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PROFESSOR: Welcome back.

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So in this session
we're going to cover

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homogeneous constant coefficient
equations for any roots.

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So we're told to
assume that z of t

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is equal to exponential minus 3t
times cosine plus i of sine t.

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And to assume that
this complex number is

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a solution to this
differential equation, which

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is second order with
constant coefficients m,

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b, and k, which are real,
and from this assumption

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give two real solutions
to this equation.

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In the second part, we're asked
to find a general solution

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for this other differential
equation, which

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is the same form as
that seen in part a,

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with now the real values for
b equals to 6, m equals to 1,

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and k equals to 10.

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And then we're asked
of this system that

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would be captured by this
differential equation is

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overdamped or underdamped.

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In the last part we
switch gears and we're

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given a series of
roots, eight roots

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to an eighth degree
polynomial, and we're

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asked to write down the general
solution for that polynomial.

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Note that here we have repeated
roots, which is basically

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where the trick is.

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So why don't you
take a few minutes,

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and we'll come back to
solve these problems.

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Welcome back.

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So for the first part,
we're given a complex number

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clearly split into its real
part and its imaginary part.

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So this is the
imaginary part of z,

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and this is the real part of z.

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Now, the differential
equation that we are given

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is constant value-- differential
equation second order, linear,

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and homogeneous.

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So there is no right-hand side.

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So clearly you can see that
if a complex number that

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can be written as real
part plus i imaginary part

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is a solution to this equation,
then both sides independently

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are also solutions
to this equation.

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So by giving us
this complex number,

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we are actually given
two types of solutions

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that we can then
write down as two

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linearly independent solutions.

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So the general solution to
this differential equation

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can then be written in the form
of-- one constant coefficient

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is undetermined, minus 3t
cos t plus another constant

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coefficient minus 3t sine of t.

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Where basically,
we're introducing

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a linear combination of two
linearly independent solutions

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that were just taken from
the original complex number

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we were given.

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So for the second
part now, we are

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asked to find a general
solution for the ODE.

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We'll just rewrite here for now.

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So again, constant coefficient,
homogeneous equation.

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And we're just asked to
find a general solution.

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So we learned from
the class notes

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that we can use the
characteristic polynomial

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method to solve this equation.

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And the characteristic
polynomial

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here will have this form.

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And basically, we
just need to find

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the roots of this
characteristic polynomial

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to be able then to
express the solutions

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of this homogeneous constant
coefficient equation in terms

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of function, exponential to
whatever roots we will find.

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So I just want to write
down the discriminant

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of this polynomial.

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So what do we have here?

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What was the question again?

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So we're asked then to
just find the discriminant

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for this equation.

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So b squared minus 40,
which is basically minus 4.

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So we have a discriminant
that is negative.

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And the roots for our polynomial
here are just going to be...

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the root absolute value
of the discriminant

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with the complex number
i out, because we

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had a negative discriminant.

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So these couple of
solutions, plus or minus,

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tell us that we are going to
have oscillations as solutions

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for this differential
equation, and that it's also

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going to be actually
damped oscillations,

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because we have the
first-- the real part

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of the number being negative.

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So let me just write down
the pair of solutions

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that we would have.

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We will have this that comes
out of the real part, cosine.

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I'll put the root
of 4 is just 2.

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And so we have
cosine t, that's one.

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6/2, of course, is just 3.

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So here I switched directly
to writing them in cosine

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and sine, in the forms.

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But we could also have
kept it in the complex form

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if we were solving the
equation in complex space.

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And then we would just
have complex exponentials

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here with the frequencies
omega-- angular frequency 1.

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So this is the
homogeneous equation,

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the homogeneous solution for
this homogeneous constant

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coefficient
differential equation.

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Now, what do we see?

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We can see that it's exactly
the same form that the solution

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that we had previously.

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And so the question here is
this oscillator overdamped

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or underdamped?

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And as we see here, we're
going to have oscillations

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that are damped by a pre-factor
of decaying exponential minus

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3t for both cases.

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So this is decaying.

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So there is definitely damping.

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And we can see that
here, because we

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had a positive
value for b, which

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is the damping term in
the oscillator system.

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But there are
still oscillations,

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because we had a discriminant
that was negative,

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which introduced complex
numbers, hence cosines

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and sines, so
basically, oscillations.

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Therefore we're in the case
of an underdamped response,

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an underdamped system.

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The discriminant was negative.

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So for the last part
of this problem,

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we now leave this
differential equation,

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which was second order, and move
to higher-order differential

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equations of eighth order.

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And we're given eight roots.

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4 minus 5i.

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We have the root of
3, which is real,

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which is repeated three times,
and a root 4 plus or minus 5i,

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which is repeated two times.

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So we saw before then
when we have repeated

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roots, that means
that we would have,

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let's say, in a second-order
differential equation.

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We would have in this
case a discriminant

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that would be equal to 0.

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And we would need to introduce
a different type of solution,

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because instead of introducing
an exponential of plus r*t,

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exponential minus r*t, now
we don't have the minus r*t.

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So we need to seek
another type of solution.

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And we'll multiply
the exponential by t.

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So one type of solution for this
unique root that would come out

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would be just 2t.

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For the first of the tripled
root, it would be a 3t.

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For the second one,
we would have t 3t.

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For the third one, we need
to go higher order, 3t.

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Now we have complex
roots, so that gives us

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solutions of the same form that
we found in questions a and b.

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So I'm going to just
write them here,

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where now we have e^(4t)
cosine 5t plus e^(4t) sine 5t.

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And for the repeated
root, it's the same game.

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Even though we're in complex
roots, it's the same idea,

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we need to multiply
all of it by t.

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So these are the families of
linearly independent solutions

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that would come
out of these roots.

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So now the general
solution is simply

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the linear combination of all
these linearly independent

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functions.

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So we will introduce
different constants.

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c_5, c_6, c_7, and c_8.

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And we have eight undetermined
coefficients to deal with,

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because these would
be eight roots

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of an eighth-order polynomial.

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And so if we were to
solve an initial value

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problems like that, we would
need eight different types

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of initial conditions
to determine

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the values of these constants.

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So to summarize, we used the
characteristic polynomial

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method to solve these
problems introducing

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higher-order equations than
the first-order differential

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equations we saw before.

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And what is
important to remember

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is that the different types
of solutions that you get,

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depending on the roots of
the characteristic polynomial

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and the sine of
the discriminant.

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So whether the roots
are real or purely

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imaginary or
imaginary numbers will

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give three types
of the behavior:

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the critically damped
behavior; if it's real

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then it's going to be
just overdamped behavior;

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and if the roots
are complex, then it

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would be underdamped behavior.

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Now when we are given
different types of roots,

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with repeated roots
it's important to know

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how to construct the
solutions as well.

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And that's by introducing
new types of functions where

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we're multiplying
by polynomial in t

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in front of the exponential.

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So that ends the
session for today.