WEBVTT

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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,

00:13:38.000 --> 00:13:44.000
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

00:14:16.000 --> 00:14:22.000
using t as the dependent
variable.

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So, T is the independent
variable.

00:14:22.000 --> 00:14:28.000
Why do I do that?
The answer is because somebody

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thought of doing it,
probably Euler,

00:14:31.000 --> 00:14:37.000
and it's been a tradition
that's handed down for the last

00:14:36.000 --> 00:14:42.000
300 or 400 years.
Some things we just know.

00:14:42.000 --> 00:14:48.000
All right, so if I do that,
as you learned from the exam,

00:14:46.000 --> 00:14:52.000
it's very easy to differentiate
exponentials.

00:14:50.000 --> 00:14:56.000
That's why people love them.
It's also very easy to

00:14:54.000 --> 00:15:00.000
integrate exponentials.
And, half of you integrated

00:14:58.000 --> 00:15:04.000
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.

00:15:09.000 --> 00:15:15.000
Substitute, in other words,
and what do we get?

00:15:12.000 --> 00:15:18.000
Well, for y double prime,
I get r squared e to the rt.

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That's y double prime

00:15:19.000 --> 00:15:25.000
because each time you
differentiate it,

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you put an extra power of r out
in front.

00:15:25.000 --> 00:15:31.000
But otherwise,
do nothing.

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The next term will be r times,
sorry, I forgot the constant.

00:15:33.000 --> 00:15:39.000
Capital A times r e to the rt,

00:15:36.000 --> 00:15:42.000
and then there's the last term,
B times y itself,

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which is B e to the rt.

00:15:41.000 --> 00:15:47.000
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

00:15:51.000 --> 00:15:57.000
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

00:16:10.000 --> 00:16:16.000
equation r squared plus ar plus
b equals zero.

00:16:16.000 --> 00:16:22.000
We are trying to find values of

00:16:19.000 --> 00:16:25.000
r that satisfy that equation.
And that, dear hearts,

00:16:24.000 --> 00:16:30.000
is why you learn to solve
quadratic equations in high

00:16:29.000 --> 00:16:35.000
school, in order that in this
moment, you would be now ready

00:16:34.000 --> 00:16:40.000
to find how spring-mass systems
behave when they are damped.

00:16:41.000 --> 00:16:47.000
This is called the
characteristic equation.

00:16:45.000 --> 00:16:51.000
The characteristic equation of
the ODE, or of the system of the

00:16:52.000 --> 00:16:58.000
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

00:17:08.000 --> 00:17:14.000
know there are several cases.
And, each of those cases

00:17:14.000 --> 00:17:20.000
corresponds to a different
behavior.

00:17:17.000 --> 00:17:23.000
And, the cases depend upon what
the roots look like.

00:17:22.000 --> 00:17:28.000
The possibilities are the roots
could be real,

00:17:25.000 --> 00:17:31.000
and distinct.
That's the easiest case to

00:17:29.000 --> 00:17:35.000
handle.
The roots might be a pair of

00:17:31.000 --> 00:17:37.000
complex conjugate numbers.
That's harder to handle,

00:17:36.000 --> 00:17:42.000
but we are ready to do it.
And, the third case,

00:17:40.000 --> 00:17:46.000
which is the one most in your
problem set is the most

00:17:45.000 --> 00:17:51.000
puzzling: when the roots are
real, and equal.

00:17:48.000 --> 00:17:54.000
And, I'm going to talk about
those three cases in that order.

00:17:53.000 --> 00:17:59.000
So, the first case is the roots
are real and unequal.

00:17:57.000 --> 00:18:03.000
If I tell you they are unequal,
and I will put down real to

00:18:02.000 --> 00:18:08.000
make that clear.
Well, that is by far the

00:18:07.000 --> 00:18:13.000
simplest case because
immediately, one sees we have

00:18:11.000 --> 00:18:17.000
two roots.
They are different,

00:18:13.000 --> 00:18:19.000
and therefore,
we get our two solutions

00:18:17.000 --> 00:18:23.000
immediately.
So, the solutions are,

00:18:19.000 --> 00:18:25.000
the general solution to the
equation, I write down without

00:18:24.000 --> 00:18:30.000
further ado as y equals c1 e to
the r1 t plus c2 e to the r2 t.

00:18:34.000 --> 00:18:40.000
There's our solution.
Now, because that was so easy,

00:18:38.000 --> 00:18:44.000
and we didn't have to do any
work, I'd like to extend this

00:18:42.000 --> 00:18:48.000
case a little bit by using it as
an example of how you put in the

00:18:48.000 --> 00:18:54.000
initial conditions,
how to put in the c.

00:18:51.000 --> 00:18:57.000
So, let me work a specific
numerical example,

00:18:54.000 --> 00:19:00.000
since we are not going to try
to do this theoretically until

00:18:59.000 --> 00:19:05.000
next Wednesday.
Let's just do a numerical

00:19:04.000 --> 00:19:10.000
example.
So, suppose I take the

00:19:07.000 --> 00:19:13.000
constants to be the damping
constant to be a four,

00:19:11.000 --> 00:19:17.000
and the spring constant,
I'll take the mass to be one,

00:19:16.000 --> 00:19:22.000
and the spring constant to be
three.

00:19:19.000 --> 00:19:25.000
So, there's more damping here,
damping force here.

00:19:24.000 --> 00:19:30.000
You can't really talk that way
since the units are different.

00:19:31.000 --> 00:19:37.000
But, this number is bigger than
that one.

00:19:33.000 --> 00:19:39.000
That seems clear,
at any rate.

00:19:35.000 --> 00:19:41.000
Okay, now, what was the
characteristic equation?

00:19:39.000 --> 00:19:45.000
Look, now watch.
Please do what I do.

00:19:41.000 --> 00:19:47.000
I've found in the past,
even by the middle of the term,

00:19:45.000 --> 00:19:51.000
there are still students who
feel that they must substitute y

00:19:50.000 --> 00:19:56.000
equals e to the rt,
and go through that whole

00:19:53.000 --> 00:19:59.000
little derivation to find that
you don't do that.

00:19:57.000 --> 00:20:03.000
It's a waste of time.
I did it that you might not

00:20:02.000 --> 00:20:08.000
ever have to do it again.
Immediately write down the

00:20:06.000 --> 00:20:12.000
characteristic equation.
That's not very hard.

00:20:10.000 --> 00:20:16.000
r squared plus 4r plus three
equals zero.

00:20:14.000 --> 00:20:20.000
And, if you can write down its

00:20:17.000 --> 00:20:23.000
roots immediately,
splendid.

00:20:20.000 --> 00:20:26.000
But, let's not assume that
level of competence.

00:20:23.000 --> 00:20:29.000
So, it's r plus three times r
plus one equals zero.

00:20:29.000 --> 00:20:35.000
Okay, you factor it.

00:20:33.000 --> 00:20:39.000
This being 18.03,
a lot of the times the roots

00:20:36.000 --> 00:20:42.000
will be integers when they are
not, God forbid,

00:20:39.000 --> 00:20:45.000
you will have to use the
quadratic formula.

00:20:42.000 --> 00:20:48.000
But here, the roots were
integers.

00:20:44.000 --> 00:20:50.000
It is, after all,
only the first example.

00:20:47.000 --> 00:20:53.000
So, the solution,
the general solution is y

00:20:50.000 --> 00:20:56.000
equals c1 e to the negative,
notice the root is minus three

00:20:54.000 --> 00:21:00.000
and minus one,
minus 3t plus c2 e to the

00:20:56.000 --> 00:21:02.000
negative t.

00:21:01.000 --> 00:21:07.000
Now, suppose it's an initial
value problem.

00:21:04.000 --> 00:21:10.000
So, I gave you an initial
condition.

00:21:06.000 --> 00:21:12.000
Suppose the initial conditions
were that y of zero were one.

00:21:11.000 --> 00:21:17.000
So, at the start,

00:21:13.000 --> 00:21:19.000
the mass has been moved over to
the position,

00:21:16.000 --> 00:21:22.000
one, here.
Well, we expected it,

00:21:18.000 --> 00:21:24.000
then, to start doing that.
But, this is fairly heavily

00:21:22.000 --> 00:21:28.000
damped.
This is heavily damped.

00:21:25.000 --> 00:21:31.000
I'm going to assume that the
mass starts at rest.

00:21:30.000 --> 00:21:36.000
So, the spring is distended.
The masses over here.

00:21:33.000 --> 00:21:39.000
But, there's no motion at times
zero this way or that way.

00:21:37.000 --> 00:21:43.000
In other words,
I'm not pushing it.

00:21:40.000 --> 00:21:46.000
I'm just releasing it and
letting it do its thing after

00:21:44.000 --> 00:21:50.000
that.
Okay, so y prime of zero,

00:21:46.000 --> 00:21:52.000
I'll assume, is zero.

00:21:49.000 --> 00:21:55.000
So, it starts at rest,
but in the extended position,

00:21:53.000 --> 00:21:59.000
one unit to the right of the
equilibrium position.

00:21:56.000 --> 00:22:02.000
Now, all you have to do is use
these two conditions.

00:22:02.000 --> 00:22:08.000
Notice I have to have two
conditions because there are two

00:22:06.000 --> 00:22:12.000
constants I have to find the
value of.

00:22:09.000 --> 00:22:15.000
All right, so,
let's substitute,

00:22:11.000 --> 00:22:17.000
well, we're going to have to
calculate the derivative.

00:22:15.000 --> 00:22:21.000
So, why don't we do that right
away?

00:22:18.000 --> 00:22:24.000
So, this is minus three c1 e to
the minus 3t minus c2 e to the

00:22:22.000 --> 00:22:28.000
negative t.

00:22:27.000 --> 00:22:33.000
And now, if I substitute in at
zero, when t equals zero,

00:22:31.000 --> 00:22:37.000
what do I get?
Well, the first equation,

00:22:34.000 --> 00:22:40.000
the left says that y of zero
should be one.

00:22:38.000 --> 00:22:44.000
And, the right says this is
one.

00:22:41.000 --> 00:22:47.000
So, it's c1 plus c2.

00:22:43.000 --> 00:22:49.000
That's the result of
substituting t equals zero.

00:22:47.000 --> 00:22:53.000
How about substituting?

00:22:49.000 --> 00:22:55.000
What should I substitute in the
second equation?

00:22:53.000 --> 00:22:59.000
Well, y prime of zero is zero.

00:22:56.000 --> 00:23:02.000
So, if the second equation,
when I put in t equals zero,

00:23:01.000 --> 00:23:07.000
the left side is zero according
to that initial value,

00:23:05.000 --> 00:23:11.000
and the right side is negative
three c1 minus c2.

00:23:12.000 --> 00:23:18.000
You see what you end up with,
therefore, is a pair of

00:23:15.000 --> 00:23:21.000
simultaneous linear equations.
And, this is why you learn to

00:23:19.000 --> 00:23:25.000
study linear set of pairs of
simultaneous linear equations in

00:23:24.000 --> 00:23:30.000
high school.
These are among the most

00:23:26.000 --> 00:23:32.000
important.
Solving problems of this type

00:23:29.000 --> 00:23:35.000
are among the most important
applications of that kind of

00:23:33.000 --> 00:23:39.000
algebra, and this kind of
algebra.

00:23:37.000 --> 00:23:43.000
All right, what's the answer
finally?

00:23:40.000 --> 00:23:46.000
Well, if I add the two of them,
I get minus 2c1 equals one.

00:23:45.000 --> 00:23:51.000
So, c1 is equal to minus one
half.

00:23:49.000 --> 00:23:55.000
And, if c1 is minus a half,

00:23:54.000 --> 00:24:00.000
then c2 is minus 3c1.

00:23:57.000 --> 00:24:03.000
So, c2 is three halves.

00:24:10.000 --> 00:24:16.000
The final question is,
what does that look like as a

00:24:13.000 --> 00:24:19.000
solution?
Well, in general,

00:24:14.000 --> 00:24:20.000
these combinations of two
exponentials aren't very easy to

00:24:17.000 --> 00:24:23.000
plot by yourself.
That's one of the reasons you

00:24:20.000 --> 00:24:26.000
are being given this little
visual which plots them for you.

00:24:24.000 --> 00:24:30.000
All you have to do is,
as you'll see,

00:24:26.000 --> 00:24:32.000
set the damping constant,
set the constants,

00:24:28.000 --> 00:24:34.000
set the initial conditions,
and by magic,

00:24:31.000 --> 00:24:37.000
the curve appears on the
screen.

00:24:34.000 --> 00:24:40.000
And, if you change either of
the constants,

00:24:40.000 --> 00:24:46.000
the curve will change nicely
right along with it.

00:24:48.000 --> 00:24:54.000
So, the solution is y equals
minus one half e to the minus 3t

00:24:57.000 --> 00:25:03.000
plus three halves e to the
negative t.

00:25:06.000 --> 00:25:12.000
How does it look?

00:25:11.000 --> 00:25:17.000
Well, I don't expect you to be
able to plot that by yourself,

00:25:16.000 --> 00:25:22.000
but you can at least get
started.

00:25:19.000 --> 00:25:25.000
It does have to satisfy the
initial conditions.

00:25:23.000 --> 00:25:29.000
That means it should start at
one, and its starting slope is

00:25:28.000 --> 00:25:34.000
zero.
So, it starts like that.

00:25:32.000 --> 00:25:38.000
These are both declining
exponentials.

00:25:35.000 --> 00:25:41.000
This declines very rapidly,
this somewhat more slowly.

00:25:40.000 --> 00:25:46.000
It does something like that.
If this term were a lot,

00:25:44.000 --> 00:25:50.000
lot more negative,
I mean, that's the way that

00:25:49.000 --> 00:25:55.000
particular solution looks.
How might other solutions look?

00:25:54.000 --> 00:26:00.000
I'll draw a few other
possibilities.

00:25:57.000 --> 00:26:03.000
If the initial term,
if, for example,

00:26:00.000 --> 00:26:06.000
the initial slope were quite
negative, well,

00:26:04.000 --> 00:26:10.000
that would have start like
this.

00:26:09.000 --> 00:26:15.000
Now, just your experience of
physics, or of the real world

00:26:12.000 --> 00:26:18.000
suggests that if I give,
if I start the thing at one,

00:26:15.000 --> 00:26:21.000
but give it a strongly negative
push, it's going to go beyond

00:26:19.000 --> 00:26:25.000
the equilibrium position,
and then come back again.

00:26:22.000 --> 00:26:28.000
But, because the damping is
big, it's not going to be able

00:26:26.000 --> 00:26:32.000
to get through that.
The equilibrium position,

00:26:29.000 --> 00:26:35.000
a second time,
is going to look something like

00:26:31.000 --> 00:26:37.000
that.
Or, if I push it in that

00:26:34.000 --> 00:26:40.000
direction, the positive
direction, that it starts off

00:26:37.000 --> 00:26:43.000
with a positive slope.
But it loses its energy because

00:26:41.000 --> 00:26:47.000
the spring is pulling it.
It comes and does something

00:26:44.000 --> 00:26:50.000
like that.
So, in other words,

00:26:46.000 --> 00:26:52.000
it might go down.
Cut across the equilibrium

00:26:49.000 --> 00:26:55.000
position, come back again,
it do that?

00:26:52.000 --> 00:26:58.000
No, that it cannot do.
I was considering giving you a

00:26:55.000 --> 00:27:01.000
problem to prove that,
but I got tired of making out

00:26:58.000 --> 00:27:04.000
the problems set,
and decided I tortured you

00:27:01.000 --> 00:27:07.000
enough already,
as you will see.

00:27:05.000 --> 00:27:11.000
So, anyway, these are different
possibilities for the way that

00:27:11.000 --> 00:27:17.000
can look.
This case, where it just

00:27:14.000 --> 00:27:20.000
returns in the long run is
called the over-damped case,

00:27:20.000 --> 00:27:26.000
over-damped.
Now, there is another case

00:27:24.000 --> 00:27:30.000
where the thing oscillates back
and forth.

00:27:30.000 --> 00:27:36.000
We would expect to get that
case if the damping is very

00:27:33.000 --> 00:27:39.000
little or nonexistent.
Then, there's very little

00:27:37.000 --> 00:27:43.000
preventing the mass from doing
that, although we do expect if

00:27:41.000 --> 00:27:47.000
there's any damping at all,
we expect it ultimately to get

00:27:45.000 --> 00:27:51.000
nearer and nearer to the
equilibrium position.

00:27:48.000 --> 00:27:54.000
Mathematically,
what does that correspond to?

00:27:51.000 --> 00:27:57.000
Well, that's going to
correspond to case two,

00:27:55.000 --> 00:28:01.000
where the roots are complex.
The roots are complex,

00:27:59.000 --> 00:28:05.000
and this is why,
let's call the roots,

00:28:03.000 --> 00:28:09.000
in that case we know that the
roots are of the form a plus or

00:28:08.000 --> 00:28:14.000
minus bi.
There are two roots,

00:28:10.000 --> 00:28:16.000
and they are a complex
conjugate.

00:28:13.000 --> 00:28:19.000
All right, let's take one of
them.

00:28:16.000 --> 00:28:22.000
What does a correspond to in
terms of the exponential?

00:28:20.000 --> 00:28:26.000
Well, remember,
the function of the r was,

00:28:24.000 --> 00:28:30.000
it's this r when we tried our
exponential solution.

00:28:30.000 --> 00:28:36.000
So, what we formally,
this means we get a complex

00:28:34.000 --> 00:28:40.000
solution.
The complex solution y equals e

00:28:38.000 --> 00:28:44.000
to this, let's use one of them,
let's say, (a plus bi) times t.

00:28:43.000 --> 00:28:49.000
The question is,

00:28:47.000 --> 00:28:53.000
what do we do that?
We are not really interested,

00:28:51.000 --> 00:28:57.000
I don't know what a complex
solution to that thing means.

00:28:56.000 --> 00:29:02.000
It doesn't have any meaning.
What I want to know is how y

00:29:01.000 --> 00:29:07.000
behaves or how x behaves in that
picture.

00:29:07.000 --> 00:29:13.000
And, that better be a real
function because otherwise I

00:29:10.000 --> 00:29:16.000
don't know what to do with it.
So, we are looking for two real

00:29:14.000 --> 00:29:20.000
functions, the y1 and the y2.
But, in fact,

00:29:17.000 --> 00:29:23.000
what we've got is one complex
function.

00:29:20.000 --> 00:29:26.000
All right, now,
a theorem to the rescue:

00:29:22.000 --> 00:29:28.000
this, I'm not going to save for
Wednesday because it's so

00:29:26.000 --> 00:29:32.000
simple.
So, the theorem is that if you

00:29:30.000 --> 00:29:36.000
have a complex solution,
u plus iv, so each of these is

00:29:36.000 --> 00:29:42.000
a function of time,
u plus iv is the complex

00:29:40.000 --> 00:29:46.000
solution to a real differential
equation with constant

00:29:45.000 --> 00:29:51.000
coefficients.
Well, it doesn't have to have

00:29:49.000 --> 00:29:55.000
constant coefficients.
It has to be linear.

00:29:53.000 --> 00:29:59.000
Let me just write it out to y
double prime plus A y prime plus

00:29:59.000 --> 00:30:05.000
B y equals zero.

00:30:04.000 --> 00:30:10.000
Suppose you got a complex
solution to that equation.

00:30:08.000 --> 00:30:14.000
These are understood to be real
numbers.

00:30:11.000 --> 00:30:17.000
They are the damping constant
and the spring constant.

00:30:15.000 --> 00:30:21.000
Then, the conclusion is that u
and v are real solutions.

00:30:20.000 --> 00:30:26.000
In other words,
having found a complex

00:30:23.000 --> 00:30:29.000
solution, all you have to do is
take its real and imaginary

00:30:28.000 --> 00:30:34.000
parts, and voila,
you've got your two solutions

00:30:31.000 --> 00:30:37.000
you were looking for for the
original equation.

00:30:37.000 --> 00:30:43.000
Now, that might seem like
magic, but it's easy.

00:30:39.000 --> 00:30:45.000
It's so easy it's the sort of
theorem I could spend one minute

00:30:43.000 --> 00:30:49.000
proving for you now.
What's the reason for it?

00:30:46.000 --> 00:30:52.000
Well, the main thing I want you
to get out of this argument is

00:30:50.000 --> 00:30:56.000
to see that it absolutely
depends upon these coefficients

00:30:54.000 --> 00:31:00.000
being real.
You have to have a real

00:30:56.000 --> 00:31:02.000
differential equation for this
to be true.

00:30:59.000 --> 00:31:05.000
Otherwise, it's certainly not.
So, the proof is,

00:31:03.000 --> 00:31:09.000
what does it mean to be a
solution?

00:31:06.000 --> 00:31:12.000
It means when you plug in A (u
plus iv) plus,

00:31:10.000 --> 00:31:16.000
prime,
plus B times u plus iv,

00:31:14.000 --> 00:31:20.000
what am I supposed to get?

00:31:17.000 --> 00:31:23.000
Zero.
Well, now, separate these into

00:31:20.000 --> 00:31:26.000
the real and imaginary parts.
What does it say?

00:31:24.000 --> 00:31:30.000
It says u double prime plus A u
prime plus B u,

00:31:29.000 --> 00:31:35.000
that's the real part of
this expression when I expand it

00:31:35.000 --> 00:31:41.000
out.
And, I've got an imaginary

00:31:39.000 --> 00:31:45.000
part, too, which all have the
coefficient i.

00:31:43.000 --> 00:31:49.000
So, from here,
I get v double prime plus i

00:31:46.000 --> 00:31:52.000
times A v prime plus
i times B v.

00:31:51.000 --> 00:31:57.000
So, this is the imaginary part.
Now, here I have something with

00:31:57.000 --> 00:32:03.000
a real part plus the imaginary
part, i, times the imaginary

00:32:02.000 --> 00:32:08.000
part is zero.
Well, the only way that can

00:32:05.000 --> 00:32:11.000
happen is if the real part is
zero, and the imaginary part is

00:32:11.000 --> 00:32:17.000
separately zero.
So, the conclusion is that

00:32:15.000 --> 00:32:21.000
therefore this part must be
zero, and therefore this part

00:32:19.000 --> 00:32:25.000
must be zero because the two of
them together make the complex

00:32:23.000 --> 00:32:29.000
number zero plus zero i.
Now, what does it mean for the

00:32:27.000 --> 00:32:33.000
real part to be zero?
It means that u is a solution.

00:32:31.000 --> 00:32:37.000
This, the imaginary part zero
means v is a solution,

00:32:34.000 --> 00:32:40.000
and therefore,
just what I said.

00:32:36.000 --> 00:32:42.000
u and v are solutions to the
real equation.

00:32:39.000 --> 00:32:45.000
Where did I use the fact that A
and B were real numbers and not

00:32:44.000 --> 00:32:50.000
complex numbers?
In knowing that this is the

00:32:47.000 --> 00:32:53.000
real part, I had to know that A
was a real number.

00:32:51.000 --> 00:32:57.000
If A were something like one
plus i,

00:32:54.000 --> 00:33:00.000
I'd be screwed,
I mean, because then I couldn't

00:32:57.000 --> 00:33:03.000
say that this was the real part
anymore.

00:33:02.000 --> 00:33:08.000
So, saying that's the real
part, and this is the imaginary

00:33:07.000 --> 00:33:13.000
part, I was using the fact that
these two numbers,

00:33:12.000 --> 00:33:18.000
constants, were real constants:
very important.

00:33:17.000 --> 00:33:23.000
So, what is the case two
solution?

00:33:20.000 --> 00:33:26.000
Well, what are the real and
imaginary parts

00:33:26.000 --> 00:33:32.000
of (a plus b i) t?
Well, y equals e to the at +

00:33:31.000 --> 00:33:37.000
ibt.
Okay, you've had experience.

00:33:37.000 --> 00:33:43.000
You know how to do this now.
That's e to the at

00:33:42.000 --> 00:33:48.000
times, well, the real part is,
well, let's write it this way.

00:33:47.000 --> 00:33:53.000
The real part is e to the at
times cosine b t.

00:33:51.000 --> 00:33:57.000
Notice how the a and b enter

00:33:54.000 --> 00:34:00.000
into the expression.
That's the real part.

00:33:58.000 --> 00:34:04.000
And, the imaginary part is e to
the at times the sine of bt.

00:34:03.000 --> 00:34:09.000
And therefore,

00:34:07.000 --> 00:34:13.000
the solution,
both of these must,

00:34:10.000 --> 00:34:16.000
therefore, be solutions to the
equation.

00:34:13.000 --> 00:34:19.000
And therefore,
the general solution to the ODE

00:34:18.000 --> 00:34:24.000
is y equals, now,
you've got to put in the

00:34:21.000 --> 00:34:27.000
arbitrary constants.
It's a nice thing to do to

00:34:26.000 --> 00:34:32.000
factor out the e to the at.

00:34:29.000 --> 00:34:35.000
It makes it look a little
better.

00:34:32.000 --> 00:34:38.000
And so, the constants are c1
cosine bt and c2 sine bt.

00:34:39.000 --> 00:34:45.000
Yeah, but what does that look
like?

00:34:41.000 --> 00:34:47.000
Well, you know that too.
This is an exponential,

00:34:45.000 --> 00:34:51.000
which controls the amplitude.
But this guy,

00:34:49.000 --> 00:34:55.000
which is a combination of two
sinusoidal oscillations with

00:34:53.000 --> 00:34:59.000
different amplitudes,
but with the same frequency,

00:34:57.000 --> 00:35:03.000
the b's are the same in both of
them, and therefore,

00:35:01.000 --> 00:35:07.000
this is, itself,
a purely sinusoidal

00:35:04.000 --> 00:35:10.000
oscillation.
So, in other words,

00:35:09.000 --> 00:35:15.000
I don't have room to write it,
but it's equal to,

00:35:15.000 --> 00:35:21.000
you know.
It's a good example of where

00:35:20.000 --> 00:35:26.000
you'd use that trigonometric
identity I spent time on before

00:35:27.000 --> 00:35:33.000
the exam.
Okay, let's work a quick

00:35:31.000 --> 00:35:37.000
example just to see how this
works out.

00:35:33.000 --> 00:35:39.000
Well, let's get rid of this.
Okay, let's now make the

00:35:36.000 --> 00:35:42.000
damping, since this is showing
oscillations,

00:35:39.000 --> 00:35:45.000
it must correspond to the case
where the damping is less strong

00:35:42.000 --> 00:35:48.000
compared with the spring
constant.

00:35:44.000 --> 00:35:50.000
So, the theorem is that if you
have a complex solution,

00:35:47.000 --> 00:35:53.000
u plus iv, so each of these is
a function of time,

00:35:50.000 --> 00:35:56.000
u plus iv is the
complex solution to a real

00:35:53.000 --> 00:35:59.000
differential equation with
constant coefficients.

00:35:56.000 --> 00:36:02.000
A stiff spring,
one that pulls with hard force

00:35:58.000 --> 00:36:04.000
is going to make that thing go
back and forth,

00:36:01.000 --> 00:36:07.000
particularly at the dipping is
weak.

00:36:05.000 --> 00:36:11.000
So, let's use almost the same
equation as I've just concealed.

00:36:09.000 --> 00:36:15.000
But, do you remember a used
four here?

00:36:12.000 --> 00:36:18.000
Okay, before we used three and
we got the solution to look like

00:36:17.000 --> 00:36:23.000
that.
Now, we will give it a little

00:36:19.000 --> 00:36:25.000
more energy by putting some
moxie in the springs.

00:36:23.000 --> 00:36:29.000
So now, the spring is pulling a
little harder,

00:36:26.000 --> 00:36:32.000
bigger force,
a stiffer spring.

00:36:30.000 --> 00:36:36.000
Okay, the characteristic
equation is now going to be r

00:36:35.000 --> 00:36:41.000
squared plus 4r plus five is
equal to zero.

00:36:40.000 --> 00:36:46.000
And therefore,
if I solve for r,

00:36:43.000 --> 00:36:49.000
I'm not going to bother trying
to factor this because I

00:36:49.000 --> 00:36:55.000
prepared for this lecture,
and I know, quadratic formula

00:36:54.000 --> 00:37:00.000
time, minus four plus or minus
the square root of b squared,

00:37:00.000 --> 00:37:06.000
16, minus four times five,
16 minus 20 is negative four

00:37:05.000 --> 00:37:11.000
all over two.
And therefore,

00:37:09.000 --> 00:37:15.000
that makes negative two plus or
minus, this makes,

00:37:13.000 --> 00:37:19.000
simply, i.
2i divided by two,

00:37:15.000 --> 00:37:21.000
which is i.
So, the exponential solution is

00:37:19.000 --> 00:37:25.000
e to the negative two,
you don't have to write this

00:37:23.000 --> 00:37:29.000
in.
You can get the thing directly.

00:37:26.000 --> 00:37:32.000
t, let's use the one with the
plus sign, and that's going to

00:37:31.000 --> 00:37:37.000
give, as the real solutions,
e to the negative two t times

00:37:36.000 --> 00:37:42.000
cosine t,
and e to the negative 2t times

00:37:41.000 --> 00:37:47.000
the sine of t.

00:37:46.000 --> 00:37:52.000
And therefore,
the solution is going to be y

00:37:51.000 --> 00:37:57.000
equals e to the negative 2t
times

00:37:58.000 --> 00:38:04.000
(c1 cosine t plus c2 sine t).

00:38:04.000 --> 00:38:10.000
If you want to put initial
conditions, you can put them in

00:38:08.000 --> 00:38:14.000
the same way I did them before.
Suppose we use the same initial

00:38:12.000 --> 00:38:18.000
conditions: y of zero
equals one,

00:38:15.000 --> 00:38:21.000
and y prime of zero equals one,
equals zero,

00:38:19.000 --> 00:38:25.000
let's say, wait,
blah, blah, blah,

00:38:22.000 --> 00:38:28.000
zero, yeah.
Okay, I'd like to take time to

00:38:24.000 --> 00:38:30.000
actually do the calculation,
but there's nothing new in it.

00:38:30.000 --> 00:38:36.000
I'd have to take,
calculate the derivative here,

00:38:35.000 --> 00:38:41.000
and then I would substitute in,
solve equations,

00:38:40.000 --> 00:38:46.000
and when you do all that,
just as before,

00:38:44.000 --> 00:38:50.000
the answer that you get is y
equals e to the negative 2t,

00:38:50.000 --> 00:38:56.000
so,
choose the constants c1 and c2

00:38:55.000 --> 00:39:01.000
by solving linear equations,
and the answer is cosine t,

00:39:01.000 --> 00:39:07.000
so, c1 turns out to be one,
and c2 turns out to be two,

00:39:07.000 --> 00:39:13.000
I hope.
Okay, I want to know,

00:39:11.000 --> 00:39:17.000
but what does that look like?
Well, use that trigonometric

00:39:15.000 --> 00:39:21.000
identity.
The e to the negative 2t

00:39:18.000 --> 00:39:24.000
is just a real
factor which is going to

00:39:21.000 --> 00:39:27.000
reproduce itself.
The question is,

00:39:24.000 --> 00:39:30.000
what is cosine t plus 2 sine t
look like?

00:39:28.000 --> 00:39:34.000
What's its amplitude as a pure
oscillation?

00:39:31.000 --> 00:39:37.000
It's the square root of one
plus two squared.

00:39:35.000 --> 00:39:41.000
Remember, it depends on looking

00:39:39.000 --> 00:39:45.000
at that little triangle,
which is one,

00:39:41.000 --> 00:39:47.000
two, this is a different scale
than that.

00:39:44.000 --> 00:39:50.000
And here is the square root of
five, right?

00:39:47.000 --> 00:39:53.000
And, here is phi,
the phase lag.

00:39:50.000 --> 00:39:56.000
So, it's equal to the square
root of five.

00:39:53.000 --> 00:39:59.000
So, it's the square root of
five times e to

00:39:57.000 --> 00:40:03.000
the negative 2t,
and the stuff inside is the

00:40:00.000 --> 00:40:06.000
cosine of, the frequency is one.
Circular frequency is one,

00:40:06.000 --> 00:40:12.000
so it's t minus phi,
where phi is this angle.

00:40:10.000 --> 00:40:16.000
How big is that,
one and two?

00:40:13.000 --> 00:40:19.000
Well, if this were the square
root of three,

00:40:16.000 --> 00:40:22.000
which is a little less than
two, it would be 60 infinity.

00:40:20.000 --> 00:40:26.000
So, this must be 70 infinity.
So, phi is 70 infinity plus or minus

00:40:24.000 --> 00:40:30.000
five, let's say.
So, it looks like a slightly

00:40:29.000 --> 00:40:35.000
delayed cosine curve,
but the amplitude is falling.

00:40:32.000 --> 00:40:38.000
So, it has to start.
So, if I draw it,

00:40:35.000 --> 00:40:41.000
here's one, here is,
let's say, the square root of

00:40:39.000 --> 00:40:45.000
five up about here.
Then, the square root of five

00:40:43.000 --> 00:40:49.000
times e to the negative 2t
looks maybe

00:40:47.000 --> 00:40:53.000
something like this.
So, that's square root of five

00:40:51.000 --> 00:40:57.000
e to the negative 2t.

00:40:54.000 --> 00:41:00.000
This is cosine t,
but shoved over by not quite pi

00:40:59.000 --> 00:41:05.000
over two.
It starts at one,

00:41:02.000 --> 00:41:08.000
and with the slope zero.
So, the solution starts like

00:41:06.000 --> 00:41:12.000
this.
It has to be guided in its

00:41:08.000 --> 00:41:14.000
amplitude by this function out
there, and in between it's the

00:41:13.000 --> 00:41:19.000
cosine curve.
But it's moved over.

00:41:15.000 --> 00:41:21.000
So, if this is pi over two,
the first time it crosses,

00:41:20.000 --> 00:41:26.000
it's 72 infinity to the right
of that. So, if this is pi

00:41:24.000 --> 00:41:30.000
over two, it's pi over two plus
70 infinity where it crosses.

00:41:27.000 --> 00:41:33.000
So, it must be doing something
like this.

00:41:32.000 --> 00:41:38.000
And now, on the other side,
it's got to stay within the

00:41:36.000 --> 00:41:42.000
same amplitude.
So, it must be doing something

00:41:40.000 --> 00:41:46.000
like this.
Okay, that gets us to,

00:41:43.000 --> 00:41:49.000
if this is the under-damped
case, because if you're trying

00:41:48.000 --> 00:41:54.000
to do this with a swinging door,
it means the door's going to be

00:41:54.000 --> 00:42:00.000
swinging back and forth.
Or, our little mass now hidden,

00:41:59.000 --> 00:42:05.000
but you could see it behind
that board, is going to be doing

00:42:04.000 --> 00:42:10.000
this.
But, it never stops.

00:42:07.000 --> 00:42:13.000
It never stops.
It doesn't realize,

00:42:11.000 --> 00:42:17.000
but not in theoretical life.
So, this is the under-damped.

00:42:16.000 --> 00:42:22.000
All right, so it's like
Goldilocks and the Three Bears.

00:42:21.000 --> 00:42:27.000
That's too hot,
and this is too cold.

00:42:25.000 --> 00:42:31.000
What is the thing which is just
right?

00:42:30.000 --> 00:42:36.000
Well, that's the thing you're
going to study on the problem

00:42:37.000 --> 00:42:43.000
set.
So, just right is called

00:42:40.000 --> 00:42:46.000
critically damped.
It's what people aim for in

00:42:46.000 --> 00:42:52.000
trying to damp motion that they
don't want.

00:42:51.000 --> 00:42:57.000
Now, what's critically damped?
It must be the case just in

00:42:58.000 --> 00:43:04.000
between these two.
Neither complex,

00:43:03.000 --> 00:43:09.000
nor the roots different.
It's the case of two equal

00:43:08.000 --> 00:43:14.000
roots.
So, r squared plus Ar plus B

00:43:11.000 --> 00:43:17.000
equals zero
has two equal roots.

00:43:17.000 --> 00:43:23.000
Now, that's a very special
equation.

00:43:20.000 --> 00:43:26.000
Suppose we call the root,
since all of these,

00:43:24.000 --> 00:43:30.000
notice these roots in this
physical case.

00:43:30.000 --> 00:43:36.000
The roots always turn out to be
negative numbers,

00:43:33.000 --> 00:43:39.000
or have a negative real part.
I'm going to call the root a.

00:43:37.000 --> 00:43:43.000
So, r equals negative a,
the root.

00:43:40.000 --> 00:43:46.000
a is understood to be a
positive number.

00:43:43.000 --> 00:43:49.000
I want that root to be really
negative.

00:43:45.000 --> 00:43:51.000
Then, the equation looks like,
the characteristic equation is

00:43:50.000 --> 00:43:56.000
going to be r plus a,
right, if the root is negative

00:43:53.000 --> 00:43:59.000
a, squared because it's a double
root.

00:43:57.000 --> 00:44:03.000
And, that means the equation is
of the form r squared plus two

00:44:01.000 --> 00:44:07.000
times a r plus a squared equals
zero.

00:44:07.000 --> 00:44:13.000
In other words,
the ODE looked like this.

00:44:10.000 --> 00:44:16.000
The ODE looked like y double
prime plus 2a y prime

00:44:15.000 --> 00:44:21.000
plus, in other words,

00:44:18.000 --> 00:44:24.000
the damping and the spring
constant were related in this

00:44:22.000 --> 00:44:28.000
special wake,
that for a given value of the

00:44:26.000 --> 00:44:32.000
spring constant,
there was exactly one value of

00:44:30.000 --> 00:44:36.000
the damping which produced this
in between case.

00:44:36.000 --> 00:44:42.000
Now, what's the problem
connected with it?

00:44:39.000 --> 00:44:45.000
Well, the problem,
unfortunately,

00:44:42.000 --> 00:44:48.000
is staring us in the face when
we want to solve it.

00:44:46.000 --> 00:44:52.000
The problem is that we have a
solution, but it is y equals e

00:44:51.000 --> 00:44:57.000
to the minus at.
I don't have another root to

00:44:56.000 --> 00:45:02.000
get another solution with.
And, the question is,

00:45:01.000 --> 00:45:07.000
where do I get that other
solution from?

00:45:04.000 --> 00:45:10.000
Now, there are three ways to
get it.

00:45:06.000 --> 00:45:12.000
Well, there are four ways to
get it.

00:45:09.000 --> 00:45:15.000
You look it up in Euler.
That's the fourth way.

00:45:12.000 --> 00:45:18.000
That's the real way to do it.
But, I've given you one way as

00:45:16.000 --> 00:45:22.000
problem number one on the
problem set.

00:45:19.000 --> 00:45:25.000
I've given you another way as
problem number two on the

00:45:23.000 --> 00:45:29.000
problem set.
And, the third way you will

00:45:26.000 --> 00:45:32.000
have to wait for about a week
and a half.

00:45:28.000 --> 00:45:34.000
And, I will give you a third
way, too.

00:45:33.000 --> 00:45:39.000
By that time,
you won't want to see any more

00:45:36.000 --> 00:45:42.000
ways.
But, I'd like to introduce you

00:45:39.000 --> 00:45:45.000
to the way on the problem set.
And, it is this,

00:45:43.000 --> 00:45:49.000
that if you know one solution
to an equation,

00:45:47.000 --> 00:45:53.000
which looks like a linear
equation, in fact,

00:45:51.000 --> 00:45:57.000
the piece can be functions of
t.

00:45:54.000 --> 00:46:00.000
They don't have to be constant,
so I'll use the books notation

00:45:59.000 --> 00:46:05.000
with p's and q's.
y prime plus q y equals zero.

00:46:05.000 --> 00:46:11.000
If you know one solution,
there's an absolute,

00:46:08.000 --> 00:46:14.000
ironclad guarantee,
if you'll know that it's true

00:46:13.000 --> 00:46:19.000
because I'm asking you to prove
it for yourself.

00:46:17.000 --> 00:46:23.000
There's another of the form,
having this as a factor,

00:46:21.000 --> 00:46:27.000
one solution y one,
let's call it,

00:46:24.000 --> 00:46:30.000
y equals y1 u is
another solution.

00:46:28.000 --> 00:46:34.000
And, you will be able to find
u, I swear.

00:46:33.000 --> 00:46:39.000
Now, let's, in the remaining
couple of minutes carry that out

00:46:37.000 --> 00:46:43.000
just for this case because I
want you to see how to arrange

00:46:41.000 --> 00:46:47.000
the work nicely.
And, I want you to arrange your

00:46:44.000 --> 00:46:50.000
work when you do the problem
sets in the same way.

00:46:48.000 --> 00:46:54.000
So, the way to do it is,
the solution we know is e to

00:46:52.000 --> 00:46:58.000
the minus at.
So, we are going to look for a

00:46:56.000 --> 00:47:02.000
solution to this differential
equation.

00:47:00.000 --> 00:47:06.000
That's the differential
equation.

00:47:02.000 --> 00:47:08.000
And, the solution we are going
to look for is of the form e to

00:47:08.000 --> 00:47:14.000
the negative at times u.

00:47:12.000 --> 00:47:18.000
Now, you're going to have to
make calculation like this

00:47:17.000 --> 00:47:23.000
several times in the course of
the term.

00:47:20.000 --> 00:47:26.000
Do it this way.
y prime equals,

00:47:23.000 --> 00:47:29.000
differentiate,
minus a e to the minus a t u

00:47:27.000 --> 00:47:33.000
plus e to the minus a
t u prime.

00:47:33.000 --> 00:47:39.000
And then, differentiate again.

00:47:37.000 --> 00:47:43.000
The answer will be a squared.

00:47:39.000 --> 00:47:45.000
You differentiate this:
a squared e to the negative at

00:47:43.000 --> 00:47:49.000
u.
I'll have to do this a little

00:47:47.000 --> 00:47:53.000
fast, but the next term will be,
okay, minus,

00:47:50.000 --> 00:47:56.000
so this times u prime,
and from this are you going to

00:47:53.000 --> 00:47:59.000
get another minus.
So, combining what you get from

00:47:57.000 --> 00:48:03.000
here, and here,
you're going to get minus 2a e

00:48:00.000 --> 00:48:06.000
to the minus a t u prime.

00:48:05.000 --> 00:48:11.000
And then, there is a final
term, which comes from this,

00:48:08.000 --> 00:48:14.000
e to the minus a t u double
prime.

00:48:11.000 --> 00:48:17.000
Two of these,
because of a piece here and a

00:48:15.000 --> 00:48:21.000
piece here combine to make that.
And now, to plug into the

00:48:19.000 --> 00:48:25.000
equation, you multiply this by
one.

00:48:21.000 --> 00:48:27.000
In other words,
you don't do anything to it.

00:48:24.000 --> 00:48:30.000
You multiply this line by 2a,
and you multiply that line by a

00:48:28.000 --> 00:48:34.000
squared, and you add them.

00:48:32.000 --> 00:48:38.000
On the left-hand side,
I get zero.

00:48:34.000 --> 00:48:40.000
What do I get on the right?
Notice how I've arrange the

00:48:39.000 --> 00:48:45.000
work so it adds nicely.
This has a squared times this,

00:48:43.000 --> 00:48:49.000
plus 2a times that,
plus one times that makes zero.

00:48:47.000 --> 00:48:53.000
2a times this plus one times
this makes zero.

00:48:50.000 --> 00:48:56.000
All that survives is e to the a
t u double prime,

00:48:56.000 --> 00:49:02.000
and therefore,
e to the minus a t u double

00:48:59.000 --> 00:49:05.000
prime is equal to zero.

00:49:04.000 --> 00:49:10.000
So, please tell me,
what's u double prime?

00:49:06.000 --> 00:49:12.000
It's zero.
So, please tell me,

00:49:09.000 --> 00:49:15.000
what's u?
It's c1 t plus c2.

00:49:11.000 --> 00:49:17.000
Now, that gives me a whole

00:49:14.000 --> 00:49:20.000
family of solutions.
Just t would be enough because

00:49:17.000 --> 00:49:23.000
all I am doing is looking for
one solution that's different

00:49:22.000 --> 00:49:28.000
from e to the minus a t.

00:49:24.000 --> 00:49:30.000
And, that solution,
therefore, is y equals e to the

00:49:28.000 --> 00:49:34.000
minus a t times t.

00:49:32.000 --> 00:49:38.000
And, there's my second
solution.

00:49:34.000 --> 00:49:40.000
So, this is a solution of the
critically damped case.

00:49:37.000 --> 00:49:43.000
And, you are going to use it in
three or four of the different

00:49:42.000 --> 00:49:48.000
problems on the problem set.
But, I think you can deal with

00:49:46.000 --> 00:49:52.000
virtually the whole problem set,
except for the last problem,

00:49:50.000 --> 00:49:56.000
now.