Hairy differential equation involving a step function that we use the Laplace Transform to solve. Created by Sal Khan.
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- Am I wrong in thinking at the very end, in speaking of simplifying, that he could actually simplify the sin(t-2pi) and sin(2(t-2pi)) into sin(t) and sin(2t) because sin(t) and sin(t-2pi) will return the same value?(28 votes)
- You're right. Then one could collect terms for sin(t) and sin(2t), and pull out the common 1 - u(t-2pi). Furthermore, one may notice that the last factor is simply 1 for t less than 2 pi and zero afterwards, and thus we could write the result as:
sin(t) / 3 - sin(2t) / 6 for t less than 2 pi and 0 otherwise
This may even give you some insight into the equation -- t = 2 pi is the moment that the forcing stops (right-hand side becomes zero), and it so happens that the system is at an equilibrium at this time, so the response disappears too.(25 votes)
- In this video you have said that you have made a whole playlist on interpretation of differential equations at 0.45 but I am not able to find it anywhere...Cananybody please post the link..
Thanks in advance(5 votes)
- He actually said that he SHOULD do a playlist on the interpretation of differential equations, not that he has made such a playlist.(11 votes)
- can Laplace transformation give general solution of differential equation?(2 votes)
- You could, I think, arbitrarily set y'(0)=c1 and y(0)=c2 and proceed from there. The hairy algebra would be a lot hairier, but you'd get an answer eventually. And no, I'm not volunteering to work it through! :-)(3 votes)
- I have a question about relating Laplace to forcing functions. I know how to solve for an initial value problem. I know what my solution and forcing function is and I can graph both, but how are they related? Is the forcing function not allowing the solution to go past certain boundaries?(3 votes)
- I'm not entirely sure I understand what you're asking, but assuming you mean in a situation similar to that in this video.
The problem in this video is, if we look just at the homogenous part, a mass on a spring problem. The angular frequency of this undriven system is 2.
The forcing function (here sin(t)-u2pi(t)sin(t)) can be seen as some additional driving force which in this case is, up to 2pi seconds (two oscillations of the undriven, homogenous system), giving the system the acceleration it would need to oscillate with an angular frequency of 1.
The driving force, is trying to force the oscillation to go at half its natural frequency.
Furthermore, because the mass starts at the origin and at rest, the driving force serves to actually get the thing moving because without it it would quite happily sit around doing nothing all day.
Because the two oscillations (the natural sin(2t) and the driven sin(t)) are in phase, and 2pi is a integer multiple of a half-period for each of them, when the step function toggles high and the driving force stops, there isn't any force to keep the mass moving and so it too will stop.(2 votes)
- Isn't sin(t-2pi) the same as sint? Sin is periodic over 2pi.(3 votes)
- is u subscript 2pi the same as u(t+2pi)?(2 votes)
- u_2pi(t) is the unit step function with the "step" (from 0 to 1) occurring at t = 2pi. If you learned that u(t) with no subscript is the unit step function that steps up at t = 0, then u_2pi(t) would be the same as u(t - 2pi) (note, minus, not plus). He discusses this function and notation at about0:40in the earlier video:
- What is a periodic function ?(1 vote)
- A function that repeats its values at regular intervals.
For example, sin(x) is a periodic function. If you start from x=y=0, the sine function fluctuates up to 1 at x=pi/2, down to 0 at pi, down to -1 at (3*pi)/2 and back to 0 at 2*pi. Since it comes back to its original point at 2pi, we say that sin(x) has period 2pi.(2 votes)
- I see why the Laplace transform gives you an exact solution (without arbitrary constants) due to the use of y(0) and y'(0). However, can we use the Laplace transform to give us the general solutions like we did in the earlier videos?(1 vote)
- I am in physics right now (in high school) and we just learned about simple harmonic motion. It's not as fun without calculus :((1 vote)
Let's apply everything we've learned to an actual differential equation. Instead of just taking Laplace transforms and taking their inverse, let's actually solve a problem. So let's say that I have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi. Let's solve this differential equation, an interpretation of it. And I actually do a whole playlist on interpretations of differential equations and how you model it, but you know, you can kind of view this is a forcing function. That it's a weird forcing function of this being applied to some weight with, you know, this is the acceleration term, right? The second derivative with respect to time is the acceleration. So the mass would be 1 whatever units, and then as a function of its position, this is probably some type of spring constant. Anyway, I won't go there. I don't want to waste your time with the interpretation of it, but let's solve it. We can do more about interpretations later. So we're going take the Laplace transform of both sides of this equation. So what's the Laplace transform of the left-hand side? So the Laplace transform of the second derivative of y is just s squared, so now I'm taking the Laplace transform of just that. The Laplace transform of s squared times the Laplace transform of y minus-- lower the degree there once-- minus s times y of 0 minus y prime of 0. So clearly, I must have to give you some initial conditions in order to do this properly. And then plus 4 times the Laplace transform of y is equal to-- what's the Laplace transform of sine of t? That should be second nature by now. It's just 1 over s squared plus 1. And then we have minus the Laplace transform of this thing. And I'll do a little side note here to figure out the Laplace transform of this thing right here. And we know, I showed it to you a couple of videos ago, we showed that the Laplace transform-- actually I could just write it out here. This is going to be the same thing as the Laplace transform of sine of t, but we're going to have to multiply it by e to the minus-- if you remember that last formula-- e to the minus cs, where c is 2 pi. Actually, let me write that down. I decided to write it down, then I decided, oh, no, I don't want to do this. But let me write that. So the Laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the Laplace transform of just the original function times the Laplace transform of f of t. So if we're taking the Laplace transform of this thing, our c is 2 pi. Our f of t is just sine of t, right? So then this is just going to be equal to-- if we just do this piece right here-- it's going to be equal to e to the minus cs-- our c is 2 pi-- e to the minus 2 pi s times the Laplace transform of f of t. f of t is just sine of t before we shifted. This is f of t minus 2 pi. So f of t is just going to be sine of t. So it's going to be times 1 over s squared plus 1. This is the Laplace transform of sine of t. So let's go back to where we had left off. So we've taken the Laplace transform of both sides of this equation. And clearly, I have some initial conditions here, so the problem must have given me some and I just forgot to write them down. So let's see, the initial conditions I'm given, and they are written kind of in the margin here, they tell us-- I'll do it in orange, they tell us that y of 0 is equal to 0, and y prime of 0 is equal to 0. That makes the math easy. That's 0 and that's 0. So let's see if I can simplify my equation. So the left-hand side, let's factor out the Laplace transform. So let's factor out this term and that term. So we get the Laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side. And what's the right-hand side? We could simplify this. Well, I'll just write it out. I don't want to do too many steps at once. It's 1 over s squared plus 1 and then plus-- or minus actually, this is a minus-- minus the Laplace transfer of this thing, which was e to the minus 2 pi s over s squared plus 1. So if we divide both sides of this equation by the s squared plus 4, then we get the Laplace transform of y is equal to-- and actually, I can just merge these two. They're the same denominator. So before I even divide by s squared plus 4, that right-hand side will look like this. It will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. And, of course, we're dividing both sides of this equation by s squared plus 4, so we're going to have to stick that s squared plus 4 over here. Now, we're at the hard part. In order to figure out why, we have to take the inverse Laplace transform of this thing. So how do we take the inverse Laplace transform of this thing? That's where the hard part is always, you know, it makes solving the differential equation's easy if you know the Laplace transforms. So it looks like we're going to have to do some partial fraction expansion. So let's see if we can do that. So we can rewrite this equation right here. Actually, let's write it as this, because this'll kind of simplify our work. Let's factor this whole thing out. So we're going to write it as 1 minus e to the minus 2 pi s, all of that times-- I'll do it in orange-- all of that times 1 over s squared plus 1 times s squared plus 4. Now, we need to do some partial fraction expansion to simplify this thing right here. We're going to do this on the side. Maybe I should do this over on the right here. This thing-- let me rewrite it-- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions, s squared plus 1 and s squared plus 4, with the numerators. This one would be As plus B. It's going to have to have degree 1, because this is degree 2. Here And then we'd have Cs plus D. And so when you add these two things up, you get As plus B times s squared plus 4 plus Cs plus D times s squared plus 1, all of that over the common denominator. We've seen this story before. We just have to do some algebra here. As you can tell, these differential equations problems, they require a lot of stamina. You kind of just have to say I will keep moving forward and do the algebra that I need to do in order to get the answer. And you kind of have to get excited about that notion that you have all this algebra to do. So let's figure it out. So this top can be simplified to As to the third plus Bs squared plus 4As plus 4B. And then this one, you end up with Cs to the third plus Ds squared plus Cs plus D. So when you add of these up together, you get-- and this is all the algebra that we have to do, for better, for worse-- A plus C over s to the third plus B plus D times s squared plus 4A plus C times s-- let's scroll over a little bit-- plus 4B plus D. And now we just have to say, OK, all of this is equal to this thing up here. This is the numerator. We just simplified the numerator. This is the numerator. That's the numerator right there. And all of this is going to be over your original s squared plus 1 times your s squared plus 4. And we established that this thing should be-- let me just write this-- that 1 over s squared plus 1 times s squared plus 4 should equal this thing. And then you just pattern match on the coefficients. This is all just intense partial fraction expansion. And you say, look, A plus C is the coefficient of the s cubed terms. I don't see any s cubed terms here. So A plus C must be equal to 0. And then you see, OK, B plus D is the coefficient of the s squared terms. I don't see any s squared terms there. So B plus D must be equal to 0. 4A plus C, the coefficient of the s terms. I don't see any s terms over here. So 4A plus C must be equal to 0. And then we're almost done. 4B plus D must be the constant terms. There is a constant term there. So 4B plus D is equal to 1. So let's see if we can do anything here. If we subtract this from that, we get minus 3A is equal to 0, or A is equal to 0. If A is equal to 0, then C is equals to 0. And let's see what we can get here. If we subtract this from that, we get minus 3B. The D's cancel out. It's equal to minus 1, or B is equal to 1/3. And then, of course, we have D is equal to minus B, if you subtract B from both sides. so D is equal to 1/3. So all of that work, and we actually have a pretty simple result. Our equation, this thing here, can be rewritten as-- the A disappeared. It's 1/3 over s squared plus 1. B was the coefficient on the-- let me make it very clear. B was the coefficient on the-- or it was a term on top of the s squared plus 1, so that's why I'm using B there. And then D is minus B, so D is minus 1. So let me make sure I have that. B is 1/3 minus-- let me make sure I get that right. D is 1/3. So, sorry, B as in boy is 1/3, so D is minus 1/3. So B, there's a term on top of the s squared plus 1. And then you have minus D over the minus 1/3 over s squared plus 4. This takes a lot of stamina to record this video. I hope you appreciate it. OK, so let me rewrite everything, just so we can get back to the problem because when you take the partial fraction detour, you forget-- not even to speak of the problem, you forget what day it is. Let's see, so you get the Laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for, times-- and I'll write it like this. 1/3 times 1 over s squared plus 1 minus 1/3 times-- actually, let me write it this way. Because I have this s squared plus 4, so I really want to have a 2 there. So I want to have a 2 in the numerator, so you want to have a 2 over s squared plus 4. So if I put a 2 in the numerator there, I have to divide this by 2 as well. So let me change this to a 6. Minus 1/6 times 2 is minus 1/3. So I did that just so I get this in the form of the Laplace transform of sine of t. Now, let's see if there's anything that I can do from here. This is an epic problem. I'll be amazed if I don't make a careless mistake while I do this. So we can rewrite everything. Let's see if we can simplify this. And by simplifying it, I'm just going to make it longer. We can write the Laplace transform of y is equal to-- I'm just going to multiply the 1 out, and then I'm going to multiply the e to the minus 2 pi s out. So if you multiply the 1 out, you get 1/3 times 1 over s squared plus 1-- I'm just multiplying the 1 out-- minus 1/6-- these are all the 1's times the 1-- times 2 over s squared plus 4. And then I'm going to multiply the minus e. Let me just switch colors, do the minus e. So then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1. And then the minus and the minus cancel out, so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4. Now, taking the inverse Laplace transform of these things are pretty straightforward. So let's do that. Let's take the inverse Laplace transform of the whole thing. And we get y is equal to the inverse Laplace transform of this guy right here, is just 1/3 sine of t-- I don't have to write a parentheses there-- sine of t, and then this is minus 1/6 times-- this is the Laplace transform of sine of 2t. That's that term right there. Now, these are almost the same, but we have this little pesky character over here. We have this e to the minus 2 pi s. And there, we just have to remind ourselves-- I'll write it here in the bottom. We just have to remind ourselves that the Laplace transform of the unit step function-- I'll put the pi there, just 2 pi times f of t minus 2 pi-- I should put as the step function of t-- is equal to e to the minus 2 pi s times the Laplace transform of just-- or let me just write it this way-- times the Laplace transform of f of t. So if we view f of t as just sine of t or sine of 2t, then we can kind of backwards pattern match. And we'll have to shift it and multiply it by the unit step function. So I want to make that clear. If you didn't have this guy here, the inverse Laplace transform of this guy would be the same thing as this guy. It'd just be sine of t. The inverse Laplace transform of this guy would be sine of 2t. But we have this pesky character here, which essentially, instead of having the inverse Laplace transform just being our f of t, it's going to be our f of t shifted by 2 pi times the unit step function, where it steps up at 2pi. So this is going to be minus 1/3 times the unit step function, where c is 2 pi of t times-- instead of sine of t-- sine of t minus 2pi. And then we're almost done. I'll do it in magenta to celebrate it. Plus this very last term, which is 1/6 times the unit step function 2 pi of t, the unit step function that steps up at 2 pi times sine of-- and we have to be careful here. Wherever we had a t before, we're going to replace it with a t minus 2 pi. So sine of, instead of 2t, is going to be 2 times t minus 2 pi. And there you have it. We finally have solved our very hairy problem. We could take some time if we want to simplify this a little bit. In fact, we might as well. At the risk of making a careless mistake at the last moment, let me see if I can make any simplifications here. Well, we could factor out this guy right here, but other than that, that seems about as simple as we can get. So this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here. This looked fairly straightforward, but we got this big mess to actually satisfy that equation, given those initial conditions that we had initially.