1 00:00:00,000 --> 00:00:00,760 2 00:00:00,760 --> 00:00:03,639 Let's say we have the following second order 3 00:00:03,640 --> 00:00:04,679 differential equation. 4 00:00:04,679 --> 00:00:11,859 We have second derivative of y, plus 4 times the first 5 00:00:11,859 --> 00:00:15,589 derivative, plus 4y is equal to 0. 6 00:00:15,589 --> 00:00:19,329 And we're asked to find the general solution to this 7 00:00:19,329 --> 00:00:20,279 differential equation. 8 00:00:20,280 --> 00:00:22,250 So the first thing we do, like we've done in the last several 9 00:00:22,250 --> 00:00:24,940 videos, we'll get the characteristic equation. 10 00:00:24,940 --> 00:00:31,590 That's r squared plus 4r plus 4 is equal to 0. 11 00:00:31,589 --> 00:00:32,829 This one is fairly easy to factor. 12 00:00:32,829 --> 00:00:34,570 We don't need the quadratic equation here. 13 00:00:34,570 --> 00:00:40,679 This is r plus 2 times r plus 2. 14 00:00:40,679 --> 00:00:43,019 And now something interesting happens, something that we 15 00:00:43,020 --> 00:00:44,830 haven't seen before. 16 00:00:44,829 --> 00:00:47,149 The two roots of our characteristic equation are 17 00:00:47,149 --> 00:00:51,149 actually the same number, r is equal to minus 2. 18 00:00:51,149 --> 00:00:53,979 So you could say we only have one solution, or one root, or 19 00:00:53,979 --> 00:00:55,000 a repeated root. 20 00:00:55,000 --> 00:00:58,560 However you want to say it, we only have one r that satisfies 21 00:00:58,560 --> 00:00:59,690 the characteristic equation. 22 00:00:59,689 --> 00:01:01,449 You might say, well that's fine. 23 00:01:01,450 --> 00:01:05,280 Maybe my general solution is just y is equal to some 24 00:01:05,280 --> 00:01:12,420 constant times e to the minus 2x, using my one solution. 25 00:01:12,420 --> 00:01:17,040 And my reply to you is, this is a solution. 26 00:01:17,040 --> 00:01:19,390 And if you don't believe me you can test it out. 27 00:01:19,390 --> 00:01:21,859 But it's not the general solution. 28 00:01:21,859 --> 00:01:22,810 And why do I say that? 29 00:01:22,810 --> 00:01:26,609 Because this is a second order differential equation. 30 00:01:26,609 --> 00:01:29,689 And if someone wanted a particular solution, they 31 00:01:29,689 --> 00:01:31,810 would have to give you two initial conditions. 32 00:01:31,810 --> 00:01:35,019 The two initial conditions we've been using so far are, 33 00:01:35,019 --> 00:01:42,859 what y of 0 equals, and what y prime of 0 equals. 34 00:01:42,859 --> 00:01:45,530 They could give you what y of 5 equals, who knows. 35 00:01:45,530 --> 00:01:48,640 But in general, when you have a second order differential 36 00:01:48,640 --> 00:01:51,790 equation, they have to give you two initial conditions. 37 00:01:51,790 --> 00:01:54,830 Now the problem with this solution, and why it's not the 38 00:01:54,829 --> 00:01:57,959 general solution, is if you use one of these initial 39 00:01:57,959 --> 00:02:01,399 conditions, you can solve for a c, right? 40 00:02:01,400 --> 00:02:02,109 You'll get an answer. 41 00:02:02,109 --> 00:02:03,469 You'll solve for that c. 42 00:02:03,469 --> 00:02:04,650 But then there's nothing to do with the 43 00:02:04,650 --> 00:02:05,880 second initial condition. 44 00:02:05,879 --> 00:02:10,299 In fact, except for only in one particular case, whatever 45 00:02:10,300 --> 00:02:13,930 c you get for the first initial condition, it won't be 46 00:02:13,930 --> 00:02:16,240 that-- this equation won't be true for the 47 00:02:16,240 --> 00:02:17,520 second initial condition. 48 00:02:17,520 --> 00:02:18,600 And you could try it out. 49 00:02:18,599 --> 00:02:24,340 I mean, if we said, y of 0 is equal to A and y prime of 0 is 50 00:02:24,340 --> 00:02:26,729 equal to 5A. 51 00:02:26,729 --> 00:02:27,949 Let's see if these work. 52 00:02:27,949 --> 00:02:34,449 If y of 0 is equal to A, that tells us that A is equal to c 53 00:02:34,449 --> 00:02:37,009 times e to the minus 2 times 0. 54 00:02:37,009 --> 00:02:38,590 So e to the 0. 55 00:02:38,590 --> 00:02:41,740 Or c is equal to A, right? 56 00:02:41,740 --> 00:02:44,790 So if you just had this first initial condition, say fine, 57 00:02:44,789 --> 00:02:50,519 my particular solution is y is equal to A times e 58 00:02:50,520 --> 00:02:52,510 to the minus 2x. 59 00:02:52,509 --> 00:02:55,639 Let's see if this particular solution satisfies the second 60 00:02:55,639 --> 00:02:57,719 initial condition. 61 00:02:57,719 --> 00:02:59,409 So what is the derivative of this? 62 00:02:59,409 --> 00:03:05,750 y prime is equal to minus 2Ae to the minus 2x. 63 00:03:05,750 --> 00:03:11,419 And it says that 5A-- this initial condition says that 5A 64 00:03:11,419 --> 00:03:15,959 is equal to minus 2A times e to the minus 2 times 65 00:03:15,960 --> 00:03:17,689 0, so e to the 0. 66 00:03:17,689 --> 00:03:20,099 Or another way of saying that, e to the 0 is just 1. 67 00:03:20,099 --> 00:03:22,150 It says that 5A is equal to minus 2A, which 68 00:03:22,150 --> 00:03:24,050 we know is not true. 69 00:03:24,050 --> 00:03:27,800 So note, when we only have this general, or 70 00:03:27,800 --> 00:03:31,580 pseudo-general, solution, it can only satisfy, generally, 71 00:03:31,580 --> 00:03:32,730 one of the initial conditions. 72 00:03:32,729 --> 00:03:35,560 And if we're really lucky, both initial conditions. 73 00:03:35,560 --> 00:03:39,060 So that at least gives you an intuitive feel of why this 74 00:03:39,060 --> 00:03:42,145 isn't the general solution. 75 00:03:42,145 --> 00:03:44,820 So let me clean that up a little bit so that I-- I have 76 00:03:44,819 --> 00:03:48,019 a feeling I'll have to use this real estate. 77 00:03:48,020 --> 00:03:50,270 So what do we do? 78 00:03:50,270 --> 00:03:53,280 We can use a technique called reduction of order. 79 00:03:53,280 --> 00:03:55,629 And it really just says, well let's just 80 00:03:55,629 --> 00:03:57,789 guess a second solution. 81 00:03:57,789 --> 00:04:01,340 In general when we first thought about these linear 82 00:04:01,340 --> 00:04:02,939 constant coefficient differential equations, we 83 00:04:02,939 --> 00:04:06,840 said, well e to rx might be a good guess. 84 00:04:06,840 --> 00:04:07,530 And why is that? 85 00:04:07,530 --> 00:04:11,650 Because all of the derivatives of e are kind of multiples of 86 00:04:11,650 --> 00:04:14,010 the original function, and that's why we used it. 87 00:04:14,009 --> 00:04:16,449 So if we're looking for a second solution, it doesn't 88 00:04:16,449 --> 00:04:18,079 hurt to kind of make the same guess. 89 00:04:18,079 --> 00:04:21,149 And in order be a little bit more general, let's make our 90 00:04:21,149 --> 00:04:23,829 guess for our second solution-- I'll call this g 91 00:04:23,829 --> 00:04:31,699 for guess-- let's say it's some function of x times our 92 00:04:31,699 --> 00:04:35,449 first solution, e to the minus 2x. 93 00:04:35,449 --> 00:04:38,589 I could say some function of x times c times e to the minus 94 00:04:38,589 --> 00:04:41,279 2x, but the c is kind of encapsulated. 95 00:04:41,279 --> 00:04:44,149 It could be part of this some random function of x. 96 00:04:44,149 --> 00:04:45,879 So let's be as general as possible. 97 00:04:45,879 --> 00:04:51,379 So let's assume that this is a solution and then substitute 98 00:04:51,379 --> 00:04:53,680 it back into our original differential equation, and see 99 00:04:53,680 --> 00:04:57,009 if we can actually solve for v that makes it all work. 100 00:04:57,009 --> 00:04:59,699 So before we do that, let's get its first and second 101 00:04:59,699 --> 00:05:00,709 derivatives. 102 00:05:00,709 --> 00:05:04,539 So the first derivative of g is equal to-- well this is 103 00:05:04,540 --> 00:05:05,790 product rule. 104 00:05:05,790 --> 00:05:08,569 And I'll drop the v of x, we know that v is a function and 105 00:05:08,569 --> 00:05:09,920 not a constant. 106 00:05:09,920 --> 00:05:14,250 So, product rule, derivative of the first, v prime times 107 00:05:14,250 --> 00:05:20,689 the second expression, e to the minus 2x, plus the first 108 00:05:20,689 --> 00:05:23,870 function, or expression, times the derivative of the second. 109 00:05:23,870 --> 00:05:25,889 So minus 2 times e to the minus 2x. 110 00:05:25,889 --> 00:05:31,629 111 00:05:31,629 --> 00:05:34,189 Or, just to write it a little bit neater, g prime is equal 112 00:05:34,189 --> 00:05:46,149 to v prime e to the minus 2x minus 2ve to the minus 2x. 113 00:05:46,149 --> 00:05:47,569 Now we have to get the second derivative. 114 00:05:47,569 --> 00:05:49,610 I'll do it in a different color, just to fight the 115 00:05:49,610 --> 00:05:51,220 monotony of it. 116 00:05:51,220 --> 00:05:52,780 So the second derivative-- we're going to have to do the 117 00:05:52,779 --> 00:05:55,769 product rule twice-- derivative of this first 118 00:05:55,769 --> 00:05:56,379 expression. 119 00:05:56,379 --> 00:06:07,259 It's going to be v prime prime e to the minus 2x, minus 2v 120 00:06:07,259 --> 00:06:09,389 prime e to the minus 2x. 121 00:06:09,389 --> 00:06:11,399 That was just the product rule again. 122 00:06:11,399 --> 00:06:15,060 And then the derivative of the second expression is going to 123 00:06:15,060 --> 00:06:17,329 be-- let's see, derivative of the first one is v prime-- so 124 00:06:17,329 --> 00:06:31,769 it's going to be minus 2v prime e to the minus 2x, plus 125 00:06:31,769 --> 00:06:36,029 4ve to the minus 2x. 126 00:06:36,029 --> 00:06:37,339 I hope I haven't made a careless mistake. 127 00:06:37,339 --> 00:06:39,009 And we can simplify this a little bit. 128 00:06:39,009 --> 00:06:42,750 So we get the second derivative of g, which is our 129 00:06:42,750 --> 00:06:47,350 guess solution, is equal to the second derivative of v 130 00:06:47,350 --> 00:06:58,129 prime, e to the minus 2x minus 2v prime-- no, minus 4, sorry, 131 00:06:58,129 --> 00:07:02,569 because we have minus 2, minus 2-- minus 4v prime e to the 132 00:07:02,569 --> 00:07:10,709 minus 2x, plus 4ve to the minus 2x. 133 00:07:10,709 --> 00:07:14,120 And now, before we substitute it into this, we can just make 134 00:07:14,120 --> 00:07:14,910 one observation. 135 00:07:14,910 --> 00:07:16,840 That will just make the algebra a little bit simpler. 136 00:07:16,839 --> 00:07:20,529 Notice that g is something times e to the minus 2x. 137 00:07:20,529 --> 00:07:23,939 G prime is-- we could factor out an e to the minus 2x. 138 00:07:23,939 --> 00:07:26,740 And g prime prime, we can factor out an e 139 00:07:26,740 --> 00:07:28,259 to the minus 2x. 140 00:07:28,259 --> 00:07:30,740 So let's factor them out, essentially. 141 00:07:30,740 --> 00:07:36,150 So when we write this, we can write-- so the second 142 00:07:36,149 --> 00:07:40,639 derivative is g prime prime, which we can write as-- and 143 00:07:40,639 --> 00:07:47,829 I'm going to try to do this-- it's e to the minus 2x times 144 00:07:47,829 --> 00:07:48,689 the second derivative. 145 00:07:48,689 --> 00:07:51,610 So now we can get rid of the e to the minus 2x terms. So 146 00:07:51,610 --> 00:08:00,580 that's v prime prime minus 4v prime plus 4v, right? 147 00:08:00,579 --> 00:08:03,129 If I just distribute this out I get the second derivative, 148 00:08:03,129 --> 00:08:04,819 which is this. 149 00:08:04,819 --> 00:08:08,219 Plus 4 times the first derivative. 150 00:08:08,220 --> 00:08:11,190 And I'm also going to factor out the e to the minus 2x. 151 00:08:11,189 --> 00:08:13,579 So, plus 4 times this. 152 00:08:13,579 --> 00:08:20,609 So it's going to be plus 4v prime minus 8v, right? 153 00:08:20,610 --> 00:08:23,490 154 00:08:23,490 --> 00:08:26,740 Once again, I factored out the e to the minus 2x, right? 155 00:08:26,740 --> 00:08:28,050 Plus 4 times y. 156 00:08:28,050 --> 00:08:31,990 157 00:08:31,990 --> 00:08:35,110 We factored out the e to the minus 2x-- so plus 4 times v. 158 00:08:35,110 --> 00:08:37,070 I did that, because if I didn't do that I'd be writing 159 00:08:37,070 --> 00:08:39,140 e to the minus 2x, and I'd probably make a careless 160 00:08:39,139 --> 00:08:41,370 mistake, and I'd run out of space, et cetera. 161 00:08:41,370 --> 00:08:44,840 But anyway, I essentially-- to get this, I just substituted 162 00:08:44,840 --> 00:08:47,350 the second derivative, the first derivative, and g back 163 00:08:47,350 --> 00:08:48,710 into the differential equation. 164 00:08:48,710 --> 00:08:51,530 And we know that that has to equal 0. 165 00:08:51,529 --> 00:08:54,009 And let's see if we can simplify this 166 00:08:54,009 --> 00:08:54,909 a little bit more. 167 00:08:54,909 --> 00:08:58,699 And then hopefully solve for v. 168 00:08:58,700 --> 00:09:01,800 So let's see, some things are popping out at me. 169 00:09:01,799 --> 00:09:07,379 So I have plus 4v plus 4v, that's plus 170 00:09:07,379 --> 00:09:11,090 8v, minus 8v, right? 171 00:09:11,090 --> 00:09:13,740 So plus 4 minus 8 plus 4, those cancel out. 172 00:09:13,740 --> 00:09:16,490 It's plus 8 minus 8, those cancel out. 173 00:09:16,490 --> 00:09:19,600 And I also have minus 4v prime plus 4v prime. 174 00:09:19,600 --> 00:09:21,500 So those cancel out. 175 00:09:21,500 --> 00:09:25,279 And lo and behold, we've done some serious simplification. 176 00:09:25,279 --> 00:09:32,839 It ends up being e to the minus 2x times v prime prime-- 177 00:09:32,840 --> 00:09:35,019 we could call that v prime prime of x, now that we've 178 00:09:35,019 --> 00:09:37,169 saved so much space-- is equal to 0. 179 00:09:37,169 --> 00:09:38,870 We know this could never equal to 0. 180 00:09:38,870 --> 00:09:42,419 So, essentially, we have now established that this 181 00:09:42,419 --> 00:09:44,120 expression has to be equal to 0. 182 00:09:44,120 --> 00:09:47,590 And we get a separable second order differential equation. 183 00:09:47,590 --> 00:09:51,585 We get that the second derivative of v with respect 184 00:09:51,585 --> 00:09:55,590 to x-- or it's a function of x-- is equal to 0. 185 00:09:55,590 --> 00:09:57,660 So now we just have to differentiate both sides of 186 00:09:57,659 --> 00:09:58,500 this equation twice. 187 00:09:58,500 --> 00:10:00,899 You differentiate once, you get what? 188 00:10:00,899 --> 00:10:10,829 v prime of x is equal to, let's call it c1. 189 00:10:10,830 --> 00:10:13,770 And if we were to take the anti-derivative of both sides 190 00:10:13,769 --> 00:10:23,329 again, we get v of x is equal to c1 x plus 191 00:10:23,330 --> 00:10:25,940 some other c2, right? 192 00:10:25,940 --> 00:10:27,940 Now remember, what was our guess? 193 00:10:27,940 --> 00:10:33,990 Our guess was that our general solution was going to be some 194 00:10:33,990 --> 00:10:39,840 arbitrary function v times that first solution we found, 195 00:10:39,840 --> 00:10:41,160 e to the minus 2x. 196 00:10:41,159 --> 00:10:43,309 And when we actually took that guess and we substituted it 197 00:10:43,309 --> 00:10:46,959 in, we actually were able to solve for that v. 198 00:10:46,960 --> 00:10:49,019 And we got that v is equal to this. 199 00:10:49,019 --> 00:10:50,350 So this is interesting. 200 00:10:50,350 --> 00:10:54,000 So what is g, or what does our guess function equal? 201 00:10:54,000 --> 00:10:54,850 And it's no longer a guess. 202 00:10:54,850 --> 00:10:57,540 We've kind of established that it works. 203 00:10:57,539 --> 00:11:03,519 g, which we can call our solution, is equal to v of x, 204 00:11:03,519 --> 00:11:05,840 times e to the minus 2x. 205 00:11:05,840 --> 00:11:15,240 Well, that equals this, c1 x plus c2 e to the minus 2x. 206 00:11:15,240 --> 00:11:25,180 That equals c1 xe to the minus 2x, plus c2 e to the minus 2x. 207 00:11:25,179 --> 00:11:28,659 And now we have a truly general solution. 208 00:11:28,659 --> 00:11:31,240 We have two constants, so we can satisfy two initial 209 00:11:31,240 --> 00:11:32,259 conditions. 210 00:11:32,259 --> 00:11:35,559 And if you were looking for a pattern, this is the pattern. 211 00:11:35,559 --> 00:11:37,949 When you have a repeated root of your characteristic 212 00:11:37,950 --> 00:11:42,250 equation, the general solution is going to be-- you're going 213 00:11:42,250 --> 00:11:45,389 to use that e to the, that whatever root is, twice. 214 00:11:45,389 --> 00:11:47,580 But one time you're going to have an x in front of it. 215 00:11:47,580 --> 00:11:52,300 And this works every time for second order homogeneous 216 00:11:52,299 --> 00:11:54,699 constant coefficient linear equations. 217 00:11:54,700 --> 00:11:57,140 I will see you in the next video. 218 00:11:57,139 --> 00:11:57,500