1 00:00:00,510 --> 00:00:01,320 ‫Welcome back. 2 00:00:01,950 --> 00:00:07,250 ‫So if this is your differential equation, then how would we solve it? 3 00:00:08,010 --> 00:00:14,640 ‫And if you remember, solving a differential equation meant that you would get rid of these derivative 4 00:00:14,640 --> 00:00:22,440 ‫terms here, and then you need to have error as a function of time and then you have some kind of equation. 5 00:00:22,740 --> 00:00:26,280 ‫And that equation only needs to contain time. 6 00:00:26,430 --> 00:00:29,340 ‫It cannot contain the derivative of the error. 7 00:00:29,880 --> 00:00:38,250 ‫So how would we find this function that would be the solution for this differential equation? 8 00:00:39,120 --> 00:00:48,240 ‫Well, the general solution for this kind of differential equation is the following error as a function 9 00:00:48,240 --> 00:00:58,590 ‫of time equals then you have some kind of random constant and then you have E field there to the power 10 00:00:58,950 --> 00:01:01,470 ‫of Lambda Times. 11 00:01:01,650 --> 00:01:07,170 ‫T note that E and E two. 12 00:01:07,170 --> 00:01:09,240 ‫There are two different things. 13 00:01:09,690 --> 00:01:11,260 ‫E is the error. 14 00:01:12,200 --> 00:01:15,220 ‫OK, this is the error of our system. 15 00:01:16,110 --> 00:01:21,010 ‫It's the reference value minus the true state. 16 00:01:21,930 --> 00:01:25,750 ‫So this one would be your error. 17 00:01:26,580 --> 00:01:29,420 ‫However, this is still there. 18 00:01:29,790 --> 00:01:30,810 ‫It's an oiler. 19 00:01:30,810 --> 00:01:36,450 ‫No, it's approximately two point seven one. 20 00:01:36,450 --> 00:01:39,660 ‫And actually you have an infinite amount of numbers here. 21 00:01:39,660 --> 00:01:48,660 ‫After the decimal point, I put a tilde here in order not to confuse those two so that you could differentiate 22 00:01:48,660 --> 00:01:51,390 ‫between the errors and the Euler number. 23 00:01:52,260 --> 00:02:02,880 ‫And again, C is just a random constant then Lambda is your power constant on top of the oilor number. 24 00:02:03,670 --> 00:02:06,960 ‫And this power constant is multiplied by time. 25 00:02:07,770 --> 00:02:18,630 ‫Let's, for example, assume that in the differential equation K one equals six and let's assume that 26 00:02:19,110 --> 00:02:23,200 ‫K two equals minus one. 27 00:02:24,030 --> 00:02:27,480 ‫So that would be your K to here then. 28 00:02:27,480 --> 00:02:30,090 ‫This would be your differential equation. 29 00:02:30,140 --> 00:02:30,420 ‫Right. 30 00:02:30,720 --> 00:02:32,730 ‫Minus and then minus one. 31 00:02:33,740 --> 00:02:43,070 ‫Which is your key to constant and then times up minus six and then six was your key one, constant times 32 00:02:43,070 --> 00:02:48,620 ‫the error equals zero or you can just write it down like this. 33 00:02:49,740 --> 00:02:53,790 ‫And so in our general solution, which is. 34 00:02:54,990 --> 00:02:56,910 ‫This one here. 35 00:02:57,980 --> 00:03:09,020 ‫We have to find the exponent power, constant lamda and then also the constancy, and for that we take 36 00:03:09,020 --> 00:03:13,390 ‫the first and second time derivatives of the general solution. 37 00:03:14,330 --> 00:03:15,470 ‫So let's do it here. 38 00:03:15,890 --> 00:03:19,540 ‫So this is your general solution. 39 00:03:20,000 --> 00:03:29,150 ‫And now if we take the first derivative of it, then you will have C, then this lambda will come down 40 00:03:29,150 --> 00:03:38,060 ‫here and then you will have the oil the no to the power of Lambda Times T and then your second time 41 00:03:38,060 --> 00:03:45,650 ‫derivative of the error will be C times lambda squared because you will have another lambda that will 42 00:03:46,190 --> 00:03:54,530 ‫come down here times the oilor number to the power of Lambda T, and this is just the chain rule from 43 00:03:54,530 --> 00:03:55,300 ‫calculus. 44 00:03:56,150 --> 00:04:01,730 ‫And now we put all of them in our differential equation. 45 00:04:02,540 --> 00:04:05,330 ‫So e double dot is this one. 46 00:04:05,480 --> 00:04:12,740 ‫So we're going to say that it's C lambda squared or the number Lambda Times T. 47 00:04:13,820 --> 00:04:27,590 ‫Then plus see Lunda, the oil no to the power of London Times T and then minus six C oil or no and then 48 00:04:27,590 --> 00:04:31,740 ‫again that would be your power here and all that equals zero. 49 00:04:32,630 --> 00:04:37,840 ‫So this is your error, double dot, then this is your error dot. 50 00:04:38,360 --> 00:04:42,910 ‫And then this here, that's for your error. 51 00:04:43,970 --> 00:04:54,680 ‫And now we can factor out the CS and then the other numbers and their powers like this C times the other 52 00:04:54,680 --> 00:04:57,410 ‫number that would be your power here. 53 00:04:57,920 --> 00:05:10,400 ‫And then here in the brackets I will have lambda squared plus lambda minus six and all that equals zero, 54 00:05:11,300 --> 00:05:22,580 ‫assuming that C does not equal to zero and well this oilor number to the power of Lunda time C cannot 55 00:05:22,580 --> 00:05:31,490 ‫equal to zero anyway, because even if Lambda is zero then you will have the oilor number to the power 56 00:05:31,490 --> 00:05:34,070 ‫of zero times t. 57 00:05:34,550 --> 00:05:41,110 ‫That means that you would have oilor number to the power of zero which would be one. 58 00:05:42,050 --> 00:05:49,100 ‫And so knowing these facts, that means that for this equation to be zero. 59 00:05:50,040 --> 00:05:53,340 ‫This thing in the brackets has to be zero. 60 00:05:54,300 --> 00:06:03,840 ‫So what you have here, you have here this quadratic equation, lambda squared, plus lambda minus six 61 00:06:04,110 --> 00:06:05,700 ‫equals zero. 62 00:06:06,720 --> 00:06:14,760 ‫And the lambdas here they are, your roots do this quadratic equation, since you have a quadratic equation, 63 00:06:14,760 --> 00:06:18,510 ‫that means that you have two roots, so you have to lambdas. 64 00:06:19,420 --> 00:06:32,410 ‫And so the formula to compute those lambdas is minus one 1/2, plus minus, and then you will have one 65 00:06:32,410 --> 00:06:37,990 ‫half here squared and then plus six. 66 00:06:39,000 --> 00:06:42,930 ‫That's a typical formula to find the roots for quadratic equation. 67 00:06:43,990 --> 00:06:47,860 ‫In a more general case, your quadratic equation looks like this. 68 00:06:48,940 --> 00:06:56,230 ‫Where you have some kind of constant in front of this square term as well, then in order to find the 69 00:06:56,230 --> 00:07:05,280 ‫roots, your formula would be minus B, so you take this constant B and you put an opposite sign here. 70 00:07:05,290 --> 00:07:06,700 ‫So here you have plus. 71 00:07:06,730 --> 00:07:15,100 ‫So that means that here you put minus if you had minus B here, then here you would have minus minus 72 00:07:15,100 --> 00:07:30,010 ‫B which would be plus B and so minus B plus minus and then square root B squared minus four times eight 73 00:07:30,010 --> 00:07:31,370 ‫times C. 74 00:07:31,420 --> 00:07:39,040 ‫So these are these constants here and then all that you divide by two times a. 75 00:07:40,130 --> 00:07:48,050 ‫Or you can also write it down like this, I divide this term by two times A and then I divide this term 76 00:07:48,050 --> 00:07:58,040 ‫by two times A, and so if you're a constant here, if it equals one, like in our case. 77 00:07:59,060 --> 00:08:00,560 ‫Because here you have one. 78 00:08:01,610 --> 00:08:09,530 ‫Well, that means that your a year will be one, so you'll just have to hear and you will also have 79 00:08:09,530 --> 00:08:15,350 ‫one here, so you will just have to hear two times one equals two. 80 00:08:16,190 --> 00:08:17,060 ‫So like this. 81 00:08:18,040 --> 00:08:24,700 ‫And that means that this a here would also be one, so that's how it will look like. 82 00:08:25,770 --> 00:08:33,840 ‫Then you can factor out this for under this square root sign, and so if you do that, then it looks 83 00:08:33,840 --> 00:08:45,300 ‫like this four times and then in the parentheses you will have B squared divided by four and then minus 84 00:08:45,300 --> 00:08:46,410 ‫C here. 85 00:08:47,460 --> 00:08:51,180 ‫And then you divide this entire thing by two. 86 00:08:52,170 --> 00:08:59,370 ‫And well, you know that if you have a square root and then you have some kind of constant here, A 87 00:08:59,520 --> 00:09:11,000 ‫and then B minus C, then you can rewrite it like this, a square root, times square root B minus C, 88 00:09:11,610 --> 00:09:21,180 ‫and in this case, this square root A would be square root four and then square root four would be two. 89 00:09:21,690 --> 00:09:29,940 ‫That means that you can cancel out this two with this four here, because once you take it out of the 90 00:09:29,940 --> 00:09:33,290 ‫square root, it becomes two and you can cancel them out. 91 00:09:34,200 --> 00:09:42,120 ‫And so in the end, you can rewrite this formula like this might just be over two plus and minus, and 92 00:09:42,120 --> 00:09:47,010 ‫that really means that you have two separate equations, one with the plus and the other one with a 93 00:09:47,010 --> 00:09:47,580 ‫minus. 94 00:09:48,560 --> 00:09:55,970 ‫And then you have your square root here and then be squared away before you can rewrite it like this, 95 00:09:56,210 --> 00:10:05,090 ‫be over two and then you take this entire thing squared and then here you have minus C, and that's 96 00:10:05,090 --> 00:10:07,400 ‫the formula that I have used here. 97 00:10:08,370 --> 00:10:17,180 ‫In this equation, A equals one, B equals one, and then C equals minus six. 98 00:10:17,940 --> 00:10:26,010 ‫And so if we go back to our original quadratic function, then our roots will be the following lambda 99 00:10:26,010 --> 00:10:31,350 ‫one will be minus one half plus. 100 00:10:31,350 --> 00:10:40,770 ‫And then here in the square root, you will have six point twenty five because one half squared will 101 00:10:40,770 --> 00:10:47,850 ‫be zero point twenty five and then plus six it will be six point twenty five and then lambda two equals 102 00:10:48,030 --> 00:10:54,890 ‫minus one half minus square root six point twenty five. 103 00:10:55,740 --> 00:11:05,520 ‫So your lambda won, your first root will be two and then your second root will be minus three. 104 00:11:06,450 --> 00:11:13,320 ‫And so these are your roots, two and minus three, and you can check it, for example, if you take 105 00:11:13,560 --> 00:11:22,440 ‫the one and you put it inside here, then, well, you will have two squared, plus two, minus six, 106 00:11:22,950 --> 00:11:24,770 ‫and then that equals to zero. 107 00:11:25,110 --> 00:11:27,780 ‫And then we can do the same thing with minus three. 108 00:11:28,320 --> 00:11:31,440 ‫So minus three squared. 109 00:11:32,690 --> 00:11:44,810 ‫Plus minus three and minus six, well, that equals nine, minus three and minus six, and that equals 110 00:11:44,810 --> 00:11:45,320 ‫zero. 111 00:11:46,350 --> 00:11:53,880 ‫So you see, you have to lambdas Lumb one and two, and it makes sense that you have two routes or two 112 00:11:54,120 --> 00:11:56,190 ‫power constants here. 113 00:11:57,180 --> 00:12:04,140 ‫It makes sense because your differential equation is a second order differential equation. 114 00:12:05,140 --> 00:12:11,980 ‫And in the next video, let's see how we formulate a general solution for this particular differential 115 00:12:11,980 --> 00:12:12,640 ‫equation. 116 00:12:12,970 --> 00:12:13,900 ‫Thank you very much.