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### Course: Differential equations>Unit 3

Lesson 2: Properties of the Laplace transform

# Laplace transform of t: L{t}

Determining the Laplace Transform of t. Created by Sal Khan.

## Want to join the conversation?

• Why use the integral?
Can't we use the derivative rule L{f'} = s(L{f}) - f(0)?
Let f = t, then f' = 1 and L{1} = 1/s.
This leads directly to L{t} = 1/s^2 without worrying about doing an integration.
• I believe he may be trying to show that it works for a simple example
• Is the content in this video any different to what we did in the video called "Laplace Transform of cos(t) and Polynomials?
• well the video quickly brushes up the concept of laplace and also makes the integration friendly by more practice!
• Laplace transform of f"(t)?
• s^2 L(y) - s y(0) - y'(0)
He's about to cover that and the formula for all the derivatives in a few videos.
• why didnt you use the l'hopital's rule ?
(to find (-t/s)*(e^(-st)) in infinity )
• Because that would complicate the process. It should be possible (in combinations with Taylor series or something), but you would have a lot more steps. That would be more of a Fourier Transform than a Laplace Transform. Laplace are better in solving linear differential functions, while Fourier is more left to digital signal processing, check out this: http://upload.wikimedia.org/math/3/2/a/32a0fcde03ba4f2ac403b07ea56cdf71.png
• how to find laplace transform of ln t ?
• laplace transform is defined for negative values of s?
• Yes, negative functions are computed in the same way as positive functions. L{s} and L{-s} will have the same result, but opposite sign.
• I dont understand when to use (int. u'v) or (int.uv') in the by parts formula. Can anybody please explain how does Sal choose u, v, u', v'?
• What is the difference between power series and Laplace transform ?
• In essence, they are extremely different concepts.
The Power series is used to determine a function in terms of an infinite series in the same variable.
Laplace transforms can be used to define a function in a different variable/dimension altogether.
• Can you obtain L{cos^2 t} ?
• You can use the formula cos^2 (t) = 1/2 (1+cos(2t)).
• F(t) = inversL{(1-e^-2s)(1+e^-4s)/s^2}
1. find F(t) in terms of Heaviside unit step function
2. find the analytical definition of F(t)
3. sketch the graph of F(t)

## Video transcript

Let's try to fill in our Laplace transform table a little bit more. And a good place to start is just to write our definition of the Laplace transform. The Laplace transform of some function f of t is equal to the integral from 0 to infinity, of e to the minus st, times our function, f of t dt. That's our definition. The very first one we solved for-- we could even do it on the side right here-- was the Laplace transform of 1. You know, we could almost view that as t to the 0, and that was equal to the integral from 0 to infinity. f of t was just 1, so it's e to the minus st, dt, which is equal to the antiderivative of e to the minus st, which is minus 1 over s, e to the minus st. And then you have to evaluate that from 0 to infinity. When you take the limit as this term approaches infinity, this e to the minus, this becomes e to the minus infinity, if we assume s is greater than 0. So if we assume s is greater than 0, this whole term goes to 0. So you end up with a 0 minus this thing evaluated at 0. So when you evaluate t is equal to 0, this term right here becomes 1, e to the 0 becomes 1, so it's minus minus 1/s, which is the same thing as plus 1/s. the? Laplace transform of 1, of just the constant function 1, is 1/s. We already solved that. Now, let's increment it a little bit. Let's see if we can figure out the Laplace transform of t. So we can view this as t to the 0. Now this is t to the 1. This is going to be equal to the integral from 0 to infinity, of e to the minus st times t dt. Now. I can tell you right now that I don't have the antiderivative of this memorized. I don't know what it is. But there's a sense that the integration by parts could be useful, because integration by parts kind of decomposes into a simpler problem. And I always forget integration by parts, so I'll rederive it here in this purple color. So if we have u times v, if we take the derivative with respect to t of that, that's equal to the derivative of the first times the second function plus the first function times the derivative of the second function. It's just the product rule. Now, if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of uv prime. Now, since we want to apply this to an integral, maybe let's make this what we want to solve for, so we can get the integral of uv prime is equal to-- we can just subtract this from that side of the equation, so it's equal-- I'm just swapping the sides, so I'm just solving for this, and to solve for this, I just subtract this from that, so it's equal to uv minus the integral of u prime v. So there you go. Even though I have trouble memorizing this formula, it's not too hard to rederive as long as you remember the product rule right there. So if we're going to do integration by parts, it's good to define our v prime to be something that's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy take the derivative of. So let's make t is equal to our u and let's make e to the minus st as being our v prime. If that's the case, then what is v? Well, v's just the antiderivative of that. In fact, we've done it before. It's minus 1/s, e to the minus st. That's v. And then if we want to figure out u prime, because we're going to have to figure out that later anyway, u prime's just the derivative of t. That's just equal to 1. So let's apply this. Let's see, so the Laplace transform of t is equal to uv. u is t, v is this right here. It's times minus 1/s, e to the minus st. e to the minus st, that's the uv term right there. And this is a definite integral, right? So we're going to evaluate this term right here from 0 to infinity. And then it's minus the integral from 0 to infinity of u prime, which is just 1 times v. v, we just figured out here, is minus-- let me write it in v's color-- times minus 1/s-- this is my v right here-- minus 1/s, e to the minus st, dt. All of that is dt. So let's see if we can simplify this. So this is equal to minus t/s, e to the minus st, evaluated from 0 to infinity. And let's see, we could take-- well, this is just 1. 1 times anything is 1. We can just not write that. And then bring the minus 1/s out. So if we bring the minus 1/s out, this becomes plus 1/s times the integral from 0 to infinity of e to the minus st, dt. And this should look familiar to you. This is exactly what we solved for right here. It was the Laplace transform of 1. So let's keep that in mind. So this right here is the Laplace transform of 1. And I want to write it that way, because we're going to see a pattern of this in the next video. I'm going to write that as a Laplace transform of 1. But what is this equal to? So we're going to evaluate this as it adds to infinity and then subtract from that evaluated at 0. You can kind of view it as a substitution, so this is equal to-- well, let me write it this way. It's the limit as A approaches infinity, of minus A/s, e to the minus sA. So that's this evaluated at infinity. And then from that, we're going to subtract this evaluated at 0. So minus all of this, but we already have a minus sign here, so we could write a plus. Plus 0/s times e to the minus s times 0. And then, of course, we have this term right here. So let me write that term. I'll do it in yellow or let me do it in blue. Plus 1/s-- that's this right there-- times the Laplace transform of 1. And what do we get? So what's the limit of this as A approaches infinity? You might say, wow, you know, as A approaches infinity right here, this becomes a really big number. There's a minus sign in there, so it would be a really big negative number. But this is an exponent. A is an exponent right here. So e to the minus infinity is going to go to zero much faster than this is going to go to infinity. This term right here is a much stronger function, I guess is the way you could see it. And you could try it out on your calculator, if you don't believe me. This term is going to overpower this term and so this whole thing is going to go to zero. Likewise, e to the minus-- e to the 0, this is 1, but you're multiplying it times a zero, so this is also going to go to zero, which is convenient because all of this stuff just disappears. And we're left with the Laplace transform of t is equal to 1/s times the Laplace transform of 1. And we know what the Laplace transform of 1 is. The Laplace transform of 1-- we just did it at beginning of the video-- was equal to 1/s, if we assume that s is greater than zero. In fact, we have to assume that s was greater than zero here in order to assume that this goes to zero. Only if s is greater than zero, when you get a minus infinity here does this approach zero. So fair enough. So the Laplace transform of t is equal to 1/s times 1/s, which is equal to 1/s squared, where s is greater than zero. So we have one more entry in our table, and then we can use this. What we're going to do in the next video is build up to the Laplace transform of t to any arbitrary exponent. And we'll do this in the next video.