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We're going to start.
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We are going to start studying
today, and for quite a while,
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the linear second-order
differential equation with
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constant coefficients.
In standard form,
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it looks like,
there are various possible
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choices for the variable,
unfortunately,
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so I hope it won't disturb you
much if I use one rather than
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another.
I'm going to write it this way
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in standard form.
I'll use y as the dependent
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variable.
Your book uses little p and
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little q.
I'll probably switch to that by
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next time.
But, for today,
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I'd like to use the most
neutral letters I can find that
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won't interfere with anything
else.
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So, of course call the constant
coefficients,
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respectively,
capital A and capital B.
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I'm going to assume for today
that the right-hand side is
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zero.
So, that means it's what we
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call homogeneous.
The left-hand side must be in
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this form for it to be linear,
it's second order because it
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involves a second derivative.
These coefficients,
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A and B, are understood to be
constant because,
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as I said, it has constant
coefficients.
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Of course, that's not the most
general linear equation there
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could be.
In general, it would be more
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general by making this a
function of the dependent
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variable, x or t,
whatever it's called.
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Similarly, this could be a
function of the dependent
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variable.
Above all, the right-hand side
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can be a function of a variable
rather than simply zero.
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In that case the equation is
called inhomogeneous.
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But it has a different physical
meaning, and therefore it's
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customary to study that after
this.
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You start with this.
This is the case we start with,
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and then by the middle of next
week we will be studying more
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general cases.
But, it's a good idea to start
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here.
Your book starts with,
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in general, some theory of a
general linear equation of
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second-order,
and even higher order.
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I'm asking you to skip that for
the time being.
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We'll come back to it next
Wednesday, it two lectures,
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in other words.
I think it's much better and
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essential for your problems at
for you to get some experience
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with a simple type of equation.
And then, you'll understand the
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general theory,
how it applies,
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a lot better,
I think.
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So, let's get experience here.
The downside of that is that
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I'm going to have to assume a
couple of things about the
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solution to this equation,
how it looks;
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I don't think that will upset
you too much.
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So, what I'm going to assume,
and we will justify it in a
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couple lectures,
that the general solution,
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that is, the solution involving
arbitrary constants,
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looks like this.
y is equal-- The arbitrary
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constants occur in a certain
special way.
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There is c one y one plus c two
y two.
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So, these are two arbitrary
constants corresponding to the
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fact that we are solving a
second-order equation.
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In general, the number of
arbitrary constants in the
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solution is the same as the
order of the equation because if
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it's a second-order equation
because if it's a second-order
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equation, that means somehow or
other, it may be concealed.
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But you're going to have to
integrate something twice to get
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the answer.
And therefore,
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there should be two arbitrary
constants.
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That's very rough,
but it sort of gives you the
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idea.
Now, what are the y1 and y2?
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Well, as you can see,
if these are arbitrary
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constants, if I take c2 to be
zero and c1 to be one,
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that means that y1 must be a
solution to the equation,
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and similarly y2.
So, where y1 and y2 are
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solutions.
Now, what that shows you is
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that the task of solving this
equation is reduced,
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in some sense,
to finding just two solutions
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of it, somehow.
All we have to do is find two
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solutions, and then we will have
solved the equation because the
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general solution is made up in
this way by multiplying those
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two solutions by arbitrary
constants and adding them.
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So, the problem is,
where do we get that solutions
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from?
But, first of all,
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or rather, second or third of
all, the initial conditions
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enter into the,
I haven't given you any initial
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conditions here,
but if you have them,
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and I will illustrate them when
I work problems,
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the initial conditions,
well, the initial values are
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satisfied by choosing c1 and c2,
are satisfied by choosing c1
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and c2 properly.
So, in other words,
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if you have an initial value
problem to solve,
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that will be taken care of by
the way those constants,
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c, enter into the solution.
Okay, without further ado,
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there is a standard example,
which I wish I had looked up in
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the physics syllabus for the
first semester.
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Did you study the
spring-mass-dashpot system in
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8.01?
I'm embarrassed having to ask
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you.
You did?
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Raise your hands if you did.
Okay, that means you all did.
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Well, just let me draw an
instant picture to remind you.
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So, this is a two second
review.
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I don't know how they draw the
picture.
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Probably they don't draw
picture at all.
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They have some elaborate system
here of the thing running back
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and forth.
Well, in the math,
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we do that all virtually.
So, here's my system.
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That's a fixed thing.
Here's a little spring.
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And, there's a little car on
the track here,
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I guess.
So, there's the mass,
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some mass in the little car,
and motion is damped by what's
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called a dashpot.
A dashpot is the sort of thing,
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you see them in everyday life
as door closers.
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They're the thing up above that
you never notice that prevent
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the door slamming shut.
So, if you take one apart,
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it looks something like this.
So, that's the dash pot.
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It's a chamber with a piston.
This is a piston moving in and
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out, and compressing the air,
releasing it,
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is what damps the motion of the
thing.
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So, this is a dashpot,
it's usually called.
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And, here's our mass in that
little truck.
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And, here's the spring.
And then, the equation which
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governs it is,
let's call this x.
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I'm already changing,
going to change the dependent
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variable from y to x,
but that's just for the sake of
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example, and because the track
is horizontal,
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it seems more natural to call
it x.
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There's some equilibrium
position somewhere,
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let's say, here.
That's the position at which
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the mass wants to be,
if the spring is not pulling on
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it or pushing on it,
and the dashpot is happy.
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I guess we'd better have a
longer dashpot here.
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So, this is the equilibrium
position where nothing is
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happening.
When you depart from that
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position, then the spring,
if you go that way,
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the spring tries to pull the
mass back.
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If it goes on the site,
the spring tries to push the
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mass away.
The dashpot,
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meanwhile, is doing its thing.
And so, the force on the,
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m x double prime.
That's by Newton's law,
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the force, comes from where?
Well, there's the spring
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pushing and pulling on it.
That force is opposed.
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If x gets to be beyond zero,
then the spring tries to pull
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it back.
If it gets to the left of zero,
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if x gets to be negative,
that that spring force is
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pushing it this way,
wants to get rid of the mass.
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So, it should be minus kx,
and this is from the spring,
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the fact that is proportional
to the amount by which x varies.
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So, that's called Hooke's Law.
Never mind that.
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This is a law.
That's a law,
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Newton's law,
okay, Newton,
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Hooke with an E,
and the dashpot damping is
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proportional to the velocity.
It's not doing anything if the
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mass is not moving,
even if it's stretched way out
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of its equilibrium position.
So, it resists the velocity.
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If the thing is trying to go
that way, the dashpot resists
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it.
It's trying to go this way,
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the dashpot resists that,
too.
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It's always opposed to the
velocity.
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And so, this is a dash pot
damping.
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I don't know whose law this is.
So, it's the force coming from
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the dashpot.
And, when you write this out,
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the final result,
therefore, is it's m x double
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prime plus c x prime,
it's important to see where the
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various terms are,
plus kx equals zero.
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And now, that's still not in
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standard form.
To put it in standard form,
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you must divide through by the
mass.
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And, it will now read like
this, plus k divided by m times
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x equals zero.
And, that's the equation
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governing the motion of the
spring.
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I'm doing this because your
problem set, problems three and
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four, ask you to look at a
little computer visual which
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illustrates a lot of things.
And, I didn't see how it would
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make, you can do it without this
interpretation of
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spring-mass-dashpot,
--
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-- but, I think thinking of it
of these constants as,
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this is the damping constant,
and this is the spring,
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the constant which represents
the force being exerted by the
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spring, the spring constant,
as it's called,
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makes it much more vivid.
So, you will note is that those
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problems are labeled Friday or
Monday.
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Make it Friday.
You can do them after today if
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you have the vaguest idea of
what I'm talking about.
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If not, go back and repeat
8.01.
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So, all this was just an
example, a typical model.
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But, by far,
the most important simple
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model.
Okay, now what I'd like to talk
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about is the solution.
What is it I have to do to
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solve the equation?
So, to solve the equation that
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I outlined in orange on the
board, the ODE,
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our task is to find two
solutions.
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Now, don't make it too trivial.
There is a condition.
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The solution should be
independent.
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All that means is that y2
should not be a constant
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multiple of y1.
I mean, if you got y1,
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then two times y1 is not an
acceptable value for this
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because, as you can see,
you really only got one there.
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You're not going to be able to
make up a two parameter family.
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So, the solutions have to be
independent, which means,
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to repeat, that neither should
be a constant multiple of the
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other.
They should look different.
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That's an adequate explanation.
Okay, now, what's the basic
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method to finding those
solutions?
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Well, that's what we're going
to use all term long,
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essentially,
studying equations of this
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type, even systems of this type,
with constant coefficients.
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The basic method is to try y
equals an exponential.
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Now, the only way you can
fiddle with an exponential is in
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the constant that you put up on
top.
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So, I'm going to try y equals e
to the rt.
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Notice you can't tell from that
what I'm using as the
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independent variable.
But, this tells you I'm using
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t.
And, I'm switching back to
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using t as the dependent
variable.
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So, T is the independent
variable.
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Why do I do that?
The answer is because somebody
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thought of doing it,
probably Euler,
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and it's been a tradition
that's handed down for the last
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300 or 400 years.
Some things we just know.
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All right, so if I do that,
as you learned from the exam,
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it's very easy to differentiate
exponentials.
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That's why people love them.
It's also very easy to
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integrate exponentials.
And, half of you integrated
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instead of differentiating.
So, we will try this and see if
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we can pick r so that it's a
solution.
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Okay, well, I will plug in,
then.
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Substitute, in other words,
and what do we get?
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Well, for y double prime,
I get r squared e to the rt.
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That's y double prime
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because each time you
differentiate it,
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you put an extra power of r out
in front.
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But otherwise,
do nothing.
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The next term will be r times,
sorry, I forgot the constant.
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Capital A times r e to the rt,
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and then there's the last term,
B times y itself,
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which is B e to the rt.
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And, that's supposed to be
equal to zero.
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So, I have to choose r so that
this becomes equal to zero.
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Now, you see,
the e to the rt
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occurs as a factor in every
term, and the e to the rt
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is never zero.
And therefore,
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you can divide it out because
it's always a positive number,
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regardless of the value of t.
So, I can cancel out from each
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term.
And, what I'm left with is the
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equation r squared plus ar plus
b equals zero.
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We are trying to find values of
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r that satisfy that equation.
And that, dear hearts,
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is why you learn to solve
quadratic equations in high
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school, in order that in this
moment, you would be now ready
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to find how spring-mass systems
behave when they are damped.
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This is called the
characteristic equation.
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The characteristic equation of
the ODE, or of the system of the
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spring mass system,
which it's modeling,
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the characteristic equation of
the system, okay?
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Okay, now, we solve it,
but now, from high school you
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know there are several cases.
And, each of those cases
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corresponds to a different
behavior.
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And, the cases depend upon what
the roots look like.
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The possibilities are the roots
could be real,
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and distinct.
That's the easiest case to
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handle.
The roots might be a pair of
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complex conjugate numbers.
That's harder to handle,
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but we are ready to do it.
And, the third case,
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which is the one most in your
problem set is the most
263
00:17:45 --> 00:17:51
puzzling: when the roots are
real, and equal.
264
00:17:48 --> 00:17:54
And, I'm going to talk about
those three cases in that order.
265
00:17:53 --> 00:17:59
So, the first case is the roots
are real and unequal.
266
00:17:57 --> 00:18:03
If I tell you they are unequal,
and I will put down real to
267
00:18:02 --> 00:18:08
make that clear.
Well, that is by far the
268
00:18:07 --> 00:18:13
simplest case because
immediately, one sees we have
269
00:18:11 --> 00:18:17
two roots.
They are different,
270
00:18:13 --> 00:18:19
and therefore,
we get our two solutions
271
00:18:17 --> 00:18:23
immediately.
So, the solutions are,
272
00:18:19 --> 00:18:25
the general solution to the
equation, I write down without
273
00:18:24 --> 00:18:30
further ado as y equals c1 e to
the r1 t plus c2 e to the r2 t.
274
00:18:29 --> 00:18:35
275
00:18:34 --> 00:18:40
There's our solution.
Now, because that was so easy,
276
00:18:38 --> 00:18:44
and we didn't have to do any
work, I'd like to extend this
277
00:18:42 --> 00:18:48
case a little bit by using it as
an example of how you put in the
278
00:18:48 --> 00:18:54
initial conditions,
how to put in the c.
279
00:18:51 --> 00:18:57
So, let me work a specific
numerical example,
280
00:18:54 --> 00:19:00
since we are not going to try
to do this theoretically until
281
00:18:59 --> 00:19:05
next Wednesday.
Let's just do a numerical
282
00:19:04 --> 00:19:10
example.
So, suppose I take the
283
00:19:07 --> 00:19:13
constants to be the damping
constant to be a four,
284
00:19:11 --> 00:19:17
and the spring constant,
I'll take the mass to be one,
285
00:19:16 --> 00:19:22
and the spring constant to be
three.
286
00:19:19 --> 00:19:25
So, there's more damping here,
damping force here.
287
00:19:24 --> 00:19:30
You can't really talk that way
since the units are different.
288
00:19:31 --> 00:19:37
But, this number is bigger than
that one.
289
00:19:33 --> 00:19:39
That seems clear,
at any rate.
290
00:19:35 --> 00:19:41
Okay, now, what was the
characteristic equation?
291
00:19:39 --> 00:19:45
Look, now watch.
Please do what I do.
292
00:19:41 --> 00:19:47
I've found in the past,
even by the middle of the term,
293
00:19:45 --> 00:19:51
there are still students who
feel that they must substitute y
294
00:19:50 --> 00:19:56
equals e to the rt,
and go through that whole
295
00:19:53 --> 00:19:59
little derivation to find that
you don't do that.
296
00:19:57 --> 00:20:03
It's a waste of time.
I did it that you might not
297
00:20:02 --> 00:20:08
ever have to do it again.
Immediately write down the
298
00:20:06 --> 00:20:12
characteristic equation.
That's not very hard.
299
00:20:10 --> 00:20:16
r squared plus 4r plus three
equals zero.
300
00:20:14 --> 00:20:20
And, if you can write down its
301
00:20:17 --> 00:20:23
roots immediately,
splendid.
302
00:20:20 --> 00:20:26
But, let's not assume that
level of competence.
303
00:20:23 --> 00:20:29
So, it's r plus three times r
plus one equals zero.
304
00:20:29 --> 00:20:35
Okay, you factor it.
305
00:20:33 --> 00:20:39
This being 18.03,
a lot of the times the roots
306
00:20:36 --> 00:20:42
will be integers when they are
not, God forbid,
307
00:20:39 --> 00:20:45
you will have to use the
quadratic formula.
308
00:20:42 --> 00:20:48
But here, the roots were
integers.
309
00:20:44 --> 00:20:50
It is, after all,
only the first example.
310
00:20:47 --> 00:20:53
So, the solution,
the general solution is y
311
00:20:50 --> 00:20:56
equals c1 e to the negative,
notice the root is minus three
312
00:20:54 --> 00:21:00
and minus one,
minus 3t plus c2 e to the
313
00:20:56 --> 00:21:02
negative t.
314
00:21:01 --> 00:21:07
Now, suppose it's an initial
value problem.
315
00:21:04 --> 00:21:10
So, I gave you an initial
condition.
316
00:21:06 --> 00:21:12
Suppose the initial conditions
were that y of zero were one.
317
00:21:11 --> 00:21:17
So, at the start,
318
00:21:13 --> 00:21:19
the mass has been moved over to
the position,
319
00:21:16 --> 00:21:22
one, here.
Well, we expected it,
320
00:21:18 --> 00:21:24
then, to start doing that.
But, this is fairly heavily
321
00:21:22 --> 00:21:28
damped.
This is heavily damped.
322
00:21:25 --> 00:21:31
I'm going to assume that the
mass starts at rest.
323
00:21:30 --> 00:21:36
So, the spring is distended.
The masses over here.
324
00:21:33 --> 00:21:39
But, there's no motion at times
zero this way or that way.
325
00:21:37 --> 00:21:43
In other words,
I'm not pushing it.
326
00:21:40 --> 00:21:46
I'm just releasing it and
letting it do its thing after
327
00:21:44 --> 00:21:50
that.
Okay, so y prime of zero,
328
00:21:46 --> 00:21:52
I'll assume, is zero.
329
00:21:49 --> 00:21:55
So, it starts at rest,
but in the extended position,
330
00:21:53 --> 00:21:59
one unit to the right of the
equilibrium position.
331
00:21:56 --> 00:22:02
Now, all you have to do is use
these two conditions.
332
00:22:02 --> 00:22:08
Notice I have to have two
conditions because there are two
333
00:22:06 --> 00:22:12
constants I have to find the
value of.
334
00:22:09 --> 00:22:15
All right, so,
let's substitute,
335
00:22:11 --> 00:22:17
well, we're going to have to
calculate the derivative.
336
00:22:15 --> 00:22:21
So, why don't we do that right
away?
337
00:22:18 --> 00:22:24
So, this is minus three c1 e to
the minus 3t minus c2 e to the
338
00:22:22 --> 00:22:28
negative t.
339
00:22:27 --> 00:22:33
And now, if I substitute in at
zero, when t equals zero,
340
00:22:31 --> 00:22:37
what do I get?
Well, the first equation,
341
00:22:34 --> 00:22:40
the left says that y of zero
should be one.
342
00:22:38 --> 00:22:44
And, the right says this is
one.
343
00:22:41 --> 00:22:47
So, it's c1 plus c2.
344
00:22:43 --> 00:22:49
That's the result of
substituting t equals zero.
345
00:22:47 --> 00:22:53
How about substituting?
346
00:22:49 --> 00:22:55
What should I substitute in the
second equation?
347
00:22:53 --> 00:22:59
Well, y prime of zero is zero.
348
00:22:56 --> 00:23:02
So, if the second equation,
when I put in t equals zero,
349
00:23:01 --> 00:23:07
the left side is zero according
to that initial value,
350
00:23:05 --> 00:23:11
and the right side is negative
three c1 minus c2.
351
00:23:12 --> 00:23:18
You see what you end up with,
therefore, is a pair of
352
00:23:15 --> 00:23:21
simultaneous linear equations.
And, this is why you learn to
353
00:23:19 --> 00:23:25
study linear set of pairs of
simultaneous linear equations in
354
00:23:24 --> 00:23:30
high school.
These are among the most
355
00:23:26 --> 00:23:32
important.
Solving problems of this type
356
00:23:29 --> 00:23:35
are among the most important
applications of that kind of
357
00:23:33 --> 00:23:39
algebra, and this kind of
algebra.
358
00:23:37 --> 00:23:43
All right, what's the answer
finally?
359
00:23:40 --> 00:23:46
Well, if I add the two of them,
I get minus 2c1 equals one.
360
00:23:45 --> 00:23:51
So, c1 is equal to minus one
half.
361
00:23:49 --> 00:23:55
And, if c1 is minus a half,
362
00:23:54 --> 00:24:00
then c2 is minus 3c1.
363
00:23:57 --> 00:24:03
So, c2 is three halves.
364
00:24:02 --> 00:24:08
365
00:24:10 --> 00:24:16
The final question is,
what does that look like as a
366
00:24:13 --> 00:24:19
solution?
Well, in general,
367
00:24:14 --> 00:24:20
these combinations of two
exponentials aren't very easy to
368
00:24:17 --> 00:24:23
plot by yourself.
That's one of the reasons you
369
00:24:20 --> 00:24:26
are being given this little
visual which plots them for you.
370
00:24:24 --> 00:24:30
All you have to do is,
as you'll see,
371
00:24:26 --> 00:24:32
set the damping constant,
set the constants,
372
00:24:28 --> 00:24:34
set the initial conditions,
and by magic,
373
00:24:31 --> 00:24:37
the curve appears on the
screen.
374
00:24:34 --> 00:24:40
And, if you change either of
the constants,
375
00:24:40 --> 00:24:46
the curve will change nicely
right along with it.
376
00:24:48 --> 00:24:54
So, the solution is y equals
minus one half e to the minus 3t
377
00:24:57 --> 00:25:03
plus three halves e to the
negative t.
378
00:25:06 --> 00:25:12
How does it look?
379
00:25:11 --> 00:25:17
Well, I don't expect you to be
able to plot that by yourself,
380
00:25:16 --> 00:25:22
but you can at least get
started.
381
00:25:19 --> 00:25:25
It does have to satisfy the
initial conditions.
382
00:25:23 --> 00:25:29
That means it should start at
one, and its starting slope is
383
00:25:28 --> 00:25:34
zero.
So, it starts like that.
384
00:25:32 --> 00:25:38
These are both declining
exponentials.
385
00:25:35 --> 00:25:41
This declines very rapidly,
this somewhat more slowly.
386
00:25:40 --> 00:25:46
It does something like that.
If this term were a lot,
387
00:25:44 --> 00:25:50
lot more negative,
I mean, that's the way that
388
00:25:49 --> 00:25:55
particular solution looks.
How might other solutions look?
389
00:25:54 --> 00:26:00
I'll draw a few other
possibilities.
390
00:25:57 --> 00:26:03
If the initial term,
if, for example,
391
00:26:00 --> 00:26:06
the initial slope were quite
negative, well,
392
00:26:04 --> 00:26:10
that would have start like
this.
393
00:26:09 --> 00:26:15
Now, just your experience of
physics, or of the real world
394
00:26:12 --> 00:26:18
suggests that if I give,
if I start the thing at one,
395
00:26:15 --> 00:26:21
but give it a strongly negative
push, it's going to go beyond
396
00:26:19 --> 00:26:25
the equilibrium position,
and then come back again.
397
00:26:22 --> 00:26:28
But, because the damping is
big, it's not going to be able
398
00:26:26 --> 00:26:32
to get through that.
The equilibrium position,
399
00:26:29 --> 00:26:35
a second time,
is going to look something like
400
00:26:31 --> 00:26:37
that.
Or, if I push it in that
401
00:26:34 --> 00:26:40
direction, the positive
direction, that it starts off
402
00:26:37 --> 00:26:43
with a positive slope.
But it loses its energy because
403
00:26:41 --> 00:26:47
the spring is pulling it.
It comes and does something
404
00:26:44 --> 00:26:50
like that.
So, in other words,
405
00:26:46 --> 00:26:52
it might go down.
Cut across the equilibrium
406
00:26:49 --> 00:26:55
position, come back again,
it do that?
407
00:26:52 --> 00:26:58
No, that it cannot do.
I was considering giving you a
408
00:26:55 --> 00:27:01
problem to prove that,
but I got tired of making out
409
00:26:58 --> 00:27:04
the problems set,
and decided I tortured you
410
00:27:01 --> 00:27:07
enough already,
as you will see.
411
00:27:05 --> 00:27:11
So, anyway, these are different
possibilities for the way that
412
00:27:11 --> 00:27:17
can look.
This case, where it just
413
00:27:14 --> 00:27:20
returns in the long run is
called the over-damped case,
414
00:27:20 --> 00:27:26
over-damped.
Now, there is another case
415
00:27:24 --> 00:27:30
where the thing oscillates back
and forth.
416
00:27:30 --> 00:27:36
We would expect to get that
case if the damping is very
417
00:27:33 --> 00:27:39
little or nonexistent.
Then, there's very little
418
00:27:37 --> 00:27:43
preventing the mass from doing
that, although we do expect if
419
00:27:41 --> 00:27:47
there's any damping at all,
we expect it ultimately to get
420
00:27:45 --> 00:27:51
nearer and nearer to the
equilibrium position.
421
00:27:48 --> 00:27:54
Mathematically,
what does that correspond to?
422
00:27:51 --> 00:27:57
Well, that's going to
correspond to case two,
423
00:27:55 --> 00:28:01
where the roots are complex.
The roots are complex,
424
00:27:59 --> 00:28:05
and this is why,
let's call the roots,
425
00:28:03 --> 00:28:09
in that case we know that the
roots are of the form a plus or
426
00:28:08 --> 00:28:14
minus bi.
There are two roots,
427
00:28:10 --> 00:28:16
and they are a complex
conjugate.
428
00:28:13 --> 00:28:19
All right, let's take one of
them.
429
00:28:16 --> 00:28:22
What does a correspond to in
terms of the exponential?
430
00:28:20 --> 00:28:26
Well, remember,
the function of the r was,
431
00:28:24 --> 00:28:30
it's this r when we tried our
exponential solution.
432
00:28:30 --> 00:28:36
So, what we formally,
this means we get a complex
433
00:28:34 --> 00:28:40
solution.
The complex solution y equals e
434
00:28:38 --> 00:28:44
to this, let's use one of them,
let's say, (a plus bi) times t.
435
00:28:43 --> 00:28:49
The question is,
436
00:28:47 --> 00:28:53
what do we do that?
We are not really interested,
437
00:28:51 --> 00:28:57
I don't know what a complex
solution to that thing means.
438
00:28:56 --> 00:29:02
It doesn't have any meaning.
What I want to know is how y
439
00:29:01 --> 00:29:07
behaves or how x behaves in that
picture.
440
00:29:07 --> 00:29:13
And, that better be a real
function because otherwise I
441
00:29:10 --> 00:29:16
don't know what to do with it.
So, we are looking for two real
442
00:29:14 --> 00:29:20
functions, the y1 and the y2.
But, in fact,
443
00:29:17 --> 00:29:23
what we've got is one complex
function.
444
00:29:20 --> 00:29:26
All right, now,
a theorem to the rescue:
445
00:29:22 --> 00:29:28
this, I'm not going to save for
Wednesday because it's so
446
00:29:26 --> 00:29:32
simple.
So, the theorem is that if you
447
00:29:30 --> 00:29:36
have a complex solution,
u plus iv, so each of these is
448
00:29:36 --> 00:29:42
a function of time,
u plus iv is the complex
449
00:29:40 --> 00:29:46
solution to a real differential
equation with constant
450
00:29:45 --> 00:29:51
coefficients.
Well, it doesn't have to have
451
00:29:49 --> 00:29:55
constant coefficients.
It has to be linear.
452
00:29:53 --> 00:29:59
Let me just write it out to y
double prime plus A y prime plus
453
00:29:59 --> 00:30:05
B y equals zero.
454
00:30:04 --> 00:30:10
Suppose you got a complex
solution to that equation.
455
00:30:08 --> 00:30:14
These are understood to be real
numbers.
456
00:30:11 --> 00:30:17
They are the damping constant
and the spring constant.
457
00:30:15 --> 00:30:21
Then, the conclusion is that u
and v are real solutions.
458
00:30:20 --> 00:30:26
In other words,
having found a complex
459
00:30:23 --> 00:30:29
solution, all you have to do is
take its real and imaginary
460
00:30:28 --> 00:30:34
parts, and voila,
you've got your two solutions
461
00:30:31 --> 00:30:37
you were looking for for the
original equation.
462
00:30:37 --> 00:30:43
Now, that might seem like
magic, but it's easy.
463
00:30:39 --> 00:30:45
It's so easy it's the sort of
theorem I could spend one minute
464
00:30:43 --> 00:30:49
proving for you now.
What's the reason for it?
465
00:30:46 --> 00:30:52
Well, the main thing I want you
to get out of this argument is
466
00:30:50 --> 00:30:56
to see that it absolutely
depends upon these coefficients
467
00:30:54 --> 00:31:00
being real.
You have to have a real
468
00:30:56 --> 00:31:02
differential equation for this
to be true.
469
00:30:59 --> 00:31:05
Otherwise, it's certainly not.
So, the proof is,
470
00:31:03 --> 00:31:09
what does it mean to be a
solution?
471
00:31:06 --> 00:31:12
It means when you plug in A (u
plus iv) plus,
472
00:31:10 --> 00:31:16
prime,
plus B times u plus iv,
473
00:31:14 --> 00:31:20
what am I supposed to get?
474
00:31:17 --> 00:31:23
Zero.
Well, now, separate these into
475
00:31:20 --> 00:31:26
the real and imaginary parts.
What does it say?
476
00:31:24 --> 00:31:30
It says u double prime plus A u
prime plus B u,
477
00:31:29 --> 00:31:35
that's the real part of
this expression when I expand it
478
00:31:35 --> 00:31:41
out.
And, I've got an imaginary
479
00:31:39 --> 00:31:45
part, too, which all have the
coefficient i.
480
00:31:43 --> 00:31:49
So, from here,
I get v double prime plus i
481
00:31:46 --> 00:31:52
times A v prime plus
i times B v.
482
00:31:51 --> 00:31:57
So, this is the imaginary part.
Now, here I have something with
483
00:31:57 --> 00:32:03
a real part plus the imaginary
part, i, times the imaginary
484
00:32:02 --> 00:32:08
part is zero.
Well, the only way that can
485
00:32:05 --> 00:32:11
happen is if the real part is
zero, and the imaginary part is
486
00:32:11 --> 00:32:17
separately zero.
So, the conclusion is that
487
00:32:15 --> 00:32:21
therefore this part must be
zero, and therefore this part
488
00:32:19 --> 00:32:25
must be zero because the two of
them together make the complex
489
00:32:23 --> 00:32:29
number zero plus zero i.
Now, what does it mean for the
490
00:32:27 --> 00:32:33
real part to be zero?
It means that u is a solution.
491
00:32:31 --> 00:32:37
This, the imaginary part zero
means v is a solution,
492
00:32:34 --> 00:32:40
and therefore,
just what I said.
493
00:32:36 --> 00:32:42
u and v are solutions to the
real equation.
494
00:32:39 --> 00:32:45
Where did I use the fact that A
and B were real numbers and not
495
00:32:44 --> 00:32:50
complex numbers?
In knowing that this is the
496
00:32:47 --> 00:32:53
real part, I had to know that A
was a real number.
497
00:32:51 --> 00:32:57
If A were something like one
plus i,
498
00:32:54 --> 00:33:00
I'd be screwed,
I mean, because then I couldn't
499
00:32:57 --> 00:33:03
say that this was the real part
anymore.
500
00:33:02 --> 00:33:08
So, saying that's the real
part, and this is the imaginary
501
00:33:07 --> 00:33:13
part, I was using the fact that
these two numbers,
502
00:33:12 --> 00:33:18
constants, were real constants:
very important.
503
00:33:17 --> 00:33:23
So, what is the case two
solution?
504
00:33:20 --> 00:33:26
Well, what are the real and
imaginary parts
505
00:33:26 --> 00:33:32
of (a plus b i) t?
Well, y equals e to the at +
506
00:33:31 --> 00:33:37
ibt.
Okay, you've had experience.
507
00:33:37 --> 00:33:43
You know how to do this now.
That's e to the at
508
00:33:42 --> 00:33:48
times, well, the real part is,
well, let's write it this way.
509
00:33:47 --> 00:33:53
The real part is e to the at
times cosine b t.
510
00:33:51 --> 00:33:57
Notice how the a and b enter
511
00:33:54 --> 00:34:00
into the expression.
That's the real part.
512
00:33:58 --> 00:34:04
And, the imaginary part is e to
the at times the sine of bt.
513
00:34:03 --> 00:34:09
And therefore,
514
00:34:07 --> 00:34:13
the solution,
both of these must,
515
00:34:10 --> 00:34:16
therefore, be solutions to the
equation.
516
00:34:13 --> 00:34:19
And therefore,
the general solution to the ODE
517
00:34:18 --> 00:34:24
is y equals, now,
you've got to put in the
518
00:34:21 --> 00:34:27
arbitrary constants.
It's a nice thing to do to
519
00:34:26 --> 00:34:32
factor out the e to the at.
520
00:34:29 --> 00:34:35
It makes it look a little
better.
521
00:34:32 --> 00:34:38
And so, the constants are c1
cosine bt and c2 sine bt.
522
00:34:39 --> 00:34:45
Yeah, but what does that look
like?
523
00:34:41 --> 00:34:47
Well, you know that too.
This is an exponential,
524
00:34:45 --> 00:34:51
which controls the amplitude.
But this guy,
525
00:34:49 --> 00:34:55
which is a combination of two
sinusoidal oscillations with
526
00:34:53 --> 00:34:59
different amplitudes,
but with the same frequency,
527
00:34:57 --> 00:35:03
the b's are the same in both of
them, and therefore,
528
00:35:01 --> 00:35:07
this is, itself,
a purely sinusoidal
529
00:35:04 --> 00:35:10
oscillation.
So, in other words,
530
00:35:09 --> 00:35:15
I don't have room to write it,
but it's equal to,
531
00:35:15 --> 00:35:21
you know.
It's a good example of where
532
00:35:20 --> 00:35:26
you'd use that trigonometric
identity I spent time on before
533
00:35:27 --> 00:35:33
the exam.
Okay, let's work a quick
534
00:35:31 --> 00:35:37
example just to see how this
works out.
535
00:35:33 --> 00:35:39
Well, let's get rid of this.
Okay, let's now make the
536
00:35:36 --> 00:35:42
damping, since this is showing
oscillations,
537
00:35:39 --> 00:35:45
it must correspond to the case
where the damping is less strong
538
00:35:42 --> 00:35:48
compared with the spring
constant.
539
00:35:44 --> 00:35:50
So, the theorem is that if you
have a complex solution,
540
00:35:47 --> 00:35:53
u plus iv, so each of these is
a function of time,
541
00:35:50 --> 00:35:56
u plus iv is the
complex solution to a real
542
00:35:53 --> 00:35:59
differential equation with
constant coefficients.
543
00:35:56 --> 00:36:02
A stiff spring,
one that pulls with hard force
544
00:35:58 --> 00:36:04
is going to make that thing go
back and forth,
545
00:36:01 --> 00:36:07
particularly at the dipping is
weak.
546
00:36:05 --> 00:36:11
So, let's use almost the same
equation as I've just concealed.
547
00:36:09 --> 00:36:15
But, do you remember a used
four here?
548
00:36:12 --> 00:36:18
Okay, before we used three and
we got the solution to look like
549
00:36:17 --> 00:36:23
that.
Now, we will give it a little
550
00:36:19 --> 00:36:25
more energy by putting some
moxie in the springs.
551
00:36:23 --> 00:36:29
So now, the spring is pulling a
little harder,
552
00:36:26 --> 00:36:32
bigger force,
a stiffer spring.
553
00:36:30 --> 00:36:36
Okay, the characteristic
equation is now going to be r
554
00:36:35 --> 00:36:41
squared plus 4r plus five is
equal to zero.
555
00:36:40 --> 00:36:46
And therefore,
if I solve for r,
556
00:36:43 --> 00:36:49
I'm not going to bother trying
to factor this because I
557
00:36:49 --> 00:36:55
prepared for this lecture,
and I know, quadratic formula
558
00:36:54 --> 00:37:00
time, minus four plus or minus
the square root of b squared,
559
00:37:00 --> 00:37:06
16, minus four times five,
16 minus 20 is negative four
560
00:37:05 --> 00:37:11
all over two.
And therefore,
561
00:37:09 --> 00:37:15
that makes negative two plus or
minus, this makes,
562
00:37:13 --> 00:37:19
simply, i.
2i divided by two,
563
00:37:15 --> 00:37:21
which is i.
So, the exponential solution is
564
00:37:19 --> 00:37:25
e to the negative two,
you don't have to write this
565
00:37:23 --> 00:37:29
in.
You can get the thing directly.
566
00:37:26 --> 00:37:32
t, let's use the one with the
plus sign, and that's going to
567
00:37:31 --> 00:37:37
give, as the real solutions,
e to the negative two t times
568
00:37:36 --> 00:37:42
cosine t,
and e to the negative 2t times
569
00:37:41 --> 00:37:47
the sine of t.
570
00:37:46 --> 00:37:52
And therefore,
the solution is going to be y
571
00:37:51 --> 00:37:57
equals e to the negative 2t
times
572
00:37:58 --> 00:38:04
(c1 cosine t plus c2 sine t).
573
00:38:04 --> 00:38:10
If you want to put initial
conditions, you can put them in
574
00:38:08 --> 00:38:14
the same way I did them before.
Suppose we use the same initial
575
00:38:12 --> 00:38:18
conditions: y of zero
equals one,
576
00:38:15 --> 00:38:21
and y prime of zero equals one,
equals zero,
577
00:38:19 --> 00:38:25
let's say, wait,
blah, blah, blah,
578
00:38:22 --> 00:38:28
zero, yeah.
Okay, I'd like to take time to
579
00:38:24 --> 00:38:30
actually do the calculation,
but there's nothing new in it.
580
00:38:30 --> 00:38:36
I'd have to take,
calculate the derivative here,
581
00:38:35 --> 00:38:41
and then I would substitute in,
solve equations,
582
00:38:40 --> 00:38:46
and when you do all that,
just as before,
583
00:38:44 --> 00:38:50
the answer that you get is y
equals e to the negative 2t,
584
00:38:50 --> 00:38:56
so,
choose the constants c1 and c2
585
00:38:55 --> 00:39:01
by solving linear equations,
and the answer is cosine t,
586
00:39:01 --> 00:39:07
so, c1 turns out to be one,
and c2 turns out to be two,
587
00:39:07 --> 00:39:13
I hope.
Okay, I want to know,
588
00:39:11 --> 00:39:17
but what does that look like?
Well, use that trigonometric
589
00:39:15 --> 00:39:21
identity.
The e to the negative 2t
590
00:39:18 --> 00:39:24
is just a real
factor which is going to
591
00:39:21 --> 00:39:27
reproduce itself.
The question is,
592
00:39:24 --> 00:39:30
what is cosine t plus 2 sine t
look like?
593
00:39:28 --> 00:39:34
What's its amplitude as a pure
oscillation?
594
00:39:31 --> 00:39:37
It's the square root of one
plus two squared.
595
00:39:35 --> 00:39:41
Remember, it depends on looking
596
00:39:39 --> 00:39:45
at that little triangle,
which is one,
597
00:39:41 --> 00:39:47
two, this is a different scale
than that.
598
00:39:44 --> 00:39:50
And here is the square root of
five, right?
599
00:39:47 --> 00:39:53
And, here is phi,
the phase lag.
600
00:39:50 --> 00:39:56
So, it's equal to the square
root of five.
601
00:39:53 --> 00:39:59
So, it's the square root of
five times e to
602
00:39:57 --> 00:40:03
the negative 2t,
and the stuff inside is the
603
00:40:00 --> 00:40:06
cosine of, the frequency is one.
Circular frequency is one,
604
00:40:06 --> 00:40:12
so it's t minus phi,
where phi is this angle.
605
00:40:10 --> 00:40:16
How big is that,
one and two?
606
00:40:13 --> 00:40:19
Well, if this were the square
root of three,
607
00:40:16 --> 00:40:22
which is a little less than
two, it would be 60 infinity.
608
00:40:20 --> 00:40:26
So, this must be 70 infinity.
So, phi is 70 infinity plus or minus
609
00:40:24 --> 00:40:30
five, let's say.
So, it looks like a slightly
610
00:40:29 --> 00:40:35
delayed cosine curve,
but the amplitude is falling.
611
00:40:32 --> 00:40:38
So, it has to start.
So, if I draw it,
612
00:40:35 --> 00:40:41
here's one, here is,
let's say, the square root of
613
00:40:39 --> 00:40:45
five up about here.
Then, the square root of five
614
00:40:43 --> 00:40:49
times e to the negative 2t
looks maybe
615
00:40:47 --> 00:40:53
something like this.
So, that's square root of five
616
00:40:51 --> 00:40:57
e to the negative 2t.
617
00:40:54 --> 00:41:00
This is cosine t,
but shoved over by not quite pi
618
00:40:59 --> 00:41:05
over two.
It starts at one,
619
00:41:02 --> 00:41:08
and with the slope zero.
So, the solution starts like
620
00:41:06 --> 00:41:12
this.
It has to be guided in its
621
00:41:08 --> 00:41:14
amplitude by this function out
there, and in between it's the
622
00:41:13 --> 00:41:19
cosine curve.
But it's moved over.
623
00:41:15 --> 00:41:21
So, if this is pi over two,
the first time it crosses,
624
00:41:20 --> 00:41:26
it's 72 infinity to the right
of that. So, if this is pi
625
00:41:24 --> 00:41:30
over two, it's pi over two plus
70 infinity where it crosses.
626
00:41:27 --> 00:41:33
So, it must be doing something
like this.
627
00:41:32 --> 00:41:38
And now, on the other side,
it's got to stay within the
628
00:41:36 --> 00:41:42
same amplitude.
So, it must be doing something
629
00:41:40 --> 00:41:46
like this.
Okay, that gets us to,
630
00:41:43 --> 00:41:49
if this is the under-damped
case, because if you're trying
631
00:41:48 --> 00:41:54
to do this with a swinging door,
it means the door's going to be
632
00:41:54 --> 00:42:00
swinging back and forth.
Or, our little mass now hidden,
633
00:41:59 --> 00:42:05
but you could see it behind
that board, is going to be doing
634
00:42:04 --> 00:42:10
this.
But, it never stops.
635
00:42:07 --> 00:42:13
It never stops.
It doesn't realize,
636
00:42:11 --> 00:42:17
but not in theoretical life.
So, this is the under-damped.
637
00:42:16 --> 00:42:22
All right, so it's like
Goldilocks and the Three Bears.
638
00:42:21 --> 00:42:27
That's too hot,
and this is too cold.
639
00:42:25 --> 00:42:31
What is the thing which is just
right?
640
00:42:30 --> 00:42:36
Well, that's the thing you're
going to study on the problem
641
00:42:37 --> 00:42:43
set.
So, just right is called
642
00:42:40 --> 00:42:46
critically damped.
It's what people aim for in
643
00:42:46 --> 00:42:52
trying to damp motion that they
don't want.
644
00:42:51 --> 00:42:57
Now, what's critically damped?
It must be the case just in
645
00:42:58 --> 00:43:04
between these two.
Neither complex,
646
00:43:03 --> 00:43:09
nor the roots different.
It's the case of two equal
647
00:43:08 --> 00:43:14
roots.
So, r squared plus Ar plus B
648
00:43:11 --> 00:43:17
equals zero
has two equal roots.
649
00:43:17 --> 00:43:23
Now, that's a very special
equation.
650
00:43:20 --> 00:43:26
Suppose we call the root,
since all of these,
651
00:43:24 --> 00:43:30
notice these roots in this
physical case.
652
00:43:30 --> 00:43:36
The roots always turn out to be
negative numbers,
653
00:43:33 --> 00:43:39
or have a negative real part.
I'm going to call the root a.
654
00:43:37 --> 00:43:43
So, r equals negative a,
the root.
655
00:43:40 --> 00:43:46
a is understood to be a
positive number.
656
00:43:43 --> 00:43:49
I want that root to be really
negative.
657
00:43:45 --> 00:43:51
Then, the equation looks like,
the characteristic equation is
658
00:43:50 --> 00:43:56
going to be r plus a,
right, if the root is negative
659
00:43:53 --> 00:43:59
a, squared because it's a double
root.
660
00:43:57 --> 00:44:03
And, that means the equation is
of the form r squared plus two
661
00:44:01 --> 00:44:07
times a r plus a squared equals
zero.
662
00:44:07 --> 00:44:13
In other words,
the ODE looked like this.
663
00:44:10 --> 00:44:16
The ODE looked like y double
prime plus 2a y prime
664
00:44:15 --> 00:44:21
plus, in other words,
665
00:44:18 --> 00:44:24
the damping and the spring
constant were related in this
666
00:44:22 --> 00:44:28
special wake,
that for a given value of the
667
00:44:26 --> 00:44:32
spring constant,
there was exactly one value of
668
00:44:30 --> 00:44:36
the damping which produced this
in between case.
669
00:44:36 --> 00:44:42
Now, what's the problem
connected with it?
670
00:44:39 --> 00:44:45
Well, the problem,
unfortunately,
671
00:44:42 --> 00:44:48
is staring us in the face when
we want to solve it.
672
00:44:46 --> 00:44:52
The problem is that we have a
solution, but it is y equals e
673
00:44:51 --> 00:44:57
to the minus at.
I don't have another root to
674
00:44:56 --> 00:45:02
get another solution with.
And, the question is,
675
00:45:01 --> 00:45:07
where do I get that other
solution from?
676
00:45:04 --> 00:45:10
Now, there are three ways to
get it.
677
00:45:06 --> 00:45:12
Well, there are four ways to
get it.
678
00:45:09 --> 00:45:15
You look it up in Euler.
That's the fourth way.
679
00:45:12 --> 00:45:18
That's the real way to do it.
But, I've given you one way as
680
00:45:16 --> 00:45:22
problem number one on the
problem set.
681
00:45:19 --> 00:45:25
I've given you another way as
problem number two on the
682
00:45:23 --> 00:45:29
problem set.
And, the third way you will
683
00:45:26 --> 00:45:32
have to wait for about a week
and a half.
684
00:45:28 --> 00:45:34
And, I will give you a third
way, too.
685
00:45:33 --> 00:45:39
By that time,
you won't want to see any more
686
00:45:36 --> 00:45:42
ways.
But, I'd like to introduce you
687
00:45:39 --> 00:45:45
to the way on the problem set.
And, it is this,
688
00:45:43 --> 00:45:49
that if you know one solution
to an equation,
689
00:45:47 --> 00:45:53
which looks like a linear
equation, in fact,
690
00:45:51 --> 00:45:57
the piece can be functions of
t.
691
00:45:54 --> 00:46:00
They don't have to be constant,
so I'll use the books notation
692
00:45:59 --> 00:46:05
with p's and q's.
y prime plus q y equals zero.
693
00:46:05 --> 00:46:11
If you know one solution,
there's an absolute,
694
00:46:08 --> 00:46:14
ironclad guarantee,
if you'll know that it's true
695
00:46:13 --> 00:46:19
because I'm asking you to prove
it for yourself.
696
00:46:17 --> 00:46:23
There's another of the form,
having this as a factor,
697
00:46:21 --> 00:46:27
one solution y one,
let's call it,
698
00:46:24 --> 00:46:30
y equals y1 u is
another solution.
699
00:46:28 --> 00:46:34
And, you will be able to find
u, I swear.
700
00:46:33 --> 00:46:39
Now, let's, in the remaining
couple of minutes carry that out
701
00:46:37 --> 00:46:43
just for this case because I
want you to see how to arrange
702
00:46:41 --> 00:46:47
the work nicely.
And, I want you to arrange your
703
00:46:44 --> 00:46:50
work when you do the problem
sets in the same way.
704
00:46:48 --> 00:46:54
So, the way to do it is,
the solution we know is e to
705
00:46:52 --> 00:46:58
the minus at.
So, we are going to look for a
706
00:46:56 --> 00:47:02
solution to this differential
equation.
707
00:47:00 --> 00:47:06
That's the differential
equation.
708
00:47:02 --> 00:47:08
And, the solution we are going
to look for is of the form e to
709
00:47:08 --> 00:47:14
the negative at times u.
710
00:47:12 --> 00:47:18
Now, you're going to have to
make calculation like this
711
00:47:17 --> 00:47:23
several times in the course of
the term.
712
00:47:20 --> 00:47:26
Do it this way.
y prime equals,
713
00:47:23 --> 00:47:29
differentiate,
minus a e to the minus a t u
714
00:47:27 --> 00:47:33
plus e to the minus a
t u prime.
715
00:47:33 --> 00:47:39
And then, differentiate again.
716
00:47:37 --> 00:47:43
The answer will be a squared.
717
00:47:39 --> 00:47:45
You differentiate this:
a squared e to the negative at
718
00:47:43 --> 00:47:49
u.
I'll have to do this a little
719
00:47:47 --> 00:47:53
fast, but the next term will be,
okay, minus,
720
00:47:50 --> 00:47:56
so this times u prime,
and from this are you going to
721
00:47:53 --> 00:47:59
get another minus.
So, combining what you get from
722
00:47:57 --> 00:48:03
here, and here,
you're going to get minus 2a e
723
00:48:00 --> 00:48:06
to the minus a t u prime.
724
00:48:05 --> 00:48:11
And then, there is a final
term, which comes from this,
725
00:48:08 --> 00:48:14
e to the minus a t u double
prime.
726
00:48:11 --> 00:48:17
Two of these,
because of a piece here and a
727
00:48:15 --> 00:48:21
piece here combine to make that.
And now, to plug into the
728
00:48:19 --> 00:48:25
equation, you multiply this by
one.
729
00:48:21 --> 00:48:27
In other words,
you don't do anything to it.
730
00:48:24 --> 00:48:30
You multiply this line by 2a,
and you multiply that line by a
731
00:48:28 --> 00:48:34
squared, and you add them.
732
00:48:32 --> 00:48:38
On the left-hand side,
I get zero.
733
00:48:34 --> 00:48:40
What do I get on the right?
Notice how I've arrange the
734
00:48:39 --> 00:48:45
work so it adds nicely.
This has a squared times this,
735
00:48:43 --> 00:48:49
plus 2a times that,
plus one times that makes zero.
736
00:48:47 --> 00:48:53
2a times this plus one times
this makes zero.
737
00:48:50 --> 00:48:56
All that survives is e to the a
t u double prime,
738
00:48:56 --> 00:49:02
and therefore,
e to the minus a t u double
739
00:48:59 --> 00:49:05
prime is equal to zero.
740
00:49:04 --> 00:49:10
So, please tell me,
what's u double prime?
741
00:49:06 --> 00:49:12
It's zero.
So, please tell me,
742
00:49:09 --> 00:49:15
what's u?
It's c1 t plus c2.
743
00:49:11 --> 00:49:17
Now, that gives me a whole
744
00:49:14 --> 00:49:20
family of solutions.
Just t would be enough because
745
00:49:17 --> 00:49:23
all I am doing is looking for
one solution that's different
746
00:49:22 --> 00:49:28
from e to the minus a t.
747
00:49:24 --> 00:49:30
And, that solution,
therefore, is y equals e to the
748
00:49:28 --> 00:49:34
minus a t times t.
749
00:49:32 --> 00:49:38
And, there's my second
solution.
750
00:49:34 --> 00:49:40
So, this is a solution of the
critically damped case.
751
00:49:37 --> 00:49:43
And, you are going to use it in
three or four of the different
752
00:49:42 --> 00:49:48
problems on the problem set.
But, I think you can deal with
753
00:49:46 --> 00:49:52
virtually the whole problem set,
except for the last problem,
754
00:49:50 --> 00:49:56
now.