# Using the convolution theorem to solve an initial valueÂ prob

## Video transcript

Now that we know a little bit
about the convolution integral and how it applies to the
Laplace transform, let's actually try to solve an actual
differential equation using what we know. So I have this equation here,
this initial value problem, where it says that the second
derivative of y plus 2 times the first derivative of y, plus
2 times y, is equal to sine of alpha t. And they give us some
initial conditions. They tell us that y of 0 is
equal to 0, and that y prime of 0 is equal to 0. And that's nice and convenient
that those initial conditions tend to make the problem
pretty clean. But let's get to the problem. So the first thing we do is we
take the Laplace transform of both sides of this equation. The Laplace transform of
the second derivative of y is just s squared. This should be a bit of second
nature to you by now. It's s squared times the Laplace
transform of Y, which I'll just write as capital Y
of s, minus s-- so we start with the same degree as the
number of derivatives we're taking, and then we decrement
that every time-- minus s times y of 0-- you kind of think
of this as the integral, and you take the derivative 1,
so this isn't exactly the derivative of that-- minus, you
decrement that 1, you just have a 1 there, y prime of o. And that's the Laplace transform
of the second derivative. Now, we have to do the Laplace
transform of 2 times the first derivative. That's just going to be equal
to plus 2, times sY of s-- s times the Laplace transform
of Y; that's that there-- minus y of 0. And we just have one left. The Laplace transform of 2Y. That's just equal to plus
2 times the Laplace transform of Y. And then that's going to be
equal to the Laplace transform of sine of alpha t. We've done that multiple
times so far. That's just alpha over s squared
plus alpha squared. Now, the next thing we want to
do is we want to separate out the Laplace transform of Y
terms, or the Y of s terms. Actually, even better, let's
get rid of these initial conditions. y of 0, and y prime of 0 is
0, so this term is 0. That term is 0, and
that term is 0. So our whole expression-- I
can get rid of the colors now-- it just becomes-- let me
pick a nice color here-- becomes s squared times Y of
s, plus 2s, Y of s-- that's that term right there-- plus
2Y of s, is equal to the right-hand side, is equal to
alpha over s squared plus alpha squared. Now let's factor out the
Y of s, or the Laplace transform of Y. And so we get s squared plus 2s,
plus 2, all of that times Y of s, is equal to this
right-hand side, is equal to alpha over s squared,
plus alpha squared. Now we can divide both sides of
this equation by this thing right here, by that
right there. And we get Y of s, the Laplace
transform of Y is equal to this thing, alpha over s
squared, plus alpha squared, times-- or, you know, I could
just say times-- 1 over s squared, plus 2s, plus 2. I could just say divided by
this, but it works out the same either way. Now, what can we do here? Remember, I was doing this in
the context of convolution, so I want to look for a Laplace
transform that looks like the product of two Laplace
transforms. I know what the inverse Laplace transform
of this is. In fact, I just took it. It's sine of alpha t. So if I can figure out the
inverse Laplace transform of this, I could at least express
our function y of t at least as a convolution integral, even
if I don't necessarily solve the integral. From there, it's just calculus,
or if it's an unsolvable integral, we could
just use a computer or something, although you could
actually use a computer to solve this so, you might
skip some steps even going through this. But anyway, let's just try
to get this in terms of a convolution integral. So what can I do with this? This is, let's see, this
isn't a perfect square. So if this isn't a perfect
square, the next best thing is to try to complete
the square here. So let's try to write this as
a, so let's see, if I write this as s squared plus 2s,
plus something, plus 2. I just rewrote it like this. And if I wrote this as s squared
plus 2s, plus 1, that becomes s plus 1 squared. But if I add a 1, I have
to also subtract a 1. I can't just add 1's arbitrarily
to things. So if I add 1 I have to subtract
a 1 to cancel out with that 1. So I really haven't changed
this at all, I just rewrote it like this. But this now, I can rewrite
this term right here as s plus 1 squared. And then this becomes plus 1. That's this term right here. This is the plus 1. So I could rewrite my whole Y
of s is now equal to alpha over s squared, plus alpha
squared, times 1 over this thing, s plus 1 squared,
plus 1. Now, I already said, I know
what the inverse Laplace transform of this thing is. Now I just have to figure out
what the inverse Laplace transform of this thing is. Of this-- let me pick a nice
color-- of this blue thing in the blue box, and then
I can express it as a convolution integral. And how do I do that? I could just do it right now. I could just immediately say
that y of t-- let me write this down-- y of t, so the
inverse is equal to the inverse Laplace transform
of, obviously of Y of s. Let me write that
down, Y of s. Which is equal to the inverse
Laplace transform of these two things. The inverse Laplace transform of
alpha over s squared, plus alpha squared, times 1 over
s plus 1 squared, plus 1. And now the convolution theorem
tells us that this is going to be equal to the inverse
Laplace transform of this first term in
the product. So the inverse Laplace transform
of that first term, alpha over s squared, plus
alpha squared, convoluted with-- I'll do a little
convolution sign there. I was about to say convulsion. They're not too different. Convoluted with the inverse
Laplace transform of this term, the inverse Laplace
transform of 1 over s plus 1 squared, plus 1. If I have the product of two
Laplace transforms, and I can take each of them independently
and I can invert them, the inverse Laplace
transform of their product is going to be the convolution of
the inverse Laplace transforms of each of them, each
of the terms. And what I just said confused
me a bit, so I don't want to confuse you. But I think you get the idea. I have these two things. I recognize these
independently. I can independently take the
inverse of each of these things, so the inverse Laplace
transform of their products is going to be the convolution
of each of their inverse transforms. Now what's
this over here? Well I had this in the beginning
of the problem? The inverse Laplace transform
of this, right here, is sine of alpha t. And then we're going to
convolute that with the inverse Laplace transform
of this right here. Let's do a little bit of work on
the side, just to make sure we get this right. So the Laplace transform of sine
of t is equal to 1 over s squared, plus 1. That looks like this, but I
was shifted by minus 1. You might remember that the
Laplace transform of e to the at sine of t, when you multiply
e to the at times anything, you're shifting
its Laplace transform. So that will be equal
to 1 over s minus a squared, plus 1. We essentially shifted
it by a. So now we have something that
looks very similar to this. If we just set our a to be equal
to negative 1, here our a is equal to negative 1, then
it fits this pattern. This is s minus negative 1. So the inverse Laplace transform
of this thing right here is just e to the a, which
is minus 1, so minus 1t, times sine of t. So this is the solution to our
differential equation, even though it's not in a pleasant
form to look at. And we can, if we want, express
it as an integral. I'm not going to actually solve
the integral in this problem, because it gets hairy,
and it's not even clear that-- well, I won't even
attempt to do it. But I just want to get into a
form, and from there it's just integral calculus. or maybe a computer. What's the convolution
of these two things? It's the integral from 0 to t,
of sine of the first function of t minus tau. Well, I could actually switch,
and I haven't shown you this, but we can switch the order
either way, but actually let me just do it this way. I could write this as
sine of [? out ?] t minus tau, times alpha-- I'm
taking the sine of all of those things-- times e to the
minus tau, sine of tau, dtau. That's one way, that if I wanted
to express the solution of this differential equation's
integral, I could write it like that. And it actually should be kind
of obvious to you that this could go either way. Because when it was a product up
here, obviously order does not matter. I could write this term first,
or I could write that term first. So regardless of which
term is written first, the same principle would apply. And I'll formally prove
it in a future video. So we could have also done
it the other way. We could have written this
expression as e to the minus t, sine of t, convoluted
with sine of alpha t. And that would be equal to the
integral from 0 to t, of e to the minus t minus tau, sine of
t, minus tau, times sine of alpha tau, dtau. So these are equivalent. Either of these would be
an acceptable answer. And normally on a test like
this, the teacher won't expect you to actually evaluate
these integrals. The teacher will just say, get
it into an integral just to kind of see whether you know
how to convolute things and get your solution to the
differential equation at least into this form, because from
here it really is just, I won't say basic calculus,
but it's non-differential equations. So hopefully, this second
example with the convolution to solve an inverse transform
clarified things up a little bit.