WEBVTT

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GILBERT STRANG: OK, this is
my second video on the Laplace

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transform, and this one will
be about solving second order

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equations.

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So let me remember the plan.

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So there's our second
order equation.

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And I'm taking
this example first,

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with the delta function
on the right-hand side.

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You remember, that's
a key example.

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And actually, we have a special
letter for the solution.

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This is an impulse.

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And the solution is
the impulse response.

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And I use a little g.

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So I should have
turned this y into a g.

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So that g is the
solution, starting

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from 0 initial
conditions, with a delta.

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So what's the plan?

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We want to take the
transform of every term.

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We have to check, what is
the transform, the Laplace

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transform of the delta function?

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You remember the definition
of the transform.

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It's this integral.

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You take your
function-- whatever

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it is, here delta-- multiply
by e to the minus st,

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and integrate.

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s equals 0 to infinity.

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Well, easy to integrate
with a delta function there.

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It's 0, except where the
impulse is, at t equals 0.

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And at t equals 0, this is 1.

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So the answer is 1.

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That's the nice Laplace
transform of the impulse.

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And now I want the transform
of the impulse response.

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So the impulse response
comes from this equation.

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So now, transform every term.

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So the transform of g
will be called capital G.

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And the transform of
the delta function is 1.

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And the derivative,
you remember,

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has an extra factor, s.

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The transform of the derivative
from the first Laplace

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transform video
was s times g of s.

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And the second
derivative, another s.

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So s squared, times g of s.

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We're not surprised to see the
very familiar quadratic, whose

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roots are the two exponents
s1 and s2, showing up here.

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We've seen this every
time, because we

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have constant coefficients.

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We always see this quantity.

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And now we see that, look,
the transform, capital G.

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I divide by that.

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It's exactly the
transfer function.

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So that's connecting in the
idea of the transfer function

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to the Laplace transform
of the impulse response,

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because I have this
in the denominator.

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OK, so I want now-- I've taken
the transform of the equation.

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I've got the transform of
little g, the impulse response.

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So now I'm ready to find G,
the impulse response by invert

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Laplace transform.

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How do I find the function
with that Laplace transform?

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Right now, it's 1
over a quadratic.

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The whole idea of
partial fractions

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is, split this so this
G of s, final step.

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It's G of s.

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Well, you remember that
this is the polynomial that

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has the two roots, s1 and s2.

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I'm going to write that
as 1 over s minus s1,

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and s minus s2.

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Those are the two roots
from the quadratic formula--

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the two poles, we could
say, poles of G of s.

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And now I want to use
partial fractions.

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So I want to separate
this into two fractions.

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And it turns out that they
are 1 over s minus s1,

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minus 1 over s minus s2.

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And it turns out
there's a factor

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there to make it correct.

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You could check.

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When you put these, this
over a common denominator,

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you get this.

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And when you put that
over a common denominator,

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you'll get a numerator,
which you have to cancel.

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OK.

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So then, now I have
two simple poles.

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And I could write here what
I know, what g of t is.

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Remember, the function
with that transform

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is just e to the s1 t.

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The one most important
of all transforms

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is the transform
of the exponential,

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is that simple pole.

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So now I invert the
Laplace transform.

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So this gives me
an e to the s1 t,

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minus the function with that
transform, is e to the s2 t.

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And I still have this
constant, s1 minus s2.

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I've re-discovered
the impulse response.

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It's a solution to my
equation with impulse force.

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And it's that
particular function

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that plays such
an important part

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in the whole subject of constant
coefficient differential

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equations.

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Because you see it's
the critical thing here.

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Here's the critical
transfer function,

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and here is the inverse
Laplace transform.

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The Laplace transform
of that is that.

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Good.

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That's that example.

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And now I just want to take
another function than delta.

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OK.

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I could go all the way to
take f of t, any f of t.

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But let me stay with examples.

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So now I'm going to do y double
prime, plus b y prime, plus cy.

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They're all the most
important example.

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The best one we could do
would be cosine of omega t.

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An oscillating problem,
with an oscillating force,

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at a frequency different
from the natural frequency,

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and we also have damping there.

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So this is the standard,
spring mass dashpot

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problem, or RLC circuit
problem, certainly

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RLC circuits, highly important.

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And you might remember the
solution gets a little messy.

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A solution gets a little messy.

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It's highly important,
but-- so I'll

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carry it through
to the last step.

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But I probably won't
take that final step.

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Just, what I want you to
see is another example

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of the inverse
Laplace transform.

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So what do we need?

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We need the Laplace
transform of that.

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I plan to take the Laplace
transform of every term.

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s squared plus bs plus c.

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We'll multiply.

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I'm transforming everything.

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And here I have to put
the transform of that.

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OK.

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So how will I get that?

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And of course, there might
be a sine omega t in there.

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I would really like to
get them both at once.

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So let me put down
them both at once.

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The cosine and the sine are the
real and imaginary parts of e

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to the i omega t, right?

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Euler's great formula.

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e to the i omega t is cosine
omega t, the real part,

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plus i sine omega t,
the imaginary part.

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But now I know the transform of
e to the a t, e to the i omega

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t.

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So that transforms
to-- so I want

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the real and the
imaginary parts of 1 over,

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you remember what it is.

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It's just that
simple pole again,

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s minus the exponent i omega.

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So I'm going to get the cosine
and the sine at the same time,

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from one calculation, finding
the real and imaginary parts

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of this, 1 over s
plus s minus i omega.

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How do you deal
with a pole, when

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if you want the real
and imaginary parts,

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you're happy to get a real
number in the denominator

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and see the real and
imaginary parts up above.

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I don't like it when
it's down there.

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So what I'm going to do is,
real and imaginary parts of,

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I'll multiply s minus i omega.

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This is the key trick
with complex numbers.

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It comes up enough,
so it's good to learn.

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I multiply that by its
conjugate, s plus i omega

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over s plus i omega.

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So I multiplied by 1.

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But you see, what's
happened now is,

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real and imaginary parts
of-- I've got what I want,

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s plus i omega is now up in the
numerator, where I can see it.

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And what do I have down below?

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s minus i omega
times s plus i omega,

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the very important quantity.

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S squared minus i omega s, plus
i omega s, minus i squared,

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that's plus 1 omega
squared plus omega squared.

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OK.

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We've done it.

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I've transformed the
cosine and the sine

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into s over this,
and omega over this.

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So two at once, two transforms.

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And of course, as always, we're
able to do, and recognize,

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and work with the transforms
of a special group

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of nice functions,
exponentials above all,

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sines and cosines coming from
exponentials, delta functions.

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It's a short list.

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Those are the ones we can
do, and fortunately those

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are the ones that we need to do.

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OK, so I'm now ready to
put in the right-- what

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did that turn out to be?

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It turned out to be
s from the cosine,

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I'm not doing the
sine now, the cosine.

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I'm taking the real part.

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It's s over that positive s
squared plus omega squared.

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Are you with me?

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Left-hand side, all normal.

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I'm starting from 0
initial conditions,

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otherwise I would see the
initial values in here.

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But I don't, because they're 0.

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And over here, I've
got the transform

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of the right-hand side.

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OK?

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Then I just bring
that down there.

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So finally, I know y of s
is s over this all-important

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quadratic, and then I have the
s squared plus omega squared.

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OK.

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Well, you see it did get harder.

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We have s squared from
a quadratic there.

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We have two quadratics.

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We have a fourth-degree
polynomial down there.

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Partial fractions will work.

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Partial fractions can simplify
this, for any polynomials,

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but the algebra
gets quickly worse

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when you get up to degree four.

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But actually, this can
be, it could be done.

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And I don't plan to do it.

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To me, this would
eventually give the solution

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to this example.

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But we have other ways
to get that solution.

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And I believe, for me at least,
the other ways are simpler.

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I know that the solution is
a combination of cos omega

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t and sine omega t.

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And this is the particular
solution I'm talking about.

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And I can figure out what
those combinations are,

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because I know the right form.

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Here, I have to deal with
partial fractions and degree

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four, down there.

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I'm going to chicken
out on that one.

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I won't completely chicken out.

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I'll say what the
pieces look like.

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But I won't figure
out all the numbers.

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OK.

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So all this thing, I see as
some constant over-- well,

00:14:39.140 --> 00:14:45.320
this factors into s
minus s1 s minus s2.

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Those are the two
roots, as we saw above.

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So I have a linear, a
linear, and a quadratic.

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And partial fraction says
that I can separate out

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the first linear,
the second linear,

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and the third quadratic,
which I could factor too,

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but-- I could factor
s squared plus omega

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squared into that
times that, but it

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brings in imaginary numbers.

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So I'll put a cs and a d.

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OK.

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Four numbers to be determined,
or rather not to be determined.

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Because I'm going to stop there.

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What I've discovered
is this part

00:15:41.800 --> 00:15:45.390
would be the null
solution, would

00:15:45.390 --> 00:15:49.540
be a null solution because
it's involves e to the s1 t

00:15:49.540 --> 00:15:51.325
and e to the s2 t.

00:15:51.325 --> 00:15:58.100
This part, when I find
the inverse transform,

00:15:58.100 --> 00:16:03.490
it must be the combination of
cos omega t and sine omega t.

00:16:03.490 --> 00:16:06.770
So that's some
combination, when I find

00:16:06.770 --> 00:16:08.460
that it's the transform of--

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