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PROFESSOR: Hi, everyone.

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Welcome back.

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In this problem, I'd like to
take a look at convolution

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and Green's formula.

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Specifically, for
part one, we're

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just asked to compute
the convolution of t

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with itself, t.

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In the second part, we're asked
to compute the convolution of e

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to the minus k*t with another
exponential, e to the a*t.

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And then in part B, we're to use
the result from part A as well

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as Green's formula to solve
x dot plus k*x equals e

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to the a*t.

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And specifically, we're
interested in the rest

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initial condition
solution to this formula.

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OK.

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So I'll let you think
about this problem,

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and I'll be back in a minute.

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Hi, everyone.

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Welcome back.

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OK, so for part one,
we're asked to compute

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the convolution of two
functions, t with itself,

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and I'll just quickly write down
the formula for a convolution.

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So if I have two
functions, f and g,

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then the convolution is
defined as, in this case,

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the integral from
zero to t of f,

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so the function f,
evaluated at, we're

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going to use tau as the
integration variable,

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multiply by g, t
minus tau, d tau.

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So notice how the
variable t appears

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in two places in this formula.

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It appears in one place
in the argument of g,

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and it appears in the
bound of integration.

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Meanwhile, tau is the variable
we're integrating over,

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and tau appears in two places
as well, one in f and one in g.

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So if we're interested
in t convolved with t,

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we have the integral from zero
to t of tau times t minus tau.

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So notice how f is
just t and g is t,

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so when we insert tau and t
minus tau into the function,

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we end up just getting tau
times t minus tau, d tau.

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And just expanding this
out, we have t times tau

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minus tau squared, d tau.

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So in the integral,
the variable t always

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appears just as a constant.

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So in this case, we
have t, tau integrated

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becomes t squared over 2.

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When I integrate from zero
to t, tau squared integrates

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to 1/3 t cubed,
and zero drops out

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from the other end
of the integral.

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So we end up with t
cubed, 1/2 minus 1/3,

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which is equal to 1/6 t cubed.

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So there's the answer
we're looking for.

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For part B, or for
question two, again we

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have another computation to do.

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So we have more
integrals to work out.

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And in this case,
it's two exponentials.

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So we have zero to t.

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The first one is going
to be e to the k*tau.

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And the second one is e to
the a t minus tau, d tau.

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So we can just
expand things out,

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and I get e to the a*t times
e to the minus a plus k tau d

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tau, zero to t.

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And again, because t is just
a dummy variable in this

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integral, we're not
integrating over t,

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I can just think of e to
the a*t as a constant.

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So I'm really just
integrating the function e

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to the minus a plus k tau.

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And when I integrate
this function,

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I get 1 over minus a plus
k, e to the minus a plus k,

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tau evaluated at the
bounds, zero and t.

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So the negative sign here
just flips the bounds.

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So we have a plus k.

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Substituting in zero
just gives us 1,

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and then substituting in t gives
us e to the minus a plus k*t.

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And I can expand out,
multiply out, e to the a*t.

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And when the dust settles,
we have e to the a*t minus e

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to the minus k*t.

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So this is just the computation
of the convolution e

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to the minus k*t
and e to the a*t.

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So note now if, for example,
k were to equal negative a,

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we would have zero
in the denominator

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and zero on the numerator.

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So one way to compute
the special case when

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k is equal to negative a
is to use L'Hopital's Rule

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and differentiate the top
and the bottom, for example,

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with respect to k.

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Another alternative way of
computing the special case when

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a is equal to negative k is
just to plug in e to the k,

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and then just work
out this integral,

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and you'll obtain
a different answer.

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And now lastly, for
part B, we're interested

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in the differential equation
x dot plus k*x equals e

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to the a*t.

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And we're interested in finding
the particular solution that

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has rest initial conditions.

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So for example, x of
zero is equal to zero.

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And we want to use
Green's formula.

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So just recall that there's
the impulse response

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formula for the
weight function w,

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which solves the same
differential equation,

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but with the delta function
on the right-hand side.

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And w, in this case, as
we've seen in lecture,

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is actually e to the negative
k*t when t is bigger than zero,

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and at zero when t
is less than zero.

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So notice how e to the minus k*t
is exactly the function that we

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convolved in part
A. Specifically,

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we convolved it
with e to the a*t.

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So for example, Green's formula
says that the particular

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solution that has rest initial
conditions is going to be

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the weight function convolved
with the right-hand side

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of the differential
equation, e to the a*t.

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In this case, on the
domain of integration

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for the convolution, w is
just e to the minus k*t.

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So we have e to the minus k*t
convolved with e to the a*t.

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So Green's formula
gives us the solution

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to this differential
equation, which

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has rest initial conditions.

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And specifically-- we've already
worked this out from part A,

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so I can just write
down what the answer is.

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It's 1 over a plus k, e to the
a*t, minus e to the minus k*t.

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And just as a quick
check to make sure

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that we've done
everything correctly,

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we can plug in t equals zero.

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And when t equals zero, we have
1 minus 1, which gives us zero.

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So indeed, this
x is the solution

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to this differential
equation, which

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has rest initial conditions.

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I'd just like to
conclude this problem.

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Quick recap.

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We worked out
several convolutions,

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and specifically we were able to
use Green's formula in addition

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to the convolution
that we worked out

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to compute the solution
with rest initial conditions

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to an ODE.

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Moreover, Green's function
is very useful because notice

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how we could have
computed the convolution

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for any right-hand
side function here.

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So in some sense, we're able to
generalize and find solutions

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to differential equations which
have arbitrary right-hand side

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forcings, not just sines and
cosines, periodic functions,

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or other simple
functions which we've

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been looking at in the past.

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So I'd like to conclude here,
and I'll see you next time.