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PROFESSOR: Hi, everyone.

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Welcome back.

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So today we're going to take
a look at solving differential

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equations using the
Laplace transforms,

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and the problem we're going to
take a look at is a simple ODE,

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x-dot plus 2x equals
3 delta of t plus 5,

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as a forcing on the
right hand side.

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We're going to look at having
rest initial conditions,

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x of 0 minus is
equal to 0, and we're

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asked to use Laplace transforms
to solve this initial value

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problem.

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For part B, we're asked
to have the initial value

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problem without any
delta function forcing

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on the right-hand side to give
an equivalent initial value

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problem without a delta function
forcing on the right-hand side,

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but yields the same
solution as in part A.

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And then in question
two, we're asked

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to solve the
second-order differential

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equation, x-dot-dot
plus 9x equals u of t

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with rest initial conditions,
so x of 0 minus is 0,

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x-dot of 0 minus is also 0.

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So I'll let you work
on this problem,

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and I'll be back in a moment.

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Hi, everyone.

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Welcome back.

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OK, so for part
A, the first step

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is to Laplace transform
both sides of the equation.

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So we take the Laplace
transform of x-dot plus 2x,

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and that's going to be equal
to the Laplace transform of 3

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delta of t plus 5.

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And we can use the distribution
properties of the Laplace

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transform, so this is going to
be Laplace transform of x-dot

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plus 2 times the
Laplace transform of x.

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On the right-hand side, we have
3 times the Laplace transform

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of the delta function plus 5
times the Laplace transform

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of 1.

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Now, we can replace the
Laplace transform of x-dot

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if we use the identity with
s times the Laplace transform

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of x minus x at 0 minus.

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And we're told that x
of 0 minus in this case

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is just 0, so this
term's going to vanish.

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And just for brevity,
I'm going to write X of s

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as the Laplace transform of x.

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So we now have s
times X of s plus 2 X

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of s equals, and on
the right-hand side,

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we have three times
the Laplace transform

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of the delta function.

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Laplace transform of the
delta function is just 1,

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so we have 3 plus 5, and
the Laplace transform of 1

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is just 1 over s.

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So I can now factor
the left-hand side,

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and I get X of s times s plus
2 equals 3 plus 5 over s.

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And note how, when we
have X of s multiplied

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by some polynomial
in s, this is always

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going to be the
characteristic polynomial.

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So if we look back, s plus 2 is
the characteristic polynomial

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of x-dot plus 2x.

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So this yields 3 divided by s
plus 2, on the right-hand side,

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plus 5 divided by
s times s plus 2.

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And for the second piece,
we can use partial fractions

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to decompose it
into a term times s

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and a term times s plus 2.

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And when we do that, we end up
getting 5/2 1 over s minus 1

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over s plus 2.

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So I can combine the 3 divided
by s plus 2 with the minus 5/2

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divided by s plus
2 into one term.

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So this gives you 1/2 1 over s
plus 2, and we also have 5/2 1

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over s.

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And now we just take the
inverse Laplace transform

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of both sides.

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So on the left-hand side, we
recover x of t, so we get 1/2 e

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to the minus 2t plus 5/2.

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The inverse Laplace
transform of 1/s is just 1.

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So we end up with x of t is
1/2 e to minus 2t plus 5/2,

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and this solution is
valid for t bigger than 0.

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Sometimes, people write
it as this quantity

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multiplied by step function.

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And x of t is also 0 for t
less than 0, for example.

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And it's just useful to quickly
sketch what x of t looks like,

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so it's an exponential
decay for t bigger than 0,

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and it's flat for t less than 0.

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So here's our x of t.

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So for part B, now, we're asked
to find a differential equation

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and new initial conditions that
reproduce the solution offered

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t bigger than 0.

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So note how we'd be looking
for a new solution, x of t,

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which would be an
exponential decay.

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And essentially, we
just grow, so we're

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looking at initial
conditions, which start at 0.

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If I were to write the original
differential equation-- so this

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is the original differential
equation from part

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A. Looks like this.

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And just quickly to note
that x-dot near the origin

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is going to be approximately e
times the delta function, which

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means in the original
differential equation,

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x is going to have a jump
of 3 about the origin.

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So the new initial
value problem--

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well, we don't want
the delta function

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on the right-hand side, so we're
going to solve x-dot plus 2x

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is equal to 5.

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But what initial
conditions do we need?

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Well, we need x of
0 minus to now be 3.

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So when we eliminate the 3
delta on the right-hand side,

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we have to introduce new initial
conditions so that the solution

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agrees for t bigger than 0.

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OK, so this concludes part one.

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For part two, we're asked
to solve a new differential

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equation, x-dot-dot
plus 9x equals u of t,

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and we're just going to follow
the same procedure where we

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Laplace transform both sides.

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So Laplace transforming
the left-hand side gives me

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x-dot-dot plus 9x equals the
Laplace transform of u of t.

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And again, I can use
the formula which

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relates derivatives of x
back to the Laplace transform

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of x, and so in this case, the
Laplace transform of x-dot-dot

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is going to be s squared times
the Laplace transform of x.

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And then I'm going to
have plus a term which

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involves x of 0 minus
and a term which

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involves x-dot of 0 minus.

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And if your initial
conditions were not 0,

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you would have to
keep these terms in.

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However in our case,
these terms are

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both 0 because we deal with
rest initial conditions,

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I'm just not going
to write them.

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Plus 9 X of s equals-- the
Laplace transform of u of t

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is 1/s.

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So again, we have X of s s
squared plus 9-- so note again

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how this is the same
characteristic polynomial

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as in our differential
equation-- is equal to 1/s.

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So X of s is 1/s
s squared plus 9,

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which we can use
partial fractions, now,

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to decompose it into A over s
plus B*s plus C divided by s

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squared plus 9.

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And if I were to take a look at
my notes, I have, in this case,

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A is 1/9, B is negative
1/9, and C is equal to 0,

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if you were to work
out these coefficients.

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So what this means is X of s
is 1/9 1 over s minus 1/9 1

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over s squared plus 9.

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And when we invert the Laplace
transform, the inverse of 1/s

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is just 1, so x of
t becomes 1 over 9.

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Sorry, this should be s up here.

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The inverse Laplace
transform of s divided by s

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squared plus 9 is
cosine 3, so we end up

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with negative 1/9 cosine
of 3t, and again, this

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is a solution for
t bigger than 0.

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So as soon as we turn on the
input, the function x of t

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starts growing
continuously from 1,

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and then achieves an
oscillation with period 3.

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So just to quickly
recap, in this problem,

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we solved several ODEs,
several initial value problem

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ODEs using Laplace transforms.

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Laplace transforms are
particularly nice because they

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convert an ODE into an
algebraic equation, which

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we can solve fairly easily.

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The drawback is
we sometimes have

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to manipulate, using
partial fractions,

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the right-hand
side into functions

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that we know how to invert using
the Laplace transform inverse.

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So I'll just conclude here,
and I'll see you next time.