:10,600 So if the slope is going to be z0, when x is equal to x0 418 00:21:10,600 --> 00:21:12,900 this equation must also be obeyed. 419 00:21:12,900 --> 00:21:15,540 Consequently, I must now solve-- 420 00:21:15,540 --> 00:21:19,260 see if I can solve these two equations for c1 and c2. 421 00:21:19,260 --> 00:21:21,690 By the way, this is very important to notice. 422 00:21:21,690 --> 00:21:25,800 Notice that this is two equations and two unknowns, 423 00:21:25,800 --> 00:21:28,470 and that my unknowns are c1 and c2-- 424 00:21:28,470 --> 00:21:31,800 that once I specify x0, y0, and z0, 425 00:21:31,800 --> 00:21:35,310 everything else is a known number in this problem. 426 00:21:35,310 --> 00:21:38,220 To see whether this has a unique solution, 427 00:21:38,220 --> 00:21:41,220 the determinant of coefficients must be unequal to 0. 428 00:21:41,220 --> 00:21:43,650 But look at what that determinant of coefficients is. 429 00:21:43,650 --> 00:21:49,980 It's e x0 e 3x0, e x0, 3e 3x0. 430 00:21:49,980 --> 00:21:51,600 That determinant is what? 431 00:21:51,600 --> 00:21:56,940 3e to the 4x0 minus e to the 4x0. 432 00:21:56,940 --> 00:21:59,460 That's twice e to the 4x0. 433 00:21:59,460 --> 00:22:02,250 And since the exponential can never be negative-- 434 00:22:02,250 --> 00:22:05,940 can never be 0, this determinant cannot be 0. 435 00:22:05,940 --> 00:22:09,960 In other words, there is a unique member of the family y 436 00:22:09,960 --> 00:22:13,380 equals c1 e to the x plus c2 e to the 3x 437 00:22:13,380 --> 00:22:18,570 that passes through the point x0, y0 with a given slope z0. 438 00:22:18,570 --> 00:22:20,310 That's what we prove over here. 439 00:22:20,310 --> 00:22:22,070 The only question that comes up is, 440 00:22:22,070 --> 00:22:24,120 is that we have now shown what? 441 00:22:24,120 --> 00:22:26,640 That at every point in space there 442 00:22:26,640 --> 00:22:30,090 is one solution from this family that 443 00:22:30,090 --> 00:22:33,600 passes through the given point with any given slope 444 00:22:33,600 --> 00:22:35,310 that you wanted to have. 445 00:22:35,310 --> 00:22:36,870 The question is that before we can 446 00:22:36,870 --> 00:22:38,970 call this the general solution, we 447 00:22:38,970 --> 00:22:41,010 have to be sure that there are no 448 00:22:41,010 --> 00:22:43,380 other solutions to the equation y 449 00:22:43,380 --> 00:22:46,920 double prime minus 4 y prime plus 3y equals 0. 450 00:22:46,920 --> 00:22:49,500 In other words, the question is are there 451 00:22:49,500 --> 00:22:52,270 other types of solutions-- 452 00:22:52,270 --> 00:22:54,300 solutions that aren't of the form e 453 00:22:54,300 --> 00:22:58,200 to the x, or e to the 3x, or linear combinations thereof? 454 00:22:58,200 --> 00:23:00,270 And the crucial theorem is this. 455 00:23:00,270 --> 00:23:04,260 Suppose we can write our second order equation in the form y 456 00:23:04,260 --> 00:23:08,250 double prime is some function of x, y, and y prime. 457 00:23:08,250 --> 00:23:10,980 And notice how analogous this is to the key theorem 458 00:23:10,980 --> 00:23:12,690 of our last lecture. 459 00:23:12,690 --> 00:23:16,020 Suppose when we treat F as a function of the three 460 00:23:16,020 --> 00:23:18,300 independent variables x, y, and z. 461 00:23:18,300 --> 00:23:22,260 It turns out that F, F sub y, F sub z, 462 00:23:22,260 --> 00:23:25,200 are all continuous in some region R. 463 00:23:25,200 --> 00:23:28,440 Then the amazing result is that for each triplet-- 464 00:23:28,440 --> 00:23:32,130 x0, y0, z0, in R-- 465 00:23:32,130 --> 00:23:33,810 see, R is three-dimensional here, 466 00:23:33,810 --> 00:23:35,940 because F is defined on three space-- 467 00:23:35,940 --> 00:23:39,480 there is a unique solution curve which passes through the point 468 00:23:39,480 --> 00:23:42,780 x0, y0 with slope equal to z0. 469 00:23:42,780 --> 00:23:45,450 Again, notice here the coding system. 470 00:23:45,450 --> 00:23:47,580 We are not talking about a solution 471 00:23:47,580 --> 00:23:50,840 passing through x0, y0, z0. 472 00:23:50,840 --> 00:23:52,740 The solution curve is in the plane. 473 00:23:52,740 --> 00:23:55,770 See, dy dx is a slope of a curve on the plane. 474 00:23:55,770 --> 00:23:57,210 We have one independent variable. 475 00:23:57,210 --> 00:23:58,590 What we're saying is what? 476 00:23:58,590 --> 00:24:02,160 That there is a unique solution that passes through the point 477 00:24:02,160 --> 00:24:06,540 x0, y0 with slope equal to z0. 478 00:24:06,540 --> 00:24:09,900 Under those conditions, the solution would be unique. 479 00:24:09,900 --> 00:24:13,080 Well, let's accept the truth of this theorem 480 00:24:13,080 --> 00:24:16,680 and apply it to a linear differential equation. 481 00:24:16,680 --> 00:24:19,590 Given the most general linear differential equation, 482 00:24:19,590 --> 00:24:22,080 to put it into the form of the crucial theorem 483 00:24:22,080 --> 00:24:25,020 I transpose everything but y double prime 484 00:24:25,020 --> 00:24:27,180 onto the right-hand side of the equation. 485 00:24:27,180 --> 00:24:29,970 I get y double prime is f of x minus p 486 00:24:29,970 --> 00:24:33,210 of x y prime minus q of xy. 487 00:24:33,210 --> 00:24:36,090 Therefore, my capital F of x, y, z 488 00:24:36,090 --> 00:24:39,920 is obtained simply by replacing y prime by z in here. 489 00:24:39,920 --> 00:24:44,520 See, I simply replace y prime by z, like I did over here. 490 00:24:44,520 --> 00:24:46,890 I wind up with what? 491 00:24:46,890 --> 00:24:52,410 Capital F of x, y, z is little f of x minus pz minus qy. 492 00:24:52,410 --> 00:24:53,745 Well look it. 493 00:24:53,745 --> 00:24:56,850 This is certainly a continuous function as soon as f, p, 494 00:24:56,850 --> 00:24:58,360 and q are continuous. 495 00:24:58,360 --> 00:25:01,290 The partial of capital F with respect to y-- 496 00:25:01,290 --> 00:25:04,150 remember, I'm treating x, y, and z as independent variables. 497 00:25:04,150 --> 00:25:06,270 If I differentiate with respect to y, 498 00:25:06,270 --> 00:25:07,980 notice that this drops out. 499 00:25:07,980 --> 00:25:10,980 The partial of capital F with respect to y is just minus q 500 00:25:10,980 --> 00:25:11,940 of x. 501 00:25:11,940 --> 00:25:15,660 The partial of capital f with respect to z is just minus p 502 00:25:15,660 --> 00:25:16,410 of x. 503 00:25:16,410 --> 00:25:20,040 Consequently, if little f, p, and q 504 00:25:20,040 --> 00:25:23,940 are all continuous functions of x, automatically capital F, 505 00:25:23,940 --> 00:25:29,350 capital F sub y, capital F sub z will be continuous functions. 506 00:25:29,350 --> 00:25:32,590 Consequently, according to that crucial theorem, 507 00:25:32,590 --> 00:25:36,260 any solution that I find will be a unique solution. 508 00:25:36,260 --> 00:25:39,230 And even more to the point, even if I can't find the solution, 509 00:25:39,230 --> 00:25:42,310 the theorem tells me that there is a unique solution. 510 00:25:42,310 --> 00:25:45,010 Well, I think that the p's and the q's sort of gets 511 00:25:45,010 --> 00:25:46,310 you a little bit messed up. 512 00:25:46,310 --> 00:25:50,530 Let's do a specific illustration of this again. 513 00:25:50,530 --> 00:25:54,540 Let's find now all solutions of the equation 514 00:25:54,540 --> 00:26:00,040 y double prime minus 4 y prime plus 3y equals e to the 2x. 515 00:26:00,040 --> 00:26:03,730 The same equation as before, only the right-hand side 516 00:26:03,730 --> 00:26:06,160 is e to the 2x rather than 0. 517 00:26:06,160 --> 00:26:08,260 The key result is going to be this. 518 00:26:08,260 --> 00:26:10,750 I already know how to solve this equation 519 00:26:10,750 --> 00:26:12,730 when the right-hand side is 0. 520 00:26:12,730 --> 00:26:16,240 Remember, one of my properties of linearity was that if I 521 00:26:16,240 --> 00:26:20,470 could find any solution of this equation, 522 00:26:20,470 --> 00:26:24,280 then by adding onto it any solution of the equation where 523 00:26:24,280 --> 00:26:25,990 the right-hand side is 0-- 524 00:26:25,990 --> 00:26:28,150 by the way, when the right hand side is 0, 525 00:26:28,150 --> 00:26:31,840 the equation is called homogeneous. 526 00:26:31,840 --> 00:26:33,780 And I don't think that's too important now, 527 00:26:33,780 --> 00:26:36,160 then, to know the language when it's mentioned, 528 00:26:36,160 --> 00:26:39,575 but this is called homogeneous if the right-hand side is 0. 529 00:26:39,575 --> 00:26:41,200 What our key theorem says is, look it-- 530 00:26:41,200 --> 00:26:42,910 we have solved this problem. 531 00:26:42,910 --> 00:26:44,920 We have found the general solution 532 00:26:44,920 --> 00:26:46,630 when the right-hand side is 0. 533 00:26:46,630 --> 00:26:49,330 Consequently, if I could find just one solution 534 00:26:49,330 --> 00:26:53,350 of this equation, by hook or by crook, just steal one, 535 00:26:53,350 --> 00:26:54,430 sneak one in. 536 00:26:54,430 --> 00:26:56,830 If I can find one solution of that equation, 537 00:26:56,830 --> 00:26:59,440 if I add that onto the general solution 538 00:26:59,440 --> 00:27:01,480 of the homogeneous equation, that 539 00:27:01,480 --> 00:27:04,720 will be the general solution of this equation. 540 00:27:04,720 --> 00:27:05,900 What does that mean? 541 00:27:05,900 --> 00:27:07,360 Let's see how we'll tackle this. 542 00:27:07,360 --> 00:27:09,910 What I'm going to do is, I look at this equation. 543 00:27:09,910 --> 00:27:11,380 And here's how I get sneaky. 544 00:27:11,380 --> 00:27:12,010 I say, look it. 545 00:27:12,010 --> 00:27:14,980 When I differentiate, and I'm all through, what I want 546 00:27:14,980 --> 00:27:17,160 to wind up with is e to the 2x. 547 00:27:17,160 --> 00:27:19,690 Well, the only function whose derivative 548 00:27:19,690 --> 00:27:22,810 gives you a factor of e to the 2x, more or less, 549 00:27:22,810 --> 00:27:26,080 is some constant times e to the 2x itself. 550 00:27:26,080 --> 00:27:29,950 So I say to myself, let me try for a solution in the form y 551 00:27:29,950 --> 00:27:33,460 equals some constant times e to the 2x. 552 00:27:33,460 --> 00:27:36,440 See, my trial solution will have this form. 553 00:27:36,440 --> 00:27:41,230 Well, yT prime would be this, yT double prime would be this. 554 00:27:41,230 --> 00:27:44,890 If I now substitute back into the original equation, 555 00:27:44,890 --> 00:27:51,970 yT double prime minus 4 yT prime plus 3yT 556 00:27:51,970 --> 00:27:54,670 must be identically equal to e to the x. 557 00:27:54,670 --> 00:27:58,090 That leads to the fact that minus A e to the 2x 558 00:27:58,090 --> 00:28:00,250 must be the same as e to the 2x. 559 00:28:00,250 --> 00:28:03,220 And that tells me that if there is a solution, 560 00:28:03,220 --> 00:28:05,560 A had better be minus 1. 561 00:28:05,560 --> 00:28:08,650 You see, if x is 0 here, this is minus A equals 1. 562 00:28:08,650 --> 00:28:13,540 A quick check will show that minus e to the 2x 563 00:28:13,540 --> 00:28:15,320 is a solution of the equation. 564 00:28:15,320 --> 00:28:19,570 So a particular solution of this equation is minus e to the 2x. 565 00:28:19,570 --> 00:28:21,100 I've found one solution. 566 00:28:21,100 --> 00:28:23,530 Now, here's the power of all this theory. 567 00:28:23,530 --> 00:28:25,870 With this one solution, I now go back 568 00:28:25,870 --> 00:28:27,850 to the homogeneous equation, which 569 00:28:27,850 --> 00:28:33,940 had as its general solution c1 e to the x plus c2 e to the 3x. 570 00:28:33,940 --> 00:28:38,680 I add these two together, and that is the general solution. 571 00:28:38,680 --> 00:28:40,720 That is the general solution of the equation 572 00:28:40,720 --> 00:28:42,820 that I started with. 573 00:28:42,820 --> 00:28:44,800 By the way, notice just as a check-- 574 00:28:44,800 --> 00:28:46,600 if this were the general solution, 575 00:28:46,600 --> 00:28:51,580 I should be able to find unique values for c1 and c2 576 00:28:51,580 --> 00:28:53,560 that allow me to make the curve-- 577 00:28:53,560 --> 00:28:59,080 a solution curve pass through x0, y0 with slope equal to z0. 578 00:28:59,080 --> 00:29:03,460 Notice that that would result in this system of equations. 579 00:29:03,460 --> 00:29:07,780 And I hope that you notice that since x0, y0, and z0 580 00:29:07,780 --> 00:29:11,860 are constants, to determine whether c1 and c2 are 581 00:29:11,860 --> 00:29:15,580 uniquely determined or not hinges on the fact 582 00:29:15,580 --> 00:29:20,020 that the coefficient matrix, the determinant of coefficients, 583 00:29:20,020 --> 00:29:22,540 is still the same determinant of coefficients 584 00:29:22,540 --> 00:29:24,550 that I had in the homogeneous case. 585 00:29:24,550 --> 00:29:26,800 By the way, let me just stress one more point 586 00:29:26,800 --> 00:29:28,660 that I forgot to mention over here. 587 00:29:28,660 --> 00:29:31,090 When I wrote down this equation, notice 588 00:29:31,090 --> 00:29:33,370 that for the fundamental theorem to be true, 589 00:29:33,370 --> 00:29:37,990 all that we required was that p, q, and little f be continuous. 590 00:29:37,990 --> 00:29:42,640 Notice that p in this problem is minus 4, q is 3, 591 00:29:42,640 --> 00:29:44,740 and f is e to the 2x. 592 00:29:44,740 --> 00:29:46,990 Certainly, the function f of-- 593 00:29:46,990 --> 00:29:49,180 the function which is identically minus 4, 594 00:29:49,180 --> 00:29:51,100 the function which is identically 3, 595 00:29:51,100 --> 00:29:53,170 and the function which is e to the 2x 596 00:29:53,170 --> 00:29:55,100 are all continuous functions. 597 00:29:55,100 --> 00:29:57,880 So what that told me was that once I find 598 00:29:57,880 --> 00:30:01,390 one family of solutions, I've found them all, 599 00:30:01,390 --> 00:30:05,150 so in the linear case, there is a general solution. 600 00:30:05,150 --> 00:30:08,710 There are no singular solutions in a problem of this type. 601 00:30:08,710 --> 00:30:10,780 There can't be any [? mongrel ?] solutions, 602 00:30:10,780 --> 00:30:15,160 because this particular equation meets the requirements 603 00:30:15,160 --> 00:30:17,110 of the crucial theorem. 604 00:30:17,110 --> 00:30:21,100 At any rate, the exercises will take care of this 605 00:30:21,100 --> 00:30:22,360 in giving you drill. 606 00:30:22,360 --> 00:30:24,430 Let me simply summarize what we've 607 00:30:24,430 --> 00:30:27,960 done today in terms of linear differential equation. 608 00:30:27,960 --> 00:30:30,160 So summary is this. 609 00:30:30,160 --> 00:30:33,097 Let's suppose that we're given a general-- 610 00:30:33,097 --> 00:30:34,930 second order is what I've been dealing with, 611 00:30:34,930 --> 00:30:36,930 but it's true for higher orders, too-- 612 00:30:36,930 --> 00:30:40,050 linear equation of the form y double prime plus p 613 00:30:40,050 --> 00:30:44,400 y prime plus qy equals f of x, where p and q are 614 00:30:44,400 --> 00:30:46,980 arbitrary functions of x. 615 00:30:46,980 --> 00:30:49,140 What I'm saying is to find a general solution 616 00:30:49,140 --> 00:30:50,680 of this equation-- 617 00:30:50,680 --> 00:30:53,370 and by the crucial theorem, this general solution 618 00:30:53,370 --> 00:30:57,030 will exist if f, p, and q are continuous-- 619 00:30:57,030 --> 00:31:01,620 first of all what I do is I find the general solution y sub 620 00:31:01,620 --> 00:31:05,550 h of the homogeneous equation L of y equals 0. 621 00:31:05,550 --> 00:31:08,550 In other words, I simply replace the right-hand side 622 00:31:08,550 --> 00:31:11,460 of this equation by 0 and find the general solution 623 00:31:11,460 --> 00:31:13,590 of this equation. 624 00:31:13,590 --> 00:31:16,620 In terms of the theory, what we're saying also is-- 625 00:31:16,620 --> 00:31:18,550 and this I'll say this as an aside-- 626 00:31:18,550 --> 00:31:21,780 if I can find two solutions of this equation which 627 00:31:21,780 --> 00:31:24,590 are not scalar multiples of one another-- 628 00:31:24,590 --> 00:31:27,600 in other words, if I can find two functions u1 and u2 629 00:31:27,600 --> 00:31:30,630 such that L of u1 and L of u2 are 0-- 630 00:31:30,630 --> 00:31:33,570 but that u2 is not a constant times u1, 631 00:31:33,570 --> 00:31:35,710 then that homogeneous solution-- in other words, 632 00:31:35,710 --> 00:31:37,170 the solution of this equation-- 633 00:31:37,170 --> 00:31:41,450 will simply be c1u1 plus c2u2. 634 00:31:41,450 --> 00:31:44,130 See, that was all in your combinations of these two. 635 00:31:44,130 --> 00:31:46,440 And the proof of that is quite simple. 636 00:31:46,440 --> 00:31:48,870 Namely, what we want is what? 637 00:31:48,870 --> 00:31:52,170 Given the point x0, y0, we want to be sure 638 00:31:52,170 --> 00:31:56,310 that we can determine c1 and c2 such that a curve passes 639 00:31:56,310 --> 00:32:00,000 through x0, y0 with slope z0. 640 00:32:00,000 --> 00:32:02,940 That means I must be able to do what? 641 00:32:02,940 --> 00:32:06,330 I must be able to solve this pair of equations. 642 00:32:06,330 --> 00:32:10,470 Namely, if I replace x by x0 and y by y0, 643 00:32:10,470 --> 00:32:12,180 and if I differentiate the equation 644 00:32:12,180 --> 00:32:16,200 and then replace x by x0 and dy dx 645 00:32:16,200 --> 00:32:19,140 by z0, because that's the derivative, the slope I 646 00:32:19,140 --> 00:32:21,830 want at this point, I again wind up with what? 647 00:32:21,830 --> 00:32:24,270 Two equations with two unknowns. 648 00:32:24,270 --> 00:32:28,860 Remember, once I specify x0, y0, and z0-- 649 00:32:28,860 --> 00:32:32,880 these are given constants, c1 and c2 are my only variables-- 650 00:32:32,880 --> 00:32:35,340 to see whether there is a unique solution here, 651 00:32:35,340 --> 00:32:37,410 it means that this determinant of coefficients 652 00:32:37,410 --> 00:32:39,730 must be unequal to 0. 653 00:32:39,730 --> 00:32:42,840 The determinant of coefficients is just u1 u2 prime 654 00:32:42,840 --> 00:32:44,880 minus u1 prime u2. 655 00:32:44,880 --> 00:32:46,650 That must be unequal to 0. 656 00:32:46,650 --> 00:32:50,040 This suggests the quotient rule again. 657 00:32:50,040 --> 00:32:52,140 This expression to be unequal to 0 658 00:32:52,140 --> 00:32:56,190 is the same as this divided by u1 squared to be unequal to 0. 659 00:32:56,190 --> 00:33:00,750 This expression here is nothing more than a derivative of u2 660 00:33:00,750 --> 00:33:03,390 divided by u1 with respect to x. 661 00:33:03,390 --> 00:33:06,690 And for that to be unequal to 0 simply says what? 662 00:33:06,690 --> 00:33:09,270 That u2/u1 is not a constant. 663 00:33:09,270 --> 00:33:12,780 And that says that u2 is not a constant times u1. 664 00:33:12,780 --> 00:33:14,760 In other words, this is just an aside 665 00:33:14,760 --> 00:33:17,310 to show you how you find the general solution 666 00:33:17,310 --> 00:33:18,870 of the homogeneous equation. 667 00:33:18,870 --> 00:33:20,040 But the point is what? 668 00:33:20,040 --> 00:33:22,320 First of all, you find the general solution 669 00:33:22,320 --> 00:33:25,110 of the homogeneous equation, meaning you take L of y 670 00:33:25,110 --> 00:33:27,960 equals f of x, replace f of x by 0, 671 00:33:27,960 --> 00:33:31,140 and find the general solution of L of y equals 0. 672 00:33:31,140 --> 00:33:35,150 After you've done that, you find any old solution, 673 00:33:35,150 --> 00:33:39,210 y sub p, of the given equation L of y equals f as x. 674 00:33:39,210 --> 00:33:41,850 Just one solution, by hook or by crook. 675 00:33:41,850 --> 00:33:44,670 Then finally, when you have the general solution 676 00:33:44,670 --> 00:33:47,130 of the reduced-- the homogeneous equation 677 00:33:47,130 --> 00:33:49,500 and the particular solution of this equation, 678 00:33:49,500 --> 00:33:51,780 the general solution of this equation 679 00:33:51,780 --> 00:33:55,110 will just be the sum of these two. 680 00:33:55,110 --> 00:33:55,925 All right? 681 00:33:55,925 --> 00:33:57,300 And that's what we're going to be 682 00:33:57,300 --> 00:34:01,500 talking about in the next few lessons, see. 683 00:34:01,500 --> 00:34:04,560 What we're going to drill on is what this stuff means. 684 00:34:04,560 --> 00:34:07,290 And I think you can now guess what the next lectures are 685 00:34:07,290 --> 00:34:08,460 going to be about. 686 00:34:08,460 --> 00:34:11,070 After all, since to find the general solution 687 00:34:11,070 --> 00:34:13,230 of the linear differential equation 688 00:34:13,230 --> 00:34:16,739 I need the general solution of the homogeneous equation 689 00:34:16,739 --> 00:34:20,100 and a particular solution of the original equation, the two 690 00:34:20,100 --> 00:34:22,199 separate topics I now have to tackle 691 00:34:22,199 --> 00:34:26,429 are, one, how do I find the general solution of the reduced 692 00:34:26,429 --> 00:34:31,260 equation; and two, how do I find a particular solution 693 00:34:31,260 --> 00:34:32,882 of the original equation? 694 00:34:32,882 --> 00:34:35,340 That, by the way, comes under the heading of [? cookbook ?] 695 00:34:35,340 --> 00:34:35,940 again. 696 00:34:35,940 --> 00:34:37,050 That's drill. 697 00:34:37,050 --> 00:34:39,060 But this is the underlying theory, 698 00:34:39,060 --> 00:34:40,770 the underlying philosophy. 699 00:34:40,770 --> 00:34:44,460 We will talk more about how to handle the techniques 700 00:34:44,460 --> 00:34:46,170 in our subsequent lectures. 701 00:34:46,170 --> 00:34:48,980 At any rate, then, until next time, goodbye. 702 00:34:53,760 --> 00:34:56,159 Funding for the publication of this video 703 00:34:56,159 --> 00:35:01,050 was provided by the Gabriella and Paul Rosenbaum Foundation. 704 00:35:01,050 --> 00:35:05,190 Help OCW continue to provide free and open access to MIT 705 00:35:05,190 --> 00:35:12,000 courses by making a donation at ocw.mit.edu/donate. 706 00:35:12,000 --> 00:35:13,250 S