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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

218
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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.

224
00:14:31 --> 00:14:37

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00:14:43 --> 00:14:49
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

229
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that circuit.
And now, you can see,

230
00:15:01 --> 00:15:07
again, what does it mean?
If this is zero,

231
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for example,
if I have a dry cell there,

232
00:15:07 --> 00:15:13
or if I have nothing at all in
the circuit, then this

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

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

235
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on the capacitor,
and waited, and of course,

236
00:15:22 --> 00:15:28
when you put a charge on there,
it's got a discharge,

237
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and discharges through the
circuit, and swings back and

238
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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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00:15:36 --> 00:15:42
But, what usually happens is
that you drive this passive

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

242
00:15:44 --> 00:15:50
know how the current behaves.
So, those are the two problems,

243
00:15:48 --> 00:15:54
the passive circuit without an
applied electromotive force,

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

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

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

247
00:16:03 --> 00:16:09
which do things because,
okay, that's why they want to

248
00:16:07 --> 00:16:13
solve inhomogeneous equations
instead of homogeneous

249
00:16:11 --> 00:16:17
equations.
But as I said,

250
00:16:13 --> 00:16:19
you have to do the homogeneous
case first.

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

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

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

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

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

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

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

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

259
00:16:49 --> 00:16:55
equations we'll be talking
about.

260
00:16:52 --> 00:16:58
I now want to tell you briefly
about the key theorem about

261
00:16:56 --> 00:17:02
solving the homogeneous
equation.

262
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.