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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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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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00:13:50,000 --> 00:13:56,000
Now, the only way you can
fiddle with an exponential is in

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00:13:54,000 --> 00:14:00,000
the constant that you put up on
top.

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00:13:57,000 --> 00:14:03,000
So, I'm going to try y equals e
to the rt.

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00:14:03,000 --> 00:14:09,000
Notice you can't tell from that
what I'm using as the

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00:14:08,000 --> 00:14:14,000
independent variable.
But, this tells you I'm using

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00:14:13,000 --> 00:14:19,000
t.
And, I'm switching back to

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using t as the dependent
variable.

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00:14:19,000 --> 00:14:25,000
So, T is the independent
variable.

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00:14:22,000 --> 00:14:28,000
Why do I do that?
The answer is because somebody

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00:14:27,000 --> 00:14:33,000
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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00:14:36,000 --> 00:14:42,000
300 or 400 years.
Some things we just know.

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00:14:42,000 --> 00:14:48,000
All right, so if I do that,
as you learned from the exam,

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00:14:46,000 --> 00:14:52,000
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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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.

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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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00:15:33,000 --> 00:15:39,000
Capital A times r e to the rt,

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00:15:36,000 --> 00:15:42,000
and then there's the last term,
B times y itself,

233
00:15:39,000 --> 00:15:45,000
which is B e to the rt.

234
00:15:41,000 --> 00:15:47,000
And, that's supposed to be
equal to zero.

235
00:15:44,000 --> 00:15:50,000
So, I have to choose r so that
this becomes equal to zero.

236
00:15:48,000 --> 00:15:54,000
Now, you see,
the e to the rt

237
00:15:51,000 --> 00:15:57,000
occurs as a factor in every
term, and the e to the rt

238
00:15:54,000 --> 00:16:00,000
is never zero.
And therefore,

239
00:15:57,000 --> 00:16:03,000
you can divide it out because
it's always a positive number,

240
00:16:01,000 --> 00:16:07,000
regardless of the value of t.
So, I can cancel out from each

241
00:16:07,000 --> 00:16:13,000
term.
And, what I'm left with is the

242
00:16:10,000 --> 00:16:16,000
equation r squared plus ar plus
b equals zero.

243
00:16:16,000 --> 00:16:22,000
We are trying to find values of

244
00:16:19,000 --> 00:16:25,000
r that satisfy that equation.
And that, dear hearts,

245
00:16:24,000 --> 00:16:30,000
is why you learn to solve
quadratic equations in high

246
00:16:29,000 --> 00:16:35,000
school, in order that in this
moment, you would be now ready

247
00:16:34,000 --> 00:16:40,000
to find how spring-mass systems
behave when they are damped.

248
00:16:41,000 --> 00:16:47,000
This is called the
characteristic equation.

249
00:16:45,000 --> 00:16:51,000
The characteristic equation of
the ODE, or of the system of the

250
00:16:52,000 --> 00:16:58,000
spring mass system,
which it's modeling,

251
00:16:57,000 --> 00:17:03,000
the characteristic equation of
the system, okay?

252
00:17:02,000 --> 00:17:08,000
Okay, now, we solve it,
but now, from high school you

253
00:17:08,000 --> 00:17:14,000
know there are several cases.
And, each of those cases

254
00:17:14,000 --> 00:17:20,000
corresponds to a different
behavior.

255
00:17:17,000 --> 00:17:23,000
And, the cases depend upon what
the roots look like.

256
00:17:22,000 --> 00:17:28,000
The possibilities are the roots
could be real,

257
00:17:25,000 --> 00:17:31,000
and distinct.
That's the easiest case to

258
00:17:29,000 --> 00:17:35,000
handle.
The roots might be a pair of

259
00:17:31,000 --> 00:17:37,000
complex conjugate numbers.
That's harder to handle,

260
00:17:36,000 --> 00:17:42,000
but we are ready to do it.
And, the third case,

261
00:17:40,000 --> 00:17:46,000
which is the one most in your
problem set is the most

262
00:17:45,000 --> 00:17:51,000
puzzling: when the roots are
real, and equal.

263
00:17:48,000 --> 00:17:54,000
And, I'm going to talk about
those three cases in that order.

264
00:17:53,000 --> 00:17:59,000
So, the first case is the roots
are real and unequal.

265
00:17:57,000 --> 00:18:03,000
If I tell you they are unequal,
and I will put down real to

266
00:18:02,000 --> 00:18:08,000
make that clear.
Well, that is by far the

267
00:18:07,000 --> 00:18:13,000
simplest case because
immediately, one sees we have

268
00:18:11,000 --> 00:18:17,000
two roots.
They are different,

269
00:18:13,000 --> 00:18:19,000
and therefore,
we get our two solutions

270
00:18:17,000 --> 00:18:23,000
immediately.
So, the solutions are,

271
00:18:19,000 --> 00:18:25,000
the general solution to the
equation, I write down without

272
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.

273
00:18:34,000 --> 00:18:40,000
There's our solution.
Now, because that was so easy,

274
00:18:38,000 --> 00:18:44,000
and we didn't have to do any
work, I'd like to extend this

275
00:18:42,000 --> 00:18:48,000
case a little bit by using it as
an example of how you put in the

276
00:18:48,000 --> 00:18:54,000
initial conditions,
how to put in the c.

277
00:18:51,000 --> 00:18:57,000
So, let me work a specific
numerical example,

278
00:18:54,000 --> 00:19:00,000
since we are not going to try
to do this theoretically until

279
00:18:59,000 --> 00:19:05,000
next Wednesday.
Let's just do a numerical

280
00:19:04,000 --> 00:19:10,000
example.
So, suppose I take the

281
00:19:07,000 --> 00:19:13,000
constants to be the damping
constant to be a four,

282
00:19:11,000 --> 00:19:17,000
and the spring constant,
I'll take the mass to be one,

283
00:19:16,000 --> 00:19:22,000
and the spring constant to be
three.

284
00:19:19,000 --> 00:19:25,000
So, there's more damping here,
damping force here.

285
00:19:24,000 --> 00:19:30,000
You can't really talk that way
since the units are different.

286
00:19:31,000 --> 00:19:37,000
But, this number is bigger than
that one.

287
00:19:33,000 --> 00:19:39,000
That seems clear,
at any rate.

288
00:19:35,000 --> 00:19:41,000
Okay, now, what was the
characteristic equation?

289
00:19:39,000 --> 00:19:45,000
Look, now watch.
Please do what I do.

290
00:19:41,000 --> 00:19:47,000
I've found in the past,
even by the middle of the term,

291
00:19:45,000 --> 00:19:51,000
there are still students who
feel that they must substitute y

292
00:19:50,000 --> 00:19:56,000
equals e to the rt,
and go through that whole

293
00:19:53,000 --> 00:19:59,000
little derivation to find that
you don't do that.

294
00:19:57,000 --> 00:20:03,000
It's a waste of time.
I did it that you might not

295
00:20:02,000 --> 00:20:08,000
ever have to do it again.
Immediately write down the

296
00:20:06,000 --> 00:20:12,000
characteristic equation.
That's not very hard.

297
00:20:10,000 --> 00:20:16,000
r squared plus 4r plus three
equals zero.

298
00:20:14,000 --> 00:20:20,000
And, if you can write down its

299
00:20:17,000 --> 00:20:23,000
roots immediately,
splendid.

300
00:20:20,000 --> 00:20:26,000
But, let's not assume that
level of competence.

301
00:20:23,000 --> 00:20:29,000
So, it's r plus three times r
plus one equals zero.

302
00:20:29,000 --> 00:20:35,000
Okay, you factor it.

303
00:20:33,000 --> 00:20:39,000
This being 18.03,
a lot of the times the roots

304
00:20:36,000 --> 00:20:42,000
will be integers when they are
not, God forbid,

305
00:20:39,000 --> 00:20:45,000
you will have to use the
quadratic formula.

306
00:20:42,000 --> 00:20:48,000
But here, the roots were
integers.

307
00:20:44,000 --> 00:20:50,000
It is, after all,
only the first example.

308
00:20:47,000 --> 00:20:53,000
So, the solution,
the general solution is y

309
00:20:50,000 --> 00:20:56,000
equals c1 e to the negative,
notice the root is minus three

310
00:20:54,000 --> 00:21:00,000
and minus one,
minus 3t plus c2 e to the

311
00:20:56,000 --> 00:21:02,000
negative t.

312
00:21:01,000 --> 00:21:07,000
Now, suppose it's an initial
value problem.

313
00:21:04,000 --> 00:21:10,000
So, I gave you an initial
condition.

314
00:21:06,000 --> 00:21:12,000
Suppose the initial conditions
were that y of zero were one.

315
00:21:11,000 --> 00:21:17,000
So, at the start,

316
00:21:13,000 --> 00:21:19,000
the mass has been moved over to
the position,

317
00:21:16,000 --> 00:21:22,000
one, here.
Well, we expected it,

318
00:21:18,000 --> 00:21:24,000
then, to start doing that.
But, this is fairly heavily

319
00:21:22,000 --> 00:21:28,000
damped.
This is heavily damped.

320
00:21:25,000 --> 00:21:31,000
I'm going to assume that the
mass starts at rest.

321
00:21:30,000 --> 00:21:36,000
So, the spring is distended.
The masses over here.

322
00:21:33,000 --> 00:21:39,000
But, there's no motion at times
zero this way or that way.

323
00:21:37,000 --> 00:21:43,000
In other words,
I'm not pushing it.

324
00:21:40,000 --> 00:21:46,000
I'm just releasing it and
letting it do its thing after

325
00:21:44,000 --> 00:21:50,000
that.
Okay, so y prime of zero,

326
00:21:46,000 --> 00:21:52,000
I'll assume, is zero.

327
00:21:49,000 --> 00:21:55,000
So, it starts at rest,
but in the extended position,

328
00:21:53,000 --> 00:21:59,000
one unit to the right of the
equilibrium position.

329
00:21:56,000 --> 00:22:02,000
Now, all you have to do is use
these two conditions.

330
00:22:02,000 --> 00:22:08,000
Notice I have to have two
conditions because there are two

331
00:22:06,000 --> 00:22:12,000
constants I have to find the
value of.

332
00:22:09,000 --> 00:22:15,000
All right, so,
let's substitute,

333
00:22:11,000 --> 00:22:17,000
well, we're going to have to
calculate the derivative.

334
00:22:15,000 --> 00:22:21,000
So, why don't we do that right
away?

335
00:22:18,000 --> 00:22:24,000
So, this is minus three c1 e to
the minus 3t minus c2 e to the

336
00:22:22,000 --> 00:22:28,000
negative t.

337
00:22:27,000 --> 00:22:33,000
And now, if I substitute in at
zero, when t equals zero,

338
00:22:31,000 --> 00:22:37,000
what do I get?
Well, the first equation,

339
00:22:34,000 --> 00:22:40,000
the left says that y of zero
should be one.

340
00:22:38,000 --> 00:22:44,000
And, the right says this is
one.

341
00:22:41,000 --> 00:22:47,000
So, it's c1 plus c2.

342
00:22:43,000 --> 00:22:49,000
That's the result of
substituting t equals zero.

343
00:22:47,000 --> 00:22:53,000
How about substituting?

344
00:22:49,000 --> 00:22:55,000
What should I substitute in the
second equation?

345
00:22:53,000 --> 00:22:59,000
Well, y prime of zero is zero.

346
00:22:56,000 --> 00:23:02,000
So, if the second equation,
when I put in t equals zero,

347
00:23:01,000 --> 00:23:07,000
the left side is zero according
to that initial value,

348
00:23:05,000 --> 00:23:11,000
and the right side is negative
three c1 minus c2.

349
00:23:12,000 --> 00:23:18,000
You see what you end up with,
therefore, is a pair of

350
00:23:15,000 --> 00:23:21,000
simultaneous linear equations.
And, this is why you learn to

351
00:23:19,000 --> 00:23:25,000
study linear set of pairs of
simultaneous linear equations in

352
00:23:24,000 --> 00:23:30,000
high school.
These are among the most

353
00:23:26,000 --> 00:23:32,000
important.
Solving problems of this type

354
00:23:29,000 --> 00:23:35,000
are among the most important
applications of that kind of

355
00:23:33,000 --> 00:23:39,000
algebra, and this kind of
algebra.

356
00:23:37,000 --> 00:23:43,000
All right, what's the answer
finally?

357
00:23:40,000 --> 00:23:46,000
Well, if I add the two of them,
I get minus 2c1 equals one.

358
00:23:45,000 --> 00:23:51,000
So, c1 is equal to minus one
half.

359
00:23:49,000 --> 00:23:55,000
And, if c1 is minus a half,

360
00:23:54,000 --> 00:24:00,000
then c2 is minus 3c1.

361
00:23:57,000 --> 00:24:03,000
So, c2 is three halves.

362
00:24:10,000 --> 00:24:16,000
The final question is,
what does that look like as a

363
00:24:13,000 --> 00:24:19,000
solution?
Well, in general,

364
00:24:14,000 --> 00:24:20,000
these combinations of two
exponentials aren't very easy to

365
00:24:17,000 --> 00:24:23,000
plot by yourself.
That's one of the reasons you

366
00:24:20,000 --> 00:24:26,000
are being given this little
visual which plots them for you.

367
00:24:24,000 --> 00:24:30,000
All you have to do is,
as you'll see,

368
00:24:26,000 --> 00:24:32,000
set the damping constant,
set the constants,

369
00:24:28,000 --> 00:24:34,000
set the initial conditions,
and by magic,

370
00:24:31,000 --> 00:24:37,000
the curve appears on the
screen.

371
00:24:34,000 --> 00:24:40,000
And, if you change either of
the constants,

372
00:24:40,000 --> 00:24:46,000
the curve will change nicely
right along with it.

373
00:24:48,000 --> 00:24:54,000
So, the solution is y equals
minus one half e to the minus 3t

374
00:24:57,000 --> 00:25:03,000
plus three halves e to the
negative t.

375
00:25:06,000 --> 00:25:12,000
How does it look?

376
00:25:11,000 --> 00:25:17,000
Well, I don't expect you to be
able to plot that by yourself,

377
00:25:16,000 --> 00:25:22,000
but you can at least get
started.

378
00:25:19,000 --> 00:25:25,000
It does have to satisfy the
initial conditions.

379
00:25:23,000 --> 00:25:29,000
That means it should start at
one, and its starting slope is

380
00:25:28,000 --> 00:25:34,000
zero.
So, it starts like that.

381
00:25:32,000 --> 00:25:38,000
These are both declining
exponentials.

382
00:25:35,000 --> 00:25:41,000
This declines very rapidly,
this somewhat more slowly.

383
00:25:40,000 --> 00:25:46,000
It does something like that.
If this term were a lot,

384
00:25:44,000 --> 00:25:50,000
lot more negative,
I mean, that's the way that

385
00:25:49,000 --> 00:25:55,000
particular solution looks.
How might other solutions look?

386
00:25:54,000 --> 00:26:00,000
I'll draw a few other
possibilities.

387
00:25:57,000 --> 00:26:03,000
If the initial term,
if, for example,

388
00:26:00,000 --> 00:26:06,000
the initial slope were quite
negative, well,

389
00:26:04,000 --> 00:26:10,000
that would have start like
this.

390
00:26:09,000 --> 00:26:15,000
Now, just your experience of
physics, or of the real world

391
00:26:12,000 --> 00:26:18,000
suggests that if I give,
if I start the thing at one,

392
00:26:15,000 --> 00:26:21,000
but give it a strongly negative
push, it's going to go beyond

393
00:26:19,000 --> 00:26:25,000
the equilibrium position,
and then come back again.

394
00:26:22,000 --> 00:26:28,000
But, because the damping is
big, it's not going to be able

395
00:26:26,000 --> 00:26:32,000
to get through that.
The equilibrium position,

396
00:26:29,000 --> 00:26:35,000
a second time,
is going to look something like

397
00:26:31,000 --> 00:26:37,000
that.
Or, if I push it in that

398
00:26:34,000 --> 00:26:40,000
direction, the positive
direction, that it starts off

399
00:26:37,000 --> 00:26:43,000
with a positive slope.
But it loses its energy because

400
00:26:41,000 --> 00:26:47,000
the spring is pulling it.
It comes and does something

401
00:26:44,000 --> 00:26:50,000
like that.
So, in other words,

402
00:26:46,000 --> 00:26:52,000
it might go down.
Cut across the equilibrium

403
00:26:49,000 --> 00:26:55,000
position, come back again,
it do that?

404
00:26:52,000 --> 00:26:58,000
No, that it cannot do.
I was considering giving you a

405
00:26:55,000 --> 00:27:01,000
problem to prove that,
but I got tired of making out

406
00:26:58,000 --> 00:27:04,000
the problems set,
and decided I tortured you

407
00:27:01,000 --> 00:27:07,000
enough already,
as you will see.

408
00:27:05,000 --> 00:27:11,000
So, anyway, these are different
possibilities for the way that

409
00:27:11,000 --> 00:27:17,000
can look.
This case, where it just

410
00:27:14,000 --> 00:27:20,000
returns in the long run is
called the over-damped case,

411
00:27:20,000 --> 00:27:26,000
over-damped.
Now, there is another case

412
00:27:24,000 --> 00:27:30,000
where the thing oscillates back
and forth.

413
00:27:30,000 --> 00:27:36,000
We would expect to get that
case if the damping is very

414
00:27:33,000 --> 00:27:39,000
little or nonexistent.
Then, there's very little

415
00:27:37,000 --> 00:27:43,000
preventing the mass from doing
that, although we do expect if

416
00:27:41,000 --> 00:27:47,000
there's any damping at all,
we expect it ultimately to get

417
00:27:45,000 --> 00:27:51,000
nearer and nearer to the
equilibrium position.

418
00:27:48,000 --> 00:27:54,000
Mathematically,
what does that correspond to?

419
00:27:51,000 --> 00:27:57,000
Well, that's going to
correspond to case two,

420
00:27:55,000 --> 00:28:01,000
where the roots are complex.
The roots are complex,

421
00:27:59,000 --> 00:28:05,000
and this is why,
let's call the roots,

422
00:28:03,000 --> 00:28:09,000
in that case we know that the
roots are of the form a plus or

423
00:28:08,000 --> 00:28:14,000
minus bi.
There are two roots,

424
00:28:10,000 --> 00:28:16,000
and they are a complex
conjugate.

425
00:28:13,000 --> 00:28:19,000
All right, let's take one of
them.

426
00:28:16,000 --> 00:28:22,000
What does a correspond to in
terms of the exponential?

427
00:28:20,000 --> 00:28:26,000
Well, remember,
the function of the r was,

428
00:28:24,000 --> 00:28:30,000
it's this r when we tried our
exponential solution.

429
00:28:30,000 --> 00:28:36,000
So, what we formally,
this means we get a complex

430
00:28:34,000 --> 00:28:40,000
solution.
The complex solution y equals e

431
00:28:38,000 --> 00:28:44,000
to this, let's use one of them,
let's say, (a plus bi) times t.

432
00:28:43,000 --> 00:28:49,000
The question is,

433
00:28:47,000 --> 00:28:53,000
what do we do that?
We are not really interested,

434
00:28:51,000 --> 00:28:57,000
I don't know what a complex
solution to that thing means.

435
00:28:56,000 --> 00:29:02,000
It doesn't have any meaning.
What I want to know is how y

436
00:29:01,000 --> 00:29:07,000
behaves or how x behaves in that
picture.

437
00:29:07,000 --> 00:29:13,000
And, that better be a real
function because otherwise I

438
00:29:10,000 --> 00:29:16,000
don't know what to do with it.
So, we are looking for two real

439
00:29:14,000 --> 00:29:20,000
functions, the y1 and the y2.
But, in fact,

440
00:29:17,000 --> 00:29:23,000
what we've got is one complex
function.

441
00:29:20,000 --> 00:29:26,000
All right, now,
a theorem to the rescue:

442
00:29:22,000 --> 00:29:28,000
this, I'm not going to save for
Wednesday because it's so

443
00:29:26,000 --> 00:29:32,000
simple.
So, the theorem is that if you

444
00:29:30,000 --> 00:29:36,000
have a complex solution,
u plus iv, so each of these is

445
00:29:36,000 --> 00:29:42,000
a function of time,
u plus iv is the complex

446
00:29:40,000 --> 00:29:46,000
solution to a real differential
equation with constant

447
00:29:45,000 --> 00:29:51,000
coefficients.
Well, it doesn't have to have

448
00:29:49,000 --> 00:29:55,000
constant coefficients.
It has to be linear.

449
00:29:53,000 --> 00:29:59,000
Let me just write it out to y
double prime plus A y prime plus

450
00:29:59,000 --> 00:30:05,000
B y equals zero.

451
00:30:04,000 --> 00:30:10,000
Suppose you got a complex
solution to that equation.

452
00:30:08,000 --> 00:30:14,000
These are understood to be real
numbers.

453
00:30:11,000 --> 00:30:17,000
They are the damping constant
and the spring constant.

454
00:30:15,000 --> 00:30:21,000
Then, the conclusion is that u
and v are real solutions.

455
00:30:20,000 --> 00:30:26,000
In other words,
having found a complex

456
00:30:23,000 --> 00:30:29,000
solution, all you have to do is
take its real and imaginary

457
00:30:28,000 --> 00:30:34,000
parts, and voila,
you've got your two solutions

458
00:30:31,000 --> 00:30:37,000
you were looking for for the
original equation.

459
00:30:37,000 --> 00:30:43,000
Now, that might seem like
magic, but it's easy.

460
00:30:39,000 --> 00:30:45,000
It's so easy it's the sort of
theorem I could spend one minute

461
00:30:43,000 --> 00:30:49,000
proving for you now.
What's the reason for it?

462
00:30:46,000 --> 00:30:52,000
Well, the main thing I want you
to get out of this argument is

463
00:30:50,000 --> 00:30:56,000
to see that it absolutely
depends upon these coefficients

464
00:30:54,000 --> 00:31:00,000
being real.
You have to have a real

465
00:30:56,000 --> 00:31:02,000
differential equation for this
to be true.

466
00:30:59,000 --> 00:31:05,000
Otherwise, it's certainly not.
So, the proof is,

467
00:31:03,000 --> 00:31:09,000
what does it mean to be a
solution?

468
00:31:06,000 --> 00:31:12,000
It means when you plug in A (u
plus iv) plus,

469
00:31:10,000 --> 00:31:16,000
prime,
plus B times u plus iv,

470
00:31:14,000 --> 00:31:20,000
what am I supposed to get?

471
00:31:17,000 --> 00:31:23,000
Zero.
Well, now, separate these into

472
00:31:20,000 --> 00:31:26,000
the real and imaginary parts.
What does it say?

473
00:31:24,000 --> 00:31:30,000
It says u double prime plus A u
prime plus B u,

474
00:31:29,000 --> 00:31:35,000
that's the real part of
this expression when I expand it

475
00:31:35,000 --> 00:31:41,000
out.
And, I've got an imaginary

476
00:31:39,000 --> 00:31:45,000
part, too, which all have the
coefficient i.

477
00:31:43,000 --> 00:31:49,000
So, from here,
I get v double prime plus i

478
00:31:46,000 --> 00:31:52,000
times A v prime plus
i times B v.

479
00:31:51,000 --> 00:31:57,000
So, this is the imaginary part.
Now, here I have something with

480
00:31:57,000 --> 00:32:03,000
a real part plus the imaginary
part, i, times the imaginary

481
00:32:02,000 --> 00:32:08,000
part is zero.
Well, the only way that can

482
00:32:05,000 --> 00:32:11,000
happen is if the real part is
zero, and the imaginary part is

483
00:32:11,000 --> 00:32:17,000
separately zero.
So, the conclusion is that

484
00:32:15,000 --> 00:32:21,000
therefore this part must be
zero, and therefore this part

485
00:32:19,000 --> 00:32:25,000
must be zero because the two of
them together make the complex

486
00:32:23,000 --> 00:32:29,000
number zero plus zero i.
Now, what does it mean for the

487
00:32:27,000 --> 00:32:33,000
real part to be zero?
It means that u is a solution.

488
00:32:31,000 --> 00:32:37,000
This, the imaginary part zero
means v is a solution,

489
00:32:34,000 --> 00:32:40,000
and therefore,
just what I said.

490
00:32:36,000 --> 00:32:42,000
u and v are solutions to the
real equation.

491
00:32:39,000 --> 00:32:45,000
Where did I use the fact that A
and B were real numbers and not

492
00:32:44,000 --> 00:32:50,000
complex numbers?
In knowing that this is the

493
00:32:47,000 --> 00:32:53,000
real part, I had to know that A
was a real number.

494
00:32:51,000 --> 00:32:57,000
If A were something like one
plus i,

495
00:32:54,000 --> 00:33:00,000
I'd be screwed,
I mean, because then I couldn't

496
00:32:57,000 --> 00:33:03,000
say that this was the real part
anymore.

497
00:33:02,000 --> 00:33:08,000
So, saying that's the real
part, and this is the imaginary

498
00:33:07,000 --> 00:33:13,000
part, I was using the fact that
these two numbers,

499
00:33:12,000 --> 00:33:18,000
constants, were real constants:
very important.

500
00:33:17,000 --> 00:33:23,000
So, what is the case two
solution?

501
00:33:20,000 --> 00:33:26,000
Well, what are the real and
imaginary parts

502
00:33:26,000 --> 00:33:32,000
of (a plus b i) t?
Well, y equals e to the at +

503
00:33:31,000 --> 00:33:37,000
ibt.
Okay, you've had experience.

504
00:33:37,000 --> 00:33:43,000
You know how to do this now.
That's e to the at

505
00:33:42,000 --> 00:33:48,000
times, well, the real part is,
well, let's write it this way.

506
00:33:47,000 --> 00:33:53,000
The real part is e to the at
times cosine b t.

507
00:33:51,000 --> 00:33:57,000
Notice how the a and b enter

508
00:33:54,000 --> 00:34:00,000
into the expression.
That's the real part.

509
00:33:58,000 --> 00:34:04,000
And, the imaginary part is e to
the at times the sine of bt.

510
00:34:03,000 --> 00:34:09,000
And therefore,

511
00:34:07,000 --> 00:34:13,000
the solution,
both of these must,

512
00:34:10,000 --> 00:34:16,000
therefore, be solutions to the
equation.

513
00:34:13,000 --> 00:34:19,000
And therefore,
the general solution to the ODE

514
00:34:18,000 --> 00:34:24,000
is y equals, now,
you've got to put in the

515
00:34:21,000 --> 00:34:27,000
arbitrary constants.
It's a nice thing to do to

516
00:34:26,000 --> 00:34:32,000
factor out the e to the at.

517
00:34:29,000 --> 00:34:35,000
It makes it look a little
better.

518
00:34:32,000 --> 00:34:38,000
And so, the constants are c1
cosine bt and c2 sine bt.

519
00:34:39,000 --> 00:34:45,000
Yeah, but what does that look
like?

520
00:34:41,000 --> 00:34:47,000
Well, you know that too.
This is an exponential,

521
00:34:45,000 --> 00:34:51,000
which controls the amplitude.
But this guy,

522
00:34:49,000 --> 00:34:55,000
which is a combination of two
sinusoidal oscillations with

523
00:34:53,000 --> 00:34:59,000
different amplitudes,
but with the same frequency,

524
00:34:57,000 --> 00:35:03,000
the b's are the same in both of
them, and therefore,

525
00:35:01,000 --> 00:35:07,000
this is, itself,
a purely sinusoidal

526
00:35:04,000 --> 00:35:10,000
oscillation.
So, in other words,

527
00:35:09,000 --> 00:35:15,000
I don't have room to write it,
but it's equal to,

528
00:35:15,000 --> 00:35:21,000
you know.
It's a good example of where

529
00:35:20,000 --> 00:35:26,000
you'd use that trigonometric
identity I spent time on before

530
00:35:27,000 --> 00:35:33,000
the exam.
Okay, let's work a quick

531
00:35:31,000 --> 00:35:37,000
example just to see how this
works out.

532
00:35:33,000 --> 00:35:39,000
Well, let's get rid of this.
Okay, let's now make the

533
00:35:36,000 --> 00:35:42,000
damping, since this is showing
oscillations,

534
00:35:39,000 --> 00:35:45,000
it must correspond to the case
where the damping is less strong

535
00:35:42,000 --> 00:35:48,000
compared with the spring
constant.

536
00:35:44,000 --> 00:35:50,000
So, the theorem is that if you
have a complex solution,

537
00:35:47,000 --> 00:35:53,000
u plus iv, so each of these is
a function of time,

538
00:35:50,000 --> 00:35:56,000
u plus iv is the
complex solution to a real

539
00:35:53,000 --> 00:35:59,000
differential equation with
constant coefficients.

540
00:35:56,000 --> 00:36:02,000
A stiff spring,
one that pulls with hard force

541
00:35:58,000 --> 00:36:04,000
is going to make that thing go
back and forth,

542
00:36:01,000 --> 00:36:07,000
particularly at the dipping is
weak.

543
00:36:05,000 --> 00:36:11,000
So, let's use almost the same
equation as I've just concealed.

544
00:36:09,000 --> 00:36:15,000
But, do you remember a used
four here?

545
00:36:12,000 --> 00:36:18,000
Okay, before we used three and
we got the solution to look like

546
00:36:17,000 --> 00:36:23,000
that.
Now, we will give it a little

547
00:36:19,000 --> 00:36:25,000
more energy by putting some
moxie in the springs.

548
00:36:23,000 --> 00:36:29,000
So now, the spring is pulling a
little harder,

549
00:36:26,000 --> 00:36:32,000
bigger force,
a stiffer spring.

550
00:36:30,000 --> 00:36:36,000
Okay, the characteristic
equation is now going to be r

551
00:36:35,000 --> 00:36:41,000
squared plus 4r plus five is
equal to zero.

552
00:36:40,000 --> 00:36:46,000
And therefore,
if I solve for r,

553
00:36:43,000 --> 00:36:49,000
I'm not going to bother trying
to factor this because I

554
00:36:49,000 --> 00:36:55,000
prepared for this lecture,
and I know, quadratic formula

555
00:36:54,000 --> 00:37:00,000
time, minus four plus or minus
the square root of b squared,

556
00:37:00,000 --> 00:37:06,000
16, minus four times five,
16 minus 20 is negative four

557
00:37:05,000 --> 00:37:11,000
all over two.
And therefore,

558
00:37:09,000 --> 00:37:15,000
that makes negative two plus or
minus, this makes,

559
00:37:13,000 --> 00:37:19,000
simply, i.
2i divided by two,

560
00:37:15,000 --> 00:37:21,000
which is i.
So, the exponential solution is

561
00:37:19,000 --> 00:37:25,000
e to the negative two,
you don't have to write this

562
00:37:23,000 --> 00:37:29,000
in.
You can get the thing directly.

563
00:37:26,000 --> 00:37:32,000
t, let's use the one with the
plus sign, and that's going to

564
00:37:31,000 --> 00:37:37,000
give, as the real solutions,
e to the negative two t times

565
00:37:36,000 --> 00:37:42,000
cosine t,
and e to the negative 2t times

566
00:37:41,000 --> 00:37:47,000
the sine of t.

567
00:37:46,000 --> 00:37:52,000
And therefore,
the solution is going to be y

568
00:37:51,000 --> 00:37:57,000
equals e to the negative 2t
times

569
00:37:58,000 --> 00:38:04,000
(c1 cosine t plus c2 sine t).

570
00:38:04,000 --> 00:38:10,000
If you want to put initial
conditions, you can put them in

571
00:38:08,000 --> 00:38:14,000
the same way I did them before.
Suppose we use the same initial

572
00:38:12,000 --> 00:38:18,000
conditions: y of zero
equals one,

573
00:38:15,000 --> 00:38:21,000
and y prime of zero equals one,
equals zero,

574
00:38:19,000 --> 00:38:25,000
let's say, wait,
blah, blah, blah,

575
00:38:22,000 --> 00:38:28,000
zero, yeah.
Okay, I'd like to take time to

576
00:38:24,000 --> 00:38:30,000
actually do the calculation,
but there's nothing new in it.

577
00:38:30,000 --> 00:38:36,000
I'd have to take,
calculate the derivative here,

578
00:38:35,000 --> 00:38:41,000
and then I would substitute in,
solve equations,

579
00:38:40,000 --> 00:38:46,000
and when you do all that,
just as before,

580
00:38:44,000 --> 00:38:50,000
the answer that you get is y
equals e to the negative 2t,

581
00:38:50,000 --> 00:38:56,000
so,
choose the constants c1 and c2

582
00:38:55,000 --> 00:39:01,000
by solving linear equations,
and the answer is cosine t,

583
00:39:01,000 --> 00:39:07,000
so, c1 turns out to be one,
and c2 turns out to be two,

584
00:39:07,000 --> 00:39:13,000
I hope.
Okay, I want to know,

585
00:39:11,000 --> 00:39:17,000
but what does that look like?
Well, use that trigonometric

586
00:39:15,000 --> 00:39:21,000
identity.
The e to the negative 2t

587
00:39:18,000 --> 00:39:24,000
is just a real
factor which is going to

588
00:39:21,000 --> 00:39:27,000
reproduce itself.
The question is,

589
00:39:24,000 --> 00:39:30,000
what is cosine t plus 2 sine t
look like?

590
00:39:28,000 --> 00:39:34,000
What's its amplitude as a pure
oscillation?

591
00:39:31,000 --> 00:39:37,000
It's the square root of one
plus two squared.

592
00:39:35,000 --> 00:39:41,000
Remember, it depends on looking

593
00:39:39,000 --> 00:39:45,000
at that little triangle,
which is one,

594
00:39:41,000 --> 00:39:47,000
two, this is a different scale
than that.

595
00:39:44,000 --> 00:39:50,000
And here is the square root of
five, right?

596
00:39:47,000 --> 00:39:53,000
And, here is phi,
the phase lag.

597
00:39:50,000 --> 00:39:56,000
So, it's equal to the square
root of five.

598
00:39:53,000 --> 00:39:59,000
So, it's the square root of
five times e to

599
00:39:57,000 --> 00:40:03,000
the negative 2t,
and the stuff inside is the

600
00:40:00,000 --> 00:40:06,000
cosine of, the frequency is one.
Circular frequency is one,

601
00:40:06,000 --> 00:40:12,000
so it's t minus phi,
where phi is this angle.

602
00:40:10,000 --> 00:40:16,000
How big is that,
one and two?

603
00:40:13,000 --> 00:40:19,000
Well, if this were the square
root of three,

604
00:40:16,000 --> 00:40:22,000
which is a little less than
two, it would be 60 infinity.

605
00:40:20,000 --> 00:40:26,000
So, this must be 70 infinity.
So, phi is 70 infinity plus or minus

606
00:40:24,000 --> 00:40:30,000
five, let's say.
So, it looks like a slightly

607
00:40:29,000 --> 00:40:35,000
delayed cosine curve,
but the amplitude is falling.

608
00:40:32,000 --> 00:40:38,000
So, it has to start.
So, if I draw it,

609
00:40:35,000 --> 00:40:41,000
here's one, here is,
let's say, the square root of

610
00:40:39,000 --> 00:40:45,000
five up about here.
Then, the square root of five

611
00:40:43,000 --> 00:40:49,000
times e to the negative 2t
looks maybe

612
00:40:47,000 --> 00:40:53,000
something like this.
So, that's square root of five

613
00:40:51,000 --> 00:40:57,000
e to the negative 2t.

614
00:40:54,000 --> 00:41:00,000
This is cosine t,
but shoved over by not quite pi

615
00:40:59,000 --> 00:41:05,000
over two.
It starts at one,

616
00:41:02,000 --> 00:41:08,000
and with the slope zero.
So, the solution starts like

617
00:41:06,000 --> 00:41:12,000
this.
It has to be guided in its

618
00:41:08,000 --> 00:41:14,000
amplitude by this function out
there, and in between it's the

619
00:41:13,000 --> 00:41:19,000
cosine curve.
But it's moved over.

620
00:41:15,000 --> 00:41:21,000
So, if this is pi over two,
the first time it crosses,

621
00:41:20,000 --> 00:41:26,000
it's 72 infinity to the right
of that. So, if this is pi

622
00:41:24,000 --> 00:41:30,000
over two, it's pi over two plus
70 infinity where it crosses.

623
00:41:27,000 --> 00:41:33,000
So, it must be doing something
like this.

624
00:41:32,000 --> 00:41:38,000
And now, on the other side,
it's got to stay within the

625
00:41:36,000 --> 00:41:42,000
same amplitude.
So, it must be doing something

626
00:41:40,000 --> 00:41:46,000
like this.
Okay, that gets us to,

627
00:41:43,000 --> 00:41:49,000
if this is the under-damped
case, because if you're trying

628
00:41:48,000 --> 00:41:54,000
to do this with a swinging door,
it means the door's going to be

629
00:41:54,000 --> 00:42:00,000
swinging back and forth.
Or, our little mass now hidden,

630
00:41:59,000 --> 00:42:05,000
but you could see it behind
that board, is going to be doing

631
00:42:04,000 --> 00:42:10,000
this.
But, it never stops.

632
00:42:07,000 --> 00:42:13,000
It never stops.
It doesn't realize,

633
00:42:11,000 --> 00:42:17,000
but not in theoretical life.
So, this is the under-damped.

634
00:42:16,000 --> 00:42:22,000
All right, so it's like
Goldilocks and the Three Bears.

635
00:42:21,000 --> 00:42:27,000
That's too hot,
and this is too cold.

636
00:42:25,000 --> 00:42:31,000
What is the thing which is just
right?

637
00:42:30,000 --> 00:42:36,000
Well, that's the thing you're
going to study on the problem

638
00:42:37,000 --> 00:42:43,000
set.
So, just right is called

639
00:42:40,000 --> 00:42:46,000
critically damped.
It's what people aim for in

640
00:42:46,000 --> 00:42:52,000
trying to damp motion that they
don't want.

641
00:42:51,000 --> 00:42:57,000
Now, what's critically damped?
It must be the case just in

642
00:42:58,000 --> 00:43:04,000
between these two.
Neither complex,

643
00:43:03,000 --> 00:43:09,000
nor the roots different.
It's the case of two equal

644
00:43:08,000 --> 00:43:14,000
roots.
So, r squared plus Ar plus B

645
00:43:11,000 --> 00:43:17,000
equals zero
has two equal roots.

646
00:43:17,000 --> 00:43:23,000
Now, that's a very special
equation.

647
00:43:20,000 --> 00:43:26,000
Suppose we call the root,
since all of these,

648
00:43:24,000 --> 00:43:30,000
notice these roots in this
physical case.

649
00:43:30,000 --> 00:43:36,000
The roots always turn out to be
negative numbers,

650
00:43:33,000 --> 00:43:39,000
or have a negative real part.
I'm going to call the root a.

651
00:43:37,000 --> 00:43:43,000
So, r equals negative a,
the root.

652
00:43:40,000 --> 00:43:46,000
a is understood to be a
positive number.

653
00:43:43,000 --> 00:43:49,000
I want that root to be really
negative.

654
00:43:45,000 --> 00:43:51,000
Then, the equation looks like,
the characteristic equation is

655
00:43:50,000 --> 00:43:56,000
going to be r plus a,
right, if the root is negative

656
00:43:53,000 --> 00:43:59,000
a, squared because it's a double
root.

657
00:43:57,000 --> 00:44:03,000
And, that means the equation is
of the form r squared plus two

658
00:44:01,000 --> 00:44:07,000
times a r plus a squared equals
zero.

659
00:44:07,000 --> 00:44:13,000
In other words,
the ODE looked like this.

660
00:44:10,000 --> 00:44:16,000
The ODE looked like y double
prime plus 2a y prime

661
00:44:15,000 --> 00:44:21,000
plus, in other words,

662
00:44:18,000 --> 00:44:24,000
the damping and the spring
constant were related in this

663
00:44:22,000 --> 00:44:28,000
special wake,
that for a given value of the

664
00:44:26,000 --> 00:44:32,000
spring constant,
there was exactly one value of

665
00:44:30,000 --> 00:44:36,000
the damping which produced this
in between case.

666
00:44:36,000 --> 00:44:42,000
Now, what's the problem
connected with it?

667
00:44:39,000 --> 00:44:45,000
Well, the problem,
unfortunately,

668
00:44:42,000 --> 00:44:48,000
is staring us in the face when
we want to solve it.

669
00:44:46,000 --> 00:44:52,000
The problem is that we have a
solution, but it is y equals e

670
00:44:51,000 --> 00:44:57,000
to the minus at.
I don't have another root to

671
00:44:56,000 --> 00:45:02,000
get another solution with.
And, the question is,

672
00:45:01,000 --> 00:45:07,000
where do I get that other
solution from?

673
00:45:04,000 --> 00:45:10,000
Now, there are three ways to
get it.

674
00:45:06,000 --> 00:45:12,000
Well, there are four ways to
get it.

675
00:45:09,000 --> 00:45:15,000
You look it up in Euler.
That's the fourth way.

676
00:45:12,000 --> 00:45:18,000
That's the real way to do it.
But, I've given you one way as

677
00:45:16,000 --> 00:45:22,000
problem number one on the
problem set.

678
00:45:19,000 --> 00:45:25,000
I've given you another way as
problem number two on the

679
00:45:23,000 --> 00:45:29,000
problem set.
And, the third way you will

680
00:45:26,000 --> 00:45:32,000
have to wait for about a week
and a half.

681
00:45:28,000 --> 00:45:34,000
And, I will give you a third
way, too.

682
00:45:33,000 --> 00:45:39,000
By that time,
you won't want to see any more

683
00:45:36,000 --> 00:45:42,000
ways.
But, I'd like to introduce you

684
00:45:39,000 --> 00:45:45,000
to the way on the problem set.
And, it is this,

685
00:45:43,000 --> 00:45:49,000
that if you know one solution
to an equation,

686
00:45:47,000 --> 00:45:53,000
which looks like a linear
equation, in fact,

687
00:45:51,000 --> 00:45:57,000
the piece can be functions of
t.

688
00:45:54,000 --> 00:46:00,000
They don't have to be constant,
so I'll use the books notation

689
00:45:59,000 --> 00:46:05,000
with p's and q's.
y prime plus q y equals zero.

690
00:46:05,000 --> 00:46:11,000
If you know one solution,
there's an absolute,

691
00:46:08,000 --> 00:46:14,000
ironclad guarantee,
if you'll know that it's true

692
00:46:13,000 --> 00:46:19,000
because I'm asking you to prove
it for yourself.

693
00:46:17,000 --> 00:46:23,000
There's another of the form,
having this as a factor,

694
00:46:21,000 --> 00:46:27,000
one solution y one,
let's call it,

695
00:46:24,000 --> 00:46:30,000
y equals y1 u is
another solution.

696
00:46:28,000 --> 00:46:34,000
And, you will be able to find
u, I swear.

697
00:46:33,000 --> 00:46:39,000
Now, let's, in the remaining
couple of minutes carry that out

698
00:46:37,000 --> 00:46:43,000
just for this case because I
want you to see how to arrange

699
00:46:41,000 --> 00:46:47,000
the work nicely.
And, I want you to arrange your

700
00:46:44,000 --> 00:46:50,000
work when you do the problem
sets in the same way.

701
00:46:48,000 --> 00:46:54,000
So, the way to do it is,
the solution we know is e to

702
00:46:52,000 --> 00:46:58,000
the minus at.
So, we are going to look for a

703
00:46:56,000 --> 00:47:02,000
solution to this differential
equation.

704
00:47:00,000 --> 00:47:06,000
That's the differential
equation.

705
00:47:02,000 --> 00:47:08,000
And, the solution we are going
to look for is of the form e to

706
00:47:08,000 --> 00:47:14,000
the negative at times u.

707
00:47:12,000 --> 00:47:18,000
Now, you're going to have to
make calculation like this

708
00:47:17,000 --> 00:47:23,000
several times in the course of
the term.

709
00:47:20,000 --> 00:47:26,000
Do it this way.
y prime equals,

710
00:47:23,000 --> 00:47:29,000
differentiate,
minus a e to the minus a t u

711
00:47:27,000 --> 00:47:33,000
plus e to the minus a
t u prime.

712
00:47:33,000 --> 00:47:39,000
And then, differentiate again.

713
00:47:37,000 --> 00:47:43,000
The answer will be a squared.

714
00:47:39,000 --> 00:47:45,000
You differentiate this:
a squared e to the negative at

715
00:47:43,000 --> 00:47:49,000
u.
I'll have to do this a little

716
00:47:47,000 --> 00:47:53,000
fast, but the next term will be,
okay, minus,

717
00:47:50,000 --> 00:47:56,000
so this times u prime,
and from this are you going to

718
00:47:53,000 --> 00:47:59,000
get another minus.
So, combining what you get from

719
00:47:57,000 --> 00:48:03,000
here, and here,
you're going to get minus 2a e

720
00:48:00,000 --> 00:48:06,000
to the minus a t u prime.

721
00:48:05,000 --> 00:48:11,000
And then, there is a final
term, which comes from this,

722
00:48:08,000 --> 00:48:14,000
e to the minus a t u double
prime.

723
00:48:11,000 --> 00:48:17,000
Two of these,
because of a piece here and a

724
00:48:15,000 --> 00:48:21,000
piece here combine to make that.
And now, to plug into the

725
00:48:19,000 --> 00:48:25,000
equation, you multiply this by
one.

726
00:48:21,000 --> 00:48:27,000
In other words,
you don't do anything to it.

727
00:48:24,000 --> 00:48:30,000
You multiply this line by 2a,
and you multiply that line by a

728
00:48:28,000 --> 00:48:34,000
squared, and you add them.

729
00:48:32,000 --> 00:48:38,000
On the left-hand side,
I get zero.

730
00:48:34,000 --> 00:48:40,000
What do I get on the right?
Notice how I've arrange the

731
00:48:39,000 --> 00:48:45,000
work so it adds nicely.
This has a squared times this,

732
00:48:43,000 --> 00:48:49,000
plus 2a times that,
plus one times that makes zero.

733
00:48:47,000 --> 00:48:53,000
2a times this plus one times
this makes zero.

734
00:48:50,000 --> 00:48:56,000
All that survives is e to the a
t u double prime,

735
00:48:56,000 --> 00:49:02,000
and therefore,
e to the minus a t u double

736
00:48:59,000 --> 00:49:05,000
prime is equal to zero.

737
00:49:04,000 --> 00:49:10,000
So, please tell me,
what's u double prime?

738
00:49:06,000 --> 00:49:12,000
It's zero.
So, please tell me,

739
00:49:09,000 --> 00:49:15,000
what's u?
It's c1 t plus c2.

740
00:49:11,000 --> 00:49:17,000
Now, that gives me a whole

741
00:49:14,000 --> 00:49:20,000
family of solutions.
Just t would be enough because

742
00:49:17,000 --> 00:49:23,000
all I am doing is looking for
one solution that's different

743
00:49:22,000 --> 00:49:28,000
from e to the minus a t.

744
00:49:24,000 --> 00:49:30,000
And, that solution,
therefore, is y equals e to the

745
00:49:28,000 --> 00:49:34,000
minus a t times t.

746
00:49:32,000 --> 00:49:38,000
And, there's my second
solution.

747
00:49:34,000 --> 00:49:40,000
So, this is a solution of the
critically damped case.

748
00:49:37,000 --> 00:49:43,000
And, you are going to use it in
three or four of the different

749
00:49:42,000 --> 00:49:48,000
problems on the problem set.
But, I think you can deal with

750
00:49:46,000 --> 00:49:52,000
virtually the whole problem set,
except for the last problem,

751
00:49:50,000 --> 00:49:56,000
now.