WEBVTT

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GILBERT STRANG: OK.

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So this is about the
world's fastest way

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to solve differential equations.

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And you'll like that method.

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First we have to
see what equations

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will we be able to solve.

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Well, linear,
constant coefficients.

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I made all the coefficients
1, but no problem

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to change those to A, B, C.
So the nice left-hand side.

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And on the right-hand side,
we also need something nice.

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We want a nice function.

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And I'll tell you which
are the nice functions.

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So I can say right away
that e to the exponentials

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are nice functions, of course.

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They're are always at the
center of this course.

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So for example,
equal e to the st.

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That would be a nice function.

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OK.

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And the key is, we're looking
for a particular solution,

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because we know how to
find null solutions.

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We're looking for particular
solution for this equation.

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One function, some
function that solves

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this equation with
right-hand side e to the st.

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And the point is, we
know what to look for.

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We just have some
coefficient to find.

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And we'll find that by
substituting in the equation.

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Now, do you remember
what we look

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for when the right-hand
side is e to the st?

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Then look for y equals some
constant times e to the st,

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right?

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When f of t-- maybe I'll put
the equal sign down there.

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If f of t is e to the st, then I
just look for a multiple of it.

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That's one coefficient to be
determined by substitute this

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into the equation.

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Do you remember the results?

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So this is our best example.

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When I put this in the equation,
I'll get the derivative

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brings an s.

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Second derivative
brings another s.

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So I get s squared and an s
and a 1 times y e to the st

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is equal to e to the st.
We've done that before.

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Here we see it as a case with
undetermined coefficient y.

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But by plugging it
in, I've discovered

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that y is 1 over that.

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So that's a nice function then.
e to the st is a nice function.

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What are the other
nice functions?

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So now, let me move to the other
board, next board, and ask,

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what other right-hand
sides could we solve?

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So I'll keep this
left-hand side equal to.

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So e to the st was 1.

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What about t?

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What about t?

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A polynomial.

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Well, that only has one term.

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So what would be a particular
solution to that equation?

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So I really have to say, what
is the-- try y particular

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equals-- now, if
I see a t there,

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then I'm going to
look for a t in y.

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And I'll also look
for a constant.

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So a plus bt would be the
correct form to look for.

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Let me just show
you how that works.

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So this now has two
undetermined coefficients.

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And we determine them by
putting that into the equation

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and making it right.

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So try yp is a plus
bt in this equation.

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OK, the second derivative
of a plus bt is 0.

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The first derivative
of that is b.

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So I get a b from that.

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And y itself is a plus bt.

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And that's supposed to give t.

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You see, I plugged it in.

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I got to this equation.

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Now I can determine a
and b by matching t.

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So then b has to be 1.

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We get b equal to 1.

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So the t equals t.

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But if b is 1, I need a to
b minus 1 to cancel that.

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So a is minus 1.

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And my answer is
minus 1 plus 1t.

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t minus 1.

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And if I put that into the
equation, it will be correct.

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So I have found a
particular solution,

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and that's my goal,
because I know

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how to find null solutions.

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And then together, that's
the complete solution.

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So we've learned what
to try with polynomials.

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With a power of t,
we want to include

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that power and all lower
powers, all the way down

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through the constants.

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OK.

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With exponentials, we just have
to include the exponential.

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What next?

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How about sine t or cosine t?

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Say sine t.

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So that case works.

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Now we want to try y double
prime plus y prime plus

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y equals, say, sine t.

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OK.

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What form do we assume for that?

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Well, I can tell you quickly.

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We assume a sine t in it.

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And we also need to
assume a cosine t.

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The rule is that the things
we try-- so I'll try y.

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y particular is what
we're always finding.

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Some c1 cos t, and
some c2 sine of t.

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That will do it.

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In fact, if I plug that in,
and I match the two sides,

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I determine c1 and
c2, I'm golden.

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Let me just comment
on that, rather

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than doing out every step.

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Again, the steps are
just substitute that in

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and make the equation correct
by choosing a good c1 and c2.

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I just noticed
that, you remember

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from Euler's great
formula that the cosine is

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a combination of e to the
it and e to the minus it.

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So in a way, we're really
using the original example.

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We're using this example,
e to the st, with two

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s's, e to the it, and
e to the minus it.

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So we have two
exponentials in a cosine.

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So I'm not surprised that there
are two constants to find.

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And now, finally, I
have to say, is this

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the end of nice functions?

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So nice functions include
exponentials, polynomials.

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These are really exponentials,
complex exponentials.

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And no, there's one
more possibility

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that we can deal with
in this simple way.

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And that possibility
is a product

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of-- so now I'll show you what
to do if it was t times sine t.

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Suppose we have the
right-hand side, the f of t,

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the forcing term,
is t times sine t.

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What is the form to assume?

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That's really all you have to
know is what form to assume?

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OK.

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Now, that t-- so we have here
a product, a polynomial times

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a sine or cosine or exponential.

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I could have done t
e to the st there.

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But what do I have to do
when the t shows up there?

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Then I have to try something
more with that t in there.

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So now I have a product
of polynomial times

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sine, cosine, or exponential.

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So what I try is at plus--
or rather, a plus bt.

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I try a product.

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Times cos t and c
plus dt times sine t.

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That's about as bad a case
as we're going to see.

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But it's still quite pleasant.

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So what do I see there?

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Because of the t here, I
needed to assume polynomials up

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to that same degree 1.

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So a plus bt.

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Had to do that, just
the way I did up there

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when there was a t.

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But now it multiplies sine t.

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So I have to allow sine
t and also cosine t.

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The pattern is, really,
we've sort of completed

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the list of nice functions.

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Exponentials, polynomials, and
polynomial times exponential.

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That's really what
a nice function is.

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A polynomial times
an exponential.

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Or we could have a
sum of those guys.

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We could have two or three
polynomial times exponential,

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like there and another one.

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And that's still
a nice function.

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And what's the real
key to nice functions?

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The key point is, why is this
such a good bunch of functions?

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Because, if I take
its derivative,

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I get a function
of the same form.

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If I take the derivative
of that right-hand side,

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and I use the
product rule, you see

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I'll get this times
the derivative of that.

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So I'll have something
looking with a sine in there.

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And I get this
times the derivative

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of that, which is just a b.

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So again, it fits the same
form, polynomial times cosine,

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polynomial times sine.

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So here I have a case
where I have actually four

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coefficients.

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But they'll all fall
out when you plug that

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into the equation.

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You just match terms
and your golden.

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So it really is a
straightforward method.

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Straightforward.

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So the key about
nice functions is--

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and they're nice for
Laplace transforms,

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they're nice at every step.

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But it's the same good functions
that we keep discovering

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as our best examples.

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The key about nice functions
is that the-- that's

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a form of a nice function
because its derivative has

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the same form.

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The derivative of that function
fits that pattern again.

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And then the second
derivative fits.

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All the derivatives fit.

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So when we put them in the
equation, everything fits.

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And always in the last
minute of a lecture,

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there's a special case.

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There's a special case.

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And let's remember what that is.

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So special case when we
have to change the form.

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And why would we
have to do that?

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Let me do y double prime
minus y, say, is e to the t.

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What is special about that?

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What's special is that
this right-hand side,

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this f function,
solves this equation.

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If I try e to the
t, it will fail.

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Try y equals some Y e to the t.

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Do you see how
that's going to fail?

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If I put that into the
equation, the second derivative

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will cancel the y and I'll
have 0 on the left side.

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Failure, because that's
the case called resonance.

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This is a case of
resonance, when

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the form of the right-hand
side is a null solution

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at the same time.

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It can't be a
particular solution.

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It won't work because
it's also a null solution.

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And do you remember how
to escape resonance?

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How to deal with resonance?

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What happens with resonance?

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The solution is a
little more complicated,

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but it fits everything here.

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We have to assume to allow a t.

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We have to allow a t.

00:13:55.560 --> 00:14:00.750
So instead of this multiple,
the and in this thing,

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we have to allow-- so I'm
going to assume-- I have

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to have-- I need a t in there.

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Oh, no.

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Actually, I don't.

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I just need a t.

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That would do it.

00:14:18.340 --> 00:14:22.210
When there's resonance,
take the form

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you would normally
assume and multiply

00:14:27.060 --> 00:14:29.750
by that extra factor t.

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Then, when I substitute that
into the differential equation,

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I'll find Y's quite safely.

00:14:37.050 --> 00:14:38.910
I'll find Y entirely safely.

00:14:38.910 --> 00:14:39.790
So I do that.

00:14:39.790 --> 00:14:44.500
So that's the resonant case,
the sort of special situation

00:14:44.500 --> 00:14:49.840
when e to the t solved this.

00:14:49.840 --> 00:14:51.960
So we need something new.

00:14:51.960 --> 00:14:55.420
And the way we get the right new
thing is to have a t in there.

00:14:55.420 --> 00:14:59.830
So when I plug that in, I take
the second derivative of that,

00:14:59.830 --> 00:15:03.490
subtract off that
itself, match e to the t.

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And that will tell
me the number Y.

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Perhaps it's 1/2 or 1.

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I won't do it.

00:15:12.870 --> 00:15:15.760
Maybe I'll leave
that as an exercise.

00:15:15.760 --> 00:15:21.040
Put that into the equation and
determine the number capital

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Y. OK.

00:15:23.730 --> 00:15:25.540
Let me pull it together.

00:15:25.540 --> 00:15:27.470
So we have certain
nice functions,

00:15:27.470 --> 00:15:29.820
which we're going to see
again, because they're nice.

00:15:29.820 --> 00:15:34.320
Every method works well
for these functions.

00:15:34.320 --> 00:15:39.800
And these functions are
exponentials, polynomials,

00:15:39.800 --> 00:15:42.640
or polynomials
times exponentials.

00:15:42.640 --> 00:15:47.070
And within exponentials,
I include sine and cosine.

00:15:47.070 --> 00:15:50.830
And for those functions,
we know the form.

00:15:50.830 --> 00:15:53.220
We plug it into the equation.

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We make it match.

00:15:55.830 --> 00:15:58.260
We choose these
undetermined coefficients.

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We determine them so that
they solve the equation.

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And then we've got a
particular solution.

00:16:05.620 --> 00:16:12.030
So this is the best
equations to solve

00:16:12.030 --> 00:16:13.800
to find particular solutions.

00:16:13.800 --> 00:16:19.100
Just by knowing the right form
and finding the constants,

00:16:19.100 --> 00:16:24.090
it did come out of the
particular equation.

00:16:24.090 --> 00:16:24.860
OK.

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All good, thanks.