1 00:00:00,000 --> 00:00:00,720 2 00:00:00,720 --> 00:00:02,700 Let's do a couple of problems where the roots of the 3 00:00:02,700 --> 00:00:04,980 characteristic equation are complex. 4 00:00:04,980 --> 00:00:07,380 And just as a little bit of a review, and we'll put this up 5 00:00:07,379 --> 00:00:09,460 here in the corner so that it's useful for us. 6 00:00:09,460 --> 00:00:12,600 We learned that if the roots of our characteristic equation 7 00:00:12,599 --> 00:00:19,289 are r is equal to lambda plus or minus mu i, that the 8 00:00:19,289 --> 00:00:26,359 general solution for our differential equation is y is 9 00:00:26,359 --> 00:00:34,740 equal to e to the lambda x times c1 with some constant 10 00:00:34,740 --> 00:00:45,120 cosine of mu x, plus c2 times sine of mu x. 11 00:00:45,119 --> 00:00:47,309 And with that said, let's do some problems. 12 00:00:47,310 --> 00:00:49,859 So let's see, this first one that our differential 13 00:00:49,859 --> 00:00:52,039 equation-- I'll do this one in blue-- our differential 14 00:00:52,039 --> 00:00:57,320 equation is the second derivative, y prime prime plus 15 00:00:57,320 --> 00:01:05,700 4y prime plus 5y is equal to 0. 16 00:01:05,700 --> 00:01:09,200 And they actually give us some initial conditions. 17 00:01:09,200 --> 00:01:15,609 They say y of 0 is equal to 1, and y prime of 18 00:01:15,609 --> 00:01:22,980 0 is equal to 0. 19 00:01:22,980 --> 00:01:25,660 So now we'll actually be able to figure out a particular 20 00:01:25,659 --> 00:01:28,000 solution, or the particular solution, for this 21 00:01:28,000 --> 00:01:29,260 differential equation. 22 00:01:29,260 --> 00:01:31,579 So let's write down the characteristic equation. 23 00:01:31,579 --> 00:01:41,700 So it's r squared plus 4r plus 5 is equal to 0. 24 00:01:41,700 --> 00:01:43,890 Let's break out our quadratic formula. 25 00:01:43,890 --> 00:01:47,599 The roots of this are going to be negative b. 26 00:01:47,599 --> 00:01:54,289 So minus 4 plus or minus the square root of b squared. 27 00:01:54,290 --> 00:01:56,210 So that's 16. 28 00:01:56,209 --> 00:02:05,239 Minus 4 times a-- well that's 1-- times 1, times c times 5. 29 00:02:05,239 --> 00:02:07,939 All of that over 2 times a. 30 00:02:07,939 --> 00:02:10,879 a is 1 so all of that over 2. 31 00:02:10,879 --> 00:02:12,240 And see this simplifies. 32 00:02:12,240 --> 00:02:19,830 This equals minus 4 plus or minus 16-- this is 20, right? 33 00:02:19,830 --> 00:02:22,310 4 times 1 times 5 is 20. 34 00:02:22,310 --> 00:02:25,990 So 16 minus 20 is minus 4. 35 00:02:25,990 --> 00:02:29,590 Minus 4 over 2. 36 00:02:29,590 --> 00:02:38,670 And that equals minus 4 plus or minus 2i, right? 37 00:02:38,669 --> 00:02:40,579 Square root of minus 4 is 2i. 38 00:02:40,580 --> 00:02:42,820 All that over 2. 39 00:02:42,819 --> 00:02:47,900 And so our roots to the characteristic equation are 40 00:02:47,900 --> 00:02:53,099 minus 2-- just dividing both by 2-- minus 2 plus or minus, 41 00:02:53,099 --> 00:02:56,609 we could say i or 1i, right? 42 00:02:56,610 --> 00:02:58,910 So if we wanted to do some pattern match, if we just 43 00:02:58,909 --> 00:03:01,049 wanted to do it really fast, what's our lambda? 44 00:03:01,050 --> 00:03:03,120 Our lambda is just minus 1. 45 00:03:03,120 --> 00:03:04,840 Let me write that down. 46 00:03:04,840 --> 00:03:07,800 That's our lambda. 47 00:03:07,800 --> 00:03:09,110 What's our mu? 48 00:03:09,110 --> 00:03:14,800 Well mu is the coefficient on the i, so mu is 1. 49 00:03:14,800 --> 00:03:16,260 mu is equal to 1. 50 00:03:16,259 --> 00:03:19,359 And now we're ready to write down our general solution. 51 00:03:19,360 --> 00:03:23,270 So the general solution to this differential equation is 52 00:03:23,270 --> 00:03:28,570 y is equal to e to the lambda x-- well lambda is minus 2-- 53 00:03:28,569 --> 00:03:36,780 minus 2x times c1 cosine of mu x-- but mu is just 1-- so c1 54 00:03:36,780 --> 00:03:42,370 cosine of x, plus c2 sine of mu x, when mu is 55 00:03:42,370 --> 00:03:44,969 1, so sine of x. 56 00:03:44,969 --> 00:03:46,069 Fair enough. 57 00:03:46,069 --> 00:03:47,889 Now let's use our initial conditions to find the 58 00:03:47,889 --> 00:03:50,989 particular solution, or a particular solution. 59 00:03:50,990 --> 00:03:53,780 So, when x is 0, y is equal to 1. 60 00:03:53,780 --> 00:03:57,439 So y is equal to 1 when x is 0. 61 00:03:57,439 --> 00:04:00,629 So 1 is equal to-- let's substitute x is 0 here. 62 00:04:00,629 --> 00:04:05,139 So e to the minus 2 times 0, that's just 1. 63 00:04:05,139 --> 00:04:07,459 So this whole thing becomes 1, so we could just ignore it. 64 00:04:07,460 --> 00:04:09,490 It's just 1 times this thing. 65 00:04:09,490 --> 00:04:11,740 So I'll write that down. 66 00:04:11,740 --> 00:04:18,970 e to the 0 is 1 times c1 times cosine of 0, plus c2 67 00:04:18,970 --> 00:04:22,300 times sine of 0. 68 00:04:22,300 --> 00:04:23,530 Now what's sine of 0? 69 00:04:23,529 --> 00:04:25,289 Sine of 0 is 0. 70 00:04:25,290 --> 00:04:28,400 So this whole term is going to be 0. 71 00:04:28,399 --> 00:04:31,479 Cosine of 0 is 1. 72 00:04:31,480 --> 00:04:34,080 So there, we already solved for c1. 73 00:04:34,079 --> 00:04:35,569 We get this, this is 1. 74 00:04:35,569 --> 00:04:39,439 So 1 times c1 times 1 is equal to 1. 75 00:04:39,439 --> 00:04:41,339 So we get our first coefficient. 76 00:04:41,339 --> 00:04:44,879 c1 is equal to 1. 77 00:04:44,879 --> 00:04:49,019 Now let's take the derivative of our general solution. 78 00:04:49,019 --> 00:04:51,180 And we could even substitute c1 in here, just so that we 79 00:04:51,180 --> 00:04:53,740 have to stop writing c1 all the time. 80 00:04:53,740 --> 00:04:56,629 And we can solve for c2. 81 00:04:56,629 --> 00:05:02,259 So right now we know that our general solution is y-- we 82 00:05:02,259 --> 00:05:03,959 could call this our pseudo-general solution, 83 00:05:03,959 --> 00:05:07,069 because we already solved for c1-- y is equal to e to the 84 00:05:07,069 --> 00:05:12,629 minus 2x times c1, but we know that c1 is 1, so I'll write 85 00:05:12,629 --> 00:05:20,129 times cosine of x, plus c2 times sine of x. 86 00:05:20,129 --> 00:05:22,180 Now let's take the derivative of this, so that we can use 87 00:05:22,180 --> 00:05:24,519 the second initial condition. 88 00:05:24,519 --> 00:05:27,839 So y prime is equal to-- we're going to have to do a little 89 00:05:27,839 --> 00:05:29,969 bit of product rule here. 90 00:05:29,970 --> 00:05:32,450 So what's the derivative of the first expression? 91 00:05:32,449 --> 00:05:37,344 It is minus 2e to the minus 2x. 92 00:05:37,345 --> 00:05:39,860 And we multiply that times the second expression. 93 00:05:39,860 --> 00:05:45,860 Cosine of x plus c2 sine of x. 94 00:05:45,860 --> 00:05:49,389 And then we add that to just the regular first expression. 95 00:05:49,389 --> 00:05:54,189 So plus e to the minus 2x times the derivative of the 96 00:05:54,189 --> 00:05:55,839 second expression. 97 00:05:55,839 --> 00:05:57,750 So what's the derivative of cosine of x? 98 00:05:57,750 --> 00:05:59,439 It's minus sine of x. 99 00:05:59,439 --> 00:06:02,360 100 00:06:02,360 --> 00:06:04,840 And then what's the derivative of c2 sine of x? 101 00:06:04,839 --> 00:06:07,750 Well , it's plus c2 cosine of x. 102 00:06:07,750 --> 00:06:11,290 103 00:06:11,290 --> 00:06:17,080 And let's see if we can do any kind of simplification here. 104 00:06:17,079 --> 00:06:18,729 Well actually, the easiest way, instead of trying to 105 00:06:18,730 --> 00:06:21,310 simplify it algebraically and everything, let's just use our 106 00:06:21,310 --> 00:06:22,530 initial condition. 107 00:06:22,529 --> 00:06:25,639 Our initial condition is y prime of 0 is equal to 0. 108 00:06:25,639 --> 00:06:27,939 Let me write that down. 109 00:06:27,939 --> 00:06:32,170 Second initial condition was y prime of 0 is equal to 0. 110 00:06:32,170 --> 00:06:36,800 So y prime, when x is equal to 0, is equal to 0. 111 00:06:36,800 --> 00:06:39,210 And let's substitute x is equal to 0 into this thing. 112 00:06:39,209 --> 00:06:41,109 We could have simplified this more, but let's not worry 113 00:06:41,110 --> 00:06:42,500 about that right now. 114 00:06:42,500 --> 00:06:46,509 So if x is 0, this is going to be 1, right? 115 00:06:46,509 --> 00:06:48,370 E to the 0. 116 00:06:48,370 --> 00:06:52,250 e to the 0 is 1, so we're left with just minus 2, right? 117 00:06:52,250 --> 00:06:58,490 Minus 2 times e to the 0, times cosine of 0, that's 1, 118 00:06:58,490 --> 00:07:01,060 plus c2 times sine of 0. 119 00:07:01,060 --> 00:07:02,550 Sine of 0 is 0. 120 00:07:02,550 --> 00:07:08,009 So that's just 1 plus 0, plus e to the minus 2 times 0. 121 00:07:08,009 --> 00:07:09,879 That's just 1. 122 00:07:09,879 --> 00:07:11,870 Times minus sine of 0. 123 00:07:11,870 --> 00:07:14,250 Sine of 0 is just 0. 124 00:07:14,250 --> 00:07:16,529 Plus c2 times cosine of 0. 125 00:07:16,529 --> 00:07:17,869 Cosine of 0 is 1. 126 00:07:17,870 --> 00:07:19,519 So plus c2. 127 00:07:19,519 --> 00:07:21,069 That simplified things, didn't it? 128 00:07:21,069 --> 00:07:25,730 So let's see, we get 0 is equal to-- this is just 1-- 129 00:07:25,730 --> 00:07:30,129 minus 2 plus c2. 130 00:07:30,129 --> 00:07:34,490 Or we get c2 is equal to 2. 131 00:07:34,490 --> 00:07:38,090 Add 2 to both sides. c2 is equal to 2. 132 00:07:38,089 --> 00:07:41,949 And then we have our particular solution. 133 00:07:41,949 --> 00:07:46,229 I know it's c2 is equal to 2, c1 is equal to 1. 134 00:07:46,230 --> 00:07:49,160 Actually, let me erase some of this, just so that we can go 135 00:07:49,160 --> 00:07:51,245 from our general solution to our particular solution. 136 00:07:51,245 --> 00:07:56,170 137 00:07:56,170 --> 00:07:59,600 So we had figured out, you can remember, c1 is 1 and c2 is 2. 138 00:07:59,600 --> 00:08:02,570 That's easy to memorize. 139 00:08:02,569 --> 00:08:05,230 So I'll just delete all of this. 140 00:08:05,230 --> 00:08:10,490 I'll write it nice and big. 141 00:08:10,490 --> 00:08:13,069 So our particular solution, given these initial 142 00:08:13,069 --> 00:08:20,980 conditions, were, or are, or is y of x is equal to-- this 143 00:08:20,980 --> 00:08:26,220 was a general solution-- e to the minus 2x times-- we solved 144 00:08:26,220 --> 00:08:27,930 for c1, it's equal to 1. 145 00:08:27,930 --> 00:08:30,889 So we can just write cosine of x. 146 00:08:30,889 --> 00:08:31,949 And then we solved for c2. 147 00:08:31,949 --> 00:08:33,090 We figured out that that was 2. 148 00:08:33,090 --> 00:08:36,860 Plus 2 sine of x. 149 00:08:36,860 --> 00:08:37,908 And there you go. 150 00:08:37,908 --> 00:08:42,710 We have our particular solution to this-- sorry where 151 00:08:42,710 --> 00:08:45,200 did I write it-- to this differential equation with 152 00:08:45,200 --> 00:08:46,290 these initial conditions. 153 00:08:46,289 --> 00:08:50,870 And what's neat is, when we originally kind of proved this 154 00:08:50,870 --> 00:08:53,070 formula-- when we originally showed this formula-- we had 155 00:08:53,070 --> 00:08:55,110 all of these i's and we simplified. 156 00:08:55,110 --> 00:08:57,210 We said c2, it was a combination of some other 157 00:08:57,210 --> 00:08:58,639 constants times some i's. 158 00:08:58,639 --> 00:09:01,379 And we said, oh we don't know whether they're imaginary or 159 00:09:01,379 --> 00:09:03,075 not, so let's just merge them into some constant. 160 00:09:03,075 --> 00:09:06,890 But what's interesting is this particular solution has no i's 161 00:09:06,889 --> 00:09:09,569 anywhere in it. 162 00:09:09,570 --> 00:09:12,530 And so, that tells us a couple of neat things. 163 00:09:12,529 --> 00:09:18,470 One that, if we had kept this c2 in terms of some multiple 164 00:09:18,470 --> 00:09:20,745 of i's, and our constants actually would have had i's 165 00:09:20,745 --> 00:09:22,440 and they would have canceled out, et cetera. 166 00:09:22,440 --> 00:09:26,860 And it also tells us that this formula is useful beyond 167 00:09:26,860 --> 00:09:29,230 formulas that just involve imaginary numbers. 168 00:09:29,230 --> 00:09:32,460 For example, this differential equation, I don't see an i 169 00:09:32,460 --> 00:09:34,070 anywhere here. 170 00:09:34,070 --> 00:09:36,790 I don't see an i anywhere here. 171 00:09:36,789 --> 00:09:40,319 And I don't see an i anywhere here. 172 00:09:40,320 --> 00:09:42,190 But given this differential equation, to get to the 173 00:09:42,190 --> 00:09:45,770 solution, we had to use imaginary numbers in between. 174 00:09:45,769 --> 00:09:48,990 And I think this is the first time-- if I'm remembering all 175 00:09:48,990 --> 00:09:51,799 my playlists correctly-- this is the first time that we used 176 00:09:51,799 --> 00:09:54,120 imaginary numbers for something useful. 177 00:09:54,120 --> 00:09:58,759 We used it as an intermediary tool where we got a real, a 178 00:09:58,759 --> 00:10:02,069 non-imaginary solution, to a real problem, a 179 00:10:02,070 --> 00:10:03,140 non-imaginary problem. 180 00:10:03,139 --> 00:10:06,100 But we used imaginary numbers as a tool to solve it. 181 00:10:06,100 --> 00:10:08,370 So, hopefully, you found that slightly interesting. 182 00:10:08,370 --> 00:10:10,960 And I'll see you in the next video. 183 00:10:10,960 --> 00:10:11,400