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GILBERT STRANG: OK.

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So speaking today about
separable equations.

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These are, in principle,
the easiest to solve.

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They include nonlinear
equations but they

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have a special feature
that makes them easy,

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makes them approachable.

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And that special feature
is that the right hand

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side of the equation
separates into some function

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of t divided by or multiplied
by some function of y.

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The t and the y have separated
on the right hand side.

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And for example, dy/dt equal y
plus t would not be separable.

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They'd be very simple
but not separable.

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Separable means that we can
keep those two separately and do

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an integral of f and an integral
of g and we're in business.

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OK.

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Examples.

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Suppose that f of y is 1.

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Then we have this simplest
differential equation of all,

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dy/dt is some function of t.

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That's what calculus is for.

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y is the integral of g.

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Suppose there was no t.

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Just a 1 over f of y,
with g of t equal one.

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Then I bring the f of y up.

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I integrate the [? f ?] dy.

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And moving the dt there,
I'm just integrating dt.

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So the right hand
side would just be t.

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And the left hand side is
an integral we have to do.

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That's the minimum
amount of work

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to solve a
differential equation.

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But the point is, with y
and t separate, we just

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have integration to do.

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And here is the case when there
is both a g of t and an f of y.

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Then let me just emphasize
what's happening here.

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The f of y I am
moving up with a dy.

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The dt I'm moving
up with a g of t dt.

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So I g of t dt equals f of y
dy and I integrate both sides.

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The left side is an integral
of y with respect to y.

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The right hand side is
an integral with respect

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to t or a dummy variable s.

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The integral going from 0 to t.

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This integral going
from y of 0 to y of t.

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Those are the two
integrations to be done.

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And you will get examples
of seperable equations.

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And what you have to
do is two integrals.

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And then there's this one
little catch at the end.

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This is some function
of y when I integrate.

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But I usually like
to have the solution

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to a differential equation
just y equal something.

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And you'll see in the examples.

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I have to solve it
for y because this

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isn't going to give me just y.

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It's going to give me some
expression involving y.

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So let me do examples.

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Let me do examples.

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You see why it's correct.

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OK.

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So here are examples.

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So let me take, what about the
equation dy/dt equals t over y.

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Clearly separable.

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The function's f.

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g of t is just t.

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f of y is just y.

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I combine those to
y dy equalling t dt.

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You see I've picked a pretty
straightforward example.

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Now I'm integrating both
sides from y of 0 to y of t

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on the left.

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And from 0 to t on the right.

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And of course that
is 1/2 t squared.

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And the left hand
side is 1/2 y squared

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between these two limits.

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So I'm getting the integral
of that is 1/2 y squared.

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So up top I have 1/2
y of t squared minus,

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at the bottom end,
1/2 y of 0 squared

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equalling the right
hand side 1/2 t squared.

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So you see, we got a function
of y equal to a function of t.

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And the equation
is solved, really.

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That differential
equation is solved.

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But I haven't found it in the
form y of t equal something.

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But I can do that.

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I just move this
to the other side.

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So that will go to the
other side with a plus.

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And then I'll cancel the 1/2.

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And then I'll take
the square root.

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So the solution y of
t is the square root

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of y of 0 squared
plus t squared.

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That's the solution to
the differential equation.

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Maybe I make a small
comment on this equation.

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Because it's essential to begin
to look for dangerous points.

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Singular points where
things are not quite right.

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Here the dangerous point
is clearly y equal zero.

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If I start at y of 0 equals
zero then I'm not sure what.

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What's the solution to that
equation if I start at y of 0

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equals 0?

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I'm starting with a 0 over 0.

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What a way to begin your life,
starting with a 0 over 0.

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This, well, actually
the solution

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would still be correct.

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If y of 0 is 0, I would get
the square root of t squared.

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I would get t.

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So y of 0 equals 0 allows
the solution y equals t.

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And that is a solution.

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That if y is equal
to t then dy/dt is 1.

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And on the right hand side
t over y is t over t is 1.

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So the equation is solved.

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But my point was,
there's got to be

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something going a little
strange when y of 0 is 0.

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And what happens strangely
is there are other solutions.

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I like, I think, y
equal negative t.

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And more, probably.

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But if y is equal to negative t,
then its derivative is minus 1.

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And on the right hand side, I
have t over negative t minus 1

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again.

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So the equation is solved.

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That's a perfectly
good solution.

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That's a second solution.

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It's an equation with
more than one solution.

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And we'll have to
think, when can we

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guarantee there is just
one solution, which

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is of course what we want.

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OK.

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I'd better do another
example going beyond this.

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And maybe the logistic
equation is a good one.

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So that's separable.

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And it's going to
be a little harder.

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So let me do that one.

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dy/dt is y minus y
squared, let's say.

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The logistic equation.

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Linear term minus
a quadratic term.

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That's separable because
the g of t part is 1.

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And what's the f of y?

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Remember f of y-- I want
to put that on the y side.

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But it's going to show
up in the denominator.

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So I have dy over y minus
y squared equaling dt.

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And I have to integrate both
sides to get the solution y.

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Now, integrating the right hand
side is of course a picnic.

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I get t.

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But integrating
the left hand side,

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I have to either know how,
or look up, or figure out

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the integral of 1 over
y minus y squared.

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So let me just make a little
comment about integrating,

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because examples often
have this problem.

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Integrating when
there is a polynomial

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a quadratic in the denominator.

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There are different
ways to do it.

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And the time that we'll really
see this type of problem

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is when we discuss
Laplace transforms.

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So I'm going to save the details
of the method until then.

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But let me give the
name of the method.

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The name is partial
fractions, which

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is a method of integration.

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Partial fractions.

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And I'll just say
here what it means.

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It means that I want to write
this 1 over y minus y squared

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in a nicer way.

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What over y minus y squared can
be split up into two fractions?

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Those are the partial fractions.

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One fraction is-- so
I'm going to factor

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that y minus y squared
factors into y and 1 minus y.

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The partial fractions will
be some number over the y

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and some other number
over the 1 minus y.

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This is just algebra now.

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Partial fractions
is just algebra.

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It's not calculus.

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So I factored the y minus y
squared into these two terms.

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You see that if I come
to a common denominator,

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if I put these two
fractions together,

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then the denominator
is going to be that.

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And the numerator, if I choose
a and b correctly, will be 1.

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So, integrating this, I can
separately integrate a over y

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dy and b dy over 1 minus y.

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And those are easy.

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So partial fractions,
after you go

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to the effort of
finding the fractions,

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then you have separate
integrations that you can do.

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That integral is just
a times the log of y.

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And this is maybe
b times-- maybe

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it's minus b times
the log of 1 minus y.

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So we've integrated.

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Just remember though.

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That with this
particular equation,

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the logistic equation, we didn't
have to use partial fractions.

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We could have done-- we've
just seen how, thinking

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of it as a separable equation.

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But that logistic equation
had the very neat approach.

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Much quicker, much nicer.

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We just introduced
z equal 1 over y.

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We looked at the unknown
1 over y, called it z,

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found the equation for
z, and it was linear.

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And we can write
down its solution.

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So when we can do that it wins.

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But if we don't
see how to do that,

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partial fractions is
the systematic way.

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One fraction, another fraction.

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Integrate those fractions.

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Put the answer together.

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And then, and then,
at the end, this

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is some integral
depending on y equal to t.

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And to finish the
problem perfectly,

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I would have to solve
for y as a function of t.

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And that was what came out
so beautifully by letting 1

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over y bz.

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We got an easy formula for z and
then we had the formula for y.

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This we would integrate
easily enough.

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But then we have to solve
to find that formula for y.

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OK.

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That's a more serious example.

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This example was
a very simple one.

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You can do other examples
of separable equations.

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y and t integrated separately.

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Good.

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Thank you.

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