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These are the user uploaded subtitles that are being translated: 1 00:00:00,800 --> 00:00:05,820 Daring to be talking about how to solve a basic initial value problem and in this particular problem 2 00:00:05,820 --> 00:00:12,720 we've been given the function D-y over DX which means the derivative of Y with respect to x is equal 3 00:00:12,720 --> 00:00:14,760 to 2 x plus 1. 4 00:00:14,760 --> 00:00:19,020 We've also been given what's called an initial condition Y of 0 equals 3. 5 00:00:19,020 --> 00:00:25,590 And what this tells us is that when we plug 0 into our original function y of X not this function. 6 00:00:25,620 --> 00:00:32,250 But the integral of this function when we plug 0 in for x into the integral of this function the value 7 00:00:32,250 --> 00:00:34,030 that we're going to get back is three. 8 00:00:34,200 --> 00:00:38,930 So we have that piece of information and we have the derivative of our original function. 9 00:00:39,000 --> 00:00:41,980 What we need to do is find the integral of this function. 10 00:00:41,990 --> 00:00:43,760 That'll give us our original function. 11 00:00:43,950 --> 00:00:48,800 Then we're going to use this initial condition to plug in and solve for an unknown constant. 12 00:00:49,080 --> 00:00:53,490 So the way that we're going to do this is by separating our variables y and x. 13 00:00:53,520 --> 00:00:58,560 So we want to get the Ys on the left hand side the X on the right hand side the way that we're going 14 00:00:58,560 --> 00:01:06,060 to do that is by multiplying both sides of this function here by DX when we do that we'll get d wine 15 00:01:06,150 --> 00:01:12,200 is equal to 2 x plus 1 times d X.. 16 00:01:12,210 --> 00:01:17,460 Now that we have our variables separated with Y's on the left and X is on the right we want to go ahead 17 00:01:17,550 --> 00:01:19,200 and integrate both sides. 18 00:01:19,200 --> 00:01:28,470 So we'll say the integral of d y is equal to the integral of 2 x plus 1 D X like this. 19 00:01:28,470 --> 00:01:31,160 Now when we take the integral of the left hand side. 20 00:01:31,290 --> 00:01:34,490 Don't think of this as just the integral of D-y. 21 00:01:34,500 --> 00:01:38,310 Think of it as the integral of one times D-y. 22 00:01:38,550 --> 00:01:44,190 Remember that when we take the integral of a constant the result is that constant multiplied by the 23 00:01:44,190 --> 00:01:49,650 variable that we're integrating with respect to in other words 1 times why or just why the integral 24 00:01:49,650 --> 00:01:53,210 of 1 is why when we're taking the integral with respect to y. 25 00:01:53,490 --> 00:01:58,950 So the result of this left hand side over here is just going to be y the integral of one is y. 26 00:01:59,130 --> 00:02:03,370 Over on the right hand side we want to take the integral term by term. 27 00:02:03,420 --> 00:02:09,900 The integral of two x remember that will add 1 to the exponent the exponent right now is 1 this is X 28 00:02:09,900 --> 00:02:11,100 to the first power. 29 00:02:11,220 --> 00:02:13,470 When we add one will get X squared. 30 00:02:13,710 --> 00:02:21,780 But then we need to divide this term here by the new exponent which is 2 2 divided by 2 our existing 31 00:02:21,780 --> 00:02:27,200 coefficient 2 divided by our new exponent 2 is just going to give us 1 which we don't need to write 32 00:02:27,210 --> 00:02:29,950 as a coefficient in front of this x squared term here. 33 00:02:29,970 --> 00:02:32,280 So we only end up with X squared. 34 00:02:32,400 --> 00:02:39,420 The integral of one is just going to be X so will get plus X but then remember that whenever we take 35 00:02:39,510 --> 00:02:45,150 an indefinite integral we always have to add c to account for the constant of integration. 36 00:02:45,150 --> 00:02:47,930 So we're going to add c to the right hand side. 37 00:02:47,940 --> 00:02:53,090 Now a lot of times people ask me why do we add c to the right hand side and not to the left hand side. 38 00:02:53,100 --> 00:02:57,380 Well really the reason that we do it is just to take a shortcut. 39 00:02:57,570 --> 00:03:03,090 But technically if we were being really technically correct we would also add c to the left hand side 40 00:03:03,090 --> 00:03:03,860 and we would do this. 41 00:03:03,870 --> 00:03:07,580 We would say plus CSUB one maybe we would call it. 42 00:03:07,620 --> 00:03:11,270 And on the right hand side we'd say plus CSUB two we call them like that. 43 00:03:11,280 --> 00:03:14,910 You'll see why in a second we just bother adding it to the right hand side. 44 00:03:14,940 --> 00:03:19,060 It's really just a shortcut because we're always going to end up with the same answer. 45 00:03:19,080 --> 00:03:23,770 But what we would do at this point is we would say why is equal to. 46 00:03:23,910 --> 00:03:25,350 We're going to go ahead and subtract. 47 00:03:25,380 --> 00:03:27,290 See someone from both sides. 48 00:03:27,300 --> 00:03:32,550 So what will be left with is y equals x squared plus X plus. 49 00:03:32,550 --> 00:03:38,070 Now we have this value here CSUB Sabtu minus see someone because we subtracted see someone from both 50 00:03:38,070 --> 00:03:38,750 sides. 51 00:03:39,030 --> 00:03:41,640 Well CSUB to minus see someone. 52 00:03:41,640 --> 00:03:43,440 We've got two constants here. 53 00:03:43,440 --> 00:03:44,870 Right now we don't know the value of each. 54 00:03:44,880 --> 00:03:51,800 But let's say that the value of CSUB 2 is equal to 5 right CSUB two is equal to five. 55 00:03:51,870 --> 00:03:56,600 And let's say see someone is equal to to see someone is equal to two. 56 00:03:56,730 --> 00:04:00,090 And here we have the value CSUB to minus see sub 1. 57 00:04:00,090 --> 00:04:04,310 So we have five minus two and the value is three. 58 00:04:04,500 --> 00:04:11,670 Well as it turns out we don't specifically need to know the value of CSUB two different from the value 59 00:04:11,670 --> 00:04:12,800 of seis of one. 60 00:04:12,810 --> 00:04:16,390 We just need to know the value of whatever constant is here. 61 00:04:16,620 --> 00:04:23,460 So what we can do is replace this whole thing with just a single value of c will say Y is equal to x 62 00:04:23,460 --> 00:04:26,200 squared plus X plus C. 63 00:04:26,520 --> 00:04:30,930 We've essentially replaced both CSUB 2 and see someone with a generic. 64 00:04:30,930 --> 00:04:32,940 See that represents a constant. 65 00:04:32,940 --> 00:04:37,080 When we solve for this constant what we'll end up with is three. 66 00:04:37,080 --> 00:04:41,290 In this example If these were the actual values we'd end up with three. 67 00:04:41,340 --> 00:04:45,450 We don't need to solve for 5 and 2 we don't need to know these values separately. 68 00:04:45,480 --> 00:04:47,490 We just need to know this value in general. 69 00:04:47,670 --> 00:04:52,250 And having just a single see here is going to lead us to this value. 70 00:04:52,290 --> 00:04:53,640 So it's plenty adequate. 71 00:04:53,640 --> 00:04:59,840 And in fact we have to do it this way because we don't have enough equations to solve for CSUB to see 72 00:04:59,850 --> 00:05:06,120 someone separately we wouldn't we wouldn't be able to go about finding separate values for these two 73 00:05:06,120 --> 00:05:08,850 constants with this one initial condition. 74 00:05:08,850 --> 00:05:14,460 We're going to be able to find a single value for the constant c so that's why we combine them and that's 75 00:05:14,460 --> 00:05:19,650 why we only have to add a single C to the right hand side and we ignore adding it to the left hand side 76 00:05:19,950 --> 00:05:25,140 because we're always just going to eventually end up in this spot where we just have A C on the right 77 00:05:25,170 --> 00:05:25,740 hand side. 78 00:05:25,830 --> 00:05:30,600 So you can skip these steps and go straight to here and only add the C on the right hand side. 79 00:05:30,690 --> 00:05:36,180 Now once we've gotten to this point we move on to our initial condition and remember that we said this 80 00:05:36,180 --> 00:05:36,930 is essentially. 81 00:05:36,960 --> 00:05:43,580 Remember the function y of x is equal to some function over here on the right in terms of x. 82 00:05:43,620 --> 00:05:50,270 So this value because this is why of X this value here 0 is what we're going to be plugging in for x 83 00:05:50,280 --> 00:05:56,360 remember that we always plug this in for x 3 is going to be our resulting y value. 84 00:05:56,370 --> 00:05:59,580 So if we plug in zero for x we get 3 for y. 85 00:05:59,580 --> 00:06:05,800 So essentially what we're going to do is plug zero in for x and 3 in for Y. 86 00:06:06,030 --> 00:06:12,380 And when we plug those in all will be left with is Seiza we'll be able to solve for that value of c.. 87 00:06:12,440 --> 00:06:12,980 OK. 88 00:06:13,080 --> 00:06:17,030 So a plug three and for y and will get three is equal to. 89 00:06:17,040 --> 00:06:20,700 Now here we plug in zero for X so we're going to get zero squared. 90 00:06:20,700 --> 00:06:29,190 So zero squared plus zero here goes in for x and then we just have our C leftovers we get plus C. 91 00:06:29,250 --> 00:06:33,530 Obviously here we're going to get zero to go away and zero to go away. 92 00:06:33,540 --> 00:06:38,200 And as you can see we're just left with a value for C of C equals 3. 93 00:06:38,370 --> 00:06:43,530 And now we just plug this value for c back into our function here that we found for Y. 94 00:06:43,530 --> 00:06:51,720 And our final answer for this initial value problem is y equals x squared plus X plus 3. 95 00:06:51,720 --> 00:06:54,060 The value here that we got for C. 10459