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0:05
- Hi, this video is going to be a first look at differential equations for a calculus class. The only kind of differential equation that we are going to look at in single variable calculus is what's called an ordinary differential equation. So, an ordinary differential equation is an equation that relates a function to some of its derivatives. It could be an equation that has y, y prime, and y double prime, for example. We'll look at several examples.
0:28
A solution to a differential equation is a function, like y equals co-sine of x plus x squared, something like that, that makes the differential equation true. The order of a differential equation is the highest level derivative that appears. For example, if you have y, y prime and y double prime, because of the y double prime, that would be a second order equation. Then, an autonomous differential equation versus a non-autonomous differential equation, I'm actually going to wait to define those
0:57
when we look at our examples, so put that on the back burner for now. An initial value problem is an ordinary differential equation, where in addition to the equation, you're given information about how the solution and maybe some of its derivatives must behave at a point in the domain. Parameters are constants in differential equation that might vary in different scientific settings. For example, parameters might encode information like birth rate and death rate,
1:24
or half-life for a chemical substance. So sometimes those look like variables because they're often letters like k or n, for example, but those are usually constants in different contexts. And then lastly, an equilibrium solution to an ordinary differential equation is a solution to that differential equation which is constant over time. Okay, that was a really quick run through of all of the terminology on this slide, but what we're going to do now
1:51
is look at several examples of differential equations. We will see first and second order examples. We will have maybe just one non-autonomous example. We'll see several autonomous examples. I'll explain what those are since I skipped that one. We'll describe what it means to say that something is a solution to a differential equation. We will identify some parameters. We'll solve some initial value problems, and also I will have a few examples where we will identify equilibrium solutions.
2:15
So we're going to work through all of this vocabulary through several examples for the rest of this video.
2:32
For this first example, consider the equation dy dx equals x plus y.
2:39
This is an equation relating the function y and its first derivative, dy dx. So here we want to think of our function as y. Typically we would write that as y, but you want to remember here that y is a function of x.
2:57
The only kind of derivative that appears is a derivative of y with respect to x. So here in my definition of an ordinary differential equation as an equation relating a function and its derivatives with respect to one single independent variable, that variable in this context is x.
3:15
I don't want to get off topic, but there's a different kind of differential equation called a partial differential equation, and that's where you might have a function that depends on, say, a spatial variable x and a temporal variable t. Then you might have derivatives with respect to x, derivatives with respect to t, you would be looking at different types of ways that your function could depend on different independent variables. That's not what we're doing in this video.
3:40
We are just looking at functions which only depend on one single independent variable. So here x is the independent variable and y is the dependent variable.
3:50
This is a first order equation
3:57
because the highest level derivative that appears is a first derivative. We have dy dx and that's it. So this is a first order equation. It is also what's called a non-autonomous differential equation because of the presence of x.
4:15
In most of the equations that we look at today, we will not actually see the independent variable appear. That's pretty common, and those are called autonomous differential equations. Because of this x, this makes it non-autonomous.
4:38
Okay. So, whoa, I wrote that really small here. But this is a first order non-autonomous
4:45
ordinary differential equation.
4:53
Now, we've classified this differential equation. I'm not going to tell you how to solve it in this video because that's kind of the point of the next few videos. I just want to present what it means to say that something is a solution to this kind of equation. So given this function, y is a function of x, in particular y is negative 5e to the x minus x minus 1. Let's show that this is a solution. So what we want to do is take this function, plug it into the differential equation
5:21
and conclude that it's a true statement. So that means that we need to check that given this function, dy dx, so if we take the derivative of this, it equals x plus y. I think for this particular example, I'm going to do the left-hand side on its own, left-hand side of the differential equation, and then the right-hand side on its own, and then we will see that they are the same. So given this function, what is dy dx?
5:46
Well, if we differentiate the left-hand side, we get dy dx. Differentiate the right-hand side and we get negative 5e to the x minus 1 minus zero.
5:57
Okay, so that was the left-hand side of the differential equation. So I computed dy dx given that y has this form. Let's do the right-hand side, which means what is x plus y? Well, that's going to be x plus this expression for y. So x plus where I had y, I'm going to put in negative 5e to the x minus x minus 1.
6:22
Notice that the x's canceled and we're left with just negative 5e to the x minus 1.
6:33
Now if you compare, when we differentiated this function to find dy dx, we had negative 5e to the x minus 1. But when we did x plus y, we also have negative 5e to the x minus 1.
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Therefore, for this function, dy dx equals x plus y.
6:54
So that's what it means to say that a function is a solution to a differential equation. It makes the differential equation true. I have here one particular solution, but it's not the only one. In fact, you might have noticed that this negative 5 never really came into
7:14
any of these computations. It never really seemed to matter. And that's because this first order differential equation has a general family of solutions. I'm going to just write down what it is now, and again, we'll learn how to solve differential equations later. So the family of solutions,
7:37
what you might think of as the general solution, is going to be y equals Ce to the x minus x minus 1
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where C is a number.
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So C is some constant.
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This is called the general solution
7:58
of which this was one particular example.
8:03
Okay, so here's what I wanted you to get out of this example. I wanted you to see that this is a first order differential equation because the highest derivative that appears is a first derivative. Because of the presence of the independent variable in the equation, this is a non-autonomous differential equation. And then what does it mean to say that a function solves a differential equation? It means that it makes it true. So that is what I wanted you to take from this first example.
8:30
Let's repeat some of this analysis now with the second example.
8:48
Our second example of an ordinary differential equation is this one. We have y double prime equals negative 4y. This is a second order differential equation.
9:02
Y prime doesn't appear, but that's okay. It's still a differential equation. It's relating information about y and its second derivative. So because of the presence of a second derivative here, this is a second order equation. Just looking at this y double prime equals negative 4y, it's not even clear what the name of the independent variable is. It's going to be x in this example, but it could be t for time or something like that. The independent variable does not appear in this equation.
9:29
There's no t or x over here on the right-hand side, so this is an autonomous differential equation.
9:41
Okay, so it's a second order autonomous differential equation. Let's show that the function y equals cosine of 2x is a solution. So in other words, we want to show that cosine of 2x, if we plug that in for y in this differential equation, we're going to get a true statement. All right, so, I think what we should do first is take its first derivative.
10:03
The derivative of cosine of 2x is negative two sine of 2x.
10:11
Now let's take its second derivative, that's going to give us the left-hand side of our second order autonomous differential equation. Y double prime is going to be negative 4 cosine of 2x.
10:26
Let's take a look at that.
10:29
Notice that cosine of 2x itself is y. You want to start recognizing these kinds of pieces whenever you want to verify that a function is a solution to a given differential equation. So when I see cosine 2x, I want to say, hey, wait, that was the function y. So, I could think of this as negative 4 times the function y itself. And this returns me the differential equation. Y double prime equals negative 4y. Therefore, the function y equals cosine of 2x is a solution to this differential equation.
11:05
Take a minute and see if you can make a little change to this to come up with another solution to this differential equation.
11:17
One thing you might've done was replace cosine with sine, and if you did that, great, you would've found another solution to this differential equation. So y equals sine of 2x also solves the second order autonomous ordinary differential equation. You also might have thrown a constant in front of this. After all, we had that constant in our first example, which we were able to generalize. You would've also found a solution to this differential equation. So for example, y equals 5 cosine of 2x
11:46
also makes this differential equation true. In fact, here's the general solution.
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The general solution is y equals a constant multiple of cosine of 2x plus a constant multiple of sine of 2x.
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So for this example, we have the multiple A is 1, B is zero.
12:18
If you tried sine of x, you would've been looking at A is zero, B is 1. But you could also try y is negative 10 cosine 2x plus 3 sine of 2x, and that will also give you a solution to this differential equation. There's one kind of special solution here, which is that if you take both A and B to be zero, you're looking at y equals zero. Now if y is the constant zero function, y double prime is zero, but also negative 4 times y is also zero. So the constant function y equals zero
12:50
is also a solution to this differential equation. So let me make a note here. Includes y equals zero.
13:02
The neat property about y equals zero that I'm highlighting is that this is a constant function which satisfies the differential equation. It does not change over time. This is what we call an equilibrium solution.
13:20
Okay, so let me identify some of the vocabulary in this example. We have a second order ordinary differential equation because the highest level derivative is a second derivative. It is autonomous because there's no dependence here on the independent variable.
13:38
A solution to this differential equation, just like before, is one that makes the differential equation true. And in this particular example, I've identified an equilibrium solution, which is a constant solution that satisfies the differential equation. So let's move on now to talking about initial value problems. For the next two examples, we are actually going to revisit what we just looked at example one and also this one. So to the examples that we've seen so far,
14:05
I'm going to add what are called initial conditions.
14:22
Now let's look at the fifth item on our vocabulary list, the initial value problem. An initial value problem is a differential equation, just like before, where you are given some extra information, an initial value, or maybe several pieces of information. So I've written here that it's a differential equation together with information about the value of the function and possibly its derivatives at a specific point in the domain. This kind of vague expression here can actually be made precise.
14:53
You need to be told as much information as the order of the differential equation. So here for our first example, we had a first order differential equation. So we need one piece of information, the value of the function at a point in its domain. For a second order differential equation, like the one that we'll look at in a minute, we will need two pieces of of information. So that possibly some of its derivatives is actually not mysterious. You need as many conditions
15:19
as the order of the differential equation. Okay, so here's what we're going to do. We have here the general solution to this differential equation. I didn't tell you how I found it because all we want to do in this video is experiment with what it means for a function to solve an ODE, or in this case solve an IVP, an initial value problem. For an initial value problem, what you're looking to do is find any constants that define the family of solutions.
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So we are going to use our initial condition
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to find a specific value of C.
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So in other words, there are infinitely many functions of this form that solve the differential equation dy dx equals x plus y. You could pick infinitely many possible values of C and they will solve just the differential equation. But there's only one that also has the property that the value of y, the dependent variable, when the independent variable x is zero is 1. So here's what we do is we take our general solution, and now everywhere I have x, I'm going to put in zero. So let x equal zero.
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And then everywhere I have y I'm going to put in 1.
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Because our initial condition is telling us that when x is zero, y is 1.
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Okay, so when x is zero, y is one. So the left-hand side is going to be 1. So 1 equals the constant we're going to find, e to the zero
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minus zero minus 1.
17:18
Okay, let's add this 1 to the other side, and we have 2 equals C times e to the zero, which is C times 1. And at this point we've actually found it. So here's our value of C. So the solution to the initial value problem
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is one particular version of this general solution. It's y equals 2e to the x minus x minus 1.
18:06
Now let's add some initial conditions to our second example. Because this was a second order differential equation, we need two pieces of information. So in particular we are told that when x is zero, y is zero, and y prime is 1.
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If you think of y as possibly a position function, that's often the case with differential equations is we might be modeling, say motion. If y is a position function, then we are being told that when we start, the initial position is zero and the initial velocity is 1. So that could be an interpretation of those two initial conditions. Initial value problems are also sometimes called boundary value problems, if the point of the domain you think of is on the boundary of the domain.
18:51
Okay, so let's return to our general solution. When we looked at this problem the first time, we said that y equals A cosine of 2x plus B sine of 2x defines a family of solutions. You could pick A and B to be any numbers you like.
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What we are going to do now is use our pair of initial conditions, a statement about y at zero and y prime at zero to solve for A and B. So there's going to be one version of the solution that makes the initial conditions true. All right, so here's what I'm going to do is I'm first going to take the derivative, let's do y prime.
19:30
Okay, y prime is going to be negative 2A sine of 2x
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plus 2B cosine of 2x.
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I wrote that kind of strangely. We can plug in x equals zero. I'm going to do that over here, sorry to work to the right.
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When x equals zero, y itself is zero. That's what this initial condition is telling us. So on the left-hand side I'm going to write zero equals, and then when x is zero, cosine of zero is 1. So we get A plus,
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and sine of zero is zero, so zero B.
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Also when x is zero, y prime is 1. So I put 1 in for the left-hand side of the second equation and then we'll get, let's see, sine of zero zero, that goes away.
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Cosine of zero is 1. So 2B.
20:35
This pair of equations is pretty ideal because we can immediately solve for both A and B just by looking at this. However, we could have ended up with something that's a little bit more interesting. We could have had like 3 equals A plus 4B, and negative 1 equals negative A plus, you could have had a system of two equations for two unknowns, okay? So that could have happened. In this particular case, it did not. We know that A equals zero.
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And B is one half.
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So therefore the solution to this initial value problem
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is y equals zero cosine 2x plus one half sine of 2x.
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So this function is the solution to this initial value problem.
21:36
For the rest of this lesson, I would like to just look at how to apply this vocabulary to a few more examples of differential equations. So I'm going to put up some differential equations like this one, the differential equation modeling exponential growth or decay. And instead of looking at general solutions or initial value problems, I just want to identify how each of these concepts comes into the differential equations. So this is the type of analysis
21:57
I might start to do every time I look at a new equation. So first of all, we have dy dt equals k times y, where the number k here, it's just some constant. So let's identify the roles that t, y, and k play in this differential equation. So dy dt tells us that we have a function y depending on t. So our independent variable is t,
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and our dependent variable is y.
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So that a solution to this differential equation would be a function y of t. This number k is what's called a parameter. So this is a constant in a differential equation, which at first glance might look like a variable, but typically what it does is it allows you to apply a differential equation to different scientific context. In this particular setting, if k is positive, you would be looking at say, exponential growth.
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You might have different values of k for say different bacteria populations, and if k is negative, you might be looking at something like chemical decay where you're looking at solving a half-life problem, and different compounds would have different values of that parameter k. So parameters allow us to adjust differential equations to different scientific settings. Okay, for the next equation, this is something that we've been looking at. What's the order of this differential equation?
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We just have a first derivative dy dt. So this is a first order of differential equation. There's no occurrence of t really in this equation. Besides seeing the dt telling us that y was differentiated with respect to t, we don't have any other instance of t. So this is an autonomous differential equation.
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And then lastly, are there any equilibrium solutions? Let me write out what I mean by that. So this is a constant solution.
23:51
We could write that as y of t equals some constant value, which is a solution to the differential equation. So this solves the differential equation.
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Okay, so are there any constants that we could plug in for y in this differential equation in order to solve the differential equation? Well, keep in mind that if y is a constant, then dy dt is zero.
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So it's really like saying is there any value of y that would make the right-hand side, k times y, equal to zero? The only constant solution there would be y equals zero itself.
24:42
So the function y of t is a constant zero is an equilibrium solution to this differential equation, and it's the only one. It's the only constant function that you could plug in that would solve this differential equation. We are going to talk about this particular model in depth, but if you think about how I alluded to that this is used for say, bacterial growth, what would that represent? Well, if y is zero, that's like having a population of zero bacteria. That would not change over time.
25:09
You would just keep having zero bacteria. If y would be any positive number, then again, just kind of looking ahead at how this equation is used, we would expect that population to change.
25:30
Let's repeat that analysis for this example. Again, this is a very famous model that we are going to look at in greater detail. So I don't want to go into the scientific context for this differential equation, but what we have here is dp dt equals some positive number k times the quantity 1 minus P divided by a positive number N times P.
25:50
So I will just say that this is used for populations. So we have dp dt represents the rate of change of a population with respect to time. So our independent variable is t, and our dependent variable is P.
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For different populations you would adjust the k and the N. So the k and N values, those aspects of our equation are parameters. Again, on the right-hand side of this differential equation, we don't have any occurrence of the independent variable t. So this is autonomous. Sorry, I jumped ahead.
26:26
It's also first order. Let me go back. This is a first order ordinary differential equation.
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And lastly, let's find the equilibrium solutions for this differential equation. In order to do that, I want to bring in just a new notation. Let me take this right-hand side,
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and recognize that it's a function of P.
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So I could rewrite this differential equation as dp dt
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equals f of P.
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Were this a non-autonomous differential equation, you could have equals f of t and P.
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So what we are looking for with our equilibrium solution is a constant value of P that makes f of P equal to zero. Because if P is constant, dp dt is zero. So we know what the left-hand side of our differential equation would have to be. Is this also equal to zero? So we're looking for the roots of f of P. So for the equilibrium solution,
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we want f of P equal to zero.
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And P constant.
27:51
All right, so are there any constant values of P that if we plugged it into this right-hand side or plugged it into this function, which I'm naming f of P, we would get zero? It's like looking for the roots of this right-hand side and there are two. The first one you might see is that if we plugged in P equals zero, then because of this final term, it would wipe the whole thing out.
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So P equals zero is an equilibrium solution, but then also in this middle term, if P were N, then the middle term would turn into 1 minus N over N, which is 1 minus 1. That would give you zero as well. So we have P equals zero and P equals the positive parameter N.
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That's all I'll say for this logistic differential equation for now. I just wanted to identify these key concepts. Let's go up now to a second order differential equation.
28:53
Our next example is another famous differential equation. This one is called the pendulum equation. So I'm not going to derive this for you in this video at least, maybe I'll make a separate one, but let me talk through what it's modeling. Consider a pendulum of length L with a mass suspended on the end. We can model the motion of that pendulum if we understand the angle that that pendulum opens with a vertical direction. In particular, let's say that the direction
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pointing straight down is like having an angle of zero, so that's theta equals zero. And then if we have a pendulum just slightly oscillating back and forth, swinging to the right would be like positive angle, swinging to the left would be like negative angle, so that we could model not only the angle that the pendulum is opening up, but also its sense of direction. Okay, so that's the context. A swinging pendulum. Then from the laws of physics, like force is mass times acceleration,
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you can get to the second order ordinary differential equation modeling the swinging of a pendulum. What it says is that the second derivative of the angle with respect to time plus the gravitational constant divided by the length of the pendulum times sine of the angle is zero. So in this second order differential equation, the independent variable is time, t. The dependent variable would be the angle theta. So that a solution to this differential equation would have the form theta of t,
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and if you plotted that solution, it would tell you how the pendulum moved over time. The main parameter here is the length of the pendulum, assuming that we have the same gravitational constant. But since you have g divided by L in one term, it's one coefficient, you could actually treat that coefficient as say a parameter for each pendulum. This is a second order differential equation, and again, we do not have any occurrence of the independent variable t in this equation,
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so this is autonomous.
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Okay, that runs through the basics. Let's wrap up this example with the discussion of the equilibrium solutions. I want to do this a little bit differently. What I would like to do is first figure out what the equilibrium should be based on our understanding of a pendulum. Then we will verify that that makes sense with the equation because this equation should be modeling the physics of this pendulum. So let me remind you that an equilibrium solution is one which is constant for all time.
31:05
So take a moment to think about a pendulum. Are there any angles which if you positioned your pendulum in that angle and then let go, the pendulum could stay in that angle forever. It would be a constant angle for all time. See if you can figure that out and then I will come back and discuss it and we will verify it with the equation.
31:30
I hope you got to two equilibrium positions for this pendulum. One is fairly immediate, and that is that if the pendulum is hanging straight down, if it's just hanging at rest, then it will stay there forever. The other one is harder to simulate in real life. But imagine we could take this pendulum and rotate it to point straight upwards. If it was perfectly positioned, then it would have no reason to go left or right, so it would also maintain that position forever.
31:54
Feels a little less stable than the first equilibrium position, but it is an equilibrium solution. So let's verify that. If our angle is constant for all time, then we have said before, that means the first derivative is zero, but also any subsequent derivatives are also zero. So theta prime prime is zero.
32:14
We take this information and plug it into our differential equation, what happens is that the first term is zero, divide out by g over L and we're left with sine of theta is zero.
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What angles make this true? Integer multiples of pi. So the solution would be theta is k times pi, where k is an integer, a whole number.
32:41
Mathematically that means there are infinitely many solutions, but each one corresponds to one of these two positions. So zero is the position hanging straight down. Pi is like pointing straight up, rotating through the right. 2pi would be rotating back down. 3pi would be rotating back up. And then negative versions just go the other direction. So negative pi means we rotate it up from the left side and then et cetera, et cetera. Even multiples of pi are ones
33:06
where the pendulum is hanging straight down. Odd multiples of pi are the ones where the pendulum is pointing straight up. Okay, let's look at one more example.
33:24
Let's end this lesson by looking at a slightly different situation. So this is the Lotka-Volterra predator-prey equations. It's two equations for a predator-prey system. So here R represents the prey. You could think of this as rabbits.
33:44
And F represents the predator. You could think of that as foxes. That helps you distinguish between these two lines. So dr dt tells us the rate of change of the rabbit population over time, and df dt tells us the rate of change of the fox population over time. And notice they must be taken together because the right-hand sides for both equations depend on both populations. You can't use just the first line to study just the rabbits and forget about the foxes because as the rabbits change,
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the foxes will change, but as the foxes will change, the rabbits will change. So these equations are coupled together. So let's go through the questions we have here for these two populations.
34:25
The first one, what is the independent variable? It's still time. So the independent variable is still t. But we are looking for two. If we wanted to solve this, we would be looking for two functions of time. We would have R and F, the two populations and how they change over time. Are there any parameters? Yes there are. I don't want to go too much into what they represent, but we've got alpha, beta, gamma, and delta. Essentially these reflect things like the growth
34:59
and death rates for the individual populations, as well as how say, the foxes eat the rabbits versus how the rabbits affect foxes. 'cause you have these interaction terms where you have R times F. Anyway, so they represent essentially rates that you would observe with these two species. What is the order of this pair of differential equations? Each one is first order. We have two of them, but they're both first order. So we would say that this is like a pair
35:24
of coupled first order differential equations.
35:32
There is once again no dependence on time. So this is an autonomous differential equation. You'll notice that was true for just about every example in this lesson.
35:44
And then what would it mean to say that we have an equilibrium solution to this Lotka-Volterra model? That would mean that we found some value of rabbits and foxes which was constant for both. So rabbits are constant, foxes are constant, and together they satisfy both of these equations at the same time. So we would want rabbits to be some constant, I'll say C sub r, and at the same time, foxes to also be a constant. They must both be constant at the same time, which means that their derivative
36:19
would simultaneously be zero for all time.
36:24
Okay, so let's actually solve this out. Let's find the equilibrium solutions for this pair of differential equations, if there is an equilibrium solution. So whatever we're looking for, the left-hand sides of this pair of differential equations would both have to be zero because dr dt and df dt would always be zero for any equilibrium pair. So we can say zero equals alpha R minus beta RF.
36:53
And also zero equals negative gamma F plus delta RF.
37:01
Okay, let's solve this system of equations where we really only have two unknowns. We're looking for R and F values that make both of these equations true.
37:13
Alpha beta gamma delta we'll just treat these as numbers. That's what they are. For a given ecosystem, they are fixed numbers. They're parameters, not variables.
37:22
What I would do to solve this differential equation is we've got multiple terms here for each line, I would try to pull out like a greatest common divisor. So R is common to both terms in this first line, let's factor R in front and write this as R times alpha minus beta F.
37:43
And similarly on the second line, we can factor F in front
37:49
and write this as F times the term negative gamma plus delta R.
37:57
Okay, take a look at that.
38:00
What we're really trying to do here is find pairings of an R value and an F value which make both of these lines both zero at the same time. So see if you can work it out and then I will come back and finish the calculation.
38:16
So what we have here is a system of two equations for two unknowns, and the right-hand sides of both of these equations have been put in a factored form. Whenever you have that a product is zero, you want to look at scenarios where each of the factors is zero. So let's look at this first line. We have two factors here.
38:34
So for the first line, either R is zero,
38:39
or alpha equals beta F, meaning alpha minus beta F equals zero, or both. Okay, so either R is zero or, I'm going to write alpha minus beta F is zero.
38:55
Okay, what I think we should do now is take one of these, put it in the second equation and see what would have to be true in the second equation in order to reach an equilibrium solution. So let's take R equals zero first and put this in the second equation.
39:14
When we do that, it's going to wipe out this delta R, and we would have zero equals negative gamma F. So this will go away and we'd just be left with that. But then this tells us that F is zero.
39:30
Okay. So when R is zero, F must be zero. So we have an equilibrium solution at zero zero
39:38
So let me just put that, whoops. B marker. Let's put that down here.
39:45
Physically, what does this mean? It means that if our rabbits don't exist and our foxes don't exist, then that will satisfy this differential equation. So does it make physical sense that they would just continue to not exist? And the answer should be yes. So if we don't have any rabbits and foxes, we never will, right? So that checks out.
40:05
Now let's take the second possibility and put that here. Okay, sorry. Hopefully the color coding is clear. So what happens if F is alpha over beta?
40:21
We plug that into the second line.
40:26
Again, sorry about my work. Trying to keep this organized, so, sort of distinguish what's going on here. Okay, hopefully I didn't just make it worse. If the fox population is the constant number alpha divided by beta and we plug that into the second line, this term out front is just going to be a positive number. We would be looking at zero equals just the number alpha over beta times the expression negative gamma plus delta R
40:57
Alpha over beta is just a positive number. This isn't going to cause this product to be zero. It would have to be the case that the second term was zero. So this would tell us
41:07
that the rabbit population is also a positive constant value. In fact, it's gamma over delta.
41:19
And those are the only ways to combine situations and have both of these equations simultaneously equal to zero. So if rabbits was zero, it had to be the case that foxes was also zero. That's what we have here. Otherwise, if foxes population was the number alpha divided by beta, it would have to be the case that the rabbits were the number gamma divided by delta. So we've got this equilibrium solution as well.
41:47
Gamma over delta for the rabbits and alpha over beta for the foxes. So I'm putting rabbits first and foxes second to match the ordering of this differential equation.
41:59
Okay, so let's wrap up this example. We said that this equilibrium corresponded to an ecosystem that had no rabbits nor foxes and never would. So that's not too interesting if we're trying to talk about rabbits and foxes. For this ecosystem, what this equilibrium tells us is that there's some number for the rabbit population and some number for the fox population, so that if those two populations were at those values, they would stay at those values forever.
42:24
It's like a sustainable balance between the prey and the predators so that there's enough food to feed the predators, but they are not so many predators that it wipes out the prey, et cetera, et cetera. So I like to think of it as like a harmonious balance between these two populations. All right, I hope you enjoyed this look at differential equations and some of the terminology and ideas that we think about when we study differential equations.
42:44
I haven't said it yet, but not every differential equation can be solved. So when we look for solutions to differential equations, sometimes we can't actually write down a nice explicit function that makes a differential equation true. However, in the next few videos, what we will do is bring in some more tools to analyze differential equations and we will look at certain scenarios where given a differential equation you can do things like find the general solution
43:08
or find the solution to the initial value problem.
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