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GILBERT STRANG: OK?

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I want to talk about a
slightly different way

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to solve a linear
first-order equation.

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And if you look at the
equation-- I'll do an example.

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That's the best.

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Do you notice what's different
from our favorite equation?

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The change is 2t.

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The interest rate a is
increasing with time,

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changing with time.

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So we still have a linear
equation, still just y.

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But the coefficient is varying.

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We have a variable
coefficient 2t.

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And if we think here of
applications to economy,

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to banks, that would
be rampant inflation,

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the interest rate 2t climbing
and climbing forever.

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But we want to see that
this is a class of equations

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that we can solve.

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OK.

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And the new method is called
an integrating factor.

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It's a magic factor that
makes the equation simple.

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So that's another
nice way to solve

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all the problems that
we've dealt with so far,

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plus this new one.

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So what is this factor?

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Well, for this 2t
problem, the right factor

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is e to the minus t squared.

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And why is that
the right factor?

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This is the factor
that I'm going

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to multiply the equation
by and make it simple.

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And the reason that's
the right choice

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is that the derivative
of this-- you

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remember how to take the
derivative by the chain rule?

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The derivative will be
the same e to the minus t

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squared, the same I, times the
derivative of the exponent.

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And the derivative of
that exponent is minus 2t.

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Minus t squared
becomes minus 2t.

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So it's that little device that
gives us an integrating factor

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that makes the equation simple.

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And now I'm going to
look at the equation.

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What I want to look at is
the derivative of I times y.

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Instead of just
dy dt, let me look

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at the derivative of I times y.

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So I have a product here.

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Got to use the product rule.

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So that will be I-- so I dy dt.

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OK.

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dy dt is-- we can take dy
dt from the equation, I

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times 2ty plus q of t.

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And now I have to
add on dI dt y.

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Good.

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So it's the product rule--
I times the derivative of y

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plus the derivative
of I times y.

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But now look.

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dI dt we know is minus 2tI.

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So that dI dt, now I'm
using the key fact about I,

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that that's minus 2tIy.

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Look, minus 2tIy cancels 2tIy.

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So I have a nice equation now.

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The derivative of Iy is Iq.

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The derivative of Iy is Iq.

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I can just integrate both sides.

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And that's the key.

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That's the key.

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If I integrate the
left-hand side--

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so I'll just move
this up-- integrate

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the derivative-- of course,
the integral of the derivative

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is the function-- at time t, Iy
at time t minus Iy at time 0,

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y of 0.

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Because notice that
I at t equals 0-- can

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I just mention
that-- I at 0 is 1.

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When t is 0, I of e to
the 0 power, which is 1.

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So that I of 0 is 1.

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So that's the integral
of the derivative.

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And on the right-hand side,
I have the integral from 0

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to t of I times q.

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So I'll put in-- yeah, e to
the minus s squared q of s ds.

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I've introduced a
variable of integration

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s going from 0 to t.

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You remember this
type of formula?

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The input is
continuous over time,

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and I'm looking at the
resulting output at time t.

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So all the inputs go in.

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They're all multiplied
by some factor

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and integrated gives the total
result from those inputs.

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OK.

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I'm almost here.

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I just want to remember I
want to divide by I of t

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so I have a formula for y.

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OK.

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So my formula for y.

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When I divide by I of t--
don't forget what I of t is.

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Let me put it again here.

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Let me remind myself.

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I of t is e to the
minus t squared.

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That was the magic
integrating factor.

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OK.

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So I'm going to
divide by that, which

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means I'll multiply
by e to the t squared.

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So that will knock
out the I here.

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I'll put this on the other
side of the equation, y of 0,

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y of 0, and it will
be multiplied by the e

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to the t squared.

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And this thing will
be multiplied by e

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to the t squared.

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The integral from 0 to t
of e to the t squared minus

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s squared q of s ds.

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That's my answer.

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Well, let's look at it.

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I have y of t.

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This is what comes
out of y of 0.

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You see that the growth factor
has changed from our old e

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to the at-- that was the
growth at constant rate,

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interest rate a-- to
e to the t squared.

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That's our growth from an
increasing interest rate.

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And over here, I'm seeing
the result, the output,

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from the input q, from all
the inputs between 0 and t.

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Each input is multiplied
by now that factor

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is the growth not from 0 to t.

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This is the growth from 0 to t.

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This is the growth from s to
t, because the input went in

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at time s, and it had the
shorter time, t minus s,

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to grow.

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So that's the formula
for the answer.

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If you give me any particular
q of s, I just do the integral,

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and I find the solution to
the differential equation.

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So that integrating factor
has made things work.

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Maybe I should say what
the integrating factor

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would be in general.

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So let me take a moment to
see-- this was an example.

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This was an example
with a of t equal to 2t.

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What's the general
integrating factor?

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So we always want the
integrating factor.

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Our construction rule
is that it should

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give us-- the derivative should
be minus a of t times I itself.

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That's how we chose the
e to the minus t squared.

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Then a of t was 2t.

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Now I want to give
the general rule.

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The general rule for
the integrating factor

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is the solution
to that equation.

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The solution to that
equation is giving us the e

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to the t squared in the example.

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This was the example.

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But now I want a formula just
to close off the entire case

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of varying interest rate.

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I want to find the
solution to that equation.

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And it is-- so here's
the integrating factor.

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It's e to the minus
because of that minus sign.

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Now I'm wanting a to come
down when I take a derivative.

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So what I'll put up here,
the integral of a of t

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dt, say, from 0 to t.

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Now, let me just do again
this example just to see.

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I have e to the minus
the integral of 2t,

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which is e to minus t squared.

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That's how I get t squared
as the right choice

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for our example.

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And the general rule is there.

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That's the integrating factor.

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And finally, finally, if
a is a constant, which

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is the most common case--
the only case we've

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had until this video--
if a is a constant,

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then the integral of a from
0 to t is just a times t.

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So number one example,
number zero example,

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would be e to the minus at.

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That would be the correct
integrating factor

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if we had constant a.

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And I'll create some
examples, some problems,

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just to go through
the steps in that best

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case of all with constant
integrating factor.

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But now we can solve it with
a varying interest rate.

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Good.

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Thank you.