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Logistic function application First order differential equations Khan Academy.mp3 | That was the whole goal, is to model population growth. Let's come up with some assumptions. Let's first think about, well, let's say that I have an island. Let's say this is my island, and I start settling it with 100 people. I'm essentially saying n naught, let me do that in the n naught color, so I'm saying n naught... |
Logistic function application First order differential equations Khan Academy.mp3 | Let's say this is my island, and I start settling it with 100 people. I'm essentially saying n naught, let me do that in the n naught color, so I'm saying n naught is equal to 100. Let's say this environment, given what current technology of farming and agriculture and the availability of water and whatever else, let's... |
Logistic function application First order differential equations Khan Academy.mp3 | So you get the idea. So we get K, capital K, is equal to 1,000. That's the limit to the population. And so now we have to think about what is r going to be. So we have to come up with some assumptions. So let's say in a generation, in a generation which is about 20 years, or I'll just assume, in 20 years, I think it's ... |
Logistic function application First order differential equations Khan Academy.mp3 | And so now we have to think about what is r going to be. So we have to come up with some assumptions. So let's say in a generation, in a generation which is about 20 years, or I'll just assume, in 20 years, I think it's reasonable that the population grows by, let's say the population grows by 50%. In 20 years, you hav... |
Logistic function application First order differential equations Khan Academy.mp3 | In 20 years, you have 50% growth, 50% increase, increase in the actual population. So what would you have to have your annual increase in order for after 20 years to grow by 50%? Well, to think about that, I'll get out my calculator. And one way to think about it, growing by 50%, that means that you are at 1.5, your or... |
Logistic function application First order differential equations Khan Academy.mp3 | And one way to think about it, growing by 50%, that means that you are at 1.5, your original population. And if I take that to the 1 20th power, so the 1 20th, well, I'll just do 1 divided by 20th, this essentially says, well, how much am I going to grow by, or what is going to, this is telling me that I'm gonna grow b... |
Logistic function application First order differential equations Khan Academy.mp3 | Let me write that. Growth each year. And we're going to assume that our T here is in years. So we're going to assume that our T is in years. So T is in years. So what would our logistic function look like, given all of these assumptions? We would have N of T, let me, N of T is equal to, is equal to N naught times K. Th... |
Logistic function application First order differential equations Khan Academy.mp3 | So we're going to assume that our T is in years. So T is in years. So what would our logistic function look like, given all of these assumptions? We would have N of T, let me, N of T is equal to, is equal to N naught times K. That's going to be 100 times 1,000. That's going to be 100, 100 times 1,000, my initial popula... |
Logistic function application First order differential equations Khan Academy.mp3 | We would have N of T, let me, N of T is equal to, is equal to N naught times K. That's going to be 100 times 1,000. That's going to be 100, 100 times 1,000, my initial population times my maximum population, divided by my initial population, my initial population, plus the difference between my final and initial, so it... |
Logistic function application First order differential equations Khan Academy.mp3 | So let me actually pause this video and then plot it. So there you go, I made a plot and I copied and pasted it here, and we see the behavior that we wanted, the behavior that we wanted to see. We see the population right over here, it's at year zero is starting at 100. Let me do this in a color that you're more likely... |
Logistic function application First order differential equations Khan Academy.mp3 | Let me do this in a color that you're more likely to see. Population starts at 100. And we can see that, let's see, after 20 years, our population looks like it's almost grown to 150. So it looks like, at least in the beginning, this term, this right over here is dominating. We are growing by this 0.0205, that would be... |
Logistic function application First order differential equations Khan Academy.mp3 | So it looks like, at least in the beginning, this term, this right over here is dominating. We are growing by this 0.0205, that would be 2.05% per year, which gets us close to 50% growth. And we see that's what's happening initially. So we go from 100 to 150 in the first 20 years, in the first generation. And then in t... |
Logistic function application First order differential equations Khan Academy.mp3 | So we go from 100 to 150 in the first 20 years, in the first generation. And then in the next generation, we should add another 75 if we weren't kind of being constrained by the environment. So 150 plus 75 would be 225. And it looks like we got, after 20 years, to about 200, so we're a little bit slower. We're a little... |
Logistic function application First order differential equations Khan Academy.mp3 | And it looks like we got, after 20 years, to about 200, so we're a little bit slower. We're a little bit slower than kind of the pure exponential growth. But we're, you know, the pure exponential growth would probably have us tracking something closer to here. But still growing pretty well. But then as our population g... |
Logistic function application First order differential equations Khan Academy.mp3 | But still growing pretty well. But then as our population gets larger and larger and larger, as we're getting closer and closer to the maximum population, our rate of growth, our rate of growth is approaching zero. So we constantly approach our maximum population, but we never quite get there. It's really an asymptote.... |
Logistic function application First order differential equations Khan Academy.mp3 | It's really an asymptote. We're just approaching it as time goes on and on and on. But you can kind of set your own threshold if you want to say, okay, when do we get to kind of 90% of maximum population? Well, that looks like that happens, 90% of maximum population happened after 210 years on this island. So on a huma... |
Logistic function application First order differential equations Khan Academy.mp3 | Well, that looks like that happens, 90% of maximum population happened after 210 years on this island. So on a human scale, that seems like a long time, many generations, but I guess on a cosmic scale, it's not that long, not even a cosmic scale, even just slightly longer than a human scale. And so it'll happen, well, ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | We have the second derivative of y plus 4 times the first derivative plus 4y is equal to 0. And we're asked to find the general solution to this differential equation. So the first thing we do, like we've done in the last several videos, we'll get the characteristic equation. That's r squared plus 4r plus 4 is equal to... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | That's r squared plus 4r plus 4 is equal to 0. This one's fairly easy to factor. We don't need the quadratic equation here. This is r plus 2 times r plus 2. And now something interesting happens, something that we haven't seen before. The two roots of our characteristic equation are actually the same number. r is equal... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | This is r plus 2 times r plus 2. And now something interesting happens, something that we haven't seen before. The two roots of our characteristic equation are actually the same number. r is equal to minus 2. So you could say we only have one solution, or one root, or a repeated root, however you want to say it. We onl... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | r is equal to minus 2. So you could say we only have one solution, or one root, or a repeated root, however you want to say it. We only have one r that satisfies the characteristic equation. So you might say, well, that's fine. Maybe my general solution is just y is equal to some constant times e to the minus 2x using ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So you might say, well, that's fine. Maybe my general solution is just y is equal to some constant times e to the minus 2x using my one solution. And my reply to you is, this is a solution. And if you don't believe me, you can test it out. But it's not the general solution. And why do I say that? Because this is a seco... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And if you don't believe me, you can test it out. But it's not the general solution. And why do I say that? Because this is a second order differential equation. And if someone wanted a particular solution, they would have to give you two initial conditions. The two initial conditions we've been using so far are what y... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Because this is a second order differential equation. And if someone wanted a particular solution, they would have to give you two initial conditions. The two initial conditions we've been using so far are what y of 0 equals and what y prime of 0 equals. They could give you what y of 5 equals, who knows. But in general... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | They could give you what y of 5 equals, who knows. But in general, when you have a second order differential equation, they have to give you two initial conditions. Now the problem with this solution, and why it's not the general solution, is if you use one of these initial conditions, you can solve for a c. You'll get... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | You'll solve for that c. But then there's nothing to do with the second initial condition. And in fact, except for only in one particular case, whatever c you get for the first initial condition, this equation won't be true for the second initial condition. And you could try it out. I mean, if we said y of 0 is equal t... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | I mean, if we said y of 0 is equal to a and y prime of 0 is equal to, I don't know, 5a, let's see if these work. If y of 0 is equal to a, so that tells us that a is equal to c times e to the minus 2 times 0. So e to the 0. Or c is equal to a. So if you just had this first initial condition, say fine, my particular solu... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Or c is equal to a. So if you just had this first initial condition, say fine, my particular solution is y is equal to a times e to the minus 2x. But let's see if this particular solution satisfies the second initial condition. So what's the derivative of this? y prime is equal to minus 2a e to the minus 2x. And it say... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So what's the derivative of this? y prime is equal to minus 2a e to the minus 2x. And it says that 5a, this initial condition says that 5a is equal to minus 2a times e to the minus 2 times 0. So e to the 0. Or another way of saying that, e to the 0 is just 1. It says that 5a is equal to minus 2a, which we know is not t... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So e to the 0. Or another way of saying that, e to the 0 is just 1. It says that 5a is equal to minus 2a, which we know is not true. So note, when we only have this general or pseudo-general solution, it can only satisfy generally one of the initial conditions, and if we're really lucky, both initial conditions. So tha... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So note, when we only have this general or pseudo-general solution, it can only satisfy generally one of the initial conditions, and if we're really lucky, both initial conditions. So that at least gives you an intuitive feel of why this isn't the general solution. So let me clean that up a little bit so that I have a ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So what do we do? So we can use a technique called reduction of order, and it really just says, well, let's just guess a second solution. In general, when we first thought about these linear constant coefficient differential equations, we said, well, e to rx might be a good guess. And why is that? Because all of the de... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And why is that? Because all of the derivatives of e are kind of multiples of the original function, and that's why we used it. So if we're looking for a second solution, it doesn't hurt to kind of make the same guess. In order to be a little bit more general, let's make our guess for our second solution. I'll call thi... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | In order to be a little bit more general, let's make our guess for our second solution. I'll call this g for guess. Let's say it's some function of x times our first solution, e to the minus 2x. I could say some function of x times c times e to the minus 2x, but this c is kind of encapsulated. It could be part of this ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | I could say some function of x times c times e to the minus 2x, but this c is kind of encapsulated. It could be part of this some random function of x. So let's be as general as possible. So let's assume that this is a solution, and then substitute it back into our original differential equation and see if we can actua... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So let's assume that this is a solution, and then substitute it back into our original differential equation and see if we can actually solve for v that makes it all work. So before we do that, let's get its first and second derivatives. So the first derivative of g is equal to, well, this is a product rule. And I'll d... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And I'll drop the v of x. We know that v is a function and not a constant. So product rule, derivative of the first, v prime times the second expression, e to the minus 2x, plus the first function or expression times the derivative of the second. So minus 2 times e to the minus 2x. So times minus 2 e to the minus 2x, o... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So minus 2 times e to the minus 2x. So times minus 2 e to the minus 2x, or just to write it a little bit neater, g prime is equal to v prime e to the minus 2x minus 2v e to the minus 2x. Now we have to get the second derivative. I'll do that in a different color just to fight the monotony of it. So the second derivativ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | I'll do that in a different color just to fight the monotony of it. So the second derivative, we're going to have to do the product rule twice. Derivative of this first expression is going to be v prime prime e to the minus 2x minus 2v prime e to the minus 2x, that was just the product rule again. And then the derivati... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And then the derivative of the second expression is going to be, let's see, the derivative of the first one is v prime, so it's going to be minus 2v prime e to the minus 2x times, or no, plus, which is the product rule, minus 2 times minus 2, plus 4v e to the minus 2x. I hope I haven't made a careless mistake. And we c... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So we get the second derivative of g, which is our guess solution, is equal to the second derivative of v prime e to the minus 2x minus 4v prime e to the minus 2x plus 4v e to the minus 2x. And now, before we substitute it into this, we can just make one observation that'll just make the algebra a little bit simpler. N... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | g prime is, we could factor out an e to the minus 2x. And g prime prime, we can factor out an e to the minus 2x. So let's factor them out, essentially. So when we write this, we can write, so the second derivative is g prime prime, which we can write as, and I'm going to try to do this, it's e to the minus 2x. Times th... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So when we write this, we can write, so the second derivative is g prime prime, which we can write as, and I'm going to try to do this, it's e to the minus 2x. Times the second derivative. So now we can get rid of the e to the minus 2x terms. So that's v prime prime minus 4v prime plus 4v. If I just distribute this out... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So that's v prime prime minus 4v prime plus 4v. If I just distribute this out, I get the second derivative, which is this. Plus 4 times the first derivative, and I'm also going to factor out the e to the minus 2x. So plus 4 times this. So it's going to be plus 4v prime minus 8v. Once again, I factored out the e to the ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So plus 4 times this. So it's going to be plus 4v prime minus 8v. Once again, I factored out the e to the minus 2x. Plus 4 times y. So plus 4 times, we factored out the e to the minus 2x, so plus 4 times v. I did that because if I didn't do that, I'd be writing e to the minus 2x, and I'd probably make a careless mistak... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Plus 4 times y. So plus 4 times, we factored out the e to the minus 2x, so plus 4 times v. I did that because if I didn't do that, I'd be writing e to the minus 2x, and I'd probably make a careless mistake, and I'd run out of space, et cetera. But anyway, I essentially, to get this, I just substituted the second deriva... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And we know that that has to equal 0. Let's see if we can simplify this a little bit more. And then hopefully solve for v. So let's see. Some things are popping out at me. So I have plus 4v, plus 4v, plus 4v. That's plus 8v minus 8v. Right? |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Some things are popping out at me. So I have plus 4v, plus 4v, plus 4v. That's plus 8v minus 8v. Right? So plus 4, minus 8, plus 4, those cancel out. It's plus 8, minus 8, those cancel out. And then I also have minus 4v prime, plus 4v prime, so those cancel out. |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Right? So plus 4, minus 8, plus 4, those cancel out. It's plus 8, minus 8, those cancel out. And then I also have minus 4v prime, plus 4v prime, so those cancel out. And lo and behold, we've done some serious simplification. It ends up being e to the minus 2x times v prime prime. We could call it v prime prime of x, no... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And then I also have minus 4v prime, plus 4v prime, so those cancel out. And lo and behold, we've done some serious simplification. It ends up being e to the minus 2x times v prime prime. We could call it v prime prime of x, now that we've saved so much space, is equal to 0. We know this can never equal to 0. So essent... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | We could call it v prime prime of x, now that we've saved so much space, is equal to 0. We know this can never equal to 0. So essentially, we have now established that this expression has to be equal to 0. And we get a separable second order differential equation. We get that the second derivative of v with respect to ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And we get a separable second order differential equation. We get that the second derivative of v with respect to x, or it's a function of x, is equal to 0. So now we just have to differentiate both sides of this equation twice. You differentiate once, you get what? v prime of x is equal to, let's call it c1. And if we... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | You differentiate once, you get what? v prime of x is equal to, let's call it c1. And if we were to take the antiderivative of both sides again, we get v of x is equal to c1x plus some other c2. Now, remember, what was our guess? Our guess was that our solution was going to be, or our general solution was going to be, ... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | Now, remember, what was our guess? Our guess was that our solution was going to be, or our general solution was going to be, some arbitrary function v times that first solution we found, e to the minus 2x. And when we actually took that guess and we substituted it in, we actually were able to solve for that v. And we g... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | So this is interesting. So what is g, or what is our guess function, equal? It's no longer a guess. We've kind of established that it works. g, which we can call our solution, is equal to v of x times e to the minus 2x. Well, that equals this, c1x plus c2 e to the minus 2x. That equals c1x e to the minus 2x plus c2 e t... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | We've kind of established that it works. g, which we can call our solution, is equal to v of x times e to the minus 2x. Well, that equals this, c1x plus c2 e to the minus 2x. That equals c1x e to the minus 2x plus c2 e to the minus 2x. This is an e. And now we have a truly general solution. We have two constants, so we... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | That equals c1x e to the minus 2x plus c2 e to the minus 2x. This is an e. And now we have a truly general solution. We have two constants, so we can satisfy two initial conditions. And if we're looking for a pattern, this is the pattern. When you have a repeated root of your characteristic equation, the general soluti... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And if we're looking for a pattern, this is the pattern. When you have a repeated root of your characteristic equation, the general solution is going to be, you're going to use that e to the, whatever root is, twice. But one time, you're going to have an x in front of it. And this works every time for second order, hom... |
Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3 | And this works every time for second order, homogenous, constant coefficient, linear equations. I will see you in the next video. |
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