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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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00:12:56,480 --> 00:12:59,040
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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00:13:07,322 --> 00:13:09,500
Do you see how
that's going to fail?
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00:13:09,500 --> 00:13:14,330
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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00:13:23,290 --> 00:13:26,480
This is a case of
resonance, when
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the form of the right-hand
side is a null solution
212
00:13:30,830 --> 00:13:32,580
at the same time.
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00:13:32,580 --> 00:13:36,410
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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00:13:39,350 --> 00:13:42,590
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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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.
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We have to allow a t.
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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.
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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.
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That would do it.
229
00:14:18,340 --> 00:14:22,210
When there's resonance,
take the form
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00:14:22,210 --> 00:14:27,060
you would normally
assume and multiply
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00:14:27,060 --> 00:14:29,750
by that extra factor t.
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00:14:29,750 --> 00:14:34,730
Then, when I substitute that
into the differential equation,
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I'll find Y's quite safely.
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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.
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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.
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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.
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00:15:07,710 --> 00:15:10,090
Perhaps it's 1/2 or 1.
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00:15:10,090 --> 00:15:12,870
I won't do it.
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00:15:12,870 --> 00:15:15,760
Maybe I'll leave
that as an exercise.
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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,
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00:15:27,470 --> 00:15:29,820
which we're going to see
again, because they're nice.
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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.