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GILBERT STRANG: OK.

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This is our last look at the
first order linear differential

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equation that you see up here.

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The dy dt is ay, that's
the interest rate growing

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in the bank example.

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y is our total balance.

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And q of t is our
deposits or withdrawals.

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Only one change.

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We allow the interest rate
a to change with time.

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This we didn't see before.

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Now we will get a formula.

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It will be a formula we had
before when a was constant.

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And now we'll see it.

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It looks a little messier, but
the point is, it can be done.

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We can solve that
equation by a new way.

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So that's really
the other point.

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Everybody in the end likes
these integrating factors.

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And I will call it m.

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And let me show you what
it is and how it works.

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What it is is the solution
to the null equation,

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with a minus sign.

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With a minus sign. dM
dt equals minus a of tM.

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No source term.

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We can solve that equation.

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If a is constant-- and I'll keep
that case going because that's

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the one with simple,
recognizable formulas.

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If a is a constant, we're
looking for the function

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M whose derivative is minus aM.

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And that function is
e to the minus at.

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The derivative brings down
the minus a that we want.

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In case a is varying, we can
still to solve this equation.

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It will still be an
exponential of minus something.

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But what we have
to put here when

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I take the derivative of
M, the derivative of that

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will come down.

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So I want the
integral of a here.

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And then the derivative of
the integral is a minus a,

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coming down as it should.

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So I want minus
the integral of a.

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And can I introduce dummy
variables, say a of T dT,

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just to make the
notation look right.

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OK.

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You see that, again,
the derivative of M

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is always with an exponential.

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It's always the exponential
times the derivative

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of the exponent.

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And the derivative of
that exponent is minus a.

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Because by the fundamental
theorem of calculus,

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if I integrate a and take its
derivative, I get a again.

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And it's that a that I want.

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Now, why do I want this M?

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How does it work?

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Here's the reason M succeeds.

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Look at the derivative
of M times y.

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That's a product.

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So I'll use the product rule.

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I get the derivative
of y times M,

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and then I get the
derivative of M times y.

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But the derivative of
M is minus a of tM,

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so I better put the derivative
of M is minus a of tM times y.

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But what have I got here?

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Factor out an M and
that's just dy dt

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minus ay, dy dt minus ay is q.

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So when I factor out
the M, I just have q.

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All together, this is M times q.

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Look, my differential
equation couldn't look nicer.

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Multiplying by M
made it just tell us

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that a derivative is
a right-hand side.

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To solve that equation, we
just integrate both sides.

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So if you'll allow me to take
that step, integrate both sides

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and see what I've
got, that will give us

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the formula we know when
we're in the constant case,

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and the formula we've never
seen when t is varying.

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And then I'll do an example.

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Let me do an example right away.

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Suppose a of t, instead of
being constant, is growing.

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The economy is really
in hyperinflation.

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Take that example if a
of t is, let's say, 2t.

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Interest rate started
low and moves up,

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then growth is going to be
faster and faster as time

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goes on.

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And what will be
the integral of 2t?

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The integral of 2t is t
squared, so M, in that case,

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will be e to the
minus a t squared.

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Sorry, there's no a anymore.

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a is just the 2t.

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e to the minus t squared.

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With a minus sign,
it's dropping fast.

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In a minute, we'll
have a plus sign there

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and we'll see the growth.

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Do you see that this is the
integrating factor when a of t

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happens to be 2t?

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OK.

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Now I come back to this
equation and integrate

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both sides to get the answer.

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OK.

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All right.

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The integral of My,
of the derivative,

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the integral of the
derivative is just

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M of t y of t minus
M of 0 y of 0.

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That's the integral
on the left side.

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And on the right side,
I have the integral

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of M times q from 0 to t.

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And again, I'm going to
put in an integration

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variable different from t
just to keep things straight.

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OK.

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So now I've got a formula for y.

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It involves the M. Actually,
the y is multiplied by M,

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I better divide by-- first of
all, do we remember what M of 0

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is?

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That's the growth factor at 0.

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It's just 1.

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Nothing's happened.

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It's the exponential of 0 in
our formulas for M. M of 0 is 1.

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That's where M starts.

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So M of 0 is 1.

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I can remove that.

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OK.

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And now-- oh, let me put that on
the other side so this will be

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equals y of 0 plus that.

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OK.

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And now if I divide by
M, I have my answer.

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So those are the steps.

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Find the integrating factor.

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Do the integration,
which is now made easy

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because I have a perfect
derivative whose integral I

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just have to integrate.

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And then put in what M is, and
divide by it so that I get y.

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OK.

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So I'm dividing by M.
So what is 1 over M?

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Well, M has this minus
sign in the exponent.

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1 over M will have a plus sign.

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M here has e to the
minus t squared.

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1 over M will be e to
the plus t squared.

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So when I divide
by M, I get y of t.

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This will be 1 over M. That will
be e to the plus the integral

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of a of t dt y of 0.

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That's the null solution.

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That's the solution that's
growing out of y of 0.

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And now I have plus the integral
from 0 to t of-- remember,

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I'm dividing by M. And
that's e to the plus

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the integral from 0 to
s of a times q of s ds.

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OK.

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Oh, just a moment.

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I'm dividing by M
and I had an M there.

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Oh, wait a minute.

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I haven't got it right here.

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So I want to know what is M at
time s divided by M at time t?

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So this was the
integral from 0 to s.

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This is an integral from 0 to t.

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And both are in the exponent.

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This is-- can I say it here?

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This is a e to the-- divided by
M is the integral from 0 to t.

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And then I'm multiplying by e
to the minus the integral from 0

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to s.

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The rule for
exponents is if I have

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a product of two exponentials,
I add the exponents.

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When I add this to
this, this knocks off

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the lower half of the integral.

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I'm left with the
integral from s to t of a.

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So this was an integral of
a minus an integral of a.

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Let me do our example.

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Our example up here.

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Example-- M of t will be--
when a is equal to 2t,

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this was the example
a equal to 2t.

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The first time we've been able
to deal with a varying interest

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rate.

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So the integral of
2t is t squared.

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From the-- is e
to the t squared.

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And I subtract the
lower limit, s squared.

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That's the growth factor.

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That's the growth factor
from time s to time t.

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When a was constant,
that exponent

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was just a times t minus s.

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That told us the time.

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But now, a is varying and
the growth factor between s

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and t is e to the t
squared minus s squared.

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So that's what goes in here.

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Let me-- that's
the growth factor.

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May I just put it in here?

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In this example, it's e to
the t squared minus s squared.

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Instead of e to the
a t minus s, I now

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have t squared minus
s squared, because I

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had an integral of a of t,
and a is not constant anymore.

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This is my example.

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And I don't know if
you like this formula.

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Can I just describe it again?

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This was an integral
from 0 to t,

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so that would be-- this part
would be e to the t squared.

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That's the growth factor that
multiplies the initial deposit.

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The growth factor
that multiplies

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the later deposit is e to the
t squared minus s squared.

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And we allow deposits all
the way from s equals 0 to t.

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So when we add those
up, we get that sum.

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We've solved an
equation that we hadn't

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been able to solve before.

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That's a small triumph in
differential equations.

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Small, admittedly.

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I'd rather move next to
non-linear equations, which

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we have not touched.

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And that's a big deal.

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Thank you.