WEBVTT
00:00:02.000 --> 00:00:08.000
We're going to start.
00:00:31.000 --> 00:00:37.000
We are going to start studying
today, and for quite a while,
00:00:37.000 --> 00:00:43.000
the linear second-order
differential equation with
00:00:42.000 --> 00:00:48.000
constant coefficients.
In standard form,
00:00:46.000 --> 00:00:52.000
it looks like,
there are various possible
00:00:51.000 --> 00:00:57.000
choices for the variable,
unfortunately,
00:00:55.000 --> 00:01:01.000
so I hope it won't disturb you
much if I use one rather than
00:01:01.000 --> 00:01:07.000
another.
I'm going to write it this way
00:01:06.000 --> 00:01:12.000
in standard form.
I'll use y as the dependent
00:01:09.000 --> 00:01:15.000
variable.
Your book uses little p and
00:01:12.000 --> 00:01:18.000
little q.
I'll probably switch to that by
00:01:16.000 --> 00:01:22.000
next time.
But, for today,
00:01:18.000 --> 00:01:24.000
I'd like to use the most
neutral letters I can find that
00:01:22.000 --> 00:01:28.000
won't interfere with anything
else.
00:01:25.000 --> 00:01:31.000
So, of course call the constant
coefficients,
00:01:28.000 --> 00:01:34.000
respectively,
capital A and capital B.
00:01:33.000 --> 00:01:39.000
I'm going to assume for today
that the right-hand side is
00:01:37.000 --> 00:01:43.000
zero.
So, that means it's what we
00:01:40.000 --> 00:01:46.000
call homogeneous.
The left-hand side must be in
00:01:44.000 --> 00:01:50.000
this form for it to be linear,
it's second order because it
00:01:49.000 --> 00:01:55.000
involves a second derivative.
These coefficients,
00:01:53.000 --> 00:01:59.000
A and B, are understood to be
constant because,
00:01:57.000 --> 00:02:03.000
as I said, it has constant
coefficients.
00:02:02.000 --> 00:02:08.000
Of course, that's not the most
general linear equation there
00:02:06.000 --> 00:02:12.000
could be.
In general, it would be more
00:02:08.000 --> 00:02:14.000
general by making this a
function of the dependent
00:02:12.000 --> 00:02:18.000
variable, x or t,
whatever it's called.
00:02:15.000 --> 00:02:21.000
Similarly, this could be a
function of the dependent
00:02:18.000 --> 00:02:24.000
variable.
Above all, the right-hand side
00:02:21.000 --> 00:02:27.000
can be a function of a variable
rather than simply zero.
00:02:25.000 --> 00:02:31.000
In that case the equation is
called inhomogeneous.
00:02:30.000 --> 00:02:36.000
But it has a different physical
meaning, and therefore it's
00:02:34.000 --> 00:02:40.000
customary to study that after
this.
00:02:36.000 --> 00:02:42.000
You start with this.
This is the case we start with,
00:02:40.000 --> 00:02:46.000
and then by the middle of next
week we will be studying more
00:02:44.000 --> 00:02:50.000
general cases.
But, it's a good idea to start
00:02:47.000 --> 00:02:53.000
here.
Your book starts with,
00:02:49.000 --> 00:02:55.000
in general, some theory of a
general linear equation of
00:02:53.000 --> 00:02:59.000
second-order,
and even higher order.
00:02:55.000 --> 00:03:01.000
I'm asking you to skip that for
the time being.
00:03:00.000 --> 00:03:06.000
We'll come back to it next
Wednesday, it two lectures,
00:03:03.000 --> 00:03:09.000
in other words.
I think it's much better and
00:03:06.000 --> 00:03:12.000
essential for your problems at
for you to get some experience
00:03:10.000 --> 00:03:16.000
with a simple type of equation.
And then, you'll understand the
00:03:14.000 --> 00:03:20.000
general theory,
how it applies,
00:03:16.000 --> 00:03:22.000
a lot better,
I think.
00:03:17.000 --> 00:03:23.000
So, let's get experience here.
The downside of that is that
00:03:21.000 --> 00:03:27.000
I'm going to have to assume a
couple of things about the
00:03:25.000 --> 00:03:31.000
solution to this equation,
how it looks;
00:03:27.000 --> 00:03:33.000
I don't think that will upset
you too much.
00:03:32.000 --> 00:03:38.000
So, what I'm going to assume,
and we will justify it in a
00:03:37.000 --> 00:03:43.000
couple lectures,
that the general solution,
00:03:42.000 --> 00:03:48.000
that is, the solution involving
arbitrary constants,
00:03:47.000 --> 00:03:53.000
looks like this.
y is equal-- The arbitrary
00:03:51.000 --> 00:03:57.000
constants occur in a certain
special way.
00:03:56.000 --> 00:04:02.000
There is c one y one plus c two
y two.
00:04:03.000 --> 00:04:09.000
So, these are two arbitrary
constants corresponding to the
00:04:06.000 --> 00:04:12.000
fact that we are solving a
second-order equation.
00:04:09.000 --> 00:04:15.000
In general, the number of
arbitrary constants in the
00:04:11.000 --> 00:04:17.000
solution is the same as the
order of the equation because if
00:04:15.000 --> 00:04:21.000
it's a second-order equation
because if it's a second-order
00:04:18.000 --> 00:04:24.000
equation, that means somehow or
other, it may be concealed.
00:04:21.000 --> 00:04:27.000
But you're going to have to
integrate something twice to get
00:04:25.000 --> 00:04:31.000
the answer.
And therefore,
00:04:26.000 --> 00:04:32.000
there should be two arbitrary
constants.
00:04:30.000 --> 00:04:36.000
That's very rough,
but it sort of gives you the
00:04:34.000 --> 00:04:40.000
idea.
Now, what are the y1 and y2?
00:04:38.000 --> 00:04:44.000
Well, as you can see,
if these are arbitrary
00:04:42.000 --> 00:04:48.000
constants, if I take c2 to be
zero and c1 to be one,
00:04:48.000 --> 00:04:54.000
that means that y1 must be a
solution to the equation,
00:04:53.000 --> 00:04:59.000
and similarly y2.
So, where y1 and y2 are
00:04:57.000 --> 00:05:03.000
solutions.
Now, what that shows you is
00:05:02.000 --> 00:05:08.000
that the task of solving this
equation is reduced,
00:05:05.000 --> 00:05:11.000
in some sense,
to finding just two solutions
00:05:09.000 --> 00:05:15.000
of it, somehow.
All we have to do is find two
00:05:12.000 --> 00:05:18.000
solutions, and then we will have
solved the equation because the
00:05:17.000 --> 00:05:23.000
general solution is made up in
this way by multiplying those
00:05:22.000 --> 00:05:28.000
two solutions by arbitrary
constants and adding them.
00:05:26.000 --> 00:05:32.000
So, the problem is,
where do we get that solutions
00:05:30.000 --> 00:05:36.000
from?
But, first of all,
00:05:33.000 --> 00:05:39.000
or rather, second or third of
all, the initial conditions
00:05:38.000 --> 00:05:44.000
enter into the,
I haven't given you any initial
00:05:42.000 --> 00:05:48.000
conditions here,
but if you have them,
00:05:45.000 --> 00:05:51.000
and I will illustrate them when
I work problems,
00:05:49.000 --> 00:05:55.000
the initial conditions,
well, the initial values are
00:05:54.000 --> 00:06:00.000
satisfied by choosing c1 and c2,
are satisfied by choosing c1
00:05:59.000 --> 00:06:05.000
and c2 properly.
So, in other words,
00:06:03.000 --> 00:06:09.000
if you have an initial value
problem to solve,
00:06:08.000 --> 00:06:14.000
that will be taken care of by
the way those constants,
00:06:14.000 --> 00:06:20.000
c, enter into the solution.
Okay, without further ado,
00:06:19.000 --> 00:06:25.000
there is a standard example,
which I wish I had looked up in
00:06:25.000 --> 00:06:31.000
the physics syllabus for the
first semester.
00:06:30.000 --> 00:06:36.000
Did you study the
spring-mass-dashpot system in
00:06:35.000 --> 00:06:41.000
8.01?
I'm embarrassed having to ask
00:06:39.000 --> 00:06:45.000
you.
You did?
00:06:40.000 --> 00:06:46.000
Raise your hands if you did.
Okay, that means you all did.
00:06:44.000 --> 00:06:50.000
Well, just let me draw an
instant picture to remind you.
00:06:49.000 --> 00:06:55.000
So, this is a two second
review.
00:06:51.000 --> 00:06:57.000
I don't know how they draw the
picture.
00:06:54.000 --> 00:07:00.000
Probably they don't draw
picture at all.
00:06:57.000 --> 00:07:03.000
They have some elaborate system
here of the thing running back
00:07:02.000 --> 00:07:08.000
and forth.
Well, in the math,
00:07:04.000 --> 00:07:10.000
we do that all virtually.
So, here's my system.
00:07:10.000 --> 00:07:16.000
That's a fixed thing.
Here's a little spring.
00:07:14.000 --> 00:07:20.000
And, there's a little car on
the track here,
00:07:18.000 --> 00:07:24.000
I guess.
So, there's the mass,
00:07:21.000 --> 00:07:27.000
some mass in the little car,
and motion is damped by what's
00:07:27.000 --> 00:07:33.000
called a dashpot.
A dashpot is the sort of thing,
00:07:32.000 --> 00:07:38.000
you see them in everyday life
as door closers.
00:07:35.000 --> 00:07:41.000
They're the thing up above that
you never notice that prevent
00:07:40.000 --> 00:07:46.000
the door slamming shut.
So, if you take one apart,
00:07:43.000 --> 00:07:49.000
it looks something like this.
So, that's the dash pot.
00:07:47.000 --> 00:07:53.000
It's a chamber with a piston.
This is a piston moving in and
00:07:51.000 --> 00:07:57.000
out, and compressing the air,
releasing it,
00:07:54.000 --> 00:08:00.000
is what damps the motion of the
thing.
00:07:57.000 --> 00:08:03.000
So, this is a dashpot,
it's usually called.
00:08:02.000 --> 00:08:08.000
And, here's our mass in that
little truck.
00:08:05.000 --> 00:08:11.000
And, here's the spring.
And then, the equation which
00:08:09.000 --> 00:08:15.000
governs it is,
let's call this x.
00:08:11.000 --> 00:08:17.000
I'm already changing,
going to change the dependent
00:08:15.000 --> 00:08:21.000
variable from y to x,
but that's just for the sake of
00:08:19.000 --> 00:08:25.000
example, and because the track
is horizontal,
00:08:23.000 --> 00:08:29.000
it seems more natural to call
it x.
00:08:27.000 --> 00:08:33.000
There's some equilibrium
position somewhere,
00:08:30.000 --> 00:08:36.000
let's say, here.
That's the position at which
00:08:34.000 --> 00:08:40.000
the mass wants to be,
if the spring is not pulling on
00:08:38.000 --> 00:08:44.000
it or pushing on it,
and the dashpot is happy.
00:08:42.000 --> 00:08:48.000
I guess we'd better have a
longer dashpot here.
00:08:46.000 --> 00:08:52.000
So, this is the equilibrium
position where nothing is
00:08:50.000 --> 00:08:56.000
happening.
When you depart from that
00:08:53.000 --> 00:08:59.000
position, then the spring,
if you go that way,
00:08:57.000 --> 00:09:03.000
the spring tries to pull the
mass back.
00:09:02.000 --> 00:09:08.000
If it goes on the site,
the spring tries to push the
00:09:07.000 --> 00:09:13.000
mass away.
The dashpot,
00:09:09.000 --> 00:09:15.000
meanwhile, is doing its thing.
And so, the force on the,
00:09:14.000 --> 00:09:20.000
m x double prime.
That's by Newton's law,
00:09:19.000 --> 00:09:25.000
the force, comes from where?
Well, there's the spring
00:09:24.000 --> 00:09:30.000
pushing and pulling on it.
That force is opposed.
00:09:29.000 --> 00:09:35.000
If x gets to be beyond zero,
then the spring tries to pull
00:09:34.000 --> 00:09:40.000
it back.
If it gets to the left of zero,
00:09:39.000 --> 00:09:45.000
if x gets to be negative,
that that spring force is
00:09:42.000 --> 00:09:48.000
pushing it this way,
wants to get rid of the mass.
00:09:46.000 --> 00:09:52.000
So, it should be minus kx,
and this is from the spring,
00:09:50.000 --> 00:09:56.000
the fact that is proportional
to the amount by which x varies.
00:09:54.000 --> 00:10:00.000
So, that's called Hooke's Law.
Never mind that.
00:09:58.000 --> 00:10:04.000
This is a law.
That's a law,
00:10:01.000 --> 00:10:07.000
Newton's law,
okay, Newton,
00:10:03.000 --> 00:10:09.000
Hooke with an E,
and the dashpot damping is
00:10:06.000 --> 00:10:12.000
proportional to the velocity.
It's not doing anything if the
00:10:11.000 --> 00:10:17.000
mass is not moving,
even if it's stretched way out
00:10:15.000 --> 00:10:21.000
of its equilibrium position.
So, it resists the velocity.
00:10:19.000 --> 00:10:25.000
If the thing is trying to go
that way, the dashpot resists
00:10:24.000 --> 00:10:30.000
it.
It's trying to go this way,
00:10:26.000 --> 00:10:32.000
the dashpot resists that,
too.
00:10:30.000 --> 00:10:36.000
It's always opposed to the
velocity.
00:10:32.000 --> 00:10:38.000
And so, this is a dash pot
damping.
00:10:35.000 --> 00:10:41.000
I don't know whose law this is.
So, it's the force coming from
00:10:39.000 --> 00:10:45.000
the dashpot.
And, when you write this out,
00:10:42.000 --> 00:10:48.000
the final result,
therefore, is it's m x double
00:10:45.000 --> 00:10:51.000
prime plus c x prime,
it's important to see where the
00:10:49.000 --> 00:10:55.000
various terms are,
plus kx equals zero.
00:10:52.000 --> 00:10:58.000
And now, that's still not in
00:10:56.000 --> 00:11:02.000
standard form.
To put it in standard form,
00:10:59.000 --> 00:11:05.000
you must divide through by the
mass.
00:11:03.000 --> 00:11:09.000
And, it will now read like
this, plus k divided by m times
00:11:07.000 --> 00:11:13.000
x equals zero.
And, that's the equation
00:11:10.000 --> 00:11:16.000
governing the motion of the
spring.
00:11:13.000 --> 00:11:19.000
I'm doing this because your
problem set, problems three and
00:11:17.000 --> 00:11:23.000
four, ask you to look at a
little computer visual which
00:11:21.000 --> 00:11:27.000
illustrates a lot of things.
And, I didn't see how it would
00:11:26.000 --> 00:11:32.000
make, you can do it without this
interpretation of
00:11:30.000 --> 00:11:36.000
spring-mass-dashpot,
--
00:11:33.000 --> 00:11:39.000
-- but, I think thinking of it
of these constants as,
00:11:37.000 --> 00:11:43.000
this is the damping constant,
and this is the spring,
00:11:41.000 --> 00:11:47.000
the constant which represents
the force being exerted by the
00:11:46.000 --> 00:11:52.000
spring, the spring constant,
as it's called,
00:11:50.000 --> 00:11:56.000
makes it much more vivid.
So, you will note is that those
00:11:54.000 --> 00:12:00.000
problems are labeled Friday or
Monday.
00:11:57.000 --> 00:12:03.000
Make it Friday.
You can do them after today if
00:12:01.000 --> 00:12:07.000
you have the vaguest idea of
what I'm talking about.
00:12:07.000 --> 00:12:13.000
If not, go back and repeat
8.01.
00:12:10.000 --> 00:12:16.000
So, all this was just an
example, a typical model.
00:12:16.000 --> 00:12:22.000
But, by far,
the most important simple
00:12:20.000 --> 00:12:26.000
model.
Okay, now what I'd like to talk
00:12:25.000 --> 00:12:31.000
about is the solution.
What is it I have to do to
00:12:30.000 --> 00:12:36.000
solve the equation?
So, to solve the equation that
00:12:37.000 --> 00:12:43.000
I outlined in orange on the
board, the ODE,
00:12:41.000 --> 00:12:47.000
our task is to find two
solutions.
00:12:44.000 --> 00:12:50.000
Now, don't make it too trivial.
There is a condition.
00:12:49.000 --> 00:12:55.000
The solution should be
independent.
00:12:53.000 --> 00:12:59.000
All that means is that y2
should not be a constant
00:12:58.000 --> 00:13:04.000
multiple of y1.
I mean, if you got y1,
00:13:02.000 --> 00:13:08.000
then two times y1 is not an
acceptable value for this
00:13:06.000 --> 00:13:12.000
because, as you can see,
you really only got one there.
00:13:10.000 --> 00:13:16.000
You're not going to be able to
make up a two parameter family.
00:13:15.000 --> 00:13:21.000
So, the solutions have to be
independent, which means,
00:13:19.000 --> 00:13:25.000
to repeat, that neither should
be a constant multiple of the
00:13:24.000 --> 00:13:30.000
other.
They should look different.
00:13:26.000 --> 00:13:32.000
That's an adequate explanation.
Okay, now, what's the basic
00:13:32.000 --> 00:13:38.000
method to finding those
solutions?
00:13:34.000 --> 00:13:40.000
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
00:13:41.000 --> 00:13:47.000
type, even systems of this type,
with constant coefficients.
00:13:46.000 --> 00:13:52.000
The basic method is to try y
equals an exponential.
00:13:50.000 --> 00:13:56.000
Now, the only way you can
fiddle with an exponential is in
00:13:54.000 --> 00:14:00.000
the constant that you put up on
top.
00:13:57.000 --> 00:14:03.000
So, I'm going to try y equals e
to the rt.
00:14:03.000 --> 00:14:09.000
Notice you can't tell from that
what I'm using as the
00:14:08.000 --> 00:14:14.000
independent variable.
But, this tells you I'm using
00:14:13.000 --> 00:14:19.000
t.
And, I'm switching back to
00:14:16.000 --> 00:14:22.000
using t as the dependent
variable.
00:14:19.000 --> 00:14:25.000
So, T is the independent
variable.
00:14:22.000 --> 00:14:28.000
Why do I do that?
The answer is because somebody
00:14:27.000 --> 00:14:33.000
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
00:15:03.000 --> 00:15:09.000
we can pick r so that it's a
solution.
00:15:07.000 --> 00:15:13.000
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.
00:15:16.000 --> 00:15:22.000
That's y double prime
00:15:19.000 --> 00:15:25.000
because each time you
differentiate it,
00:15:22.000 --> 00:15:28.000
you put an extra power of r out
in front.
00:15:25.000 --> 00:15:31.000
But otherwise,
do nothing.
00:15:27.000 --> 00:15:33.000
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,
00:15:39.000 --> 00:15:45.000
which is B e to the rt.
00:15:41.000 --> 00:15:47.000
And, that's supposed to be
equal to zero.
00:15:44.000 --> 00:15:50.000
So, I have to choose r so that
this becomes equal to zero.
00:15:48.000 --> 00:15:54.000
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
00:15:54.000 --> 00:16:00.000
is never zero.
And therefore,
00:15:57.000 --> 00:16:03.000
you can divide it out because
it's always a positive number,
00:16:01.000 --> 00:16:07.000
regardless of the value of t.
So, I can cancel out from each
00:16:07.000 --> 00:16:13.000
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,
00:16:57.000 --> 00:17:03.000
the characteristic equation of
the system, okay?
00:17:02.000 --> 00:17:08.000
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.