In the last video, we showed
that the Laplace transform of f prime of t is equal to s times
the Laplace transform of our function f minus f of 0. Now, what we're going to do
here is actually use this property that we
showed is true. And use it to fill in some more
of the entries in our Laplace transform table, that
you'll probably have to memorize, sooner or later,
if you use Laplace transforms a lot. But we already learned that the
Laplace transform of sine of a t is equal-- and we did
a very hairy integration by parts problems to show that
that is equal to a over s squared plus a squared. So let's use these two things we
know to figure out what the Laplace transform of
cosine of a t is. So the Laplace transform
of cosine of a t is equal to what? Well, if we assume that the
Laplace transform of cosine of a t is the derivative of some
function, what is it the derivative of? Right? Let me do it on the side. If f prime of t is equal to
cosine of a t, what is a potential f of t? Well, it's the antiderivative. And we can just forget about the
constant, because we just have to know nf of t for
which this is true. So what's the antiderivative
cosine of it? It's 1/a sine of a t. Right? So if this is f prime of t, then
that is equal to s times the Laplace transform of its
antiderivative, or 1/a sine of a t minus the antiderivative
evaluated at 0. Minus 1/a sine of-- well,
a times 0 is 0. Well, sine of 0 is 0. So this whole term goes away. This is a constant, right? This 1/a. And we showed that the Laplace transform is a linear operator. So we can take it out. So this is equal to s/a times
the Laplace transform of sine of a t. And that is equal to
s/a times a over f squared plus a squared. And the a's cancel out. And that was much simpler than
the integration by parts we had to do to figure this out. So then we get that the Laplace
transform of cosine of a t is equal to s over s
squared plus a squared. And in three minutes, we filled
in another table in our Laplace transform table. And now we have the two most
important trig functions. Let's keep going. We haven't really done much
with polynomials. We know a couple of things. We did this already. We know that the Laplace
transform of 1 is equal to 1/s. So let's see if we could use
this and the fact that the Laplace transform of f prime
is equal to s times the Laplace transform of
f minus f of 0. Or another way. Let's rearrange this. If we know f, how can we
figure out some Laplace transforms in terms of
f prime and f of 0? So let's just rearrange
this equation. So we get the Laplace transform
of f prime. I could write of t, but
that gets monotonous. Plus f of 0 is equal to
s times the Laplace transform of f. Divide both sides by s. Let me put the Laplace transform
of-- and I'm also going to the sides. So I guess the Laplace
transform-- my l's are getting funky. The Laplace transform of
f is equal to 1/s. I'm just dividing both sides
by s, so 1/s times this. Times the Laplace transform of
my derivative plus my function evaluated at 0. And let's see if we can use this
and this to figure out some more useful Laplace
transforms. Well what is the Laplace
transform of f of t is equal to t? Well, just use this property. This is going to be equal
to 1/s times the Laplace transform of the derivative. Well, what's the derivative
of t? The derivative of t is 1. So it's the Laplace transform
of 1 minus f of 0. When t equals 0,
this becomes 0. Minus 0. So the Laplace transform of t
is equal to 1/s times the Laplace transform of 1. Well that's just 1/s. So it's 1 over s squared
minus 0. Interesting. The Laplace transform of 1 is
1/s, Laplace transform of t is 1/s squared. Let's figure out what the
Laplace transform of t squared is. And I'll do this one in green. And maybe we'll see
a pattern emerge. The Laplace transform
of t squared. Well, it equals 1/s times the
Laplace transform of it's derivative. So what's it's derivative? Times the Laplace transform of
2t plus this evaluated at 0. Well, that's just 0. So this is equal to-- well
we can just take this constant out. This is equal to 2/s times the
Laplace transform of t. Well, what does that equal? We just solved it. 1/s squared. So it's 2/s times 1/s squared. So it's equal to 2/s
to the third. Fascinating. Well, let me just
do t the third. And I think then you'll
see the pattern. The pattern will emerge. The Laplace transform. And this is actually
kind of fun. I recommend you do it. It's somehow satisfying. It's much more satisfying than
integration by parts. So the Laplace transform of t to
the third is 1/s times the Laplace transform of it's derivative, which is 3t squared. Which is-- take the constant
out because it's a linear operator. 3/s times the Laplace transform
of t squared. So it equals what? What's the Laplace transform
of t squared? It's 2/s to the third. So this equals 3 times 2 over
what? s to the fourth. And you can put a t/n here and
use an inductive argument to figure out a general formula. And that general formula
is-- I think you see the pattern here. Whatever my exponent is, the
Laplace transform has an s in the denominator with one
larger exponent. And then the numerator is the
factorial of my exponent. So in general, and this is one
more entry in our Laplace transform table. The Laplace transform of t to
the nth power is equal to n factorial over s to
the n plus 1. That's a parenthesis. I guess I didn't have to write
those parenthesis. That just confuses it. But anyway, when you see this in
a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n
factorials and all of that. But it's just saying with this
pattern we showed, t to the third increase it by 1, so
s the fourth, put another denominator and take three
factorial on the numerator, which is 6, right? And that's all it is. So using the derivative property
of Laplace transform, we figured out the Laplace
transform of cosine of a t and the Laplace transform of really
any polynomial, right? Because it's a linear
operator. So now we know t to the
nth power, t to the whatever power. And we can multiply
it by a constant. So we know the basic
trig functions. We know the exponential
function. And we know how to take the
Laplace transform of polynomials. See you in the next video.