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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,
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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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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.
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And, the voltage shop across
the capacitance is q
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divided by C.
And so, that's equal to,
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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.
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E could be a battery.
It could be a source of
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alternating current,
something like that.
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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
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question of the signs,
which I know I'll never
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understand.
But, let's assume you've chosen
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the sign convention so that this
comes out nicely on the
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right-hand side.
So, this might be varying
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sinusoidally,
in which case you'd have source
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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,
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so don't do it.
Now, this doesn't look like
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what it's supposed to look like
because it's got q and i.
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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
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little soul mate on the other
side is the current that's
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flowing in the circuit.
That's why current flows,
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except nothing really happens.
Electrons just push on each
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other, and they stay where they
are.
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I don't understand this at all.
So, if I differentiate this,
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you can do two things.
Either you could integrate i,
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and expressed the thing
entirely in terms of q,
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or you can differentiate it,
and express everything in terms
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of i.
Your book does nicely both,
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does not take sides.
So, let's differentiate it,
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and then it will look like L i
double prime plus R i prime plus
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i divided by C equals,
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and now, watch out,
you have now not the
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electromotive force,
but its derivative.
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So, if you were so unfortunate
as to put a little dry cell
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there, now you've got nothing,
and you've got the homogeneous
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case.
That's okay.
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Where are the erasers?
One eraser?
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I don't believe this.
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So, there's the equation.
There are our two equations.
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Why don't we put them up in
colored chalk.
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There's the spring equation.
And, here's the equation that
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governs the current,
for how the current flows in
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that circuit.
And now, you can see,
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again, what does it mean?
If this is zero,
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for example,
if I have a dry cell there,
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or if I have nothing at all in
the circuit, then this
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represents the passive circuit.
It's just sitting there.
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It wouldn't do anything at all,
except that you've put a charge
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on the capacitor,
and waited, and of course,
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when you put a charge on there,
it's got a discharge,
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and discharges through the
circuit, and swings back and
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forth a little bit if it's
under-damped until finally
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towards the end the current dies
away to zero.
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But, what usually happens is
that you drive this passive
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circuit by putting an effective
E in it, and then you want to
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know how the current behaves.
So, those are the two problems,
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the passive circuit without an
applied electromotive force,
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or plugging it into the wall,
and wanting it to do things.
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That's the normal state of
affairs.
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People don't want passive
circuits, they want circuits
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which do things because,
okay, that's why they want to
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solve inhomogeneous equations
instead of homogeneous
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equations.
But as I said,
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you have to do the homogeneous
case first.
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Okay, you are now officially
responsible for this,
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and I don't care that you
haven't had it in physics yet.
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You will before the next exam.
So, I don't even feel guilty.
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But, you're going to start
using it on the problem set
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right away.
So, it's never too soon to
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start learning it.
Okay, now, the main theorem,
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I now want to go,
so that was just examples to
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give you some physical feeling
for the sorts of differential
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equations we'll be talking
about.
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I now want to tell you briefly
about the key theorem about
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solving the homogeneous
equation.
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00:16:59 --> 00:17:05
So, the main theorem about
solving the homogeneous equation
263
00:17:04 --> 00:17:10
is, the inhomogeneous equation.
So, I'm going to write the
264
00:17:09 --> 00:17:15
inhomogeneous equation out.
I'm going to make the left-hand
265
00:17:14 --> 00:17:20
side a linear operator,
and am going to write the
266
00:17:18 --> 00:17:24
equation as Ly equals f of x.
267
00:17:21 --> 00:17:27
That's the inhomogeneous
equation.
268
00:17:23 --> 00:17:29
So, L is the linear operator,
second order because I'm only
269
00:17:28 --> 00:17:34
talking about second-order
equations.
270
00:17:32 --> 00:17:38
L is a linear operator,
and then this is the
271
00:17:37 --> 00:17:43
differential equation.
So, here's our differential
272
00:17:43 --> 00:17:49
equation.
It's inhomogeneous because it's
273
00:17:48 --> 00:17:54
go the f of x on the
right hand side.
274
00:17:53 --> 00:17:59
And, what the theorem says is
that the solution has the
275
00:18:00 --> 00:18:06
following form,
y sub p, I'll explain what that
276
00:18:05 --> 00:18:11
is in just a moment,
plus y sub c.
277
00:18:13 --> 00:18:19
So, the hypothesis is we've got
the linear equation,
278
00:18:17 --> 00:18:23
and the conclusion is that
that's what its solution looks
279
00:18:21 --> 00:18:27
like.
Now, you already know what y
280
00:18:24 --> 00:18:30
sub c looks like.
In other words,
281
00:18:27 --> 00:18:33
if I write this out in more
detail, it would be i.e.,
282
00:18:31 --> 00:18:37
department of fuller
explanation, --
283
00:18:35 --> 00:18:41
-- the general solution looks
like y equals yp,
284
00:18:38 --> 00:18:44
and then this thing is going to
look like an arbitrary constant
285
00:18:43 --> 00:18:49
times y1 plus an arbitrary
constant times y2,
286
00:18:47 --> 00:18:53
where these are solutions of
the homogeneous equation.
287
00:18:51 --> 00:18:57
So, Yc looks like this part,
and the yp, what's yp?
288
00:18:55 --> 00:19:01
p stands for particular,
the most confusing word in this
289
00:19:00 --> 00:19:06
subject.
But, you've got at least four
290
00:19:04 --> 00:19:10
weeks to learn what it means.
Okay, yp is a particular
291
00:19:10 --> 00:19:16
solution to Ly equals f of x.
292
00:19:14 --> 00:19:20
Now, I'm not going to explain
what particular means.
293
00:19:19 --> 00:19:25
First, I'll chat as if you knew
what it meant,
294
00:19:24 --> 00:19:30
and then we'll see if you have
picked it up.
295
00:19:30 --> 00:19:36
In other words,
the procedure for solving this
296
00:19:33 --> 00:19:39
equation is composed of two
steps.
297
00:19:36 --> 00:19:42
First, to find this part.
In other words,
298
00:19:40 --> 00:19:46
to find the complementary
solution, in other words,
299
00:19:44 --> 00:19:50
to do what we've been doing for
the last week,
300
00:19:48 --> 00:19:54
solve not the equation you are
given, but the reduced equation.
301
00:19:53 --> 00:19:59
So, the first step is to find
this.
302
00:19:56 --> 00:20:02
The second step is to find yp.
Now, what's yp?
303
00:20:01 --> 00:20:07
yp is a particular solution to
the whole equation.
304
00:20:05 --> 00:20:11
Yeah, but which one?
Well, if it's any one,
305
00:20:09 --> 00:20:15
then it's not a particular
solution, yeah.
306
00:20:12 --> 00:20:18
I say, unfortunately the word
particular here is not being
307
00:20:17 --> 00:20:23
used in exactly the same sense
in which most people use it in
308
00:20:22 --> 00:20:28
ordinary English.
It's a perfectly valid way to
309
00:20:26 --> 00:20:32
use it.
It's just confusing,
310
00:20:29 --> 00:20:35
and no one has ever come up
with a better word.
311
00:20:33 --> 00:20:39
So, particular means any one
solution.
312
00:20:38 --> 00:20:44
Any one will do.
Okay, even these have slightly
313
00:20:45 --> 00:20:51
different meanings.
Any questions about this?
314
00:20:51 --> 00:20:57
I refuse to answer them.
[LAUGHTER]
315
00:20:58 --> 00:21:04
Now, well, examples of course
will make it all clear.
316
00:21:03 --> 00:21:09
But I'd like,
first, to prove the theorem,
317
00:21:08 --> 00:21:14
to show you how simple it is.
It's extremely simple if you
318
00:21:15 --> 00:21:21
just use the fact that L is a
linear operator.
319
00:21:20 --> 00:21:26
We've got two things to prove.
What have we got to prove?
320
00:21:26 --> 00:21:32
Well, I have to prove two
statements, first of all,
321
00:21:32 --> 00:21:38
that all the yp plus c1 y1 plus
c2 y2 are
322
00:21:39 --> 00:21:45
solutions.
How are we going to prove that?
323
00:21:43 --> 00:21:49
Well, how do you know if
something is a solution?
324
00:21:46 --> 00:21:52
Well, you plug it into the
equation, and you see if it
325
00:21:49 --> 00:21:55
satisfies the equation.
Good, let's do it,
326
00:21:52 --> 00:21:58
proof.
L, I'm going to plug it into
327
00:21:54 --> 00:22:00
the equation.
That means I calculate L of yp
328
00:21:56 --> 00:22:02
plus c1 y1 plus c2 y2.
329
00:22:00 --> 00:22:06
Now, what's the answer?
Because this is a linear
330
00:22:04 --> 00:22:10
operator, and notice,
the argument doesn't use the
331
00:22:08 --> 00:22:14
fact that the equation is second
order.
332
00:22:12 --> 00:22:18
It immediately generalizes to a
linear equation of any order,
333
00:22:17 --> 00:22:23
whatever-- 47.
Okay, this is L of yp plus L of
334
00:22:21 --> 00:22:27
c1 y1 plus c2 y2.
335
00:22:25 --> 00:22:31
Well, what's that?
What's L of the complementary
336
00:22:29 --> 00:22:35
solution?
What does it mean to be the
337
00:22:34 --> 00:22:40
complementary solution?
It means when you apply the
338
00:22:37 --> 00:22:43
operator L to it,
you get zero because this
339
00:22:40 --> 00:22:46
satisfies the homogeneous
equation.
340
00:22:43 --> 00:22:49
So, this is zero.
What's L of yp?
341
00:22:46 --> 00:22:52
Well, it was a particular
solution to the equation.
342
00:22:50 --> 00:22:56
Therefore, when I plugged it
into the equation,
343
00:22:53 --> 00:22:59
I must have gotten out on the
right-hand side,
344
00:22:57 --> 00:23:03
f of x.
So, this is since yp is a
345
00:23:00 --> 00:23:06
solution to the whole equation.
So, what's the conclusion?
346
00:23:06 --> 00:23:12
That, if I take any one of
these guys, no matter what c1
347
00:23:10 --> 00:23:16
and c2 are, apply the linear
operator, L to it,
348
00:23:13 --> 00:23:19
the answer comes out
to be f of.
349
00:23:17 --> 00:23:23
Therefore, this proves that
this shows that these are all
350
00:23:21 --> 00:23:27
solutions because that's what it
means.
351
00:23:24 --> 00:23:30
Therefore, they satisfy L of y
equals f of x.
352
00:23:30 --> 00:23:36
They satisfy the whole
inhomogeneous differential
353
00:23:34 --> 00:23:40
equation, and that's it.
Well, that's only half the
354
00:23:38 --> 00:23:44
story.
The other half of the story is
355
00:23:41 --> 00:23:47
to show that there are no other
solutions.
356
00:23:45 --> 00:23:51
Okay, so we got our little u of
x coming up again,
357
00:23:50 --> 00:23:56
and he thinks he's a solution.
Okay, so, to prove there are no
358
00:23:55 --> 00:24:01
other solutions,
it almost sounds biblical,
359
00:23:59 --> 00:24:05
thou shalt have no other
solutions before me,
360
00:24:03 --> 00:24:09
okay.
There are no other solutions
361
00:24:07 --> 00:24:13
accept these guys for different
values of c1 and c2.
362
00:24:10 --> 00:24:16
Okay, so, u of x is a
solution.
363
00:24:13 --> 00:24:19
I have to show that u of x is
one of these guys.
364
00:24:16 --> 00:24:22
How am I going to do that?
Easy.
365
00:24:19 --> 00:24:25
If it's a solution that,
L of u,
366
00:24:21 --> 00:24:27
okay, I'm going to drop the x,
okay, just to make the,
367
00:24:25 --> 00:24:31
like I dropped the x over
there.
368
00:24:29 --> 00:24:35
If it's a solution to the whole
inhomogeneous equation,
369
00:24:33 --> 00:24:39
then this must come out to be f
of x.
370
00:24:37 --> 00:24:43
Now, what's L of yp?
That's f of x too,
371
00:24:41 --> 00:24:47
by secret little particular
solution I've got in my pocket.
372
00:24:47 --> 00:24:53
Okay, I pull it out,
ah-ha, L of yp,
373
00:24:50 --> 00:24:56
that's f of x,
too.
374
00:24:51 --> 00:24:57
Now, I'm going to not add them.
I'm going to subtract them.
375
00:24:56 --> 00:25:02
What is L of u minus yp?
376
00:25:00 --> 00:25:06
Well, it's zero.
It's zero because this is a
377
00:25:05 --> 00:25:11
linear operator.
This would be L of u minus L of
378
00:25:08 --> 00:25:14
yp.
I guess the answer is zero on
379
00:25:12 --> 00:25:18
the right-hand side.
And therefore,
380
00:25:14 --> 00:25:20
what is the conclusion?
If that's zero,
381
00:25:17 --> 00:25:23
it must be a solution to the
homogeneous equation.
382
00:25:21 --> 00:25:27
Therefore, u minus yp
is equal to,
383
00:25:24 --> 00:25:30
there must be c1 and c2.
I won't give them the generic
384
00:25:28 --> 00:25:34
names.
I'll give them a name,
385
00:25:30 --> 00:25:36
a particular one.
I'll put a tilde to indicate
386
00:25:34 --> 00:25:40
it's a particular one.
c1 plus c2 y2 tilde,
387
00:25:37 --> 00:25:43
so, in other words,
for some choice of these
388
00:25:41 --> 00:25:47
constants, and I'll call those
particular choices c1 tilde and
389
00:25:45 --> 00:25:51
c2 tilde, it must be that these
are equal.
390
00:25:48 --> 00:25:54
Well, what does that say?
It says that u is equal to yp
391
00:25:52 --> 00:25:58
plus c1 tilde,
blah, blah, blah,
392
00:25:54 --> 00:26:00
blah, plus c2 tilde,
blah, blah, blah,
393
00:25:57 --> 00:26:03
blah, and therefore chose that
u wasn't a new solution.
394
00:26:02 --> 00:26:08
It was one of these.
So, u isn't new.
395
00:26:08 --> 00:26:14
So, I should write it down.
Otherwise some of you will have
396
00:26:18 --> 00:26:24
missed the punch line.
Okay, therefore,
397
00:26:25 --> 00:26:31
u is equal to yp plus c1 tilde
y1 plus c2 tilde y2.
398
00:26:36 --> 00:26:42
And, it shows.
This guy who thought he was new
399
00:26:39 --> 00:26:45
was not new at all.
It was just one of the other
400
00:26:43 --> 00:26:49
solutions.
Okay, well, now,
401
00:26:45 --> 00:26:51
since the coefficient's a
constant, apparently we've done
402
00:26:49 --> 00:26:55
half the work.
We know what the complementary
403
00:26:52 --> 00:26:58
solution is because you know how
to do those in terms of
404
00:26:56 --> 00:27:02
exponentials and complex
exponentials,
405
00:26:59 --> 00:27:05
signs and cosines,
and so on.
406
00:27:03 --> 00:27:09
So, what's left to do?
All we have to do is find to
407
00:27:07 --> 00:27:13
solve equations,
which are inhomogeneous.
408
00:27:10 --> 00:27:16
All we have to do is find a
particular solution,
409
00:27:14 --> 00:27:20
find one solution.
It doesn't matter which one,
410
00:27:18 --> 00:27:24
any one.
Just find one,
411
00:27:20 --> 00:27:26
okay?
Now, we're going to spend the
412
00:27:23 --> 00:27:29
next two weeks trying to do
this.
413
00:27:26 --> 00:27:32
I'll give you various methods.
I'll give you a general method
414
00:27:32 --> 00:27:38
involving Fourier series because
it's a good excuse for learning
415
00:27:37 --> 00:27:43
what Fourier series are.
But, the answer is that in
416
00:27:41 --> 00:27:47
general, for a few standard
functions, it's known how to do
417
00:27:45 --> 00:27:51
this.
You will learn those methods
418
00:27:48 --> 00:27:54
for finding those using
operators.
419
00:27:50 --> 00:27:56
For all the others,
it's done by a series,
420
00:27:54 --> 00:28:00
or a method involving
approximation.
421
00:27:58 --> 00:28:04
Or, the worse comes to worst,
you throw it on a computer and
422
00:28:02 --> 00:28:08
just take a graph and the
numerical output of answers as
423
00:28:07 --> 00:28:13
the particular solution.
Okay, now before,
424
00:28:10 --> 00:28:16
we are going to start that
work, not today.
425
00:28:14 --> 00:28:20
We'll start it next Monday,
and it will last,
426
00:28:18 --> 00:28:24
as I say the next two weeks.
And, we will be up to spring
427
00:28:22 --> 00:28:28
break.
But, before we do that,
428
00:28:25 --> 00:28:31
I'd like to relate this to what
we did for first order equations
429
00:28:30 --> 00:28:36
because there is something to be
learned from that.
430
00:28:36 --> 00:28:42
Think back to the linear
first-order equation,
431
00:28:38 --> 00:28:44
and I'm going to,
since from now on for the rest
432
00:28:41 --> 00:28:47
of the period,
I'm going to be considering the
433
00:28:44 --> 00:28:50
case for constant coefficients.
In other words,
434
00:28:47 --> 00:28:53
this case of springs or
circuits or simple systems which
435
00:28:51 --> 00:28:57
behave like those and have
constant coefficients.
436
00:28:54 --> 00:29:00
So, for the linear,
first-order equation,
437
00:28:56 --> 00:29:02
there, too, I'm going to think
of constant coefficients.
438
00:29:01 --> 00:29:07
We talked quite a bit about
this equation.
439
00:29:04 --> 00:29:10
What did I call the right-hand
side?
440
00:29:06 --> 00:29:12
I think we usually called it q
of t,
441
00:29:09 --> 00:29:15
right?
This is in ancient history.
442
00:29:12 --> 00:29:18
The definition of ancient
history was before the first
443
00:29:16 --> 00:29:22
exam.
Okay, now how does that fit
444
00:29:18 --> 00:29:24
into this theorem that I've
given you?
445
00:29:21 --> 00:29:27
Remember what the solution
looked like.
446
00:29:25 --> 00:29:31
The solution looked like,
remember, you took the
447
00:29:28 --> 00:29:34
integrating factor was e to the
kt,
448
00:29:32 --> 00:29:38
and then after you integrated
both sides, multiplied through,
449
00:29:37 --> 00:29:43
and then the final answer
looked like this,
450
00:29:41 --> 00:29:47
y equaled, it was e to the
negative kt times
451
00:29:45 --> 00:29:51
either an indefinite integral,
or a definite integral
452
00:29:50 --> 00:29:56
depending on your preference,
q of t,
453
00:29:54 --> 00:30:00
so, x is metamorphosed into t.
I gather you've got that,
454
00:29:58 --> 00:30:04
e to the kt plus,
what was the other term?
455
00:30:02 --> 00:30:08
A constant times e to the
negative kt
456
00:30:08 --> 00:30:14
How does this fit into the
paradigm I've given you over
457
00:30:12 --> 00:30:18
there for solving the second
order equation?
458
00:30:15 --> 00:30:21
Which term is which?
Well, this has the arbitrary
459
00:30:19 --> 00:30:25
constant in it.
So, this must be the
460
00:30:22 --> 00:30:28
complementary solution.
Is it?
461
00:30:24 --> 00:30:30
Is this the solution to the
associated homogeneous equation?
462
00:30:30 --> 00:30:36
What's the associated
homogeneous equation?
463
00:30:32 --> 00:30:38
Put zero here.
Okay, if you put zero there,
464
00:30:35 --> 00:30:41
what's the solution?
Now, this you ought to know.
465
00:30:38 --> 00:30:44
y prime equals negative ky.
466
00:30:40 --> 00:30:46
What's the solution?
e to the negative kt.
467
00:30:43 --> 00:30:49
You are supposed to come into
468
00:30:46 --> 00:30:52
this course knowing that,
except there's an arbitrary
469
00:30:49 --> 00:30:55
constant in front.
So, right, this is exactly the
470
00:30:52 --> 00:30:58
solution to the associated
homogeneous equation,
471
00:30:55 --> 00:31:01
where there is zero here.
Then, what's this thing?
472
00:31:00 --> 00:31:06
This is a particular solution.
This is my yp.
473
00:31:02 --> 00:31:08
But that's not a particular
solution because this indefinite
474
00:31:06 --> 00:31:12
integral, you know,
has an arbitrary constant in
475
00:31:09 --> 00:31:15
it.
In fact, it's just that
476
00:31:10 --> 00:31:16
arbitrary constant.
So, it's totally confusing.
477
00:31:13 --> 00:31:19
But, this symbol,
you know when you actually
478
00:31:16 --> 00:31:22
solve the equation this way,
all you did was you found one
479
00:31:20 --> 00:31:26
function here.
You didn't throw in the
480
00:31:22 --> 00:31:28
arbitrary constant right away.
All you needed to do was find
481
00:31:26 --> 00:31:32
one function.
And, even if you really are
482
00:31:29 --> 00:31:35
bothered by the fact that this
is so indefinite,
483
00:31:32 --> 00:31:38
and therefore,
make it a particular solution
484
00:31:35 --> 00:31:41
by making this zero,
make it a definite integral,
485
00:31:38 --> 00:31:44
zero, here, t there,
and then change those t's to
486
00:31:42 --> 00:31:48
dummy t's, t1's or t tildes,
or something like that.
487
00:31:45 --> 00:31:51
So, this fits into that thing.
In other words,
488
00:31:48 --> 00:31:54
I could have done it at that
time, but I didn't the point
489
00:31:52 --> 00:31:58
because this can be solved
directly, whereas,
490
00:31:55 --> 00:32:01
of course, the general second
order equation in homogeneous
491
00:31:58 --> 00:32:04
cannot be solved directly,
and therefore you have to be
492
00:32:02 --> 00:32:08
willing to talk about what its
solutions look like in advance.
493
00:32:08 --> 00:32:14
Now, remember I said,
we talked, I said there was two
494
00:32:12 --> 00:32:18
different cases,
although both of them had the
495
00:32:15 --> 00:32:21
identical looking solution.
Their meaning in the physical
496
00:32:19 --> 00:32:25
world was so different that they
really should be considered as
497
00:32:24 --> 00:32:30
solving the same equation.
And, one of these was the case.
498
00:32:30 --> 00:32:36
Of the two, perhaps the more
important was the case when k
499
00:32:34 --> 00:32:40
was positive,
and of course the other is when
500
00:32:38 --> 00:32:44
k is negative.
When k is positive,
501
00:32:41 --> 00:32:47
that had the effect of
separating that solution into
502
00:32:45 --> 00:32:51
this part, which was a
transient, and the other part,
503
00:32:49 --> 00:32:55
which was a steady state.
The steady state solution,
504
00:32:53 --> 00:32:59
that was the yp part of it in
that terminology.
505
00:32:57 --> 00:33:03
And, the transient part,
it was trangent because it went
506
00:33:02 --> 00:33:08
to zero.
If k is positive,
507
00:33:05 --> 00:33:11
the exponential dies regardless
of what c is.
508
00:33:09 --> 00:33:15
So, the transient,
that's the yc part.
509
00:33:12 --> 00:33:18
It goes to zero as Ttgoes to
infinity.
510
00:33:16 --> 00:33:22
The transient depends on,
uses, the initial condition,
511
00:33:21 --> 00:33:27
whatever it is,
because that's what determines
512
00:33:25 --> 00:33:31
the value of c.
On the other hand,
513
00:33:28 --> 00:33:34
this initial condition makes no
difference as t goes towards
514
00:33:33 --> 00:33:39
infinity.
All that's left is this steady
515
00:33:38 --> 00:33:44
state solution.
And, all solutions tend to the
516
00:33:42 --> 00:33:48
steady state solution.
So, if k is positive,
517
00:33:46 --> 00:33:52
one gets this analysis of the
solutions into the sum of one
518
00:33:52 --> 00:33:58
basic solution,
and the others,
519
00:33:55 --> 00:34:01
which just die away,
have no influence on this,
520
00:33:59 --> 00:34:05
less and less influence as time
goes to infinity.
521
00:34:05 --> 00:34:11
For k less than zero,
this analysis does not work
522
00:34:09 --> 00:34:15
because this term,
if k is less than zero,
523
00:34:12 --> 00:34:18
this term goes to infinity or
negative infinity,
524
00:34:16 --> 00:34:22
and typically tends to dominate
that.
525
00:34:19 --> 00:34:25
So, it's the start that the
important one.
526
00:34:23 --> 00:34:29
It depends on the initial
conditions, and the analysis is
527
00:34:28 --> 00:34:34
meaningless.
So, the above is meaningless.
528
00:34:33 --> 00:34:39
And now, what I'd like to do is
try to see what the analog of
529
00:34:39 --> 00:34:45
that is for second order
equations, and higher order
530
00:34:45 --> 00:34:51
equations.
If you understand second-order,
531
00:34:49 --> 00:34:55
that's good enough.
Higher order goes exactly the
532
00:34:54 --> 00:35:00
same way.
So, the question is,
533
00:34:58 --> 00:35:04
for second-order,
let's make it with constant
534
00:35:02 --> 00:35:08
coefficients plus,
I could call it b and k,
535
00:35:07 --> 00:35:13
oh, no, b k,
or p.
536
00:35:11 --> 00:35:17
The trouble is,
that wouldn't take care of the
537
00:35:14 --> 00:35:20
electrical circuits.
So, I just want to use neutral
538
00:35:17 --> 00:35:23
letters, which suggest nothing.
And, you can make them turn it
539
00:35:22 --> 00:35:28
into a circuit,
so springs, or yet other
540
00:35:25 --> 00:35:31
examples undreamt of.
But these are constants.
541
00:35:28 --> 00:35:34
And I'm going to think of it as
time.
542
00:35:32 --> 00:35:38
I think I'll switch back to
time, let x be the time.
543
00:35:37 --> 00:35:43
So, B y equals f of t.
So, there is our equation.
544
00:35:43 --> 00:35:49
A and B are constants.
And, the question is,
545
00:35:47 --> 00:35:53
the question I'm asking,
can think of either of these
546
00:35:53 --> 00:35:59
two models or others,
the question I'm asking is,
547
00:35:58 --> 00:36:04
under what circumstances can I
make that same type of analysis
548
00:36:04 --> 00:36:10
into steady-state and transient?
Well, what does the solution
549
00:36:11 --> 00:36:17
look like?
The solution looks like y
550
00:36:15 --> 00:36:21
equals a particular solution
plus c1 y1 plus c2 y2.
551
00:36:20 --> 00:36:26
Therefore, to make that look
552
00:36:24 --> 00:36:30
like this, the c1 and c2 contain
the initial conditions.
553
00:36:31 --> 00:36:37
This part does not.
Therefore, if I want to say
554
00:36:35 --> 00:36:41
that the solutions look like a
steady state solution plus
555
00:36:41 --> 00:36:47
something that dies away,
which becomes less and less
556
00:36:47 --> 00:36:53
important as time goes on,
what I'm really asking is,
557
00:36:52 --> 00:36:58
under what circumstances is
this part guaranteed to go to
558
00:36:58 --> 00:37:04
zero?
So, the question is,
559
00:37:01 --> 00:37:07
when, in other words,
under what conditions on the
560
00:37:06 --> 00:37:12
equation A and B,
in effect, is what we are
561
00:37:10 --> 00:37:16
asking.
When does c1 y1 plus c2 y2 go
562
00:37:14 --> 00:37:20
to zero as t goes to infinity,
563
00:37:19 --> 00:37:25
regardless of what
c1 and c2 are for all c1 c2.
564
00:37:25 --> 00:37:31
Now, here there was no
difficulty.
565
00:37:30 --> 00:37:36
We had the thing very
explicitly, and you could see k
566
00:37:34 --> 00:37:40
is positive: this goes to zero.
And if k is negative,
567
00:37:38 --> 00:37:44
it doesn't go to zero.
It goes to infinity.
568
00:37:41 --> 00:37:47
Here, I want to make the same
kind of analysis,
569
00:37:44 --> 00:37:50
except it's just going to take,
it's a little more trouble.
570
00:37:49 --> 00:37:55
But the answer,
when it finally comes out is
571
00:37:52 --> 00:37:58
very beautiful.
So, when are all these guys
572
00:37:55 --> 00:38:01
going to go to zero?
First of all,
573
00:37:58 --> 00:38:04
you might as well just have the
definition.
574
00:38:01 --> 00:38:07
So, all the good things that
this is going to imply,
575
00:38:05 --> 00:38:11
if this is so,
in other words,
576
00:38:07 --> 00:38:13
if they all go to zero,
everything in the complementary
577
00:38:12 --> 00:38:18
solution, then the ODE is called
stable.
578
00:38:17 --> 00:38:23
Some people call it
asymptotically stable.
579
00:38:21 --> 00:38:27
I don't know what to call it.
I can make the analysis,
580
00:38:28 --> 00:38:34
and then I use the identical
terminology, c1 y1 plus c2 y2.
581
00:38:36 --> 00:38:42
This is called the transient
because it goes to zero.
582
00:38:40 --> 00:38:46
This is called the particular
solution now that we labored so
583
00:38:45 --> 00:38:51
hard to get for the next two
weeks.
584
00:38:47 --> 00:38:53
It's the important part.
It's the steady-state part.
585
00:38:52 --> 00:38:58
It's what lasts out to infinity
after the other stuff has
586
00:38:56 --> 00:39:02
disappeared.
So, this is the steady-state
587
00:38:59 --> 00:39:05
solution, steady-state solution,
okay?
588
00:39:04 --> 00:39:10
And, the differential equation
is called stable.
589
00:39:07 --> 00:39:13
Now, it's of the highest
interest to know when a
590
00:39:10 --> 00:39:16
differential equation is stable,
linear differential equation is
591
00:39:14 --> 00:39:20
stable in this sense because you
have a control.
592
00:39:17 --> 00:39:23
You know what its solutions
look like.
593
00:39:20 --> 00:39:26
You have some feeling for how
it's behaving in the long term.
594
00:39:24 --> 00:39:30
If this is not so,
each equation is a law unto
595
00:39:27 --> 00:39:33
itself if you don't know.
So, let's do the work.
596
00:39:31 --> 00:39:37
For the rest of the period,
what I'd like to do is to find
597
00:39:36 --> 00:39:42
out what the conditions are,
which make this true.
598
00:39:40 --> 00:39:46
Those were the equations which
we will have a right to call
599
00:39:45 --> 00:39:51
stable.
So, when does this happen,
600
00:39:48 --> 00:39:54
and where is it going to
happen?
601
00:39:50 --> 00:39:56
I don't know.
I guess, here.
602
00:39:54 --> 00:40:00
603
00:40:05 --> 00:40:11
Now, I think the first step is
fairly easy, and it will give
604
00:40:10 --> 00:40:16
you a good review of what we've
been doing up until now.
605
00:40:14 --> 00:40:20
So, I'm simply going to make a
case-by-case analysis.
606
00:40:19 --> 00:40:25
Don't worry,
it won't take very long.
607
00:40:22 --> 00:40:28
What are the cases we've been
studying?
608
00:40:25 --> 00:40:31
Well, what do the
characteristic roots look like?
609
00:40:29 --> 00:40:35
The roots of the characteristic
equation, in other words,
610
00:40:34 --> 00:40:40
remember, there are cases.
The first case is they are real
611
00:40:41 --> 00:40:47
and distinct,
r1 not equal to r2,
612
00:40:45 --> 00:40:51
real and distinct.
What are the other cases?
613
00:40:50 --> 00:40:56
Well, r1 equals r2.
And then, there's the case
614
00:40:56 --> 00:41:02
where there are complex.
So, I will write it r equals a
615
00:41:02 --> 00:41:08
plus or minus b i.
What do the solutions look
616
00:41:07 --> 00:41:13
like?
So, my ham-handed approach to
617
00:41:09 --> 00:41:15
this problem is going to be,
in each case,
618
00:41:12 --> 00:41:18
I'll look at the solutions,
and first get the condition on
619
00:41:16 --> 00:41:22
the roots.
So, in other words,
620
00:41:18 --> 00:41:24
I'm not going to worry right
away about the a and the b.
621
00:41:21 --> 00:41:27
I'm going, instead,
to worry about expressing this
622
00:41:24 --> 00:41:30
condition of stability in terms
of the characteristic roots.
623
00:41:28 --> 00:41:34
In fact, that's the only way in
which many people know the
624
00:41:32 --> 00:41:38
conditions.
You're going to be smarter.
625
00:41:35 --> 00:41:41
Okay, what do the solutions
look like?
626
00:41:38 --> 00:41:44
Well, the general solution
looks like e to the r1 t plus c2
627
00:41:42 --> 00:41:48
e to the r2 t.
628
00:41:45 --> 00:41:51
Okay, so, what's the stability
condition?
629
00:41:48 --> 00:41:54
In other words,
if equation happened to have
630
00:41:51 --> 00:41:57
its characteristic roots,
real and distinct,
631
00:41:54 --> 00:42:00
under what circumstances would
it be stable?
632
00:41:57 --> 00:42:03
Would it, in other words,
all its solutions go to zero?
633
00:42:01 --> 00:42:07
So, I'm talking about the
homogeneous equation,
634
00:42:04 --> 00:42:10
the reduced equation,
the associated homogeneous
635
00:42:07 --> 00:42:13
equation.
Why?
636
00:42:10 --> 00:42:16
Because that's all that's
involved in this.
637
00:42:13 --> 00:42:19
In other words,
when I write that,
638
00:42:15 --> 00:42:21
I am no longer interested in
the whole equation.
639
00:42:19 --> 00:42:25
All I'm interested in is the
reduced equation,
640
00:42:22 --> 00:42:28
the equation where you turn the
f of t on the
641
00:42:26 --> 00:42:32
right-hand side into zero.
So, what's the stability
642
00:42:31 --> 00:42:37
condition?
Well, let's write it out.
643
00:42:35 --> 00:42:41
Under what circumstances will
all these guys go to zero?
644
00:42:41 --> 00:42:47
If r1 and r2 should be
negative, can they be zero?
645
00:42:46 --> 00:42:52
No, because then it will be a
constant and it will go to zero.
646
00:42:52 --> 00:42:58
How about this one?
Well, in this one,
647
00:42:56 --> 00:43:02
it's (c1 plus c2 times t)
multiplied by e to the r1 t.
648
00:43:01 --> 00:43:07
Of course, both of these are
649
00:43:08 --> 00:43:14
the same.
I'll just arbitrarily pick one
650
00:43:11 --> 00:43:17
of them.
What happens to this as things
651
00:43:14 --> 00:43:20
go to zero?
Well, this part is rising,
652
00:43:16 --> 00:43:22
at least if c2 is positive.
This part is either helping or
653
00:43:21 --> 00:43:27
it's hindering.
But, I hope you know what these
654
00:43:24 --> 00:43:30
functions look like,
and you know which of them go
655
00:43:28 --> 00:43:34
to zero.
They go to zero if r1 is
656
00:43:31 --> 00:43:37
negative.
It might rise in the beginning,
657
00:43:35 --> 00:43:41
but after a while they lose the
energy.
658
00:43:39 --> 00:43:45
Of course, if r1 is equal to
zero, what do these guys do?
659
00:43:44 --> 00:43:50
Linear, go to infinity.
Well, we are doing okay.
660
00:43:49 --> 00:43:55
How about here?
Well, here, it's a little more
661
00:43:53 --> 00:43:59
complicated.
The solutions look like e to
662
00:43:57 --> 00:44:03
the at times (c1 cosine bt plus
c2 sine bt).
663
00:44:03 --> 00:44:09
Now, this part is a pure
664
00:44:07 --> 00:44:13
oscillation.
You know that.
665
00:44:09 --> 00:44:15
It might have a big amplitude,
but whatever it does,
666
00:44:13 --> 00:44:19
it does the same thing all the
time.
667
00:44:16 --> 00:44:22
So, whether this goes to zero
depends entirely upon what that
668
00:44:20 --> 00:44:26
exponential is doing.
And, that exponential goes to
669
00:44:24 --> 00:44:30
zero if a is negative.
So here, the condition is
670
00:44:28 --> 00:44:34
negative.
And now, the only thing left to
671
00:44:33 --> 00:44:39
do is to say it nicely.
I've got three cases,
672
00:44:38 --> 00:44:44
and I want to say them all in
one breath.
673
00:44:43 --> 00:44:49
So, the stability condition is,
the ODE is stable.
674
00:44:49 --> 00:44:55
So, this is,
or f of t.
675
00:44:53 --> 00:44:59
It doesn't matter.
But, psychologically,
676
00:44:57 --> 00:45:03
you can put this as zero there,
is stable if what?
677
00:45:05 --> 00:45:11
In case one,
this is true.
678
00:45:07 --> 00:45:13
In case two,
that's true.
679
00:45:10 --> 00:45:16
In case three,
that's true.
680
00:45:13 --> 00:45:19
But that's ugly.
Make it beautiful.
681
00:45:17 --> 00:45:23
The beautiful way of saying it
is if all the characteristic
682
00:45:23 --> 00:45:29
roots have negative real parts.
If the characteristic roots,
683
00:45:31 --> 00:45:37
the r's or the a plus or minus
b i, have negative real part.
684
00:45:38 --> 00:45:44
That's the form in which the
electrical engineers will nod
685
00:45:44 --> 00:45:50
their head, tell you,
yeah, that's right,
686
00:45:49 --> 00:45:55
negative real part,
sorry.
687
00:45:52 --> 00:45:58
Isn't it right?
Is that right here?
688
00:45:56 --> 00:46:02
Yeah.
What's the real part of these
689
00:45:59 --> 00:46:05
guys?
They themselves,
690
00:46:03 --> 00:46:09
because they are real.
What's the real part of this?
691
00:46:09 --> 00:46:15
Yeah.
The only case in which I really
692
00:46:13 --> 00:46:19
had to use real part is when I
talk about the complex case
693
00:46:20 --> 00:46:26
because a is just the real part
of a complex number.
694
00:46:26 --> 00:46:32
It's not the whole thing.