1 00:00:02,350 --> 00:00:06,886 Welcome back, this is Doctor Ferri. Today's lesson is on First-Order 2 00:00:06,886 --> 00:00:10,512 Differential Equations. Previous lessons, we've seen that 3 00:00:10,512 --> 00:00:15,230 inductors and capacitors have i-v characteristics that involve derivatives. 4 00:00:16,520 --> 00:00:20,740 So, looking at our outline here, we have to be able to handle those derivatives. 5 00:00:20,740 --> 00:00:25,460 And what we're going to find is that we need to be able to solve first-order 6 00:00:25,460 --> 00:00:30,740 differential equations in order to be able to be able to apply the methods to 7 00:00:30,740 --> 00:00:37,358 solving RC and RL circuits. The specific lesson objectives are to 8 00:00:37,358 --> 00:00:40,543 examine first-order differential equations with a constant input and write 9 00:00:40,543 --> 00:00:47,084 the solution and sketch the solution. Ordinary differential equations are 10 00:00:47,084 --> 00:00:50,520 equations that involve functions of variables and their derivatives. 11 00:00:51,580 --> 00:00:55,460 Let's look at some examples. This is a first order differential 12 00:00:55,460 --> 00:00:59,480 equation that, it has a first derivative as the highest term, and that's why it's 13 00:00:59,480 --> 00:01:04,590 first order. Let's look at a different one. 14 00:01:04,590 --> 00:01:06,740 This one has a different forcing term in it. 15 00:01:09,700 --> 00:01:12,430 So if I look at these two terms, this is s forcing term that's constant, and this 16 00:01:12,430 --> 00:01:15,280 one is a forcing term that contains a sine wave. 17 00:01:15,280 --> 00:01:19,250 And these are two very common forcing terms that we see in electrical circuits. 18 00:01:20,940 --> 00:01:23,358 This is a second order differential equation because the highest order is 19 00:01:23,358 --> 00:01:26,311 second derivative. And here, I gave the forcing term as just 20 00:01:26,311 --> 00:01:32,543 a little bit more generic. In circuits, the equations that we will 21 00:01:32,543 --> 00:01:36,325 be looking at involve voltages and currents, so it makes sense to write our 22 00:01:36,325 --> 00:01:41,240 differential equation in terms of i or v as our variables. 23 00:01:43,800 --> 00:01:47,794 Now these three differential equations. Our first order, because their highest 24 00:01:47,794 --> 00:01:51,140 derivative is first derivative. In particular, we will be looking at 25 00:01:51,140 --> 00:01:55,295 differential equations of this form where it's a constant forcing term. 26 00:01:55,295 --> 00:02:03,800 And a circuit is a physical system. So it makes sense to look at models of 27 00:02:03,800 --> 00:02:09,858 physical systems that. have first order of differential 28 00:02:09,858 --> 00:02:15,376 equations. A system model typically has an input 29 00:02:15,376 --> 00:02:19,072 that we'll show here, which is our forcing term to our differential 30 00:02:19,072 --> 00:02:25,820 equation, and the output is the solution to the differential equation. 31 00:02:25,820 --> 00:02:30,020 Where a is just a coefficient that depends on the physical properties. 32 00:02:30,020 --> 00:02:34,460 Now, let's look at some common examples of other systems. 33 00:02:34,460 --> 00:02:37,305 Physical symptoms that involve differential equations. 34 00:02:37,305 --> 00:02:41,109 Particularly, let's look at a, a thermal system. 35 00:02:41,109 --> 00:02:44,590 A heat source is the input. And the temperature at a particular 36 00:02:44,590 --> 00:02:48,121 point's the output. So, when you have heat source, it takes a 37 00:02:48,121 --> 00:02:52,500 little bit of time for the heat to permeate through the, the area. 38 00:02:52,500 --> 00:02:55,762 And raise the temperature at a, a point. So it's modeled as a differential 39 00:02:55,762 --> 00:02:59,330 equation. Typically, another example that we often 40 00:02:59,330 --> 00:03:04,140 see is biological systems. Where we're looking the population growth 41 00:03:04,140 --> 00:03:07,520 or decay of a system given a production rate. 42 00:03:07,520 --> 00:03:11,396 Now in this course, we are interested in linear circuits, and they are solved, are 43 00:03:11,396 --> 00:03:15,320 written, modeled as differential equations. 44 00:03:15,320 --> 00:03:19,412 Particularly we have voltage or current as our input terms, and voltage or 45 00:03:19,412 --> 00:03:23,760 current across a particular component as our output. 46 00:03:26,070 --> 00:03:29,630 So, let's look at a solution to first order differential equations. 47 00:03:29,630 --> 00:03:32,854 Now our approach here is not to show you how to find the solution, but just to 48 00:03:32,854 --> 00:03:37,510 give you the solution and have you use it as a formula. 49 00:03:37,510 --> 00:03:42,160 So given a general, generic sort of differential equation here, where you 50 00:03:42,160 --> 00:03:46,885 have an initial condition, and this is given that time equals zero, and k is a 51 00:03:46,885 --> 00:03:54,720 constant, and a is also a constant, and that's our coefficient right there. 52 00:03:54,720 --> 00:03:58,482 The solution is given right her for time greater than or equal to zero. 53 00:03:58,482 --> 00:04:02,122 Notice that we have an exponential term in there that relates to that constant 54 00:04:02,122 --> 00:04:07,003 coefficient. And this K over a comes from this K over 55 00:04:07,003 --> 00:04:14,684 that A. It helps for us to understand the 56 00:04:14,684 --> 00:04:18,960 behavior as time gets very large. If a is greater than or equal to zero, 57 00:04:18,960 --> 00:04:23,622 this exponential decays to zero. So this term k is out, and this term k is 58 00:04:23,622 --> 00:04:29,986 out and we're left with K over a. I'm going to call that my steady state 59 00:04:29,986 --> 00:04:39,220 value, K over a. We want to be able to graph the response. 60 00:04:39,220 --> 00:04:42,302 Given the same formula for the solution to a first order differential equation, 61 00:04:42,302 --> 00:04:45,890 we've already identified the steady state value. 62 00:04:45,890 --> 00:04:49,630 And I'm going to call everything else in my solution as a transient. 63 00:04:49,630 --> 00:04:51,410 The transients the part that the k is out. 64 00:04:54,280 --> 00:04:57,790 In a time constant is something that's important to determine its a measure of 65 00:04:57,790 --> 00:05:01,408 how fast the transient decays now these are two plots of solutions to first order 66 00:05:01,408 --> 00:05:05,945 of each differential equations now in this first case. 67 00:05:05,945 --> 00:05:11,600 The initial condition starts at 1, and then it decays, the transient decays to 68 00:05:11,600 --> 00:05:16,595 steady-state value of 0. In this particular case, this solution to 69 00:05:16,595 --> 00:05:20,406 this differential equation, we start out with the initial condition of 0. 70 00:05:20,406 --> 00:05:29,890 And the steady-state value is 1. The time constant It's defined. 71 00:05:29,890 --> 00:05:33,940 We are going to called it Tau. It's defined as a time, the exponential 72 00:05:33,940 --> 00:05:40,650 transient to decay to e to the minus 1, which is 0.37 of its initial value. 73 00:05:40,650 --> 00:05:45,750 Another way of saying it is 63% of it's final value. 74 00:05:45,750 --> 00:05:50,793 So looking at this case over here. We want the transient de, decay to 0.37, 75 00:05:50,793 --> 00:05:55,390 which is about right here, of its initial value. 76 00:05:55,390 --> 00:05:57,910 And I go over and look at the time that it takes, and say that's my time 77 00:05:57,910 --> 00:06:06,940 constant. So in this case the time constant is 0.5. 78 00:06:06,940 --> 00:06:12,682 In this case, we want the transient to the k to you know, 63% of its final 79 00:06:12,682 --> 00:06:17,263 value. So from here to here, we want it to be 80 00:06:17,263 --> 00:06:22,992 about 63% of the way. So that's about 0.63 right there, and I 81 00:06:22,992 --> 00:06:29,038 go over here, and this would be the time constant right there. 82 00:06:29,038 --> 00:06:34,519 In both cases the time constant is 0.5, so I'm looking to reach go about 2 3rds 83 00:06:34,519 --> 00:06:39,739 away, 0.63 is about 2 3rds, so about 2 3rds of the way to the final value and 84 00:06:39,739 --> 00:06:47,735 over here about 2 3rds of the way to the final value. 85 00:06:47,735 --> 00:07:04,080 I'd be do some sample problems dy/dt plus 10y is equal to 20. 86 00:07:04,080 --> 00:07:06,712 Given an initial value I'll call it equal to one. 87 00:07:06,712 --> 00:07:14,840 The solution directly from that formula where k is equal to 20 and a is equal to 88 00:07:14,840 --> 00:07:22,460 10, I have y of t is equal to k over a or 2 times 1 minus e to the minus 10t plus e 89 00:07:22,460 --> 00:07:34,300 to the minus 10t, for t greater than or equal to 0. 90 00:07:34,300 --> 00:07:43,710 I can identify this steady state solution and that's k over a which is 2. 91 00:07:43,710 --> 00:07:49,693 We can see that right here, because these are there terms will be k to 0. 92 00:07:50,710 --> 00:07:53,680 And then the transient is going to be the value of time that makes this term, this 93 00:07:53,680 --> 00:08:00,795 exponential term e to the minus one. Well that value of time is going to be 1 94 00:08:00,795 --> 00:08:10,914 over 10, and that's always the case. The time concept is always one over this 95 00:08:10,914 --> 00:08:18,688 coefficient. Next I want to plot this, versus time. 96 00:08:18,688 --> 00:08:26,829 Show my scales here one, two, and I am plotting this solution y of t. 97 00:08:27,840 --> 00:08:31,988 It start out, it starts out with initial condition of 1, and let me show in a 98 00:08:31,988 --> 00:08:36,871 different color here. It starts out with initial condition of 99 00:08:36,871 --> 00:08:41,324 1, and then it's going to reach a steady state value of 2. 100 00:08:41,324 --> 00:08:49,710 Let me kind of highlight in green, what that steady state value is? 101 00:08:49,710 --> 00:08:55,122 So, my solution should start out with the initial condition and exponentially 102 00:08:55,122 --> 00:09:01,872 approach the steady state value. The time comes into tell us how to scale 103 00:09:01,872 --> 00:09:07,300 this axis here. So, at the time constant is when I get 104 00:09:07,300 --> 00:09:11,265 about 2 3rds of the way, from the initial condition, to the final value, that's 105 00:09:11,265 --> 00:09:17,930 about 2 3rds right there. So, I come down here and mark that as 1 106 00:09:17,930 --> 00:09:24,110 over 10 or 0.1. Then I can go ahead and just scale out 107 00:09:24,110 --> 00:09:28,502 the rest of my axis in a proportional way, by time I get to 4 time constant or 108 00:09:28,502 --> 00:09:32,966 four times a time constant add very very close to the final solution, so the 109 00:09:32,966 --> 00:09:37,430 method again was to write down a solution, identify the steady state value 110 00:09:37,430 --> 00:09:45,406 in the time constant. And then to draw your sketch from those 111 00:09:45,406 --> 00:09:50,720 quantities. We're going to do one more example. 112 00:09:50,720 --> 00:09:54,710 This one I want you to be a little bit more interactive in the way you do this. 113 00:09:54,710 --> 00:09:57,936 So, part of this I'm going to have you do this with me. 114 00:09:57,936 --> 00:10:09,247 So, dy dt plus 20y is equal to minus 2, y of zero is 2. 115 00:10:09,247 --> 00:10:19,615 The solution, y of 2 is equal to minus 2 over 20, which is minus 0.1 times 1 minus 116 00:10:19,615 --> 00:10:28,687 e to the minus a, or a is 20 ,so minus 20t plus y of 0 is 2e to the minus 20t 117 00:10:28,687 --> 00:10:43,286 and this is for time greater than or equal to 0. 118 00:10:43,286 --> 00:10:47,652 At this point I'm going to want you to pause this and I want you to find the 119 00:10:47,652 --> 00:10:52,854 steady state value. The time constant, and then we'll come 120 00:10:52,854 --> 00:11:03,840 back to sketch this. So hopefully, you've paused it and you're 121 00:11:03,840 --> 00:11:09,130 joining me back with the, the solution. The steady state value. 122 00:11:13,250 --> 00:11:18,080 Is going to be minus 0.01, which comes from K over A. 123 00:11:18,080 --> 00:11:28,348 The time constant is 1 over the coefficient, which is 1 over 20. 124 00:11:28,348 --> 00:11:38,540 Which is 0.05. So then if I want to sketch this, let's 125 00:11:38,540 --> 00:11:55,060 call this 1, 2, this is y of t. and my steady state values minus point 1. 126 00:11:55,060 --> 00:12:02,370 Let me go ahead and mark that. So minus point 1 is about right there. 127 00:12:08,550 --> 00:12:13,185 And my solution, I'll draw in blue here. Let's see, I start with initial condition 128 00:12:13,185 --> 00:12:19,040 of 2 and I'm going to exponentially approach my steady state value. 129 00:12:21,450 --> 00:12:23,811 The time constant is 0.05. That again is the time that it's going to 130 00:12:23,811 --> 00:12:25,226 take to start, to get 2 3rds of the way there. 131 00:12:25,226 --> 00:12:31,014 It's about right here. The time constant is 0.05 and just 132 00:12:31,014 --> 00:12:43,190 proportioning out my scale I've got 0.1. .15, and so on. 133 00:12:46,270 --> 00:12:51,115 So, you should be able to work sample problems like this. 134 00:12:51,115 --> 00:12:54,714 The important thing is to be able to identify the city state response, the 135 00:12:54,714 --> 00:13:01,697 time constant, and to sketch it. In summary. 136 00:13:03,290 --> 00:13:06,215 We've discussed how various physical phenomena are modeled by differential 137 00:13:06,215 --> 00:13:09,796 equations. We've showed the solution to a generic 138 00:13:09,796 --> 00:13:13,940 first-order differential equation with a constant input and initial condition. 139 00:13:13,940 --> 00:13:18,050 We're using it like a formula. We've introduced a transient and 140 00:13:18,050 --> 00:13:21,235 steady-state responses. And we've showed how to the sketch the 141 00:13:21,235 --> 00:13:23,430 response. And plot the time constant. 142 00:13:24,670 --> 00:13:29,979 In our next lesson we want to solve RC circuit equations and plot the responses. 143 00:13:29,979 --> 00:13:32,971 And we're going to be doing that by using this generic method that was developed in 144 00:13:32,971 --> 00:13:36,246 this lesson. Now please go to the forum to ask any 145 00:13:36,246 --> 00:13:40,860 questions that you have and please work on your homework. 146 00:13:40,860 --> 00:13:41,250 Thank you.