GILBERT STRANG: OK. This is the start of
Laplace transforms. And that's going to take
more than one short video. But I'll devote this video to
first order equations, where the steps are easy
and pretty quick. Then will come second
order equations. So Laplace transforms
starting now. So let me tell you
what-- I use a capital letter for the Laplace transform
of little f, a function of t. The transform is capital
F, a function of s. And you'll see where s comes in. Or if it's the solution
I'm looking at, y of t, its transform
is naturally called capital Y of s. So that's what we want-- we
want to find y, and we know f. OK. So can I do an example? Well, first tell you what
the Laplace transform is. Suppose the function is f of t. Here is the transform. I multiply by e to the minus
st, and I integrate from 0 to infinity. 0 to infinity. Very important. The function doesn't
start until t equals 0, but it goes on to
t equal infinity. I integrate, and
when I integrate, t disappears but
s is still there. So I have a function of s. Well, I have to do an example. So to find the Laplace transform
is to do an integration. And you won't be surprised
that the good functions we know are the ones where we
can do the integration and discover the transform,
and make a little table of nice transforms. And the number one function
we know is the exponential. So can I find--
for that function, I'll compute its transform. So what do I have to do? I have to integrate
from 0 to infinity-- you might say 0 to
infinity is hard, but it's actually the
best-- of my function, which is e to the at. So that's my function
times e to the minus st dt. OK. I can do that integral,
because those combine into e to the a minus st. I
can put those together into e to the a
minus st. I integrate so I get e to the a minus
st divided by a minus s. That's the integral of that. Because what I have
in here is just that. To integrate the exponential,
I just divide by the exponent there. And I have just substitute t
equal infinity and t equal 0. So t equal infinity,
starting at 0 to infinity. OK. Infinity is the nice one. It's the easy one. I will look only at s's
that are bigger than a. s larger than a means that this
exponential is decreasing to 0. It gets to 0 at
t equal infinity. So at t equal infinity, that
upper limit of the integral ends up with a 0. So I just have to
subtract the lower limit. And look how nice. Now I put in t equal 0. Well, then that becomes 1. And it's a lower limit, so
it comes with a minus sign. So it's just the 1
over, the minus sign will flip that s minus a. The most important Laplace
transform in the world. Remember, the function
was in to the at. The transform is
a function of s. The original function depended
on t and a parameter a. The result depends on
s and a parameter a. And an engineer would say,
here we have the exponent. The growth rate is a. And over in the transform--
so this is the transform, remember. This is the transform f of x. In the transform,
I see blow up-- a pole, that's called
a pole-- at s equal a. 1/0 is a pole. And I'm not surprised. So the answer is
blowing up at s equal a. Well, of course. If s equals a, then this
is the integral of 1 from 0 to infinity, and it's infinite. So I'm not surprised to
see the pole showing up. The blow up showing up
exactly at the exponent a. But this is a nice transform. OK. I need to do one other-- oh, no. I could already
solve the equation. So let me start
with the equation dy dt minus ay equal 0. Oh, well, I can take the
Laplace transform of 0 is 0, safe enough. The Laplace transform of
y is capital Y. But what's the transform of this? Oh, I have to do one
more transform for you. I'm hoping that the transform
of the derivative, dy dt, connects to the transform of y. So the transform of this
guy is the integral from 0 to infinity of that function,
whatever it is, times e to the minus st dt. This is the transform. So this Laplace transform. Now what can I do
with that integral? This is a step that goes back
to the beginning of calculus. But it's easy to forget. When you see a derivative
there inside that integral, you think, I could
integrate by parts. I could integrate
that term and take the derivative of that term. That's what integration
by parts does. It moves the derivative
away from that and onto that where it's no problem. And do you remember that a minus
sign comes in when I do this? So I have the integral
from 0 to infinity of-- now the derivative is coming off
of that, so that's just y of t. And the derivative
is going onto that, so that's minus se
to the minus st dt. Good. And then do you remember
in integration by parts, there's also another term
that comes from y times e to the minus st? This is ye to the minus
st at 0 and infinity. OK. I've integrated by parts. A very useful, powerful
thing, not just a trick. OK. Now, can I recognize
some of this? That is minus minus, no problem. I bring out-- that
s is a constant. Bring it out, s. Now, what do I have left
when I bring out that s? I have the integral of
ye to the minus st dt. That is exactly the
Laplace transform of y. It's exactly capital Y. Put the equal sign here. I'll make that 0 a little
smaller, get it out of the way. OK. sY of s. So that whole term
has a nice form. When you take the
derivative of a function, you multiply its
Laplace transform by s. That's the rule. Take the derivative
of the function, multiply the Laplace
transform by s. If we have two derivatives,
we'll multiply by s twice. Easy. That's why the Laplace
transform works. But now, here is a final term. y at infinity-- well,
and e to the minus st at t equal infinity, 0. Forget it. So I just have to subtract off
y at 0 times e to the minus st at 0, which is 1. e to the 0 is 1. So do you see that
the initial condition comes into the transform? It's like, great. We have the transform of Y.
Now, all this is the transform. This is the transform
of dy dt that we found. Now, why did I want that? Because I plan to
take the transform of every term in my equation. So like there are two steps to
using the Laplace transform. One is to compute some
transforms like this one, and some rules like this one. That's the preparation step. That comes from just
looking at these integrals. And then to use them, I'm
going to take the Laplace transform of every term. So I have an equation. I take the Laplace
transform of every term. I've got another equation. So the Laplace transform of
this is sY of s minus y of 0. That was a Laplace
transform of this part. Now the Laplace transform
of this is minus a, a constant, Y of x. And the Laplace
transform of 0 is 0. Do you realize what we've done? I've taken a
differential equation and I've produced
an algebra equation. That's the point of
the Laplace transform, to turn differential
equations-- derivatives turn into multiplications, algebra. So all the terms
turn into that one. And now comes-- so
that's big step one. Transform every term. Get an algebra
problem for each s. We've changed from t, time
in the differential equation, to s in Laplace transform. Now solve this. So how am I going to solve that? I'm going to put y of 0
on to the right-hand side. And then I have Y of
s times s minus a. So I will divide by s minus a. And that gives me Y of s. So that was easy to do. The algebra problem
was easy to solve. The differential
equation more serious. OK. The algebra problem is easy. Are we finished? Got the answer, but we're in
the s variable, the s domain. I've got to get back
to-- so now this is going to be an inverse
Laplace transform. That's the inverse transform. To give me back y
of t equals what? How am I going to do
the inverse transform? So now I have the
transform of the answer, and I want the answer. I have to invert that transform
and get out of s and back to t. Well, y of 0 is a constant. Laplace transform is
linear, no problem. So have y of 0 from that. And now I have 1 over s minus a. So I'm asking myself, what is
the function whose transform is 1 over s minus a? Then it's that function
that I want to put in there. And what is the function whose
transform is 1 over s minus a? It's the one we did. It's this one up here. 1 over s minus a came from
the function e to the at. So that 1 over s minus
a, when I transform back, is the e to the at. And I'm golden. And that you recognize, of
course, as the correct answer, correct solution to this
differential equation. The initial value y of 0
takes off with exponential e to the at. No problem. OK. Can I do one more example
of a first order equation? Now I'm going to
put it in an f of t. I'm going to put
in a source term . So I'll do all the
same stuff, but I'm going to have an f of t. And what shall I take for-- I'll
take an exponential again, e to the ct. So that's my right-hand side. Can I do the same
idea, the central idea? Take my differential equation,
transform every term. I've started with
a time equation and I'm going to
get an s equation. So again, dy dt minus ay, that
transformed to-- what did that transform to? sY of s minus y of 0. Came from there. Minus aY of 0-- minus aY of s. Minus aY of s. And on the right-hand side,
I have the transform of e to the ct. We're getting good
at this transform. 1 over s minus c,
instead of a at c. OK. That's our equation transform. Now algebra. I just pull Y of s out of that. How am I going to pull Y
of s out of this equation? Well, I'll move y of
0 to the other side. And I'll divide by s minus a. Look at it. Y of s is-- OK. I have 1 over s minus c. And I have an s minus
a that I'm dividing by. S minus a. And then I have the y
of 0 over s minus a. I've transformed the
differential equation to an s equation. I've just done simple algebra
to solve that equation. And I've got two terms. Two terms. You see that term? That's what I had before. That's what I had just there. The inverse transform was this. No problem. That's the null solution that's
coming out of the initial value . The new term that's coming
from the e to the ct, coming from the
force, is this one. And I have to do its
inverse transform. I have to figure out what
function has that transform. And you may say,
that's completely new. But we can connect it
to the one we know. OK. So that will give me the same
inverse transform, the growing exponential. But what does this one give? That's a key question. We have to be able to do--
invert, figure out what function has that transform? The function will involve
a and c and t, the time. And s, the transform variable,
will become t, the time period. So that's the big question. What do I do with this? And notice, it has two poles. It blows up at s equal a,
and it blows up at s equal c. And I have to figure out--
well, actually, by good luck, I want to separate
those two poles. Because if I separate
the two poles, I know what to do with a blow up
at s equal a and a blow up at s equal c. The problem is, right now I
have both blow ups at once. So I'm going to separate that. And that's called
partial fractions. So I will have to say more
about partial fractions. Right now, let me just do it. That expression there,
I'll take this guy away. Because it gives that
term that we know. It's this one. Is that one. It's this two poles
thing that I want to separate those two poles. So this is algebra again. Partial fractions
is just algebra. No calculus. No derivatives are in here. I just want to write
that as 1 over s minus c. It turns out-- look. There's a way to
remember the answer. s minus c times c minus
a and 1 over s minus a. And now a minus c. Do you see that that has
only one pole at s equals c? This is just a number. This has one pole at s equals a. That's just a number. In fact, those numbers
are the opposite. So now, are we golden? I can take the inverse
transform with just one pole. So now that gives me the
solution y from-- so this is just a constant,
1 over c minus a. And what is the inverse
transform of this? That's the simple pole at c. It came from a pure
exponential, e to the ct. Right? And now this guy, this one. OK. Well, this has a minus c, which
is the opposite of c minus a. So if I put a minus sign, I can
put them all over c minus a. Look at that. Look at this. C minus a is in both of these. Here it is with a plus sign. And that transform came
from that function. Here it is with a minus sign. So I want a minus there. And what function gave
me that transform? e to the at, right? That's the one we know. The unforgettable transform
of a simple exponential, e to the at. That is the particular solution. So the Laplace transform, we
transformed the differential equation. We got an algebra equation. We solved that algebra
equation, and then we had to go backwards to find what
function had this transform y. And to see that, clearly we had
to use this partial fraction idea which separated these
two poles into one pole there, when s is c, and
another pole, when s is a. We've got two easy fractions. The easy fractions each
gave me an exponential. And the final
result was this one. And I don't know if
you remember that. That is the correct solution
to the first order linear constant coefficient
equation, the simple equation there, when the right-hand
side is e to the ct. So our final solution
then is the null solution with the initial value in it. And that function comes from
the right-hand side, comes from the force, e to the ct. And so that's how
Laplace transforms work. Take the Laplace
transform of every term. Solve for y of s, and try your
best to invert that transform. OK. More of that coming in the next
lecture on Laplace transforms. Thank you.
