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.