# "Shifting" transform by multiplying function byÂ exponential

## Video transcript

Now I think is a good time
to add some notation and techniques to our Laplace
Transform tool kit. So the first thing I want to
introduce is just kind of a quick way of doing something. And that is, if I had the
Laplace Transform, let's say I want to take the Laplace
Transform of the second derivative of y. Well, we proved several videos
ago that if I wanted to take the Laplace Transform of the
first derivative of y, that is equal to s times the Laplace
Transform of y minus y of 0. And we used this property in the
last couple of videos to actually figure out the Laplace
Transform of the second derivative. Because if you just, you know,
if you say this is y prime, this is the anti-derivative
of it, then you could just pattern match. You could say, well, the Laplace
Transform of y prime prime, that's just equal to s
times the Laplace Transform of y prime minus y prime of 0. This is the derivative of this,
just like this is the derivative of this. I'll draw a line here just so
you don't get confused. So the Laplace Transform of y
prime prime is this thing. And now we can use this, which
we proved several videos ago, to resubstitute it and get it
in terms of the Laplace Transform of y. So we can expand this part. The Laplace Transform of the
derivative of y, that's just equal to s times the Laplace
Transform of y minus y of 0. And then we have the
outside, right? We have s minus y prime of 0. And then when you expand it all
out, and we've done this before, you get s squared times
the Laplace Transform of y minus s times y of 0
minus y prime of 0. Now there's something
interesting to note here, and if you learn this it'll
make it a lot faster. You won't have to go through
all of this and risk making careless mistakes when you have
scarce time and paper on your test. Just notice that
when you take the Laplace Transform of the second
derivative, what do we end up? We end up with s
squared, right? This was the second
derivative. So I end up with s squared times
the Laplace Transform of y, minus s times y of 0 minus
1 times y prime of 0. So every term, we started with s
squared, and then every term we lower the degree of s one,
and then everything except the first term is a negative sign. And then we started with the
Laplace Transform of y, and then you can almost view the
Laplace Transform as a kind of integral, so we kind of take
the derivatives, so then you get y. And then you take
the derivative again, you get y prime. And of course every other
term is negative. And these aren't the
actual functions. These are those functions
evaluated at 0. But that's a good way to help
you, hopefully, remember how to do these. And once you get the hang of it,
you can take the Laplace Transform of any arbitrary
function very, very quickly. Or any arbitrary derivatives. So let's say we wanted to take
the Laplace Transform of, I don't know, this should hit
the point home, the fourth derivative of y. That 4 in parentheses means
the fourth derivative. I could have drawn four prime
marks, but either way. So what is this equal to? If we use this technique and
substitute it, we're bound to make some form of careless
mistake or other, and it would take us forever and it would
waste a lot of paper. But now we see the pattern, and
so we can just say, well, the Laplace Transform of this,
in terms of the Laplace Transform of y, right, that's
what we want to get to, is going to be s to the fourth
times the Laplace Transform of y-- now every other term is
going to have a minus in front of it-- minus-- lower the
degree on the s-- minus s to the third. And then you could kind of say,
let's take the, you know, so form of derivative, so you
get y of 0-- it's not a real derivative. The Laplace Transform really
isn't the anti-derivative of y of 0, but anyway, I think
you get the idea. And then we lower the degree on
s again, minus s squared, take the derivative. And of course these
aren't functions. But we're evaluating the
derivative of that function now of 0. So y prime of 0, minus-- now we
lower the degree one more-- minus s, times-- this is an s--
times y prime prime of 0. We have one more term. Lower the degree on the
s one more time. Then you get s to the
0, which is just 1. So minus-- and 1 is a
coefficient-- and then you have y, the third derivative
of y-- let me scroll over a little bit-- the third
derivative of y evaluated at 0. So I think you see
the pattern now. And this is a much faster way
of evaluating the Laplace Transform of an arbitrary
derivative of y, as opposed to keep going through that pattern
over and over again. Another thing I want to
introduce you to is just a notational savings. And it's just something that
you'll see, so you might as well get used to it. And it actually saves time over,
you know, keep writing this curly L in this bracket. If I have the Laplace Transform
of y of t, I can write as, and people tend to
write it as-- well, it's going to be a function of s, and what
they use is a capital Y to denote the function of s. It makes sense, because normally
when we're doing antiderivatives, you just take--
you know, when you learn the fundamental theorem
of calculus, you learn that the integral of f with respect
to dx, you know, from 0 to x, is equal to capital F of x. So it's kind of borrowing that
notation, because this function of s is kind of
an integral of y of t. The Laplace Transform, to some
degree, is like a special type of integral where you have a
little exponential function in there to mess around with
things a little bit. Anyway, I just wanted you to
get used to this notation. When you see capital Y of s,
that's the same thing as a Laplace Transform of y of t. And you might also
see it this way. The Laplace Transform of f of t
is equal to capital F of s. And the clue that tells you that
this isn't just a normal antiderivative, is the fact that
they're using that s as the independent variable. Because in general, s represents
the frequency domain, and if people were to
use s with just a general antiderivative, people
would get confused, et cetera, et cetera. Anyway, I'm trying to think
whether I have time to teach you more fascinating concepts
of Laplace Transform. Well, sure, I think we do. So my next question for you--
and now we'll teach you a couple more properties, and
this'll be helpful in taking Laplace Transforms. What is the
Laplace Transform of e to the at times f of t? Fascinating. Well, let's just should go back
to our definition of the Laplace Transform. It is the integral from 0 to
infinity of e to the minus st times whatever we have between
the curly brackets. So, with the curly brackets we
have e to the at f of t dt. And now we can add
these exponents. We have a similar base, so
this is equal to what? This is equal to the integral
from 0 to infinity. And let's see, I want to write
it as, I could write it minus s plus a, but I'm going to write
it as minus s minus a t. And you could expand this out. It becomes minus s plus a, which
is exactly what we have here, times f of t dt. Now let me show you something. if I were to just take the
Laplace Transform of f of t, that is equal to some
function of s. Whatever we essentially have
right here for s, it becomes some function of that. So this is interesting. This is some function of s. Here, all we did to go
from-- well actually let me rewrite this. The Laplace, which is equal to
0 to infinity e to the minus st f of t dt. The Laplace Transform of just f
of t is equal to this, which is some function of s. Well, the Laplace Transform of
e to the at, times f of t, it equals this. And what's the difference
between this and this? What's the difference
between the two? Well, it's not much. Here, wherever I have an s,
I have an s minus a here. So if this is a function of s,
what's this going to be? It's going to be that
same function. Whatever the Laplace Transform
of f was, it's going to be that same function, but instead
of s, it's going to be a function of s minus a. And once again, how
did I get that? Well I said the Laplace
Transform of f is a function of s, and it's equal to this. Well if I just replace an s with
an s minus a, I get this, which is a function
of s minus a. Which was the Laplace Transform
of e to the at times f of t. Maybe that's a little
confusing. Let me show you an example. Let's just take the Laplace
Transform of cosine of 2t. We've shown is equal to-- well
I'll write the notation-- it's equal to some function of s. And that function of s is
s over s squared plus 4. We've shown that already. And so the Laplace Transform of
e to the, I don't know, 3t times cosine of 2t, is going
to be equal to the same function, but instead of
s, it's going to be a function of s minus a. So s minus 3, which is equal
to s minus 3 over s minus 3 squared plus 4. Notice, when you just multiply
something by this, either the 3t and then or either the at,
you take the Laplace Transform of it, you just-- it's the
same thing as the Laplace Transform of this function, but
everywhere where you had an s, you replace it with
an s minus this a. Anyway, I hope I didn't
confuse you too much with that last part. I think my power adaptor
actually just went on. I hope the video keeps
recording. I'll see you in the next one.