video_title stringlengths 68 104 | Sentence stringlengths 129 2.65k |
|---|---|
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out. Because if you think about it, no matter the solution for all homogenous equations, or when you kind of apply, when you kind of solve the equation, they always equal 0. So they're homogenized, I guess, is... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So we could call this a second order linear, because a, b, and c definitely are functions just of, well, they're not even functions of x or y, they're just constants. So second order linear, homogenous, because they equal 0, differential equations. And I think you'll see that these are, in some ways, the most fun diffe... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And actually, often the most useful, because a lot of the applications in classical mechanics, this is all you need to solve. But they're the most fun to solve because they all boil down to algebra 2 problems. And I'll touch on that in a second. But let's just think about this a little bit. Think about what the propert... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | But let's just think about this a little bit. Think about what the properties of these solutions might be. Let me just throw out something. Let's say that g of x is a solution. So that means that a times g prime prime plus b times g prime plus c times g is equal to 0. That means the same thing. Now my question to you i... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Let's say that g of x is a solution. So that means that a times g prime prime plus b times g prime plus c times g is equal to 0. That means the same thing. Now my question to you is, what if I have some constant times g? Is that still a solution? So let's say, so my question is, let's say some constant, c1 gx, c1 times... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Now my question to you is, what if I have some constant times g? Is that still a solution? So let's say, so my question is, let's say some constant, c1 gx, c1 times g. Is this a solution? Let's try it out. Let's substitute this into our original equation. So a times the second derivative of this would just be, and I'll... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Let's try it out. Let's substitute this into our original equation. So a times the second derivative of this would just be, and I'll switch colors here. I'll switch maybe, let me switch to brown. So a times the second derivative of this would be the constant, every time you take a derivative, the constant just carries ... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | I'll switch maybe, let me switch to brown. So a times the second derivative of this would be the constant, every time you take a derivative, the constant just carries over, right? So that'll just be a times c1 g prime prime plus the same thing for the first derivative, b times c1 g prime plus c times, and this c is dif... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So we could factor out that c1 constant and we get c1 times a g prime prime plus b g prime plus cg. And lo and behold, we already know, because we know that g of x is a solution, we know that this is true, so this is going to be equal to 0, right? Because g is a solution. So if this is 0, c1 times 0 is going to be equa... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogenous differential equation, then some constant times g is also a solution. So this is also a solution to the differential equation.... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So this is also a solution to the differential equation. And then the next property I want to show you, and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our o... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Well, let's substitute this whole thing into our original differential equation, right? So we'll have g of x plus h of x. a times the second derivative of this thing. Well, that's straightforward enough. That's just g prime prime plus h prime prime. Plus b times the first derivative of this thing, g prime plus h prime.... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | That's just g prime prime plus h prime prime. Plus b times the first derivative of this thing, g prime plus h prime. Plus c times this function, g plus h. And now what can we do? Let's distribute all of these constants. We get a times g prime prime plus a times h prime prime plus b times the first derivative of g plus ... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Let's distribute all of these constants. We get a times g prime prime plus a times h prime prime plus b times the first derivative of g plus b times the first derivative of h plus c times g plus c times h. And now we can rearrange them. We get a, let's take this one, let's take all the g terms, a times the second deriv... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Plus a times the second derivative of h plus b times first derivative plus c times h. And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, this was the left-hand side of that differential equation, this is goin... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation, of this second-order linear homogenous differential equation, and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in gen... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x plus some constant times h of x is also a solution. And maybe the constant in one of the cases is 0 or somethi... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second-order homogenous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see ... |
Creating a slope field First order differential equations Khan Academy.mp3 | Let's say we don't know how to find the solutions to this, but we at least want to get a sense of what the solutions might look like. And to do that, what we could do is we could look at a coordinate plane. So, let me draw some axes here. So, let me draw a relatively straight line. Alright, so that's my y-axis, and thi... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and ... |
Creating a slope field First order differential equations Khan Academy.mp3 | And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could v... |
Creating a slope field First order differential equations Khan Academy.mp3 | Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by do... |
Creating a slope field First order differential equations Khan Academy.mp3 | And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes thr... |
Creating a slope field First order differential equations Khan Academy.mp3 | Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x i... |
Creating a slope field First order differential equations Khan Academy.mp3 | And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent lin... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had so... |
Creating a slope field First order differential equations Khan Academy.mp3 | Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we ... |
Creating a slope field First order differential equations Khan Academy.mp3 | Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one ne... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. |
Creating a slope field First order differential equations Khan Academy.mp3 | Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well. |
Creating a slope field First order differential equations Khan Academy.mp3 | If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, ... |
Creating a slope field First order differential equations Khan Academy.mp3 | That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table. |
Creating a slope field First order differential equations Khan Academy.mp3 | And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. |
Creating a slope field First order differential equations Khan Academy.mp3 | I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. |
Creating a slope field First order differential equations Khan Academy.mp3 | It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, the... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And... |
Creating a slope field First order differential equations Khan Academy.mp3 | And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something li... |
Creating a slope field First order differential equations Khan Academy.mp3 | So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do... |
Finding particular linear solution to differential equation Khan Academy.mp3 | And so here we have a differential equation, and we haven't started exploring how we find the solutions for differential equations yet, but let's just say you saw this and someone just walked up to you on the street and says, hey, I will give you a clue, that there's a solution to this differential equation that is ess... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So if we know that this kind of a solution can be described in this way, we have to figure out some m's and b's here, this is telling us that if we were to take the derivative of this with respect to x, if we take the derivative of mx plus b with respect to x, that that should be equal to negative two times x plus thre... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So let's work through it. Let's figure out first what our dy dx is. So dy dx, we'll just take the derivative here with respect to x, dy dx is derivative of mx with respect to x is just going to be m, and of course derivative of b with respect to x is just a constant, so it's just going to be zero. So dy dx is m. So we ... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So dy dx is m. So we could write m is equal to negative two x, is equal to negative two x plus three times, instead of putting y there, I could write mx plus b. Remember, y is equal to mx plus b. And just as a repeated reminder, this has to be true for all x's. mx plus b, and then of course we have the minus five. And ... |
Finding particular linear solution to differential equation Khan Academy.mp3 | mx plus b, and then of course we have the minus five. And so if you weren't able to solve it the first time, I encourage you to start from here and now figure out what m and b needs to be in order for this equation right over here, in order for this to be true for all x's. In order for this to be true for all x's. So I... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So I'm assuming you have paused again and had a go at it, so let's just keep algebraically manipulating this. And I'll just switch to one color here. So we have m, m is equal to negative two x plus, if we distribute this three, we're going to have three mx plus three b, and then of course we're going to have minus five... |
Finding particular linear solution to differential equation Khan Academy.mp3 | And now we can group the x terms. So if we were to group, if we were to group, let me find a new color here, maybe this blue. So if we were to take these two and add them together, that's going to be negative two plus three m times x, or we could write this as three m minus two times x, and then you have your constant ... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So you have these terms right over here. So plus three b minus five, and of course that's all going to be equal to m. That's going to be equal to m. Now remember, this needs to be true, this needs to be true for all x's. So notice, over here I have some, I have some coefficient times x on the right-hand side, but on th... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So somehow this thing must disappear. This is a constant, so it's completely reasonable, it's completely reasonable that this constant could be equal to m, but the only way that I get these x's to disappear, so all I'm left with is an m, is if this thing is equal to zero. Let me say that again, because I think it might... |
Finding particular linear solution to differential equation Khan Academy.mp3 | We're saying that m, some constant value, is equal to some coefficient times x plus some other constant value. Well, in order for a constant value to be equal to a coefficient times x plus some other constant value, the coefficient on x must be equal to zero. Another way to think about it is, this should be, you could ... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So zero must be equal to three m minus two, and m is equal to three b minus five. m is equal to three b minus five. So let's use that knowledge, that information, to solve for m and b. So we could use this first one. So three m minus two must be equal to zero. So let's write that. Three m minus two is equal to zero, or... |
Finding particular linear solution to differential equation Khan Academy.mp3 | So we could use this first one. So three m minus two must be equal to zero. So let's write that. Three m minus two is equal to zero, or three m is equal to two, or m is equal to 2 3rds. So we figured out what m is. And then we could use that information because we know that m is equal to three b minus five. m is 2 3rds... |
Finding particular linear solution to differential equation Khan Academy.mp3 | Three m minus two is equal to zero, or three m is equal to two, or m is equal to 2 3rds. So we figured out what m is. And then we could use that information because we know that m is equal to three b minus five. m is 2 3rds, so we get 2 3rds is equal to three b minus five. We could add five to both sides, which is the ... |
Finding particular linear solution to differential equation Khan Academy.mp3 | m is 2 3rds, so we get 2 3rds is equal to three b minus five. We could add five to both sides, which is the same thing as adding 15 3rds to both sides. Did I do that right? Yeah, adding five to both sides is the same thing as adding 15 3rds to both sides. So let's do that, 15 over three plus 15 over three. These cancel... |
Finding particular linear solution to differential equation Khan Academy.mp3 | Yeah, adding five to both sides is the same thing as adding 15 3rds to both sides. So let's do that, 15 over three plus 15 over three. These cancel out, that's just five right over there on the left-hand side. We have 17 over three is equal to three b. Or if you divide both sides by three, you get b is equal to 17 over... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Now, I'll introduce you to the concept of exact equations. And it's just another method for solving a certain type of differential equations. Let me write that down. Exact equations. Before I show you what an exact equation is, I'm just going to give you a little bit of the building blocks, just so that when I later pr... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Exact equations. Before I show you what an exact equation is, I'm just going to give you a little bit of the building blocks, just so that when I later prove it, or at least give you the intuition behind it, it doesn't seem like it's coming out of the blue. So let's say I had some function of x and y, and we'll call it... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So psi is a function of x and y. So you're probably not familiar with taking the chain rule onto partial derivatives, but I'll show it to you now. And I'll give you a little intuition, although I won't prove it. So if I were to take the derivative of this with respect to x, where y is also a function of x. So maybe I c... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So if I were to take the derivative of this with respect to x, where y is also a function of x. So maybe I could also write this as y, sorry, not y, psi. So I could also write this as psi as x and y, which is a function of x. I could write it just like that. These are just two different ways of writing the same thing. ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | These are just two different ways of writing the same thing. Now, if I were to take the derivative of psi with respect to x, and these are just the building blocks. If I were to take the derivative of psi with respect to x, it is equal to, and this is the chain rule using partial derivatives, and I won't prove it, but ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So this is going to be equal to the partial derivative of psi with respect to x plus the partial derivative of psi with respect to y times dy dx. And this should make a little bit of intuition. I'm kind of taking the derivative with respect to x, and then if you could say, and I know you can't, because this partial wit... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | And if you were to kind of add them up, then you would get the full derivative with respect to x. That's not even in the intuition. That's just to kind of show you that even this should make a little bit of intuitive sense. Now the intuition here, let's just say, and I'm not, let's just say psi, and psi doesn't always ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Now the intuition here, let's just say, and I'm not, let's just say psi, and psi doesn't always have to take this form, but you could use the same methodology to take psi to kind of more complex notations. But let's say that psi, and I won't write this, it's a function of x and y. We know it's a function of x and y. Le... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Let's say it's equal to some function of x, we'll call that f1 of x, times some function of y. And let's say there's a bunch of terms like this. So there's n terms like this. Plus all the way, and the n-th term is the n-th function of x times the n-th function of y. I just define psi like this, just so I can give you t... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Plus all the way, and the n-th term is the n-th function of x times the n-th function of y. I just define psi like this, just so I can give you the intuition that when I use implicit differentiation on this, when I take the derivative of this with respect to x, I actually get something that looks just like that. So wha... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | That's equal, we just do the product rule, right? So the first expression, you take the derivative of that with respect to x, well that's just going to be f1 prime of x times the second function, well that's just g1 of y. Now you add that to the derivative of the second function times the first function. So plus f1 of ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So plus f1 of x, that's just the first function, times the derivative of the second function. Now the derivative of the second function, it's going to be this function with respect to y, so you could write that as g1 prime of y. But of course we're doing the chain rule, so it's that times dy dx, and you might want to r... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | But this right here, what I just did, this expression right here, this is the derivative with respect to x of this, right? And we have n terms like that, so if we keep adding them, I'll do them vertically down. So plus, and then you'd have a bunch of them, and then the last one's going to look the same, it's just the n... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So fn prime of x times the second function, gn of y, plus the first function, fn of x, times the derivative of the second function. The derivative of the second function with respect to y is just g prime of y times dy dx. That's just the chain rule. dy dx. Now we have two n terms. We had n terms here, right? Where each... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | dy dx. Now we have two n terms. We had n terms here, right? Where each term was f of x times g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y. Now we have, for each of those, we got two of them when we did the product rule. If we group the terms, so if we group all the terms that don't ha... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Where each term was f of x times g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y. Now we have, for each of those, we got two of them when we did the product rule. If we group the terms, so if we group all the terms that don't have a dy dx on them, what do we get? If we add all of these, ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | If we add all of these, I guess you could call them on the left-hand side, you get, I'm just rearranging, it all equals f1 prime of x times g1 of y plus f2, g2, all the way to fn prime of x, gn of y. That's just all of these added up. Plus all of these added up. All of the terms that have the dy dx in them. So those ar... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | All of the terms that have the dy dx in them. So those are, and I'll do them in a different color. So all of these terms, which are going to be the different color, I'll do it in different parentheses, plus f1 of x g1 prime of y, and I'll do the dy dx later. I'll distribute it out. Plus, and we have n terms, plus fn of... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | I'll distribute it out. Plus, and we have n terms, plus fn of x gn prime of y. And then all of these terms are multiplied by dy dx. Now something looks interesting here. We originally defined our xi up here as this right here. But what is this green term? Well, what we did is we took all of these individual terms and t... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Now something looks interesting here. We originally defined our xi up here as this right here. But what is this green term? Well, what we did is we took all of these individual terms and these green terms here are just taking the derivative with respect to just x on each of these terms. Because if you take the derivati... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | Well, what we did is we took all of these individual terms and these green terms here are just taking the derivative with respect to just x on each of these terms. Because if you take the derivative just with respect to x of this, then the function of y is just a constant, right? If you were to take just the partial de... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So if you take the partial derivative with respect to x of this term, you treat a function of y as a constant. So the derivative of this would just be f prime of x g1 of y, right? Because g1 of y is just a constant. And so forth and so on. All of these green terms you can view as the partial derivative of xi with respe... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | And so forth and so on. All of these green terms you can view as the partial derivative of xi with respect to x. We just pretended like y is a constant. And that same logic, if you just look at this part right here, what is this? We took xi up here. We treated the functions of x as a constant. And we just took the part... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | And that same logic, if you just look at this part right here, what is this? We took xi up here. We treated the functions of x as a constant. And we just took the partial derivative with respect to y. And that's why the primes are on all the g's. And then we multiply that times dy dx. So you could write this. |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | And we just took the partial derivative with respect to y. And that's why the primes are on all the g's. And then we multiply that times dy dx. So you could write this. This is equal to, I'll do this green. This green is the same thing as the partial of xi with respect to x plus, what's this purple? This part of the pu... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | So you could write this. This is equal to, I'll do this green. This green is the same thing as the partial of xi with respect to x plus, what's this purple? This part of the purple. Let me do it in a different color, in a magenta. This right here is the partial of xi with respect to y. And then times dy dx. |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | This part of the purple. Let me do it in a different color, in a magenta. This right here is the partial of xi with respect to y. And then times dy dx. So that's essentially all I wanted to show you right now in this video, because I realize I'm almost running out of time, that the chain rule, when you're taking, with ... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | And then times dy dx. So that's essentially all I wanted to show you right now in this video, because I realize I'm almost running out of time, that the chain rule, when you're taking, with respect to one of the variables, but the second variable of the function is also a function of x, the chain rule is this. If xi is... |
Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3 | If y wasn't a function of x, or if y in no way was independent of x, then dy dx would be 0. And this term would be 0. And then the derivative of xi with respect to x would be just the partial of xi with respect to x. But anyway, I want you to just keep this in mind. And in this video, I didn't prove it, but I hopefully... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | We're now ready to solve non-homogenous, second-order, linear differential equations with constant coefficients. So what does all that mean? Well, it means an equation that looks like this. a times the second derivative plus b times the first derivative plus c times the function is equal to g of x. Before I show you an... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | a times the second derivative plus b times the first derivative plus c times the function is equal to g of x. Before I show you an actual example, I want to show you something interesting. That the general solution of this non-homogenous equation is actually the general solution of the homogenous equation plus a partic... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | I'll explain what that means in a second. So let's say that h is a solution of the homogenous equation. h is homogenous. And that worked out well, because h for homogenous. h is solution for homogenous. There should be some shorthand notation for homogenous. So what does that mean? |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | And that worked out well, because h for homogenous. h is solution for homogenous. There should be some shorthand notation for homogenous. So what does that mean? That means that a times the second derivative of h plus b times h prime plus c times h is equal to 0. That's what I mean when I say that h is a solution. And ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | So what does that mean? That means that a times the second derivative of h plus b times h prime plus c times h is equal to 0. That's what I mean when I say that h is a solution. And actually, let's just say that h is the general solution for this homogenous equation. And we know how to solve that. Take the characterist... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | And actually, let's just say that h is the general solution for this homogenous equation. And we know how to solve that. Take the characteristic equation, depending on how many roots it has and whether they're real or complex. You can figure out a general equation, a general solution. And then if you have initial condi... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | You can figure out a general equation, a general solution. And then if you have initial conditions, you can substitute them and get the values of the constants. Fair enough. Now let's say that I were to say that g is a solution. Let's say it's a. Well, I already used g up here. Let's say, what am I? |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | Now let's say that I were to say that g is a solution. Let's say it's a. Well, I already used g up here. Let's say, what am I? Well, I don't like using vowels. Let's say j. Let's say j is a particular solution to this differential equation. |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | Let's say, what am I? Well, I don't like using vowels. Let's say j. Let's say j is a particular solution to this differential equation. So what does that mean? That means that a times j prime prime plus b times j prime plus c times j is equal to g of x. So that means we're just defining j of x to be a particular soluti... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | Let's say j is a particular solution to this differential equation. So what does that mean? That means that a times j prime prime plus b times j prime plus c times j is equal to g of x. So that means we're just defining j of x to be a particular solution. Now what I want to show you is that j of x plus h of x is also g... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | So that means we're just defining j of x to be a particular solution. Now what I want to show you is that j of x plus h of x is also going to be a solution to this original equation. And that it's the general solution for this non- homogenous equation. And before I just do it mathematically, what's the intuition? Well,... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | And before I just do it mathematically, what's the intuition? Well, when you substitute h here, you get 0. When you substitute j here, you get g of x. So when you add them together, you're going to get 0 plus g of x here. So you're going to get g of x here. And I'll show you that right now. So let's say I wanted to sub... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | So when you add them together, you're going to get 0 plus g of x here. So you're going to get g of x here. And I'll show you that right now. So let's say I wanted to substitute h plus j here. So what is, and I'll do it in a different color, a. So the second derivative of the sum of those two functions is going to be th... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | So let's say I wanted to substitute h plus j here. So what is, and I'll do it in a different color, a. So the second derivative of the sum of those two functions is going to be the second derivative of both of them summed up. Plus b times the first derivative of the sum. Plus c times the sum of the functions. And my go... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | Plus b times the first derivative of the sum. Plus c times the sum of the functions. And my goal is to show that this is equal to g of x. So what does this simplify to? Well, if we take all the h terms, we get a h prime prime plus b h prime plus c h plus, let's do all the j terms, a j prime prime plus b j prime plus c ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | So what does this simplify to? Well, if we take all the h terms, we get a h prime prime plus b h prime plus c h plus, let's do all the j terms, a j prime prime plus b j prime plus c j. Well, by definition of how we defined h and j, what is this equal to? Well, we said that h is a solution for the homogeneous equation o... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.