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# Laplace transform solves an equation 2

Video transcript

Welcome back. We're finally using the Laplace
Transform to do something useful. In the first part of this
problem, we just had this fairly straightforward
differential equation. And I know it's a little bit
frustrating right now, because you're like, this is such an
easy one to solve using the characteristic equation. Why are we doing Laplace
Transforms? Well I just want to show you
that they can solve even these problems. But later on there
are going to be classes of problems that, frankly, our
traditional methods aren't as good as the Laplace Transform. But anyway, how did
we solve this? We just took the Laplace
Transform of both sides of this equation. We got all of this hairy mess. We used the property of the
derivative of functions, where you take the Laplace Transform,
and we ended up, after doing a lot of algebra
essentially, we got this. We got the Laplace Transform of
y is equal to this thing. We just took the Laplace
Transform of both sides and manipulated algebraically. So now our task in this video
is to figure out what y's Laplace Transform
is this thing? And essentially what we're
trying to do, is we're trying to take the inverse Laplace
Transform of both sides of this equation. So another way to say it, we
could say that y-- if we take the inverse Laplace Transform
of both sides-- we could say that y is equal to the
inverse Laplace Transform of this thing. 2s plus 13, over s squared
plus 5s plus 6. Now we'll eventually actually
learn the formal definition of the inverse Laplace Transform. How do you go from the s
domain to the t domain? Or how do you go from
the frequency domain to the time domain? We're not going to worry
about that right now. What we're going to do is we're
going to get this into a form that we recognize,
and say, oh, I know those functions. That's the Laplace Transform
of whatever and whatever. And then we'll know what y is. So let's try to do that. So what we're going to use is
something that you probably haven't used since Algebra two,
which is I think when it's taught in, you know,
eighth, ninth, or 10th grade, depending. And you finally see it now in
differential equations that it actually has some use. Let me write it. We're going to use partial
fraction expansion. And I'll do a little primer
on that, in case you don't remember it. So anyway, let's just factor
the bottom part right here. And you'll see where I'm
going with this. So if I factor the bottom, I get
s plus 2 times s plus 3. And what we want to do, is we
want to rewrite this fraction as the sum of 2-- I
guess you could call it partial fractions. I think that's why it's called
partial fraction expansion. So we want to write this as a
sum of A over s plus 2, plus B over s plus 3. And if we can do this, then--
and bells might already be ringing in your head-- we know
that these things that look like this are the Laplace
Transform of functions that we've already solved for. And I'll do a little review
on that in a second. But anyway, how do we
figure out A and B? Well if we were to actually add
A and B, if we were to-- let's do a little aside right
here-- so if we said that A-- so if we were to give them a
common denominator, which is this, s plus 2 times s plus 3. Then what would A become? We'd have to multiply A
times s plus 3, right? So we'd get As plus 3A. This, as I've written it right
now, is the same thing as A over s plus 2. You could cancel out an s plus
3 in the top and the bottom. And now we're going to
add the B to it. So plus-- I'll do that in a
different color-- plus-- well, if we have this as the
denominator, we could multiply the numerator and
the denominator by s plus 2, right? To get B times s, plus
2B, and that's going to equal this thing. And all I did is I added
these two fractions. Nothing fancier than there. That was Algebra two. Actually, I think I should
do an actual video on that as well. But that's going to
equal this thing. 2s plus 13, all of that over
s plus 2 times s plus 3. Notice in all differential
equations, the hairiest part's always the algebra. So now what we do
is we match up. We say, well, let's add
the s terms here. And we could say that the
numerators have to equal each other, because the denominators
are equal. So we have A plus Bs plus 3A
plus 2B is equal to 2s plus B. So the coefficient on s, on
the right-hand side, is 2. The coefficient on the left-hand
side is A plus B, so we know that A plus
B is equal to 2. And then on the right-hand side,
we see 3A plus 2B must be equal to-- oh,
this is a 13. Did I say B? This is a 13. That's a 13. It looks just like a B, right? That was 2s plus 13. Anyway, so on the right-hand
side I get, it was 3A plus 2B is equal to 13. Now we have two equations
with two unknowns, and what do we get? I know this is very tiresome,
but it'll be satisfying in the end. Because you'll actually
solve something with the Laplace Transform. So let's multiply the top
equation by 2, or let's just say minus 2. So we get minus 2A minus
2B equals minus 4. And then we get-- add the two
equations-- you get A is equal to-- these cancel out--
A is equal to 9. Great. If A is equal to 9, what
is B equal to? B is equal to 9 plus
what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious
simplification. Because now we can rewrite this
whole expression as the Laplace Transform of y is equal
to A over s plus 2, is equal to 9 over s plus 2,
minus 7 over s plus 3. Or another way of writing it, we
could write it as equal to 9 times 1 over s plus 2, minus
7 times 1 over s plus 3. Why did I take the trouble
to do this? Well hopefully, you'll recognize
this was actually the second Laplace Transform
we figured out. What was that? I'll write it down here just
so you remember it. It was the Laplace Transform of
e to the at, was equal to 1 over s minus a. That was the second Laplace
Transform we figured out. So this is interesting. This is the Laplace
Transform of what? So if we were to take the
inverse Laplace Transform-- actually let me just
stay consistent. So that means that this is the
Laplace Transform of y, is equal to 9 times the Laplace
Transform of what? If we just do pattern matching,
if this is s minus a, then a is minus 2. So 9 times the Laplace
Transform of e to the minus 2t. Does that make sense? Take this, put it in this one,
which we figured out, and you get 1 over s plus 2. And let me clean this up a
little bit, because I'm going to need that real estate. I'll write this. I'll leave that there, because
we'll still use that. And then we have minus 7 times--
this is the Laplace Transform of what? This is the Laplace Transform
of e to the minus 3t. This pattern matching, you're
like, wow, if you saw this, you would go to your Laplace
Transform table, if you didn't remember it, you'd see this. You're like, wow, that looks
a lot like that. I just have to figure
out what a is. I have s plus 3. I have s minus a. So in this case, a is
equal to minus 3. So if a is equal to minus 3,
this is the Laplace Transform of e to the minus 3t. So now we can take the inverse
Laplace-- actually, before we do that. We know that because the Laplace
Transform is a linear operator-- and actually now I
can delete this down here-- we know that the Laplace Transform
is a linear operator, so we can
write this. And you normally wouldn't go
through all of these steps. I just really want to make you
understand what we're doing. So we could say that this is the
same thing as the Laplace Transform of 9e to the minus 2t,
minus 7e to the minus 3t. Now we have something
interesting. The Laplace Transform of y
is equal to the Laplace Transform of this. Well if that's the case, then
y must be equal to 9e to the minus 2t, minus 7e
to the minus 3t. And I never proved to you, but
the Laplace Transform is actually a 1:1 Transformation. That if a function's Laplace
Transform, if I take a function against the Laplace
Transform, and then if I were take the inverse Laplace
Transform, the only function whose Laplace Transform
that that is, is that original function. It's not like two different
functions can have the same Laplace Transform. Anyway, a couple of things
to think about here. Notice, we had that thing that
kind of looked like a characteristic equation
pop up here and there. And we still have to solve a
system of two equations with two unknowns. Those are both things that we
had to do when we solve an initial value problem, when we
use just traditional, the characteristic equation. But here it happened
all at once. And frankly it was a little bit
hairier because we had to do all this partial fraction
expansion. But it's pretty neat. The Laplace Transform got
us something useful. In the next video I'll actually
do a non-homogeneous equation, and show you that the
Laplace Transform applies equally well there. So it's kind of a more
consistent theory of solving differential equations, instead
of kind of guessing solutions, and solving for
coefficients and all of that. See you in the next video.