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### Course: Differential equations > Unit 3

Lesson 1: Laplace transform# Laplace transform 1

Introduction to the Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

- At4:29of the video Sal begins integration. He starts with -1/s times e to the -st but it gets hairy for me because what happened to adding 1 to the exponent??(14 votes)
- It involves integration by substitution, wherein:

Let -st=u

=> du = -s.dt

Thus int e^-st = int (-1/s) e^u du

= -1/s e^u

Substituting back

int e^-st = -1/s.e^-st(40 votes)

- I really didn't get how int. e^-st = -1/s e^-st..Please help!!(7 votes)
- This is a classic case of u-substitution.

∫ e^(-st) dt. u=-st. du/dt=-s. dt=-du/s

∫ e^(u) * -du/s = -1/s * ∫ e^u du = -1/s * e^u = -1/s * e^(-st) + c

With enough practice, you will be able to do u-sub problems like this in your head.(32 votes)

- Would you do a video about the proof of the Laplace Transform definition ?(3 votes)
- Huh? How can you prove a definition?(14 votes)

- at "4:10" Sal mentions anti derivitive. what is that?(0 votes)
- It's another name for the integral (without any bounds).(30 votes)

- What if s < 0?

Then this integral would diverge.(4 votes)- Yes it will diverge.

Remember that a laplace transform is essentially telling you how close the function is to e^(st). If the integral diverges that just means the function gets farther and farther away from e^(st) as s gets closer to 0 (in this case).

Really what it means is you need to pick an s much larger than 0.(7 votes)

- why does this transform actually work in helping to solve differential equations?(4 votes)
- It's a property of Laplace transform that solves differential equations without using integration,called"Laplace transform of derivatives".

Laplace transform of derivatives: {f'(t)}= S* L{f(t)}-f(0).

This property converts derivatives into just function of f(S),that can be seen from eq. above.

Next inverse laplace transform converts again function F(S) into f(t).

If my ans. looks confusing .Just observe am example of solving D.E. using laplace,i hope droughts will disappear.(0 votes)

- and what is actually the difference between laplace transform and laplace operator ?(3 votes)
- The Laplace operator involves partial derivatives.

The Laplace transform involves an improper integral to transform to a function of a different variable.(2 votes)

- If I am asked to use the Laplace transform on f(x)=x, would that be the same as using f(t)=t?(2 votes)
- A laplace transform on a polynomial with degree n is n!/s^(n+1).

Given that you have f(x) = x, your Laplace transform should be:

L(s) = 1/s^2(1 vote)

- Hi,

May you please do the Derivation of the Fourier Transform Pairs?

I tried to search online, but I can NEVER find any.

Please help, very please(2 votes) - Sir ,I read laplace transform is used to change from time domain to frequency domain.For Simple harmonic functions,T=1/F, sin at is also simple harmonic function, but laplace transform of sin at is a/(s2+a2),HOW? please explain me physically sir,Thank you..(2 votes)

## Video transcript

I'll now introduce you
to the concept of the Laplace Transform. And this is truly one of the
most useful concepts that you'll learn, not just in
differential equations, but really in mathematics. And especially if you're going
to go into engineering, you'll find that the Laplace Transform,
besides helping you solve differential equations,
also helps you transform functions or waveforms from
the time domain to this frequency domain, and study
and understand a whole set of phenomena. But I won't get into
all of that yet. Now I'll just teach
you what it is. Laplace Transform. I'll teach you what it is, make
you comfortable with the mathematics of it and then in
a couple of videos from now, I'll actually show you how it
is useful to use it to solve differential equations. We'll actually solve some of the
differential equations we did before, using the
previous methods. But we'll keep doing it, and
we'll solve more and more difficult problems. So what is the Laplace
Transform? Well, the Laplace Transform,
the notation is the L like Laverne from Laverne
and Shirley. That might be before many
of your times, but I grew up on that. Actually, I think it was even
reruns when I was a kid. So Laplace Transform
of some function. And here, the convention,
instead of saying f of x, people say f of t. And the reason is because in
a lot of the differential equations or a lot of
engineering you actually are converting from a function
of time to a function of frequency. And don't worry about
that right now. If it confuses you. But the Laplace Transform
of a function of t. It transforms that function into
some other function of s. and And does it do that? Well actually, let me just do
some mathematical notation that probably won't
mean much to you. So what does it transform? Well, the way I think of
it is it's kind of a function of functions. A function will take you from
one set of-- well, in what we've been dealing with-- one
set of numbers to another set of numbers. A transform will take you from
one set of functions to another set of functions. So let me just define this. The Laplace Transform for our
purposes is defined as the improper integral. I know I haven't actually done
improper integrals just yet, but I'll explain them
in a few seconds. The improper integral from 0 to
infinity of e to the minus st times f of t-- so whatever's
between the Laplace Transform brackets-- dt. Now that might seem very
daunting to you and very confusing, but I'll now do
a couple of examples. So what is the Laplace
Transform? Well let's say that f
of t is equal to 1. So what is the Laplace
Transform of 1? So if f of t is equal to 1--
it's just a constant function of time-- well actually, let me
just substitute exactly the way I wrote it here. So that's the improper integral
from 0 to infinity of e to the minus st
times 1 here. I don't have to rewrite it here,
but there's a times 1dt. And I know that infinity is
probably bugging you right now, but we'll deal
with that shortly. Actually, let's deal with
that right now. This is the same thing
as the limit. And let's say as A approaches
infinity of the integral from 0 to Ae to the minus st. dt. Just so you feel a little bit
more comfortable with it, you might have guessed that this
is the same thing. Because obviously you can't
evaluate infinity, but you could take the limit as
something approaches infinity. So anyway, let's take the
anti-derivative and evaluate this improper definite
integral, or this improper integral. So what's anti-derivative
of e to the minus st with respect to dt? Well it's equal to minus 1/s
e to the minus st, right? If you don't believe me, take
the derivative of this. You'd take minus s times that. That would all cancel out, and
you'd just be left with e to the minus st. Fair enough. Let me delete this here,
this equal sign. Because I could actually use
some of that real estate. We are going to take the limit
as A approaches infinity. You don't always have to do
this, but this is the first time we're dealing with
improper intergrals. So I figured I might as
well remind you that we're taking a limit. Now we took the anti-derivative. Now we have to value it at A
minus the anti-derivative valued at 0. And then take the limit of
whatever that ends up being as A approaches infinity. So this is equal to the limit
as A approaches infinity. OK. If we substitute A in here
first, we get minus 1/s. Remember we're, dealing
with t. We took the integral
with respect to t. e to the minus sA, right? That's what happens when
I put A in here. Now what happens when I put
t equals 0 in here? So when t equals 0, it becomes
e to the minus s times 0. This whole thing becomes 1. And I'm just left
with minus 1/s. Fair enough. And then let me scroll
down a little bit. I wrote a little bit bigger
than I wanted to, but that's OK. So this is going to be the limit
as A approaches infinity of minus 1/s e to the
minus sA minus 1/s. So plus 1/s. So what's the limit as A
approaches infinity? Well what's this term
going to do? As A approaches infinity, if
we assume that s is greater than 0-- and we'll make that
assumption for now. Actually, let me write
that down explicitly. Let's assume that s
is greater than 0. So if we assume that s is
greater than 0, then as A approaches infinity, what's
going to happen? Well this term is going to go to
0, right? e to the minus-- a googol is a very,
very small number. And an e to the minus googol
is an even smaller number. So then this e to the minus
infinity approaches 0, so this term approaches 0. This term isn't affected because
it has no A in it, so we're just left with 1/s. So there you go. This is a significant to
moment in your life. You have just been exposed to
your first Laplace Transform. I'll show you in a few videos,
there are whole tables of Laplace Transforms, and
eventually we'll prove all of them. But for now, we'll just
work through some of the more basic ones. But this can be our
first entry in our Laplace Transform table. The Laplace Transform of
f of t is equal to 1 is equal to 1/s. Notice we went from a function
of t-- although obviously this one wasn't really dependent
on t-- to a function of s. I have about 3 minutes left,
but I don't think that's enough time to do another
Laplace Transform. So I will save that for
the next video. See you soon.