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PROFESSOR: Hi everyone.
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Welcome back.
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So today, I'd like
to tackle a problem
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in undetermined
coefficients, specifically
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find a particular
solution to each
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of the following equations
using undetermined coefficients.
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So for part A, we have
x dot plus 3x equals
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t squared plus t.
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And for B, we have x dot
dot plus x dot equals t
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to the four.
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So I'll let you work
this problem out.
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And I'll come back in a minute.
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Hi Everyone.
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Welcome back.
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So we're asked to solve this
problem using the method
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of undetermined coefficients.
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And specifically, the
observation is if we have
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a differential equation
with constant coefficients,
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and we have a forcing on
the right-hand side which is
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a polynomial, then there's
always going to be a particular
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solution, which is a polynomial,
that has the form of some
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constant times t to the power
of r plus constant t^(r-1) plus
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a constant c_(r-2) t^(r-2)
plus dot, dot, dot,
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plus c_1*t and then
possibly plus c_0.
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And typically, the
problem is to find out
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what's the highest-order
polynomial that we
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should guess.
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These constant c's,
they're referred to as
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the undetermined coefficients.
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And these are the constants
that we seek to solve.
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So for our first
differential equation,
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we have x dot plus 3x
equals t squared plus t.
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And the power that we
should guess for a solution
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are, it's always going to be
at least as high as the highest
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power on the right-hand side.
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So the right-hand side tells
us at least the largest
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that we should at least
start our guess at.
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And then, sometimes, we have
to knock it up a few orders
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depending on what the
lowest-order derivative
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on the left-hand side is.
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So in this case, the term
on the left-hand side,
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there's no derivative term.
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And so what we can
do for this case
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is we can just take
r is equal to 2,
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the power of the polynomial
on the right-hand side.
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So we're going to seek
a guess or an ansatz,
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which is some constant
A times t squared
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plus B times t plus
some constant C.
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And again, I've taken
the highest polynomial
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that we should guess to be 2,
because the right-hand side is
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2 in this case.
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So now what we do is
we just substitute this
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in into the
differential equation.
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And we choose our A, B, and C
to construct this as a solution.
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So let's go ahead and do that.
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So taking its derivative, we
have 2A*t plus B. And then,
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we have 3 times A*t
squared plus B*t plus C.
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And we want this to be
equal to t squared plus t.
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And now, the only way
that the left-hand side
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is going to equal
the right-hand side
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is if the coefficients in
front of each polynomial
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are the same.
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So in the left-hand
side, we only
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have one term that
has a t squared in it.
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And that must, therefore,
equal the t squared term
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on the right-hand side.
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So these two terms have
to be equal to each other.
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So we end up getting that
3A has to be equal 1.
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And so A is equal to one third.
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And now what we do is we
collect terms with a power of t.
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This gives us 3B plus 2A
on the left-hand side,
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so that's this
term and this term.
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And we want that t equal t on
the right-hand side or just 1.
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Now, we already know that
A is equal to one third.
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So we end up getting that
3B is equal to 1 minus 2/3,
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which is equal to one third.
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Put 3B as one third, which
means that B is equal to 1/9.
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And now lastly,
to determine C, we
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see that B plus 3C must be 0.
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So B plus 3C is 0, or C
is equal to negative 1/27.
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So notice how we always
start with the highest power.
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In this case, A
is the coefficient
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in front of t squared.
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That lets us solve
for A. And then
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the rest of the
undetermined coefficients,
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we can solve for, almost
like a giant zipper.
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So the particular solution
that we just constructed,
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when the dust settles is 1/3 t
squared plus 1/9 t minus 1/27.
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So this concludes part A.
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For part B, we have the
differential equation
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x dot dot plus x dot
equals t to the four.
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And in this case, we know
that the right-hand side
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is a power of four.
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So we should guess at least
a fourth-order polynomial.
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However, we see that there's
no constant multiple of x.
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In fact, the lowest-order
derivative is 1.
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So what this means
is we should actually
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knock the solution that we
guess up one order polynomial.
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So we should try and guess x is
equal to A t to the fifth plus
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B t to the four plus C t cubed
plus D t squared plus E*t.
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And we can leave out an F term.
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We can leave out a constant.
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Because if we just substitute
that in the left-hand side,
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we see that the x dot dot is
going to vanish, and x dot
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is going to vanish.
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So we can omit
any constant term.
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So let's seek this
ansatz, and plug it
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into the differential equation.
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And if we do that,
we get 20A t cubed.
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And this is plugging
into the x dot dot term.
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This term is going to give us
12B t squared plus 6C*t plus D.
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And then the x dot term is going
to give us 5A t to the four
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plus 4B t cubed plus 3C t
squared plus 2D*t plus E.
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And we want the sum of these two
terms to equal t to the four.
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So again, what we do is we
start with the highest power.
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We see that on the
left-hand side,
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we have 5A must be
equated with just 1.
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So we have 5A is equal to
1, or A is equal to 1/5.
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Secondly, we're
going to have 20A
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plus 4B equals the 0
polynomial, or the 0 which
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is the coefficient of t cubed.
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And this gives us B is
equal to negative 5A, which
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is negative 1.
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And now for the quadratic
terms, we have 12B plus 3C is 0.
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So C is going to equal
negative 4B, which
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is negative 4 times negative
1, which just gives us 4.
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The linear term is going
to give us 6C plus 2D
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equals 0, which gives us
D is equal to negative 3C,
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which gives us negative 12.
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And then the last
term is D plus E
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is 0, which gives
us E is equal to 12.
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So at the end of
the day, we end up
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with a particular
solution, which
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is a polynomial of 1/5 t to
the five minus 2 to the four
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plus C, which is 4, t cubed
plus D, which is negative 12,
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t squared plus E,
which is 12, times t.
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And by construction,
this polynomial
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solves the differential
equation with a forcing of t
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to the four on the
right-hand side.
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So just to recap,
when we're faced
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with an undetermined coefficient
problem, what we have to do
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is we just guess a solution,
which is a polynomial.
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And the main difficulty is
just guessing the highest power
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of the polynomial.
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And that's always going to be at
least as big as the polynomial
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on the right-hand side.
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But as we saw in
part B, we sometimes
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have to knock it up a
few orders depending
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on what the differential
equation actually is.
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So I'd like to conclude here.
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And I'll see you next time.