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Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3 | So this is an a. That's an a. And so this is an a. And so this is an a. And so this is an a. This was an a. And so we're left with, and this is the correct answer, a over s squared plus a squared. |
Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3 | And so this is an a. And so this is an a. This was an a. And so we're left with, and this is the correct answer, a over s squared plus a squared. So I hope those careless mistakes didn't throw you off too much. These things happen when you do integration by parts twice with a bunch of variables. But anyway, now we are ... |
Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3 | And so we're left with, and this is the correct answer, a over s squared plus a squared. So I hope those careless mistakes didn't throw you off too much. These things happen when you do integration by parts twice with a bunch of variables. But anyway, now we are ready to add a significant entry into our table of Laplac... |
Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3 | But anyway, now we are ready to add a significant entry into our table of Laplace transforms. And that is that the Laplace transform of sine of at is equal to a over s squared plus a squared. And that's a significant entry. And maybe a good exercise for you, just to see how fun it is to do these integration by parts pr... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | I've spoken a lot about second order linear homogenous differential equations in abstract terms and how if g is a solution, then some constant times g is also a solution. Or if g and h are solutions, then g plus h is also a solution. Let's actually do problems because I think that will actually help you learn as oppose... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | So let's say I have this differential equation. The second derivative of y with respect to x plus 5 times the first derivative of y with respect to x plus 6 times y is equal to 0. So we need to find a y where 1 times its second derivative plus 5 times its first derivative plus 6 times itself is equal to 0. And now let'... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And now let's just do a little bit of, take a step back and think about what kind of function. Most functions, if I have the function and I take its derivative and then I take its second derivative, most times I get something completely different. Like if I have y was x squared, then y prime would be 2x and y prime pri... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And then to add them together, you'd say, well, how would my x terms cancel out so that you get 0 in the end? So draw back into your brain and think, is there some function that when I take its first and second derivatives and third and fourth derivatives, it essentially becomes the same function. Maybe the constant in... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And if you've listened to a lot of my videos, you'd realize that it probably is what I consider to be the most amazing function in mathematics. And that is the function e to the x. And in particular, maybe e to the x won't work here. You could even try it out. If you did e to the x, it won't satisfy this equation. e to... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | You could even try it out. If you did e to the x, it won't satisfy this equation. e to the x, you'd get e to the x plus 5e to the x plus 6e to the x, that would not equal to 0. But maybe y is equal to e to some constant r times x. Let's just make the assumption that y is equal to some constant r times x, substitute it ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | But maybe y is equal to e to some constant r times x. Let's just make the assumption that y is equal to some constant r times x, substitute it back into this, and then see if we can actually solve for an r that makes this equation true. And if we can, we've found the solution. Or maybe we found several solutions. So le... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | Or maybe we found several solutions. So let's try it out. Let's try y is equal to e to the rx into this differential equation. So what is the first derivative of it, first of all? It's always useful to. So y prime is equal to what? Derivative chain rule. |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | So what is the first derivative of it, first of all? It's always useful to. So y prime is equal to what? Derivative chain rule. Derivative of the inside is r, and then derivative of the outside is still just e to the rx. And what's the second derivative? y prime prime is equal to derivative. |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | Derivative chain rule. Derivative of the inside is r, and then derivative of the outside is still just e to the rx. And what's the second derivative? y prime prime is equal to derivative. r is just a constant, so derivative of the inside is r times r on the outside, that's r squared times e to the rx. And now we're rea... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | y prime prime is equal to derivative. r is just a constant, so derivative of the inside is r times r on the outside, that's r squared times e to the rx. And now we're ready to substitute back in. And I will switch colors. So the second derivative, that's r squared times e to the rx plus 5 times the first derivative, so... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And I will switch colors. So the second derivative, that's r squared times e to the rx plus 5 times the first derivative, so that's 5r e to the rx plus 6 times our function, 6 times e to the rx is equal to 0. And something might already be surfacing to you as something we can do to this equation to solve for r. All of ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | So let's factor that out. So this is equal to e to the rx times r squared plus 5r plus 6 is equal to 0. And our goal, remember, was to solve for the r or the r's that will make this true. And in order for this side of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to som... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And in order for this side of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to some exponent and get 0? Well, we'll know. So this cannot equal 0. So in order for this left-hand side of the equation to be 0, this term, this expression right here has to be 0. And I'll do ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | Well, we'll know. So this cannot equal 0. So in order for this left-hand side of the equation to be 0, this term, this expression right here has to be 0. And I'll do that in a different color. So we know if we want to solve for r that this, r squared plus 5r plus 6, that has to be 0. And this is called the characterist... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And I'll do that in a different color. So we know if we want to solve for r that this, r squared plus 5r plus 6, that has to be 0. And this is called the characteristic equation. This or this, the r squared plus 5r plus 6 is called the characteristic equation. And it should be obvious to you that now this is no longer ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | This or this, the r squared plus 5r plus 6 is called the characteristic equation. And it should be obvious to you that now this is no longer calculus. This is just factoring a quadratic. And this one actually is fairly straightforward to factor. So what is this? This is r plus 2 times r plus 3 is equal to 0. And so the... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And this one actually is fairly straightforward to factor. So what is this? This is r plus 2 times r plus 3 is equal to 0. And so the solutions of the characteristic equation, or actually the solutions to this original equation, are r is equal to negative 2 and r is equal to minus 3. So you say, hey, we found two solut... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And so the solutions of the characteristic equation, or actually the solutions to this original equation, are r is equal to negative 2 and r is equal to minus 3. So you say, hey, we found two solutions, because we found two suitable r's that make this equation true, this differential equation true. And what are those? ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | Well, the first one is y is equal to e to the minus 2x. r is minus 2. And then we could call that y1. And then the second solution we found, y2, is e to the, what is this, r is minus 3x. Now my question to you is, is this the most general solution? Well, in the last video, in kind of our introductory video, we learned ... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And then the second solution we found, y2, is e to the, what is this, r is minus 3x. Now my question to you is, is this the most general solution? Well, in the last video, in kind of our introductory video, we learned that a constant times a solution is still a solution. So if y1 is a solution, we also know that we can... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | So if y1 is a solution, we also know that we can multiply y1 times any constant. Let's do that. Let's multiply it by c1. That's a c1 there. This is also going to be a solution. Now it's a little bit more general, right? It's a whole class of functions. |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | That's a c1 there. This is also going to be a solution. Now it's a little bit more general, right? It's a whole class of functions. The c doesn't have to just be a 1. It can be any constant. And then when you use your initial values, you actually can figure out what that constant is. |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | It's a whole class of functions. The c doesn't have to just be a 1. It can be any constant. And then when you use your initial values, you actually can figure out what that constant is. And same for y2. y2 doesn't have to be 1 times e to the minus 3x. It has to be some, any constant. |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And then when you use your initial values, you actually can figure out what that constant is. And same for y2. y2 doesn't have to be 1 times e to the minus 3x. It has to be some, any constant. And we learned that in the last video, that if something's a solution, some constant times that is also a solution. And we also... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | It has to be some, any constant. And we learned that in the last video, that if something's a solution, some constant times that is also a solution. And we also learned that if we have two different solutions, that if you add them together, you also get a solution. So the most general solution to this differential equa... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | So the most general solution to this differential equation is y of x. Just to hit it home that this is definitely a function of x. y of x is equal to c1 e to the minus 2x plus c2 e to the minus 3x. And this is the general solution of this differential equation. And I won't prove it, because the proof is fairly involved... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And I won't prove it, because the proof is fairly involved. I mean, we just tried out e to the rx. Maybe there's some other wacko function that would have worked here. But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution. There isn't some crazy outside functi... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution. There isn't some crazy outside function there that would have also worked. And so the other question that might be popping in your brain is, Sal, in previous, when we did first order differential equations... |
2nd order linear homogeneous differential equations 2 Khan Academy.mp3 | And that was OK, because we had one set of initial conditions, and we solved for our constants. But here I have two constants. So if I wanted a particular solution, how can I solve for two variables if I'm only given one initial condition? And if that's what you actually thought, your intuition would be correct. You ac... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And now I'll show you, at least in the context of differential equations. And I've gotten, actually, a bunch of letters on the Laplace transform, what does it really mean, and all of that, and those are excellent questions and you should strive for that. It's hard to really have an intuition of the Laplace transform in... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | But I'll give you a hint, and if you want a path to learn it in, you should learn about Fourier series and Fourier transforms, which are very similar to Laplace transforms, and that'll actually build up the intuition on what the frequency domain is all about. Anyway, let's actually use a Laplace transform to solve a di... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | OK, so let's say the differential equation is y prime prime plus 5 times the first derivative plus 6y is equal to 0. And you know how to solve this one, but I just want to show you, with a fairly straightforward differential equation, that you could solve it with a Laplace transform, and actually you end up kind of hav... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Now, to use a Laplace transform here, we essentially just take the Laplace transform of both sides of this equation. So we get, let me use a more vibrant color, so we get the Laplace transform of y, the second derivative, plus, well we could say Laplace transform of 5 times y prime, but that's the same thing as 5 times... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | What's the Laplace transform of 0? Let me do that in a, so the Laplace transform of 0 would be the integral from 0 to infinity of 0 times e to the minus st dt. So this is a 0 in here, so this is equal to 0. So Laplace transform of 0 is 0, and that's good, because I didn't have space to draw another curly L. So what are... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So Laplace transform of 0 is 0, and that's good, because I didn't have space to draw another curly L. So what are the Laplace transforms of these things? Well, this is where we break out one of the useful properties that we learned, and that useful property, let me write it over here, because I think that's going to be... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So we learned that the Laplace transform, I'll do it here, actually I'll do it down here. The Laplace transform of f prime, or we could even say y prime, is equal to s times the Laplace transform of y minus y of 0. We proved that to you, and this is extremely important to know. So let's see if we can apply that. So the... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So let's see if we can apply that. So the Laplace transform of y prime prime, if we apply that, that's equal to s times the Laplace transform of, well, to go from y prime to y, you're just taking the antiderivative. So if we're taking the antiderivative of the second derivative, we just end up with the first derivative... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Minus the first derivative at 0. Notice, we're already using our initial conditions. I won't substitute it just yet. And then we end up with plus 5 times, I'll write it every time, just so we can, so plus 5 times the Laplace transform of y prime plus 6 times the Laplace transform of y. All of that is equal to 0. So jus... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And then we end up with plus 5 times, I'll write it every time, just so we can, so plus 5 times the Laplace transform of y prime plus 6 times the Laplace transform of y. All of that is equal to 0. So just to be clear, all I did is I expanded this into this. Using this. So how can we rewrite the Laplace transform of y p... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Using this. So how can we rewrite the Laplace transform of y prime? Well, we could use this once again. So let's do that. So this over here, I'll do it in magenta, this is equal to s times what? s times the Laplace transform of y prime. Well, that's s times the Laplace transform of y minus y of 0, right? |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So let's do that. So this over here, I'll do it in magenta, this is equal to s times what? s times the Laplace transform of y prime. Well, that's s times the Laplace transform of y minus y of 0, right? I took this part and replaced it with what I have in parentheses. So minus y prime of 0. And now I'll switch colors. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Well, that's s times the Laplace transform of y minus y of 0, right? I took this part and replaced it with what I have in parentheses. So minus y prime of 0. And now I'll switch colors. Plus 5 times, once again, the Laplace transform of y prime. Well, we could use this again. So 5 times s times the Laplace transform of... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And now I'll switch colors. Plus 5 times, once again, the Laplace transform of y prime. Well, we could use this again. So 5 times s times the Laplace transform of y minus y of 0 plus 6 times the Laplace transform of y. All of that is equal to 0. I know this looks really confusing, but we'll simplify right now. We could... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So 5 times s times the Laplace transform of y minus y of 0 plus 6 times the Laplace transform of y. All of that is equal to 0. I know this looks really confusing, but we'll simplify right now. We could get rid of this right here, because we've used it as much as we need to. So now we just simplify. And notice, using th... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | We could get rid of this right here, because we've used it as much as we need to. So now we just simplify. And notice, using the Laplace transform, we didn't have to guess at a general solution or anything like that. Even when we did characteristic equation, we guessed what the original general solution was. Now we're ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Even when we did characteristic equation, we guessed what the original general solution was. Now we're just taking Laplace transforms. And let's see where this gets us. So simplifying, actually I just want to make it clear, because I know it's very confusing. So I rewrote this part as this, and I rewrote this thing as ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So simplifying, actually I just want to make it clear, because I know it's very confusing. So I rewrote this part as this, and I rewrote this thing as this. And everything else is the same. But now let's simplify the math. So we get s squared times the Laplace transform of y. I'm going to write smaller. I've learned my... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | But now let's simplify the math. So we get s squared times the Laplace transform of y. I'm going to write smaller. I've learned my lesson. Minus s times y of 0. Let's substitute y of 0 here. y of 0 is 2. So s times y of 0 is 2 times s. So 2s. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Minus s times y of 0. Let's substitute y of 0 here. y of 0 is 2. So s times y of 0 is 2 times s. So 2s. Minus y prime of 0. y prime of 0 is 3. So minus 3. So we have 5 times s times the Laplace transform of y. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So s times y of 0 is 2 times s. So 2s. Minus y prime of 0. y prime of 0 is 3. So minus 3. So we have 5 times s times the Laplace transform of y. So plus 5s times the Laplace transform of y. Minus 5 times y of 0. y of 0 is 2. So minus 10. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So we have 5 times s times the Laplace transform of y. So plus 5s times the Laplace transform of y. Minus 5 times y of 0. y of 0 is 2. So minus 10. 5 times, this is 2 right here. So 5 times 2. Plus 6 times the Laplace transform of y. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So minus 10. 5 times, this is 2 right here. So 5 times 2. Plus 6 times the Laplace transform of y. All of that is equal to 0. Now let's group our Laplace transform of y terms and our constant terms. And we should be hopefully getting someplace. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Plus 6 times the Laplace transform of y. All of that is equal to 0. Now let's group our Laplace transform of y terms and our constant terms. And we should be hopefully getting someplace. So let's see. My Laplace transform of y terms. I have this one. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And we should be hopefully getting someplace. So let's see. My Laplace transform of y terms. I have this one. I have this one. And I have that one. So what am I left with? |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | I have this one. I have this one. And I have that one. So what am I left with? Well let me factor out the Laplace transform of y part. So I get the Laplace transform of y. And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So what am I left with? Well let me factor out the Laplace transform of y part. So I get the Laplace transform of y. And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6. So those are all my Laplace transform terms. And then I have my constant terms. So minus 2s minus 3 m... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6. So those are all my Laplace transform terms. And then I have my constant terms. So minus 2s minus 3 minus 10 is equal to 0. And what can we do here? Well this is interesting, first of all. Notice that the coefficients ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So minus 2s minus 3 minus 10 is equal to 0. And what can we do here? Well this is interesting, first of all. Notice that the coefficients on the Laplace transform of y terms, that those are that characteristic equation that we dealt with so much. And that is hopefully, to some degree, second nature to you. So that's a ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Notice that the coefficients on the Laplace transform of y terms, that those are that characteristic equation that we dealt with so much. And that is hopefully, to some degree, second nature to you. So that's a little bit of a clue. And just if you want some very tenuous connections, well that makes a lot of sense. Bec... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And just if you want some very tenuous connections, well that makes a lot of sense. Because the characteristic equation, to get that we substituted e to the rt. And the Laplace transform involves a very similar function. But anyway, let's go back to the problem. So how do we solve this? So actually, let me just give yo... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | But anyway, let's go back to the problem. So how do we solve this? So actually, let me just give you the big picture here, because this is a good point. What I'm going to do is I'm going to solve this. I'm going to say the Laplace transform of y is equal to something. And then I'm going to say, boy, what functions Lapl... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | What I'm going to do is I'm going to solve this. I'm going to say the Laplace transform of y is equal to something. And then I'm going to say, boy, what functions Laplace transform is at something? And then I'll have the solution. If that confuses you, just wait and hopefully it'll make some sense. From here until that... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | And then I'll have the solution. If that confuses you, just wait and hopefully it'll make some sense. From here until that point, it's just some fairly hairy algebra. So let's scroll down a little bit, just so we have some breathing room. And so I get the Laplace transform of y times s squared plus 5s plus 6 is equal t... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | So let's scroll down a little bit, just so we have some breathing room. And so I get the Laplace transform of y times s squared plus 5s plus 6 is equal to, let's add these terms to both sides of this equation, is equal to 2s plus 3 plus 10. Oh, that's silly, plus 13. This is minus 13 here. A phone call. Who's calling? ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | This is minus 13 here. A phone call. Who's calling? I think it's some kind of marketing phone call. Anyway, 2s plus 13. And now what can I do? Well, let's divide both sides by this s squared plus 5s plus 6. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | I think it's some kind of marketing phone call. Anyway, 2s plus 13. And now what can I do? Well, let's divide both sides by this s squared plus 5s plus 6. So I get the Laplace transform of y is equal to 2s plus 13 over s squared plus 5s plus 6. Now we're almost done. Everything here is just a little bit of algebra. |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Well, let's divide both sides by this s squared plus 5s plus 6. So I get the Laplace transform of y is equal to 2s plus 13 over s squared plus 5s plus 6. Now we're almost done. Everything here is just a little bit of algebra. So now we're almost done. We haven't solved for y yet, but we know that the Laplace transform ... |
Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3 | Everything here is just a little bit of algebra. So now we're almost done. We haven't solved for y yet, but we know that the Laplace transform of y is equal to this. Now if we just had this in our table of our Laplace transforms, we would immediately know what y was. But I don't see something, or I don't remember anyth... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | Derivative of y with respect to x is equal to y cosine of x divided by 1 plus 2y squared. And they give us an initial condition that y of 0 is equal to 1, or when x is equal to 0, y is equal to 1. And I know we did a couple already, but another way to think about separable differential equations is really all you're do... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | Or another way to think about it is, whenever you took an implicit derivative, the end product was a separable differential equation. And so that just hopefully forms a little bit of a connection. But anyway, let's just do this. We have to separate the y's from the x's. Let's multiply both sides times 1 plus 2y squared... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | We have to separate the y's from the x's. Let's multiply both sides times 1 plus 2y squared. We get 1 plus 2y squared times dy dx is equal to y cosine of x. We still haven't fully separated the y's and the x's. Let's divide both sides of this by y. And then let's see. We get 1 over y plus 2y squared divided by y. |
Old separable differential equations example First order differential equations Khan Academy.mp3 | We still haven't fully separated the y's and the x's. Let's divide both sides of this by y. And then let's see. We get 1 over y plus 2y squared divided by y. That's just 2y times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can int... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | We get 1 over y plus 2y squared divided by y. That's just 2y times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides. Let's integrate both sides. So what's the integral of 1 over y with respect to y? I know your... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | And now we can integrate both sides. Let's integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that whose derivative is actually 1 over y. And that's the natural log of the a... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that whose derivative is actually 1 over y. And that's the natural log of the absolute value of y. And this is just a slightly broader function because its domain includes positive and negative numb... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | It just excludes 0, while natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice. And it's actually true that at all points other than 0, its derivative is 1 over y. So it's just a slightly broader function. So that's the antiderivative of 1 over y. We proved that, or at lea... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | So it's just a slightly broader function. So that's the antiderivative of 1 over y. We proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared. So 2y is equal to, I'll do the plus c on this side, whose derivati... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared. So 2y is equal to, I'll do the plus c on this side, whose derivative is cosine of x? What's sine of x? And then we could add the plus c. We can add that plus c there. And what was our initial condition? y of 0 is equal to 1. |
Old separable differential equations example First order differential equations Khan Academy.mp3 | What's sine of x? And then we could add the plus c. We can add that plus c there. And what was our initial condition? y of 0 is equal to 1. So when x is equal to 0, y is equal to 1. So when x is equal to 0, y is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | y of 0 is equal to 1. So when x is equal to 0, y is equal to 1. So when x is equal to 0, y is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of 1, e to the what power is 1? Well, 0. Plus 1. Sine of 0 is 0. |
Old separable differential equations example First order differential equations Khan Academy.mp3 | So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of 1, e to the what power is 1? Well, 0. Plus 1. Sine of 0 is 0. It's equal to c. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it. We figured out c is equal ... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | Sine of 0 is 0. It's equal to c. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it. We figured out c is equal to 1. So we could just scratch this out. We could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | We figured out c is equal to 1. So we could just scratch this out. We could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you could actually get rid of t... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you could actually get rid of that absolute value function there. But anyway, that's just a little technicality. But... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | But anyway, that's just a little technicality. But this is the implicit form of the solution to this differential equation. That makes sense, because these separable differential equations are really just implicit derivatives backwards. And in general, one thing that's kind of fun about differential equations, but kind... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | And in general, one thing that's kind of fun about differential equations, but kind of not as satisfying about differential equations, is it really is just a whole hodgepodge of tools to solve different types of equations. There isn't just one tool or one theory that will solve all differential equations. There are a f... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | But there's not just one consistent way to solve all of them. And even today, there are unsolved differential equations, where the only way that we know how to get solutions is using a computer and numerically. And one day I'll do videos on that. And actually, you'll find that in most applications, that's what you end ... |
Old separable differential equations example First order differential equations Khan Academy.mp3 | And actually, you'll find that in most applications, that's what you end up doing anyway, because most differential equations you encounter in science or in any kind of science, whether it's economics or physics or engineering, they often are unsolvable, because they're going to have a second or third derivatives invol... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | We'll now move from the world of first order differential equations to the world of second order differential equations. What does that mean? That means that we're now going to start involving the second derivative. And the first class that I'm going to show you, and this is probably the most useful class when you're s... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And the first class that I'm going to show you, and this is probably the most useful class when you're studying classical physics, are linear second order differential equations. So what is a linear second order differential equation? So I think I touched on it a little bit in our very first intro video, but it's somet... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | If I have a of x, so some function only of x, times the second derivative of y with respect to x, plus b of x times the first derivative of y with respect to x, plus c of x times y is equal to some function that's only a function of x. So just to review our terminology, y is the second order because the highest derivat... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | Well, all of the coefficients on, and I want to be careful with the term coefficients because traditionally we view coefficients as always being constants, but here we have functions of x as coefficients. So in order for this to be a linear differential equation, a of x, b of x, c of x, and d of x, they all have to be ... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | So what will that look like? So I could just rewrite that as a, so now a is not a function anymore, it's just a number. a times the second derivative of y with respect to x plus b times the first derivative plus c times y. And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal to 0. And by setting this equal to 0, I have now introduced you to the other form of homogenous differential equation, and this one is called homogenous. And I haven't made the connection yet on how these second order diff... |
2nd order linear homogeneous differential equations 1 Khan Academy.mp3 | I think they just happen to have the same name, even though they're not that related. So the reason why this one is called homogenous is because you have it equal to 0. So this is what makes it homogenous. And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out... |
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