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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.

173
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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.

185
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But it's the same good functions
that we keep discovering

186
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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

189
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the same form.

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The derivative of that function
fits that pattern again.

191
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And then the second
derivative fits.

192
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All the derivatives fit.

193
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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,

195
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there's a special case.

196
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There's a special case.

197
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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.

199
00:12:31,960 --> 00:12:33,980
And why would we
have to do that?

200
00:12:33,980 --> 00:12:42,580
Let me do y double prime
minus y, say, is e to the t.

201
00:12:42,580 --> 00:12:45,250
What is special about that?

202
00:12:45,250 --> 00:12:49,200
What's special is that
this right-hand side,

203
00:12:49,200 --> 00:12:52,905
this f function,
solves this equation.

204
00:12:56,480 --> 00:12:59,040
If I try e to the
t, it will fail.

205
00:12:59,040 --> 00:13:07,322
Try y equals some Y e to the t.

206
00:13:07,322 --> 00:13:09,500
Do you see how
that's going to fail?

207
00:13:09,500 --> 00:13:14,330
If I put that into the
equation, the second derivative

208
00:13:14,330 --> 00:13:17,870
will cancel the y and I'll
have 0 on the left side.

209
00:13:17,870 --> 00:13:23,290
Failure, because that's
the case called resonance.

210
00:13:23,290 --> 00:13:26,480
This is a case of
resonance, when

211
00:13:26,480 --> 00:13:30,830
the form of the right-hand
side is a null solution

212
00:13:30,830 --> 00:13:32,580
at the same time.

213
00:13:32,580 --> 00:13:36,410
It can't be a
particular solution.

214
00:13:36,410 --> 00:13:39,350
It won't work because
it's also a null solution.

215
00:13:39,350 --> 00:13:42,590
And do you remember how
to escape resonance?

216
00:13:42,590 --> 00:13:44,570
How to deal with resonance?

217
00:13:44,570 --> 00:13:46,590
What happens with resonance?

218
00:13:46,590 --> 00:13:48,990
The solution is a
little more complicated,

219
00:13:48,990 --> 00:13:50,670
but it fits everything here.

220
00:13:50,670 --> 00:13:53,930
We have to assume to allow a t.

221
00:13:53,930 --> 00:13:55,560
We have to allow a t.

222
00:13:55,560 --> 00:14:00,750
So instead of this multiple,
the and in this thing,

223
00:14:00,750 --> 00:14:08,060
we have to allow-- so I'm
going to assume-- I have

224
00:14:08,060 --> 00:14:10,370
to have-- I need a t in there.

225
00:14:10,370 --> 00:14:10,870
Oh, no.

226
00:14:10,870 --> 00:14:12,510
Actually, I don't.

227
00:14:12,510 --> 00:14:15,340
I just need a t.

228
00:14:15,340 --> 00:14:18,340
That would do it.

229
00:14:18,340 --> 00:14:22,210
When there's resonance,
take the form

230
00:14:22,210 --> 00:14:27,060
you would normally
assume and multiply

231
00:14:27,060 --> 00:14:29,750
by that extra factor t.

232
00:14:29,750 --> 00:14:34,730
Then, when I substitute that
into the differential equation,

233
00:14:34,730 --> 00:14:37,050
I'll find Y's quite safely.

234
00:14:37,050 --> 00:14:38,910
I'll find Y entirely safely.

235
00:14:38,910 --> 00:14:39,790
So I do that.

236
00:14:39,790 --> 00:14:44,500
So that's the resonant case,
the sort of special situation

237
00:14:44,500 --> 00:14:49,840
when e to the t solved this.

238
00:14:49,840 --> 00:14:51,960
So we need something new.

239
00:14:51,960 --> 00:14:55,420
And the way we get the right new
thing is to have a t in there.

240
00:14:55,420 --> 00:14:59,830
So when I plug that in, I take
the second derivative of that,

241
00:14:59,830 --> 00:15:03,490
subtract off that
itself, match e to the t.

242
00:15:03,490 --> 00:15:07,710
And that will tell
me the number Y.

243
00:15:07,710 --> 00:15:10,090
Perhaps it's 1/2 or 1.

244
00:15:10,090 --> 00:15:12,870
I won't do it.

245
00:15:12,870 --> 00:15:15,760
Maybe I'll leave
that as an exercise.

246
00:15:15,760 --> 00:15:21,040
Put that into the equation and
determine the number capital

247
00:15:21,040 --> 00:15:23,730
Y. OK.

248
00:15:23,730 --> 00:15:25,540
Let me pull it together.

249
00:15:25,540 --> 00:15:27,470
So we have certain
nice functions,

250
00:15:27,470 --> 00:15:29,820
which we're going to see
again, because they're nice.

251
00:15:29,820 --> 00:15:34,320
Every method works well
for these functions.

252
00:15:34,320 --> 00:15:39,800
And these functions are
exponentials, polynomials,

253
00:15:39,800 --> 00:15:42,640
or polynomials
times exponentials.

254
00:15:42,640 --> 00:15:47,070
And within exponentials,
I include sine and cosine.

255
00:15:47,070 --> 00:15:50,830
And for those functions,
we know the form.

256
00:15:50,830 --> 00:15:53,220
We plug it into the equation.

257
00:15:53,220 --> 00:15:55,830
We make it match.

258
00:15:55,830 --> 00:15:58,260
We choose these
undetermined coefficients.

259
00:15:58,260 --> 00:16:02,090
We determine them so that
they solve the equation.

260
00:16:02,090 --> 00:16:05,620
And then we've got a
particular solution.

261
00:16:05,620 --> 00:16:12,030
So this is the best
equations to solve

262
00:16:12,030 --> 00:16:13,800
to find particular solutions.

263
00:16:13,800 --> 00:16:19,100
Just by knowing the right form
and finding the constants,

264
00:16:19,100 --> 00:16:24,090
it did come out of the
particular equation.

265
00:16:24,090 --> 00:16:24,860
OK.

266
00:16:24,860 --> 00:16:26,910
All good, thanks.