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00:00:07,000 --> 00:00:13,000
We are going to start today in
a serious way on the

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inhomogenous equation,
second-order linear

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00:00:17,000 --> 00:00:23,000
differential,
I'll simply write it out

4
00:00:22,000 --> 00:00:28,000
instead of writing out all the
words which go with it.

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00:00:28,000 --> 00:00:34,000
So, such an equation looks
like, the second-order equation

6
00:00:35,000 --> 00:00:41,000
is going to look like y double
prime plus p of x,

7
00:00:40,000 --> 00:00:46,000
t, x plus q of x times y.
Now, up to now the right-hand

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00:00:47,000 --> 00:00:53,000
side has been zero.
So, now we are going to make it

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00:00:53,000 --> 00:00:59,000
not be zero.
So, this is going to be f of x.

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In the most frequent

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applications,
x is time.

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x is usually time,
often, but not always.

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So, maybe just for today,
I will use X in talking about

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00:01:13,000 --> 00:01:19,000
the general theory.
And, from now on,

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I'll probably make X equal time
because that's what is most of

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00:01:22,000 --> 00:01:28,000
the time in the applications.
So, this is the part we've been

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studying up until now.
It has a lot of names.

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It's input, signal,
commas between those,

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a driving term,
or sometimes it's called the

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forcing term.
You'll see all of these in the

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literature, and it pretty much
depends upon what course you're

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00:01:48,000 --> 00:01:54,000
sitting at, what the professor
habitually calls it.

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I will try to use all these
terms now and then,

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probably most often I will
lapse into input as the most

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00:01:59,000 --> 00:02:05,000
generic term,
suggesting nothing in

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particular, and therefore,
equally acceptable or

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unacceptable to everybody.
The response,

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00:02:09,000 --> 00:02:15,000
the solution,
then, the solution as you know

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is then called the response.
The response,

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sometimes it's called the
output.

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I think I'll stick pretty much
with response.

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So, I'm using pretty much the
same terminology we use for

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00:02:35,000 --> 00:02:41,000
studying first-order equations.
Now, as you will see,

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the reason we had to study the
homogeneous case first was

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because you cannot solve this
without knowing the homogeneous

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solutions.
So, that's the inhomogeneous

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00:02:53,000 --> 00:02:59,000
case.
But the homogeneous one,

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00:02:56,000 --> 00:03:02,000
the corresponding homogeneous
thing, y double prime plus p of

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00:03:01,000 --> 00:03:07,000
x y prime plus q of x times y
equals zero

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00:03:06,000 --> 00:03:12,000
is an essential
part of the solution to this

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equation.
That's called,

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therefore, it has names.
Now, unfortunately,

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it doesn't have a single name.
I don't know what to call it,

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00:03:24,000 --> 00:03:30,000
but I think I'll probably call
it the associated homogeneous

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00:03:30,000 --> 00:03:36,000
equation, or ODE,
the associated homogeneous

46
00:03:34,000 --> 00:03:40,000
equation, the one associated to
the guy on the left.

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00:03:40,000 --> 00:03:46,000
It's also called the reduced
equation by some people.

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00:03:44,000 --> 00:03:50,000
There is some other term for
it, which escapes me totally,

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00:03:48,000 --> 00:03:54,000
but what the heck.
Now, its solution has a name.

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00:03:52,000 --> 00:03:58,000
So, its solution,
of course, doesn't depend on

51
00:03:55,000 --> 00:04:01,000
anything in particular,
the general solution,

52
00:03:58,000 --> 00:04:04,000
because the right-hand side is
always zero.

53
00:04:03,000 --> 00:04:09,000
So, its solution,
we know can be written as y

54
00:04:06,000 --> 00:04:12,000
equals in the form c1 y1 plus c2
y2,

55
00:04:10,000 --> 00:04:16,000
where y1 and y2 are any two
independent solutions of that,

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00:04:14,000 --> 00:04:20,000
and then c1's and c2's are
arbitrary constants.

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00:04:17,000 --> 00:04:23,000
Now, what you are looking at
this equation,

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00:04:20,000 --> 00:04:26,000
you're going to need this also.
And therefore,

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00:04:24,000 --> 00:04:30,000
it has a name.
It has various names.

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00:04:26,000 --> 00:04:32,000
Sometimes there is a subscript,
c, there.

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00:04:31,000 --> 00:04:37,000
Sometimes there's a subscript,
h.

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00:04:34,000 --> 00:04:40,000
Sometimes there's no subscript
at all, which is the most

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00:04:39,000 --> 00:04:45,000
confusing of all.
But, anyway,

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00:04:42,000 --> 00:04:48,000
what's the name given to it?
Well, there is no name.

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00:04:47,000 --> 00:04:53,000
Many books call it the solution
to the associated homogeneous

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equation.
That's maximally long.

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00:04:56,000 --> 00:05:02,000
Your book calls it the
complementary solution.

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00:05:02,000 --> 00:05:08,000
Many people call it that,
and many will look at you with

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00:05:06,000 --> 00:05:12,000
a blank, who know differential
equations very well,

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00:05:10,000 --> 00:05:16,000
and will not have the faintest
idea what you're talking about.

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00:05:14,000 --> 00:05:20,000
If you call it (y)h,
then you are thinking of it as

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the solution;
the h is for homogeneous to

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indicate it's the solution.
So, it's the solution to the,

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I'm not going to write that.
You put it in her books if you

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like writing.
Write solution to the

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associated homogeneous equation,
y(h).

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00:05:36,000 --> 00:05:42,000
But, it's all the same thing.
Now, or the solution to the

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reduced equation,
I see I have in my notes.

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Okay, good, the solution to the
reduced equation,

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too.
Okay, now, the examples,

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there are, of course,
two classical examples,

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of which you know one.
But, use them as the model for

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00:05:57,000 --> 00:06:03,000
what solutions of these things
should look like and how they

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00:06:02,000 --> 00:06:08,000
should behave.
So, the model you know already

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00:06:06,000 --> 00:06:12,000
is the one, I won't make the
leading coefficient one because

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00:06:11,000 --> 00:06:17,000
it usually isn't,
is the one, m x double prime,

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00:06:15,000 --> 00:06:21,000
so t is the
independent variable,

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00:06:18,000 --> 00:06:24,000
plus b x prime plus k x equals
f of t.

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00:06:23,000 --> 00:06:29,000
That's the spring-mass system,

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00:06:26,000 --> 00:06:32,000
the spring-mass-dashpot system.
Mass, the damping constant and

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00:06:32,000 --> 00:06:38,000
the spring constant,
except up to now,

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00:06:35,000 --> 00:06:41,000
it's always been zero here.
What does this f of t

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00:06:40,000 --> 00:06:46,000
represent?
Well, if you think of the way

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00:06:43,000 --> 00:06:49,000
in which I derived the equation,
the mx, that was the Newton's

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00:06:49,000 --> 00:06:55,000
law.
That's the acceleration.

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00:06:51,000 --> 00:06:57,000
So, it's the acceleration,
the mass times the

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00:06:55,000 --> 00:07:01,000
acceleration.
By Newton's law,

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00:06:57,000 --> 00:07:03,000
this is equal to the imposed
force on the little mass truck.

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Okay, you got that truck,
there.

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00:07:06,000 --> 00:07:12,000
I'm not going to draw the truck
for the nth time.

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00:07:09,000 --> 00:07:15,000
You'll have to imagine it.
So, here's our truck.

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Okay, forces are acting on it.
Remember, the forces were

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minus kx.
That came from the spring.

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00:07:20,000 --> 00:07:26,000
There was a force,
minus b x prime.

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00:07:23,000 --> 00:07:29,000
That came from the dashpot,
the damping force.

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So, this other guy is f of t.

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00:07:30,000 --> 00:07:36,000
What's this?
This is the external force,

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which is acting out.
In other words,

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instead of the little truck
going back and forth and doing

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its own thing all by itself,
here's someone with an

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00:07:41,000 --> 00:07:47,000
electromagnet,
and the mass it's carrying is a

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00:07:44,000 --> 00:07:50,000
big pile of iron ore.
You're turning it on and off,

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00:07:47,000 --> 00:07:53,000
and pulling that thing from
afar where nobody can see it.

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00:07:51,000 --> 00:07:57,000
So, this is the external force.
Now, think, that is the model

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00:07:55,000 --> 00:08:01,000
you must have in your mind of
how these equations are treated.

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00:08:00,000 --> 00:08:06,000
In other words,
when f of t is zero,

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the system is passive.
There is no external force on

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it when this is zero.
The system is sitting,

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and just doing what it wants to
do, all by itself.

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You wanted up by giving it an
initial push,

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00:08:17,000 --> 00:08:23,000
and putting its initial
position somewhere.

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But after that,
you lay your hands off.

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The system then just passively
responds to its initial

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conditions and does what it
wants.

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The other model is that you
don't let it respond the way it

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wants to.
You force it from the outside

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by pushing it with an external
force.

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Now, those are clearly two
entirely different problems:

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what it does by itself,
or what it does when it's acted

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on from outside.
And, when I explained to you

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how the thing is to be solved,
you have to keep in mind those

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two models.
So, this is the forced system.

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I'll just use the word,
forced system,

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that's where f of t is
not zero, versus the passive

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system where there is no
external applied force.

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The passive system,
the forced system,

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00:09:14,000 --> 00:09:20,000
now, you have to both,
even if you wanted to solve the

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00:09:18,000 --> 00:09:24,000
forced system,
the way the system would behave

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00:09:22,000 --> 00:09:28,000
if nothing would be done to it
from the outside is nonetheless

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going to be an important part of
the solution.

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00:09:32,000 --> 00:09:38,000
And, I won't be able to give
you that solution without

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knowing this also.
Now, I'd like to give you the

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other model very rapidly because
it's in your book.

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It's in the problems I have to
give you.

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You know, it's part of
everybody's culture,

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whether they like it or not.
So, that's example number one.

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Example number two,
which follows the differential

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00:09:55,000 --> 00:10:01,000
equation just as perfectly as
the spring-mass-dashpot system

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is the simple electric circuit.
The inductance,

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you don't know yet what an
inductance is,

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00:10:08,000 --> 00:10:14,000
officially, but you will,
a resistance,

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00:10:11,000 --> 00:10:17,000
sorry, that's okay,
put the capacitance up there,

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00:10:15,000 --> 00:10:21,000
resistance, and then maybe a
thing.

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So, this is a resistance.
I think you know these symbols.

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By now, you certainly know the
system for capacitance.

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What I mean when I say C is the
capacitance, you may not know

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yet what L is.
That's called the inductance.

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So, this is something called a
coil because it looks like one.

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L is what's called its
inductance.

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00:10:45,000 --> 00:10:51,000
And, the differential equation,
there are two differential

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equations which can be used in
this.

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They are essentially the same.
One is simply the derivative of

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the other.
Both differential equations

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come from Kirchhoff's voltage
law, that the sum of the voltage

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00:11:04,000 --> 00:11:10,000
drops as you move around the
circuit --

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-- has to be zero because
otherwise, I don't have to,

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that's because of somebody's
law, Kirchhoff,

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00:11:14,000 --> 00:11:20,000
with two h's.
The sum of the voltage drops to

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zero, and now you know the
voltage drop across this,

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and you know the voltage drop
across that because you learned

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in 8.02.
You will, one day,

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00:11:27,000 --> 00:11:33,000
learn the voltage drop across
this.

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00:11:29,000 --> 00:11:35,000
But, I already know it.
It's Li.

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00:11:32,000 --> 00:11:38,000
So, i is the current.
I'll write this thing in its

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00:11:37,000 --> 00:11:43,000
primitive form first.
So, i is the current that's

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00:11:42,000 --> 00:11:48,000
flowing in the circuit.
q is the charge on the

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capacitance.
So, the voltage drop across the

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coil is L times i.
The voltage shop across the,

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00:11:57,000 --> 00:12:03,000
Li prime,
the voltage drop across the

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00:12:02,000 --> 00:12:08,000
resistance is,
well, you know that.

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00:12:07,000 --> 00:12:13,000
And, the voltage shop across
the capacitance is q

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00:12:10,000 --> 00:12:16,000
divided by C.
And so, that's equal to,

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00:12:13,000 --> 00:12:19,000
well, it's equal to zero,
except if there's a battery

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00:12:16,000 --> 00:12:22,000
here or something generating a
voltage drop,

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00:12:19,000 --> 00:12:25,000
so, let's call that E is a
generic word.

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00:12:21,000 --> 00:12:27,000
E could be a battery.
It could be a source of

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00:12:24,000 --> 00:12:30,000
alternating current,
something like that.

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00:12:27,000 --> 00:12:33,000
But, there's a voltage drop
across it, and I'm giving E the

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00:12:31,000 --> 00:12:37,000
name of the voltage drop.
So, and then there's the

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00:12:35,000 --> 00:12:41,000
question of the signs,
which I know I'll never

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00:12:38,000 --> 00:12:44,000
understand.
But, let's assume you've chosen

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00:12:41,000 --> 00:12:47,000
the sign convention so that this
comes out nicely on the

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00:12:44,000 --> 00:12:50,000
right-hand side.
So, this might be varying

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00:12:47,000 --> 00:12:53,000
sinusoidally,
in which case you'd have source

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00:12:50,000 --> 00:12:56,000
of alternating current.
Or, it might be constant,

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00:12:53,000 --> 00:12:59,000
in which that would be a
battery, a little dry cell

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00:12:56,000 --> 00:13:02,000
giving you direct current of a
constant voltage,

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00:12:59,000 --> 00:13:05,000
stuff like that.
So, you could make this minus

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00:13:03,000 --> 00:13:09,000
if you want, but everything will
have the wrong signs,

200
00:13:07,000 --> 00:13:13,000
so don't do it.
Now, this doesn't look like

201
00:13:09,000 --> 00:13:15,000
what it's supposed to look like
because it's got q and i.

202
00:13:13,000 --> 00:13:19,000
So, the final thing you have to
know is that q prime is

203
00:13:17,000 --> 00:13:23,000
equal to i.
The rate at which that charge

204
00:13:19,000 --> 00:13:25,000
leaves the condenser and hurries
around the circuit to find its

205
00:13:23,000 --> 00:13:29,000
little soul mate on the other
side is the current that's

206
00:13:27,000 --> 00:13:33,000
flowing in the circuit.
That's why current flows,

207
00:13:30,000 --> 00:13:36,000
except nothing really happens.
Electrons just push on each

208
00:13:35,000 --> 00:13:41,000
other, and they stay where they
are.

209
00:13:37,000 --> 00:13:43,000
I don't understand this at all.
So, if I differentiate this,

210
00:13:42,000 --> 00:13:48,000
you can do two things.
Either you could integrate i,

211
00:13:46,000 --> 00:13:52,000
and expressed the thing
entirely in terms of q,

212
00:13:50,000 --> 00:13:56,000
or you can differentiate it,
and express everything in terms

213
00:13:54,000 --> 00:14:00,000
of i.
Your book does nicely both,

214
00:13:57,000 --> 00:14:03,000
does not take sides.
So, let's differentiate it,

215
00:14:02,000 --> 00:14:08,000
and then it will look like L i
double prime plus R i prime plus

216
00:14:06,000 --> 00:14:12,000
i divided by C equals,

217
00:14:09,000 --> 00:14:15,000
and now, watch out,
you have now not the

218
00:14:12,000 --> 00:14:18,000
electromotive force,
but its derivative.

219
00:14:15,000 --> 00:14:21,000
So, if you were so unfortunate
as to put a little dry cell

220
00:14:19,000 --> 00:14:25,000
there, now you've got nothing,
and you've got the homogeneous

221
00:14:24,000 --> 00:14:30,000
case.
That's okay.

222
00:14:25,000 --> 00:14:31,000
Where are the erasers?
One eraser?

223
00:14:28,000 --> 00:14:34,000
I don't believe this.

224
00:14:43,000 --> 00:14:49,000
So, there's the equation.
There are our two equations.

225
00:14:47,000 --> 00:14:53,000
Why don't we put them up in
colored chalk.

226
00:14:50,000 --> 00:14:56,000
There's the spring equation.
And, here's the equation that

227
00:14:54,000 --> 00:15:00,000
governs the current,
for how the current flows in

228
00:14:58,000 --> 00:15:04,000
that circuit.
And now, you can see,

229
00:15:01,000 --> 00:15:07,000
again, what does it mean?
If this is zero,

230
00:15:04,000 --> 00:15:10,000
for example,
if I have a dry cell there,

231
00:15:07,000 --> 00:15:13,000
or if I have nothing at all in
the circuit, then this

232
00:15:11,000 --> 00:15:17,000
represents the passive circuit.
It's just sitting there.

233
00:15:16,000 --> 00:15:22,000
It wouldn't do anything at all,
except that you've put a charge

234
00:15:20,000 --> 00:15:26,000
on the capacitor,
and waited, and of course,

235
00:15:22,000 --> 00:15:28,000
when you put a charge on there,
it's got a discharge,

236
00:15:26,000 --> 00:15:32,000
and discharges through the
circuit, and swings back and

237
00:15:29,000 --> 00:15:35,000
forth a little bit if it's
under-damped until finally

238
00:15:32,000 --> 00:15:38,000
towards the end the current dies
away to zero.

239
00:15:36,000 --> 00:15:42,000
But, what usually happens is
that you drive this passive

240
00:15:39,000 --> 00:15:45,000
circuit by putting an effective
E in it, and then you want to

241
00:15:44,000 --> 00:15:50,000
know how the current behaves.
So, those are the two problems,

242
00:15:48,000 --> 00:15:54,000
the passive circuit without an
applied electromotive force,

243
00:15:52,000 --> 00:15:58,000
or plugging it into the wall,
and wanting it to do things.

244
00:15:56,000 --> 00:16:02,000
That's the normal state of
affairs.

245
00:16:00,000 --> 00:16:06,000
People don't want passive
circuits, they want circuits

246
00:16:03,000 --> 00:16:09,000
which do things because,
okay, that's why they want to

247
00:16:07,000 --> 00:16:13,000
solve inhomogeneous equations
instead of homogeneous

248
00:16:11,000 --> 00:16:17,000
equations.
But as I said,

249
00:16:13,000 --> 00:16:19,000
you have to do the homogeneous
case first.

250
00:16:16,000 --> 00:16:22,000
Okay, you are now officially
responsible for this,

251
00:16:20,000 --> 00:16:26,000
and I don't care that you
haven't had it in physics yet.

252
00:16:24,000 --> 00:16:30,000
You will before the next exam.
So, I don't even feel guilty.

253
00:16:30,000 --> 00:16:36,000
But, you're going to start
using it on the problem set

254
00:16:34,000 --> 00:16:40,000
right away.
So, it's never too soon to

255
00:16:37,000 --> 00:16:43,000
start learning it.
Okay, now, the main theorem,

256
00:16:41,000 --> 00:16:47,000
I now want to go,
so that was just examples to

257
00:16:44,000 --> 00:16:50,000
give you some physical feeling
for the sorts of differential

258
00:16:49,000 --> 00:16:55,000
equations we'll be talking
about.

259
00:16:52,000 --> 00:16:58,000
I now want to tell you briefly
about the key theorem about

260
00:16:56,000 --> 00:17:02,000
solving the homogeneous
equation.

261
00:16:59,000 --> 00:17:05,000
So, the main theorem about
solving the homogeneous equation

262
00:17:04,000 --> 00:17:10,000
is, the inhomogeneous equation.
So, I'm going to write the

263
00:17:09,000 --> 00:17:15,000
inhomogeneous equation out.
I'm going to make the left-hand

264
00:17:14,000 --> 00:17:20,000
side a linear operator,
and am going to write the

265
00:17:18,000 --> 00:17:24,000
equation as Ly equals f of x.

266
00:17:21,000 --> 00:17:27,000
That's the inhomogeneous
equation.

267
00:17:23,000 --> 00:17:29,000
So, L is the linear operator,
second order because I'm only

268
00:17:28,000 --> 00:17:34,000
talking about second-order
equations.

269
00:17:32,000 --> 00:17:38,000
L is a linear operator,
and then this is the

270
00:17:37,000 --> 00:17:43,000
differential equation.
So, here's our differential

271
00:17:43,000 --> 00:17:49,000
equation.
It's inhomogeneous because it's

272
00:17:48,000 --> 00:17:54,000
go the f of x on the
right hand side.

273
00:17:53,000 --> 00:17:59,000
And, what the theorem says is
that the solution has the

274
00:18:00,000 --> 00:18:06,000
following form,
y sub p, I'll explain what that

275
00:18:05,000 --> 00:18:11,000
is in just a moment,
plus y sub c.

276
00:18:13,000 --> 00:18:19,000
So, the hypothesis is we've got
the linear equation,

277
00:18:17,000 --> 00:18:23,000
and the conclusion is that
that's what its solution looks

278
00:18:21,000 --> 00:18:27,000
like.
Now, you already know what y

279
00:18:24,000 --> 00:18:30,000
sub c looks like.
In other words,

280
00:18:27,000 --> 00:18:33,000
if I write this out in more
detail, it would be i.e.,

281
00:18:31,000 --> 00:18:37,000
department of fuller
explanation, --

282
00:18:35,000 --> 00:18:41,000
-- the general solution looks
like y equals yp,

283
00:18:38,000 --> 00:18:44,000
and then this thing is going to
look like an arbitrary constant

284
00:18:43,000 --> 00:18:49,000
times y1 plus an arbitrary
constant times y2,

285
00:18:47,000 --> 00:18:53,000
where these are solutions of
the homogeneous equation.

286
00:18:51,000 --> 00:18:57,000
So, Yc looks like this part,
and the yp, what's yp?

287
00:18:55,000 --> 00:19:01,000
p stands for particular,
the most confusing word in this

288
00:19:00,000 --> 00:19:06,000
subject.
But, you've got at least four

289
00:19:04,000 --> 00:19:10,000
weeks to learn what it means.
Okay, yp is a particular

290
00:19:10,000 --> 00:19:16,000
solution to Ly equals f of x.

291
00:19:14,000 --> 00:19:20,000
Now, I'm not going to explain
what particular means.

292
00:19:19,000 --> 00:19:25,000
First, I'll chat as if you knew
what it meant,

293
00:19:24,000 --> 00:19:30,000
and then we'll see if you have
picked it up.

294
00:19:30,000 --> 00:19:36,000
In other words,
the procedure for solving this

295
00:19:33,000 --> 00:19:39,000
equation is composed of two
steps.

296
00:19:36,000 --> 00:19:42,000
First, to find this part.
In other words,

297
00:19:40,000 --> 00:19:46,000
to find the complementary
solution, in other words,

298
00:19:44,000 --> 00:19:50,000
to do what we've been doing for
the last week,

299
00:19:48,000 --> 00:19:54,000
solve not the equation you are
given, but the reduced equation.

300
00:19:53,000 --> 00:19:59,000
So, the first step is to find
this.

301
00:19:56,000 --> 00:20:02,000
The second step is to find yp.
Now, what's yp?

302
00:20:01,000 --> 00:20:07,000
yp is a particular solution to
the whole equation.

303
00:20:05,000 --> 00:20:11,000
Yeah, but which one?
Well, if it's any one,

304
00:20:09,000 --> 00:20:15,000
then it's not a particular
solution, yeah.

305
00:20:12,000 --> 00:20:18,000
I say, unfortunately the word
particular here is not being

306
00:20:17,000 --> 00:20:23,000
used in exactly the same sense
in which most people use it in

307
00:20:22,000 --> 00:20:28,000
ordinary English.
It's a perfectly valid way to

308
00:20:26,000 --> 00:20:32,000
use it.
It's just confusing,

309
00:20:29,000 --> 00:20:35,000
and no one has ever come up
with a better word.

310
00:20:33,000 --> 00:20:39,000
So, particular means any one
solution.

311
00:20:38,000 --> 00:20:44,000
Any one will do.
Okay, even these have slightly

312
00:20:45,000 --> 00:20:51,000
different meanings.
Any questions about this?

313
00:20:51,000 --> 00:20:57,000
I refuse to answer them.
[LAUGHTER]

314
00:20:58,000 --> 00:21:04,000
Now, well, examples of course
will make it all clear.

315
00:21:03,000 --> 00:21:09,000
But I'd like,
first, to prove the theorem,

316
00:21:08,000 --> 00:21:14,000
to show you how simple it is.
It's extremely simple if you

317
00:21:15,000 --> 00:21:21,000
just use the fact that L is a
linear operator.

318
00:21:20,000 --> 00:21:26,000
We've got two things to prove.
What have we got to prove?

319
00:21:26,000 --> 00:21:32,000
Well, I have to prove two
statements, first of all,

320
00:21:32,000 --> 00:21:38,000
that all the yp plus c1 y1 plus
c2 y2 are

321
00:21:39,000 --> 00:21:45,000
solutions.
How are we going to prove that?

322
00:21:43,000 --> 00:21:49,000
Well, how do you know if
something is a solution?

323
00:21:46,000 --> 00:21:52,000
Well, you plug it into the
equation, and you see if it

324
00:21:49,000 --> 00:21:55,000
satisfies the equation.
Good, let's do it,

325
00:21:52,000 --> 00:21:58,000
proof.
L, I'm going to plug it into

326
00:21:54,000 --> 00:22:00,000
the equation.
That means I calculate L of yp

327
00:21:56,000 --> 00:22:02,000
plus c1 y1 plus c2 y2.

328
00:22:00,000 --> 00:22:06,000
Now, what's the answer?
Because this is a linear

329
00:22:04,000 --> 00:22:10,000
operator, and notice,
the argument doesn't use the

330
00:22:08,000 --> 00:22:14,000
fact that the equation is second
order.

331
00:22:12,000 --> 00:22:18,000
It immediately generalizes to a
linear equation of any order,

332
00:22:17,000 --> 00:22:23,000
whatever-- 47.
Okay, this is L of yp plus L of

333
00:22:21,000 --> 00:22:27,000
c1 y1 plus c2 y2.

334
00:22:25,000 --> 00:22:31,000
Well, what's that?
What's L of the complementary

335
00:22:29,000 --> 00:22:35,000
solution?
What does it mean to be the

336
00:22:34,000 --> 00:22:40,000
complementary solution?
It means when you apply the

337
00:22:37,000 --> 00:22:43,000
operator L to it,
you get zero because this

338
00:22:40,000 --> 00:22:46,000
satisfies the homogeneous
equation.

339
00:22:43,000 --> 00:22:49,000
So, this is zero.
What's L of yp?

340
00:22:46,000 --> 00:22:52,000
Well, it was a particular
solution to the equation.

341
00:22:50,000 --> 00:22:56,000
Therefore, when I plugged it
into the equation,

342
00:22:53,000 --> 00:22:59,000
I must have gotten out on the
right-hand side,

343
00:22:57,000 --> 00:23:03,000
f of x.
So, this is since yp is a

344
00:23:00,000 --> 00:23:06,000
solution to the whole equation.
So, what's the conclusion?

345
00:23:06,000 --> 00:23:12,000
That, if I take any one of
these guys, no matter what c1

346
00:23:10,000 --> 00:23:16,000
and c2 are, apply the linear
operator, L to it,

347
00:23:13,000 --> 00:23:19,000
the answer comes out
to be f of.

348
00:23:17,000 --> 00:23:23,000
Therefore, this proves that
this shows that these are all

349
00:23:21,000 --> 00:23:27,000
solutions because that's what it
means.

350
00:23:24,000 --> 00:23:30,000
Therefore, they satisfy L of y
equals f of x.

351
00:23:30,000 --> 00:23:36,000
They satisfy the whole
inhomogeneous differential

352
00:23:34,000 --> 00:23:40,000
equation, and that's it.
Well, that's only half the

353
00:23:38,000 --> 00:23:44,000
story.
The other half of the story is

354
00:23:41,000 --> 00:23:47,000
to show that there are no other
solutions.

355
00:23:45,000 --> 00:23:51,000
Okay, so we got our little u of
x coming up again,

356
00:23:50,000 --> 00:23:56,000
and he thinks he's a solution.
Okay, so, to prove there are no

357
00:23:55,000 --> 00:24:01,000
other solutions,
it almost sounds biblical,

358
00:23:59,000 --> 00:24:05,000
thou shalt have no other
solutions before me,

359
00:24:03,000 --> 00:24:09,000
okay.
There are no other solutions

360
00:24:07,000 --> 00:24:13,000
accept these guys for different
values of c1 and c2.

361
00:24:10,000 --> 00:24:16,000
Okay, so, u of x is a
solution.

362
00:24:13,000 --> 00:24:19,000
I have to show that u of x is
one of these guys.

363
00:24:16,000 --> 00:24:22,000
How am I going to do that?
Easy.

364
00:24:19,000 --> 00:24:25,000
If it's a solution that,
L of u,

365
00:24:21,000 --> 00:24:27,000
okay, I'm going to drop the x,
okay, just to make the,

366
00:24:25,000 --> 00:24:31,000
like I dropped the x over
there.

367
00:24:29,000 --> 00:24:35,000
If it's a solution to the whole
inhomogeneous equation,

368
00:24:33,000 --> 00:24:39,000
then this must come out to be f
of x.

369
00:24:37,000 --> 00:24:43,000
Now, what's L of yp?
That's f of x too,

370
00:24:41,000 --> 00:24:47,000
by secret little particular
solution I've got in my pocket.

371
00:24:47,000 --> 00:24:53,000
Okay, I pull it out,
ah-ha, L of yp,

372
00:24:50,000 --> 00:24:56,000
that's f of x,
too.

373
00:24:51,000 --> 00:24:57,000
Now, I'm going to not add them.
I'm going to subtract them.

374
00:24:56,000 --> 00:25:02,000
What is L of u minus yp?

375
00:25:00,000 --> 00:25:06,000
Well, it's zero.
It's zero because this is a

376
00:25:05,000 --> 00:25:11,000
linear operator.
This would be L of u minus L of

377
00:25:08,000 --> 00:25:14,000
yp.
I guess the answer is zero on

378
00:25:12,000 --> 00:25:18,000
the right-hand side.
And therefore,

379
00:25:14,000 --> 00:25:20,000
what is the conclusion?
If that's zero,

380
00:25:17,000 --> 00:25:23,000
it must be a solution to the
homogeneous equation.

381
00:25:21,000 --> 00:25:27,000
Therefore, u minus yp
is equal to,

382
00:25:24,000 --> 00:25:30,000
there must be c1 and c2.
I won't give them the generic

383
00:25:28,000 --> 00:25:34,000
names.
I'll give them a name,

384
00:25:30,000 --> 00:25:36,000
a particular one.
I'll put a tilde to indicate

385
00:25:34,000 --> 00:25:40,000
it's a particular one.
c1 plus c2 y2 tilde,

386
00:25:37,000 --> 00:25:43,000
so, in other words,
for some choice of these

387
00:25:41,000 --> 00:25:47,000
constants, and I'll call those
particular choices c1 tilde and

388
00:25:45,000 --> 00:25:51,000
c2 tilde, it must be that these
are equal.

389
00:25:48,000 --> 00:25:54,000
Well, what does that say?
It says that u is equal to yp

390
00:25:52,000 --> 00:25:58,000
plus c1 tilde,
blah, blah, blah,

391
00:25:54,000 --> 00:26:00,000
blah, plus c2 tilde,
blah, blah, blah,

392
00:25:57,000 --> 00:26:03,000
blah, and therefore chose that
u wasn't a new solution.

393
00:26:02,000 --> 00:26:08,000
It was one of these.
So, u isn't new.

394
00:26:08,000 --> 00:26:14,000
So, I should write it down.
Otherwise some of you will have

395
00:26:18,000 --> 00:26:24,000
missed the punch line.
Okay, therefore,

396
00:26:25,000 --> 00:26:31,000
u is equal to yp plus c1 tilde
y1 plus c2 tilde y2.

397
00:26:36,000 --> 00:26:42,000
And, it shows.
This guy who thought he was new

398
00:26:39,000 --> 00:26:45,000
was not new at all.
It was just one of the other

399
00:26:43,000 --> 00:26:49,000
solutions.
Okay, well, now,

400
00:26:45,000 --> 00:26:51,000
since the coefficient's a
constant, apparently we've done

401
00:26:49,000 --> 00:26:55,000
half the work.
We know what the complementary

402
00:26:52,000 --> 00:26:58,000
solution is because you know how
to do those in terms of

403
00:26:56,000 --> 00:27:02,000
exponentials and complex
exponentials,

404
00:26:59,000 --> 00:27:05,000
signs and cosines,
and so on.

405
00:27:03,000 --> 00:27:09,000
So, what's left to do?
All we have to do is find to

406
00:27:07,000 --> 00:27:13,000
solve equations,
which are inhomogeneous.

407
00:27:10,000 --> 00:27:16,000
All we have to do is find a
particular solution,

408
00:27:14,000 --> 00:27:20,000
find one solution.
It doesn't matter which one,

409
00:27:18,000 --> 00:27:24,000
any one.
Just find one,

410
00:27:20,000 --> 00:27:26,000
okay?
Now, we're going to spend the

411
00:27:23,000 --> 00:27:29,000
next two weeks trying to do
this.

412
00:27:26,000 --> 00:27:32,000
I'll give you various methods.
I'll give you a general method

413
00:27:32,000 --> 00:27:38,000
involving Fourier series because
it's a good excuse for learning

414
00:27:37,000 --> 00:27:43,000
what Fourier series are.
But, the answer is that in

415
00:27:41,000 --> 00:27:47,000
general, for a few standard
functions, it's known how to do

416
00:27:45,000 --> 00:27:51,000
this.
You will learn those methods

417
00:27:48,000 --> 00:27:54,000
for finding those using
operators.

418
00:27:50,000 --> 00:27:56,000
For all the others,
it's done by a series,

419
00:27:54,000 --> 00:28:00,000
or a method involving
approximation.

420
00:27:58,000 --> 00:28:04,000
Or, the worse comes to worst,
you throw it on a computer and

421
00:28:02,000 --> 00:28:08,000
just take a graph and the
numerical output of answers as

422
00:28:07,000 --> 00:28:13,000
the particular solution.
Okay, now before,

423
00:28:10,000 --> 00:28:16,000
we are going to start that
work, not today.

424
00:28:14,000 --> 00:28:20,000
We'll start it next Monday,
and it will last,

425
00:28:18,000 --> 00:28:24,000
as I say the next two weeks.
And, we will be up to spring

426
00:28:22,000 --> 00:28:28,000
break.
But, before we do that,

427
00:28:25,000 --> 00:28:31,000
I'd like to relate this to what
we did for first order equations

428
00:28:30,000 --> 00:28:36,000
because there is something to be
learned from that.

429
00:28:36,000 --> 00:28:42,000
Think back to the linear
first-order equation,

430
00:28:38,000 --> 00:28:44,000
and I'm going to,
since from now on for the rest

431
00:28:41,000 --> 00:28:47,000
of the period,
I'm going to be considering the

432
00:28:44,000 --> 00:28:50,000
case for constant coefficients.
In other words,

433
00:28:47,000 --> 00:28:53,000
this case of springs or
circuits or simple systems which

434
00:28:51,000 --> 00:28:57,000
behave like those and have
constant coefficients.

435
00:28:54,000 --> 00:29:00,000
So, for the linear,
first-order equation,

436
00:28:56,000 --> 00:29:02,000
there, too, I'm going to think
of constant coefficients.

437
00:29:01,000 --> 00:29:07,000
We talked quite a bit about
this equation.

438
00:29:04,000 --> 00:29:10,000
What did I call the right-hand
side?

439
00:29:06,000 --> 00:29:12,000
I think we usually called it q
of t,

440
00:29:09,000 --> 00:29:15,000
right?
This is in ancient history.

441
00:29:12,000 --> 00:29:18,000
The definition of ancient
history was before the first

442
00:29:16,000 --> 00:29:22,000
exam.
Okay, now how does that fit

443
00:29:18,000 --> 00:29:24,000
into this theorem that I've
given you?

444
00:29:21,000 --> 00:29:27,000
Remember what the solution
looked like.

445
00:29:25,000 --> 00:29:31,000
The solution looked like,
remember, you took the

446
00:29:28,000 --> 00:29:34,000
integrating factor was e to the
kt,

447
00:29:32,000 --> 00:29:38,000
and then after you integrated
both sides, multiplied through,

448
00:29:37,000 --> 00:29:43,000
and then the final answer
looked like this,

449
00:29:41,000 --> 00:29:47,000
y equaled, it was e to the
negative kt times

450
00:29:45,000 --> 00:29:51,000
either an indefinite integral,
or a definite integral

451
00:29:50,000 --> 00:29:56,000
depending on your preference,
q of t,

452
00:29:54,000 --> 00:30:00,000
so, x is metamorphosed into t.
I gather you've got that,

453
00:29:58,000 --> 00:30:04,000
e to the kt plus,
what was the other term?

454
00:30:02,000 --> 00:30:08,000
A constant times e to the
negative kt

455
00:30:08,000 --> 00:30:14,000
How does this fit into the
paradigm I've given you over

456
00:30:12,000 --> 00:30:18,000
there for solving the second
order equation?

457
00:30:15,000 --> 00:30:21,000
Which term is which?
Well, this has the arbitrary

458
00:30:19,000 --> 00:30:25,000
constant in it.
So, this must be the

459
00:30:22,000 --> 00:30:28,000
complementary solution.
Is it?

460
00:30:24,000 --> 00:30:30,000
Is this the solution to the
associated homogeneous equation?

461
00:30:30,000 --> 00:30:36,000
What's the associated
homogeneous equation?

462
00:30:32,000 --> 00:30:38,000
Put zero here.
Okay, if you put zero there,

463
00:30:35,000 --> 00:30:41,000
what's the solution?
Now, this you ought to know.

464
00:30:38,000 --> 00:30:44,000
y prime equals negative ky.

465
00:30:40,000 --> 00:30:46,000
What's the solution?
e to the negative kt.

466
00:30:43,000 --> 00:30:49,000
You are supposed to come into

467
00:30:46,000 --> 00:30:52,000
this course knowing that,
except there's an arbitrary

468
00:30:49,000 --> 00:30:55,000
constant in front.
So, right, this is exactly the

469
00:30:52,000 --> 00:30:58,000
solution to the associated
homogeneous equation,

470
00:30:55,000 --> 00:31:01,000
where there is zero here.
Then, what's this thing?

471
00:31:00,000 --> 00:31:06,000
This is a particular solution.
This is my yp.

472
00:31:02,000 --> 00:31:08,000
But that's not a particular
solution because this indefinite

473
00:31:06,000 --> 00:31:12,000
integral, you know,
has an arbitrary constant in

474
00:31:09,000 --> 00:31:15,000
it.
In fact, it's just that

475
00:31:10,000 --> 00:31:16,000
arbitrary constant.
So, it's totally confusing.

476
00:31:13,000 --> 00:31:19,000
But, this symbol,
you know when you actually

477
00:31:16,000 --> 00:31:22,000
solve the equation this way,
all you did was you found one

478
00:31:20,000 --> 00:31:26,000
function here.
You didn't throw in the

479
00:31:22,000 --> 00:31:28,000
arbitrary constant right away.
All you needed to do was find

480
00:31:26,000 --> 00:31:32,000
one function.
And, even if you really are

481
00:31:29,000 --> 00:31:35,000
bothered by the fact that this
is so indefinite,

482
00:31:32,000 --> 00:31:38,000
and therefore,
make it a particular solution

483
00:31:35,000 --> 00:31:41,000
by making this zero,
make it a definite integral,

484
00:31:38,000 --> 00:31:44,000
zero, here, t there,
and then change those t's to

485
00:31:42,000 --> 00:31:48,000
dummy t's, t1's or t tildes,
or something like that.

486
00:31:45,000 --> 00:31:51,000
So, this fits into that thing.
In other words,

487
00:31:48,000 --> 00:31:54,000
I could have done it at that
time, but I didn't the point

488
00:31:52,000 --> 00:31:58,000
because this can be solved
directly, whereas,

489
00:31:55,000 --> 00:32:01,000
of course, the general second
order equation in homogeneous

490
00:31:58,000 --> 00:32:04,000
cannot be solved directly,
and therefore you have to be

491
00:32:02,000 --> 00:32:08,000
willing to talk about what its
solutions look like in advance.

492
00:32:08,000 --> 00:32:14,000
Now, remember I said,
we talked, I said there was two

493
00:32:12,000 --> 00:32:18,000
different cases,
although both of them had the

494
00:32:15,000 --> 00:32:21,000
identical looking solution.
Their meaning in the physical

495
00:32:19,000 --> 00:32:25,000
world was so different that they
really should be considered as

496
00:32:24,000 --> 00:32:30,000
solving the same equation.
And, one of these was the case.

497
00:32:30,000 --> 00:32:36,000
Of the two, perhaps the more
important was the case when k

498
00:32:34,000 --> 00:32:40,000
was positive,
and of course the other is when

499
00:32:38,000 --> 00:32:44,000
k is negative.
When k is positive,

500
00:32:41,000 --> 00:32:47,000
that had the effect of
separating that solution into

501
00:32:45,000 --> 00:32:51,000
this part, which was a
transient, and the other part,

502
00:32:49,000 --> 00:32:55,000
which was a steady state.
The steady state solution,

503
00:32:53,000 --> 00:32:59,000
that was the yp part of it in
that terminology.

504
00:32:57,000 --> 00:33:03,000
And, the transient part,
it was trangent because it went

505
00:33:02,000 --> 00:33:08,000
to zero.
If k is positive,

506
00:33:05,000 --> 00:33:11,000
the exponential dies regardless
of what c is.

507
00:33:09,000 --> 00:33:15,000
So, the transient,
that's the yc part.

508
00:33:12,000 --> 00:33:18,000
It goes to zero as Ttgoes to
infinity.

509
00:33:16,000 --> 00:33:22,000
The transient depends on,
uses, the initial condition,

510
00:33:21,000 --> 00:33:27,000
whatever it is,
because that's what determines

511
00:33:25,000 --> 00:33:31,000
the value of c.
On the other hand,

512
00:33:28,000 --> 00:33:34,000
this initial condition makes no
difference as t goes towards

513
00:33:33,000 --> 00:33:39,000
infinity.
All that's left is this steady

514
00:33:38,000 --> 00:33:44,000
state solution.
And, all solutions tend to the

515
00:33:42,000 --> 00:33:48,000
steady state solution.
So, if k is positive,

516
00:33:46,000 --> 00:33:52,000
one gets this analysis of the
solutions into the sum of one

517
00:33:52,000 --> 00:33:58,000
basic solution,
and the others,

518
00:33:55,000 --> 00:34:01,000
which just die away,
have no influence on this,

519
00:33:59,000 --> 00:34:05,000
less and less influence as time
goes to infinity.

520
00:34:05,000 --> 00:34:11,000
For k less than zero,
this analysis does not work

521
00:34:09,000 --> 00:34:15,000
because this term,
if k is less than zero,

522
00:34:12,000 --> 00:34:18,000
this term goes to infinity or
negative infinity,

523
00:34:16,000 --> 00:34:22,000
and typically tends to dominate
that.

524
00:34:19,000 --> 00:34:25,000
So, it's the start that the
important one.

525
00:34:23,000 --> 00:34:29,000
It depends on the initial
conditions, and the analysis is

526
00:34:28,000 --> 00:34:34,000
meaningless.
So, the above is meaningless.

527
00:34:33,000 --> 00:34:39,000
And now, what I'd like to do is
try to see what the analog of

528
00:34:39,000 --> 00:34:45,000
that is for second order
equations, and higher order

529
00:34:45,000 --> 00:34:51,000
equations.
If you understand second-order,

530
00:34:49,000 --> 00:34:55,000
that's good enough.
Higher order goes exactly the

531
00:34:54,000 --> 00:35:00,000
same way.
So, the question is,

532
00:34:58,000 --> 00:35:04,000
for second-order,
let's make it with constant

533
00:35:02,000 --> 00:35:08,000
coefficients plus,
I could call it b and k,

534
00:35:07,000 --> 00:35:13,000
oh, no, b k,
or p.

535
00:35:11,000 --> 00:35:17,000
The trouble is,
that wouldn't take care of the

536
00:35:14,000 --> 00:35:20,000
electrical circuits.
So, I just want to use neutral

537
00:35:17,000 --> 00:35:23,000
letters, which suggest nothing.
And, you can make them turn it

538
00:35:22,000 --> 00:35:28,000
into a circuit,
so springs, or yet other

539
00:35:25,000 --> 00:35:31,000
examples undreamt of.
But these are constants.

540
00:35:28,000 --> 00:35:34,000
And I'm going to think of it as
time.

541
00:35:32,000 --> 00:35:38,000
I think I'll switch back to
time, let x be the time.

542
00:35:37,000 --> 00:35:43,000
So, B y equals f of t.
So, there is our equation.

543
00:35:43,000 --> 00:35:49,000
A and B are constants.
And, the question is,

544
00:35:47,000 --> 00:35:53,000
the question I'm asking,
can think of either of these

545
00:35:53,000 --> 00:35:59,000
two models or others,
the question I'm asking is,

546
00:35:58,000 --> 00:36:04,000
under what circumstances can I
make that same type of analysis

547
00:36:04,000 --> 00:36:10,000
into steady-state and transient?
Well, what does the solution

548
00:36:11,000 --> 00:36:17,000
look like?
The solution looks like y

549
00:36:15,000 --> 00:36:21,000
equals a particular solution
plus c1 y1 plus c2 y2.

550
00:36:20,000 --> 00:36:26,000
Therefore, to make that look

551
00:36:24,000 --> 00:36:30,000
like this, the c1 and c2 contain
the initial conditions.

552
00:36:31,000 --> 00:36:37,000
This part does not.
Therefore, if I want to say

553
00:36:35,000 --> 00:36:41,000
that the solutions look like a
steady state solution plus

554
00:36:41,000 --> 00:36:47,000
something that dies away,
which becomes less and less

555
00:36:47,000 --> 00:36:53,000
important as time goes on,
what I'm really asking is,

556
00:36:52,000 --> 00:36:58,000
under what circumstances is
this part guaranteed to go to

557
00:36:58,000 --> 00:37:04,000
zero?
So, the question is,

558
00:37:01,000 --> 00:37:07,000
when, in other words,
under what conditions on the

559
00:37:06,000 --> 00:37:12,000
equation A and B,
in effect, is what we are

560
00:37:10,000 --> 00:37:16,000
asking.
When does c1 y1 plus c2 y2 go

561
00:37:14,000 --> 00:37:20,000
to zero as t goes to infinity,

562
00:37:19,000 --> 00:37:25,000
regardless of what
c1 and c2 are for all c1 c2.

563
00:37:25,000 --> 00:37:31,000
Now, here there was no
difficulty.

564
00:37:30,000 --> 00:37:36,000
We had the thing very
explicitly, and you could see k

565
00:37:34,000 --> 00:37:40,000
is positive: this goes to zero.
And if k is negative,

566
00:37:38,000 --> 00:37:44,000
it doesn't go to zero.
It goes to infinity.

567
00:37:41,000 --> 00:37:47,000
Here, I want to make the same
kind of analysis,

568
00:37:44,000 --> 00:37:50,000
except it's just going to take,
it's a little more trouble.

569
00:37:49,000 --> 00:37:55,000
But the answer,
when it finally comes out is

570
00:37:52,000 --> 00:37:58,000
very beautiful.
So, when are all these guys

571
00:37:55,000 --> 00:38:01,000
going to go to zero?
First of all,

572
00:37:58,000 --> 00:38:04,000
you might as well just have the
definition.

573
00:38:01,000 --> 00:38:07,000
So, all the good things that
this is going to imply,

574
00:38:05,000 --> 00:38:11,000
if this is so,
in other words,

575
00:38:07,000 --> 00:38:13,000
if they all go to zero,
everything in the complementary

576
00:38:12,000 --> 00:38:18,000
solution, then the ODE is called
stable.

577
00:38:17,000 --> 00:38:23,000
Some people call it
asymptotically stable.

578
00:38:21,000 --> 00:38:27,000
I don't know what to call it.
I can make the analysis,

579
00:38:28,000 --> 00:38:34,000
and then I use the identical
terminology, c1 y1 plus c2 y2.

580
00:38:36,000 --> 00:38:42,000
This is called the transient
because it goes to zero.

581
00:38:40,000 --> 00:38:46,000
This is called the particular
solution now that we labored so

582
00:38:45,000 --> 00:38:51,000
hard to get for the next two
weeks.

583
00:38:47,000 --> 00:38:53,000
It's the important part.
It's the steady-state part.

584
00:38:52,000 --> 00:38:58,000
It's what lasts out to infinity
after the other stuff has

585
00:38:56,000 --> 00:39:02,000
disappeared.
So, this is the steady-state

586
00:38:59,000 --> 00:39:05,000
solution, steady-state solution,
okay?

587
00:39:04,000 --> 00:39:10,000
And, the differential equation
is called stable.

588
00:39:07,000 --> 00:39:13,000
Now, it's of the highest
interest to know when a

589
00:39:10,000 --> 00:39:16,000
differential equation is stable,
linear differential equation is

590
00:39:14,000 --> 00:39:20,000
stable in this sense because you
have a control.

591
00:39:17,000 --> 00:39:23,000
You know what its solutions
look like.

592
00:39:20,000 --> 00:39:26,000
You have some feeling for how
it's behaving in the long term.

593
00:39:24,000 --> 00:39:30,000
If this is not so,
each equation is a law unto

594
00:39:27,000 --> 00:39:33,000
itself if you don't know.
So, let's do the work.

595
00:39:31,000 --> 00:39:37,000
For the rest of the period,
what I'd like to do is to find

596
00:39:36,000 --> 00:39:42,000
out what the conditions are,
which make this true.

597
00:39:40,000 --> 00:39:46,000
Those were the equations which
we will have a right to call

598
00:39:45,000 --> 00:39:51,000
stable.
So, when does this happen,

599
00:39:48,000 --> 00:39:54,000
and where is it going to
happen?

600
00:39:50,000 --> 00:39:56,000
I don't know.
I guess, here.

601
00:40:05,000 --> 00:40:11,000
Now, I think the first step is
fairly easy, and it will give

602
00:40:10,000 --> 00:40:16,000
you a good review of what we've
been doing up until now.

603
00:40:14,000 --> 00:40:20,000
So, I'm simply going to make a
case-by-case analysis.

604
00:40:19,000 --> 00:40:25,000
Don't worry,
it won't take very long.

605
00:40:22,000 --> 00:40:28,000
What are the cases we've been
studying?

606
00:40:25,000 --> 00:40:31,000
Well, what do the
characteristic roots look like?

607
00:40:29,000 --> 00:40:35,000
The roots of the characteristic
equation, in other words,

608
00:40:34,000 --> 00:40:40,000
remember, there are cases.
The first case is they are real

609
00:40:41,000 --> 00:40:47,000
and distinct,
r1 not equal to r2,

610
00:40:45,000 --> 00:40:51,000
real and distinct.
What are the other cases?

611
00:40:50,000 --> 00:40:56,000
Well, r1 equals r2.
And then, there's the case

612
00:40:56,000 --> 00:41:02,000
where there are complex.
So, I will write it r equals a

613
00:41:02,000 --> 00:41:08,000
plus or minus b i.
What do the solutions look

614
00:41:07,000 --> 00:41:13,000
like?
So, my ham-handed approach to

615
00:41:09,000 --> 00:41:15,000
this problem is going to be,
in each case,

616
00:41:12,000 --> 00:41:18,000
I'll look at the solutions,
and first get the condition on

617
00:41:16,000 --> 00:41:22,000
the roots.
So, in other words,

618
00:41:18,000 --> 00:41:24,000
I'm not going to worry right
away about the a and the b.

619
00:41:21,000 --> 00:41:27,000
I'm going, instead,
to worry about expressing this

620
00:41:24,000 --> 00:41:30,000
condition of stability in terms
of the characteristic roots.

621
00:41:28,000 --> 00:41:34,000
In fact, that's the only way in
which many people know the

622
00:41:32,000 --> 00:41:38,000
conditions.
You're going to be smarter.

623
00:41:35,000 --> 00:41:41,000
Okay, what do the solutions
look like?

624
00:41:38,000 --> 00:41:44,000
Well, the general solution
looks like e to the r1 t plus c2

625
00:41:42,000 --> 00:41:48,000
e to the r2 t.

626
00:41:45,000 --> 00:41:51,000
Okay, so, what's the stability
condition?

627
00:41:48,000 --> 00:41:54,000
In other words,
if equation happened to have

628
00:41:51,000 --> 00:41:57,000
its characteristic roots,
real and distinct,

629
00:41:54,000 --> 00:42:00,000
under what circumstances would
it be stable?

630
00:41:57,000 --> 00:42:03,000
Would it, in other words,
all its solutions go to zero?

631
00:42:01,000 --> 00:42:07,000
So, I'm talking about the
homogeneous equation,

632
00:42:04,000 --> 00:42:10,000
the reduced equation,
the associated homogeneous

633
00:42:07,000 --> 00:42:13,000
equation.
Why?

634
00:42:10,000 --> 00:42:16,000
Because that's all that's
involved in this.

635
00:42:13,000 --> 00:42:19,000
In other words,
when I write that,

636
00:42:15,000 --> 00:42:21,000
I am no longer interested in
the whole equation.

637
00:42:19,000 --> 00:42:25,000
All I'm interested in is the
reduced equation,

638
00:42:22,000 --> 00:42:28,000
the equation where you turn the
f of t on the

639
00:42:26,000 --> 00:42:32,000
right-hand side into zero.
So, what's the stability

640
00:42:31,000 --> 00:42:37,000
condition?
Well, let's write it out.

641
00:42:35,000 --> 00:42:41,000
Under what circumstances will
all these guys go to zero?

642
00:42:41,000 --> 00:42:47,000
If r1 and r2 should be
negative, can they be zero?

643
00:42:46,000 --> 00:42:52,000
No, because then it will be a
constant and it will go to zero.

644
00:42:52,000 --> 00:42:58,000
How about this one?
Well, in this one,

645
00:42:56,000 --> 00:43:02,000
it's (c1 plus c2 times t)
multiplied by e to the r1 t.

646
00:43:01,000 --> 00:43:07,000
Of course, both of these are

647
00:43:08,000 --> 00:43:14,000
the same.
I'll just arbitrarily pick one

648
00:43:11,000 --> 00:43:17,000
of them.
What happens to this as things

649
00:43:14,000 --> 00:43:20,000
go to zero?
Well, this part is rising,

650
00:43:16,000 --> 00:43:22,000
at least if c2 is positive.
This part is either helping or

651
00:43:21,000 --> 00:43:27,000
it's hindering.
But, I hope you know what these

652
00:43:24,000 --> 00:43:30,000
functions look like,
and you know which of them go

653
00:43:28,000 --> 00:43:34,000
to zero.
They go to zero if r1 is

654
00:43:31,000 --> 00:43:37,000
negative.
It might rise in the beginning,

655
00:43:35,000 --> 00:43:41,000
but after a while they lose the
energy.

656
00:43:39,000 --> 00:43:45,000
Of course, if r1 is equal to
zero, what do these guys do?

657
00:43:44,000 --> 00:43:50,000
Linear, go to infinity.
Well, we are doing okay.

658
00:43:49,000 --> 00:43:55,000
How about here?
Well, here, it's a little more

659
00:43:53,000 --> 00:43:59,000
complicated.
The solutions look like e to

660
00:43:57,000 --> 00:44:03,000
the at times (c1 cosine bt plus
c2 sine bt).

661
00:44:03,000 --> 00:44:09,000
Now, this part is a pure

662
00:44:07,000 --> 00:44:13,000
oscillation.
You know that.

663
00:44:09,000 --> 00:44:15,000
It might have a big amplitude,
but whatever it does,

664
00:44:13,000 --> 00:44:19,000
it does the same thing all the
time.

665
00:44:16,000 --> 00:44:22,000
So, whether this goes to zero
depends entirely upon what that

666
00:44:20,000 --> 00:44:26,000
exponential is doing.
And, that exponential goes to

667
00:44:24,000 --> 00:44:30,000
zero if a is negative.
So here, the condition is

668
00:44:28,000 --> 00:44:34,000
negative.
And now, the only thing left to

669
00:44:33,000 --> 00:44:39,000
do is to say it nicely.
I've got three cases,

670
00:44:38,000 --> 00:44:44,000
and I want to say them all in
one breath.

671
00:44:43,000 --> 00:44:49,000
So, the stability condition is,
the ODE is stable.

672
00:44:49,000 --> 00:44:55,000
So, this is,
or f of t.

673
00:44:53,000 --> 00:44:59,000
It doesn't matter.
But, psychologically,

674
00:44:57,000 --> 00:45:03,000
you can put this as zero there,
is stable if what?

675
00:45:05,000 --> 00:45:11,000
In case one,
this is true.

676
00:45:07,000 --> 00:45:13,000
In case two,
that's true.

677
00:45:10,000 --> 00:45:16,000
In case three,
that's true.

678
00:45:13,000 --> 00:45:19,000
But that's ugly.
Make it beautiful.

679
00:45:17,000 --> 00:45:23,000
The beautiful way of saying it
is if all the characteristic

680
00:45:23,000 --> 00:45:29,000
roots have negative real parts.
If the characteristic roots,

681
00:45:31,000 --> 00:45:37,000
the r's or the a plus or minus
b i, have negative real part.

682
00:45:38,000 --> 00:45:44,000
That's the form in which the
electrical engineers will nod

683
00:45:44,000 --> 00:45:50,000
their head, tell you,
yeah, that's right,

684
00:45:49,000 --> 00:45:55,000
negative real part,
sorry.

685
00:45:52,000 --> 00:45:58,000
Isn't it right?
Is that right here?

686
00:45:56,000 --> 00:46:02,000
Yeah.
What's the real part of these

687
00:45:59,000 --> 00:46:05,000
guys?
They themselves,

688
00:46:03,000 --> 00:46:09,000
because they are real.
What's the real part of this?

689
00:46:09,000 --> 00:46:15,000
Yeah.
The only case in which I really

690
00:46:13,000 --> 00:46:19,000
had to use real part is when I
talk about the complex case

691
00:46:20,000 --> 00:46:26,000
because a is just the real part
of a complex number.

692
00:46:26,000 --> 00:46:32,000
It's not the whole thing.