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2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | This differential equations problem was literally just a problem in using the quadratic equation, and once you figure out the r's, you have your general solution. Now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do th... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3 halves times c1 e to the 3 halves x plus, derivative of the inside, 1 half times c2 e to the 1 half x. And now let's use our actual initial conditions. I don't want to lose them. Let me rewrite them down here so I can scroll ... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And now let's use our actual initial conditions. I don't want to lose them. Let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1 half. Those are our initial conditions. So let's use that information. So y of 0, what happens when you substitute x is equ... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1 half. Those are our initial conditions. So let's use that information. So y of 0, what happens when you substitute x is equal to 0 here? You get c1 times e to the 0, essentially, so that's just 1, plus c2, well that's just e to the 0 again because x i... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | So y of 0, what happens when you substitute x is equal to 0 here? You get c1 times e to the 0, essentially, so that's just 1, plus c2, well that's just e to the 0 again because x is 0, is equal to, so this is when x is equal to 0, what is y? y is equal to 2. y of 0 is equal to 2. And then let's use the second equation.... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And then let's use the second equation. So when we substitute x is equal to 0 in the derivative, so when x is 0, we get 3 halves c1, this goes to 1 again, plus 1 half c2, this is 1 again, e to the 1 half times 0 is e to the 0, which is 1, is equal to, so when x is 0 for the derivative, y is equal to 1 half, or the deri... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Let's multiply the top equation, I don't know, let's multiply it by 3 halves, and what do we get? We get, I'll do it in a different color, we get 3 halves c1 plus 3 halves c2 is equal to, what's 3 halves times 2? It's equal to 3. And now, let's subtract, well, I don't want to confuse you, so let's just subtract the bot... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And now, let's subtract, well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1 half minus 3 halves? 1 half minus 1 and 1 half. Well, that's just minus 1, right? So minus c2 is equal to, what's 1 half minus 3? That's minus 2 and 1 half, or minus 5 halves. And so... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | Well, that's just minus 1, right? So minus c2 is equal to, what's 1 half minus 3? That's minus 2 and 1 half, or minus 5 halves. And so we get c2 is equal to 5 halves, and we could substitute back to this top equation, c1 plus 5 halves is equal to 2, or c1 is equal to 2, which is the same thing as 4 halves minus 5 halve... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And so we get c2 is equal to 5 halves, and we could substitute back to this top equation, c1 plus 5 halves is equal to 2, or c1 is equal to 2, which is the same thing as 4 halves minus 5 halves, which is equal to minus 1 half. And now we can just substitute c1 and c2 back into our general solution, and we have found th... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And it might seem really fancy. We're solving a differential equation, our solution has e in it, and we're taking derivatives, and we're doing all sorts of things. But really, the meat of this problem was solving a quadratic, which was our characteristic equation. And watch the previous video just to see why this chara... |
2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3 | And watch the previous video just to see why this characteristic equation works. But it's very easy to come up with the characteristic equation, right? I think you obviously see that y prime turns into r squared, y prime turns into r, and then y just turns into 1, essentially. So you solve a quadratic, and then after d... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | This is a scenario where we take an object that is hotter or cooler than the ambient room temperature and we want to model how fast it cools or heats up. And the way that we'll think about it is the way that Newton thought about it. And it is, Newton's described as Newton's law of cooling. Newton's law of cooling. And ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | Newton's law of cooling. And in a lot of ways, it's common sense. It states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature. So let me write that in mathematical terms. So Newton's law of cooling tells us that the rate of ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So let me write that in mathematical terms. So Newton's law of cooling tells us that the rate of change of temperature, I'll use that with a capital T, with respect to time, lowercase t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | And once again, it's common sense. If something is much, much, much, much hotter than the ambient temperature, the rate of change should be pretty steep. It should be declining in temperature quickly. If something is much, much, much, much cooler, it should be increasing in temperature quickly. And if something is clos... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | If something is much, much, much, much cooler, it should be increasing in temperature quickly. And if something is close, well, maybe the rate, if these two things are pretty close, well, maybe this rate of change shouldn't be so big. Now, I know one thing that you're thinking. It's like, okay, if the temperature is ho... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | It's like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our temperature of our object is greater than our am... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature? And the way that that would happen is you would have to have a negative K. Now, if you don't like thinking in terms of a negative K, you could just put a negative right over here, and now you wo... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | We'll be getting cooler. If it was the other way around, if our temperature of our object is cooler than our ambient temperature, then this thing is going to be a negative. Well, then the negative of that is going to be a positive. We're assuming a positive K, and our temperature will be increasing. So hopefully, this ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | We're assuming a positive K, and our temperature will be increasing. So hopefully, this makes some intuitive sense, and our constant K could depend on the specific heat of the object, how much surface area is exposed to it, or whatever else. But now, given this, let's see if we can solve this differential equation for ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So we just have to algebraically manipulate this, so all my Ts and DTs are on one side, or I should say, so all my capital Ts and D capital Ts are on one side, and so this is going to be a little bit mor... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So one thing I could do is I could divide both sides by capital T minus ambient temperature, minus T sub a. Remember, this is just going to be a constant based on what our ambient temperature is. We're going to assume our ambient temperature doesn't change as a function of time, that there's just, it's just such a big ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So if we do that, if we divide both sides by this, we are going to have, so I'm going to divide both sides, let me do this in a new color. If I divide both sides by that, I get one over capital T minus T sub a, and let me multiply both sides times the time differential. So I'm going to have, so that DT, DT, so our temp... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So once again, to separate the variables, all I did is divide both sides by this, and multiply both sides by that. Now I can integrate both sides. We've seen this show before. So I can integrate both sides, and the integral of this is going to be the natural log of the absolute value of what we have in the denominator.... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So I can integrate both sides, and the integral of this is going to be the natural log of the absolute value of what we have in the denominator. You could do U substitution if you want. If we said U is equal to capital T minus T sub a, then DU is just going to be one DT, and so this is essentially, you could say the in... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So this is the natural log of the absolute value of capital T minus T sub a is equal to, and once again, I could put a constant here, but I'm going to end up with a constant on the right-hand side too, so I'm just going to merge them into the constant on the right-hand side. So that is going to be equal to, now here, t... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | Let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to E, something about E I always think of the color green, E to the negative KT plus C, plus C. This, of course, is the same thing as this is equal to E to the negative KT. We've done this multiple times before. ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | Negative KT times E to the C, times E, let me do that in that same, so times E to the C power, and we could just call this, well, we could just call this another arbitrary constant. Let's see, if we call this C1, then we could just call this whole thing C. So this, we could say, is CE, CE to the negative KT, negative K... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | Instead of just temperature on this left-hand side, we have temperature minus our ambient temperature. And so we can do a couple of things. If in a world, in a world, say, where we're dealing with a hot cup of tea, something that's hotter than the ambient temperature, so we could imagine a world where T is, let's say, ... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive, or the thing inside the absolute value is going to be positive. So we don't need the absolute value, or the absolute value of it's go... |
Newton's Law of Cooling First order differential equations Khan Academy.mp3 | And then we could just add TA to both sides, T sub A to both sides, and then we would have our temperature, and I could even write this, as a function of time, is going to be equal to this business, is going to be equal to CE, let me do that in the same color, CE to the negative KT, negative KT plus T sub A, plus T sub... |
Differential equation introduction First order differential equations Khan Academy.mp3 | And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now, let's just think about, or at least look at, what a differential equation actually is. So if I were to write, so here's an example of a differential... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could wr... |
Differential equation introduction First order differential equations Khan Academy.mp3 | We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first der... |
Differential equation introduction First order differential equations Khan Academy.mp3 | They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the so... |
Differential equation introduction First order differential equations Khan Academy.mp3 | It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation, I guess I could say, may... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set... |
Differential equation introduction First order differential equations Khan Academy.mp3 | We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the... |
Differential equation introduction First order differential equations Khan Academy.mp3 | Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution ac... |
Differential equation introduction First order differential equations Khan Academy.mp3 | And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here. We'll verify that these indee... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution o... |
Differential equation introduction First order differential equations Khan Academy.mp3 | One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this tog... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the sec... |
Differential equation introduction First order differential equations Khan Academy.mp3 | And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have ... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equ... |
Differential equation introduction First order differential equations Khan Academy.mp3 | Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. T... |
Differential equation introduction First order differential equations Khan Academy.mp3 | Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this different... |
Differential equation introduction First order differential equations Khan Academy.mp3 | So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a... |
Differential equation introduction First order differential equations Khan Academy.mp3 | And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x. And... |
Differential equation introduction First order differential equations Khan Academy.mp3 | The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to ... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | We're given a differential equation right over here, cosine of y plus two, this whole thing, times the derivative of y with respect to x, is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero. And we're asked, what is x when y is equal to pi? So the first t... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So ... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with r... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't for... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we ca... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, ... |
Worked example separable equation with an implicit solution Khan Academy.mp3 | So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one. And now what... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | Before we move on past the method of undetermined coefficients, I want to make an interesting and actually a useful point. Let's say that I had the following non-homogenous differential equation. Second derivative of y minus 3 times the first derivative minus 4y is equal to, now this is where it gets interesting, 3e to... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | So you might say, wow, this is a tremendously complicated problem. I have the three types of functions I've been exposed to. I would have so many undetermined coefficients, it would get really unwieldy. And this is where you need to make a simplifying realization. We know the three particular solutions to the following... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | And this is where you need to make a simplifying realization. We know the three particular solutions to the following differential equations. We know the solution to second derivative minus 3 times the first derivative minus 4y. Well, this is a homogenous, right? And we know that the solution to the homogenous equation... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | Well, this is a homogenous, right? And we know that the solution to the homogenous equation, we did this a bunch of times, is c1 e to the 4x plus c2 e to the minus x. We know the solution to, and I'll switch colors just for variety, y prime prime minus 3y prime minus 4y is equal to just this alone, 3e to the 2x. And we... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | And we saw that that solution, that particular solution there, y particular, was minus 1 half e to the 2x. And we did that using undetermined coefficients. We did that a couple of videos ago. And then, let me just write this out a couple of times. We know the solution to this one as well. This was another particular so... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | And then, let me just write this out a couple of times. We know the solution to this one as well. This was another particular solution we found. I think it was two videos ago. And we found that the particular solution in this case, and this was a fairly hairy problem, was minus 5 over 17x plus 3 over 17. Sorry. The par... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | I think it was two videos ago. And we found that the particular solution in this case, and this was a fairly hairy problem, was minus 5 over 17x plus 3 over 17. Sorry. The particular solution was minus 5 over 17 sine of x plus 3 over 17 cosine of x. And then finally, this last polynomial, we could call it. We know the ... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | The particular solution was minus 5 over 17 sine of x plus 3 over 17 cosine of x. And then finally, this last polynomial, we could call it. We know the solution when that was just the right-hand side. That was this equation. And there, we figured out, and this was in the last video, we figured out that the particular s... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | That was this equation. And there, we figured out, and this was in the last video, we figured out that the particular solution in this case was minus x squared plus 3 halves x minus 13 over 8. So we know the particular solution when 0 is on the right-hand side. We know it when just 3 e to the 2x is on the right-hand si... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | We know it when just 3 e to the 2x is on the right-hand side. We know it just when 2 sine of x is on the right-hand side. And we know it just when 4x squared is on the right-hand side. So if we want, first of all, the particular solution to this non-homogeneous equation, we could just take the sum of the three particul... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | So if we want, first of all, the particular solution to this non-homogeneous equation, we could just take the sum of the three particular solutions. And that makes sense, right? Because one of the particular solutions, like this one, when you put it on the left-hand side, it'll just equal this term. This particular sol... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | This particular solution, when you put it on the left-hand side, will equal this term. And finally, this particular solution, when you put it on the left-hand side, will equal the 4x squared. And then you could add this homogenous solution to that. You put it on this side, and you'll get 0. So it won't change the right... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | You put it on this side, and you'll get 0. So it won't change the right-hand side. And then you will have the most general solution, because you have these two constants that you can solve for depending on your initial condition. So the solution to this seemingly hairy differential equation is really just the sum of th... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | So the solution to this seemingly hairy differential equation is really just the sum of these four solutions. And let me clean up some space, because I want everything to be on the board at the same time. So the solution is going to be, well, I don't want that to be deleted. The solution is going to be, I'll do it in b... |
Undetermined coefficients 4 Second order differential equations Khan Academy.mp3 | The solution is going to be, I'll do it in baby blue, it's going to be the solution to the homogenous c1 e to the 4x plus c2 e to the minus x minus 1 half e to the 2x. And I'll continue this line. Minus 5 over 17 sine of x plus 3 over 17 cosine of x minus x squared plus 3 halves x minus 13 over 8. And it seems daunting... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | For one of the first times, a mathematician's actually named something similar to what it's actually doing. You're actually convoluting the functions. And in this video, I'm not going to dive into the intuition of the convolution. Because a convolution, well, there's a lot of different ways you can look at it. It has a... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | Because a convolution, well, there's a lot of different ways you can look at it. It has a lot of different applications. And if you become an engineer, really of any kind, you're going to see the convolution in kind of a discrete form and a continuous form and a bunch of different ways. But in this video, I just want t... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | But in this video, I just want to make you comfortable with the idea of a convolution, especially in the context of taking Laplace transforms. So the convolution theorem. Actually, before I even go to the convolution theorem, let me define what a convolution is. So let's say that I have some function f of t. So if I co... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So let's say that I have some function f of t. So if I convolute f with g. So this means that I'm going to take the convolution of f and g, and this is going to be a function of t. And so far, nothing I've written should make any sense to you, because I haven't defined what this means. This is like those SAT problems w... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So let me undo this silliness that I just wrote there. The definition of the convolution, we're going to do it over a, well, there's several definitions you'll see, but the definition we're going to use in this context, there's actually one other definition you'll see in the continuous case, is the integral from 0 to t... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | And to kind of give you that comfort, let's actually compute a convolution. Let's say that, and it's actually hard to find some functions that are very easy to analytically compute, and you're going to find that we're going to go into a lot of trig identities to actually compute this. But if I say that f of t, if I def... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So the convolution of f with g, and this is going to be a function of t, it equals this. I'm just going to show you how to apply this integral. So it equals the integral, I'll do it in purple, the integral from 0 to t of f of t minus tau. This is my f of t, so it's going to be sine of t minus tau times g of tau. Well, ... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | This is my f of t, so it's going to be sine of t minus tau times g of tau. Well, this is my g of t, so g of tau is cosine of tau. So that's the integral. And now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So ... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | And now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So let's evaluate this. But I want to evaluate this in this video, because I want to show you that this isn't just some abstract thing, that you can actually... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So let's evaluate this. But I want to evaluate this in this video, because I want to show you that this isn't just some abstract thing, that you can actually evaluate these functions. So the first thing I want to do, I mean, I don't know what the antiderivative of this is. It's tempting. You see a sine and a cosine, ma... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | It's tempting. You see a sine and a cosine, maybe they're the derivatives of each other, but this is a sine of t minus tau. So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is just equal to the sine of t times the cosine of tau minus the sine of tau times th... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So if we make this substitution, this you'll find it on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to, I'll just write that, f star g, I'll just write it with that, is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this th... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | And let's see, t and tau and tau and t, everything's working so far. So let's see, so then that's dt. I have to be very careful here. Now let's distribute this cosine of tau out. And what do we get? We get this is equal to, so f convoluted with g, I guess we'll call it f star g, it's equal to the integral from 0 to t o... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | Now let's distribute this cosine of tau out. And what do we get? We get this is equal to, so f convoluted with g, I guess we'll call it f star g, it's equal to the integral from 0 to t of sine of t times cosine of tau times cosine of tau. I'm just distributing this cosine of tau. So it's cosine squared of tau. And then... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | I'm just distributing this cosine of tau. So it's cosine squared of tau. And then minus, let's rewrite the cosine of t first. And I'm doing that because we're integrating with respect to tau. So I'm just going to write my cosine of t first. So cosine of t times sine of tau times the cosine of tau, d tau. And now since ... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | And I'm doing that because we're integrating with respect to tau. So I'm just going to write my cosine of t first. So cosine of t times sine of tau times the cosine of tau, d tau. And now since we're taking the integral of really two things subtracting from each other, let's just turn this into two separate integrals. ... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | And now since we're taking the integral of really two things subtracting from each other, let's just turn this into two separate integrals. So this is equal to the integral from 0 to t of sine of t times the cosine squared of tau, d tau, minus the integral from 0 to t of cosine of t times sine of tau, cosine of tau, d ... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | Well to simplify it more, remember we're integrating with respect to, let me be careful here, we're integrating with respect to tau. I wrote a t there. We're integrating with respect to tau. So all of these, this cosine of t right here, that's a constant, the sine of t is a constant. For all I know, t could be equal to... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So all of these, this cosine of t right here, that's a constant, the sine of t is a constant. For all I know, t could be equal to 5. It doesn't matter that one of the boundaries of our integration is also a t. That t would be a 5. In which case, these are all just constants. We're integrating only with respect to the t... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | In which case, these are all just constants. We're integrating only with respect to the tau. So if cosine of 5, that's a constant. We can take it out of the integral. So this is equal to sine of t times the integral from 0 to t of cosine squared of tau, d tau, and then minus cosine of t, that's just a constant, I'm bri... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | We can take it out of the integral. So this is equal to sine of t times the integral from 0 to t of cosine squared of tau, d tau, and then minus cosine of t, that's just a constant, I'm bringing it out, times the integral from 0 to t of sine of tau, cosine of tau, d tau. Now this antiderivative is pretty straightforwar... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | You could do u substitution. And let me do it here. Instead of doing it in our head, this is a complicated problem, so we don't want to skip steps. If we set u is equal to sine of tau, then du d tau is equal to the cosine of tau, just the derivative of sine. Or we could write that du is equal to cosine of tau d tau. An... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | If we set u is equal to sine of tau, then du d tau is equal to the cosine of tau, just the derivative of sine. Or we could write that du is equal to cosine of tau d tau. And then of course, we'll undo the substitution before we evaluate the endpoints here. But this one's a little bit more of a conundrum. I don't know h... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | But this one's a little bit more of a conundrum. I don't know how to take the antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going to break out some more trigonometric identities. And in the video I just recorded, it might not be the last video in the playlist, I showed tha... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So to do this, we're going to break out some more trigonometric identities. And in the video I just recorded, it might not be the last video in the playlist, I showed that the cosine squared of tau, I'm just using tau as an example, is equal to 1 half times 1 plus the cosine of 2 tau. And once again, this is just a tri... |
Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3 | So we can make this substitution here. And then let's see what our integrals become. So this first one over here, let me just write it here. We get sine of t times the integral from 0 to t of this thing here. I could just take the 1 half out. So let me just, to keep things simple, so I'll put the 1 half out here, that'... |
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