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PROFESSOR: Hi everyone.

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Welcome back.

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So today, I'd like to take a
look at solving a linear ODE

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but with a periodic input.

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And specifically,
we're asked to find

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one solution, a particular
solution, which is also

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the periodic solution of
the differential equation

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x dot dot plus 2x dot plus
4x equals the square wave

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function.

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So the square wave function
is a periodic function

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with period 2pi.

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It's defined as minus 1 and
1 on the intervals minus pi

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to 0 and 0 to pi.

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And we know that the
square wave function has

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the following Fourier series.

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So I'll let you think about
this problem for a moment.

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And I'll come back in a second.

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Hi everyone.

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Welcome back.

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So the reason we've been
studying Fourier series

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is to essentially solve
differential equations

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with complicated forcing inputs
on the right-hand side which

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are periodic.

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And the reason we've been
studying Fourier series

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is because we know that
differential equations

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with sines and cosines
as forcing terms

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on the right-hand side are
relatively easy to solve.

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And we want to be able to
solve the same differential

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equation with a more
complicated periodic function

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on the right-hand side.

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So the general approach
is to first decompose

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the right-hand side
into a Fourier series.

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And this step is essentially
already done for us.

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We're told what the
Fourier series is.

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And what we do is
we solve the ODE x

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dot dot plus 2x dot plus 4x.

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And I'm just going to take one
term of the Fourier series,

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sine n*t.

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So if we know the
right-hand side is

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a sum of a whole bunch of
sines, what we're going to do

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is we're going to solve for any
specific one of those sines.

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So we want to solve this
differential equation.

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And then, we use superposition.

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So if we know what the
solution is to one sine n*t,

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and we know that the right-hand
side forcing is a sum of many

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sine n*t's with
appropriate weight factors,

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then we can use superposition
to construct a final solution.

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So this is the method
to attack this problem.

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So we were given step one,
and we want to solve step two.

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And we can solve this just
using the exponential response

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formula.

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So what we're going
to do is we're just

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going to complexify
the right-hand side.

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So if I want to solve the
differential equation x dot dot

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plus 2x dot plus
4x equals sine n*t,

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I'm going to denote the
solution with the subscript n.

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And it's going to be
the imaginary part

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of-- and I'm just using the
exponential response formula--

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1 over the characteristic
polynomial evaluated at i*n,

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e to the i*n*t.

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You might ask how do I get that.

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Well, I basically just took
the complexified equation,

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and noted that sine n*t
was the imaginary part of e

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to the i*n*t, and then used the
exponential response formula.

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And I want to take the imaginary
part at the end of the day.

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So what is the
characteristic polynomial?

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In this case, it's s
squared plus 2s plus 4,

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which means that p of i*n is
going to be negative n squared

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plus 4-- so I'll just group
the real terms together--

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plus 2i*n, so the 2i*n
comes from the 2s term.

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And then x of n is going to be
the imaginary part of 1 over 4

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minus n squared plus
2i*n e to the i*n*t.

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And I'm going to use the
amplitude-phase form.

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I'm going to convert this
Cartesian complex number

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into an amplitude-phase
form just because it's

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going to make taking the
imaginary part of the solution

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very easy at the end of the day.

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So the amplitude of this complex
number is 4 minus n squared,

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quantity squared, plus the
imaginary part squared,

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so that's 4n squared,
square rooted.

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And then, upstairs is e^(i*n*t).

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And downstairs, it's
going to be e^(i*phi).

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And I'm going to put a
subscript n on the phi

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just because for
each complex number,

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we're going to have a
different phase phi.

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And that phase
phi_n is going to be

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the arctangent of 2n divided by
4 minus n squared, like that.

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So now, I can combine phi of n
with the upstairs term, i*n*t,

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in the exponent.

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And when I take
the imaginary part,

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I'm only going to be left
with sine of n*t minus phi_n.

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So x of n is going to be 1 over
4 minus n squared squared plus

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4n squared quantity
squared-- square rooted--

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times sine of n*t minus phi_n,
where phi_n was given using

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the arctangent formula.

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And this gives us a
solution, which note

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is periodic with period 2t--

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AUDIENCE: [INAUDIBLE].

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PROFESSOR: Right.

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I should also note that phi
of n is between 0 and pi.

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And this gives us this one
solution to the differential

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equation with a forcing of sine
n*t on the right-hand side.

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So now what we want to
do is we want sum up

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many of these solutions using
the superposition principle.

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So if I go back now,
I'm going to write

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the original
differential equation.

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So I'm just going to go
back and just rewrite this.

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And I'm going to write the
right-hand side using its

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Fourier series, 1 over n
sine n*t, where n is odd.

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And this is the problem we
originally wanted to solve.

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And essentially what we've done
is we've solved the problem

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for each individual sine n*t.

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So how do we get
the full solution?

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Well, what we have to do is we
have to multiply the solution

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for each sine n*t by a factor
of 4 divided by pi and 1 over n.

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And then, we have to add
all of these solutions

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up for all odd values of n.

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So for example, x is just going
to be 4 over pi sum of n odd,

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1 over n times the
solution to every sine n*t,

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which we've already computed,
and I've denoted it as x sub n.

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And x sub n is up here.

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So if we want to be
explicit about it,

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I'll write the whole thing out
as 4 over pi, n odd, 1 over 4

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minus n squared,
quantity squared,

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plus 4n squared, square rooted.

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We also have a factor
of n out front.

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We have a sine n*t minus phi_n.

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So there's the final answer
for a particular solution

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in its full glorious detail.

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We can also check that
this particular solution

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is periodic.

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Note how each sine
n*t is periodic.

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And each sine n*t has
period of at least 2pi.

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So when we sum up a
whole bunch of functions

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which all have a
period of at least 2pi,

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the sum is also going to be
periodic function with at least

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2pi.

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And as a result, this gives us
the answer we're looking for.

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Also, in addition, if we want
the full general solution

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to the differential equation
to this particular solution,

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we also have to add
the homogeneous piece.

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So this concludes the problem.

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And I'll just quickly recap.

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When solving a
differential equation

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with a periodic forcing
function on the right-hand side,

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again, to iterate
the steps, you first

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Fourier decompose
the right-hand side

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into summation of
sines and cosines.

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You then solve the differential
equation for sine n*t,

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cosine n*t individually.

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This gives you a
solution for each term

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on the right-hand side of
the differential equation.

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And then at the
end of the day, you

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use superposition to sum
up all of the solutions.

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And that gives you one
final big solution.

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So I hope you
enjoy this problem.

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And I'll see you next time.