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GILBERT STRANG: OK.
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So today is a specific way
to solve linear differential
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equations.
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I'll take second order
equations as a good example.
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This way is called
variation of parameters,
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and it will lead us to a formula
for the answer, an integral.
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So that's the big step, to get
from the differential equation
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to y of t equal a
certain integral.
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That integral will
involve the right hand
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side, of course, the source.
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And if we can do
the integration,
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then we get a complete answer.
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But in any case, we get a
nice form for the answer.
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OK.
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So what's the idea?
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So we're looking for
a particular solution.
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Today is about a
particular solution.
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We have to know two null
solutions to get started.
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So we must know these, y1
and y2, null solutions,
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with f equals 0.
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And of course we do
know two null solutions
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when those coefficients
b and c are constants.
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And we'll do it as a good
example, the most important
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example.
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But maybe sometimes we can
find the null solutions
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when B and t are changing
in time, time varying,
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and that's all to the good.
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Those problems are
not easy to solve.
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But it's really
constant coefficients
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that we know it works.
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OK.
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What's the idea?
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So the idea is to use y1
and y2, the null solutions.
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If we multiply by constants,
we get another null solution
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by linearity.
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But the idea is to
multiply them by functions.
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The varying, those
constants are not constant,
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they are varying parameters.
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So this would be the
form of the solution.
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And we want to find a c1
and a c2 depending on t,
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so that the differential
equation above it is solved.
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So of course, I'm going to
plug that into the differential
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equation and find out the
conditions on c1 and c2.
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May I just jump to the result?
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This will satisfy that equation.
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We know that y1
and y2 satisfy it
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with a 0 on the right-hand side.
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So we just have to--
So it's c1 of t and c2
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of t that are going
to deal with the f.
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So here is the result
after putting it in there.
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I discover that c1 prime,
the derivative of t,
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comes into it times y1 plus
c2 prime times y2 equaling 0
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will give me one equation
for c1 prime and c2 prime.
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The other equation
for c1 prime times--
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We'll multiply y1 prime
plus c2 prime y2 prime.
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Of course, prime
means the derivative.
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And that, when I'm plugging
all this into the equation,
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I'm going to see an f of
t on the right-hand side.
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I have two equations
at each time.
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At each time instant I have
two ordinary linear equations.
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They're straight lines
in the c1, c2 plane.
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They intersect.
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We know how to solve--
The most basic problem
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of linear algebra, solve two
equations and two unknowns.
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And we do that for each
t, and we get an answer.
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So this leads us to c1 and
c2 and they depend on t.
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Because-- Well,
actually it leads us
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to c1 prime and c2
prime, the derivatives.
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It just happens that when we
plug it in, the c1 and the c2
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themselves disappear because
these were null solutions.
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So we know that c1
constant would be good,
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but when we put it in
there, we get equations that
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involve c1 prime at each time.
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So I'll take that away.
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At each time.
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All right.
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Two equations, two
unknowns, we solved it.
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OK.
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And then when we solve them,
we put them back in to y of t
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and can now just
write the answer.
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I'll just write.
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So I'm not doing all
the gory calculations.
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I'm just going to
write the answer.
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So y of t is going to be-- Just
to be sure, I'm looking here.
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That's what I'm going to write,
and I'm going to part c1 and c2
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into that-- c1 and c2 come
from these two equations.
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OK.
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So I have a y1 of t times c1.
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Now, the c1 that comes out
of that happens to be-- Well,
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c1 prime comes out of it,
so I have to integrate.
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I have to integrate c1 prime.
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So c1 prime that comes out of
that is a minus y2 times the f,
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dt, and then there
is a denominator,
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because if I have two
equations and two unknowns,
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there's a little 2
by 2 determinant.
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I'll just call it W.
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It has a famous name in
differential equations.
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And I'll tell you that name.
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W of t is the determinant
of what I have there,
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y1, y2, y1 prime, y2 prime.
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Two equations, they
have to be independent.
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They have to be
invertible to give me--
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and it's this determinant
that is the critical thing.
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So that's y1, y2 prime
minus y2, y1 prime.
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That's the function.
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Remember we know y1 and y2.
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The whole deal is starting
with y1 and y2, null solutions,
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and combining them.
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And you see there is a first.
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This is the c1 and now
I have to add in the c2.
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It looks a little messy,
but it's just an integral.
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It's just an integral.
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So now what is y2
of t multiplied by?
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Well, it's multiplied by c2,
and since this equation gives me
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c2 prime, I'm going to have
to integrate to get c2.
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And it turns out to
be y1 of t, f of t,
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dt, divided by that
same determinant W of t.
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This is-- Let me
tell you its name.
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It's named after a
guy named Wronski,
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so it's called the Wronskian.
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You can call it that or not.
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It's this.
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So we know y1 and y2.
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That means we've plugged those
in, we find the Wronskian.
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We put that in these integrals.
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We have the y2s and the
y1s and we're given the f
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and we integrate.
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Well, if we can integrate.
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I don't plan to go
probably much beyond that.
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Just I'll stop there.
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That's a formula for the answer.
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Well, I won't
completely stop there.
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Let me do an example.
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Let me do an example.
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So let's see.
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I guess my example will be a
constant coefficient equation.
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So constant
coefficient when-- What
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are the null solutions for a
constant coefficient equation?
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You remember what happens
with constant coefficients.
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You plug in.
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You get s squared plus Bs
plus C times-- You're looking
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for-- you're trying for an
exponential, and it'll work.
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So I plugged-- This is
looking for null solutions,
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because we have to
have those to start.
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They are the y1 and
y2 in my formula.
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So my variation and parameters,
the whole new thing,
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is completed there.
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Just if I want an example, I
have to solve the differential
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equation, and for
that, believe me,
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I'm going to make
B and C constant.
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So I solve this equation.
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So this gives me-- Of course,
I can cancel the e to the st,
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because it's never 0.
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So this has to be 0.
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And that has two roots, and
the quadratic formula tells me
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those two roots s1 and s2.
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And then my null solutions
are y1 is e to s1t
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and y2 is e to the s2t.
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Now, I'm ready to go.
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I'm going to put those
into this formula.
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I have to do the Wronskian,
and I need another blackboard.
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Sorry.
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This is the only video so far
that went to a third board.
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Can you remember I need
the Wronskian and then
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I'll reproduce the-- So
there's-- Let me copy again.
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Y1 is e to the s1t
y2 is e to the s2t.
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And now that Wronskian is y1,
y2 prime minus y2, y1 prime.
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And what does that
come out to be?
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y1 is e to the s1t.
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y2 prime is the derivative
of that, so it's an s2,
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e to the s2t.
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That's this term gave me that.
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And now I subtract
y2, which is there.
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y1 prime, which is the
derivative of this,
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brings down an s1.
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You see the beauty of
all these formulas.
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Well, beautiful if you like,
formulas, not everybody does.
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OK.
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So now I know all
the terms, and I'm
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prepared to write
again my formula for y
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of t, my solution.
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This is the climax now.
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I'm going to use the
formula that came
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from variation of parameters.
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And I'm going to put in that
function, that function,
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and that W, and I'm
going to see what I have.
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OK.
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So do you remember it
was y1 times an integral.
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Oh, I had better use a
variable of integration here.
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I don't want to put
t in there, because t
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is the limit of integration.
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So y1, do you remember, it
was a minus y2 e to s2--
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I'll use just capital
T-- time divided by W.
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Remember, there's an f
of t and there's a dT,
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and I'm using a capital
T as the dummy variable.
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And in the bottom goes W, which
is s2 minus s1 e to the s1T,
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e to the s2T.
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That was the first term.
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This was the c1
that multiplied y1.
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Now, I have to remember the
y2, the second null solution,
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times its c2.
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And that was the integral.
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And again, now I have
a plus sign, I think,
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so it was-- put in
parentheses there.
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That was the W on the bottom.
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So it was e to the s1T, f of
T, dT, dT, and that same W
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down here, s2 minus s1,
e to s1T, e to the s2T.
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OK.
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That's the variation
of parameters formula
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for these nice null solutions.
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It doesn't get better than this.
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And in fact, I guess I can
cancel e to the s2T there.
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I can cancel the e
to the s1T there.
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I can put this up
with a minus exponent.
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Oh yeah, it's going to be good.
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It's going to be good.
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So here is a constant s2 minus
s1, the same for both terms.
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Then I'll put this up here
as a negative exponent,
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so there's an e to
the s1t, I'll get--
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You might even see this coming.
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I get this 1 over s2 minus s1,
and then I have the integral.
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And here I have an f
of T in both integrals.
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Let me just put
that f of T down.
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And I have a dT.
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And then what do I--
What else do I have?
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I have this thing which I can
put inside but with a small t.
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I'm not integrating that.
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I think if you look there,
you have e to the s1t,
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and then this will come
up with a minus sign.
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Does that look good to
you for the first term?
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And then the second
term will have
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that same 1 over s2 minus s1.
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Now, what do I have?
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e to the s2T is going to
come up with a minus sign,
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so I have an integral of
e to the s2, t minus T.
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I think there's a minus.
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Haha.
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Don't forget that
minus, Professor Strang.
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And f of T, dT.
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Now, right there I'm
going to put the answer.
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I'm going to put the
answer, and for me, this
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was the highlight
of the whole thing.
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I didn't know what would come
out of variation of parameters.
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00:16:20,200 --> 00:16:23,790
I wasn't-- I'm not at
all an expert in that.
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00:16:23,790 --> 00:16:30,970
But I just followed the
rules, put in these two
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00:16:30,970 --> 00:16:35,350
null solutions, computed
W, put it all in,
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00:16:35,350 --> 00:16:39,250
ended up with this
answer, and then I
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00:16:39,250 --> 00:16:43,080
was very happy to recognize
what this answer was.
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00:16:43,080 --> 00:16:48,600
This answer is the integral
of something times--
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00:16:48,600 --> 00:16:56,060
This is the integral from 0 to
t of something times f of T, dT.
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00:16:56,060 --> 00:16:58,530
And what is that something?
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00:16:58,530 --> 00:17:04,010
It comes from that term divided
by this with a minus sign,
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00:17:04,010 --> 00:17:07,930
and this term divided by
that with a minus sign,
254
00:17:07,930 --> 00:17:11,770
and when you put
those in, what do you
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00:17:11,770 --> 00:17:15,660
get but the impulse response.
256
00:17:15,660 --> 00:17:18,119
What I've called g of t.
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00:17:18,119 --> 00:17:20,470
And it's that t minus T.
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00:17:20,470 --> 00:17:21,940
So here you go.
259
00:17:21,940 --> 00:17:23,619
This is the big moment.
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00:17:23,619 --> 00:17:31,220
g of t minus capital
T, f of T, dT.
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00:17:31,220 --> 00:17:36,140
Focus camera and attention
on that last result.
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00:17:36,140 --> 00:17:38,150
So that's the formula
that we end up
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00:17:38,150 --> 00:17:43,820
with from the
variation of parameters
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00:17:43,820 --> 00:17:46,970
applied to the constant
coefficient problem.
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00:17:46,970 --> 00:17:50,800
It's given us what
we already knew.
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00:17:50,800 --> 00:17:57,240
It's told us that the solution,
the particular solution, y of t
267
00:17:57,240 --> 00:18:05,630
is the integral of the
inputs, the right-hand side,
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00:18:05,630 --> 00:18:13,520
the forcing term, f of T, times
the growth term, the impulse
269
00:18:13,520 --> 00:18:14,710
response.
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00:18:14,710 --> 00:18:18,450
We could imagine
that at every time T,
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00:18:18,450 --> 00:18:21,990
we have an impulse
of size f of T.
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00:18:21,990 --> 00:18:25,060
And that impulse
grows by the impulse
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00:18:25,060 --> 00:18:31,750
response, g, over the remaining
time until it gets us to here.
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00:18:31,750 --> 00:18:35,770
Let me just clean that up.
275
00:18:35,770 --> 00:18:40,630
And then we have to
take all those inputs,
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00:18:40,630 --> 00:18:45,940
so we integrate over all those
inputs, and we get that answer.
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00:18:45,940 --> 00:18:50,070
That's the ultimate
formula for the solution
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00:18:50,070 --> 00:18:53,240
to our differential
equations, to
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00:18:53,240 --> 00:18:56,730
our linear constant coefficient
differential equation.
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00:18:56,730 --> 00:19:00,940
So we've reproduced that
formula in the one case,
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00:19:00,940 --> 00:19:06,200
constant coefficient case, when
we can find the null solutions
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00:19:06,200 --> 00:19:11,450
and run this variation of
parameters formula right
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00:19:11,450 --> 00:19:15,090
through to the end,
and that's the end.
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00:19:15,090 --> 00:19:16,880
Thank you.