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### Course: Differential equations > Unit 3

Lesson 2: Properties of the Laplace transform- Laplace as linear operator and Laplace of derivatives
- Laplace transform of cos t and polynomials
- "Shifting" transform by multiplying function by exponential
- Laplace transform of t: L{t}
- Laplace transform of t^n: L{t^n}
- Laplace transform of the unit step function
- Inverse Laplace examples
- Dirac delta function
- Laplace transform of the dirac delta function

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# 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.(18 votes)- I believe he may be trying to show that it works for a simple example(9 votes)

- Is the content in this video any different to what we did in the video called "Laplace Transform of cos(t) and Polynomials?(14 votes)
- well the video quickly brushes up the concept of laplace and also makes the integration friendly by more practice!(7 votes)

- Laplace transform of f"(t)?(4 votes)
- 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.(4 votes)

- why didnt you use the l'hopital's rule ?

(to find (-t/s)*(e^(-st)) in infinity )(4 votes)- 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(3 votes)

- how to find laplace transform of ln t ?(4 votes)
- laplace transform is defined for negative values of s?(3 votes)
- 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.(3 votes)

- 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'?(3 votes)
- https://www.youtube.com/watch?v=NXH7p8NwE70

here some explanation(3 votes)

- What is the difference between power series and Laplace transform ?(3 votes)
- 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.(2 votes)

- Can you obtain L{cos^2 t} ?(2 votes)
- You can use the formula cos^2 (t) = 1/2 (1+cos(2t)).(3 votes)

- 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)(2 votes)

## 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.