WEBVTT

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GILBERT STRANG: OK.

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This is the start of
Laplace transforms.

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And that's going to take
more than one short video.

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But I'll devote this video to
first order equations, where

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the steps are easy
and pretty quick.

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Then will come second
order equations.

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So Laplace transforms
starting now.

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So let me tell you
what-- I use a capital

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letter for the Laplace transform
of little f, a function of t.

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The transform is capital
F, a function of s.

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And you'll see where s comes in.

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Or if it's the solution
I'm looking at,

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y of t, its transform
is naturally

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called capital Y of s.

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So that's what we want-- we
want to find y, and we know f.

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OK.

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So can I do an example?

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Well, first tell you what
the Laplace transform is.

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Suppose the function is f of t.

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Here is the transform.

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I multiply by e to the minus
st, and I integrate from 0

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to infinity.

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0 to infinity.

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Very important.

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The function doesn't
start until t equals 0,

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but it goes on to
t equal infinity.

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I integrate, and
when I integrate,

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t disappears but
s is still there.

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So I have a function of s.

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Well, I have to do an example.

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So to find the Laplace transform
is to do an integration.

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And you won't be surprised
that the good functions we know

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are the ones where we
can do the integration

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and discover the transform,
and make a little table

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of nice transforms.

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And the number one function
we know is the exponential.

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So can I find--
for that function,

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I'll compute its transform.

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So what do I have to do?

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I have to integrate
from 0 to infinity--

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you might say 0 to
infinity is hard,

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but it's actually the
best-- of my function, which

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is e to the at.

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So that's my function
times e to the minus st dt.

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OK.

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I can do that integral,
because those combine into e

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to the a minus st. I
can put those together

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into e to the a
minus st. I integrate

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so I get e to the a minus
st divided by a minus s.

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That's the integral of that.

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Because what I have
in here is just that.

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To integrate the exponential,
I just divide by the exponent

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there.

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And I have just substitute t
equal infinity and t equal 0.

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So t equal infinity,
starting at 0 to infinity.

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OK.

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Infinity is the nice one.

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It's the easy one.

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I will look only at s's
that are bigger than a.

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s larger than a means that this
exponential is decreasing to 0.

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It gets to 0 at
t equal infinity.

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So at t equal infinity, that
upper limit of the integral

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ends up with a 0.

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So I just have to
subtract the lower limit.

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And look how nice.

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Now I put in t equal 0.

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Well, then that becomes 1.

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And it's a lower limit, so
it comes with a minus sign.

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So it's just the 1
over, the minus sign

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will flip that s minus a.

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The most important Laplace
transform in the world.

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Remember, the function
was in to the at.

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The transform is
a function of s.

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The original function depended
on t and a parameter a.

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The result depends on
s and a parameter a.

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And an engineer would say,
here we have the exponent.

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The growth rate is a.

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And over in the transform--
so this is the transform,

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remember.

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This is the transform f of x.

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In the transform,
I see blow up--

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a pole, that's called
a pole-- at s equal a.

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1/0 is a pole.

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And I'm not surprised.

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So the answer is
blowing up at s equal a.

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Well, of course.

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If s equals a, then this
is the integral of 1 from 0

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to infinity, and it's infinite.

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So I'm not surprised to
see the pole showing up.

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The blow up showing up
exactly at the exponent a.

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But this is a nice transform.

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OK.

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I need to do one other-- oh, no.

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I could already
solve the equation.

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So let me start
with the equation dy

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dt minus ay equal 0.

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Oh, well, I can take the
Laplace transform of 0

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is 0, safe enough.

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The Laplace transform of
y is capital Y. But what's

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the transform of this?

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Oh, I have to do one
more transform for you.

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I'm hoping that the transform
of the derivative, dy dt,

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connects to the transform of y.

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So the transform of this
guy is the integral from 0

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to infinity of that function,
whatever it is, times e

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to the minus st dt.

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This is the transform.

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So this Laplace transform.

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Now what can I do
with that integral?

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This is a step that goes back
to the beginning of calculus.

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But it's easy to forget.

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When you see a derivative
there inside that integral,

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you think, I could
integrate by parts.

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I could integrate
that term and take

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the derivative of that term.

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That's what integration
by parts does.

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It moves the derivative
away from that and onto that

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where it's no problem.

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And do you remember that a minus
sign comes in when I do this?

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So I have the integral
from 0 to infinity of-- now

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the derivative is coming off
of that, so that's just y of t.

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And the derivative
is going onto that,

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so that's minus se
to the minus st dt.

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Good.

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And then do you remember
in integration by parts,

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there's also another term
that comes from y times e

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to the minus st?

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This is ye to the minus
st at 0 and infinity.

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OK.

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I've integrated by parts.

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A very useful, powerful
thing, not just a trick.

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OK.

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Now, can I recognize
some of this?

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That is minus minus, no problem.

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I bring out-- that
s is a constant.

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Bring it out, s.

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Now, what do I have left
when I bring out that s?

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I have the integral of
ye to the minus st dt.

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That is exactly the
Laplace transform of y.

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It's exactly capital Y.

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Put the equal sign here.

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I'll make that 0 a little
smaller, get it out of the way.

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OK. sY of s.

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So that whole term
has a nice form.

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When you take the
derivative of a function,

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you multiply its
Laplace transform by s.

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That's the rule.

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Take the derivative
of the function,

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multiply the Laplace
transform by s.

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If we have two derivatives,
we'll multiply by s twice.

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Easy.

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That's why the Laplace
transform works.

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But now, here is a final term.

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y at infinity-- well,
and e to the minus

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st at t equal infinity, 0.

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Forget it.

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So I just have to subtract off
y at 0 times e to the minus st

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at 0, which is 1.

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e to the 0 is 1.

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So do you see that
the initial condition

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comes into the transform?

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It's like, great.

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We have the transform of Y.
Now, all this is the transform.

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This is the transform
of dy dt that we found.

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Now, why did I want that?

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Because I plan to
take the transform

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of every term in my equation.

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So like there are two steps to
using the Laplace transform.

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One is to compute some
transforms like this one,

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and some rules like this one.

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That's the preparation step.

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That comes from just
looking at these integrals.

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And then to use them, I'm
going to take the Laplace

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transform of every term.

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So I have an equation.

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I take the Laplace
transform of every term.

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I've got another equation.

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So the Laplace transform of
this is sY of s minus y of 0.

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That was a Laplace
transform of this part.

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Now the Laplace transform
of this is minus a,

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a constant, Y of x.

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And the Laplace
transform of 0 is 0.

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Do you realize what we've done?

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I've taken a
differential equation

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and I've produced
an algebra equation.

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That's the point of
the Laplace transform,

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to turn differential
equations-- derivatives turn

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into multiplications, algebra.

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So all the terms
turn into that one.

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And now comes-- so
that's big step one.

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Transform every term.

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Get an algebra
problem for each s.

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We've changed from t, time
in the differential equation,

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to s in Laplace transform.

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Now solve this.

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So how am I going to solve that?

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I'm going to put y of 0
on to the right-hand side.

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And then I have Y of
s times s minus a.

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So I will divide by s minus a.

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And that gives me Y of s.

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So that was easy to do.

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The algebra problem
was easy to solve.

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The differential
equation more serious.

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OK.

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The algebra problem is easy.

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Are we finished?

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Got the answer, but we're in
the s variable, the s domain.

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I've got to get back
to-- so now this

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is going to be an inverse
Laplace transform.

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That's the inverse transform.

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To give me back y
of t equals what?

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How am I going to do
the inverse transform?

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So now I have the
transform of the answer,

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and I want the answer.

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I have to invert that transform
and get out of s and back to t.

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Well, y of 0 is a constant.

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Laplace transform is
linear, no problem.

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So have y of 0 from that.

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And now I have 1 over s minus a.

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So I'm asking myself, what is
the function whose transform is

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1 over s minus a?

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Then it's that function
that I want to put in there.

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And what is the function whose
transform is 1 over s minus a?

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It's the one we did.

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It's this one up here.

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1 over s minus a came from
the function e to the at.

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So that 1 over s minus
a, when I transform back,

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is the e to the at.

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And I'm golden.

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And that you recognize, of
course, as the correct answer,

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correct solution to this
differential equation.

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The initial value y of 0
takes off with exponential e

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to the at.

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No problem.

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OK.

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Can I do one more example
of a first order equation?

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Now I'm going to
put it in an f of t.

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I'm going to put
in a source term .

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So I'll do all the
same stuff, but I'm

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going to have an f of t.

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And what shall I take for-- I'll
take an exponential again, e

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to the ct.

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So that's my right-hand side.

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Can I do the same
idea, the central idea?

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Take my differential equation,
transform every term.

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I've started with
a time equation

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and I'm going to
get an s equation.

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So again, dy dt minus ay, that
transformed to-- what did that

00:14:29.470 --> 00:14:30.890
transform to?

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sY of s minus y of 0.

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Came from there.

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Minus aY of 0-- minus aY of s.

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Minus aY of s.

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And on the right-hand side,
I have the transform of e

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to the ct.

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We're getting good
at this transform.

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1 over s minus c,
instead of a at c.

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OK.

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That's our equation transform.

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Now algebra.

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I just pull Y of s out of that.

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How am I going to pull Y
of s out of this equation?

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Well, I'll move y of
0 to the other side.

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And I'll divide by s minus a.

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Look at it.

00:15:22.710 --> 00:15:26.960
Y of s is-- OK.

00:15:26.960 --> 00:15:31.320
I have 1 over s minus c.

00:15:31.320 --> 00:15:34.170
And I have an s minus
a that I'm dividing by.

00:15:34.170 --> 00:15:37.790
S minus a.

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And then I have the y
of 0 over s minus a.

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I've transformed the
differential equation

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to an s equation.

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I've just done simple algebra
to solve that equation.

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And I've got two terms.

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Two terms.

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You see that term?

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That's what I had before.

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That's what I had just there.

00:16:07.650 --> 00:16:10.580
The inverse transform was this.

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No problem.

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That's the null solution that's
coming out of the initial value

00:16:17.780 --> 00:16:18.460
.

00:16:18.460 --> 00:16:24.180
The new term that's coming
from the e to the ct,

00:16:24.180 --> 00:16:28.030
coming from the
force, is this one.

00:16:28.030 --> 00:16:33.320
And I have to do its
inverse transform.

00:16:33.320 --> 00:16:38.020
I have to figure out what
function has that transform.

00:16:38.020 --> 00:16:40.890
And you may say,
that's completely new.

00:16:40.890 --> 00:16:44.580
But we can connect it
to the one we know.

00:16:44.580 --> 00:16:45.080
OK.

00:16:45.080 --> 00:16:54.340
So that will give me the same
inverse transform, the growing

00:16:54.340 --> 00:16:55.490
exponential.

00:16:55.490 --> 00:16:58.360
But what does this one give?

00:16:58.360 --> 00:17:01.190
That's a key question.

00:17:01.190 --> 00:17:07.200
We have to be able to do--
invert, figure out what

00:17:07.200 --> 00:17:09.800
function has that transform?

00:17:09.800 --> 00:17:14.810
The function will involve
a and c and t, the time.

00:17:14.810 --> 00:17:21.260
And s, the transform variable,
will become t, the time period.

00:17:21.260 --> 00:17:23.170
So that's the big question.

00:17:23.170 --> 00:17:27.440
What do I do with this?

00:17:27.440 --> 00:17:29.840
And notice, it has two poles.

00:17:29.840 --> 00:17:34.390
It blows up at s equal a,
and it blows up at s equal c.

00:17:34.390 --> 00:17:43.020
And I have to figure out--
well, actually, by good luck,

00:17:43.020 --> 00:17:46.030
I want to separate
those two poles.

00:17:46.030 --> 00:17:47.540
Because if I separate
the two poles,

00:17:47.540 --> 00:17:51.430
I know what to do with a blow up
at s equal a and a blow up at s

00:17:51.430 --> 00:17:52.210
equal c.

00:17:52.210 --> 00:17:56.190
The problem is, right now I
have both blow ups at once.

00:17:56.190 --> 00:17:57.800
So I'm going to separate that.

00:17:57.800 --> 00:18:01.480
And that's called
partial fractions.

00:18:01.480 --> 00:18:05.150
So I will have to say more
about partial fractions.

00:18:05.150 --> 00:18:07.580
Right now, let me just do it.

00:18:07.580 --> 00:18:10.860
That expression there,
I'll take this guy away.

00:18:10.860 --> 00:18:14.330
Because it gives that
term that we know.

00:18:18.520 --> 00:18:20.610
It's this one.

00:18:20.610 --> 00:18:21.910
Is that one.

00:18:21.910 --> 00:18:25.420
It's this two poles
thing that I want

00:18:25.420 --> 00:18:28.430
to separate those two poles.

00:18:28.430 --> 00:18:30.450
So this is algebra again.

00:18:30.450 --> 00:18:32.400
Partial fractions
is just algebra.

00:18:32.400 --> 00:18:33.900
No calculus.

00:18:33.900 --> 00:18:36.080
No derivatives are in here.

00:18:36.080 --> 00:18:42.190
I just want to write
that as 1 over s minus c.

00:18:42.190 --> 00:18:44.250
It turns out-- look.

00:18:44.250 --> 00:18:47.020
There's a way to
remember the answer.

00:18:47.020 --> 00:19:00.730
s minus c times c minus
a and 1 over s minus a.

00:19:00.730 --> 00:19:03.700
And now a minus c.

00:19:06.448 --> 00:19:11.070
Do you see that that has
only one pole at s equals c?

00:19:11.070 --> 00:19:12.870
This is just a number.

00:19:12.870 --> 00:19:15.260
This has one pole at s equals a.

00:19:15.260 --> 00:19:16.250
That's just a number.

00:19:16.250 --> 00:19:19.660
In fact, those numbers
are the opposite.

00:19:19.660 --> 00:19:24.150
So now, are we golden?

00:19:24.150 --> 00:19:27.910
I can take the inverse
transform with just one pole.

00:19:27.910 --> 00:19:35.050
So now that gives me the
solution y from-- so this

00:19:35.050 --> 00:19:40.310
is just a constant,
1 over c minus a.

00:19:40.310 --> 00:19:44.720
And what is the inverse
transform of this?

00:19:44.720 --> 00:19:49.230
That's the simple pole at c.

00:19:49.230 --> 00:19:52.805
It came from a pure
exponential, e to the ct.

00:19:56.350 --> 00:19:57.375
Right?

00:19:57.375 --> 00:20:01.070
And now this guy, this one.

00:20:01.070 --> 00:20:01.970
OK.

00:20:01.970 --> 00:20:05.730
Well, this has a minus c, which
is the opposite of c minus a.

00:20:05.730 --> 00:20:10.410
So if I put a minus sign, I can
put them all over c minus a.

00:20:10.410 --> 00:20:11.090
Look at that.

00:20:11.090 --> 00:20:12.160
Look at this.

00:20:12.160 --> 00:20:15.280
C minus a is in both of these.

00:20:15.280 --> 00:20:17.570
Here it is with a plus sign.

00:20:17.570 --> 00:20:24.390
And that transform came
from that function.

00:20:24.390 --> 00:20:26.550
Here it is with a minus sign.

00:20:26.550 --> 00:20:28.850
So I want a minus there.

00:20:28.850 --> 00:20:34.180
And what function gave
me that transform?

00:20:34.180 --> 00:20:36.780
e to the at, right?

00:20:36.780 --> 00:20:39.040
That's the one we know.

00:20:39.040 --> 00:20:44.680
The unforgettable transform
of a simple exponential,

00:20:44.680 --> 00:20:47.660
e to the at.

00:20:47.660 --> 00:20:53.360
That is the particular solution.

00:20:53.360 --> 00:21:01.440
So the Laplace transform, we
transformed the differential

00:21:01.440 --> 00:21:02.980
equation.

00:21:02.980 --> 00:21:04.900
We got an algebra equation.

00:21:04.900 --> 00:21:08.060
We solved that algebra
equation, and then we

00:21:08.060 --> 00:21:15.240
had to go backwards to find what
function had this transform y.

00:21:15.240 --> 00:21:19.680
And to see that, clearly we had
to use this partial fraction

00:21:19.680 --> 00:21:23.980
idea which separated these
two poles into one pole

00:21:23.980 --> 00:21:29.430
there, when s is c, and
another pole, when s is a.

00:21:29.430 --> 00:21:31.520
We've got two easy fractions.

00:21:31.520 --> 00:21:36.110
The easy fractions each
gave me an exponential.

00:21:36.110 --> 00:21:38.970
And the final
result was this one.

00:21:38.970 --> 00:21:41.380
And I don't know if
you remember that.

00:21:41.380 --> 00:21:47.670
That is the correct solution
to the first order linear

00:21:47.670 --> 00:21:51.100
constant coefficient
equation, the simple equation

00:21:51.100 --> 00:21:56.070
there, when the right-hand
side is e to the ct.

00:21:56.070 --> 00:21:59.480
So our final solution
then is the null solution

00:21:59.480 --> 00:22:02.620
with the initial value in it.

00:22:02.620 --> 00:22:08.280
And that function comes from
the right-hand side, comes

00:22:08.280 --> 00:22:11.500
from the force, e to the ct.

00:22:11.500 --> 00:22:16.210
And so that's how
Laplace transforms work.

00:22:16.210 --> 00:22:19.490
Take the Laplace
transform of every term.

00:22:19.490 --> 00:22:25.421
Solve for y of s, and try your
best to invert that transform.

00:22:25.421 --> 00:22:25.920
OK.

00:22:25.920 --> 00:22:30.830
More of that coming in the next
lecture on Laplace transforms.

00:22:30.830 --> 00:22:32.630
Thank you.