WEBVTT
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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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differential,
I'll simply write it out
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instead of writing out all the
words which go with it.
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So, such an equation looks
like, the second-order equation
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is going to look like y double
prime plus p of x,
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t, x plus q of x times y.
Now, up to now the right-hand
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side has been zero.
So, now we are going to make it
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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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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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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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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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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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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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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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case.
But the homogeneous one,
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the corresponding homogeneous
thing, y double prime plus p of
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x y prime plus q of x times y
equals zero
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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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but I think I'll probably call
it the associated homogeneous
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equation, or ODE,
the associated homogeneous
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equation, the one associated to
the guy on the left.
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It's also called the reduced
equation by some people.
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There is some other term for
it, which escapes me totally,
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but what the heck.
Now, its solution has a name.
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So, its solution,
of course, doesn't depend on
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anything in particular,
the general solution,
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because the right-hand side is
always zero.
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So, its solution,
we know can be written as y
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equals in the form c1 y1 plus c2
y2,
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where y1 and y2 are any two
independent solutions of that,
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and then c1's and c2's are
arbitrary constants.
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Now, what you are looking at
this equation,
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you're going to need this also.
And therefore,
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it has a name.
It has various names.
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Sometimes there is a subscript,
c, there.
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Sometimes there's a subscript,
h.
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Sometimes there's no subscript
at all, which is the most
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confusing of all.
But, anyway,
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what's the name given to it?
Well, there is no name.
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Many books call it the solution
to the associated homogeneous
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equation.
That's maximally long.
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Your book calls it the
complementary solution.
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Many people call it that,
and many will look at you with
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a blank, who know differential
equations very well,
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and will not have the faintest
idea what you're talking about.
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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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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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what solutions of these things
should look like and how they
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should behave.
So, the model you know already
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is the one, I won't make the
leading coefficient one because
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it usually isn't,
is the one, m x double prime,
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so t is the
independent variable,
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plus b x prime plus k x equals
f of t.
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That's the spring-mass system,
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the spring-mass-dashpot system.
Mass, the damping constant and
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the spring constant,
except up to now,
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it's always been zero here.
What does this f of t
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represent?
Well, if you think of the way
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in which I derived the equation,
the mx, that was the Newton's
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law.
That's the acceleration.
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So, it's the acceleration,
the mass times the
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acceleration.
By Newton's law,
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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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I'm not going to draw the truck
for the nth time.
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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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There was a force,
minus b x prime.
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That came from the dashpot,
the damping force.
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So, this other guy is f of t.
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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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electromagnet,
and the mass it's carrying is a
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big pile of iron ore.
You're turning it on and off,
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and pulling that thing from
afar where nobody can see it.
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So, this is the external force.
Now, think, that is the model
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you must have in your mind of
how these equations are treated.
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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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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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now, you have to both,
even if you wanted to solve the
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forced system,
the way the system would behave
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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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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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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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officially, but you will,
a resistance,
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sorry, that's okay,
put the capacitance up there,
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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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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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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,
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
00:11:25.000 --> 00:11:31.000
in 8.02.
You will, one day,
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learn the voltage drop across
this.
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But, I already know it.
It's Li.
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So, i is the current.
I'll write this thing in its
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primitive form first.
So, i is the current that's
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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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Li prime,
the voltage drop across the
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resistance is,
well, you know that.
00:12:07.000 --> 00:12:13.000
And, the voltage shop across
the capacitance is q
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divided by C.
And so, that's equal to,
00:12:13.000 --> 00:12:19.000
well, it's equal to zero,
except if there's a battery
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here or something generating a
voltage drop,
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so, let's call that E is a
generic word.
00:12:21.000 --> 00:12:27.000
E could be a battery.
It could be a source of
00:12:24.000 --> 00:12:30.000
alternating current,
something like that.
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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name of the voltage drop.
So, and then there's the
00:12:35.000 --> 00:12:41.000
question of the signs,
which I know I'll never
00:12:38.000 --> 00:12:44.000
understand.
But, let's assume you've chosen
00:12:41.000 --> 00:12:47.000
the sign convention so that this
comes out nicely on the
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right-hand side.
So, this might be varying
00:12:47.000 --> 00:12:53.000
sinusoidally,
in which case you'd have source
00:12:50.000 --> 00:12:56.000
of alternating current.
Or, it might be constant,
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in which that would be a
battery, a little dry cell
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giving you direct current of a
constant voltage,
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stuff like that.
So, you could make this minus
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if you want, but everything will
have the wrong signs,
00:13:07.000 --> 00:13:13.000
so don't do it.
Now, this doesn't look like
00:13:09.000 --> 00:13:15.000
what it's supposed to look like
because it's got q and i.
00:13:13.000 --> 00:13:19.000
So, the final thing you have to
know is that q prime is
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equal to i.
The rate at which that charge
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leaves the condenser and hurries
around the circuit to find its
00:13:23.000 --> 00:13:29.000
little soul mate on the other
side is the current that's
00:13:27.000 --> 00:13:33.000
flowing in the circuit.
That's why current flows,
00:13:30.000 --> 00:13:36.000
except nothing really happens.
Electrons just push on each
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other, and they stay where they
are.
00:13:37.000 --> 00:13:43.000
I don't understand this at all.
So, if I differentiate this,
00:13:42.000 --> 00:13:48.000
you can do two things.
Either you could integrate i,
00:13:46.000 --> 00:13:52.000
and expressed the thing
entirely in terms of q,
00:13:50.000 --> 00:13:56.000
or you can differentiate it,
and express everything in terms
00:13:54.000 --> 00:14:00.000
of i.
Your book does nicely both,
00:13:57.000 --> 00:14:03.000
does not take sides.
So, let's differentiate it,
00:14:02.000 --> 00:14:08.000
and then it will look like L i
double prime plus R i prime plus
00:14:06.000 --> 00:14:12.000
i divided by C equals,
00:14:09.000 --> 00:14:15.000
and now, watch out,
you have now not the
00:14:12.000 --> 00:14:18.000
electromotive force,
but its derivative.
00:14:15.000 --> 00:14:21.000
So, if you were so unfortunate
as to put a little dry cell
00:14:19.000 --> 00:14:25.000
there, now you've got nothing,
and you've got the homogeneous
00:14:24.000 --> 00:14:30.000
case.
That's okay.
00:14:25.000 --> 00:14:31.000
Where are the erasers?
One eraser?
00:14:28.000 --> 00:14:34.000
I don't believe this.
00:14:43.000 --> 00:14:49.000
So, there's the equation.
There are our two equations.
00:14:47.000 --> 00:14:53.000
Why don't we put them up in
colored chalk.
00:14:50.000 --> 00:14:56.000
There's the spring equation.
And, here's the equation that
00:14:54.000 --> 00:15:00.000
governs the current,
for how the current flows in
00:14:58.000 --> 00:15:04.000
that circuit.
And now, you can see,
00:15:01.000 --> 00:15:07.000
again, what does it mean?
If this is zero,
00:15:04.000 --> 00:15:10.000
for example,
if I have a dry cell there,
00:15:07.000 --> 00:15:13.000
or if I have nothing at all in
the circuit, then this
00:15:11.000 --> 00:15:17.000
represents the passive circuit.
It's just sitting there.
00:15:16.000 --> 00:15:22.000
It wouldn't do anything at all,
except that you've put a charge
00:15:20.000 --> 00:15:26.000
on the capacitor,
and waited, and of course,
00:15:22.000 --> 00:15:28.000
when you put a charge on there,
it's got a discharge,
00:15:26.000 --> 00:15:32.000
and discharges through the
circuit, and swings back and
00:15:29.000 --> 00:15:35.000
forth a little bit if it's
under-damped until finally
00:15:32.000 --> 00:15:38.000
towards the end the current dies
away to zero.
00:15:36.000 --> 00:15:42.000
But, what usually happens is
that you drive this passive
00:15:39.000 --> 00:15:45.000
circuit by putting an effective
E in it, and then you want to
00:15:44.000 --> 00:15:50.000
know how the current behaves.
So, those are the two problems,
00:15:48.000 --> 00:15:54.000
the passive circuit without an
applied electromotive force,
00:15:52.000 --> 00:15:58.000
or plugging it into the wall,
and wanting it to do things.
00:15:56.000 --> 00:16:02.000
That's the normal state of
affairs.
00:16:00.000 --> 00:16:06.000
People don't want passive
circuits, they want circuits
00:16:03.000 --> 00:16:09.000
which do things because,
okay, that's why they want to
00:16:07.000 --> 00:16:13.000
solve inhomogeneous equations
instead of homogeneous
00:16:11.000 --> 00:16:17.000
equations.
But as I said,
00:16:13.000 --> 00:16:19.000
you have to do the homogeneous
case first.
00:16:16.000 --> 00:16:22.000
Okay, you are now officially
responsible for this,
00:16:20.000 --> 00:16:26.000
and I don't care that you
haven't had it in physics yet.
00:16:24.000 --> 00:16:30.000
You will before the next exam.
So, I don't even feel guilty.
00:16:30.000 --> 00:16:36.000
But, you're going to start
using it on the problem set
00:16:34.000 --> 00:16:40.000
right away.
So, it's never too soon to
00:16:37.000 --> 00:16:43.000
start learning it.
Okay, now, the main theorem,
00:16:41.000 --> 00:16:47.000
I now want to go,
so that was just examples to
00:16:44.000 --> 00:16:50.000
give you some physical feeling
for the sorts of differential
00:16:49.000 --> 00:16:55.000
equations we'll be talking
about.
00:16:52.000 --> 00:16:58.000
I now want to tell you briefly
about the key theorem about
00:16:56.000 --> 00:17:02.000
solving the homogeneous
equation.
00:16:59.000 --> 00:17:05.000
So, the main theorem about
solving the homogeneous equation
00:17:04.000 --> 00:17:10.000
is, the inhomogeneous equation.
So, I'm going to write the
00:17:09.000 --> 00:17:15.000
inhomogeneous equation out.
I'm going to make the left-hand
00:17:14.000 --> 00:17:20.000
side a linear operator,
and am going to write the
00:17:18.000 --> 00:17:24.000
equation as Ly equals f of x.
00:17:21.000 --> 00:17:27.000
That's the inhomogeneous
equation.
00:17:23.000 --> 00:17:29.000
So, L is the linear operator,
second order because I'm only
00:17:28.000 --> 00:17:34.000
talking about second-order
equations.
00:17:32.000 --> 00:17:38.000
L is a linear operator,
and then this is the
00:17:37.000 --> 00:17:43.000
differential equation.
So, here's our differential
00:17:43.000 --> 00:17:49.000
equation.
It's inhomogeneous because it's
00:17:48.000 --> 00:17:54.000
go the f of x on the
right hand side.
00:17:53.000 --> 00:17:59.000
And, what the theorem says is
that the solution has the
00:18:00.000 --> 00:18:06.000
following form,
y sub p, I'll explain what that
00:18:05.000 --> 00:18:11.000
is in just a moment,
plus y sub c.
00:18:13.000 --> 00:18:19.000
So, the hypothesis is we've got
the linear equation,
00:18:17.000 --> 00:18:23.000
and the conclusion is that
that's what its solution looks
00:18:21.000 --> 00:18:27.000
like.
Now, you already know what y
00:18:24.000 --> 00:18:30.000
sub c looks like.
In other words,
00:18:27.000 --> 00:18:33.000
if I write this out in more
detail, it would be i.e.,
00:18:31.000 --> 00:18:37.000
department of fuller
explanation, --
00:18:35.000 --> 00:18:41.000
-- the general solution looks
like y equals yp,
00:18:38.000 --> 00:18:44.000
and then this thing is going to
look like an arbitrary constant
00:18:43.000 --> 00:18:49.000
times y1 plus an arbitrary
constant times y2,
00:18:47.000 --> 00:18:53.000
where these are solutions of
the homogeneous equation.
00:18:51.000 --> 00:18:57.000
So, Yc looks like this part,
and the yp, what's yp?
00:18:55.000 --> 00:19:01.000
p stands for particular,
the most confusing word in this
00:19:00.000 --> 00:19:06.000
subject.
But, you've got at least four
00:19:04.000 --> 00:19:10.000
weeks to learn what it means.
Okay, yp is a particular
00:19:10.000 --> 00:19:16.000
solution to Ly equals f of x.
00:19:14.000 --> 00:19:20.000
Now, I'm not going to explain
what particular means.
00:19:19.000 --> 00:19:25.000
First, I'll chat as if you knew
what it meant,
00:19:24.000 --> 00:19:30.000
and then we'll see if you have
picked it up.
00:19:30.000 --> 00:19:36.000
In other words,
the procedure for solving this
00:19:33.000 --> 00:19:39.000
equation is composed of two
steps.
00:19:36.000 --> 00:19:42.000
First, to find this part.
In other words,
00:19:40.000 --> 00:19:46.000
to find the complementary
solution, in other words,
00:19:44.000 --> 00:19:50.000
to do what we've been doing for
the last week,
00:19:48.000 --> 00:19:54.000
solve not the equation you are
given, but the reduced equation.
00:19:53.000 --> 00:19:59.000
So, the first step is to find
this.
00:19:56.000 --> 00:20:02.000
The second step is to find yp.
Now, what's yp?
00:20:01.000 --> 00:20:07.000
yp is a particular solution to
the whole equation.
00:20:05.000 --> 00:20:11.000
Yeah, but which one?
Well, if it's any one,
00:20:09.000 --> 00:20:15.000
then it's not a particular
solution, yeah.
00:20:12.000 --> 00:20:18.000
I say, unfortunately the word
particular here is not being
00:20:17.000 --> 00:20:23.000
used in exactly the same sense
in which most people use it in
00:20:22.000 --> 00:20:28.000
ordinary English.
It's a perfectly valid way to
00:20:26.000 --> 00:20:32.000
use it.
It's just confusing,
00:20:29.000 --> 00:20:35.000
and no one has ever come up
with a better word.
00:20:33.000 --> 00:20:39.000
So, particular means any one
solution.
00:20:38.000 --> 00:20:44.000
Any one will do.
Okay, even these have slightly
00:20:45.000 --> 00:20:51.000
different meanings.
Any questions about this?
00:20:51.000 --> 00:20:57.000
I refuse to answer them.
[LAUGHTER]
00:20:58.000 --> 00:21:04.000
Now, well, examples of course
will make it all clear.
00:21:03.000 --> 00:21:09.000
But I'd like,
first, to prove the theorem,
00:21:08.000 --> 00:21:14.000
to show you how simple it is.
It's extremely simple if you
00:21:15.000 --> 00:21:21.000
just use the fact that L is a
linear operator.
00:21:20.000 --> 00:21:26.000
We've got two things to prove.
What have we got to prove?
00:21:26.000 --> 00:21:32.000
Well, I have to prove two
statements, first of all,
00:21:32.000 --> 00:21:38.000
that all the yp plus c1 y1 plus
c2 y2 are
00:21:39.000 --> 00:21:45.000
solutions.
How are we going to prove that?
00:21:43.000 --> 00:21:49.000
Well, how do you know if
something is a solution?
00:21:46.000 --> 00:21:52.000
Well, you plug it into the
equation, and you see if it
00:21:49.000 --> 00:21:55.000
satisfies the equation.
Good, let's do it,
00:21:52.000 --> 00:21:58.000
proof.
L, I'm going to plug it into
00:21:54.000 --> 00:22:00.000
the equation.
That means I calculate L of yp
00:21:56.000 --> 00:22:02.000
plus c1 y1 plus c2 y2.
00:22:00.000 --> 00:22:06.000
Now, what's the answer?
Because this is a linear
00:22:04.000 --> 00:22:10.000
operator, and notice,
the argument doesn't use the
00:22:08.000 --> 00:22:14.000
fact that the equation is second
order.
00:22:12.000 --> 00:22:18.000
It immediately generalizes to a
linear equation of any order,
00:22:17.000 --> 00:22:23.000
whatever-- 47.
Okay, this is L of yp plus L of
00:22:21.000 --> 00:22:27.000
c1 y1 plus c2 y2.
00:22:25.000 --> 00:22:31.000
Well, what's that?
What's L of the complementary
00:22:29.000 --> 00:22:35.000
solution?
What does it mean to be the
00:22:34.000 --> 00:22:40.000
complementary solution?
It means when you apply the
00:22:37.000 --> 00:22:43.000
operator L to it,
you get zero because this
00:22:40.000 --> 00:22:46.000
satisfies the homogeneous
equation.
00:22:43.000 --> 00:22:49.000
So, this is zero.
What's L of yp?
00:22:46.000 --> 00:22:52.000
Well, it was a particular
solution to the equation.
00:22:50.000 --> 00:22:56.000
Therefore, when I plugged it
into the equation,
00:22:53.000 --> 00:22:59.000
I must have gotten out on the
right-hand side,
00:22:57.000 --> 00:23:03.000
f of x.
So, this is since yp is a
00:23:00.000 --> 00:23:06.000
solution to the whole equation.
So, what's the conclusion?
00:23:06.000 --> 00:23:12.000
That, if I take any one of
these guys, no matter what c1
00:23:10.000 --> 00:23:16.000
and c2 are, apply the linear
operator, L to it,
00:23:13.000 --> 00:23:19.000
the answer comes out
to be f of.
00:23:17.000 --> 00:23:23.000
Therefore, this proves that
this shows that these are all
00:23:21.000 --> 00:23:27.000
solutions because that's what it
means.
00:23:24.000 --> 00:23:30.000
Therefore, they satisfy L of y
equals f of x.
00:23:30.000 --> 00:23:36.000
They satisfy the whole
inhomogeneous differential
00:23:34.000 --> 00:23:40.000
equation, and that's it.
Well, that's only half the
00:23:38.000 --> 00:23:44.000
story.
The other half of the story is
00:23:41.000 --> 00:23:47.000
to show that there are no other
solutions.
00:23:45.000 --> 00:23:51.000
Okay, so we got our little u of
x coming up again,
00:23:50.000 --> 00:23:56.000
and he thinks he's a solution.
Okay, so, to prove there are no
00:23:55.000 --> 00:24:01.000
other solutions,
it almost sounds biblical,
00:23:59.000 --> 00:24:05.000
thou shalt have no other
solutions before me,
00:24:03.000 --> 00:24:09.000
okay.
There are no other solutions
00:24:07.000 --> 00:24:13.000
accept these guys for different
values of c1 and c2.
00:24:10.000 --> 00:24:16.000
Okay, so, u of x is a
solution.
00:24:13.000 --> 00:24:19.000
I have to show that u of x is
one of these guys.
00:24:16.000 --> 00:24:22.000
How am I going to do that?
Easy.
00:24:19.000 --> 00:24:25.000
If it's a solution that,
L of u,
00:24:21.000 --> 00:24:27.000
okay, I'm going to drop the x,
okay, just to make the,
00:24:25.000 --> 00:24:31.000
like I dropped the x over
there.
00:24:29.000 --> 00:24:35.000
If it's a solution to the whole
inhomogeneous equation,
00:24:33.000 --> 00:24:39.000
then this must come out to be f
of x.
00:24:37.000 --> 00:24:43.000
Now, what's L of yp?
That's f of x too,
00:24:41.000 --> 00:24:47.000
by secret little particular
solution I've got in my pocket.
00:24:47.000 --> 00:24:53.000
Okay, I pull it out,
ah-ha, L of yp,
00:24:50.000 --> 00:24:56.000
that's f of x,
too.
00:24:51.000 --> 00:24:57.000
Now, I'm going to not add them.
I'm going to subtract them.
00:24:56.000 --> 00:25:02.000
What is L of u minus yp?
00:25:00.000 --> 00:25:06.000
Well, it's zero.
It's zero because this is a
00:25:05.000 --> 00:25:11.000
linear operator.
This would be L of u minus L of
00:25:08.000 --> 00:25:14.000
yp.
I guess the answer is zero on
00:25:12.000 --> 00:25:18.000
the right-hand side.
And therefore,
00:25:14.000 --> 00:25:20.000
what is the conclusion?
If that's zero,
00:25:17.000 --> 00:25:23.000
it must be a solution to the
homogeneous equation.
00:25:21.000 --> 00:25:27.000
Therefore, u minus yp
is equal to,
00:25:24.000 --> 00:25:30.000
there must be c1 and c2.
I won't give them the generic
00:25:28.000 --> 00:25:34.000
names.
I'll give them a name,
00:25:30.000 --> 00:25:36.000
a particular one.
I'll put a tilde to indicate
00:25:34.000 --> 00:25:40.000
it's a particular one.
c1 plus c2 y2 tilde,
00:25:37.000 --> 00:25:43.000
so, in other words,
for some choice of these
00:25:41.000 --> 00:25:47.000
constants, and I'll call those
particular choices c1 tilde and
00:25:45.000 --> 00:25:51.000
c2 tilde, it must be that these
are equal.
00:25:48.000 --> 00:25:54.000
Well, what does that say?
It says that u is equal to yp
00:25:52.000 --> 00:25:58.000
plus c1 tilde,
blah, blah, blah,
00:25:54.000 --> 00:26:00.000
blah, plus c2 tilde,
blah, blah, blah,
00:25:57.000 --> 00:26:03.000
blah, and therefore chose that
u wasn't a new solution.
00:26:02.000 --> 00:26:08.000
It was one of these.
So, u isn't new.
00:26:08.000 --> 00:26:14.000
So, I should write it down.
Otherwise some of you will have
00:26:18.000 --> 00:26:24.000
missed the punch line.
Okay, therefore,
00:26:25.000 --> 00:26:31.000
u is equal to yp plus c1 tilde
y1 plus c2 tilde y2.
00:26:36.000 --> 00:26:42.000
And, it shows.
This guy who thought he was new
00:26:39.000 --> 00:26:45.000
was not new at all.
It was just one of the other
00:26:43.000 --> 00:26:49.000
solutions.
Okay, well, now,
00:26:45.000 --> 00:26:51.000
since the coefficient's a
constant, apparently we've done
00:26:49.000 --> 00:26:55.000
half the work.
We know what the complementary
00:26:52.000 --> 00:26:58.000
solution is because you know how
to do those in terms of
00:26:56.000 --> 00:27:02.000
exponentials and complex
exponentials,
00:26:59.000 --> 00:27:05.000
signs and cosines,
and so on.
00:27:03.000 --> 00:27:09.000
So, what's left to do?
All we have to do is find to
00:27:07.000 --> 00:27:13.000
solve equations,
which are inhomogeneous.
00:27:10.000 --> 00:27:16.000
All we have to do is find a
particular solution,
00:27:14.000 --> 00:27:20.000
find one solution.
It doesn't matter which one,
00:27:18.000 --> 00:27:24.000
any one.
Just find one,
00:27:20.000 --> 00:27:26.000
okay?
Now, we're going to spend the
00:27:23.000 --> 00:27:29.000
next two weeks trying to do
this.
00:27:26.000 --> 00:27:32.000
I'll give you various methods.
I'll give you a general method
00:27:32.000 --> 00:27:38.000
involving Fourier series because
it's a good excuse for learning
00:27:37.000 --> 00:27:43.000
what Fourier series are.
But, the answer is that in
00:27:41.000 --> 00:27:47.000
general, for a few standard
functions, it's known how to do
00:27:45.000 --> 00:27:51.000
this.
You will learn those methods
00:27:48.000 --> 00:27:54.000
for finding those using
operators.
00:27:50.000 --> 00:27:56.000
For all the others,
it's done by a series,
00:27:54.000 --> 00:28:00.000
or a method involving
approximation.
00:27:58.000 --> 00:28:04.000
Or, the worse comes to worst,
you throw it on a computer and
00:28:02.000 --> 00:28:08.000
just take a graph and the
numerical output of answers as
00:28:07.000 --> 00:28:13.000
the particular solution.
Okay, now before,
00:28:10.000 --> 00:28:16.000
we are going to start that
work, not today.
00:28:14.000 --> 00:28:20.000
We'll start it next Monday,
and it will last,
00:28:18.000 --> 00:28:24.000
as I say the next two weeks.
And, we will be up to spring
00:28:22.000 --> 00:28:28.000
break.
But, before we do that,
00:28:25.000 --> 00:28:31.000
I'd like to relate this to what
we did for first order equations
00:28:30.000 --> 00:28:36.000
because there is something to be
learned from that.
00:28:36.000 --> 00:28:42.000
Think back to the linear
first-order equation,
00:28:38.000 --> 00:28:44.000
and I'm going to,
since from now on for the rest
00:28:41.000 --> 00:28:47.000
of the period,
I'm going to be considering the
00:28:44.000 --> 00:28:50.000
case for constant coefficients.
In other words,
00:28:47.000 --> 00:28:53.000
this case of springs or
circuits or simple systems which
00:28:51.000 --> 00:28:57.000
behave like those and have
constant coefficients.
00:28:54.000 --> 00:29:00.000
So, for the linear,
first-order equation,
00:28:56.000 --> 00:29:02.000
there, too, I'm going to think
of constant coefficients.
00:29:01.000 --> 00:29:07.000
We talked quite a bit about
this equation.
00:29:04.000 --> 00:29:10.000
What did I call the right-hand
side?
00:29:06.000 --> 00:29:12.000
I think we usually called it q
of t,
00:29:09.000 --> 00:29:15.000
right?
This is in ancient history.
00:29:12.000 --> 00:29:18.000
The definition of ancient
history was before the first
00:29:16.000 --> 00:29:22.000
exam.
Okay, now how does that fit
00:29:18.000 --> 00:29:24.000
into this theorem that I've
given you?
00:29:21.000 --> 00:29:27.000
Remember what the solution
looked like.
00:29:25.000 --> 00:29:31.000
The solution looked like,
remember, you took the
00:29:28.000 --> 00:29:34.000
integrating factor was e to the
kt,
00:29:32.000 --> 00:29:38.000
and then after you integrated
both sides, multiplied through,
00:29:37.000 --> 00:29:43.000
and then the final answer
looked like this,
00:29:41.000 --> 00:29:47.000
y equaled, it was e to the
negative kt times
00:29:45.000 --> 00:29:51.000
either an indefinite integral,
or a definite integral
00:29:50.000 --> 00:29:56.000
depending on your preference,
q of t,
00:29:54.000 --> 00:30:00.000
so, x is metamorphosed into t.
I gather you've got that,
00:29:58.000 --> 00:30:04.000
e to the kt plus,
what was the other term?
00:30:02.000 --> 00:30:08.000
A constant times e to the
negative kt
00:30:08.000 --> 00:30:14.000
How does this fit into the
paradigm I've given you over
00:30:12.000 --> 00:30:18.000
there for solving the second
order equation?
00:30:15.000 --> 00:30:21.000
Which term is which?
Well, this has the arbitrary
00:30:19.000 --> 00:30:25.000
constant in it.
So, this must be the
00:30:22.000 --> 00:30:28.000
complementary solution.
Is it?
00:30:24.000 --> 00:30:30.000
Is this the solution to the
associated homogeneous equation?
00:30:30.000 --> 00:30:36.000
What's the associated
homogeneous equation?
00:30:32.000 --> 00:30:38.000
Put zero here.
Okay, if you put zero there,
00:30:35.000 --> 00:30:41.000
what's the solution?
Now, this you ought to know.
00:30:38.000 --> 00:30:44.000
y prime equals negative ky.
00:30:40.000 --> 00:30:46.000
What's the solution?
e to the negative kt.
00:30:43.000 --> 00:30:49.000
You are supposed to come into
00:30:46.000 --> 00:30:52.000
this course knowing that,
except there's an arbitrary
00:30:49.000 --> 00:30:55.000
constant in front.
So, right, this is exactly the
00:30:52.000 --> 00:30:58.000
solution to the associated
homogeneous equation,
00:30:55.000 --> 00:31:01.000
where there is zero here.
Then, what's this thing?
00:31:00.000 --> 00:31:06.000
This is a particular solution.
This is my yp.
00:31:02.000 --> 00:31:08.000
But that's not a particular
solution because this indefinite
00:31:06.000 --> 00:31:12.000
integral, you know,
has an arbitrary constant in
00:31:09.000 --> 00:31:15.000
it.
In fact, it's just that
00:31:10.000 --> 00:31:16.000
arbitrary constant.
So, it's totally confusing.
00:31:13.000 --> 00:31:19.000
But, this symbol,
you know when you actually
00:31:16.000 --> 00:31:22.000
solve the equation this way,
all you did was you found one
00:31:20.000 --> 00:31:26.000
function here.
You didn't throw in the
00:31:22.000 --> 00:31:28.000
arbitrary constant right away.
All you needed to do was find
00:31:26.000 --> 00:31:32.000
one function.
And, even if you really are
00:31:29.000 --> 00:31:35.000
bothered by the fact that this
is so indefinite,
00:31:32.000 --> 00:31:38.000
and therefore,
make it a particular solution
00:31:35.000 --> 00:31:41.000
by making this zero,
make it a definite integral,
00:31:38.000 --> 00:31:44.000
zero, here, t there,
and then change those t's to
00:31:42.000 --> 00:31:48.000
dummy t's, t1's or t tildes,
or something like that.
00:31:45.000 --> 00:31:51.000
So, this fits into that thing.
In other words,
00:31:48.000 --> 00:31:54.000
I could have done it at that
time, but I didn't the point
00:31:52.000 --> 00:31:58.000
because this can be solved
directly, whereas,
00:31:55.000 --> 00:32:01.000
of course, the general second
order equation in homogeneous
00:31:58.000 --> 00:32:04.000
cannot be solved directly,
and therefore you have to be
00:32:02.000 --> 00:32:08.000
willing to talk about what its
solutions look like in advance.
00:32:08.000 --> 00:32:14.000
Now, remember I said,
we talked, I said there was two
00:32:12.000 --> 00:32:18.000
different cases,
although both of them had the
00:32:15.000 --> 00:32:21.000
identical looking solution.
Their meaning in the physical
00:32:19.000 --> 00:32:25.000
world was so different that they
really should be considered as
00:32:24.000 --> 00:32:30.000
solving the same equation.
And, one of these was the case.
00:32:30.000 --> 00:32:36.000
Of the two, perhaps the more
important was the case when k
00:32:34.000 --> 00:32:40.000
was positive,
and of course the other is when
00:32:38.000 --> 00:32:44.000
k is negative.
When k is positive,
00:32:41.000 --> 00:32:47.000
that had the effect of
separating that solution into
00:32:45.000 --> 00:32:51.000
this part, which was a
transient, and the other part,
00:32:49.000 --> 00:32:55.000
which was a steady state.
The steady state solution,
00:32:53.000 --> 00:32:59.000
that was the yp part of it in
that terminology.
00:32:57.000 --> 00:33:03.000
And, the transient part,
it was trangent because it went
00:33:02.000 --> 00:33:08.000
to zero.
If k is positive,
00:33:05.000 --> 00:33:11.000
the exponential dies regardless
of what c is.
00:33:09.000 --> 00:33:15.000
So, the transient,
that's the yc part.
00:33:12.000 --> 00:33:18.000
It goes to zero as Ttgoes to
infinity.
00:33:16.000 --> 00:33:22.000
The transient depends on,
uses, the initial condition,
00:33:21.000 --> 00:33:27.000
whatever it is,
because that's what determines
00:33:25.000 --> 00:33:31.000
the value of c.
On the other hand,
00:33:28.000 --> 00:33:34.000
this initial condition makes no
difference as t goes towards
00:33:33.000 --> 00:33:39.000
infinity.
All that's left is this steady
00:33:38.000 --> 00:33:44.000
state solution.
And, all solutions tend to the
00:33:42.000 --> 00:33:48.000
steady state solution.
So, if k is positive,
00:33:46.000 --> 00:33:52.000
one gets this analysis of the
solutions into the sum of one
00:33:52.000 --> 00:33:58.000
basic solution,
and the others,
00:33:55.000 --> 00:34:01.000
which just die away,
have no influence on this,
00:33:59.000 --> 00:34:05.000
less and less influence as time
goes to infinity.
00:34:05.000 --> 00:34:11.000
For k less than zero,
this analysis does not work
00:34:09.000 --> 00:34:15.000
because this term,
if k is less than zero,
00:34:12.000 --> 00:34:18.000
this term goes to infinity or
negative infinity,
00:34:16.000 --> 00:34:22.000
and typically tends to dominate
that.
00:34:19.000 --> 00:34:25.000
So, it's the start that the
important one.
00:34:23.000 --> 00:34:29.000
It depends on the initial
conditions, and the analysis is
00:34:28.000 --> 00:34:34.000
meaningless.
So, the above is meaningless.
00:34:33.000 --> 00:34:39.000
And now, what I'd like to do is
try to see what the analog of
00:34:39.000 --> 00:34:45.000
that is for second order
equations, and higher order
00:34:45.000 --> 00:34:51.000
equations.
If you understand second-order,
00:34:49.000 --> 00:34:55.000
that's good enough.
Higher order goes exactly the
00:34:54.000 --> 00:35:00.000
same way.
So, the question is,
00:34:58.000 --> 00:35:04.000
for second-order,
let's make it with constant
00:35:02.000 --> 00:35:08.000
coefficients plus,
I could call it b and k,
00:35:07.000 --> 00:35:13.000
oh, no, b k,
or p.
00:35:11.000 --> 00:35:17.000
The trouble is,
that wouldn't take care of the
00:35:14.000 --> 00:35:20.000
electrical circuits.
So, I just want to use neutral
00:35:17.000 --> 00:35:23.000
letters, which suggest nothing.
And, you can make them turn it
00:35:22.000 --> 00:35:28.000
into a circuit,
so springs, or yet other
00:35:25.000 --> 00:35:31.000
examples undreamt of.
But these are constants.
00:35:28.000 --> 00:35:34.000
And I'm going to think of it as
time.
00:35:32.000 --> 00:35:38.000
I think I'll switch back to
time, let x be the time.
00:35:37.000 --> 00:35:43.000
So, B y equals f of t.
So, there is our equation.
00:35:43.000 --> 00:35:49.000
A and B are constants.
And, the question is,
00:35:47.000 --> 00:35:53.000
the question I'm asking,
can think of either of these
00:35:53.000 --> 00:35:59.000
two models or others,
the question I'm asking is,
00:35:58.000 --> 00:36:04.000
under what circumstances can I
make that same type of analysis
00:36:04.000 --> 00:36:10.000
into steady-state and transient?
Well, what does the solution
00:36:11.000 --> 00:36:17.000
look like?
The solution looks like y
00:36:15.000 --> 00:36:21.000
equals a particular solution
plus c1 y1 plus c2 y2.
00:36:20.000 --> 00:36:26.000
Therefore, to make that look
00:36:24.000 --> 00:36:30.000
like this, the c1 and c2 contain
the initial conditions.
00:36:31.000 --> 00:36:37.000
This part does not.
Therefore, if I want to say
00:36:35.000 --> 00:36:41.000
that the solutions look like a
steady state solution plus
00:36:41.000 --> 00:36:47.000
something that dies away,
which becomes less and less
00:36:47.000 --> 00:36:53.000
important as time goes on,
what I'm really asking is,
00:36:52.000 --> 00:36:58.000
under what circumstances is
this part guaranteed to go to
00:36:58.000 --> 00:37:04.000
zero?
So, the question is,
00:37:01.000 --> 00:37:07.000
when, in other words,
under what conditions on the
00:37:06.000 --> 00:37:12.000
equation A and B,
in effect, is what we are
00:37:10.000 --> 00:37:16.000
asking.
When does c1 y1 plus c2 y2 go
00:37:14.000 --> 00:37:20.000
to zero as t goes to infinity,
00:37:19.000 --> 00:37:25.000
regardless of what
c1 and c2 are for all c1 c2.
00:37:25.000 --> 00:37:31.000
Now, here there was no
difficulty.
00:37:30.000 --> 00:37:36.000
We had the thing very
explicitly, and you could see k
00:37:34.000 --> 00:37:40.000
is positive: this goes to zero.
And if k is negative,
00:37:38.000 --> 00:37:44.000
it doesn't go to zero.
It goes to infinity.
00:37:41.000 --> 00:37:47.000
Here, I want to make the same
kind of analysis,
00:37:44.000 --> 00:37:50.000
except it's just going to take,
it's a little more trouble.
00:37:49.000 --> 00:37:55.000
But the answer,
when it finally comes out is
00:37:52.000 --> 00:37:58.000
very beautiful.
So, when are all these guys
00:37:55.000 --> 00:38:01.000
going to go to zero?
First of all,
00:37:58.000 --> 00:38:04.000
you might as well just have the
definition.
00:38:01.000 --> 00:38:07.000
So, all the good things that
this is going to imply,
00:38:05.000 --> 00:38:11.000
if this is so,
in other words,
00:38:07.000 --> 00:38:13.000
if they all go to zero,
everything in the complementary
00:38:12.000 --> 00:38:18.000
solution, then the ODE is called
stable.
00:38:17.000 --> 00:38:23.000
Some people call it
asymptotically stable.
00:38:21.000 --> 00:38:27.000
I don't know what to call it.
I can make the analysis,
00:38:28.000 --> 00:38:34.000
and then I use the identical
terminology, c1 y1 plus c2 y2.
00:38:36.000 --> 00:38:42.000
This is called the transient
because it goes to zero.
00:38:40.000 --> 00:38:46.000
This is called the particular
solution now that we labored so
00:38:45.000 --> 00:38:51.000
hard to get for the next two
weeks.
00:38:47.000 --> 00:38:53.000
It's the important part.
It's the steady-state part.
00:38:52.000 --> 00:38:58.000
It's what lasts out to infinity
after the other stuff has
00:38:56.000 --> 00:39:02.000
disappeared.
So, this is the steady-state
00:38:59.000 --> 00:39:05.000
solution, steady-state solution,
okay?
00:39:04.000 --> 00:39:10.000
And, the differential equation
is called stable.
00:39:07.000 --> 00:39:13.000
Now, it's of the highest
interest to know when a
00:39:10.000 --> 00:39:16.000
differential equation is stable,
linear differential equation is
00:39:14.000 --> 00:39:20.000
stable in this sense because you
have a control.
00:39:17.000 --> 00:39:23.000
You know what its solutions
look like.
00:39:20.000 --> 00:39:26.000
You have some feeling for how
it's behaving in the long term.
00:39:24.000 --> 00:39:30.000
If this is not so,
each equation is a law unto
00:39:27.000 --> 00:39:33.000
itself if you don't know.
So, let's do the work.
00:39:31.000 --> 00:39:37.000
For the rest of the period,
what I'd like to do is to find
00:39:36.000 --> 00:39:42.000
out what the conditions are,
which make this true.
00:39:40.000 --> 00:39:46.000
Those were the equations which
we will have a right to call
00:39:45.000 --> 00:39:51.000
stable.
So, when does this happen,
00:39:48.000 --> 00:39:54.000
and where is it going to
happen?
00:39:50.000 --> 00:39:56.000
I don't know.
I guess, here.
00:40:05.000 --> 00:40:11.000
Now, I think the first step is
fairly easy, and it will give
00:40:10.000 --> 00:40:16.000
you a good review of what we've
been doing up until now.
00:40:14.000 --> 00:40:20.000
So, I'm simply going to make a
case-by-case analysis.
00:40:19.000 --> 00:40:25.000
Don't worry,
it won't take very long.
00:40:22.000 --> 00:40:28.000
What are the cases we've been
studying?
00:40:25.000 --> 00:40:31.000
Well, what do the
characteristic roots look like?
00:40:29.000 --> 00:40:35.000
The roots of the characteristic
equation, in other words,
00:40:34.000 --> 00:40:40.000
remember, there are cases.
The first case is they are real
00:40:41.000 --> 00:40:47.000
and distinct,
r1 not equal to r2,
00:40:45.000 --> 00:40:51.000
real and distinct.
What are the other cases?
00:40:50.000 --> 00:40:56.000
Well, r1 equals r2.
And then, there's the case
00:40:56.000 --> 00:41:02.000
where there are complex.
So, I will write it r equals a
00:41:02.000 --> 00:41:08.000
plus or minus b i.
What do the solutions look
00:41:07.000 --> 00:41:13.000
like?
So, my ham-handed approach to
00:41:09.000 --> 00:41:15.000
this problem is going to be,
in each case,
00:41:12.000 --> 00:41:18.000
I'll look at the solutions,
and first get the condition on
00:41:16.000 --> 00:41:22.000
the roots.
So, in other words,
00:41:18.000 --> 00:41:24.000
I'm not going to worry right
away about the a and the b.
00:41:21.000 --> 00:41:27.000
I'm going, instead,
to worry about expressing this
00:41:24.000 --> 00:41:30.000
condition of stability in terms
of the characteristic roots.
00:41:28.000 --> 00:41:34.000
In fact, that's the only way in
which many people know the
00:41:32.000 --> 00:41:38.000
conditions.
You're going to be smarter.
00:41:35.000 --> 00:41:41.000
Okay, what do the solutions
look like?
00:41:38.000 --> 00:41:44.000
Well, the general solution
looks like e to the r1 t plus c2
00:41:42.000 --> 00:41:48.000
e to the r2 t.
00:41:45.000 --> 00:41:51.000
Okay, so, what's the stability
condition?
00:41:48.000 --> 00:41:54.000
In other words,
if equation happened to have
00:41:51.000 --> 00:41:57.000
its characteristic roots,
real and distinct,
00:41:54.000 --> 00:42:00.000
under what circumstances would
it be stable?
00:41:57.000 --> 00:42:03.000
Would it, in other words,
all its solutions go to zero?
00:42:01.000 --> 00:42:07.000
So, I'm talking about the
homogeneous equation,
00:42:04.000 --> 00:42:10.000
the reduced equation,
the associated homogeneous
00:42:07.000 --> 00:42:13.000
equation.
Why?
00:42:10.000 --> 00:42:16.000
Because that's all that's
involved in this.
00:42:13.000 --> 00:42:19.000
In other words,
when I write that,
00:42:15.000 --> 00:42:21.000
I am no longer interested in
the whole equation.
00:42:19.000 --> 00:42:25.000
All I'm interested in is the
reduced equation,
00:42:22.000 --> 00:42:28.000
the equation where you turn the
f of t on the
00:42:26.000 --> 00:42:32.000
right-hand side into zero.
So, what's the stability
00:42:31.000 --> 00:42:37.000
condition?
Well, let's write it out.
00:42:35.000 --> 00:42:41.000
Under what circumstances will
all these guys go to zero?
00:42:41.000 --> 00:42:47.000
If r1 and r2 should be
negative, can they be zero?
00:42:46.000 --> 00:42:52.000
No, because then it will be a
constant and it will go to zero.
00:42:52.000 --> 00:42:58.000
How about this one?
Well, in this one,
00:42:56.000 --> 00:43:02.000
it's (c1 plus c2 times t)
multiplied by e to the r1 t.
00:43:01.000 --> 00:43:07.000
Of course, both of these are
00:43:08.000 --> 00:43:14.000
the same.
I'll just arbitrarily pick one
00:43:11.000 --> 00:43:17.000
of them.
What happens to this as things
00:43:14.000 --> 00:43:20.000
go to zero?
Well, this part is rising,
00:43:16.000 --> 00:43:22.000
at least if c2 is positive.
This part is either helping or
00:43:21.000 --> 00:43:27.000
it's hindering.
But, I hope you know what these
00:43:24.000 --> 00:43:30.000
functions look like,
and you know which of them go
00:43:28.000 --> 00:43:34.000
to zero.
They go to zero if r1 is
00:43:31.000 --> 00:43:37.000
negative.
It might rise in the beginning,
00:43:35.000 --> 00:43:41.000
but after a while they lose the
energy.
00:43:39.000 --> 00:43:45.000
Of course, if r1 is equal to
zero, what do these guys do?
00:43:44.000 --> 00:43:50.000
Linear, go to infinity.
Well, we are doing okay.
00:43:49.000 --> 00:43:55.000
How about here?
Well, here, it's a little more
00:43:53.000 --> 00:43:59.000
complicated.
The solutions look like e to
00:43:57.000 --> 00:44:03.000
the at times (c1 cosine bt plus
c2 sine bt).
00:44:03.000 --> 00:44:09.000
Now, this part is a pure
00:44:07.000 --> 00:44:13.000
oscillation.
You know that.
00:44:09.000 --> 00:44:15.000
It might have a big amplitude,
but whatever it does,
00:44:13.000 --> 00:44:19.000
it does the same thing all the
time.
00:44:16.000 --> 00:44:22.000
So, whether this goes to zero
depends entirely upon what that
00:44:20.000 --> 00:44:26.000
exponential is doing.
And, that exponential goes to
00:44:24.000 --> 00:44:30.000
zero if a is negative.
So here, the condition is
00:44:28.000 --> 00:44:34.000
negative.
And now, the only thing left to
00:44:33.000 --> 00:44:39.000
do is to say it nicely.
I've got three cases,
00:44:38.000 --> 00:44:44.000
and I want to say them all in
one breath.
00:44:43.000 --> 00:44:49.000
So, the stability condition is,
the ODE is stable.
00:44:49.000 --> 00:44:55.000
So, this is,
or f of t.
00:44:53.000 --> 00:44:59.000
It doesn't matter.
But, psychologically,
00:44:57.000 --> 00:45:03.000
you can put this as zero there,
is stable if what?
00:45:05.000 --> 00:45:11.000
In case one,
this is true.
00:45:07.000 --> 00:45:13.000
In case two,
that's true.
00:45:10.000 --> 00:45:16.000
In case three,
that's true.
00:45:13.000 --> 00:45:19.000
But that's ugly.
Make it beautiful.
00:45:17.000 --> 00:45:23.000
The beautiful way of saying it
is if all the characteristic
00:45:23.000 --> 00:45:29.000
roots have negative real parts.
If the characteristic roots,
00:45:31.000 --> 00:45:37.000
the r's or the a plus or minus
b i, have negative real part.
00:45:38.000 --> 00:45:44.000
That's the form in which the
electrical engineers will nod
00:45:44.000 --> 00:45:50.000
their head, tell you,
yeah, that's right,
00:45:49.000 --> 00:45:55.000
negative real part,
sorry.
00:45:52.000 --> 00:45:58.000
Isn't it right?
Is that right here?
00:45:56.000 --> 00:46:02.000
Yeah.
What's the real part of these
00:45:59.000 --> 00:46:05.000
guys?
They themselves,
00:46:03.000 --> 00:46:09.000
because they are real.
What's the real part of this?
00:46:09.000 --> 00:46:15.000
Yeah.
The only case in which I really
00:46:13.000 --> 00:46:19.000
had to use real part is when I
talk about the complex case
00:46:20.000 --> 00:46:26.000
because a is just the real part
of a complex number.
00:46:26.000 --> 00:46:32.000
It's not the whole thing.