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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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# Laplace as linear operator and Laplace of derivatives

Useful properties of the Laplace Transform. Created by Sal Khan.

## Want to join the conversation?

- All other videos except this one work for me. Anyone else having this issue?(15 votes)
- yup, it works after skipping ahead in the video and then going back to the beginning. It should play normally after that.(10 votes)

- laplace transform of the laplace transform of f(x) equals what?(6 votes)
- at6:08why do we evaluate e^(-st)* f(t) from 0 to infinity?(5 votes)
- I believe it's because it's the "uv" part of the solution for integration by parts. The integral of udv (evaluated from 0 to infinity) is equal to uv (evaluated from 0 to infinity) plus integral of vdu (evaluated from 0 to infinity).(3 votes)

- So this explains what happens when taking the Laplace transform of a derivative, but what happens when one takes the derivative of a Laplace Transform?(2 votes)
- Take the derivative as you normally would after taking the Laplace transform, I believe.(5 votes)

- Is there a problem with this video, it has been loading for several minutes, and still won't play.(2 votes)
- try to skip the beginning, and it will start to work.(4 votes)

- hello, can someone explain to me where is the constant C when we have done the antiderivative of f'(t) at5:37!

thank you for your answer(3 votes)- The integral is evaluated at the limits 0 to infinity6:10, so the constant C is not necessary.(2 votes)

- at0:41and3:51, sal switches from integrating from negative infinity to positive infinity as he has done in prior videos and as he did in the definition of the laplace operation to only integrating from zero to positive infinity. i understand the necessity of this change as the evaluation of e^(-st) within the evaluation of uv at8:46as t approaches negative infinity would yield e raised to the positive infinity, which would render the laplace transform of f'(t) infinity, destroying the proof of the laplace transform of f'(t), but why can he make this change?(2 votes)
- The Laplace transform starts at some arbitrary fixed time and proceeds ad infinitum. Generally it is computationally simpler to start at zero than at any other time. If you need to start at a time other than zero, you can use a switch (or Heaviside or unit step) function to rectify it. But this tool is not really capable of handling times from eternity past.(3 votes)

- Why does e^-st when t is given the values of infinite and 0 approaches 0? I thought e^-st when t=0 would be 1, and when t = infinite would be 0, thus 0 - 1 = -1. I am not getting it(3 votes)
- what did he do at5:14in the video(1 vote)
- Are you referring to the point where he takes the derivative of both sides of the equation u=e^(-st)? He does this by using chain rule. He takes the derivative of e^(-st) with respect to -st, which results in e^(-st). He then takes the derivative of -st with respect to t, which results in -s. He lastly multiplied the two answers,which resulted in -s[e^(-st)]. This was the new right side of the equation. The derivative of u was simply u'. The resulting equation from all of this was u'=-s[e^(-st)].(2 votes)

- I do not get why Laplace transform is called a linear operator. Also did not understand the point of deriving the final result of this video where Sal says it turns derivatives into multiplication by s, because we were still left with a derivative term and a transform term so what was the point?(1 vote)

## Video transcript

Well, now's as good a time
as any to go over some interesting and very
useful properties of the Laplace transform. And the first is to show that
it is a linear operator. And what does that mean? Well, let's say I wanted to take
the Laplace transform of the sum of the-- we call it the weighted sum of two functions. So say some constant, c1, times
my first function, f of t, plus some constant,
c2, times my second function, g of t. Well, by the definition of the
Laplace transform, this would be equal to the improper
integral from 0 to infinity of e to the minus st, times
whatever our function that we're taking the Laplace
transform of, so times c1, f of t, plus c2, g of t-- I think
you know where this is going-- all of that dt. And then that is equal to the
integral from 0 to infinity. Let's just distribute
the e the minus st. That is equal to what? That is equal to c1e to the
minus st, f of t, plus c2e to the minus st, g of t, and
all of that times dt. And just by the definition
of how the properties of integrals work, we know that we
can split this up into two integrals, right? If the integral of the sum of
two functions is equal to the sum of their integrals. And these are just constant. So this is going to be equal to
c1 times the integral from 0 to infinity of e to the minus
st, times f of t, d of t, plus c2 times the integral
from 0 to infinity of e to the minus st, g of t, dt. And this was just a very
long-winded way of saying, what is this? This is the Laplace transform
of f of t. This is the Laplace transform
of g of t. So this is equal to c1 times the
Laplace transform of f of t, plus c2 times-- this is the
Laplace transform-- the Laplace transform of g of t. And so, we have just shown that
the Laplace transform is a linear operator, right? The Laplace transform of
this is equal to this. So essentially, you can kind of
break up the sum and take out the constants, and just take
the Laplace transform. That's something useful to
know, and you might have guessed that was the
case anyway. But now you know for sure. Now we'll do something which
I consider even more interesting. And this is actually going to
be a big clue as to why Laplace transforms are extremely
useful for solving differential equations. So let's say I want to find
the Laplace transform of f prime of t. So I have some f of t, I take
its derivative, and then I want the Laplace transform
of that. Let's see if we can find a
relationship between the Laplace transform of the
derivative of a function, and the Laplace transform
of the function. So we're going to use some
integration by parts here. Let me just say what this
is, first of all. This is equal to the integral
from 0 to infinity of e to the minus st, times f
prime of t, dt. And to solve this, we're going
to use integration by parts. Let me write it in the
corner, just so you remember what it is. So I think I memorized it,
because I recorded that last video not too long ago. I'm just going to write
this shorthand. The integral of u-- well, let's
say uv prime, because that will match what we have up
here better-- is equal to both functions without the
derivitives, uv minus the integral of the opposite. So the opposite is u prime v. So here, the substitution
is pretty clear, right? Because we want to end up
with f of x, right? So let's make v prime is f
prime, and let's make u e to the minus st. So
let's do that. u is going to be e to the minus
st, and v is going to equal what? v is going to equal
f prime of t. And then u prime would be minus
se to the minus st. And then, v prime-- oh, sorry,
this is v prime, right? v prime is f prime of t, so
v is just going to be equal to f of t. I hope I didn't say that
wrong the first time. But you see what I'm saying. This is u, that's u, and
this is v prime. And if this is v prime, then
if you were to take the antiderivative of both sides,
then v is equal to f of t. So let's apply integration
by parts. So this Laplace transform, which
is this, is equal to uv, which is equal to e to the minus
st, times v, f of t, minus the integral-- and, of
course, we're going to have to evaluate this from
0 to infinity. I'll keep the improper
integral with us the whole time. I won't switch back and forth
between the definite and indefinite integral. So minus this part. So the integral from 0 to
infinity of u prime. u prime is minus se to the
minus st times v-- v is f of t-- dt. Now, let's see. We have a minus and a
minus, let's make both of these pluses. This s is just a constant,
so we can bring it out. So that is equal to e to the
minus st, f of t, evaluated from 0 to infinity, or as we
approach infinity, plus s times the integral from 0 to
infinity of e to the minus st, f of t, dt. And here, we see,
what is this? This is the Laplace transform
of f of t, right? Let's evaluate this part. So when we evaluated in
infinity, as we approach infinity, e to the minus
infinity approaches 0. f of infinity-- now this is
an interesting question. f of infinity-- I don't know. That could be large, that
could be small, that approaches some value, right? This approach 0, so
we're not sure. If this increases faster than
this approaches 0, then this will diverge. I won't go into the mathematics
of whether this converges or diverges, but let's
just say, in very rough terms, that this will converge
to 0 if f of t grows slower than e to the minus
st shrinks. And maybe later on we'll do some
more rigorous definitions of under what conditions
will this expression actually converge. But let's assume that f of t
grows slower than e to the st, or it diverges slower than this
converges, is another way to view it. Or this grows slower
than this shrinks. So if this grows slower than
this shrinks, then this whole expression will approach 0. And then you want to subtract
this whole expression evaluated at 0. So e to the 0 is 1 times f of
0-- so that's just f of 0-- plus s times-- we said, this is
the Laplace transform of f of t, that's our definition--
so the Laplace transform of f of t. And now we have an interesting
property. What was the left-hand side of
everything we were doing? The Laplace transform
of f prime of t. So let me just write
all over again. And I'll switch colors. The Laplace transform of f prime
of t is equal to s times the Laplace transform of
f of t minus f of 0. And now, let's just extend
this further. What is the Laplace transform--
and this is a really useful thing to know--
what is the Laplace transform of f prime prime of t? Well, we can do a little pattern
matching here, right? That's going to be s times the
Laplace transform of its antiderivative, times the
Laplace transform of f prime of t, right? This goes to this, that's
an antiderivative. This goes to this, that's
one antiderivative. Minus f prime of 0, right? But then what's the Laplace
transform of this? This is going to be equal to s
times the Laplace transform of f prime of t, but what's that? That's this, right? That's s times the Laplace
transform of f of t, minus f of 0, right? I just substituted
this with this. Minus f prime of 0. And we get the Laplace transform
of the second derivative is equal to s squared
times the Laplace transform of our function, f
of t, minus s times f of 0, minus f prime of 0. And I think you're starting
to see a pattern here. This is the Laplace transform
of f prime prime of t. And I think you're starting
to see why the Laplace transform is useful. It turns derivatives into
multiplications by f. And actually, as you'll see
later, it turns integration to divisions by s. And you can take arbitrary
derivatives and just keep multiplying by s. And you see this pattern. And I'm running out of time. But I'll leave it up to you to
figure out what the Laplace transform of the third
derivative of f is. See you in the next video.