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Logistic differential equation intuition First order differential equations Khan Academy.mp3 | But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero. And what that does is as N approaches the natural limit, the ceiling to population, then no matter what this is doing, if this thing is approaching zero, that's goi... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | It's going to be harder to find things. And so what can I construct here dealing with N and K that will have these properties? And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an exp... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Well, let's see. What if we start with a one, we start with a one, and we subtract N over K? We subtract N over, my K is in pink, over K. Does this have those properties? Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fractio... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fraction, then this whole thing is going to be pretty close to one. It's going to be a little bit less than one. And when N approaches K, as N gets closer and closer and closer t... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted. And this thing right over here is actually, and this is used in tons of applications, not just in ... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | It's called the logistic differential equation. Logistic differential equation. Logistic differential equation. And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration. It's a little bit hairier tha... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration. It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way. But before we actually solve for it, let's jus... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like. And to do that, actually, let me, it's nice to see the faces. So let me draw some axes here. Let me draw some axes here. So that's my time axis. That is my populat... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Let me draw some axes here. So that's my time axis. That is my population axis. Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our ... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population. And that's good, because if your population is zero, how are you going to actually be able to add population? There's no one there to have children. So there's a... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero. And that's neat that this satisfies the logistic differential equation. Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever. Well, th... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Well, that's actually what would happen in real life. There's no one there to have kids. Now let's think about another situation. What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals ... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals zero, this is our population? Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of po... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero. So essentially, if my population is zero, then after a little bit of time, my population is still the same k. If my rate of change of population is zero, that means my population is s... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around. But for our purposes, you can never model anything perfectly, for our purposes, that's pretty good. You'... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important. If you start at zero, you're going to stay at zero. If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food,... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a ... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is. And if this is kind of dictating it, we're kind of looking more of a, well, let's just think of it this way. As the population grows, the rate of change is going to grow. So it's going to look, it's going to look som... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | So it's going to look, it's going to look something, it's going to look something like this. As our population gets larger, our slope is getting higher, and it's getting steeper and steeper. But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this... |
Logistic differential equation intuition First order differential equations Khan Academy.mp3 | It's going to make this whole thing approach zero. So as n approaches k, the whole thing, the rate of change, is going to flatten out, and we're going to asymptote towards k. And so the solution, the solution to the logistic differential equation should look something like this, depending on what your initial condition... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | At the point negative one comma one, I would draw a short segment of slope blank. And like always, pause this video and see if you can fill out these three blanks. Well, when you're, the short segments that you're trying to draw to construct this slope field, you figure out their slope based on the differential equatio... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two. Well, let's see, when x is zero and ... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do. Now, if we actuall... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so fo... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how ... |
Worked example forming a slope field AP Calculus AB Khan Academy.mp3 | So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that. And then at the point two comma three, at two comma... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | It's been over a year since I last did a video in the differential equations playlist, and I thought I would start taking up making a couple of videos, and I think where I left off, I said that I would do a non-homogeneous linear equation using the Laplace transform. So let's do one as a bit of a warmup now that we've ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So if we have the equation, the second derivative of y plus y is equal to sine of two t, and we're given some initial conditions here. The initial conditions are y of zero is equal to two, and y prime of zero is equal to one. And where we left off, and you probably remember this, you probably recently watched the last ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | To solve these, we take the Laplace transforms of all the sides. We solve for the Laplace transform of the function, and then we take the inverse Laplace transform. If that doesn't make sense, then let's just do it in this video, and hopefully the example will clarify all confusion. So in the last video, it was either ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So in the last video, it was either the last one or the previous one, I showed you that the Laplace transform, the Laplace transform of the second derivative of y is equal to s squared times the Laplace transform of y, and we keep lowering the degree on s, so minus s times y of zero. You can kind of think of it as taki... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | It's not exactly the antiderivative of this, but the Laplace transform, it is an integral. The transform is an integral. So y of zero is kind of a one derivative away from that, and then minus y prime of zero, and then we could also rewrite this, and this is just a purely notational issue. I could write this instead of... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | I could write this instead of writing the Laplace transform of y all the time. I could write this as s squared times capital Y of s, because this is going to be a function of s, not a function of y, minus s times y of zero minus y prime y of zero. These are going to be numbers, right? These aren't functions. These are ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | These aren't functions. These are the function evaluated at zero or the function evaluated, or the derivative of the function evaluated at zero, and we know what these values are. Y of zero right here is two, and y prime of zero is one. It was given to us. So if we take the Laplace transforms of both sides of this equa... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | It was given to us. So if we take the Laplace transforms of both sides of this equation, first we're going to want to take the Laplace transform of this term right there, which we've really just done. The Laplace transform of the second derivative is s squared times Laplace transform of the function, which we write as ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | They gave us that initial condition. Minus 2s and then minus one, right? This term right here is just one, so minus one. So that's the term right there. Then we want to take the Laplace transform of y by itself. So this is just plus y of s, right? The Laplace transform of y, so I'll just rewrite Laplace transform of y.... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So that's the term right there. Then we want to take the Laplace transform of y by itself. So this is just plus y of s, right? The Laplace transform of y, so I'll just rewrite Laplace transform of y. I'm just rewriting it in this notation. Y of s. It's good to get used to either one. And then we want to take, this is g... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | The Laplace transform of y, so I'll just rewrite Laplace transform of y. I'm just rewriting it in this notation. Y of s. It's good to get used to either one. And then we want to take, this is going to be equal to the Laplace transform of sine of 2t. And I showed you in a video last year of what we showed what the Lapla... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | And I showed you in a video last year of what we showed what the Laplace transform of sine of at is, but I'll write it down here just so you remember it. Laplace transform of the sine of at is equal to a over s squared plus a squared. Right? a over s squared plus a squared. So the Laplace transform of sine of 2t, here ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | a over s squared plus a squared. So the Laplace transform of sine of 2t, here a is two. This is going to be two over s squared plus four. So if we take the Laplace transform of both sides of this, the right-hand side is going to be two over s squared plus four. Now what we can do is we can separate out all the y of s t... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So if we take the Laplace transform of both sides of this, the right-hand side is going to be two over s squared plus four. Now what we can do is we can separate out all the y of s terms. And so we can factor, well, I guess we could say factor out their coefficients. So that's a y of s term. That's a y of s term. And s... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So that's a y of s term. That's a y of s term. And so we could write the left-hand side here as s squared, that's that term, plus one, the coefficient on that term, s squared plus one times y of s. Let me do it in green. So this is a y of s and this is a y of s times y of s. And then we have the non-y of s terms, these... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So this is a y of s and this is a y of s times y of s. And then we have the non-y of s terms, these two right here. So minus 2s minus one is equal to two over s squared plus four. We can add 2s plus one to both sides to essentially move this to the right-hand side. And we're left with, we are left with, s squared plus ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | And we're left with, we are left with, s squared plus one times y of s, times y of s, is equal to two over s squared plus four plus 2s plus one. Now we can divide both sides of this equation by s squared plus one and we get the Laplace transform of y, y of s, is equal to two, let me switch colors, it's equal to two ove... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Divided by s squared plus one, divided by s squared plus one. Now, in order to be able to take the inverse Laplace transform of this, I need to get it in some type of simple fraction form, these are actually easier to do, but this one's a little bit difficult. I want to do some partial fraction decomposition to break t... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | And since both of these, so what I want to do, I'm going to do a little bit of an aside here. And this really is the hardest part of these problems, is the algebra of breaking this thing up. So, since we're going to break this up, I'm going to break this up, so let me write this this way, two over s squared plus four, ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | I'm going to break this up into two fractions, this is the partial fraction decomposition, one fraction is s squared plus four, and then the other fraction is s squared plus one. And since both of these, both of these denominators are of degree two, the numerators are going to have, they're going to be of degree one, s... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | I would have to multiply these times, so my denominator, my common denominator would be this thing again, it would be s squared plus four, times s squared plus one, and now I'm going to have to multiply the a s plus b, a s plus b, times this s squared plus one, times s squared plus one, this has its right now, if these... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So a s times s squared is a s to the third, a s times one is plus a s, a s times s squared, so plus b s squared, and then you have b times one is plus b, and then you have c s times s squared, that's c s to the third, and then c s times four, so it's plus four c s, and then let's see, these problems are tiring, and I a... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So if I were to add the entire numerator, I get, and I'll just switch colors somewhat arbitrarily, I get a plus c times s to the third, plus, let me write the s squared term next, plus b plus d times s squared, and I'll write this s term, plus a plus four c times s, plus b plus four d. This is just the numerator, this ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Of course I have to show that this is a fraction, and this is going to be equal to this thing over here, two over s squared plus four times s squared plus one. Now, why did I go through this whole mess right here? Well, the reason why I went through it is because we should be able to solve for a, b, c, and d. So let's ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Do we see any s cubed terms here? No, we see no s cubed terms here, so a plus c, let me write this down, a plus c must be equal to zero, because we see nothing here that has an s to the third. b plus d is the coefficient on the s squared term. Do we see any s squared terms here? No, so b plus d must be equal to zero. a... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Do we see any s squared terms here? No, so b plus d must be equal to zero. a plus four c are the coefficient on the s term. I see no s term over here, so a plus four c must also be equal to zero. And then finally, we look at just the constant terms, and we do have a constant term on the left-hand side of this equation.... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | I see no s term over here, so a plus four c must also be equal to zero. And then finally, we look at just the constant terms, and we do have a constant term on the left-hand side of this equation. We have two, so b plus four d, didn't want to make it that thick, b plus four d must be equal to two. So let's see if we ca... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So let's see if we can do any, this seems like these linear equations are pretty easy to solve for. Let's subtract this from this. So a, or let me subtract the bottom one from the top one. So a minus a, that's zero a, and then c minus four c minus three c is equal to zero, and so you get c is equal to zero. If c is equ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So a minus a, that's zero a, and then c minus four c minus three c is equal to zero, and so you get c is equal to zero. If c is equal to zero, a plus c is equal to zero, a must be equal to zero. Let's do the same thing here. Let's subtract this from that. So you get b minus b is zero, and then minus three d, that's jus... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Let's subtract this from that. So you get b minus b is zero, and then minus three d, that's just d minus four d, and then zero minus two is equal to minus two, and then you get d is equal to two thirds. Minus two divided by minus three is two thirds, and then this isn't a minus here, I wrote that there later, we said b... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Let's remember all of this and go back to our original problem, because we've kind of, so actually let me just be clear. So we can rewrite two over s squared plus four times s squared plus one. We can rewrite this as, well a is zero, b is minus two thirds, so this is equal to minus two thirds over s squared plus four, ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So all of that work that I just did, that was just to break up this piece right here. That was just to break up that piece right there, and of course we have these other two pieces here that we forgot about. So after all of this work, what do we have? I'm going to make sure I don't make a careless mistake here. We get ... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | I'm going to make sure I don't make a careless mistake here. We get the Laplace transform of y. As you can see, the algebra is the hardest part here. Is equal to this first term, I'm just going back, this first term which I've now decomposed into this. So it's minus, let me write it this way, minus one third, and I thi... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Is equal to this first term, I'm just going back, this first term which I've now decomposed into this. So it's minus, let me write it this way, minus one third, and I think you're going to see in a second why I'm writing this way, minus one third times two over s squared plus four, and then plus two thirds times one ov... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | So I wanted to write it in this form. This was just the first term. We had two more terms to worry about. We don't want to make a careless mistake. I have two s over s squared plus one, so let me write that down. So plus two times s over s squared plus one, plus last one, plus one over s squared plus one, plus one over... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | We don't want to make a careless mistake. I have two s over s squared plus one, so let me write that down. So plus two times s over s squared plus one, plus last one, plus one over s squared plus one, plus one over s squared plus one. Now we just take the inverse Laplace transform of the whole thing, and then we'll kno... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Now we just take the inverse Laplace transform of the whole thing, and then we'll know what y is. Let me just write, just to remember, the Laplace transform, so this is going to be a little inverse, this is going to be sine of two t. Let me just write, just so we have it here, so you know I'm not doing some type of voo... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | Let's just remember those two things when we take the inverse Laplace transform of both sides of this equation. The inverse Laplace transform of the Laplace transform of y, well, that's just y. This y, maybe we'll write it as a function of t, is equal to, well, this is the Laplace transform of sine of two t. You can ju... |
Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3 | If a is equal to two, then this would be the Laplace transform of sine of two t, so it's minus 1 3rd times sine of two t plus 2 3rds times, this is the Laplace transform of sine of t. If you just make a is equal to one, sine of t's Laplace transform is one over s squared plus one, so plus 2 3rds times the sine of t. Le... |
Old separable differential equations introduction Khan Academy.mp3 | So now let's try to solve some. And this first class of differential equations I'll introduce you to, they're called separable equations. And I think what you'll find is that we're not learning really anything new. Using just your first year calculus, derivative and integrating skills, you can solve a separable equatio... |
Old separable differential equations introduction Khan Academy.mp3 | Using just your first year calculus, derivative and integrating skills, you can solve a separable equation. And the reason why they're called separable is because you can actually separate the x and y terms and integrate them separately to get the solution of the differential equation. So let's separable equations. So ... |
Old separable differential equations introduction Khan Academy.mp3 | So let's do a couple, and I think you'll get the point. These often are really more of exercises in algebra than anything else. So the first separable differential equation is dy over dx is equal to x squared over 1 minus y squared. And actually this is a good time to just review our terminology. So first of all, what ... |
Old separable differential equations introduction Khan Academy.mp3 | And actually this is a good time to just review our terminology. So first of all, what is the order of this differential equation? Well, the highest derivative in it is just the first derivative, so the order is equal to 1. So it's first order. It's ordinary because we only have a regular derivative, no partial derivat... |
Old separable differential equations introduction Khan Academy.mp3 | So it's first order. It's ordinary because we only have a regular derivative, no partial derivatives here. And then is this linear or nonlinear? Well, at first you say, oh, well, you know, this looks linear. I'm not multiplying the derivative times anything else. But if you look carefully, something interesting is goin... |
Old separable differential equations introduction Khan Academy.mp3 | Well, at first you say, oh, well, you know, this looks linear. I'm not multiplying the derivative times anything else. But if you look carefully, something interesting is going on. First of all, you have a y squared, and y is a dependent variable. y is a function of x. So to have the y squared, that makes it nonlinear.... |
Old separable differential equations introduction Khan Academy.mp3 | First of all, you have a y squared, and y is a dependent variable. y is a function of x. So to have the y squared, that makes it nonlinear. And even if this was a y, if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation, you would ... |
Old separable differential equations introduction Khan Academy.mp3 | And even if this was a y, if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation, you would have 1 minus y squared. Actually, this is actually the first step of what we have to do anyway, so I'll write it down. So if I'm just multip... |
Old separable differential equations introduction Khan Academy.mp3 | And then you immediately see that you're actually, even if this wasn't a squared here, you'd be multiplying the y times dy dx, and that also makes it nonlinear because you're multiplying the dependent variable times the derivative of itself. So that also makes this a nonlinear equation. But anyway, let's get back to so... |
Old separable differential equations introduction Khan Academy.mp3 | So this was the first step. Let's multiply both sides by 1 minus y squared. And the real end goal is just to separate the y's and the x's and then integrate both sides. So I'm almost there. So now what I want to do is I want to multiply both sides of this equation times dx. So I have a dx here and get rid of this dx th... |
Old separable differential equations introduction Khan Academy.mp3 | So I'm almost there. So now what I want to do is I want to multiply both sides of this equation times dx. So I have a dx here and get rid of this dx there. I'm going to go here. I don't want to waste too much space. So you get 1 minus y squared dy is equal to x squared dx. I have separated the x and y variables and the... |
Old separable differential equations introduction Khan Academy.mp3 | I'm going to go here. I don't want to waste too much space. So you get 1 minus y squared dy is equal to x squared dx. I have separated the x and y variables and the differentials. All I did is I multiplied both sides of this equation times dx to get here. Now I can just integrate both sides. So let's do that. |
Old separable differential equations introduction Khan Academy.mp3 | I have separated the x and y variables and the differentials. All I did is I multiplied both sides of this equation times dx to get here. Now I can just integrate both sides. So let's do that. So whatever you do to one side of the equation, you have to do to the other. That's true with regular equations or differential... |
Old separable differential equations introduction Khan Academy.mp3 | So let's do that. So whatever you do to one side of the equation, you have to do to the other. That's true with regular equations or differential equations. So we're going to integrate both sides. So what's the integral of this expression with respect to y? The integral of 1 is y. The integral of y squared, well that's... |
Old separable differential equations introduction Khan Academy.mp3 | So we're going to integrate both sides. So what's the integral of this expression with respect to y? The integral of 1 is y. The integral of y squared, well that's minus y to the third over 3. And I'll write the plus c here just to kind of show you something. But you really don't have to write a plus c on both sides. I... |
Old separable differential equations introduction Khan Academy.mp3 | The integral of y squared, well that's minus y to the third over 3. And I'll write the plus c here just to kind of show you something. But you really don't have to write a plus c on both sides. I'll call that plus the constant due to y, the y integration. You'll never see this in a calculus class, but I just want to ma... |
Old separable differential equations introduction Khan Academy.mp3 | I'll call that plus the constant due to y, the y integration. You'll never see this in a calculus class, but I just want to make a point here. Is equal to, I just want to show you this, our plus c's never disappeared from when we were taking our traditional antiderivatives. And what's the derivative of this? Well that'... |
Old separable differential equations introduction Khan Academy.mp3 | And what's the derivative of this? Well that's x to the third over 3. And this is also going to have a plus c due to the x variable. Now the reason why I did this magenta one in magenta, I labeled it like that, is because you really just have to write a plus c on one side of the equation. And if that doesn't make a lot... |
Old separable differential equations introduction Khan Academy.mp3 | Now the reason why I did this magenta one in magenta, I labeled it like that, is because you really just have to write a plus c on one side of the equation. And if that doesn't make a lot of sense, let's subtract this c from both sides and we get y minus, let me scroll down a little bit, y, my y looks like a g, y minus... |
Old separable differential equations introduction Khan Academy.mp3 | So you could have just, you have to have a constant, but it doesn't have to be on both sides of this equation because they're arbitrary. Cx minus Cy, well that's still just another constant. And then if we wanted to simplify this equation more, we can multiply both sides of this by 3, just to make it look nicer. And yo... |
Old separable differential equations introduction Khan Academy.mp3 | And you get 3y minus y to the third is equal to x to the third plus, well I could write 3c here, but once again, c is an arbitrary constant, right? So 3 times an arbitrary constant, that's just another arbitrary constant. So I'll write the c there. And there you have it. We have solved this differential equation, altho... |
Old separable differential equations introduction Khan Academy.mp3 | And there you have it. We have solved this differential equation, although it's in implicit form right now, and it's fairly hard to get it out of implicit form. We could put the c on one side, so the solution could be 3y minus y to the third minus x to the third is equal to c. Some people might like that a little bit b... |
Old separable differential equations introduction Khan Academy.mp3 | And notice, the solution, just like when you take an antiderivative, the solution is a class of implicit functions in this case. And why is it a class? Because we have that constant there. Depending on what number you pick there, it will be another solution. But any constant there will satisfy the original differential... |
Old separable differential equations introduction Khan Academy.mp3 | Depending on what number you pick there, it will be another solution. But any constant there will satisfy the original differential equation. Which was up here. This was the original differential equation. And if you want to solve for that constant, someone has to give you an initial condition. Someone has to say, well... |
Old separable differential equations introduction Khan Academy.mp3 | This was the original differential equation. And if you want to solve for that constant, someone has to give you an initial condition. Someone has to say, well, when x is 2, y is 3. And then you could solve for c. Anyway, let's do another one that gives us an initial condition. So this one's a little bit... I don't wan... |
Old separable differential equations introduction Khan Academy.mp3 | And then you could solve for c. Anyway, let's do another one that gives us an initial condition. So this one's a little bit... I don't want to... I'll start over. Clear image, invert colors. So I have optimal space. So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 ... |
Old separable differential equations introduction Khan Academy.mp3 | I'll start over. Clear image, invert colors. So I have optimal space. So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 times y minus 1. This is a parentheses, not an absolute value. And they give us initial conditions. They say that y of 0 is equal to negative 1. |
Old separable differential equations introduction Khan Academy.mp3 | So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 times y minus 1. This is a parentheses, not an absolute value. And they give us initial conditions. They say that y of 0 is equal to negative 1. So once we solve this differential equation, and this is a separable di... |
Old separable differential equations introduction Khan Academy.mp3 | They say that y of 0 is equal to negative 1. So once we solve this differential equation, and this is a separable differential equation, then we can use this initial condition, when x is 0, y is 1, to figure out the constant. So let's first separate this equation. So let's multiply both sides by 2 times y minus 1. And ... |
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