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PROFESSOR: Welcome back.

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So today we're
going to take a look

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at a problem with
partial fractions,

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and specifically how to
use these partial fractions

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to compute Laplace inverses.

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So just as a
warm-up, we're asked

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to recall what the formula is
for the Laplace transform of f

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prime, in terms of the
Laplace transform of f.

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In the second part,
we're asked to find

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the inverse Laplace transform
for three different functions.

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The first one is 1
over s squared minus 4,

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s squared divided by
s squared plus 4, e

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to the minus 5s divided
by s squared minus 4.

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And notice how none
of these functions

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appear in the look-up table
for Laplace transform.

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So in each case, we have
to use partial fractions

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to convert it or massage
these functions into something

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which we do know the Laplace
transform inverse for.

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And then lastly, for
the third problem

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we're asked to write
down the partial fraction

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decomposition of this
function, 1 over s

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squared times s squared plus 4
times s plus 1 times s plus 3.

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And specifically, when we do the
partial fraction decomposition,

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we're just going to leave
the undetermined coefficients

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unknown.

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We're not actually
going to solve for them.

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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 moment.

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Hi everyone.

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Welcome back.

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OK.

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So for part one,
we're just asked

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to recall the Laplace
transform of f prime in terms

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of the Laplace transform of f.

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So the Laplace
transform of f prime

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we can write down as s times the
Laplace transform of F, which

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I'll use capital F, minus
f evaluated at the lower

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bound of the Laplace transform
integral, which in this case

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is just zero minus.

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OK.

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So this is just part one.

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This is a warm-up problem.

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For part two,
we're asked to find

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the inverse Laplace transform
for three different functions.

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The first one is 1
over s squared minus 4.

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And we see here that we can
factor the denominator into 1

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over s minus 2 times 1 over s
plus 2, which means that we can

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use a partial fraction
decomposition which

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has the form of A
divided by s minus 2

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plus B divided by s plus 2.

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And then we need to solve
for the coefficients A and B.

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So one way to solve
for the coefficients

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A and B is just to multiply
both sides of this equation

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through by the factors
s minus 2 and s plus 2.

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We can then plug in values
of s and solve for A and B.

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There's another way,
which is sometimes

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referred to a cover-up method.

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And in this case,
what we do is we pick,

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for example, say
one of the places

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where the denominator blows up.

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So for example, if we look
at this factor s minus 2,

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this factor diverges
when s approaches 2.

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So then what we do is we go
back to our original function,

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we cover up the term s minus
2 where it diverges, and then

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in the remaining term, we
plug in the value of s which

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causes the factor s
minus 2 to diverge.

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So in this case, s
minus 2 diverges at 2.

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So we would cover this
factor up and then plug

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in s is equal to 2.

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And this would give us the
value of A. So for this problem,

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A is equal to 1 over 2 plus
2, which is just 1 over 4.

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For B, we look at plugging
in s equals negative 2.

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So we cover up the
factor that diverges

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in the original function.

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Plugging in s equals negative
2 gives me negative 2

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minus negative 2.

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So B is just negative 4.

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So this function is 1/4 1
over s minus 2 minus 1/4 1

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over s plus 2.

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And when we take the
inverse Laplace transform,

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so the inverse Laplace transform
of 1 over s squared minus 4

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is going to be 1 over 4-- the
inverse Laplace transform of s

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minus 2 is e to the plus 2t.

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The second factor is negative
1/4 e to the negative 2t.

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And this concludes
the first problem.

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For the second function,
we have s squared

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divided by s squared plus 4.

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So again, this function's
not in the correct form

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to use a Laplace
transform look-up table.

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But what we can do
is we can divide--

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long division of polynomials--
the numerator out

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by the denominator.

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So when we do that we end
up with 1 minus 4 divided by

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s squared plus 4.

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So when we take the inverse
Laplace transform of s

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squared divided by
s squared plus 4,

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we're left with the inverse
Laplace transform of 1

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minus the inverse
Laplace transform of 4

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divided by s squared plus 4.

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And if we use our
look-up table, we

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know that the inverse
Laplace transform of 1

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is the delta function.

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Meanwhile, the inverse
Laplace transform

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of some number
divided by s squared

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plus 4 looks like, in
this case, sine of 2t.

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Now we need a 2 upstairs,
so this is going

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to give us 2 times sine of 2t.

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And I obtain this by noting that
the Laplace transform of sine

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omega*t is equal to omega
divided by s squared plus omega

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squared.

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So that's how I went from
this function, this Laplace

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transform inverse, to 2 sine 2t.

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Lastly, for part three,
the third problem,

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we're asked for the inverse
Laplace transform of e

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to the minus 5s divided
by s squared minus 4.

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Now note that we already know
the inverse Laplace transform

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to 1 over s squared minus 4.

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In this case, we're
just multiplying 1

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over s squared minus 4
by e to the minus 5s.

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So we can use the shift
formula in addition

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to what we already computed,
the inverse Laplace transform

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to 1 over s squared
minus 4, to compute

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this inverse Laplace transform.

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So the shift
formula says that we

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need to multiply by the
step function t minus 5.

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So we're shifting with
the step function.

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And then what we do
is we want to shift

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the inverse Laplace transform
of 1 over s squared minus 4.

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So the inverse Laplace transform
to 1 over s squared minus 4

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is 2 t minus 5.

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So if we shift t by 5,
we end up with 1/4 e

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to the 2 t minus 5 minus 1/4
e to the minus 2 t minus 5.

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So again, just to
reiterate, when

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we have an exponential
e to the minus, say, 5s

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and we already know the inverse
Laplace transform of everything

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else, we just multiply
by the step function u

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t minus 5, the same shift.

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And then wherever we see
the t in the inverse Laplace

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transform, we just replace it
with t minus 5, in this case.

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OK?

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And now lastly, for
part three we're

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just asked to write down the
decomposition for this very

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large function, the partial
fraction decomposition

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of this large function.

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So it's s squared times
s squared plus 1 times

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s plus 1 times s plus 3.

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Let me just double
check. s squared plus 4.

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OK?

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And unfortunately
this function's

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going to be a little ugly.

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We're going to have a constant,
A, divided by s plus B divided

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by s squared.

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So these first two terms come
from this term right here.

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For the s squared
plus 4 term, we're

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going to seek a C
times s plus D--

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so we need both factors,
Cs plus D-- divided by

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s squared plus 4.

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We have E divided by s plus 1.

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And then plus F
divided by s plus 3.

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So just to associate each
factor in the partial fraction

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decomposition with
the original function,

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I've just drawn some arrows.

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OK?

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And then lastly,
although I didn't

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state this in the
original problem,

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we can also just write down
the inverse Laplace transform

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of this entire mess
without actually

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determining the constants.

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We can just keep
the constants in.

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So if we wanted to compute
the inverse Laplace transform,

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well, it would just be A--
because inverse Laplace

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transform of 1 over s is
1-- plus B-- inverse Laplace

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transform of s
squared is t-- plus,

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now this part has two terms,
Cs divided by s squared plus 4.

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That's going to give us
C times cosine of 2t.

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The D divided by s squared plus
4 is going to give us, well,

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we need to multiply upstairs by
2 and divide downstairs by 2.

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So we get D divided
by 2 of sine 2t.

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This term, E, is going
to give us E times

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the exponential of negative t.

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And then the last term is
going to give us F times e

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to the minus 3t.

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OK?

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So just as a quick
wrap-up, in this problem

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we've computed partial
fraction decompositions

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of several functions.

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And the reason we've done
this is because we often

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end up in a situation when
we want to solve an ODE using

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Laplace transforms where we
have some complicated function

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and we need to use partial
fraction decomposition to write

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down simpler functions,
in which case

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we can then use the Laplace
inverse formula on each

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of the simpler functions.

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OK?

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