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PROFESSOR: OK, this lecture,
this day, is differential
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equations day.
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I just feel even though these
are not on the BC exams, that
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we've got everything
we need to actually
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see calculus in use.
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We've got the derivatives of
the key functions and ready
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for a differential equation.
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And there it is.
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When I look at that equation--
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so it's a differential equation
because it has the
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derivatives of y as well as
y itself in the equation.
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And when I look at it, I see
it's a second order equation
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because there's a second
derivative.
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It's a linear equation because
second derivative, first
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derivative, and y itself are
separate, no multiplying of y
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times y prime.
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In fact, the only
multiplications are by these
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numbers m, r and k, and those
are constant numbers coming
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from the application.
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So I have a constant
coefficient, linear, second
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order, differential equation,
and I'd like to solve it.
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And we can do it because it uses
the very functions that
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we know how to find
derivatives of.
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Let me take two or three special
cases where those
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functions appear purely.
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So one special case is if
I knock out the second
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derivative term, and let
me just choose--
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rewrite it in the way it looks
easiest and best. It's just
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the derivative of y as
some multiple of y.
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It's now first order, and we
know the function that solves
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that equation.
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When the derivative equals the
function itself, that's the
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exponential.
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If I want to have an extra
factor a here, then I
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need e to the at.
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And usually, in fact, we expect
that with a first order
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equation, in the solution
there'll be some constant that
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we can set later to match
the starting condition.
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And that constant
shows up here.
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It's just if e to the at solves
that equation, as it
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does, because when I take the
derivative, down comes on a,
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so does c times e to the at.
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That's because the equation
is linear.
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So that's a nice one.
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Pure exponential.
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OK, ready for this one.
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So this one, I don't have
this middle term.
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I just have the second
derivative and the function.
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And let me again change to
letters I like, putting the ky
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on the other side with
a minus sign.
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So this omega squared
will be k/m when I
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reorganize that equation.
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My point is we can solve this
equation, the second
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derivative equaling minus
the function.
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We've met that.
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That's this equation
that the sine and
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the cosine both solve.
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There we get two solutions for
this second order equation.
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And just as with the a in this
problem, so with this number
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here, I'll just jiggle the sine
and the cosine a little
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bit so that we get omega to come
down twice when I take
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two derivatives.
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So the solution here will be--
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one solution will be the cosine
of omega t because two
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derivatives of the cosine
is minus the cosine--
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that's what this asks for--
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with the factor omega
coming out twice.
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And another solution will
be sine of omega t
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for the same reason.
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And again, now with a second
order equation, I'm expecting
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a couple of constants to be
able to choose later.
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And here they are: c cosine
omega t and d sine omega t.
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That's the general solution
to that equation.
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So we know that.
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And these are the two
important ones.
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There's another less important
one and an
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extremely simple one.
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Suppose all that went away, and
I just had as a third very
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special case d second y
dt squared equals 0.
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We sure know the solution
to that.
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What functions have second
derivative equals 0?
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Well, a constant function
does, certainly.
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A constant, even its first
derivative is 0
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much less its second.
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And then t does.
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Its first derivative is
1 and then the second
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derivative is 0.
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So there show up the
powers of t.
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Well, the first two powers, t to
the 0 and t to the 1, show
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up in that very special case.
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Sine and cosine show up here.
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e to the at shows up here.
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And now let me tell you the
good part of this lecture.
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The solution to this equation
and in fact to equations of
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third, fourth, all orders, are
products of these ones that we
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know: exponentials times sines
and cosines times powers of t.
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That's all we need to solve
constant coefficient
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differential equations.
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So I plan now to go ahead
and solve the--
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and move toward this equation.
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I should have said that's a
fundamental equation of
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engineering.
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m stands for some mass.
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Oh yeah, let me draw a picture,
and you'll see why I
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chose to choose t rather than
x, because things are
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happening in time.
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And what is happening?
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Let me show you.
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What's happening in time is
typically this would be a
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problem with us some kind of a
spring hanging down, and on
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that spring is a mass m.
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OK, so what happens if I--
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I've pulled that mass down, so
I've stretched the spring, and
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then I let go.
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Then what does the spring do?
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Well, the spring will pull the
mass back upwards, and it will
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pull it back up to the point
where it squeezes, compresses
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the spring.
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The spring will be--
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like instead of being
stretched out,
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it'll be the opposite.
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It'll be compressed in.
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And then when compressed in,
it'll push the mass.
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Being compressed, the
spring will push.
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It'll push the mass down again
and up again, and I get
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oscillation.
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Oscillation is what
I'm seeing here.
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And what are examples of
oscillation in real life?
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The spring or a clock,
especially a grandfather clock
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that's swinging back and forth,
back and forth, so I'll
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just put a clock.
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Music, a violin string
is oscillating.
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That's where the beautiful
sound comes from.
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Our heart is in and out,
in and out, a regular
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oscillation.
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I could add molecules.
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They oscillate extremely
quickly.
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So this equation that I'm
aiming for comes up in
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biology, in chemistry--
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for molecules, it's a very
important equation--
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in physics and mechanics and
engineering for springs.
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It comes up in economics.
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It's everywhere.
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And by choosing constant
coefficients, I have the basic
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model and the simplest model.
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OK, so I'll talk in this
language of springs, but it's
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all these oscillations that lead
to equations like that.
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Actually, this model often
would have r equals 0.
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So let me take that case
r equals 0 again.
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So where does the equation
come from?
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Can I do two cents worth of
physics and then go back to
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the math, solving
the equation.
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The physics is just remembering
Newton's
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Law: f equals ma.
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So there is the m, the mass.
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The a is the acceleration.
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That's the second derivative,
right?
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You don't mind if I write that
as second derivative.
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Acceleration, the mass
is constant.
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We're not going at the speed of
light here so we can assume
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that mass is not being
converted to energy.
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It's mass.
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And then what's the force?
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Well, for this spring
force, for this
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spring, what did we say?
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So y is going this way.
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It's the disposition, the
displacement of the spring.
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When it's down like that, when
y is positive, the spring is
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pulling back.
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The force from the spring
is pulling back.
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And the force from the spring
is proportional to y.
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And the proportionality constant
is my number k.
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That fact that I
just said is--
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and I guess it's pulling
opposite to a positive y.
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When y is positive, and this
spring is way down, the force
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is pulling it up,
pulling in the
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negative direction, upwards.
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So I need that minus sign.
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So that k is the stiffness
of the spring.
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This is Hooke's Law, that the
force coming from a spring is
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proportional to the stretch.
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And the constant in there is
Hooke's constant, the spring
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constant k.
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So now that if I put the minus
ky over there as plus ky, you
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see my equation again.
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But this is the one that we were
able to solve right here.
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Do you see that this is the
case where r is 0 in this
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first model?
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r is 0 because r involves
resistance: r for resistance,
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air resistance, damping.
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And right now, I don't
have that.
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I just have a little spring
that'll oscillate forever.
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It'll oscillate forever
following sine and cosine.
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That's exactly what the
spring will do.
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It will just go on forever,
and the c and the d, their
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constants, will depend
on how it started.
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Did it start from rest?
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If it started from rest, there
would be no sine term.
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It's easy to find
c and d later.
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The real problem is to solve the
equation, and we've done
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it for this equation.
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OK, and let's just
remember this.
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Let me repeat that when I put
that onto the other side and
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divide by k, then I see the--
oh, divided by m, sorry--
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then I see the omega squared.
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So omega squared is k/m.
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Right.
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Oh, it is k/m, right.
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OK, that's the simple case,
pure sines and cosines.
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Now I'm letting in
some resistance.
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So now I'm coming back
to my equation.
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Let me write it again.
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m y double prime--
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second derivative--
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plus 2r y prime--
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first derivative--
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plus ky--
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0-th derivative equals 0.
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I want to solve that equation
now for any
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numbers m and r and k.
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OK, well, the nice thing is that
the exponential function
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takes us right to the answer,
the best plan here.
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So this is the most important
equation you would see in a
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differential equations course.
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And then the course kind of goes
past, and then you easily
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forget this is the most
important and the simplest.
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Why is it simple?
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Because if the key idea--
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this is the key idea: try
y equals e to the--
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an exponential e to the
something times t.
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Let me call that something
lambda.
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You might have preferred c.
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I'm happy with c or any
other number there.
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Yeah, OK, I'll call that lambda,
just because it gives
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it a little Greek importance.
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All right, so I try this.
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I substitute that into the
equation, and I'm going to
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choose lambda to make
things work.
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OK, but here's the key.
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The key idea is so easy.
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Now, put it into the equation.
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So what happens when I put--
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when this is y, take
two derivatives.
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Well, let me start with
taking no derivatives.
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So I have down here the k
e to the lambda t, and
258
00:15:23,500 --> 00:15:25,170
over here is a 0.
259
00:15:25,170 --> 00:15:29,160
And now let me back up to
the first derivative.
260
00:15:29,160 --> 00:15:33,650
So that's 2r times
the derivative
261
00:15:33,650 --> 00:15:35,460
of this guy y prime.
262
00:15:35,460 --> 00:15:38,940
I'm just substituting this
into the equation.
263
00:15:38,940 --> 00:15:41,480
So what's the derivative?
264
00:15:41,480 --> 00:15:45,360
We know that the derivative of
this brings down the lambda.
265
00:15:45,360 --> 00:15:46,680
We already did it.
266
00:15:46,680 --> 00:15:50,130
lambda e to the lambda
t, right?
267
00:15:50,130 --> 00:15:51,770
That's the derivative.
268
00:15:51,770 --> 00:15:53,550
And what about this one?
269
00:15:53,550 --> 00:15:57,980
This one is going to be
an m y double prime.
270
00:15:57,980 --> 00:16:00,520
What happens with
two derivatives?
271
00:16:00,520 --> 00:16:05,160
Bring down lambda twice, two
times, so I have lambda
272
00:16:05,160 --> 00:16:09,140
squared e to the lambda t.
273
00:16:09,140 --> 00:16:14,430
And then in a minute, you know
what I'm going to do.
274
00:16:14,430 --> 00:16:18,900
I'm going to cancel that common
factor e to the lambda
275
00:16:18,900 --> 00:16:24,140
t, which is never 0 so I can
safely divide it out, and then
276
00:16:24,140 --> 00:16:26,690
write the equation we get.
277
00:16:26,690 --> 00:16:29,470
OK, let me write that out
more with more space.
278
00:16:29,470 --> 00:16:36,260
m lambda squared plus
2r lambda--
279
00:16:36,260 --> 00:16:37,230
taking that--
280
00:16:37,230 --> 00:16:41,435
plus k is 0.
281
00:16:41,435 --> 00:16:44,140
282
00:16:44,140 --> 00:16:48,120
This is the equation.
283
00:16:48,120 --> 00:16:51,640
It's just an ordinary
quadratic equation.
284
00:16:51,640 --> 00:16:55,730
It's a high school
algebra equation.
285
00:16:55,730 --> 00:16:59,740
Lambda appears squared because
we had two derivatives.
286
00:16:59,740 --> 00:17:03,920
We had a second derivative, and
so I need the quadratic
287
00:17:03,920 --> 00:17:06,130
formula to know--
288
00:17:06,130 --> 00:17:09,410
I expect two answers,
two lambdas.
289
00:17:09,410 --> 00:17:14,220
And that's normal for a second
order equation, and I'll get
290
00:17:14,220 --> 00:17:19,329
two solutions: e to the lambda
1t and e to the lambda 2t.
291
00:17:19,329 --> 00:17:23,750
Two different exponentials will
both solve the problem.
292
00:17:23,750 --> 00:17:27,560
All right, what's the lambda?
293
00:17:27,560 --> 00:17:31,380
Well, can I just recall
the quadratic formula?
294
00:17:31,380 --> 00:17:35,160
Well, it's a little messy, but
it's not too bad here, just to
295
00:17:35,160 --> 00:17:39,280
show that I remember it.
296
00:17:39,280 --> 00:17:42,770
And the 2 there is kind of
handy with the quadratic
297
00:17:42,770 --> 00:17:47,730
formula because then I just get
minus an r plus or minus
298
00:17:47,730 --> 00:17:51,370
the square root of r squared.
299
00:17:51,370 --> 00:17:55,830
And it's not minus 4km, but
because of the 2 there, it's
300
00:17:55,830 --> 00:17:57,080
just minus km.
301
00:17:57,080 --> 00:17:59,440
302
00:17:59,440 --> 00:18:04,680
And then I divide by m.
303
00:18:04,680 --> 00:18:10,280
OK, well, big deal.
304
00:18:10,280 --> 00:18:12,710
I get two roots.
305
00:18:12,710 --> 00:18:14,280
Let me use numbers.
306
00:18:14,280 --> 00:18:19,020
So what you see there
is like solving the
307
00:18:19,020 --> 00:18:20,860
differential equation.
308
00:18:20,860 --> 00:18:22,410
Not too bad.
309
00:18:22,410 --> 00:18:27,990
Now let me put in numbers to
show what's typical, and, of
310
00:18:27,990 --> 00:18:31,200
course, as those numbers
change, we'll
311
00:18:31,200 --> 00:18:35,710
see different lambdas.
312
00:18:35,710 --> 00:18:38,820
And actually, as the numbers
change, that will take us
313
00:18:38,820 --> 00:18:44,330
between the exponential stuff
and the oscillating stuff.
314
00:18:44,330 --> 00:18:48,270
All right, let me take one
where I think it start--
315
00:18:48,270 --> 00:18:50,770
I think this will be-- so
this is example one.
316
00:18:50,770 --> 00:18:54,310
317
00:18:54,310 --> 00:19:01,640
I'll choose m equal 1 y
double prime plus--
318
00:19:01,640 --> 00:19:07,110
let me take r to be 3, so
then I have 6y prime.
319
00:19:07,110 --> 00:19:13,260
And let me choose k
to be 8y equals 0.
320
00:19:13,260 --> 00:19:15,370
All right, now we've
got numbers.
321
00:19:15,370 --> 00:19:19,050
322
00:19:19,050 --> 00:19:25,690
So the numbers are m equals
1, r equals 3, k equals 8.
323
00:19:25,690 --> 00:19:30,740
And I claim we can write the
solution to that equation, or
324
00:19:30,740 --> 00:19:33,850
the two solutions, because there
will be two lambdas.
325
00:19:33,850 --> 00:19:40,400
OK, so when I try e to the
lambda t, plug it in, I'll get
326
00:19:40,400 --> 00:19:46,220
lambda twice, and then I'll get
6 lambda once, and then
327
00:19:46,220 --> 00:19:50,510
I'll get 8 without a lambda
coming down, all multiplied by
328
00:19:50,510 --> 00:19:53,540
e to the lambda t, which I'm
canceling, equals 0.
329
00:19:53,540 --> 00:19:56,090
330
00:19:56,090 --> 00:20:03,000
So I solve that equation either
directly by recognizing
331
00:20:03,000 --> 00:20:07,960
that it factors into lambda
plus 2 times lambda plus 4
332
00:20:07,960 --> 00:20:13,310
equals 0 or by plugging
in r and k and m in
333
00:20:13,310 --> 00:20:14,640
the quadratic formula.
334
00:20:14,640 --> 00:20:22,450
Either way, I'm learning that
lambda is minus 2 or minus 4.
335
00:20:22,450 --> 00:20:24,720
Those are the two solutions.
336
00:20:24,720 --> 00:20:29,330
The two decay rates, you could
call them, because they're up
337
00:20:29,330 --> 00:20:30,820
in the exponent.
338
00:20:30,820 --> 00:20:32,140
So what's the solution?
339
00:20:32,140 --> 00:20:39,670
y of t, the solution to that
equation, the general solution
340
00:20:39,670 --> 00:20:45,370
with a constant c and a constant
d is an e to the
341
00:20:45,370 --> 00:20:52,680
minus 2t and an e
to the minus 4t.
342
00:20:52,680 --> 00:20:58,960
The two lambdas are in the
exponent, and we've solved it.
343
00:20:58,960 --> 00:21:01,190
So that's the point.
344
00:21:01,190 --> 00:21:07,380
We have the ability to solve
differential equations based
345
00:21:07,380 --> 00:21:12,560
on the three most important
derivatives we know:
346
00:21:12,560 --> 00:21:16,590
exponential, sines-cosines,
powers of t.
347
00:21:16,590 --> 00:21:19,070
OK, ready for example two?
348
00:21:19,070 --> 00:21:23,370
Example two, I'm just going to
change that 8 to a 10, so
349
00:21:23,370 --> 00:21:26,280
you're going to see
a 10 show up here.
350
00:21:26,280 --> 00:21:27,915
All right, but that will
make a difference.
351
00:21:27,915 --> 00:21:30,460
It won't just be some
new numbers.
352
00:21:30,460 --> 00:21:33,410
There'll be a definite
difference here.
353
00:21:33,410 --> 00:21:38,630
OK, let me go over across
here to the one with 10.
354
00:21:38,630 --> 00:21:46,210
OK, so now my equation is 1y
double prime, 6y prime still,
355
00:21:46,210 --> 00:21:51,660
and now 10y is equals to 0,
remembering prime means
356
00:21:51,660 --> 00:21:53,220
derivative.
357
00:21:53,220 --> 00:21:58,470
OK, so again, I try y is
e to the lambda t.
358
00:21:58,470 --> 00:21:59,940
I plug it in.
359
00:21:59,940 --> 00:22:03,290
When I have two derivatives,
bring down lambda squared.
360
00:22:03,290 --> 00:22:06,140
One derivative brings
down lambda.
361
00:22:06,140 --> 00:22:09,930
No derivatives leaves
the 10, equals 0.
362
00:22:09,930 --> 00:22:13,860
That's my equation for lambda.
363
00:22:13,860 --> 00:22:14,870
Ha!
364
00:22:14,870 --> 00:22:17,730
I don't know how to
factor that one.
365
00:22:17,730 --> 00:22:24,930
And in fact, I better use the
quadratic formula just to show
366
00:22:24,930 --> 00:22:26,420
what happens here.
367
00:22:26,420 --> 00:22:31,350
So the quadratic formula will
be the two roots lambda.
368
00:22:31,350 --> 00:22:35,580
369
00:22:35,580 --> 00:22:38,500
Can I remember that
dumb formula?
370
00:22:38,500 --> 00:22:43,290
Minus r plus or minus the square
root of r squared minus
371
00:22:43,290 --> 00:22:47,010
km, all divided by m.
372
00:22:47,010 --> 00:22:51,690
I got the 2's and the
4's out of it by
373
00:22:51,690 --> 00:22:54,080
taking r to be 3 here.
374
00:22:54,080 --> 00:23:01,850
OK, so it's minus 3 plus or
minus the square root of r
375
00:23:01,850 --> 00:23:07,820
squared is 9 minus 1 times 10.
376
00:23:07,820 --> 00:23:13,394
k and m is 10 divided by 1.
377
00:23:13,394 --> 00:23:15,270
Ha!
378
00:23:15,270 --> 00:23:18,570
You see something different's
going on here.
379
00:23:18,570 --> 00:23:22,240
I have the square root
of a negative number.
380
00:23:22,240 --> 00:23:27,430
Over there, if I wrote out that
square root, you would
381
00:23:27,430 --> 00:23:30,910
have seen the square
root of plus 1.
382
00:23:30,910 --> 00:23:34,900
That gave me minus 3 plus
1 or minus 3 minus 1.
383
00:23:34,900 --> 00:23:37,870
That was the minus
2 and minus 4.
384
00:23:37,870 --> 00:23:39,870
Now I'm different.
385
00:23:39,870 --> 00:23:45,020
Now seeing the square root of
minus 1, so this is minus 3
386
00:23:45,020 --> 00:23:46,920
plus or minus i.
387
00:23:46,920 --> 00:23:52,460
So I see the solution y of t.
388
00:23:52,460 --> 00:23:57,645
You see, this is i here, the
square root of minus 1.
389
00:23:57,645 --> 00:24:02,110
390
00:24:02,110 --> 00:24:04,610
We can deal with that.
391
00:24:04,610 --> 00:24:07,800
It's a complex number,
an imaginary number.
392
00:24:07,800 --> 00:24:11,600
And the combination minus 3 plus
i is a complex number,
393
00:24:11,600 --> 00:24:14,690
and we have to accept that
that's our lambda.
394
00:24:14,690 --> 00:24:20,760
So I have any multiple of c,
and the lambda here is
395
00:24:20,760 --> 00:24:26,060
minus 3 plus i t.
396
00:24:26,060 --> 00:24:33,980
And the second solution is
e minus 3 minus i t.
397
00:24:33,980 --> 00:24:38,290
398
00:24:38,290 --> 00:24:39,560
I found the general solution.
399
00:24:39,560 --> 00:24:42,880
400
00:24:42,880 --> 00:24:48,770
We could say done, except you
might feel, well, how did
401
00:24:48,770 --> 00:24:49,660
imaginary--
402
00:24:49,660 --> 00:24:53,040
what are we going to do with
these imaginary numbers here?
403
00:24:53,040 --> 00:24:55,460
How did they get in this
perfectly real
404
00:24:55,460 --> 00:24:57,210
differential equation?
405
00:24:57,210 --> 00:25:01,360
Well, they slipped in because
the solutions
406
00:25:01,360 --> 00:25:03,110
were not real numbers.
407
00:25:03,110 --> 00:25:06,560
The solutions were minus
3 plus or minus i.
408
00:25:06,560 --> 00:25:10,290
But we can get real again.
409
00:25:10,290 --> 00:25:14,330
So this is one way to write the
answer, but I just want to
410
00:25:14,330 --> 00:25:20,870
show you using the earlier
lecture, using the beautiful
411
00:25:20,870 --> 00:25:24,660
fact that Euler discovered.
412
00:25:24,660 --> 00:25:29,350
So now let me complete this
example by remembering Euler's
413
00:25:29,350 --> 00:25:32,870
great formula for e to the it.
414
00:25:32,870 --> 00:25:37,790
Because you see we have an e
to the minus 3t, perfectly
415
00:25:37,790 --> 00:25:39,650
real, decaying.
416
00:25:39,650 --> 00:25:43,950
The spring is slowing down
because of air resistance.
417
00:25:43,950 --> 00:25:47,910
But we also have
an e to the it.
418
00:25:47,910 --> 00:25:51,840
That's what Euler's formulas
about and Euler's formula says
419
00:25:51,840 --> 00:25:57,550
that e to the it is the
cosine of t plus i
420
00:25:57,550 --> 00:26:01,220
times the sine of t.
421
00:26:01,220 --> 00:26:04,620
So it's through Euler's
formula that these
422
00:26:04,620 --> 00:26:06,500
oscillations are coming in.
423
00:26:06,500 --> 00:26:10,800
The direct method led
to an e to the it.
424
00:26:10,800 --> 00:26:18,640
But the next day, Euler
realized that e
425
00:26:18,640 --> 00:26:20,420
to the minus it--
426
00:26:20,420 --> 00:26:24,970
or probably being Euler, it
didn't take till next day--
427
00:26:24,970 --> 00:26:29,580
will be minus i sine t.
428
00:26:29,580 --> 00:26:36,080
So both e to the it and e to
the minus it, they both can
429
00:26:36,080 --> 00:26:39,780
get replaced by sines
and cosines.
430
00:26:39,780 --> 00:26:43,460
So in place of e to the
it, I'll put that.
431
00:26:43,460 --> 00:26:46,080
In place of e to the minus
it, I put that one.
432
00:26:46,080 --> 00:26:48,160
The final result is--
433
00:26:48,160 --> 00:26:50,200
can I just jump to that?
434
00:26:50,200 --> 00:26:54,730
The final result is that with
some different constants, we
435
00:26:54,730 --> 00:26:56,080
have the cosine--
436
00:26:56,080 --> 00:26:57,140
oh!
437
00:26:57,140 --> 00:27:00,700
let me not forget e
to the minus 3t.
438
00:27:00,700 --> 00:27:03,510
That's part of this answer.
439
00:27:03,510 --> 00:27:08,010
I'm damping this out by e to the
minus 3t, this resistance
440
00:27:08,010 --> 00:27:18,990
r, times cosine of t and the e
to the minus 3t sine of t.
441
00:27:18,990 --> 00:27:21,055
OK, that's good.
442
00:27:21,055 --> 00:27:23,630
443
00:27:23,630 --> 00:27:29,080
General solution, back
to a real numbers.
444
00:27:29,080 --> 00:27:32,310
It describes a damped
oscillation.
445
00:27:32,310 --> 00:27:33,650
It's damped out.
446
00:27:33,650 --> 00:27:35,290
It's slowing down.
447
00:27:35,290 --> 00:27:37,780
Rather, it's decaying.
448
00:27:37,780 --> 00:27:40,940
The amplitude is-- the
spring is like--
449
00:27:40,940 --> 00:27:43,250
it's like having a shock
absorber or something.
450
00:27:43,250 --> 00:27:50,760
It's settling down to the center
point pretty fast. But
451
00:27:50,760 --> 00:27:54,170
as it settles, it's goes back
across that center point,
452
00:27:54,170 --> 00:27:55,680
oscillates across.
453
00:27:55,680 --> 00:28:01,250
OK, now you might finally ask,
the last step of this lecture,
454
00:28:01,250 --> 00:28:06,010
where do powers of t come in?
455
00:28:06,010 --> 00:28:07,580
Where does t come in?
456
00:28:07,580 --> 00:28:11,470
So far we've seen exponentials
come in.
457
00:28:11,470 --> 00:28:14,180
We've seen sines and
cosines come in.
458
00:28:14,180 --> 00:28:19,090
Can I do a last example just
here in the corner, which will
459
00:28:19,090 --> 00:28:25,980
be y double prime, 6 y prime,
and now this time instead of 8
460
00:28:25,980 --> 00:28:31,850
or 10, I'm going to
take 9y equals 0.
461
00:28:31,850 --> 00:28:38,610
OK, can we use this as example
3, which we can solve?
462
00:28:38,610 --> 00:28:42,880
You know that I'm going to
try y equals to lambda t.
463
00:28:42,880 --> 00:28:45,360
Let me substitute that.
464
00:28:45,360 --> 00:28:49,790
I'll get lambda squared coming
from two derivatives, lambda
465
00:28:49,790 --> 00:28:55,930
coming from one derivative, 9
coming from no derivatives.
466
00:28:55,930 --> 00:28:59,400
I've got my quadratic equation
that's supposed
467
00:28:59,400 --> 00:29:02,210
to give me two lambdas.
468
00:29:02,210 --> 00:29:03,790
Little problem here.
469
00:29:03,790 --> 00:29:08,100
When I factor this, it factors
into lambda plus 3
470
00:29:08,100 --> 00:29:10,900
squared equals 0.
471
00:29:10,900 --> 00:29:16,440
So the answer, the lambda,
is minus 3 twice.
472
00:29:16,440 --> 00:29:19,620
473
00:29:19,620 --> 00:29:21,400
Twice!
474
00:29:21,400 --> 00:29:28,810
The two lambdas happen to hit
the same value: minus 3.
475
00:29:28,810 --> 00:29:32,730
OK, we don't have any
complex stuff here.
476
00:29:32,730 --> 00:29:36,790
It's two real values that
happen to coincide.
477
00:29:36,790 --> 00:29:42,120
And when that happens, well,
minus 3 tells us that a
478
00:29:42,120 --> 00:29:47,240
solution e to the
minus 3t works.
479
00:29:47,240 --> 00:29:50,360
Where do we get the
other solution?
480
00:29:50,360 --> 00:29:54,260
We want two solutions to a
second order equation.
481
00:29:54,260 --> 00:29:58,040
I can't just use e to
the minus 3t again.
482
00:29:58,040 --> 00:30:02,580
I need a second solution, and
it shows up at this--
483
00:30:02,580 --> 00:30:05,560
it's typical of math
that it shows--
484
00:30:05,560 --> 00:30:08,960
something special happens at
this special situation of a
485
00:30:08,960 --> 00:30:16,440
double root, and the solutions
will be e to the minus 3t and
486
00:30:16,440 --> 00:30:19,170
t e to the minus 3t.
487
00:30:19,170 --> 00:30:22,490
That's the last step, and
maybe I just put it--
488
00:30:22,490 --> 00:30:25,670
well, where am I going
to put it?
489
00:30:25,670 --> 00:30:29,120
I'll bring this down and
just put it here.
490
00:30:29,120 --> 00:30:34,030
So now I'm solving this
particular problem y is a
491
00:30:34,030 --> 00:30:37,330
multiple of e to the minus 3t.
492
00:30:37,330 --> 00:30:42,340
Good, but I can't just repeat
it for the second one.
493
00:30:42,340 --> 00:30:46,160
So the second one, it just turns
out that then is when a
494
00:30:46,160 --> 00:30:47,490
little factor t appears.
495
00:30:47,490 --> 00:30:50,030
496
00:30:50,030 --> 00:30:54,760
You might like to substitute
that for practice with the
497
00:30:54,760 --> 00:30:56,220
product rule.
498
00:30:56,220 --> 00:30:59,280
If you substitute that in the
differential equation, you'll
499
00:30:59,280 --> 00:31:04,360
find everything cancels,
and it works.
500
00:31:04,360 --> 00:31:11,480
So the conclusion is linear
constant coefficient
501
00:31:11,480 --> 00:31:17,030
differential equations are
completely solved by trying e
502
00:31:17,030 --> 00:31:22,140
to the lambda t and finding
that number lambda.
503
00:31:22,140 --> 00:31:25,950
If it's a real number,
we have exponentials.
504
00:31:25,950 --> 00:31:29,810
If it's an imaginary number,
we have sines and cosines.
505
00:31:29,810 --> 00:31:33,950
If it's a repeated number,
we have an extra
506
00:31:33,950 --> 00:31:36,650
factor t showing up.
507
00:31:36,650 --> 00:31:42,690
That's the exceptional event
in that particular case.
508
00:31:42,690 --> 00:31:43,890
So there you go.
509
00:31:43,890 --> 00:31:46,950
Differential equations
with constant
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coefficients we can handle.
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You can handle.
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Thank you.
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This has been a production
of MIT
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OpenCourseWare and Gilbert Strang.
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