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# Laplace transform to solve an equation

Video transcript

I've been doing a ton of videos
on the mechanics of taking the Laplace Transform,
but you've been sitting through them always wondering,
what am I learning this for? And now I'll show you, at
least in the context of differential equations. And I've gotten a
bunch of letters on the Laplace Transform. What does it really mean? And and all that. And those are excellent
questions and you should strive for that. It's hard to really have an
intuition of the Laplace Transform in the differential
equations context, other than it being a very useful tool that
converts differential or integral problems into algebra
problems. But I'll give you a hint, and if you want a path
to learn it in, you should learn about Fourier series and
Fourier Transforms, which are very similar to Laplace
Transforms. And that'll actually build up the intuition
on what the frequency domain is all about. Well anyway, let's actually use
the Laplace Transform to solve a differential equation. And this is one we've
seen before. So let me see. So let's say the differential
equation is y prime prime, plus 5, times the first
derivative, plus 6y, is equal to 0. And you know how to solve this
one, but I just want to show you, with a fairly
straightforward differential equation, that you could solve
it with the Laplace Transform. And actually, you
end up having a characteristic equation. And the initial conditions are
y of 0 is equal to 2, and y prime of 0 is equal to 3. Now, to use the Laplace
Transform here, we essentially just take the Laplace Transform
of both sides of this equation. Let me use a more
vibrant color. So we get the Laplace Transform
of y the second derivative, plus-- well we
could say the Laplace Transform of 5 times y prime,
but that's the same thing as 5 times the Laplace Transform--
y prime. y prime plus 6 times the
Laplace Transform of y. And let me ask you a question. What's the Laplace
Transform of 0? Let me do that. So the Laplace Transform of 0
would be be the integral from 0 to infinity, of 0 times
e to the minus stdt. So this is a 0 in here. So this is equal to 0. So the Laplace Transform
of 0 is 0. And that's good, because I
didn't have space to do another curly L. So what are the Laplace
Transforms of these things? Well this is where we break
out one of the useful properties that we learned. Let me write it over here. I think that's going
to need as much real estate as possible. Let me erase this. So we learned that the Laplace
Transform-- I'll do it here. Actually, I'll do
it down here. The Laplace Transform of f
prime, or we could even say y prime, is equal to s times
the Laplace Transform of y, minus y of 0. We proved that to you. And this is extremely
important to know. So let's see if we
can apply that. So the Laplace Transform of y
prime prime, if we apply that, that's equal to s times the
Laplace Transform of-- well if we go from y prime to y,
you're just taking the anti-derivative, so if you're
taking the anti-derivative of y, of the second derivative, we
just end up with the first derivative-- minus the first
derivative at 0. Notice, we're already using
our initial conditions. I won't substitute
it just yet. And then we end up with plus 5,
times-- I'll write it every time-- so plus 5 times the
Laplace Transform of y prime, plus 6 times the Laplace
Transform of y. All of that is equal to 0. So just to be clear, all I did
is I expanded this into this using this. So how can we rewrite the
Laplace Transform of y prime? Well, we could use this once
again, so let's do that. So this over here-- I'll do it
in magenta-- this is equal to s times what? s times the Laplace Transform
of y prime. Well that's s times the Laplace
Transform of y, minus y of 0, right? I took this part and replaced
it with what I have in parentheses. So minus y prime of 0-- and now
I'll switch colors-- plus 5 times-- once again the Laplace
Transform of y prime. Well we can use this again. So 5 times s times Laplace
Transform of y, minus y of 0, plus 6 times the Laplace
Transform-- oh I ran out of space, I'll do it in another
line-- plus 6 times the Laplace Transform of y. All of that is equal to 0. I know this looks really
confusing but we'll simplify right now. And we could get rid of this
right here, because we've used it as much as we need to. So now we just simplify. And notice, using the Laplace
Transform, we didn't have to guess at a general solution
or anything like that. Even when we did a
characteristic equation, we guessed what the original
general solution was. Now we're just taking Laplace
Transforms, and let's see where this gets us. And actually I just want to
make clear, because I know it's very confusing, so I
rewrote this part as this. And I rewrote this
thing as this. And everything else
is the same. But now let's simplify
the math. So we get s squared, times the
Laplace Transform of y-- I'm going to write smaller, I've
learned my lesson-- minus s times y of 0. Let's substitute y of 0 here. y
of 0 is 2, so s times y of 0 is 2 times s, so 2s, distribute
that s, minus y prime of 0. Y prime of 0 is 3. So minus 3, plus-- so we have
5 times s times the Laplace Transform of y, so plus 5s times
the Laplace Transform of y, minus 5 times y of 0. y
of 0 is 2, so minus 10. Minus 10, right? 5 times-- this is 2 right here--
so 5 times 2, plus 6 times the Laplace
Transform of y. All of that is equal to 0. Now, let's group our Laplace
Transform of y terms and our constant terms, and we
should be hopefully getting some place. So let's see, my Laplace
Transform of y terms, I have this one, I have this one,
and I have that one. So what am I left with? Well let me factor out the
Laplace Transform of y part. So I get the Laplace Transform
of y-- and that's good because it's a pain to keep writing
it over and over-- times s squared plus 5s plus 6. So those are all my Laplace
Transform terms. And then I have my constant terms. So let's
see, I have 1s, so minus 2s, minus 3, minus 10,
is equal to 0. And what can we do here? Well, this is interesting,
first of all. Notice that the coefficients on
the Laplace Transform of y terms, that those are that
characteristic equation that we dealt with so much, and that
is hopefully, to some degree, second nature to you. So that's a little bit of a
clue, and if you want some very tenuous connections, well
that makes a lot of sense. Because the characteristic
equation to get that, we substituted e to the rt, and the
Laplace Transform involves very similar function. But anyway, let's go back
to the problem. So how do we solve this? And actually, let me just give
you the big picture here, because this is a good point. What I'm going to do is I'm
going to solve this. I'm going to say the Laplace
Transform of y is equal to something. And then I'm going to say,
boy, what functions the Laplace Transform
is at something? And then I'll have
the solution. If that confuses you, just wait
and hopefully it'll make some sense. From here until that point
it's just some fairly hairy algebra. So let's scroll down a little
bit, just so we have some breathing room. And so I get the Laplace
Transform of y, times s squared, plus 5s, plus 6, is
equal to-- let's add these terms to both sides of this
equation-- is equal to 2s plus 3 plus 10-- oh, that's
silly-- plus 13. This is minus 13 here. A phone call. Who's calling? I think it's some kind of
marketing phone call. Anyway, 2s plus 13, and
now what can I do? Well. Let's divide both sides by this
s squared plus 5s plus 6. So I get the Laplace Transform
of y is equal to 2s plus 13, over s squared plus 5s plus 6. Now we're almost done. Everything here is just a
little bit of algebra. So now we're almost done. We haven't solved for y yet, but
we know that the Laplace Transform of y is
equal to this. Now, if we just had this in
our table of our Laplace Transforms, we would immediately
know what y was, but I don't see something, or
I don't remember anything we did in our table that looks
like this expression of s. I'm essentially out of time, so
the next video we're going to figure out what functions
Laplace Transform is this. And it actually turns out it's
a sum of things we already know, and we just have to
manipulate this a little bit algebraically. See you in the next video.