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Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So let's just rearrange this equation so we get the Laplace transform of f prime. I could write of t, but that gets monotonous. Plus f of 0 is equal to s times the Laplace transform of f, divide both sides by s. Let me put the Laplace transform of f. I'm also going to switch the sides so I get the Laplace transform of ... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | Times the Laplace transform of my derivative plus my function evaluated at 0. And let's see if we can use this and this to figure out some more useful Laplace transforms. Well, what is the Laplace transform of f of t is equal to t? Well, let's just use this property. This is going to be equal to 1 over s times the Lapl... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | Well, let's just use this property. This is going to be equal to 1 over s times the Laplace transform of the derivative. Well, what's the derivative of t? The derivative of t is 1. So it's the Laplace transform of 1 minus f of 0. When t equals 0, this becomes 0. Minus 0. |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | The derivative of t is 1. So it's the Laplace transform of 1 minus f of 0. When t equals 0, this becomes 0. Minus 0. So the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. Well, that's just 1 over s. So it's 1 over s squared minus 0. Interesting. |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | Minus 0. So the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. Well, that's just 1 over s. So it's 1 over s squared minus 0. Interesting. Laplace transform of 1 is 1 over s. Laplace transform of t is 1 over s squared. Let's figure out what the Laplace transform of t squared is. And I'll d... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | Interesting. Laplace transform of 1 is 1 over s. Laplace transform of t is 1 over s squared. Let's figure out what the Laplace transform of t squared is. And I'll do this one in green. Maybe we'll see a pattern emerge. The Laplace transform of t squared, well, it equals 1 over s times the Laplace transform of its deriv... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | And I'll do this one in green. Maybe we'll see a pattern emerge. The Laplace transform of t squared, well, it equals 1 over s times the Laplace transform of its derivative. So what's its derivative? Times the Laplace transform of 2t plus this evaluated 0. Well, that's just 0. So this is equal to, well, we could just ta... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So what's its derivative? Times the Laplace transform of 2t plus this evaluated 0. Well, that's just 0. So this is equal to, well, we could just take this constant out. This is equal to 2 over s times the Laplace transform of t. Well, what does that equal? That is equal to, we just solved it, 1 over s squared. So it's ... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So this is equal to, well, we could just take this constant out. This is equal to 2 over s times the Laplace transform of t. Well, what does that equal? That is equal to, we just solved it, 1 over s squared. So it's 2 over s times 1 over s squared. So it's equal to 2 over s to the third. Fascinating. Well, let me ask y... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So it's 2 over s times 1 over s squared. So it's equal to 2 over s to the third. Fascinating. Well, let me ask you a, well, let me just do t to the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform. |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | Well, let me ask you a, well, let me just do t to the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying. |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of 2 to the third is 1 over s times the Laplace transform of its derivative, which is 3t squared, which is, take the constant out, becaus... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of 2 to the third is 1 over s times the Laplace transform of its derivative, which is 3t squared, which is, take the constant out, because it's a linear operator, 3 over s times the Laplace transform of t squared. So ... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So it equals what? What's the Laplace transform of t squared? It's 2 over s to the third. So this equals 3 times 2 over what? s to the fourth. And you could put a t over n here and use an inductive argument to figure out a general formula. And that general formula is, and I think you see the pattern here, whatever my e... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So this equals 3 times 2 over what? s to the fourth. And you could put a t over n here and use an inductive argument to figure out a general formula. And that general formula is, and I think you see the pattern here, whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. An... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | And that general formula is, and I think you see the pattern here, whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. And then the numerator is the factorial of my exponent. So in general, and this is one more entry in our Laplace transform table, the Laplace transform ... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | That's a parentheses. I guess I didn't have to write those parentheses. It just confuses it. But anyway, this sometimes looks like a fairly, you know when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all that. But it's just saying with this pattern we ... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | But anyway, this sometimes looks like a fairly, you know when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all that. But it's just saying with this pattern we showed, whatever, you know, t to the third, increase it by 1. So s to the fourth, put it in t... |
Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3 | So s to the fourth, put it in the denominator and take 3 factorial on the numerator, which is 6. And that's all it is. So we have, in using the derivative property of a Laplace transform, we figured out the Laplace transform of cosine of at and the Laplace transform of any, well really any polynomial, right? Because it... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | So I encourage you to pause this video and see if you can figure this out on your own. All right, now let's work through it together. So some of you might have immediately said, hey, this is the form of a differential equation where the solution is going to be an exponential and you just got right to it. But I'm not go... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide bot... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable diffe... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I ... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus ... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be so... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one. And so to solve for... |
Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3 | And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it b... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | But anyway, the problem we have here, it's the derivative of y with respect to x is equal to x squared plus 3y squared, I'm writing it a little bit small so that I'll run a space, divided by 2xy. So with all of these homogenous equations, and obviously we don't know if it's homogenous yet, so we have to try to write it... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So if we just multiply, we do it in a different color. 1 over x squared, or x to the negative 2, over 1 over x squared, right? We're essentially just multiplying by 1. And then we get that is equal to what? 1 plus 3y squared over x squared, divided by 2. If you divide x divided by x squared, you just get a 1 over x, so... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And then we get that is equal to what? 1 plus 3y squared over x squared, divided by 2. If you divide x divided by x squared, you just get a 1 over x, so 2 times y over x. Or we could just rewrite the whole thing, and we get this is just equal to 1 plus 3y over x squared, divided by 2 times y over x. So yes, this is a h... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Or we could just rewrite the whole thing, and we get this is just equal to 1 plus 3y over x squared, divided by 2 times y over x. So yes, this is a homogenous equation, because we were able to write it as a function of y divided by x. So now we can do the substitution with v, and hopefully this is starting to become a ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v. This is just the produc... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And now we can substitute. The derivative of y with respect to x is just this. And then the right-hand side of the equation is this, but we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv, dvx, I'll write v prime for now, just so that I don't take up too much space. v prime is equ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So let's do that. And so we get v plus x. Instead of dv, dvx, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared. We're making the substitution v is equal to y over x. 3v squared, all of that over 2v. Now let's see what we can do. |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | v prime is equal to 1 plus 3v squared. We're making the substitution v is equal to y over x. 3v squared, all of that over 2v. Now let's see what we can do. This is where we just put our algebra hat on and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equati... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Now let's see what we can do. This is where we just put our algebra hat on and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. Let's see, we'll get 2v squared plus 2xvv prime. Write 2v times that. Yep, that's 2xvv prime. Is equal to 1 plus 3v ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Let's see, we'll get 2v squared plus 2xvv prime. Write 2v times that. Yep, that's 2xvv prime. Is equal to 1 plus 3v squared. Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xvv prime is equal to 1 plus, let's see, we're subtracting 2v squared from both sides, so we just left ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Is equal to 1 plus 3v squared. Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xvv prime is equal to 1 plus, let's see, we're subtracting 2v squared from both sides, so we just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left-hand side. So we get 2xvv prime divided by 1 plus v squared is equal to 1, and let's divide both sides by x. So we get the x's on the other side. So then we get 2v, and now I'll switch back to ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So we get the x's on the other side. So then we get 2v, and now I'll switch back to the other notation, instead of v prime I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to, I'm dividing both sides by x. Notice I didn't write the x on this side, so that is equal to... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Notice I didn't write the x on this side, so that is equal to 1 over x. And then if we just multiply both sides of this times dx, we've separated the two variables and we can integrate both sides, so let's do that. Let's go up here, I'll switch to a different color so that you know I'm working on a different column now... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So multiply both sides by dx, I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first you might think, oh boy, this is complicated, this is difficult, maybe some type... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And what's the integral of this? At first you might think, oh boy, this is complicated, this is difficult, maybe some type of trig function, but you'll see that it's kind of just the reverse chain rule. We have a function here, 1 plus v squared, an expression here, we have its derivative sitting right there. So the ant... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So the antiderivative of this, and you could actually make a substitution if you like, you could say u is equal to 1 plus v squared, then du is equal to 2v dv, and then you'd end up saying that the antiderivative is just the natural log of u, or in this case, the antiderivative of this is just the natural log of 1 plus... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And I hope I didn't confuse you. That's how I think about it. I say if I have an expression and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression, and I don't have to worry about what's inside of it. So if this was a 1 over an x or 1 over u, it's just the natural lo... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So if this was a 1 over an x or 1 over u, it's just the natural log of it. So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully it'll make a little bit more sense. But anyway, that's the left-hand side,... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And hopefully it'll make a little bit more sense. But anyway, that's the left-hand side, and then that equals, well, this one's easy. That's the natural log, say the absolute value of x, plus, we could say plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that a... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | I mean, this is still some arbitrary constant c. So we could rewrite this whole equation as the natural log of 1 plus v squared is equal to, when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of. The natural log of, we could say, the absolute value of cx. And... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So we could say that 1 plus v squared is equal to cx. And now we can unsubstitute. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's see. Let's multiply both sides of this equation times x squared. |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's see. Let's multiply both sides of this equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And w... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | Let's multiply both sides of this equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on the left hand side, we could ... |
First order homogeneous equations 2 First order differential equations Khan Academy.mp3 | And we're essentially done. If we want to put all of the variable terms on the left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function or curve or whatever you want to call it is the solution to our original homogenous first o... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And we already found some constant solutions. We can think through that a little bit, just as a little bit of a review from the last few videos. So this is the t-axis and this is the n-axis. We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this different... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this differential equation is consistent with that because if n is zero, this thing is going to be zero, and so our rate of change is going to be zero with respect to time, and so our population just won't... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | That's one solution to this differential equation. Not that interesting. A zero population will never grow or change. The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One min... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One minus one is zero, and so in there, the population would also not change. It would just stay at k, and so the rate of cha... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | It would just stay at k, and so the rate of change would just stay at zero. So that's another constant solution that we start at the maximum population and then this differential equation tells us a scenario that never changes. But we said, hey, look, well, there could be something interesting that happens if our initi... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that. As n increases, our rate of change increases, but then as n approaches k, this thing is going to a... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | This is what could be used to model populations that would be more consistent with a Malthusian mindset. So let's see if we can do that. And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that. I'm going to leave the r on the right-hand side. That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that. So this is going to be equal to one over n times one m... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way. So we get that, and now what I want to do is take the antiderivative of both s... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Do a search for partial fraction expansion on Khan Academy. So how do we do it? Well, let's just add these two right over here. So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n times one minus n over k. One way to think about it, I just multiply the numerator and the denominator of this one by one minus ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And I multiply the numerator and denominator of this both by n, bn, and n times that. And then I added the two now that I had the same denominator. So that's just adding fractions with unlike denominators. It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to? Or what can they be equal to? Well, I have a constant term here. I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a l... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a little bit. Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero. And that this, which is our a, is going to be equal to one. And that'... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And that this, which is our a, is going to be equal to one. And that's pretty nice. So we could say that a is equal to one. a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be? Well, b would be equal to one, one o... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be? Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | You can rewind the video and review that if you find it necessary. So let's do that. And so how does this help us? Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of o... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can se... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And we could do the same thing over here. Notice, I have, I have an expression. What would be the derivative of this expression right over here? It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative. I can turn this, I can clear this out of the way. Instead of having a positive there, I could have a, I guess I could say a double ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So a double negative, I haven't changed the value of it. And notice, now this is the derivative of this, which makes it very nice for you substitution, or you might be already used to doing this thing in your head. And so we know that the derivative with respect to n, let me write this in a new color. We know that the ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So copy and paste. It's going to be that. It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt. Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not? Let's just do that just to make it clear, because this is really,... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not? Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the a... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps. So if I distribute this, edit, paste, whoops. I wanted to copy a... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So copy and paste. So I have that. And then I have, and then I have, I can copy and paste. So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit. So I have that right over here. |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit. So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this. I am just going to get this. Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let's do that. Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. A... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | I wanna find a pretty general solution. Let's call it c one, is equal to the antiderivative of this with respect to t is r times t, maybe plus some other constant, plus some other constant, just like that. And now I'm going to assume that my n of t meets this assumption right over here. And so I'm going to assume, I'm ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero. That means that this thing, this n is always going to be positive. And if n is always between zero and k, that means that this thing is always going to be positive. And so that helps ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And so that helps me clear things up a little bit. Actually, let me do it in this color. That helps me get rid of that, helps me get rid of, helps me get rid of that. I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here. So edit, cut, an... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here. So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little b... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little bit. And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this ... |
Solving the logistic differential equation part 1 Khan Academy.mp3 | And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this some other constant. Let me just call it a general c. So let me just call this, let me just call this c. So clear, and I'm just going to call that c. I am just going to call that c. And I... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | And if you have your initial conditions, you can solve for c1 and c2. But the question I'm asking is, what happens when you have two complex roots? Or essentially, when you're trying to solve the characteristic equation, when you're trying to solve that quadratic, the b squared minus 4ac, that's negative. So you get th... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | So you get the two roots end up being complex conjugates. And we said, OK, let's say that our two roots are lambda plus or minus mu i. And we just did a bunch of algebra to, we said, well, if those are the roots and we substitute it back into this formula for the general solution, we get all of this. And we kept simpli... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | And we kept simplifying it all the way until we got here, where we said y is equal to e to the lambda x plus c1, et cetera, et cetera. And we said, can we simplify this further? And that's where we took out Euler's equation, or Euler's formula, or Euler's definition, depending on what you want, which I'm always in awe ... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | But we've talked a lot about that in the calculus playlist. We could use this to maybe further simplify it. So I wrote e to the mu xi as cosine mu x plus i sine mu x. And I wrote e to the minus mu xi as cosine minus mu x plus i sine minus mu x. And now we could use a little bit about what we know about trigonometry. Co... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | And I wrote e to the minus mu xi as cosine minus mu x plus i sine minus mu x. And now we could use a little bit about what we know about trigonometry. Cosine of minus theta is equal to cosine of theta. And we also know that sine of minus theta is equal to minus sine of theta. So let's use these identities to simplify t... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | And we also know that sine of minus theta is equal to minus sine of theta. So let's use these identities to simplify this a little bit more. So we get y is equal to e to the lambda x times, and we could actually distribute the c1 too, so times c1 cosine of mu x plus i times c1 sine of mu x plus, all of this is in this ... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | Plus c i times c2 sine of minus mu x, that's the same thing as minus sine of x. So actually, let's take the minus sign out there. So minus sine of mu x. And let's see, it seems like we're getting to a point that we can simplify it even more. We could add the two cosine terms. So we get the general solution, and I know ... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | And let's see, it seems like we're getting to a point that we can simplify it even more. We could add the two cosine terms. So we get the general solution, and I know this is a high, this problem requires a lot of algebraic stamina, but as long as you don't make careless mistakes, you'll find it reasonably rewarding, b... |
Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3 | So we get y, the general solution is y is equal to e to the lambda x times, let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x, and let's add the two sine of mu x terms. So plus i, we could call that c1i, minus c2i times sine of mu x, and we're almost done simplifying. And the last thing we... |
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