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GILBERT STRANG: OK.
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We're still talking about first
order differential equations
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with a dy dt.
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And there is still a gross term
proportional to the balance.
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This might be the interest
that's added on to y.
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And then there is
input, a source term,
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of deposits being
made all the time.
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So I'm looking to solve
that differential equation.
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And this is the best
possible function
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for differential equations.
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Exponentials.
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Because the derivative of
exponential is exponential.
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It's just easiest to work with.
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And the output, the
solution is called
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the exponential response.
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That word response says what
comes out when e to the st
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goes in.
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And as before, we have some
starting deposit, y of 0.
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Initial condition at time 0.
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OK.
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Here is the key point, that
with this nice source function,
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the solution, or one solution,
a particular solution,
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will be just a multiple of e
to the st. So all I have to do
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is find that number, capital
Y, and I've got a solution
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to that equation.
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How do I do it?
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Substitute this
into the equation
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and solve for Y.
So let's do that.
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The derivative of this, the
derivative of exponential
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will bring down a factor s.
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So there'll be a ys e to
the st from the derivative.
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And that will have
to equal a times Y
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e to the st, plus the source
term e to the st. good?
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I just substituted it in.
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Now, the nice thing, I
cancel, I divide by e
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to the st, which is never 0.
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So I divide by-- factor out
e to the st, factor that out.
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It just leaves me with a 1.
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So I have Y times s,
Y times a, plus 1.
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So let me write that
equation so you see it.
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That's just s minus
a times Y is 1.
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Right?
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I took a times Y and put it on
the left side of the equation.
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So I've discovered the
exponential response.
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1 over s minus a.
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OK.
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So I have a solution
to the equation.
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That's not the end,
because that solution won't
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match the initial condition.
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So how do I match an
initial condition?
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What I've found is a
particular solution,
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and I need also the null
solution, the homogeneous
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solution.
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So the full solution, y of
t, is this Y particular.
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So capital Y, I
now know is that.
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So I have an e to
the st over-- I'm
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putting in Y, the
right value of Y.
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So that's the particular
solution that I've found.
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Plus any null solution.
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So remember, the null
solutions, that term is gone.
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So the source is 0.
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That's why the word null.
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So I'm looking for the
solution to dy dt equal ay.
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And the solutions
to dy dt equal ay
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are e to the at,
times any number.
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Because the right-hand
side is 0 now.
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This is y particular.
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And let me write that.
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This is y particular, and this
is y null, or y homogeneous.
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So that's the general solution.
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The complete solution
has that form.
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And now I can match y
equal y of 0 at t equal 0.
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I put in t equal 0, I get y of 0
equals-- t equals 0, this is 1.
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So 1 over s minus a.
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And when t is 0,
this is 1, so plus C.
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So now I know what C is.
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And notice, C is not just y of
0, as sometimes in the past.
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C is y of 0 minus this.
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Are you ready now for
the complete solution
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with satisfying the
initial conditions?
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So now I'm going to--
this in the correct form.
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This tells me what C has to be.
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So I put it in and I
have this solution.
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y of t is e the st over s minus
a, the easy one, plus C. Now,
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C is y of 0 minus
1 over s minus a.
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That's what we needed.
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Times e to the at.
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That's our answer.
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That's our answer.
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I can make it look
a little nicer.
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I want to.
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I want to separate out the y
of 0 part, the part that's just
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growing from the
initial condition,
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from the part that is
coming from the source term.
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So I just want to put
that together with this.
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So I have the same
s minus a below.
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Here is an e to the st above.
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And I have a minus, that 1 over
s minus a, times e to the at.
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And then I have this
term, which is growing.
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Well, this is really good.
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This is the part growing
out of the initial deposit.
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I'm using, again, money in a
bank with additional deposits,
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e to the st. And this
is the part coming
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from those later deposits.
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Initial part, and the
part coming from there.
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So this is, again,
a null solution.
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A multiple of e to the at.
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This is another
particular solution.
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Remember, there isn't just
one particular solution.
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Any solution is a
particular solution.
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And this is, I call that the
very particular solution.
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Because it has the nice
property that it starts from 0.
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So a t equals 0, that's
1 minus 1, I'm getting 0.
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So I would call that y vp.
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I'll introduce those letters,
not standard, for that part.
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And then y homogeneous,
or y null, is this part.
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OK.
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Problem solved.
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The exponential
grows in this way,
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and the initial condition
grows directly that way.
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OK.
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The problem is solved
with one exception.
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And now I have to take a
minute with that exception.
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The exception is the formula
breaks down if s equals a.
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If s equals a,
I'm dividing by 0.
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My formula is falling apart.
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And that's the
case of resonance.
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So let me put over
here, s equal a.
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That resonance.
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And we always have
to expect that that's
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a possibility, that
we're putting money
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with the same exponential as
the natural growth of the money,
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or whatever we're growing.
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And our formula has to change.
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You what you might
say it's infinite,
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because I'm dividing by 0.
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But notice the part
above is also 0.
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If s equals a, this is e to
the st minus e to the at.
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Those are the same.
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So I have a 0/0 situation.
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My formula is breaking
down, but it isn't dying.
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It needs more thinking.
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The case of resonance
needs to-- I
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have to understand what
this is when s equals a.
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Let me tell you what it is.
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And then show you why.
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So this is the case s equal a.
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So if s equal a, then y very
particular plus y null space.
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So it's this very
particular solution that
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has to have a different form.
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And here's the form it gets.
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A factor t appears.
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You just learn to recognize
resonance by that factor t.
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So it'll be a t e to the at.
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a is the same as s now,
so s doesn't appear.
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So that's the solution
which starts from 0,
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and it comes from the input.
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And this is the part that
starts from y of 0 and grows.
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So you see, eventually, this
is going to be the bigger one.
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The resonant case,
it grows like e
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to the at, with that little
bit extra growth of t.
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OK.
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So now I have a solution
in that special case
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also, when s equals a.
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All right.
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Do you want to know how
this comes out of this
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as s approaches a?
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Let me take three minutes
to tell you about that.
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It's L'Hopital's rule.
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Do you remember
from calculus, 0/0,
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the way to deal with that was
called by this guy's name,
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L'Hopital?
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Hospital, I guess.
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Probably hospital in French.
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And you this 0/0 expression.
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It's a ratio of two things.
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The top going to 0
when s goes to a,
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because these become the same.
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The bottom going to
0 when s goes to a.
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And the right--
L'Hopital's cool idea
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was you get the same
answer if you take
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the ratio of the derivatives.
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So L'Hopital says,
take the ratio
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of the derivatives of the top
minus the derivative-- oh,
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divided by the
derivative of the bottom.
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And then let s go
to a in the end.
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OK.
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So I have to take
this derivative,
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I have to take that
derivative with respect to s.
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Often in calculus, it was an x.
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Here it's an s.
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No big deal.
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So that derivative is--
You take the derivative
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with respect to s.
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We'll bring down--
ah, here comes the t.
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The derivative of
s is t e to the st.
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And the derivative
of this thing is 1.
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And now I let s go to a.
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Well, it's easy to
let s go to a now.
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This thing approaches that.
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I have the t e of
the at in the limit.
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So L'Hopital's
rule was the reason
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behind this formula
for resonance.
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But again, I emphasize
that expect a factor t
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when you have this resonance.
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OK.
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So that's the solution for the
best possible right-hand side
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e to the st. Well, maybe
the best is a constant.
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Second simplest
is an exponential.
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Next, will come
sines and cosigns.
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That's the next lecture.
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Thanks.