1 00:00:00,135 --> 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 00:00:30,504 --> 00:00:31,350 HERBERT GROSS: Hi. 9 00:00:31,350 --> 00:00:35,250 Today we begin our discussion of differential equations. 10 00:00:35,250 --> 00:00:38,430 It has been said that the language of the universe 11 00:00:38,430 --> 00:00:40,410 is differential equations. 12 00:00:40,410 --> 00:00:42,570 And for this reason, I would suspect 13 00:00:42,570 --> 00:00:45,510 from a practical point of view differential 14 00:00:45,510 --> 00:00:48,040 equations is most important. 15 00:00:48,040 --> 00:00:52,140 For example, one is used to stating physical principles 16 00:00:52,140 --> 00:00:54,720 in terms of rates of change. 17 00:00:54,720 --> 00:00:57,180 To say, for example, that the rate of change 18 00:00:57,180 --> 00:00:59,090 is proportional to the amount of present 19 00:00:59,090 --> 00:01:02,070 is a rather simple sounding physical principle. 20 00:01:02,070 --> 00:01:03,990 The idea is knowing that the rate of change 21 00:01:03,990 --> 00:01:06,570 is proportional to the amount present. 22 00:01:06,570 --> 00:01:09,300 We sometimes like to find out explicitly 23 00:01:09,300 --> 00:01:13,740 what the amount looks like as a function of time. 24 00:01:13,740 --> 00:01:17,070 And that's what we mean by a solution or a general solution 25 00:01:17,070 --> 00:01:18,990 to differential equations. 26 00:01:18,990 --> 00:01:21,690 And that will be our subject for our first lesson 27 00:01:21,690 --> 00:01:24,900 in our study of differential equations-- the concept 28 00:01:24,900 --> 00:01:27,150 of a general solution. 29 00:01:27,150 --> 00:01:29,340 The first and simplest type of equation 30 00:01:29,340 --> 00:01:32,670 that we're going to talk about is the differential equation 31 00:01:32,670 --> 00:01:34,860 which we say has order 1. 32 00:01:34,860 --> 00:01:37,890 In other words, the greatest derivative that appears 33 00:01:37,890 --> 00:01:39,210 is the first derivative. 34 00:01:39,210 --> 00:01:41,680 There are no second derivatives involved-- in other words, 35 00:01:41,680 --> 00:01:44,070 a problem where, in terms of cannot kinematics, 36 00:01:44,070 --> 00:01:47,250 you are told what the velocity looks like in terms 37 00:01:47,250 --> 00:01:49,170 of the time-- 38 00:01:49,170 --> 00:01:51,750 in other words-- and you want to find the answer in terms 39 00:01:51,750 --> 00:01:54,240 of displacement. dv dt equals-- 40 00:01:54,240 --> 00:01:56,100 dx dt equals something. 41 00:01:56,100 --> 00:01:59,640 Find what x looks like explicitly in terms of t. 42 00:01:59,640 --> 00:02:01,890 Now, in general, what we're saying 43 00:02:01,890 --> 00:02:05,070 is all we know about a first order differential equation is 44 00:02:05,070 --> 00:02:07,080 that there is some relationship between 45 00:02:07,080 --> 00:02:11,250 the independent variable x, the dependent variable y, and dy 46 00:02:11,250 --> 00:02:11,820 dx. 47 00:02:11,820 --> 00:02:14,610 And the symbolic way of writing that is to say that some 48 00:02:14,610 --> 00:02:16,320 function-- say, capital F-- 49 00:02:16,320 --> 00:02:19,380 of x, y, and dy dx is 0. 50 00:02:19,380 --> 00:02:21,240 That's just a mathematical symbolism 51 00:02:21,240 --> 00:02:23,760 for writing down the most general first order 52 00:02:23,760 --> 00:02:25,420 differential equation. 53 00:02:25,420 --> 00:02:28,810 Now the question that comes up is twofold. 54 00:02:28,810 --> 00:02:31,110 One is, first of all, does equation 55 00:02:31,110 --> 00:02:33,370 1 even have a solution? 56 00:02:33,370 --> 00:02:36,000 How do we know that just because we can state the differential 57 00:02:36,000 --> 00:02:39,960 equation that there is a specific function-- 58 00:02:39,960 --> 00:02:43,680 say, y as a function of x-- that satisfies the equation? 59 00:02:43,680 --> 00:02:46,410 And the second is, assuming that there is a function 60 00:02:46,410 --> 00:02:48,750 y as a function of x-- 61 00:02:48,750 --> 00:02:51,900 that y as a function of x does satisfy this equation, 62 00:02:51,900 --> 00:02:54,810 how do we know that that solution is unique? 63 00:02:54,810 --> 00:02:55,810 Two questions. 64 00:02:55,810 --> 00:02:57,480 Does the equation have a solution? 65 00:02:57,480 --> 00:02:58,743 Is the solution unique? 66 00:02:58,743 --> 00:03:00,660 And I think that the easiest way to illustrate 67 00:03:00,660 --> 00:03:04,290 this is in terms of part one of our course, where we already 68 00:03:04,290 --> 00:03:07,800 solve a rather simple type of differential equation, 69 00:03:07,800 --> 00:03:09,540 namely that differential equation that 70 00:03:09,540 --> 00:03:13,260 had the specific form dy dx is some function of x. 71 00:03:13,260 --> 00:03:18,180 In particular, suppose I were to say find all solutions of dy dx 72 00:03:18,180 --> 00:03:19,980 equals 2x-- 73 00:03:19,980 --> 00:03:21,870 and I'll just add on here-- which 74 00:03:21,870 --> 00:03:24,430 pass through a particular point x0, y0. 75 00:03:24,430 --> 00:03:26,130 I'll come back to that in a moment. 76 00:03:26,130 --> 00:03:29,640 By the way, notice what I mean by an explicit solution. 77 00:03:29,640 --> 00:03:33,570 If somebody says to me, find all functions y such 78 00:03:33,570 --> 00:03:37,150 that dy dx equals 2x, in the language of sets, 79 00:03:37,150 --> 00:03:40,320 notice that somebody could get kind of cute with us and say, 80 00:03:40,320 --> 00:03:42,870 gee, isn't that just a set of all functions 81 00:03:42,870 --> 00:03:48,750 y such that dy dx equals 2x? 82 00:03:48,750 --> 00:03:50,700 And the answer is, that is correct. 83 00:03:50,700 --> 00:03:52,540 Of course, that's the answer. 84 00:03:52,540 --> 00:03:54,690 But what we mean is we would like to know 85 00:03:54,690 --> 00:03:58,020 what y looks like explicitly in terms of x. 86 00:03:58,020 --> 00:04:00,690 In other words, given x, what is y? 87 00:04:00,690 --> 00:04:03,960 This only tells me what dy d looks like explicitly 88 00:04:03,960 --> 00:04:05,400 in terms of x. 89 00:04:05,400 --> 00:04:08,790 So what I am really interested in in problems of this type 90 00:04:08,790 --> 00:04:11,580 is to convert from this set builder-- 91 00:04:11,580 --> 00:04:14,010 this implicit solution set-- 92 00:04:14,010 --> 00:04:15,960 to an explicit solution set. 93 00:04:15,960 --> 00:04:19,390 And as a trivial review, recall that, in this case, 94 00:04:19,390 --> 00:04:23,430 we knew that if dy dx was 2x then y, 95 00:04:23,430 --> 00:04:26,040 had to be x squared plus c. 96 00:04:26,040 --> 00:04:32,100 In other words, the solution set S written in explicit form 97 00:04:32,100 --> 00:04:36,180 is the set of all y such that y equals x squared plus c. 98 00:04:36,180 --> 00:04:39,480 And I suppose if you wanted a definition of what 99 00:04:39,480 --> 00:04:42,660 it means to solve a differential equation, 100 00:04:42,660 --> 00:04:44,280 why couldn't we just say-- 101 00:04:44,280 --> 00:04:46,430 just like we do an algebra in a sense-- 102 00:04:46,430 --> 00:04:50,160 that solving a differential equation essentially means 103 00:04:50,160 --> 00:04:55,410 to convert from the set builder-- the implicit form-- 104 00:04:55,410 --> 00:04:57,760 to the explicit form. 105 00:04:57,760 --> 00:05:00,750 So in other words, explicitly to express y 106 00:05:00,750 --> 00:05:04,560 in terms of x, the set of all y such that y equals 107 00:05:04,560 --> 00:05:06,540 x squared plus c. 108 00:05:06,540 --> 00:05:08,730 So now I know what y looks like in terms of x, 109 00:05:08,730 --> 00:05:10,040 up to an arbitrary constant. 110 00:05:10,040 --> 00:05:12,840 It is the solution set of the differential equation 111 00:05:12,840 --> 00:05:14,760 dy dx equals 2x. 112 00:05:14,760 --> 00:05:18,100 Now you may have noticed I said it as part of the problem, 113 00:05:18,100 --> 00:05:20,220 let's see if there is a solution which passes 114 00:05:20,220 --> 00:05:23,070 to a particular point x0, y0. 115 00:05:23,070 --> 00:05:25,800 What I do is as I look at this particular equation 116 00:05:25,800 --> 00:05:31,100 and I replace x by x0 and y by y0 and solve for see. 117 00:05:31,100 --> 00:05:33,530 And by the way, I would intuitively 118 00:05:33,530 --> 00:05:36,530 expect to be able to find that value of c. 119 00:05:36,530 --> 00:05:38,780 Because after all, I have an arbitrary constant 120 00:05:38,780 --> 00:05:41,640 in here, which means I have a degree of freedom. 121 00:05:41,640 --> 00:05:45,710 And therefore, I would suspect that by replacing x by x0 and y 122 00:05:45,710 --> 00:05:50,390 by y0, I should be able to find a specific value of c. 123 00:05:50,390 --> 00:05:54,980 Now notice what this leads to if I replace x by x0, y by y0, 124 00:05:54,980 --> 00:05:57,740 I get y0 equals x0 squared plus c. 125 00:05:57,740 --> 00:06:01,250 Consequently, c is y0 minus x0 squared. 126 00:06:01,250 --> 00:06:04,610 x0 and y0 were specific numbers, real numbers. 127 00:06:04,610 --> 00:06:08,660 Consequently, y0 minus x0 squared is a specific number. 128 00:06:08,660 --> 00:06:11,330 And consequently, the specific choice 129 00:06:11,330 --> 00:06:13,700 of c, such that the equation becomes 130 00:06:13,700 --> 00:06:19,940 what y equals x squared plus the constant y0 minus x0 squared, 131 00:06:19,940 --> 00:06:24,500 is the only solution of what the equation dy 132 00:06:24,500 --> 00:06:28,490 dx equals 2x that passes through the point x0, y0. 133 00:06:29,280 --> 00:06:31,670 This is kind of baby stuff in terms of the fact 134 00:06:31,670 --> 00:06:34,170 that it was elementary in part 1 of our course. 135 00:06:34,170 --> 00:06:36,380 But remember, this solution set is what? 136 00:06:36,380 --> 00:06:40,870 It's a set of parabolas which are, essentially, 137 00:06:40,870 --> 00:06:42,883 a parallel family of parabolas. 138 00:06:42,883 --> 00:06:44,300 And what you're sort of saying is, 139 00:06:44,300 --> 00:06:46,940 gee, it seems obvious, geometrically, 140 00:06:46,940 --> 00:06:50,780 that since I have this infinite family of parallel parabolas 141 00:06:50,780 --> 00:06:53,030 that's specifying a point that has 142 00:06:53,030 --> 00:06:55,580 to be on the parabola uniquely determines 143 00:06:55,580 --> 00:06:57,050 a member of that family. 144 00:06:57,050 --> 00:06:59,720 At any rate, I don't want to belabor this point. 145 00:06:59,720 --> 00:07:02,840 I want to get to the greater subtleties that 146 00:07:02,840 --> 00:07:05,730 come up in handling first order differential equations. 147 00:07:05,730 --> 00:07:09,830 So I've picked a slightly tougher problem for example 2. 148 00:07:09,830 --> 00:07:13,180 Example 2-- and by the way, this is a special equation 149 00:07:13,180 --> 00:07:13,680 I've picked. 150 00:07:13,680 --> 00:07:16,100 It's known as Clairaut's Equation. 151 00:07:16,100 --> 00:07:18,380 And the equation is this. 152 00:07:18,380 --> 00:07:23,030 It's y equals x dy dx minus dy dx squared. 153 00:07:23,030 --> 00:07:26,460 Clairaut's Equation is any equation 154 00:07:26,460 --> 00:07:31,280 which has the form y equals x times dy dx 155 00:07:31,280 --> 00:07:34,400 plus something which depends only on dy dx. 156 00:07:34,400 --> 00:07:38,300 In other words, rather than write this in an abstract form, 157 00:07:38,300 --> 00:07:41,750 I decided to pick a particular application of this. 158 00:07:41,750 --> 00:07:45,450 And that is, let's just pick this particular example. 159 00:07:45,450 --> 00:07:47,480 That's one special form of Clairaut's Equation. 160 00:07:47,480 --> 00:07:52,340 It's y equals x dy dx minus dy dx squared. 161 00:07:52,340 --> 00:07:57,200 x times dy dx, and what's left is a function of dy dx alone. 162 00:07:57,200 --> 00:08:01,730 At any rate, it turns out that one explicit solution 163 00:08:01,730 --> 00:08:04,640 of Clairaut's Equation-- it's a very interesting thing-- 164 00:08:04,640 --> 00:08:08,090 is you simply replace dy dx by c. 165 00:08:08,090 --> 00:08:09,000 Very amazing thing. 166 00:08:09,000 --> 00:08:11,660 In other words, if you have a Clairaut's Equation, 167 00:08:11,660 --> 00:08:15,620 and you want a solution, every place you see dy dx, 168 00:08:15,620 --> 00:08:17,120 replace it by c. 169 00:08:17,120 --> 00:08:19,040 For example, in this case, I would get what? 170 00:08:19,040 --> 00:08:21,533 y is equal to xc, or cx-- 171 00:08:21,533 --> 00:08:23,450 I guess you like to write the constant first-- 172 00:08:23,450 --> 00:08:25,760 cx minus c squared, which, by the way, 173 00:08:25,760 --> 00:08:29,360 is what a straight line whose slope is c and whose 174 00:08:29,360 --> 00:08:32,580 y-intercept is minus c squared. 175 00:08:32,580 --> 00:08:35,240 By the way, the proof that this is a solution 176 00:08:35,240 --> 00:08:37,100 is simply differentiate this. 177 00:08:37,100 --> 00:08:39,530 You get that dy dx equals c. 178 00:08:39,530 --> 00:08:43,070 Since c is dy dx, every place you see a c, 179 00:08:43,070 --> 00:08:45,470 replace it by dy dx. 180 00:08:45,470 --> 00:08:46,640 And that will give you what? 181 00:08:46,640 --> 00:08:51,350 y equals x dy dx minus dy dx squared. 182 00:08:51,350 --> 00:08:54,680 I'll give you more on that in the exercises on how one 183 00:08:54,680 --> 00:08:56,640 solves Clairaut's Equations. 184 00:08:56,640 --> 00:08:59,910 I want it for illustrative purposes here. 185 00:08:59,910 --> 00:09:04,190 So anyway, this is a solution of this particular equation. 186 00:09:04,190 --> 00:09:05,400 And it's only one solution. 187 00:09:05,400 --> 00:09:08,070 I don't know if there are other solutions, or what have you. 188 00:09:08,070 --> 00:09:10,070 The next question that I would investigate here, 189 00:09:10,070 --> 00:09:11,720 just as in the previous case-- 190 00:09:11,720 --> 00:09:14,030 I would say, gee, I wonder what c 191 00:09:14,030 --> 00:09:18,410 must be if I want a solution curve to pass through the point 192 00:09:18,410 --> 00:09:22,230 x0, y0 in the xy-plane. 193 00:09:22,230 --> 00:09:24,650 So what I do, purely algebraically, 194 00:09:24,650 --> 00:09:29,360 is I replace x by x0, y by y0, and solve 195 00:09:29,360 --> 00:09:31,340 the resulting equation for c-- 196 00:09:31,340 --> 00:09:34,880 namely y0 equals x0 c minus c squared. 197 00:09:34,880 --> 00:09:36,050 And I now observe-- 198 00:09:36,050 --> 00:09:37,910 remember, x and y look like variables. 199 00:09:37,910 --> 00:09:41,210 With a subscript 0, they represent fixed x and y 200 00:09:41,210 --> 00:09:42,650 coordinates of a point. 201 00:09:42,650 --> 00:09:44,660 c is the only variable in here. 202 00:09:44,660 --> 00:09:48,230 In other words, this is a quadratic equation in c. 203 00:09:48,230 --> 00:09:50,300 And solving this quadratic equation 204 00:09:50,300 --> 00:09:53,240 by the quadratic formula, I find that c is 205 00:09:53,240 --> 00:09:55,610 equal to this expression here. 206 00:09:55,610 --> 00:09:59,870 Remembering that we're dealing with real variables, 207 00:09:59,870 --> 00:10:03,230 I now know that I'm in trouble if this thing here 208 00:10:03,230 --> 00:10:05,490 happens to be negative. 209 00:10:05,490 --> 00:10:08,060 In other words, if this is negative, 210 00:10:08,060 --> 00:10:11,160 I have an imaginary number when I take the square root. 211 00:10:11,160 --> 00:10:13,760 In other words, there will be no solution, at least 212 00:10:13,760 --> 00:10:15,440 of this particular type. 213 00:10:15,440 --> 00:10:19,310 There'll be no solution to this equation if y happens to be-- 214 00:10:19,310 --> 00:10:23,360 if y0 is greater than 1/4 x0 squared. 215 00:10:23,360 --> 00:10:26,360 And that means there will be no solution if you're 216 00:10:26,360 --> 00:10:29,940 above the parabola y equals 1/4 x squared. 217 00:10:29,940 --> 00:10:32,340 Remember, the set of all points x 218 00:10:32,340 --> 00:10:35,940 comma y for which y is greater than 1/4 x squared 219 00:10:35,940 --> 00:10:39,870 means you're above the parabola y equals 1/4 x squared, 220 00:10:39,870 --> 00:10:43,770 because on that parabola, since y is equal to 1/4 x squared, 221 00:10:43,770 --> 00:10:47,520 you would have to go above that to make y greater than 1/4 222 00:10:47,520 --> 00:10:48,880 x squared. 223 00:10:48,880 --> 00:10:50,820 By the way, you see, all this proves 224 00:10:50,820 --> 00:10:54,540 is that there is no solution of this type, 225 00:10:54,540 --> 00:10:56,880 to this particular equation, if you're 226 00:10:56,880 --> 00:11:00,060 above the parabola y equals 1/4 x squared. 227 00:11:00,060 --> 00:11:03,090 The question is, what if I didn't give you this as a hint? 228 00:11:03,090 --> 00:11:05,460 Suppose I just gave you this equation, 229 00:11:05,460 --> 00:11:08,330 didn't even give you a hint as to what the solution should be, 230 00:11:08,330 --> 00:11:11,160 and just said, tell me where there will be solutions? 231 00:11:11,160 --> 00:11:14,700 Notice that you did not need this piece of information 232 00:11:14,700 --> 00:11:16,230 to get started on this problem. 233 00:11:16,230 --> 00:11:18,210 Namely, all you had to say was, gee, 234 00:11:18,210 --> 00:11:21,330 this is a quadratic equation in dy dx. 235 00:11:21,330 --> 00:11:24,460 Let me solve this quadratic equation. 236 00:11:24,460 --> 00:11:26,790 See, what I could have done was what? 237 00:11:26,790 --> 00:11:31,290 I could have said, dy dx squared minus x dy dx plus y 238 00:11:31,290 --> 00:11:35,490 equals 0, and then solved that equation for dy dx. 239 00:11:35,490 --> 00:11:38,880 Without boring you with the details of a quadratic formula 240 00:11:38,880 --> 00:11:42,270 all over again, it would have turned out, more generally, 241 00:11:42,270 --> 00:11:46,290 that dy dx is equal to x plus or minus the square root of x 242 00:11:46,290 --> 00:11:48,630 squared minus 4y over 2. 243 00:11:48,630 --> 00:11:52,418 And then I would have seen, right from this, 244 00:11:52,418 --> 00:11:54,210 without even knowing what the solution was, 245 00:11:54,210 --> 00:11:56,230 that dy dx isn't even a real number. 246 00:11:56,230 --> 00:11:58,860 Remember, dy dx, geometrically, is slope. 247 00:11:58,860 --> 00:12:00,570 And slope is real. 248 00:12:00,570 --> 00:12:02,550 So this wouldn't even be a real number 249 00:12:02,550 --> 00:12:05,580 if y was greater than 1/4 x squared. 250 00:12:05,580 --> 00:12:07,530 So no matter what the solution is, 251 00:12:07,530 --> 00:12:09,330 the geometry of this problem tells 252 00:12:09,330 --> 00:12:14,670 me there can be no solution if y is greater than 1/4 x squared. 253 00:12:14,670 --> 00:12:17,210 And what that means is, if I draw the graph now-- 254 00:12:17,210 --> 00:12:21,300 see, in other words, this is the x-axis, the y-axis. 255 00:12:21,300 --> 00:12:24,360 This is the parabola y equals 1/4 x squared. 256 00:12:24,360 --> 00:12:27,630 There are no solutions that pass through any points in here. 257 00:12:27,630 --> 00:12:29,460 No solutions in here. 258 00:12:29,460 --> 00:12:32,070 By the way, the plus or minus sign here 259 00:12:32,070 --> 00:12:33,840 is relatively unimportant. 260 00:12:33,840 --> 00:12:38,320 What it means is, dy dx is a double-valued function. 261 00:12:38,320 --> 00:12:42,570 We would solve this in real life as two separate problems-- 262 00:12:42,570 --> 00:12:46,020 namely, dy dx equals x plus the square root of x squared 263 00:12:46,020 --> 00:12:49,830 minus 4y over 2, and dy dx is x minus the square root 264 00:12:49,830 --> 00:12:52,120 of x squared minus 4y over 2. 265 00:12:52,120 --> 00:12:54,060 Notice that the plus or minus sign 266 00:12:54,060 --> 00:12:57,090 has no bearing on the fact that the crucial point is 267 00:12:57,090 --> 00:13:01,980 that y must be no greater than 1/4 x squared, all right? 268 00:13:01,980 --> 00:13:04,320 By the way, if you want to see this geometrically, 269 00:13:04,320 --> 00:13:09,480 notice that y equals cx minus c squared is what kind of a line? 270 00:13:09,480 --> 00:13:13,780 It's a straight line whose y-intercept is minus c squared. 271 00:13:13,780 --> 00:13:16,330 Notice that minus c squared is negative 272 00:13:16,330 --> 00:13:20,520 whether c is positive or negative, because c squared 273 00:13:20,520 --> 00:13:21,760 can't be negative. 274 00:13:21,760 --> 00:13:24,360 So minus c squared can't be positive. 275 00:13:24,360 --> 00:13:26,550 Notice, however, if c is positive-- 276 00:13:26,550 --> 00:13:28,020 c is the slope of this line-- 277 00:13:28,020 --> 00:13:32,160 if c is positive, the straight line goes through this point, 278 00:13:32,160 --> 00:13:33,150 all right? 279 00:13:33,150 --> 00:13:34,350 And has this slope. 280 00:13:34,350 --> 00:13:36,990 If c is negative, it goes through this point 281 00:13:36,990 --> 00:13:37,860 with this slope. 282 00:13:37,860 --> 00:13:41,310 There's a very amazing property that the parabola 1/4 x 283 00:13:41,310 --> 00:13:44,720 squared has with respect to this family of straight lines. 284 00:13:44,720 --> 00:13:47,490 And I'll also talk about that more in the exercises. 285 00:13:47,490 --> 00:13:50,390 The parabola y equals 1/4 x squared 286 00:13:50,390 --> 00:13:54,120 is called the envelope of this family of straight lines. 287 00:13:54,120 --> 00:13:59,370 The amazing thing is that every straight line in this family 288 00:13:59,370 --> 00:14:01,890 is tangent to this parabola. 289 00:14:01,890 --> 00:14:07,740 And conversely, the tangent line at any point of this parabola 290 00:14:07,740 --> 00:14:11,040 is a line that belongs to this family. 291 00:14:11,040 --> 00:14:13,230 And that leads to a very remarkable thing, 292 00:14:13,230 --> 00:14:15,750 which is not immediately, intuitively obvious. 293 00:14:15,750 --> 00:14:18,450 And that is, in particular, it must 294 00:14:18,450 --> 00:14:20,700 mean that the parabola itself must 295 00:14:20,700 --> 00:14:23,760 be a solution of my differential equation. 296 00:14:23,760 --> 00:14:25,950 You see, after all, if I pick a point, now, 297 00:14:25,950 --> 00:14:29,160 that's on the parabola and draw the tangent line in here, 298 00:14:29,160 --> 00:14:31,680 that tangent line belongs to this family. 299 00:14:31,680 --> 00:14:34,680 Because this point is on this tangent line, 300 00:14:34,680 --> 00:14:37,680 it must be a solution to the original differential equation. 301 00:14:37,680 --> 00:14:39,928 But that point is also on the parabola. 302 00:14:39,928 --> 00:14:41,970 The point doesn't know whether I'm considering it 303 00:14:41,970 --> 00:14:45,330 as being part of the parabola or a part of the family 304 00:14:45,330 --> 00:14:46,290 of straight lines. 305 00:14:46,290 --> 00:14:49,260 Consequently, what it means is that this parabola must also 306 00:14:49,260 --> 00:14:51,270 satisfy the differential equation. 307 00:14:51,270 --> 00:14:52,860 And the best way of proving that is 308 00:14:52,860 --> 00:14:55,050 to show you that it does satisfy the equation. 309 00:14:55,050 --> 00:14:57,760 Namely, if y equals 1/4 x squared, 310 00:14:57,760 --> 00:15:00,210 notice that dy dx is x over 2. 311 00:15:00,210 --> 00:15:03,750 Consequently, dy dx squared is x squared over 4. 312 00:15:03,750 --> 00:15:07,530 The differential equation I'm trying to satisfy is this one. 313 00:15:07,530 --> 00:15:13,800 If I replace y by 1/4 x squared, dy dx by x over 2-- 314 00:15:13,800 --> 00:15:16,270 and look at what this says. 315 00:15:16,270 --> 00:15:19,410 It says x squared over 4 is equal to x squared 316 00:15:19,410 --> 00:15:22,260 over 2 minus x squared over 4, which, 317 00:15:22,260 --> 00:15:24,460 of course, is x squared over 4. 318 00:15:24,460 --> 00:15:28,330 And indeed, the parabola does satisfy this equation. 319 00:15:28,330 --> 00:15:32,600 And now you see we've run into a very interesting situation 320 00:15:32,600 --> 00:15:33,110 here. 321 00:15:33,110 --> 00:15:37,220 Namely, on the parabola y equals 1/4 x squared, 322 00:15:37,220 --> 00:15:41,090 y equals d dy dx minus dy dx squared has 323 00:15:41,090 --> 00:15:42,740 at least two solutions-- 324 00:15:42,740 --> 00:15:45,440 namely, a member of the family of straight lines y 325 00:15:45,440 --> 00:15:49,490 equals cx minus c squared, and also the parabola y 326 00:15:49,490 --> 00:15:51,860 equals 1/4 x squared. 327 00:15:51,860 --> 00:15:53,960 Remember, at the beginning of our lecture 328 00:15:53,960 --> 00:15:56,430 today, we mentioned two questions. 329 00:15:56,430 --> 00:15:58,970 First of all, does the differential equation 330 00:15:58,970 --> 00:16:00,180 have a solution? 331 00:16:00,180 --> 00:16:03,410 And if it does have a solution, is the solution unique? 332 00:16:03,410 --> 00:16:06,050 What we've shown so far is that on the parabola, 333 00:16:06,050 --> 00:16:08,120 the solution is not unique. 334 00:16:08,120 --> 00:16:10,430 Namely, there were two solutions on the parabola. 335 00:16:10,430 --> 00:16:12,770 And we've shown that, above the parabola, 336 00:16:12,770 --> 00:16:14,870 the equation has no solutions. 337 00:16:14,870 --> 00:16:18,260 Consequently, the answers to both questions, 1 and 2, 338 00:16:18,260 --> 00:16:20,630 can be in the negative. 339 00:16:20,630 --> 00:16:22,730 And this is what makes differential equations 340 00:16:22,730 --> 00:16:24,410 a very tough subject, and why it's 341 00:16:24,410 --> 00:16:28,010 difficult to talk about the solution, or a solution, 342 00:16:28,010 --> 00:16:29,180 or what have you. 343 00:16:29,180 --> 00:16:31,640 But fortunately, there is a key theorem 344 00:16:31,640 --> 00:16:33,500 which we have at our disposal-- 345 00:16:33,500 --> 00:16:36,680 a theorem which is far more difficult to prove than it 346 00:16:36,680 --> 00:16:38,600 is to state and memorize. 347 00:16:38,600 --> 00:16:41,360 And I will settle right now for just stating the theorem 348 00:16:41,360 --> 00:16:43,340 and having you see what it means. 349 00:16:43,340 --> 00:16:44,960 The key theorem is this. 350 00:16:44,960 --> 00:16:47,060 Let's assume, for the sake of argument, 351 00:16:47,060 --> 00:16:49,670 that we can write our differential equation 352 00:16:49,670 --> 00:16:52,400 explicitly in terms of dy dx-- 353 00:16:52,400 --> 00:16:55,100 in other words, that dy dx is explicitly 354 00:16:55,100 --> 00:16:57,380 some function of x and y. 355 00:16:57,380 --> 00:17:03,200 If it turns out that f and its partial with respect to y 356 00:17:03,200 --> 00:17:08,930 are continuous in some region R, then at each point in R, 357 00:17:08,930 --> 00:17:12,050 there is a unique solution of dy dx 358 00:17:12,050 --> 00:17:16,970 equals f of x, y which passes through x0 y0. 359 00:17:16,970 --> 00:17:20,270 You see, notice that in the particular problem 360 00:17:20,270 --> 00:17:24,005 that we were just dealing with, what was f of x, y? 361 00:17:24,005 --> 00:17:28,400 f of x, y was this function here. 362 00:17:28,400 --> 00:17:32,630 Notice that f of x, y will be continuous as long 363 00:17:32,630 --> 00:17:34,130 as this expression exists. 364 00:17:34,130 --> 00:17:36,950 In other words, f of x, y will be continuous 365 00:17:36,950 --> 00:17:40,430 as long as y is less than or equal to 1/4 x squared. 366 00:17:40,430 --> 00:17:43,040 What about the partial of f with respect to y? 367 00:17:43,040 --> 00:17:45,110 To find the partial of f with respect to y, 368 00:17:45,110 --> 00:17:47,030 I have to differentiate this thing. 369 00:17:47,030 --> 00:17:50,720 Notice that y is under the square root sign. 370 00:17:50,720 --> 00:17:53,240 When I differentiate the square root, 371 00:17:53,240 --> 00:17:55,970 the square root comes down into the denominator. 372 00:17:55,970 --> 00:17:58,310 I can't allow a 0 denominator. 373 00:17:58,310 --> 00:18:02,300 So that tells me that not only must y be less than 374 00:18:02,300 --> 00:18:10,160 or equal to 1/4 x squared, but rather, y must be less than 1/4 375 00:18:10,160 --> 00:18:12,920 x squared, because if y equaled 1/4 376 00:18:12,920 --> 00:18:14,990 x squared, when this factor comes down 377 00:18:14,990 --> 00:18:16,950 into the denominator, I'm in trouble. 378 00:18:16,950 --> 00:18:18,820 I have a 0 denominator. 379 00:18:18,820 --> 00:18:21,980 Notice, by the way, the bad solution that we got. 380 00:18:21,980 --> 00:18:23,390 In other words, notice that where 381 00:18:23,390 --> 00:18:26,990 we got the two solutions occurred where 382 00:18:26,990 --> 00:18:29,720 y was equal to 1/4 x squared. 383 00:18:29,720 --> 00:18:31,850 And our key theorem doesn't apply 384 00:18:31,850 --> 00:18:35,240 in that case, because notice that our key theorem says 385 00:18:35,240 --> 00:18:39,290 that the solution will be unique only in that region R 386 00:18:39,290 --> 00:18:43,280 where f and f sub y are continuous. 387 00:18:43,280 --> 00:18:46,130 You see, notice that, below the parabola, 388 00:18:46,130 --> 00:18:48,040 f and f sub y were continuous. 389 00:18:48,040 --> 00:18:51,410 In other words, these functions couldn't go bad in our problem. 390 00:18:51,410 --> 00:18:54,920 Consequently, since we had one solution of the form y 391 00:18:54,920 --> 00:18:58,730 equals cx minus c squared that passed through every point that 392 00:18:58,730 --> 00:19:02,570 was below the parabola, the fact that this theorem applies 393 00:19:02,570 --> 00:19:05,960 says there can't be any other solution, because once 394 00:19:05,960 --> 00:19:08,900 you've found one solution, that's all there are. 395 00:19:08,900 --> 00:19:12,210 That's exactly what you mean by unique. 396 00:19:12,210 --> 00:19:14,900 See, in other words, if the solution is unique, 397 00:19:14,900 --> 00:19:17,930 then we can talk about a general solution-- 398 00:19:17,930 --> 00:19:21,170 namely, that solution that has one arbitrary constant. 399 00:19:21,170 --> 00:19:25,100 We can find what point it passes through by solving 400 00:19:25,100 --> 00:19:26,518 for the arbitrary constant. 401 00:19:26,518 --> 00:19:28,560 And that's what we mean by a particular solution. 402 00:19:28,560 --> 00:19:30,500 In fact, let me just summarize that for you 403 00:19:30,500 --> 00:19:32,420 in terms of our two examples. 404 00:19:32,420 --> 00:19:35,120 By a general solution, you mean what? 405 00:19:35,120 --> 00:19:39,290 You mean a solution that has one arbitrary constant 406 00:19:39,290 --> 00:19:43,010 such that, through any point in your region, 407 00:19:43,010 --> 00:19:46,110 you can get one and only one solution. 408 00:19:46,110 --> 00:19:48,470 Notice that the differential equation dy 409 00:19:48,470 --> 00:19:53,315 dx equals 2x has one solution, y equals x squared plus c-- one 410 00:19:53,315 --> 00:19:55,220 family of solutions, all right? 411 00:19:55,220 --> 00:19:57,000 That's the general solution. 412 00:19:57,000 --> 00:20:00,110 Notice that 2x-- you see, if f of xy is 2x, 413 00:20:00,110 --> 00:20:01,780 f is certainly continuous. 414 00:20:01,780 --> 00:20:06,650 The partial of f with respect to y, since f of xy is just 2x 415 00:20:06,650 --> 00:20:09,200 is a function of x alone, the partial with respect to y 416 00:20:09,200 --> 00:20:11,960 is 0, which is certainly a continuous function. 417 00:20:11,960 --> 00:20:13,460 In other words, that's another way 418 00:20:13,460 --> 00:20:15,732 of seeing what we had in part one-- 419 00:20:15,732 --> 00:20:16,940 another way of looking at it. 420 00:20:16,940 --> 00:20:18,950 That's, in terms of our key theorem 421 00:20:18,950 --> 00:20:22,370 here, why we have a unique family of solutions-- 422 00:20:22,370 --> 00:20:26,570 that in the whole xy-plane, 2x and its partial with respect to 423 00:20:26,570 --> 00:20:28,730 y happen to be continuous. 424 00:20:28,730 --> 00:20:31,960 On the other hand, in example 2, the general solution 425 00:20:31,960 --> 00:20:37,180 was y equals cx minus c squared provided that our region 426 00:20:37,180 --> 00:20:39,700 R was restricted to being what? 427 00:20:39,700 --> 00:20:44,230 That we were below the parabola y equals 1/4 x squared. 428 00:20:44,230 --> 00:20:47,110 By a particular solution, we mean a solution 429 00:20:47,110 --> 00:20:50,300 that you can get to pass through a particular point. 430 00:20:50,300 --> 00:20:52,610 In other words, by arbitrarily specifying 431 00:20:52,610 --> 00:20:54,100 a value of the constant. 432 00:20:54,100 --> 00:20:56,740 For example, referring to example 1 433 00:20:56,740 --> 00:21:01,360 again, if I pick c to be 7 or minus pi [INAUDIBLE] y equals 434 00:21:01,360 --> 00:21:04,720 x squared plus 7, y equals x squared minus pi, 435 00:21:04,720 --> 00:21:07,780 these are particular solutions of the equation 436 00:21:07,780 --> 00:21:09,160 that we're talking about. 437 00:21:09,160 --> 00:21:11,590 In particular, what we're saying is, if you 438 00:21:11,590 --> 00:21:14,470 want the particular solution-- 439 00:21:14,470 --> 00:21:16,420 see, maybe I should emphasize that-- 440 00:21:16,420 --> 00:21:19,030 the particular solution that passes through the point 441 00:21:19,030 --> 00:21:22,300 x0, y0, your constant must be chosen 442 00:21:22,300 --> 00:21:25,840 to be y0 minus x0 squared. 443 00:21:25,840 --> 00:21:28,960 In example 2, the general solution, 444 00:21:28,960 --> 00:21:30,610 if we were below the parabola, was 445 00:21:30,610 --> 00:21:32,950 y equals cx minus c squared. 446 00:21:32,950 --> 00:21:35,380 In particular, if I were to pick c to be 3, 447 00:21:35,380 --> 00:21:37,270 a particular solution would be what? 448 00:21:37,270 --> 00:21:40,570 y equals 3x minus 9. 449 00:21:40,570 --> 00:21:45,160 Or to do this more generally, given the point x0, y0, 450 00:21:45,160 --> 00:21:49,700 the particular solution of y equals cx minus c squared, 451 00:21:49,700 --> 00:21:52,690 which passes through this point, is determined 452 00:21:52,690 --> 00:21:57,520 by c being this expression, where the plus or minus sign is 453 00:21:57,520 --> 00:21:59,260 not really ambiguous here. 454 00:21:59,260 --> 00:22:02,050 What we meant was, is that when we have the plus or minus sign, 455 00:22:02,050 --> 00:22:04,300 we look at this as two separate problems. 456 00:22:04,300 --> 00:22:07,090 Finally, we come to a third concept that's 457 00:22:07,090 --> 00:22:08,920 called a singular solution. 458 00:22:08,920 --> 00:22:11,560 A singular solution occurs only when 459 00:22:11,560 --> 00:22:15,300 you're in the region where f or its partial with respect to y 460 00:22:15,300 --> 00:22:16,600 are not continuous. 461 00:22:16,600 --> 00:22:19,120 For example, there are no singular solutions 462 00:22:19,120 --> 00:22:21,040 to example 1. 463 00:22:21,040 --> 00:22:24,460 On the other hand, a singular solution to example 2 464 00:22:24,460 --> 00:22:27,460 was y equals 1/4 x squared. 465 00:22:27,460 --> 00:22:29,380 Why do I call it a singular solution? 466 00:22:29,380 --> 00:22:33,550 Let me point out that there is no way of getting 467 00:22:33,550 --> 00:22:38,680 y equals 1/4 x squared by judiciously choosing 468 00:22:38,680 --> 00:22:40,060 a constant c. 469 00:22:40,060 --> 00:22:43,240 As long as c is a constant, observe 470 00:22:43,240 --> 00:22:47,380 that, for every choice of c, y equals cx minus c squared 471 00:22:47,380 --> 00:22:50,050 is a straight line-- in particular, the straight line 472 00:22:50,050 --> 00:22:54,200 whose slope is c and whose y-intercept is minus c squared. 473 00:22:54,200 --> 00:22:58,450 In other words, I cannot get the curve y equals 1/4 x squared 474 00:22:58,450 --> 00:23:00,400 by specifying a constant here. 475 00:23:00,400 --> 00:23:03,340 That's why it's called a singular solution. 476 00:23:03,340 --> 00:23:06,130 It cannot be obtained from the general solution. 477 00:23:06,130 --> 00:23:09,070 But notice that that singular solution only 478 00:23:09,070 --> 00:23:13,750 existed because we violated the condition that our region be 479 00:23:13,750 --> 00:23:15,190 below the parabola. 480 00:23:15,190 --> 00:23:17,260 In other words, once we get outside of the region 481 00:23:17,260 --> 00:23:19,570 where f and f sub y were continuous, 482 00:23:19,570 --> 00:23:23,200 then mongrel-type solutions could have snuck in. 483 00:23:23,200 --> 00:23:25,420 And those are what are called singular solutions. 484 00:23:25,420 --> 00:23:27,730 But I'll talk about those more. 485 00:23:27,730 --> 00:23:29,650 In the exercises, what I thought I 486 00:23:29,650 --> 00:23:32,860 would like to do for the remainder of this lecture 487 00:23:32,860 --> 00:23:38,470 is to talk specifically about what the textbook is all about 488 00:23:38,470 --> 00:23:42,150 and what we'll be dealing with for the next few lessons. 489 00:23:42,150 --> 00:23:45,980 You see, the next few units will not have any lectures. 490 00:23:45,980 --> 00:23:48,340 This is the lecture that will launch you 491 00:23:48,340 --> 00:23:51,580 into solving first order, first degree equations. 492 00:23:51,580 --> 00:23:55,780 From this point on, until we get to second order equations, 493 00:23:55,780 --> 00:23:58,060 there will be no lectures-- simply 494 00:23:58,060 --> 00:24:01,450 reading assignments in the texts and learning exercises, where 495 00:24:01,450 --> 00:24:04,300 I'll try to give you the so-called cookbook 496 00:24:04,300 --> 00:24:06,550 part of differential equations. 497 00:24:06,550 --> 00:24:09,100 But I thought that, to lead into that, what I thought 498 00:24:09,100 --> 00:24:10,670 I wanted to bring out was this. 499 00:24:10,670 --> 00:24:13,090 Let's suppose that we restrict our attention 500 00:24:13,090 --> 00:24:15,160 to not only first order equations, 501 00:24:15,160 --> 00:24:16,700 but first degree equations. 502 00:24:16,700 --> 00:24:19,330 In other words, not only does the only derivative that 503 00:24:19,330 --> 00:24:20,620 appears is-- 504 00:24:20,620 --> 00:24:22,690 be the first derivative, but it also 505 00:24:22,690 --> 00:24:26,800 happens to appear only to the first power, all right? 506 00:24:26,800 --> 00:24:29,020 Notice that, in that situation, as I 507 00:24:29,020 --> 00:24:31,720 look at every term in my differential equation, 508 00:24:31,720 --> 00:24:33,370 they fall into two types. 509 00:24:33,370 --> 00:24:37,630 There'll be a term in which dy dx is a factor or a term where 510 00:24:37,630 --> 00:24:39,760 dy dx is not a factor. 511 00:24:39,760 --> 00:24:42,940 What I can therefore do is collect 512 00:24:42,940 --> 00:24:46,030 all the terms which have dy dx as a factor 513 00:24:46,030 --> 00:24:48,430 and factor out dy dx. 514 00:24:48,430 --> 00:24:50,020 So I have a term of the form what? 515 00:24:50,020 --> 00:24:56,140 Some function N of x and y times dy dx plus what? 516 00:24:56,140 --> 00:24:58,420 A bunch of terms involving only x and y, 517 00:24:58,420 --> 00:25:00,370 which have no dy dx in them-- 518 00:25:00,370 --> 00:25:02,070 equals 0. 519 00:25:02,070 --> 00:25:05,050 In other words, this is the most general form of my first order, 520 00:25:05,050 --> 00:25:06,640 first degree equation. 521 00:25:06,640 --> 00:25:10,180 If I now treat these as differentials and multiply 522 00:25:10,180 --> 00:25:14,440 through by dx, notice that I'm back to the standard 523 00:25:14,440 --> 00:25:17,327 differential equation form that we talked about-- in fact, 524 00:25:17,327 --> 00:25:18,910 I'll bring this up again in a moment-- 525 00:25:18,910 --> 00:25:21,790 in block 3 when we introduced exact differentials. 526 00:25:21,790 --> 00:25:24,910 Namely, I have an equation of what form? 527 00:25:24,910 --> 00:25:30,330 M dx plus N dy equals 0. 528 00:25:30,330 --> 00:25:33,690 By the way, to correlate this with our key theorem, 529 00:25:33,690 --> 00:25:35,790 notice that, from here, I could have 530 00:25:35,790 --> 00:25:41,670 written that dy dx was minus M of x, y over N of x, y. 531 00:25:41,670 --> 00:25:44,370 And that's the f of x, y that I'm 532 00:25:44,370 --> 00:25:48,250 talking about in our key theorem that we talked about earlier. 533 00:25:48,250 --> 00:25:51,720 See, in other words, f of x, y is the right-hand side 534 00:25:51,720 --> 00:25:54,630 of the equation when the left-hand side is specifically 535 00:25:54,630 --> 00:25:55,830 dy dx. 536 00:25:55,830 --> 00:25:59,340 So in other words, what I'm guaranteed of is this. 537 00:25:59,340 --> 00:26:03,720 As long as this function on the right-hand side is continuous, 538 00:26:03,720 --> 00:26:08,150 and its partial with respect to y is continuous, 539 00:26:08,150 --> 00:26:10,110 I'm guaranteed that this equation 540 00:26:10,110 --> 00:26:12,330 has a general solution. 541 00:26:12,330 --> 00:26:16,320 In particular, if M and N happen to be continuously differential 542 00:26:16,320 --> 00:26:19,410 functions, the general solution will exist, 543 00:26:19,410 --> 00:26:23,590 provided only that N and its partial with respect to y 544 00:26:23,590 --> 00:26:26,580 are not 0, because, after all, to differentiate 545 00:26:26,580 --> 00:26:30,510 this with respect to y, this is a quotient. 546 00:26:30,510 --> 00:26:32,640 And the only place that I'm going to be in trouble 547 00:26:32,640 --> 00:26:34,590 is if the square of the denominator 548 00:26:34,590 --> 00:26:36,770 appears when I use the quotient rule. 549 00:26:36,770 --> 00:26:41,640 So I have to be careful where either the partial with N 550 00:26:41,640 --> 00:26:44,480 respect to y or N itself happen to be 0. 551 00:26:44,480 --> 00:26:46,470 But if that doesn't happen, it means 552 00:26:46,470 --> 00:26:48,430 there is a general solution. 553 00:26:48,430 --> 00:26:52,620 Now, you may remember, back in block 3, 554 00:26:52,620 --> 00:26:56,400 we said, if this differential happens to be exact-- 555 00:26:56,400 --> 00:26:59,430 in other words, if there happens to be a function w 556 00:26:59,430 --> 00:27:03,570 such that dw is M dx plus N dy, then 557 00:27:03,570 --> 00:27:06,390 it's trivial to solve this equation. 558 00:27:06,390 --> 00:27:10,050 In particular, what we mean by an exact differential 559 00:27:10,050 --> 00:27:11,530 equation is simply this. 560 00:27:11,530 --> 00:27:15,090 If M dx plus N dy is exact, then we 561 00:27:15,090 --> 00:27:18,840 call the equation M dx plus N dy equals 0 562 00:27:18,840 --> 00:27:21,030 an exact differential equation. 563 00:27:21,030 --> 00:27:25,380 And the solution of that equation is simply f of x, y 564 00:27:25,380 --> 00:27:28,830 equals c, where f is any function 565 00:27:28,830 --> 00:27:32,082 such that df is M dx plus N dy. 566 00:27:32,082 --> 00:27:33,540 And we're sure that such a function 567 00:27:33,540 --> 00:27:38,470 f exists by definition of M dx plus N dy being exact. 568 00:27:38,470 --> 00:27:42,075 See, by way of example, suppose I were given the equation y 569 00:27:42,075 --> 00:27:44,153 dx plus x dy equals 0. 570 00:27:44,153 --> 00:27:45,570 Forget about the fact that I could 571 00:27:45,570 --> 00:27:49,240 have solved this easier by just separating the variables. 572 00:27:49,240 --> 00:27:52,860 Notice, by way of illustration, that the left-hand side here 573 00:27:52,860 --> 00:27:55,620 is just the differential of x times y. 574 00:27:55,620 --> 00:27:57,570 In other words, y dx plus x dy equals 575 00:27:57,570 --> 00:28:00,090 0 says that the differential of xy 576 00:28:00,090 --> 00:28:03,810 is 0, where y is implicitly, now, some function of x. 577 00:28:03,810 --> 00:28:06,660 Now, if the differential of some function of x is 0, 578 00:28:06,660 --> 00:28:09,270 then that function itself must be a constant. 579 00:28:09,270 --> 00:28:13,050 Consequently, xy equals a constant is a solution 580 00:28:13,050 --> 00:28:15,960 of this differential equation. 581 00:28:15,960 --> 00:28:19,812 The problem that comes up is, wouldn't it be nice-- 582 00:28:19,812 --> 00:28:20,770 that's not the problem. 583 00:28:20,770 --> 00:28:22,500 The question is, wouldn't it be nice 584 00:28:22,500 --> 00:28:25,560 if every first order, first degree differential 585 00:28:25,560 --> 00:28:28,000 equation happened to be exact? 586 00:28:28,000 --> 00:28:29,730 See, if it were, we'd be all done. 587 00:28:29,730 --> 00:28:32,400 We talked about that, as I say, back in block 3. 588 00:28:32,400 --> 00:28:35,460 If this were exact, this is how we solve it. 589 00:28:35,460 --> 00:28:38,500 But we also saw, in block 3, that it's 590 00:28:38,500 --> 00:28:41,450 very unique if the thing happens to be exact. 591 00:28:41,450 --> 00:28:45,090 See, in general, the differential won't be exact. 592 00:28:45,090 --> 00:28:48,240 And that's why I look now at non-exact equations. 593 00:28:48,240 --> 00:28:50,010 See, non-exact means what? 594 00:28:50,010 --> 00:28:51,780 Not that the equation isn't exact, 595 00:28:51,780 --> 00:28:54,240 but the differential isn't an exact differential. 596 00:28:54,240 --> 00:28:56,370 Look at y dx minus x dy. 597 00:28:56,370 --> 00:28:59,190 The partial of y with respect to y is 1. 598 00:28:59,190 --> 00:29:02,250 The partial of minus x with respect to x is minus 1. 599 00:29:02,250 --> 00:29:03,750 This thing is not exact. 600 00:29:03,750 --> 00:29:06,190 Consequently, this is not the differential 601 00:29:06,190 --> 00:29:08,200 of any particular function. 602 00:29:08,200 --> 00:29:10,440 Let me show you a little trick over here, though. 603 00:29:10,440 --> 00:29:12,870 This sort of suggests the quotient rule, see? 604 00:29:12,870 --> 00:29:14,580 The denominator times the differential 605 00:29:14,580 --> 00:29:16,740 of the numerator minus the numerator 606 00:29:16,740 --> 00:29:19,650 times the differential of the denominator-- except there 607 00:29:19,650 --> 00:29:23,110 should be a denominator here appearing as y squared, 608 00:29:23,110 --> 00:29:23,610 you see? 609 00:29:23,610 --> 00:29:25,410 And what I do is I say, OK. 610 00:29:25,410 --> 00:29:28,050 Why don't I just divide both sides of this equation 611 00:29:28,050 --> 00:29:29,940 by y squared? 612 00:29:29,940 --> 00:29:33,510 If I do that and invoke the rule that equals divided by equals 613 00:29:33,510 --> 00:29:36,360 are equal, being careful to remember that I'm 614 00:29:36,360 --> 00:29:37,740 in trouble when y equals 0-- 615 00:29:37,740 --> 00:29:39,620 I've got to keep that in the back of my mind, 616 00:29:39,620 --> 00:29:41,040 remembering that y is 0-- 617 00:29:41,040 --> 00:29:43,650 at any rate, this equation transforms 618 00:29:43,650 --> 00:29:47,910 into y dx minus x dy over y squared equals 0. 619 00:29:47,910 --> 00:29:51,540 The left-hand side-- this differential is now exact, 620 00:29:51,540 --> 00:29:52,920 believe it or not. 621 00:29:52,920 --> 00:29:54,540 It wasn't exact to begin with. 622 00:29:54,540 --> 00:29:57,510 But by dividing it by y squared, it became exact. 623 00:29:57,510 --> 00:29:59,140 In fact, the left-hand side is now 624 00:29:59,140 --> 00:30:01,125 the differential of x over y. 625 00:30:01,125 --> 00:30:03,870 In other words, the differential of x over y is 0. 626 00:30:03,870 --> 00:30:06,000 Therefore, x over y is a constant. 627 00:30:06,000 --> 00:30:09,240 Or, in particular, y is some constant times x. 628 00:30:09,240 --> 00:30:13,680 By the way, remembering that y had to be unequal to 0 for this 629 00:30:13,680 --> 00:30:17,730 to be true, notice that we have to check y equals 0 separately. 630 00:30:17,730 --> 00:30:19,650 Notice, by the way, that y equals 631 00:30:19,650 --> 00:30:23,280 0 is a particular solution of this equation obtained 632 00:30:23,280 --> 00:30:25,800 by choosing c equal to 0. 633 00:30:25,800 --> 00:30:28,730 So y equals 0 gives us no great hardship here. 634 00:30:28,730 --> 00:30:31,200 But the thing I'm leading up to-- and I am sorry for all 635 00:30:31,200 --> 00:30:32,190 these asides-- 636 00:30:32,190 --> 00:30:33,990 but what I'm leading up to is something 637 00:30:33,990 --> 00:30:36,280 called an integrating factor. 638 00:30:36,280 --> 00:30:38,400 And that's what we just used over here. 639 00:30:38,400 --> 00:30:42,390 What we say is, if M dx plus N dy is not 0-- 640 00:30:42,390 --> 00:30:46,260 equals 0-- is not exact, make it exact. 641 00:30:46,260 --> 00:30:47,580 What does that mean? 642 00:30:47,580 --> 00:30:50,730 Find a function u of x and y such 643 00:30:50,730 --> 00:30:52,680 that when I multiply or divide-- it makes 644 00:30:52,680 --> 00:30:56,010 no difference-- both sides of this equation by it, 645 00:30:56,010 --> 00:30:59,910 the new equation, which you see, has the same solution set 646 00:30:59,910 --> 00:31:01,120 as this one. 647 00:31:01,120 --> 00:31:05,340 See, in other words, something is a solution 648 00:31:05,340 --> 00:31:06,888 of this equation if and only if it's 649 00:31:06,888 --> 00:31:08,055 a solution of this equation. 650 00:31:08,055 --> 00:31:11,250 See, it's, equals multiplied by equals are equal. 651 00:31:11,250 --> 00:31:12,810 What if this is exact? 652 00:31:12,810 --> 00:31:15,420 See, that's precisely what we did up here. 653 00:31:15,420 --> 00:31:18,360 y dx minus x dy was not exact. 654 00:31:18,360 --> 00:31:21,810 The u of xy, in that case, was just 1 over y squared. 655 00:31:21,810 --> 00:31:23,940 We multiplied both sides of the equation 656 00:31:23,940 --> 00:31:26,820 by 1 over y squared and made this exact. 657 00:31:26,820 --> 00:31:28,740 The problem is this, though. 658 00:31:28,740 --> 00:31:31,990 It turns out, theoretically, that if an equation 659 00:31:31,990 --> 00:31:34,183 of first order, first degree is not exact, 660 00:31:34,183 --> 00:31:36,600 there is an integrating factor-- in other words, something 661 00:31:36,600 --> 00:31:39,540 that you can multiply it by to make it exact. 662 00:31:39,540 --> 00:31:41,940 The problem is that, in real life, 663 00:31:41,940 --> 00:31:44,850 that factor is harder to find than the solution 664 00:31:44,850 --> 00:31:47,130 of the original equation, in most cases. 665 00:31:47,130 --> 00:31:51,180 Namely, how would I find a u such that this was exact? 666 00:31:51,180 --> 00:31:53,430 Remember, the condition for exactness 667 00:31:53,430 --> 00:31:56,280 is that the partial of this with respect to y 668 00:31:56,280 --> 00:32:00,210 has to equal the partial of this with respect to x. 669 00:32:00,210 --> 00:32:02,940 Remembering that u is a function of x and y, 670 00:32:02,940 --> 00:32:05,370 I have to use the product rule over here. 671 00:32:05,370 --> 00:32:07,740 I get u times the partial of M with respect 672 00:32:07,740 --> 00:32:10,800 to y plus the partial of u with respect to y times 673 00:32:10,800 --> 00:32:13,620 M equals u times the partial of N with respect 674 00:32:13,620 --> 00:32:17,160 to x plus the partial of u with respect to x times N. 675 00:32:17,160 --> 00:32:18,810 And look at this equation. 676 00:32:18,810 --> 00:32:20,890 I'm trying to solve this for u. 677 00:32:20,890 --> 00:32:23,700 And notice that this is still a differential equation, 678 00:32:23,700 --> 00:32:25,830 but it involves even partial derivatives. 679 00:32:25,830 --> 00:32:28,260 In other words, to solve this equation 680 00:32:28,260 --> 00:32:30,960 involves having to solve a partial differential 681 00:32:30,960 --> 00:32:34,050 equation, which is even harder to do than the equation 682 00:32:34,050 --> 00:32:36,170 that we're trying to solve now. 683 00:32:36,170 --> 00:32:38,460 Well, at any rate, let me show you now, 684 00:32:38,460 --> 00:32:42,180 in summary, why people refer to differential equations 685 00:32:42,180 --> 00:32:43,710 as a cookbook course. 686 00:32:43,710 --> 00:32:47,130 Philosophically, there's nothing to solving first order, 687 00:32:47,130 --> 00:32:48,780 first degree differential equations. 688 00:32:48,780 --> 00:32:50,880 Namely, look at the differential. 689 00:32:50,880 --> 00:32:52,920 If it's exact, bang. 690 00:32:52,920 --> 00:32:54,990 You just integrate it directly. 691 00:32:54,990 --> 00:32:58,470 Find the w such that dw is M dx plus N dy. 692 00:32:58,470 --> 00:33:01,090 Then w equals a constant is a solution. 693 00:33:01,090 --> 00:33:04,500 If it's not exact, find an integrating factor. 694 00:33:04,500 --> 00:33:06,060 Make it exact. 695 00:33:06,060 --> 00:33:07,650 The trouble is that, in real life, 696 00:33:07,650 --> 00:33:11,160 given a particular equation, it's very, very difficult 697 00:33:11,160 --> 00:33:13,260 to find the integrating factor. 698 00:33:13,260 --> 00:33:18,350 Consequently, what one does is one categorizes various types 699 00:33:18,350 --> 00:33:20,340 of first order, first degree differential 700 00:33:20,340 --> 00:33:23,160 equations according to their structure, 701 00:33:23,160 --> 00:33:25,740 and shows little tricks for solving 702 00:33:25,740 --> 00:33:27,630 special kinds of equations-- 703 00:33:27,630 --> 00:33:31,740 special types where we can cut through the red tape and either 704 00:33:31,740 --> 00:33:37,380 find our integrating factor or a direct solution rather easily. 705 00:33:37,380 --> 00:33:41,310 At any rate, that part is taken care of magnificently 706 00:33:41,310 --> 00:33:42,480 in the textbook. 707 00:33:42,480 --> 00:33:45,540 And coupled with my unique learning exercises, 708 00:33:45,540 --> 00:33:48,720 you will get an excellent amount of drill 709 00:33:48,720 --> 00:33:50,370 in how to do the mechanics. 710 00:33:50,370 --> 00:33:52,560 But what I wanted this lecture to do 711 00:33:52,560 --> 00:33:54,930 was to have you at least understand, for sure, what 712 00:33:54,930 --> 00:33:57,480 you meant by a solution to an equation-- what you meant 713 00:33:57,480 --> 00:34:00,540 by a general solution, a particular solution, a singular 714 00:34:00,540 --> 00:34:02,100 solution, et cetera. 715 00:34:02,100 --> 00:34:04,950 At any rate, then, until we meet again 716 00:34:04,950 --> 00:34:08,570 for second and higher order differential equations, 717 00:34:08,570 --> 00:34:12,420 let's just say so long for now. 718 00:34:12,420 --> 00:34:14,820 Funding for the publication of this video 719 00:34:14,820 --> 00:34:19,710 was provided by the Gabriella and Paul Rosenbaum Foundation. 720 00:34:19,710 --> 00:34:23,850 Help OCW continue to provide free and open access to MIT 721 00:34:23,850 --> 00:34:29,284 courses by making a donation at ocw.mit.edu/donate.