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## Differential equations

### 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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# Inverse Laplace examples

Using our toolkit to take some inverse Laplace Transforms. Created by Sal Khan.

## Want to join the conversation?

- Mr. Sal,

I tried to solve the same problem using MATLAB and I got the following answer:

2*cos(t - 2)*heaviside(t - 2)*exp(t - 2)

where the heaviside is the unit step function. Would you explain the reason how MATLAB did it with (t-2) in the unit step function?(4 votes)- Also notice how Sal has the answer as u₂(t)
**with**a sub-scripted 2.

This is the same as "the step function starting from two".

Which is also the same as "shifting the step function two to the right" (the step function starts at 0), or u(t-2). Notice**no**subscripts with this answer.(14 votes)

- How can I do the inverse laplace transform of 1/(s-a)?(1 vote)
- That's 1/s shifted by a. And you can shift by a by multiplying your function f(t) with e^-at. Since L{1} = 1/s and therefore f(t) = 1, L{ e^-at*f(t) } = 1/(s-a). Of course if s is equal or greater than a.(7 votes)

- At10:58why not cancel out (s-1) from numerator & denominator before we continue? (which would leave ( 2 e^-2s ) / s )(2 votes)
- You can only cancel factors if they are actually factors both in the numerator and in the denominator.

In this case`(s-1)`

is a factor of the numerator (it's multiplying the whole numerator), but in the denominator you have`(s-1)² + 1`

; since you have 2 terms on the denominator, you don't have any factors that you could cancel out.(4 votes)

- Was the fact that there was a 2 constant, and also an e^(-2s) just a coincidence that they were both 2's?(2 votes)
- Yes it is a coincidence. The 2 constant in the question is the 2 constant in the answer. The 2s from u2(t)*e^(t-2)*cos(t-2) is a result of the 2 in e^(-2s).(4 votes)

- what is the inverse laplace of 1 ?

L^-1{1}(3 votes)- The inverse laplace of 1 is the dirac delta function d(t). The inverse laplace transform of any number (K) is K*d(t).(2 votes)

- How did it become e^(t-2) ?(1 vote)
- Okay, so in the video, Sal establishes that f(t) = e^t * cos(t).

Whenever you change or shift the variable in the "f" portion of the equation, you equivalently change the other side. Do you remember, back in algebra/algebra II that you might be given something like "f(x) = 2x^2 + 3x, find f(2x+3)"? Those kinds of things show what Sal is doing here. We know f(t), but we want to find f(t-2): therefore, we substitute in t-2 for every t in the f(t) function.(2 votes)

- If I understand correctly,

ℒ{ μ_c(t) f(t-c) } = e^(cs) F(s)

Shifting by f(t) to the right by c in the time domain shrinks F in the frequency domain by e^(-cs)

ℒ{ e^(at) f(t)} = F(s-a)

Multiplying f(t) by e^(at) in the time domain shifts F in the frequency domain to the right by a, resulting in F(s-a)(1 vote)- I think you have a typo in second line. Should be e^(-cs) F(s), but other than that I'd say you summed it up well.(1 vote)

- Hi to the khan academy team, would you please practice section in this lesson?(1 vote)
- Isn't there a 1/s factor missing in the original equiation for the time domain answer to contain a step function? I believe that a lone e^-ts factor is an indication of the impulse function not the step function.(1 vote)
- Why do these inverses work so well with s^(-n) but not with s^(n)?(1 vote)
- Because for functions that are polynomials, the Laplace transform function, F(s), has the variable ("s") part in the denominator, which yields s^(-n). However, there's no restriction on whether we have/use "+n" or "-n" so just make sure you pay attention to your (-) signs!(1 vote)

## Video transcript

A lot of what we do with Laplace
transforms, taking them and taking their
inverse, it's a lot of pattern matching. And it shouldn't just be a
mechanical thing, and that's why I've gone through the
exercise of showing you why they work. But in order to just kind of
make sure we don't get confused, I think it might be
useful to review a little bit of everything that we've
learned so far. So in the last video, we saw
that the Laplace transform-- well, let me just
write something. The Laplace transform of f of
t, let me just get some notation down, and we can write
that as big capital F of s, and I've told you
that before. And so given that, in the last
video I showed you that if we have to deal with the unit step
function, so if I said, look, the Laplace transform of
the unit step function, it becomes 1 at some value c times
some shifted function f of t minus c, in the last video,
we saw that this is just equal to e to the minus cs
times the Laplace transform of just this function right
there, so times the F of s. And it's really important not
to get this confused with another Laplace transform
property or rule, or whatever you want to call it, that
we figured out. I think it was one of the videos
that I made last year, but if you're just following
these in order I think it's three or four videos ago. And that one told us that the
Laplace transform of e to the at, times f of t, that this is
equal to-- and I want to make this distinction very clear. Here we shifted the f of t
and we got just kind of a regular F of s. In this situation, when we
multiply it times a e to the positive at, we end up shifting
the actual transform. So this becomes F
of s minus a. And these two rules, or
properties, or whatever you want to call them, they're
very easy to confuse with each other. So we're going to do a couple
of examples that we're going to have to figure out which
one of these two apply. Let's write all the other stuff
that we learned as well. The very first thing we learned
was that the Laplace transform of 1 was
equal to 1/s. We know that that's a pretty
straightforward one, easy to prove to yourself. And more generally, we learned
that the Laplace transform of t to the n, where n is a
positive integer, it equaled n factorial over s to
the n plus 1. And then we had our
trig functions that we've gone over. Let me do this in a
different color. I'll do it right here. The Laplace transform of sine
of at is equal to a over s squared, plus a squared. And the Laplace transform of the
cosine of at is equal to s over s squared plus a squared. And you'll be amazed by how far
we can go with just what I've written here. In future videos, we're going
to broaden our toolkit even further, but just these right
here, you can already do a whole set of Laplace transforms
and inverse Laplace transforms. So let's
try to do a few. So let's say I were to give
you the Laplace transform. And you know, this is
just the hard part. I think you know how to solve
a differential equation, if you know how to take the
Laplace transforms and go back and forth. The hard part is just
recognizing or inverting your Laplace transforms. So let's
say we had the Laplace transform of some
function F of s. Let's say it's 3 factorial over
s minus 2 to the fourth. Now, your pattern matching, or
your pattern recognition part of your brain, should
immediately say, look, I have a Laplace transform of something
that has a factorial in it, and it's over
an exponent. This must be something
related to this thing right here, right? If I just had the Laplace
transform-- let me write that down-- the Laplace transform
of-- you see a 3 factorial and a fourth power, so it looks
like n is equal to 3. So if you write the Laplace
transform of t to the 3, this rule that we showed right here,
this means that it would be equal to 3 factorial
over s to the fourth. Now, this thing isn't
exactly this thing. They're not quite
the same thing. You know, I'm doing this to
instruct you, but I find these, when I'm actually doing
them on an exam-- I remember when I did them when I first
learned this, I would actually go through this step because you
definitely don't want to make a careless mistake and you
definitely want to make sure you have a good handle
on what you're doing. So you're like, OK, it's
something related to this, but what's the difference between
this expression right here and the expression that we're trying
to take the inverse Laplace transform of,
and this one here? Well, we've shifted our s. If we call this expression right
here F of s, then what's this expression? This expression right here
is F of s minus 2. So what are we dealing
with here? So you see here, you have
a shifted F of s. So in this case, a would
be equal to 2. So this is the Laplace transform
of e to the at times our f of t. So let me write this down. This is the Laplace transform
of e to the-- and what's a? a is what we shifted by. It's what we shifted by minus
a, so you have a positive a, so e to the 2t times the
actual function. If this was just an F of s,
what would f of t be? Well, we figured out, it's t the
3, t to the third power. So the Laplace transform of
this is equal to that. Or we could write that the
inverse Laplace transform of 3 factorial over s minus 2 to the
fourth is equal to e to the 2t times t to the third. Now, if that seemed confusing
to you, you can kind of go forward. Let's go the other direction,
and maybe this will make it a little bit clearer for you. So let's go from
this direction. If I have to take the Laplace
transform of this thing, I'd say, OK, well, the Laplace
transform of t to the third is easy. I think the tool isn't working
right there properly. Let me scroll up a little bit. So I could write
it right here. So if I wanted to figure out the
Laplace transform of e to the 2t times t to the third,
I'll say, well, you know, this e to the 2t, I remember that
it shifts something. So if I know that the Laplace
transform of t the third, this is an easy one. It's equal to 3 factorial
over s to the fourth. That's 3 plus 1. Then the Laplace transform of e
to the 2t times t the third is going to be this shifted. This is equal to F of s. Then this is going to
be f of s minus 2. So what's F of s minus 2? It's going to be equal to 3
factorial over s minus 2 to the fourth. I think you're already getting
an appreciation that the hardest thing about these
Laplace transform problems are really kind of all of these
shifts and kind of recognizing the patterns and recognizing
what's your a, and what's your c, and being very careful about
it so you don't make a careless mistake. And I think doing a lot of
examples probably helps a lot, so let's do a couple of more
to kind of make sure things really get hammered home
in your brain. So let's try this
one right here. This looks a little bit
more complicated. They give us that the Laplace
transform of some function is equal to 2 times s minus 1 times
e to the minus 2s, all of that over s squared
minus 2s plus 2. Now this looks very daunting. How do you do this? I have an e here. I have something shifted here. I have this polynomial in
the denominator here. What can I do with this? So the first thing, when I look
at these polynomials in the denominator, I say can
I factor it somehow? Can I factor it fairly simply? And actually, in the exams
that you'll find in differential equation class,
they'll never give you something that's factorable
into these weird numbers. It tends to be integers. So you see, OK, what
two numbers? They have to be positive. When you give their product,
you get 2. And then when you add them, you
get negative 2, or they could both be negative. But there's no two easy
numbers, not 1 and 2. None of those work. So if you can't factor this
outright, the next idea is maybe we could complete the
square and maybe this will match one of the cosine
or the sine formulas. So how can we complete the
square in this denominator? Well, this can be rewritten
as s squared minus 2s. And I'm going to put
a plus 2 out here. And you can watch, I have
a bunch of videos on the completing of the square,
if all of this looks foreign to you. And to complete the square, we
just want to turn this into a perfect square. So to turn this into a perfect
square-- so something when I add it to itself twice becomes
minus 2, and so that when I square it, when I add it to
itself twice, it becomes minus 2, it's minus 1. And when I square it,
it'll become plus 1. I can't just add plus 1
arbitrarily to some expression, I have to
make it neutral. So let me subtract 1. I haven't changed this. I added 1 and I subtracted 1. A little bit of a primer on
completing the square. But by doing this, I now can
call this expression right here, I can now say that this
thing is s minus 1 squared. And then this stuff out here,
this out here is 2 minus 1. This is just plus 1. So I can rewrite my entire
expression now as 2 times s minus 1 times e to the minus
2s-- make sure I'm not clipping off at the top-- e to
the minus 2s, all of that over s minus 1 squared plus 1. So a couple of interesting
things seem to be happening here. Let's just do a couple of test
Laplace transforms. So if a Laplace transform of cosine of
t, we know that this is equal to s over s squared plus 1,
which this kind of looks like if this was an s and this
was an s squared plus 1. If this was F of s,
then what is this? Well, let's ignore this guy
right here for a little bit. So what is it? We know, actually, from
the last video. We saw, well, what if we took
the Laplace transform of e to the-- I'll call it 1t. But let's say e to the-- yeah
I'll just write it e to the 1t times cosine of t? Well, then this will just
shift this Laplace transform by 1. It will shift it by
1 to the right. Wherever you see an s, you
would put an s minus a 1. So this will be equal to s
minus 1 over s minus 1 squared plus 1. We're getting close. We now figured out this
part right here. Now, in the previous video, I
think it was two videos ago, or maybe it was the last
video, I forget. Memory fails me. I showed you that if you have
the Laplace transform of the unit step function of t times
some f of t shifted by some value of c, then that this is
equal to e to the minus cs times F of s. OK, And this can get
very confusing. This can get very confusing,
so I want to be very careful here. Let's ignore all of this. I called this F of s before, but
now I'm going to backtrack a little bit. And let's just ignore this,
because I'm going to redefine our F of s. So let's just ignore
that for a second. Let's define our new
f of t to be this. Let's say that that is f of t. Let's say f of t is equal to
e to the t cosine of t. Then if you take the Laplace
transform of that, that means that F of s is equal to s
minus 1 over s minus 1 squared plus 1. Nothing fancy there. I just defined our f of
t as this, and then our F of s is that. Now, we have a situation here. Let's ignore the 2 here. The 2 is just kind of
a scaling factor. This expression right here, we
can rewrite as that expression is equal to-- this
is our F of s. This expression right here is
equal to 2 times our F of s times e to the minus 2s. Or let me just write it. Let me switch the order, just
so we make it look right. 2 times e to the minus
2s times F of s. Well, that looks just like
this if our 2 was equal to our c. So what does that tell us? That tells us that the inverse
Laplace transform, if we take the inverse Laplace
transform-- and let's ignore the 2. Let's do the inverse Laplace
transform of the whole thing. The inverse Laplace transform
of this thing is going to be equal to-- we can just write
the 2 there as a scaling factor, 2 there times
this thing times the unit step function. What's our c? You can just pattern match. You have a 2 here. You have a c, a minus c,
a minus 2, so c is 2. The unit step function is zero
until it gets to 2 times t, or of t, so, then it becomes 1
after t is equal to 2, times our function shifted by 2. So this is our inverse
Laplace transform. Now, what was our function? Our function was this
thing right here. So if our inverse Laplace
transform of that thing that I had written is this thing, an
f of t, f of t is equal to e to the t cosine of t. Then our inverse-- let me
write all of this down. Let me write our big result. We established that the inverse
Laplace transform of that big thing that I had
written before, 2 times s minus 1 times e to the minus 2--
sorry, e to the minus 2s over s squared minus 2s plus 2
is equal to this thing where f of t is this. Or we could just rewrite this
as 2 times the unit step function starting at 2, where
that's when it becomes non-zero of t times f of t minus
2. f of t minus 2 is this with t being replaced
by t minus 2. I'll do it in another color,
just to ease the monotony. So it would be e to the t minus
2 cosine of t minus 2. Now, you might be thinking, Sal,
you know, he must have taken all these baby steps with
this problem, because he's trying to explain
it to me. But I'm taking baby steps with
this problem so that I myself don't get confused. And I think it's essential
that you do take these baby steps. And let's just think about
what baby steps we took. And I really want
to review this. This is actually a surprisingly
good problem. I didn't realize it when I
first decided to do it. We solved this thing. We wanted to get this
denominator into some form that is vaguely useful to us,
so I completed the square there and then we rewrote
our Laplace transform, our f of s like this. And then we used a little
pattern recognition. We said, look, if I take the
Laplace transform of cosine of t, I'd get s over s
squared plus 1. But this isn't s over
s squared plus 1. It's s minus 1 over s minus
1 squared plus 1. So we said, oh, well, that means
that we're multiplying our original time
domain function. We're multiplying our f of
t times e to the 1t. And that's what we got there. So the Laplace transform of e to
the t cosine of t became s minus 1 over s minus
1 squared plus 1. And then we had this e to the
minus 2s this entire time. And that's where we said, hey,
if we have e to the minus 2s in our Laplace transform, when
you take the inverse Laplace transform, it must be the step
function times the shifted version of that function. And that's why I was
very careful. And you had this 2 hanging out
the whole time, and I could have used that any time. But the simple constants
just scale. A function is equal to two times
the Laplace transform of that function and vice versa. So the 2's are very easy to
deal with, so I kind of ignored that most of the time. But that's why I was
very careful. I redefined f of t to be this,
F of s to be this, and said, gee, if F of s is this, and if
I'm multiplying it times e to the minus 2s, then what I'm
essentially doing, I'm fitting this pattern right here. And so the answer to my problem
is going to be the unit step function-- I just
throw the 2 out there-- the 2 times the unit step function
times my f of t shifted by c. And we established this was
our f of t, so we just shifted it by c. We shifted it by 2, and we
got our final answer. So this is about as hard up to
this point as you'll see an inverse Laplace transform
problem. So, hopefully, you found that
pretty interesting.