 We are at lecture 21 today, math 241. We are hopefully going to wrap up slope fields, trying to model differential equations, or at least what the solutions to those might look like in the plane, depending on which points they happen to go through or initial conditions. So I want to, first of all, explain the red vest today. That's our stat keeping uniform. I keep stats for the basketball games, men's and women's. And we have a men's game tonight against Wake. And I don't have time to go home and change before the game. But I like red. I'm a big state fan. State fan that actually is a true fan, not one that gets bent out of shape and completely distraught when we get beat and let's fire the coach and do all this crazy kind of stuff. And we need new players. I'm happy with the players we have and want to see them perform well. And I'm all about us beating the devil out of Duke and Wake and Carolina especially. So I'm a true state fan. But I do stats. Have I mentioned that in here? I sit down at the scores table on the court. And I get to hear some interesting language and interaction between coaches and players. I sit about, I don't know, 15 feet away from the visiting coach in the visiting bench. And it's interesting. Gary Williams has a very unique vocabulary at University of Maryland. Very unique. And he begins right away in the game with his unique vocabulary. But most of the coaches are well-behaved in their interaction with their players. It's just kind of fun to listen in to. But if you ever see me down there, that's what I'm doing. I don't get paid for that. It's a volunteer job. I'd probably pay them to do it because it's fun. But I probably shouldn't have said that. And they give you free stuff, like free vests and stuff. But it is fun. And it is kind of work. It gets intense when there's a shot and somebody blocks a shot and then there's an offensive rebound and a tip and another block. And sometimes during the game, stuff is happening too fast to actually kind of record in a computer. And usually somebody will get behind and I'll have to start writing. And then I'm looking down to write. And when I'm looking down to write, something else happened. And it's like, what happened there? Who turned it over? Who stole it? And somebody's dunking at the other end. And it's like, so it gets a little wild at times. But it is a lot of fun. And occasionally I get to travel with the team and do stats for the radio crew when their radio guy goes on cruises, the radio stat guy. When he's gone on cruises, I'm the backup stat guy for the radio too, which is also kind of fun. Are you going to ACC tournament? Have I ever done it? I've done it before. I don't think we're on schedule to do it this year. But we have done it before. That was fun. And that was wild, like game after game after game after game. Yeah, yes. So let's pick up with 7.2. I do want to put a little disclaimer on the fact that I am by no means an expert in what we're doing today when we look at modeling differential equations with electrical circuits. And we're going to take a look at a couple of them. We don't have to at this point in time because I know we're going to see them later in this book. But I think it makes it easier when we see them later if we have examined them in terms of how the differential equation goes and what are the laws that govern voltage drops and all that stuff. So if we take an initial look at them today, I think it'll be easier when we see them again. Before we get there, I made a statement yesterday that I want to qualify. When we're trying to model what the exact solution curve is by using Euler's method, let's take a look at something that I said yesterday. And let me clarify it just a little bit. When I drive somewhere in a car, I sometimes will turn the music off. It's pretty pleasant, actually. I listen to a lot of sports radio, which is just a bunch of chatter. But sometimes I'll turn it off. And a lot of times when I walk from my office at Harrelson over to here, I'll intentionally not use my cell phone because I like time to contemplate. You're going to laugh at me again. But a lot of times I'll think about what I did in this class or my other class that I teach and is that what I really wanted to say? And I thought about something I said in here. And I said, it's kind of like this. And I said, you can kind of reduce this with this. And I want to go back and revisit that because it isn't kind of like that. It is exactly like that. So let me revisit my verbiage. So here's the solution curve, what it's really supposed to look like. So this is our initial x value. And in fact, this is the exact point. And that kind of reminds me of my daughter. She's 21 now. But when she was a little girl, we'd see a formation in the clouds. And we say, hey, that kind of looks like a giraffe. No, it is a giraffe. And look at that. It does look like it. No, it is. It is a giraffe. So I said, you can kind of reduce these. No, you can reduce this. So how are we doing this? We're using this with a tangent line. The slope of the tangent line is the first derivative. So instead of, and we do want to make delta x a lot smaller than I'm making it in this diagram, but let's say we go over here delta x to get to our next x value. We want it to be a lot smaller. So we come up here. And there's a distinct difference between what we are finding. Here's what we're finding right here. Here's x1, y1. We're finding a point along the tangent to the curve. What we would really like to be finding is the point on the curve itself. So there is a difference. Let me try to show you on the diagram what we are finding. We're really finding this distance, which is dy, which is the change in y along the tangent to the curve. What in the equation that I wrote down yesterday shows what dy really is. Well, let me clarify one more thing. This change in x that we undergo from going from x0 to x1, it is exactly the same on the tangent to the curve as it is along the curve itself. So the delta x that we're changing, whether we're on the tangent line or on the curve, this distance is still delta x. So we say, how do we get from x0 to x1? Sorry, we want x1 is x0 plus delta x. That's not what I want to correct or clarify. How do we get to y1? y1 being the point along the tangent to the curve, it's y0 plus dy over dx, which is the derivative. That's what we're using here, differential equations. So we're given some derivative. We throw it in here at the prior x value and y value. So here's what I want to clarify or correct. I said, if it helps to remember this, think of this dx and this delta x knocking each other out. They are the same. The change in y along the tangent to the curve, which is really what this is, is exactly the same as the change in y along the curve itself. So the exact change in x and the change in x along the tangent to the curve are 100% equal in this diagram. So what we have found then is actually this. The new y is the old y plus dy is the change in y along the tangent to the curve. So dy is not delta y. Delta y would be the exact y in the change in the curve itself. We don't have the solution curve. All we have is the differential equation. We're trying to approximate it. So this is the best we can do at this point as far as our approximation. So I said, it's almost like you can reduce those. It is totally that you can reduce those. Then you end up with old y plus change in y along the tangent to the curve. So here's where we're going to end up. Hopefully, it's fairly close to the exact point along the curve itself. What will make this process better? Certainly better than my diagram illustrates. Small delta x. Small delta x's. Smaller delta x's. I mean, you can see it on this. If delta x is small, it's pretty hard to distinguish between the tangent to the curve and the curve itself. So the smaller delta x is, the better. Now, it is relative to the curve or to the description of the derivative to the curve, what small actually is. But clearly, the smaller, the better. So is it kind of like you can reduce those? It is exactly that you can reduce those as long as we clarify what dy is. All right, I chose a new color. Actually, pink was suggested to me today because it's nearing Valentine's Day, but I opted for another color. I don't know how that looks. Salmon? Let's see how it goes. I realize we're on at the five o'clock hour on cable TV, and there's a lot of other choices that people could be making, like sports center. So we've got to spice this thing up a little bit. We've got to get some color to this thing. So let's see how these circuits fly on salmon paper. This is exactly something from the book. It is page 506 in the book, the bottom of the page. Got a couple of different circuits once again. I'm not a circuit expert, but let's see how differential equations can model, according to some properties that circuits possess, how they can model a circuit and what's going on in the circuit with a differential equation. And we will see these again, and we'll do more with them the next time we see them. So let's see how direction fields give insight into physical situations. This simple electric circuit shown in figure nine, this is figure nine, contains an electromotive force, usually a battery or a generator, that produce a voltage. So we do want to pay attention to the units so that if we're given information, it goes along with these same units. So the battery or generator is going to be voltage in volts and a current of I, in terms of T, amps at time T. This circuit contains a resistor. Resistors ought to have a resistance, which they do, of r ohms. And an inductor, inductors ought to have inductance, which they do, of L, Henrys. Some unique units, too. That's another reason to look at this. So you're used to all these other units, and here's some new ones. So here's what the circuit looks like. We have a couple of principles that are going to guide how these things are interrelated on the circuit. Ohm's law gives the drop in voltage due to the resistor as ri, which is going to be part of our equation. The voltage drop, so we're going to have some kind of an equation that has the voltage drops in it, obviously. Drop in voltage due to the resistor is r times i. Voltage drop due to the inductor is L times di over dt. Thus, we've got the derivative in the equation, so it's a differential equation. One of Kirchhoff's laws says that the sum of the voltage drops is equal to the supplied voltage. That's where this equation comes from. So here's a voltage drop based on the inductor. Here is the voltage drop based on the resistor. The sum of the voltage drops is equal to the supplied voltage. So there's the supplied voltage. I don't know a lot about that other than that this is the principle that tells us what to do. And here's what those voltage drops are across the different things that we see on this particular circuit. So what is it? It's a first order. First order because it has a first derivative in it. It's a first order differential equation that models current as a function of time. i is current. So you're not going to be asked to memorize this. This will be provided for you if we have a current problem. But we need to be able to plug the numbers into the appropriate places. So let's use that in this example, which is in the book. So we have this equation. And I don't know time permitting today we'll do our first actual exact solution technique, separable differential equations. This equation that we're going to come up with will end up being a separable differential equation. And we can actually solve it for i in terms of t. Suppose that in the simple circuit that we just looked at, the resistance is 12 ohms. The inductance is 4 Henrys. And a battery gives a constant voltage of 60 volts. So if that is the equation written above this particular slope field, one of the questions is to draw the slope field. So I went ahead and included that. We could draw it, but that takes a long time. And we could put it into maple and it would construct it. But this is what it's going to look like. So our equation, tell me what to put where with these numbers. So the next time you see a circuit problem, you will have at least kind of plugged in some numbers in the right positions. L is what? 4. 4 plus 12 equals 60. So the units are exactly what we want, volts, ohms, and Henrys. So if we solve 4 di over dt, divide all the way through by 4, there's our differential equation that models that circuit problem. So we could, at individual points, we could come out here to 0, 2 and figure out what the slope is at this point by plugging in those values into the equation. So the first question is draw a direction field for equation 1 with these values. This is what it looks like. What can you say about the limiting value of the current? And I hope once we get to the equation for i in terms of t, you'll be able to see this also in the equation, not just from the slope field. So let's say we want a solution that goes through here. Can you visualize this solution coming up here and doing this, right? Or maybe it starts here at this point. So we've got a solution that does something like that. So what appears to be the limiting value? Appears that it's 5, right? And I think we'll be able to show this with the equation once we separate the variables and integrate both sides and go from there. Identify any equilibrium solutions. So we see a limiting value of 5. What if the initial value is, in fact, 5? Plug it in here. What's the rate of change of i with respect to t? 0. It would be 0. And if the rate of change is 0, then it's an equilibrium solution. So it would be stuck right here. If the switch is closed when t equals 0, so the current starts with i of 0 equals 0. I've already done that one. i of 0 equals 0 would be this solution. Sketch the solution curve. I don't know if there are other parts to this problem that I did not include. I think that's it. I think those are the parts to this example problem. So what we'll want to do before we finish today, I hope, is take this equation and solve it for i in a technique that's called separable, not that all equations are this type, separable differential equations. So for how many of you is this your first circuit problem? That's your first circuit problem. Several of you, a third of you. So the rest of you have dealt with circuits in some math or science related course. All right, let's look at another one. So whichever one we see later down the road will at least have had some example with it. This is a problem at the end of this section, problem 27. It's another circuit, a different circuit. This figure, down at the bottom, the figure shows a circuit containing an electromotive force, a capacitor with a capacitance of, I don't think we had one of these in the last one, C ferrids. Am I saying that right? Aren't these named after people that have done a lot of the research with electricity, right? And a resistor with a resistance of r ohms. The voltage drop across the capacitor is q over c, or you'll see it in the equation, 1 over c times q, where q is the charge in coulombs. So as this case with Kirchhoff's law, it gives us this equation. So the sum of the voltage drops is equal to the voltage. So we have r times i. We have q over c, which is going to become 1 over q. 1 over, excuse me, 1 over c times q. And then we've got the voltage over here on the right side. So we've seemingly got too many different letters in this differential equation. So how do we get things related to one another? i is the rate of change of q with respect to t. So let's see what these things are. q is the charge. So apparently if we take the instantaneous rate of change of charge with respect to t, we get i. And what was i in the previous problem? i was the current. So the current is the first derivative of the charge. So we get this equation, which is a first order differential equation. It's got a first derivative in it. This time, instead of being a first order in terms of i, or i in terms of t, it's q. So there's a derivative of q. There is q itself. So we've got a first order differential equation in terms of q. Suppose the resistance is 5. The capacitance is 0.05, and the battery gives a constant voltage of 60 volts. So what's the equation going to look like here? With a capacitor, as opposed to what we have in the other one. What do we have? So give me some numbers to plug in here. R is 5. First order. 0.05. q. q equals 60. So there's our first order differential equation with this particular circuit. And again, we'll revisit those later in this chapter. We could solve for dq over dt, which is what you normally do by moving everything to the other side and dividing everything by 5. But that's really as far as I wanted to go with this example problem, is that there are other things in these circuit problems that we're going to encounter. As long as we follow the right law and we know the interrelationships of the i's and q's and c's and ohms and ferrids and all the different things. And we can, you don't have to be an expert, I guess, is what I'm saying, at circuits to be able to do the math that's associated with it. OK, this is about right as far as where I thought we'd be time wise. I think it'd be to our advantage to actually work a little bit ahead. And one of the things I want to work ahead with is one of the circuit problems. But let's just take, so we're kind of dabbling now into the beginning of 7.3, which are separable differential equations. So this is the first type, not that all types that we're going to deal with are, in fact, separable, but some of them are separable. So we want to, whatever we're given, we want to be able to separate the x's and the dx's if those are our letters in the problem from the y's and the d y's. Once we get things separated, and there's a little bit of, we're going to have to do something we really haven't done before as far as how we treat a first derivative. But that's where the differentials part. We're going to treat a dy over dx as two separate things, the differential of y over the differential of x. So it is one thing. It's a derivative, but we're going to treat it as two separate entities. That way we can separate the x's and dx's from the y's and the dy's. So let's start with a real simple one. So let's start with this. dy over dx equals x. So if we think this is a separable differential equation, we'll find out if we are indeed able to separate the x's and dx's from the y's and dy's. So again, we're going to treat dy over dx as if it were two separate entities. We'll multiply both sides by dx. And now things are separated. We don't have any y's, but we do have a dy. So it's on the opposite side of the equation from the x's and the dx's. By the way, we should know our original equation was this. We ought to know without this technique what the solution is going to be. If the derivative of y with respect to x is x, what was y to begin with? 1 1⁄2x squared plus some constant, right? So we know what we're going to get on this one. But it's a good first example because we can separate things. It's pretty easy to separate. The next step, once we have the x's and the dx's separate from the y's and the dy's would be to integrate both sides. So the integral of dy equals the integral of x dx. Now we're opening up a little problem here is that when we start with two things that are equal and we integrate indefinite integral of both sides, then we introduce the possibility of some arbitrary constant here and some arbitrary constant here. So they're not necessarily going to be equal anymore, but they're going to at least differ by a constant. So what's the integral of the derivative of y? That would be y. Or if it suits you better that this is 1 dy. We're integrating 1 with respect to y. So in terms of y, what has 1 for its derivative? It would be y. I probably went a little too quickly there. Actually, we have to allow for a constant, some arbitrary constant because it's an indefinite integral. What's the integral of x with respect to x? And that constant is potentially different, so I'll call it c2. Are we getting there? We know what the solution is. We're getting there. So how do we rectify the fact that we've got two distinct constants? Can we put them together into a single constant? So we could subtract c1 from here and subtract c1 from here. What is c2 minus c1? Is that another constant? It's a constant. So solve for y. And we've got what we thought we were going to get 1 half x squared plus a constant. So you don't have to show c1 on one side and c2 on the other to show the fact that there is a constant involved when we integrated both sides, some arbitrary constant. So this is the exact set of curves that has this for its differential equation. Every one of these curves, and it's a whole bunch of curves, if you took the derivative, you'd get this. Questions on the technique? And that's exact. There's no approximation to that at all. It is the exact family of curves that we were looking for. Trying to think of ones that we looked at in terms of pictures. So let's say that the rate of change of y with respect to x is directly proportional to y itself. Now, I think we looked at this in terms of population. dp over dt, the rate of change of population is directly proportional to the population. This is the same thing. So if this is a separable differential equation, what should we be able to do? Separate x's and dx's from y's and dy's. So we can multiply both sides by dx. We better get the y to the other side. What's going to move the y to the other side? Divide by y or multiply by 1 over y. The k, it doesn't matter where it is. I mean, so it's there on one side. We could divide both sides by k. That's not going to really matter that much. So we've got k dx. And over here, we've got 1 over y dy. Does that appear to be separated properly? OK, what's the next step? Now that we've separated x's and dx's from y's and dy's. Integrate both sides. What's the integral of 1 over y dy? Natural log of y. There'd be a constant over here. I'm not going to write the constant over here. I'm going to put the two constants together on the right side. Now that would be absolute value of y. But we're assuming that we're only going to choose values for which the natural log function would be defined. If that bothers you, we'll revisit that topic. But we'll just be careful with the y's that we choose when we're done with the problem. How about the integral of k, where k is a number with respect to x? kx. And then we've got a constant on the left, a constant on the right. We'll put them together into a single constant on the right. I usually put the constants together opposite the y term because now the goal would be to solve for y. We didn't have to do that in the other problem. When we integrated both sides, we integrated the derivative of y and we got y. Now we got something else other than y, so let's solve for y. What e to the k is? OK, good. So three of you are saying what we need to do. Actually, that idea has a name, which I didn't know till like five years ago, that if these two things are equal, then e to those powers ought to be equal as well. It's called exponentiating both sides. So e to the natural log of y equals e to the kx plus c. So if these two things are equal, they can both be the exponent of the base e, and they still ought to be equal. And what's the advantage of having done that? So e to the natural log of y, this is a composition of two functions that are inverses of each other. So what do you start with here? You start with y, you take the natural log of it, then you raise e to that power. You ought to be back to where you started, which in fact is y. e to the natural log of y. e to the kx plus c, isn't that the same thing as e to the kx times e to the c? Is that correct? Kind of an algebra process in reverse. Normally, we would see two things that are multiplied with like bases, and we would add their exponents, right? Well, what if you have a sum in the exponent? Is it OK to kind of go backward algebraically to write them as a product? It is OK. And what is e as a number? c is some arbitrary number, so what is e to the c? It's some constant. So this equation becomes y equals some constant. I don't know, I'll just put it capital B. You can call this anything you want to call it. But you do need to split it up? Yes. Which kind of leads to my next question. Does this look at all familiar with anything else you've had in a math class prior to today? Isn't it exponential growth? That's just exponential growth, right? You might have seen it written like this, something like that, where it's a function in terms of time. This is the initial amount of whatever it is we're watching grow in this problem. The fact is that it's growing exponentially. If k is positive, it is actually growing. If k is negative, it's decaying. Is that those things ringing a bell at all? So y0, what is y0? Well, that's really just b. And isn't that what b is in this problem? Isn't b, when we put in x equals 0, wouldn't that be the initial y value? If x is 0, that's e to the k0, which is e to the 0, which is 1, right? So when x is 0, b is the initial y value. Now, you've probably seen this in another form. Sorry. Putting two things together there. Isn't that the same animal? Isn't the p value the initial amount of money, the principal investment, right? That's the initial amount of whatever it is illustrating here. This is the initial y value. e to a power, what is r in that mathematical model? That's the rate, kind of the growth rate. That's what this is here. k is positive, it's growing. k is positive, it's growing. It's the same thing. So this mathematical model is exactly the same as this mathematical model. It's something that is growing exponentially. What is that? Do you remember what this is normally used for? Interest. Right. What kind of interest? Not compound. OK, it is compounded, but not just compounded. It's compounded continuously. That's what this model is. Don't the rate and time just cancel out? Do you just get a number? There's no rate up there anymore? No? Because it's per time, and you multiply it by time. Right, so if you knew, let's say, district, look at that model. Let's say you had $10,000, and you want to invest it so that it's in an account that's not compounded one time a year, or twice a year, or four times a year, or monthly, or daily, or hourly, or every minute. It's more than that. It's compounded continuously, by the way. You'll never see a bank pay that. So this is kind of the maximum that this money could make at this interest rate. So let's say the interest rate is, I don't know what would be good now, let's say 3%. And your time that you leave it in there is 10 years. So you would punch the buttons on your calculator, and this would be the most that this amount of money could make at that interest rate for that length of time. You'll never get this. You might get a higher interest rate, but it's compounded quarterly or monthly. But this would be the most that it could make at 3% for 10 years. Somebody run that through. About $13,500. What is it? $13,500 almost. $13,500, roughly. Does that seem like a realistic answer? You started with $10,000. It's a fairly low interest rate, 3%. It's only in there for 10 years. Picked up $3,500. What if it were just simple interest? And if it were just 3%, what would this be? You're going to get 3% on your money for this 10-year period. Well, 3% of 10 grand is what? $300. $300? Would be $3,000, right? Just flat for that 10-year period I'm talking about. So it'd be $300 per year for 10 years. So $300 per year for 10 years is $3,000. So how did we come up with the extra $500 to explain how that came about? You're just all kind of, you're Pittsburgh, aren't you? Pittsburgh penguins? Man. You got to support the teams. I like that. This is the year to support them, too. Steelers. How's it going really bad, though? They are. Well, we won't talk about that. Let's just talk about the Steelers. So where did the extra $500 come from? They just liked you? Well, it's 3%. But you know, you're a college student. Struggling a little bit. Just give you $500. So if it's compounded continuously, which it is in this model, doesn't the interest you make today make interest for you tomorrow? Yes. Right? And doesn't the interest that you just made while I made my previous sentence? So the interest that you made, because it's being compounded continuously, the interest that you just made 10 seconds ago, isn't it thrown into the mix, and it makes interest for you all the way through this whole 10 years? So that's where that extra $500 comes from. That's interest making interest for you. It's money you've already made in interest being reinvested, and it makes interest for you. But that's the most. You can't compound something more than continuously. That's the most that amount of money could make for that length of time at that interest rate, which is exponential growth. All right, one more problem, and that would be to revisit the circuit problem that we had earlier. Let me put this picture back up, because we said, can you see a limiting value of the current when it looks like the limiting value from the slope field? It looks like it's 5, right, from the picture, that things begin to flatten out as we get close to 5, and in fact, at 5, they are flat, and that's the equilibrium solution. So here's our equation. This is a separable differential equation. So let's get the i's and the d i's separate from the t's and the d t's. So if we multiply both sides by d t, that's a start. So we've got the d t's. So we need t's and d t's on the right, i's and d i's on the left. So we better get rid of this term, similar to, but a little more complicated than the one we just looked at. So we've got i's and d i's separated from t's, which there aren't any, and d t's. What's next? Integrate both sides. Integrate both sides. The integral of the derivative of t ought to be t. I'll go ahead and put both constants together on the right side, call that c. How about this left side? Negative 1 third times the l n of 15. Good, good. So we've got u, if you want to think about this, equaling 15 minus 3 i d u would be negative 3 d i, right? So we need a negative 3 and a negative 1 third to be able to do the integration with a what? 1 over u, that's what this is, 1 over u. And this circled part is d u. So what's the integral of 1 over u d u? Natural log of u. Again, we're only going to choose values that make sense, so we don't need the natural log. So similar to the one we just looked at, we had 1 over y d y with a little bit more work in there. What's next? Solve for i. Exponentiate both sides? That's a nice word, isn't it? So you want to do what first before we exponentiate? We're the 1 third. Multiply both sides by negative 3? Yeah. You could do something else with that negative 1 third. Couldn't we put it up here in the exponent position? We wanted to? Let's not do that. Actually, a little bit easier to multiply both sides by negative 3. So this side by negative 3, so negative 3 times this gets rid of the coefficient of negative 1 third. Negative 3 times t is that. What's negative 3 times some constant that we don't even know? Another constant that we don't even know, right? Would it bother you if I called that c? Negative 3 times c is c? I won't. We'll call it something else. C1. Now c1. Now we can exponentiate both sides. What's e to the natural log of something? That's something. Here we've got e to the negative 3t. And I'm going to go ahead and write this as times e to the c1. We've got a sum in the exponent position that translates backward algebraically to the product of two things with like bases. So e is a number raised to some c1, which is also a number. There's nothing variable about that. That's just a number. So e to the c1 is a number. And typically then we solve for i, right? So if we solve for negative 3i, we would do what? Subtract 15 from both sides. Divide both sides by negative 3. So negative 15 divided by negative 3 is 5. And then we've got b over negative 3. Well, b is a number divided by negative 3 is a number. Plus or minus? Does it really matter? Let's call it minus. So b over 3 is a number. I'll call it something. Let's not call it q. Let's call it theta. I was going to call it q. We'll change it there with our mark. q means something else in this problem. E to the negative 3t. Now, I said we would get to a point where the equation would back up something we already saw in the diagram. Didn't we see a limiting value of 5, right? What happens to this equation as we let it run out to the right? If we let t approach 0, what happens to this term right here? t approaches infinity. I already gave it away, right? This term goes to 0. Because it's e to the negative 3t. So it's in the denominator. That's the thing that's exploding. So that term gradually disappears. Is it clear to see that the limiting value from the equation is also 5? So we can see it from the slope field that it's 5. We can see it from the equation that it's also 5. We're out of time. We will continue with separable differential equations on Friday.