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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 homogeneous equations
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with constant coefficients,
and specifically, the case
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where we have real roots.
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And we'll start the
problem off by looking
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at the equation x dot dot
plus 8x dot plus 7x equals 0.
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And we're asked to find
the general solution
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to this differential equation.
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And then we also
have the question,
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do all the solutions go to
0 as t goes to infinity?
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And then for part B, we're
going to take a look at just
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the differential equation
y dot equals negative k*y.
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So this is the same
equation that we've
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seen in past recitations.
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And we're just going to show
that we can use this method
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to solve this
differential equation
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and obtain the same result.
And then lastly, we're asked,
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or we're told that
we have eight roots
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to an eighth-order differential
equation, negative 4,
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negative 3, negative 2,
negative 1, 0, 1, 2, and 3.
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And we're asked, what
is the general solution.
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So why don't you take a moment
and try and work these problems
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out, 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 we're asked to find the
general solution to x double
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dot plus 8x dot
plus 7x equals 0.
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And we see that this is
a differential equation,
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it's linear, and it has
constant coefficients.
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And whenever we have a
differential equation
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that's linear with
constant coefficients,
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one of the standard ways
to generate the solution is
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to seek what sometimes
mathematicians call an ansatz,
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but it's to try a solution
of the form x is equal
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to a constant
times e to the s*t.
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And if we substitute a
solution n of this form,
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we see that taking the second
derivative of this function
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pulls down two s's.
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One derivative pulls down one s.
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We have no derivatives here.
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And we also have, on each
term, a factor of c times e
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to the s*t.
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And we want this to be 0.
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So specifically, c e to the
s*t can't be 0 for all time.
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So the only way
that this can hold
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is if s squared plus
8s plus 7 equals 0.
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So what this means is if
we choose s to solve this
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polynomial, then x equals c e
to the s*t will be the solution.
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And this will be the
solution for any constant c.
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OK, so what are the roots
to this algebraic equation.
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Well, we can factorize it.
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The roots are going to be
negative 7 and negative 1.
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And notice how
this whole process
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has turned a differential
equation into a simpler
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algebraic equation.
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So if we can solve the
algebraic equation,
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then we can solve the
differential equation.
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OK, so the general solution.
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Well, we've just shown that we
can take any constant times e
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to the s*t, provided s is equal
to negative 1 or negative 7.
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So the general solution is going
to be some constant, c_1, times
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e to the minus t, plus c_2,
can be a different constant,
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e to the minus 7t.
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So notice how
there's two constants
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in the final solution.
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And the reason
there's two constants
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is because we started out with
a second-order differential
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equation.
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So, in some sense,
for each order
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of the differential equation,
we always have one constant.
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It's almost as if for
each time we integrate,
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we have a constant
of integration.
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So at the end of the day,
we have two constants
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in our general solution.
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As part of part A, we're
also asked for any solution
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to this differential
equation, does the solution
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go to 0 as t goes to infinity?
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Well, the general
solution has this form.
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So for any constant c_1
and c_2, the solution
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is c_1 e to the minus t
plus c_2 e to the minus 7t.
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And we see that no matter
what c_1 and c_2 are,
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this term, as t
goes to infinity,
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is multiplied by e to the
minus t, which goes to 0.
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And the second term
also goes to 0.
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So as t goes to infinity,
both e to the minus t
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and e to the minus
7t both go to 0.
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So that means that
any constant times
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e to the minus t plus
any constant times
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e to the minus 7t
must also go to 0.
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So hence, x of t goes to
0 as t goes to infinity.
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OK.
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For part B, we have the
differential equation y dot
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equals negative k*y.
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And this is the first-order
linear differential equation
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with constant coefficients.
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And we're going to
use the same trick.
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We let y is equal to
c times e to the s*t.
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And we see that the
characteristic equation
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in this case, it's
not a polynomial.
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It's just s, s is
equal to negative k.
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So we get y is equal to c e to
the negative k*t is the general
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solution.
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And this is exactly what we
had in previous recitations,
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when we used, for example,
integrating factors
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to solve this very same
differential equation.
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So this just shows that
we can use the same method
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to solve first-order linear
differential equations.
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OK.
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Now, lastly, we're
given eight roots
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to an eighth-order
differential equation.
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An eighth-order
differential equations
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with constant coefficients.
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So I'll just write
out the roots again.
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So we're told the
roots are negative 4,
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negative 3, negative 2,
negative 1, 0, 1, 2, and 3.
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And in general, the solution
to an eighth-order differential
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equation whose roots to the
characteristic polynomial
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are negative 4 through 3,
the general solution, x of t,
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is going to be a
constant c_1 times e
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to the power of the first
root, which will be minus 4t,
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plus c_2 e to the minus 3t.
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And of course, we take different
constants for each term.
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c_3 e to the minus 2t, plus
c_4 e to the minus t, plus c_5.
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And now for this term,
it should be e to the 0t,
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but e to the 0t is just 1.
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So the zero root is just going
to give us a constant c_5.
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And we have c_6 to the
t, plus c_7 e to the 2t.
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And then plus c_8 e to the 3t.
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So the solution has eight
terms and eight constants.
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And just for fun, we can
ask, does every solution
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to this differential equation
go to 0 as t goes to infinity.
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And the answer is no.
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In fact, although each
term with a negative root
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does go to zero as
t goes to infinity,
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there are three terms that
go to positive infinity
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as t goes to infinity,
and there's one term
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that just stays constant.
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So in general, as t
grows, goes to infinity,
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these terms will
become very large
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and won't necessarily go to 0.
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Well, they'll never go to 0.
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So I'd just like
to conclude there,
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and I'll see you next time.