WEBVTT
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>> Welcome to the Cypress
College Math Review
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on Laplace Transforms and
Differential Equations.
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First let's review how to
take the Laplace Transform
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of derivatives of functions.
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Let's take the Laplace Transform
of the first derivative.
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So n is equal to 1.
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So we get s to the first power,
times the Laplace Transform
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of F, minus -- since n is equal
to 1, this term just becomes f
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of 0 and there are
no other terms.
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Now if n is equal to 2, we're
taking the Laplace Transform
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of the second derivative.
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So now we have s to the
second power times the Laplace
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Transform of F, minus
the sum, n is equal to 2.
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So this first term becomes s to
the first power, times f of 0,
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plus -- now we'll have
s to the 2 minus 2 power
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so now we have no s's, we
just have f prime of 0.
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And then that's the
last of our sum.
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If n is equal to 3, we're
taking the Laplace Transform
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of the third derivative,
which should be enough.
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So now we have s
to the third power,
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times the Laplace
Transform of F minus the sum.
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So now we have s to the
second power times f of 0,
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plus s to the first, times
the derivative of f, plus --
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no s's anymore -- and now we
have the second derivative
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evaluated at 0.
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Let's solve this initial value
differential equation using
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Laplace Transform.
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First let's transform
the equation
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so the Laplace Transform applied
to both sides of the equation.
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First we have the
Laplace Transform
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of the second derivative.
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So that's s squared times
the Laplace Transform
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of the function x,
minus s times little x
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at 0, minus x prime at 0.
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Then we have 4 times
the quantity --
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then we have the
Laplace Transform
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of the first derivative, which
is s times the Laplace Transform
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of X minus little x at 0.
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Then we have plus 5 times
the Laplace Transform
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of X. That's equal to the
Laplace Transform of 25t.
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Looking at a table
of transforms,
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we notice that the
Laplace Transform of t
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to f power is the
quotient of n factorial
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over s to the n plus 1.
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Well, n in this case would be 1,
so we would have the constant,
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which is 25 times one factorial,
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over s to the 1 plus
1 or s squared.
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There we go.
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Now let's substitute in
the values that were given.
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x of 0 is negative 5.
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And x prime at 0 is 7.
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Multiplying both sides of
the equation by s squared
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and factoring out cap X of s.
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Over on the left
hand side we have s
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to the fourth plus 4s
cubed, plus 5s squared.
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Moving all terms that do
not involve cap X of s
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to the other side -- and
once again we're calling
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that we're multiplying
by s squared.
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We have 25 minus 13s
squared minus 5s cubed.
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Solving for cap X of s we have
25 minus 13s squared minus 5s
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cubed, factoring out an s
squared in the denominator,
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we have s squared
plus 4s plus 5.
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Next we set ourselves up for a
partial fraction decomposition.
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So we have A over s, plus B
over s squared, plus C s plus D
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over s squared plus 4s plus 5.
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OK, multiply by the
original denominator
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and we have 25 minus 13 s
squared minus 5 s cubed equals A
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times s, times the quantity
s squared plus 4s plus 5.
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Plus B times the quantity
s squared plus 4s plus 5.
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Plus the quantity Cs
plus D times s squared.
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We can let s be 0 and
we immediately get
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that 25 is equal
to 5B, so B is 5.
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But that's the only easy value
we're going to get that way.
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So next I think we'll
equate coefficients.
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So negative 5 s cubed, minus 13s
squared plus 0s plus 25 is equal
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to As cubed plus 4As squared,
plus 5As, plus B s squared,
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plus 4Bs, plus 5B, plus
Cs cubed, plus Ds squared.
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Now let's group like terms.
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For s cubed, we have A plus C.
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For s squared we have
4A plus B plus D. For s
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to the first power
we have 5A plus 4B.
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And our constant is simply 5B.
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We already know that B is
5 so we don't need that.
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The next one that's easiest
to look at would be one
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that involves B and
only one other unknown,
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which would be this equation.
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So 5A plus 4B is equal to
0, by equating coefficients,
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so 5A plus 4 times 5 is
equal to 0, which gives us
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that A is equal to negative 4.
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Since A is negative 4, the next
one that we should probably look
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at would be the s cubes.
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A plus C is equal to negative 5.
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Since A is negative 4, we
get that C is negative 1.
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The final equation
will give us D.
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This is equal to negative
13 by equating coefficients.
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So D is equal to negative
13 minus B minus 4A,
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so that's equal to
negative 13 minus 5 plus 16,
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so we get a value of negative
2 for D. So cap X of s is equal
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to A, which is negative
4 over s. Plus B,
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which is 5 over s squared.
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Plus C, which is negative
1 times s. Plus D,
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which is negative 2
and all of that is
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over s squared plus 4s plus 5.
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Let's rewrite this.
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Put the coefficients out front.
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So negative 4 times 1
over s, plus 5 times 1
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over s squared minus 1
times quantity s plus 2
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and then we'll complete
the square and write this
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as s plus 2 quantities
squared plus 1.
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Now we want to take the
inverse Laplace Transform
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of each of these three terms.
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Here's a table of
transforms that will help us
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with the first two terms.
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So we want the Laplace
Transform of 1 over s.
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So in this case, n would be 0.
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So the inverse Laplace Transform
of 1 over s, would simply be t
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to the 0 power and the Laplace
Transform of 1 over s squared --
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in this case n would be 1.
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The Laplace Transform
would be t to the first.
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So, so far we have
for our answer --
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x of t is equal to negative
4 plus 5 times t. Now we look
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at our next term.
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Our final term is a
composition of two functions.
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So since we have
the composition,
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we're going to have e to the
ct times f of t. So we're going
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to have e to the -- what is c?
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c is negative 2 because we
have s minus negative 2, yes?
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So that's going to
be negative 2t
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and that's being
multiplied times f of t. f
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of is the inverse
Laplace Transform of s
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over the quantity s
squared plus a squared
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and the inverse Laplace
Transform of that is cosine
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of a times t. a is
this value here, 1.
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So I have cosine of just 1t.
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And there's the answer
to our problem.