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The last time I spent solving a
system of equations dealing with
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the chilling of this hardboiled
egg being put in an ice bath.
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We called T1 the temperature of
the yoke and T2 the temperature
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of the white.
What I am going to do is
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revisit that same system of
equations, but basically the
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topic for today is to learn to
solve that system of equations
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by a completely different
method.
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It is the method that is
normally used in practice.
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Elimination is used mostly by
people who have forgotten how to
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do it any other way.
Now, in order to make it a
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little more general,
I am not going to use the
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dependent variables T1 and T2
because they suggest temperature
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a little too closely.
Let's change them to neutral
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variables.
I will use x equals T1,
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and for T2 I will just use
y.
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I am not going to re-derive
anything.
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I am not going to resolve
anything.
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I am not going to repeat
anything of what I did last
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time, except to write down to
remind you what the system was
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in terms of these variables,
the system we derived using the
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particular conductivity
constants, two and three,
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respectively.
The system was this one,
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minus 2x plus 2y.
And the y prime was
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2x minus 5y.
And so we solved this by
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elimination.
We got a single second-order
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equation with constant
coefficients,
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which we solved in the usual
way.
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From that I derived what the x
was, from that we derived what
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the y was, and then I put them
all together.
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I will just remind you what the
final solution was when written
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out in terms of arbitrary
constants.
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It was c1 times e to the
negative t plus c2 e to the
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negative 6t,
and y was c1 over 2 e
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to the negative t minus 2c2 e to
the negative 6t.
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That was the solution we got.
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And then I went on to put in
initial conditions,
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but we are not going to explore
that aspect of it today.
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We will in a week or so.
This was the general solution
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because it had two arbitrary
constants in it.
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What I want to do now is
revisit this and do it by a
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different method,
which makes heavy use of
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matrices.
That is a prerequisite for this
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course, so I am assuming that
you reviewed a little bit about
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matrices.
And it is in your book.
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Your book puts in a nice little
review section.
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Two-by-two and three-by-three
will be good enough for 18.03
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mostly because I don't want you
to calculate all night on bigger
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matrices, bigger systems.
So nothing serious,
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matrix multiplication,
solving systems of linear
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equations, end-by-end systems.
I will remind you at the
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appropriate places today of what
it is you need to remember.
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The very first thing we are
going to do is,
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let's see.
I haven't figured out the color
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coding for this lecture yet,
but let's make this system in
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green and the solution can be in
purple.
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Invisible purple,
but I have a lot of it.
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Let's abbreviate,
first of all,
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the system using matrices.
I am going to make a column
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vector out of (x,
y).
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Then you differentiate a column
vector by differentiating each
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component.
I can write the left-hand side
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of the system as (x,
y) prime.
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How about the right-hand side?
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Well, I say I can just write
the matrix of coefficients to
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negative 2, 2,
2, negative 5 times x,y.
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And I say that this matrix
equation says exactly the same
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thing as that green equation
and, therefore,
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it is legitimate to put it up
in green, too.
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The top here is x prime.
What is the top here?
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After I multiply these two I
get a column vector.
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And what is its top entry?
It is negative 2x plus 2y.
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There it is.
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And the bottom entry the same
way is 2x minus 5y,
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just as it is down there.
Now, what I want to do is,
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well, maybe I should translate
the solution.
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What does the solution look
like?
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We got that,
too.
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How am I going to write this as
a matrix equation?
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Actually, if I told you to use
matrices, use vectors,
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the point at which you might be
most hesitant is this one right
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here, the very next step.
Because how you should write it
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is extremely well-concealed in
this notation.
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But the point is,
this is a column vector and I
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am adding together two column
vectors.
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And what is in each one of the
column vectors?
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Think of these two things as a
column vector.
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Pull out all the scalars from
them that you can.
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Well, you see that c1 is a
common factor of both entries
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and so is e to the negative t,
that function.
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Now, if I pull both of those
out of the vector,
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what is left of the vector?
Well, you cannot even see it.
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What is left is a 1 up here and
a one-half there.
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So I am going to write that in
the following form.
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I will put out the c1,
it's the common factor in both,
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and put that out front.
Then I will put in the guts of
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the vector, even though you
cannot see it,
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the column vector 1,
one-half.
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And then I will put the other
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scalar function in back.
The only reason for putting one
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of these in front and one in
back is visual so to make it
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easy to read.
There is no other reason.
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You could put the c1 here,
you could put it here,
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you could put the e negative t
in front if you want
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to, but people will fire you.
Don't do that.
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Write it the standard way
because that is the way that it
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is easiest to read.
The constants out front,
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the functions behind,
and the column vector of
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numbers in the middle.
And so the other one will be
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written how?
Well, here, that one is a
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little more transparent.
c2, 1, 2 and the other thing is
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e to the negative 6t.
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There is our solution.
That is going to need a lot of
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purple, but I have it.
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And now I want to talk about
how the new method of solving
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the equation.
It is based just on the same
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idea as the way we solve
second-order equations.
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Yes, question.
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Oh, here.
Sorry.
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This should be negative two.
Thanks very much.
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What I am going to use is a
trial solution.
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Remember when we had a
second-order equation with
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constant coefficients the very
first thing I did was I said we
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are going to try a solution of
the form e to the rt.
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Why that?
Well, because Oiler thought of
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it and it has been known for
or 300 years that that is the
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thing you should do.
Well, this has not been known
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nearly as long because matrices
were only invented around
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or so, and people did not really
use them to solve systems of
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differential equations until the
middle of the last century,
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1950-1960.
If you look at books written in
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1950, they won't even talk about
systems of differential
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equations, or talk very little
anyway and they won't solve them
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using matrices.
This is only 50 years old.
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I mean, my God,
in mathematics that is very up
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to date, particularly elementary
mathematics.
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Anyway, the method of solving
is going to use as a trial
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solution.
Now, if you were left to your
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own devices you might say,
well, let's try x equals some
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constant times e to the lambda1
t and y
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equals some other constant times
e to the lambda2 t.
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Now, if you try that,
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it is a sensible thing to try,
but it will turn out not to
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work.
And that is the reason I have
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written out this particular
solution, so we can see what
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solutions look like.
The essential point is here is
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the basic solution I am trying
to find.
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Here is another one.
Their form is a column vector
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of constants.
But they both use the same
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exponential factor,
which is the point.
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In other words,
I should not use here,
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in my trial solution,
two different lambdas,
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I should use the same lambda.
And so the way to write the
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trial solution is (x,
y) equals two unknown numbers,
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that or that or whatever,
times e to a single unknown
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exponent factor.
Let's call it lambda t.
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It is called lambda.
It is called r.
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It is called m.
I have never seen it called
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anything but one of those three
things.
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I am using lambda.
Your book uses lambda.
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It is a common choice.
Let's stick with it.
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Now what is the next step?
Well, we plug into the system.
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Substitute into the system.
What are we going to get?
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Well, let's do it.
First of all,
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I have to differentiate.
The left-hand side asks me to
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differentiate this.
How do I differentiate this?
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Column vector times a function.
Well, the column vector acts as
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a constant.
And I differentiate that.
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That is lambda e to the lambda
t.
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So the (x, y) prime is (a1,
a2) times e to the lambda t
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times lambda.
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Now, it is ugly to put the
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lambda afterwards because it is
a number so you should put it in
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front, again,
to make things easier to read.
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But this lambda comes from
differentiating e to the lambda
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t and using the chain rule.
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This much is the left-hand
side.
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That is the derivative (x,
y) prime.
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I differentiate the x and I
differentiated the y.
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How about the right-hand side.
Well, the right-hand side is
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negative 2, 2,
2, negative 5 times what?
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Well, times (x,
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y), which is (a1,
a2) e to the lambda t.
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Now, the same thing that
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happened a month or a month and
a half ago happens now.
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The whole point of making that
substitution is that the e to
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the lambda t,
the function part of it drops
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out completely.
And one is left with what?
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An algebraic equation to be
solved for lambda a1 and a2.
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In other words,
by means of that substitution,
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and it basically uses the fact
that the coefficients are
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constant, what you have done is
reduced the problem of calculus,
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of solving differential
equations, to solving algebraic
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equations.
In some sense that is the only
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method there is,
unless you do numerical stuff.
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You reduce the calculus to
algebra.
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The Laplace transform is
exactly the same thing.
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All the work is algebra.
You turn the original
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differential equation into an
algebraic equation for Y of s,
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you solve it,
and then you use more algebra
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to find out what the original
little y of t was.
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It is not different here.
So let's solve this system of
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equations.
Now, the whole problem with
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solving this system,
first of all,
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what is the system?
Let's write it out explicitly.
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Well, it is really two
equations, isn't it?
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The first one says lambda a1 is
equal to negative 2 a1
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plus 2 a2.
That is the first one.
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The other one says lambda a2 is
equal to 2 a1 minus 5 a2.
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Now, purely,
if you want to classify that,
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that is two equations and three
variables, three unknowns.
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The a1, a2, and lambda are all
unknown.
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And, unfortunately,
if you want to classify them
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correctly, they are nonlinear
equations because they are made
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nonlinear by the fact that you
have multiplied two of the
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variables.
Well, if you sit down and try
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to hack away at solving those
without a plan,
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you are not going to get
anywhere.
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It is going to be a mess.
Also, two equations and three
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unknowns is indeterminate.
You can solve three equations
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and three unknowns and get a
definite answer,
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but two equations and three
unknowns usually have an
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infinity of solutions.
Well, at this point it is the
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only idea that is required.
Well, this was a little idea,
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but I assume one would think of
that.
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And the idea that is required
here is, I think,
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not so unnatural,
it is not to view these a1,
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a2, and lambda as equal.
Not all variables are created
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equal.
Some are more equal than
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others.
a1 and a2 are definitely equal
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to each other,
and let's relegate lambda to
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the background.
In other words,
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I am going to think of lambda
as just a parameter.
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I am going to demote it from
the status of variable to
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parameter.
If I demoted it further it
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would just be an unknown
constant.
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That is as bad as you can be.
I am going to focus my
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attention on the a1,
a2 and sort of view the lambda
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as a nuisance.
Now, as soon as I do that,
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I see that these equations are
linear if I just look at them as
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equations in a1 and a2.
And moreover,
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they are not just linear,
they are homogenous.
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Because if I think of lambda
just as a parameter,
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I should rewrite the equations
this way.
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I am going to subtract this and
move the left-hand side to the
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00:16:19 --> 00:16:25
right side, and it is going to
look like (minus 2 minus lambda)
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times a1 plus 2 a2 is equal to
zero.
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And the same way for the other
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one.
It is going to be 2a1 plus,
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what is the coefficient,
(minus 5 minus lambda) a2
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equals zero.
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That is a pair of simultaneous
linear equations for determining
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a1 and a2, and the coefficients
involved are parameter lambda.
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Now, what is the point of doing
that?
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Well, now the point is whatever
you learned about linear
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equations, you should have
learned the most fundamental
265
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theorem of linear equations.
The main theorem is that you
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00:17:10 --> 00:17:16
have a square system of
homogeneous equations,
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this is a two-by-two system so
it is square,
268
00:17:16 --> 00:17:22
it always has the trivial
solution, of course,
269
00:17:20 --> 00:17:26
a1, a2 equals zero.
Now, we don't want that trivial
270
00:17:24 --> 00:17:30
solution because if a1 and a2
are zero, then so are x and y
271
00:17:28 --> 00:17:34
zero.
Now that is a solution.
272
00:17:32 --> 00:17:38
Unfortunately,
it is of no interest.
273
00:17:35 --> 00:17:41
If the solution were x,
y zero, it corresponds to the
274
00:17:39 --> 00:17:45
fact that this is an ice bath.
The yoke is at zero,
275
00:17:44 --> 00:17:50
the white is at zero and it
stays that way for all time
276
00:17:48 --> 00:17:54
until the ice melts.
So that is the solution we
277
00:17:52 --> 00:17:58
don't want.
We don't want the trivial
278
00:17:56 --> 00:18:02
solution.
Well, when does it have a
279
00:17:59 --> 00:18:05
nontrivial solution?
Nontrivial means non-zero,
280
00:18:03 --> 00:18:09
in other words.
If and only if this determinant
281
00:18:09 --> 00:18:15
is zero.
282
00:18:10 --> 00:18:16
283
00:18:18 --> 00:18:24
In other words,
by using that theorem on linear
284
00:18:22 --> 00:18:28
equations, what we find is there
is a condition that lambda must
285
00:18:28 --> 00:18:34
satisfy, an equation in lambda
in order that we would be able
286
00:18:33 --> 00:18:39
to find non-zero values for a1
and a2.
287
00:18:38 --> 00:18:44
Let's write it out.
I will recopy it over here.
288
00:18:41 --> 00:18:47
What was it?
Negative 2 minus lambda,
289
00:18:44 --> 00:18:50
two, here it was 2 and minus 5
minus lambda.
290
00:18:49 --> 00:18:55
All right.
You have to expand the
291
00:18:51 --> 00:18:57
determinant.
In other words,
292
00:18:54 --> 00:19:00
we are trying to find out for
what values of lambda is this
293
00:18:58 --> 00:19:04
determinant zero.
Those will be the good values
294
00:19:03 --> 00:19:09
which lead to nontrivial
solutions for the a's.
295
00:19:06 --> 00:19:12
This is the equation lambda
plus 2.
296
00:19:09 --> 00:19:15
See, this is minus that and
minus that, the product of the
297
00:19:14 --> 00:19:20
two minus ones is plus one.
So it is lambda plus 2 times
298
00:19:18 --> 00:19:24
lambda plus 5,
299
00:19:21 --> 00:19:27
which is the product of the two
diagonal elements,
300
00:19:25 --> 00:19:31
minus the product of the two
anti-diagonal elements,
301
00:19:29 --> 00:19:35
which is 4, is equal to zero.
And if I write that out,
302
00:19:35 --> 00:19:41
what is that,
that is the equation lambda
303
00:19:39 --> 00:19:45
squared plus 7 lambda,
304
00:19:43 --> 00:19:49
5 lambda plus 2 lambda,
and then the constant term is
305
00:19:48 --> 00:19:54
10 minus 4 which is 6.
How many of you have long
306
00:19:52 --> 00:19:58
enough memories,
two-day memories that you
307
00:19:56 --> 00:20:02
remember that equation?
When I did the method of
308
00:20:02 --> 00:20:08
elimination, it led to exactly
the same equation except it had
309
00:20:10 --> 00:20:16
r's in it instead of lambda.
And this equation,
310
00:20:15 --> 00:20:21
therefore, is given the same
name and another color.
311
00:20:21 --> 00:20:27
Let's make it salmon.
312
00:20:25 --> 00:20:31
313
00:20:30 --> 00:20:36
And it is called the
characteristic equation for this
314
00:20:34 --> 00:20:40
method.
All right.
315
00:20:36 --> 00:20:42
Now I am going to use now the
word from last time.
316
00:20:40 --> 00:20:46
You factor this.
From the factorization we get
317
00:20:44 --> 00:20:50
its root easily enough.
The roots are lambda equals
318
00:20:48 --> 00:20:54
negative 1 and
lambda equals negative 6
319
00:20:54 --> 00:21:00
by factoring the
equation.
320
00:20:57 --> 00:21:03
Now what I am supposed to do?
You have to keep the different
321
00:21:03 --> 00:21:09
parts of the method together.
Now I have found the only
322
00:21:08 --> 00:21:14
values of lambda for which I
will be able to find nonzero
323
00:21:12 --> 00:21:18
values for the a1 and a2.
For each of those values of
324
00:21:17 --> 00:21:23
lambda, I now have to find the
corresponding a1 and a2.
325
00:21:21 --> 00:21:27
Let's do them one at a time.
Let's take first lambda equals
326
00:21:26 --> 00:21:32
negative one.
My problem is now to find a1
327
00:21:32 --> 00:21:38
and a2.
Where am I going to find them
328
00:21:35 --> 00:21:41
from?
Well, from that system of
329
00:21:38 --> 00:21:44
equations over there.
I will recopy it over here.
330
00:21:42 --> 00:21:48
What is the system?
The hardest part of this is
331
00:21:47 --> 00:21:53
dealing with multiple minus
signs, but you had experience
332
00:21:52 --> 00:21:58
with that in determinants so you
know all about that.
333
00:21:58 --> 00:22:04
In other words,
there is the system of
334
00:22:01 --> 00:22:07
equations over there.
Let's recopy them here.
335
00:22:06 --> 00:22:12
Minus 2, minus minus 1 makes
minus 1.
336
00:22:09 --> 00:22:15
What's the other coefficient?
It is just plain old 2.
337
00:22:15 --> 00:22:21
Good.
There is my first equation.
338
00:22:18 --> 00:22:24
And when I substitute lambda
equals negative one
339
00:22:24 --> 00:22:30
for the second equation,
what do you get?
340
00:22:30 --> 00:22:36
2 a1 plus negative 5 minus
negative 1 makes negative 4.
341
00:22:37 --> 00:22:43
There is my system that will
find me a1 and a2.
342
00:22:43 --> 00:22:49
What is the first thing you
notice about it?
343
00:22:48 --> 00:22:54
You immediately notice that
this system is fake because this
344
00:22:56 --> 00:23:02
second equation is twice the
first one.
345
00:23:03 --> 00:23:09
Something is wrong.
No, something is right.
346
00:23:06 --> 00:23:12
If that did not happen,
if the second equation were not
347
00:23:12 --> 00:23:18
a constant multiple of the first
one then the only solution of
348
00:23:17 --> 00:23:23
the system would be a1 equals
zero, a2 equals zero because the
349
00:23:23 --> 00:23:29
determinant of the coefficients
would not be zero.
350
00:23:29 --> 00:23:35
The whole function of this
exercise was to find the value
351
00:23:33 --> 00:23:39
of lambda, negative 1,
for which the system would be
352
00:23:37 --> 00:23:43
redundant and,
therefore, would have a
353
00:23:40 --> 00:23:46
nontrivial solution.
Do you get that?
354
00:23:43 --> 00:23:49
In other words,
calculate the system out,
355
00:23:47 --> 00:23:53
just as I have done here,
you have an automatic check on
356
00:23:51 --> 00:23:57
the method.
If one equation is not a
357
00:23:54 --> 00:24:00
constant multiple of the other
you made a mistake.
358
00:24:00 --> 00:24:06
You don't have the right value
of lambda or you substituted
359
00:24:05 --> 00:24:11
into the system wrong,
which is frankly a more common
360
00:24:10 --> 00:24:16
error.
Go back, recheck first the
361
00:24:13 --> 00:24:19
substitution,
and if convinced that is right
362
00:24:17 --> 00:24:23
then recheck where you got
lambda from.
363
00:24:20 --> 00:24:26
But here everything is going
fine so we can now find out what
364
00:24:26 --> 00:24:32
the value of a1 and a2 are.
You don't have to go through a
365
00:24:32 --> 00:24:38
big song and dance for this
since most of the time you will
366
00:24:36 --> 00:24:42
have two-by-two equations and
now and then three-by-three.
367
00:24:40 --> 00:24:46
For two-by-two all you do is,
since we really have the same
368
00:24:44 --> 00:24:50
equation twice,
to get a solution I can assign
369
00:24:47 --> 00:24:53
one of the variables any value
and then simply solve for the
370
00:24:51 --> 00:24:57
other.
The natural thing to do is to
371
00:24:54 --> 00:25:00
make a2 equal one,
then I won't need fractions and
372
00:24:57 --> 00:25:03
then a1 will be a2.
So the solution is (2,
373
00:25:01 --> 00:25:07
1).
I am only trying to find one
374
00:25:04 --> 00:25:10
solution.
Any constant multiple of this
375
00:25:07 --> 00:25:13
would also be a solution,
as long as it wasn't zero,
376
00:25:12 --> 00:25:18
zero which is the trivial one.
And, therefore,
377
00:25:15 --> 00:25:21
this is a solution to this
system of algebraic equations.
378
00:25:20 --> 00:25:26
And the solution to the whole
system of differential equations
379
00:25:25 --> 00:25:31
is, this is only the (a1,
a2) part.
380
00:25:30 --> 00:25:36
I have to add to it,
as a factor,
381
00:25:32 --> 00:25:38
lambda is negative,
therefore, e to the minus t.
382
00:25:36 --> 00:25:42
There is our purple thing.
383
00:25:40 --> 00:25:46
384
00:25:50 --> 00:25:56
See how I got it?
Starting with the trial
385
00:25:52 --> 00:25:58
solution, I first found out
through this procedure what the
386
00:25:57 --> 00:26:03
lambda's have to be.
Then I took the lambda and
387
00:26:00 --> 00:26:06
found what the corresponding a1
and a2 that went with it and
388
00:26:04 --> 00:26:10
then made up my solution out of
that.
389
00:26:06 --> 00:26:12
Now, quickly I will do the same
thing for lambda
390
00:26:11 --> 00:26:17
equals negative 6.
Each one of these must be
391
00:26:13 --> 00:26:19
treated separately.
They are separate problems and
392
00:26:17 --> 00:26:23
you are looking for separate
solutions.
393
00:26:19 --> 00:26:25
Lambda equals negative 6.
What do I do?
394
00:26:22 --> 00:26:28
How do my equations look now?
Well, the first one is minus 2
395
00:26:25 --> 00:26:31
minus negative 6 makes plus 4.
It is 4a1 plus 2a2 equals zero.
396
00:26:32 --> 00:26:38
Then I hold my breath while I
397
00:26:37 --> 00:26:43
calculate the second one to see
if it comes out to be a constant
398
00:26:44 --> 00:26:50
multiple.
I get 2a1 plus negative 5 minus
399
00:26:49 --> 00:26:55
negative 6, which makes plus 1.
And, indeed,
400
00:26:54 --> 00:27:00
one is a constant multiple of
the other.
401
00:27:00 --> 00:27:06
I really only have on equation
there.
402
00:27:04 --> 00:27:10
I will just write down
immediately now what the
403
00:27:09 --> 00:27:15
solution is to the system.
Well, the (a1,
404
00:27:13 --> 00:27:19
a2) will be what?
Now, it is more natural to make
405
00:27:19 --> 00:27:25
a1 equal 1 and then solve to get
an integer for a2.
406
00:27:25 --> 00:27:31
If a1 is 1, then a2 is negative
2.
407
00:27:30 --> 00:27:36
And I should multiply that by e
to the negative 6t
408
00:27:34 --> 00:27:40
because negative 6 is the
corresponding value.
409
00:27:38 --> 00:27:44
There is my other one.
And now there is a
410
00:27:41 --> 00:27:47
superposition principle,
which if I get a chance will
411
00:27:44 --> 00:27:50
prove for you at the end of the
hour.
412
00:27:47 --> 00:27:53
If not, you will have to do it
yourself for homework.
413
00:27:51 --> 00:27:57
Since this is a linear system
of equations,
414
00:27:54 --> 00:28:00
once you have two separate
solutions, neither a constant
415
00:27:58 --> 00:28:04
multiple of the other,
you can multiply each one of
416
00:28:02 --> 00:28:08
these by a constant and it will
still be a solution.
417
00:28:08 --> 00:28:14
You can add them together and
that will still be a solution,
418
00:28:11 --> 00:28:17
and that gives the general
solution.
419
00:28:13 --> 00:28:19
The general solution is the sum
of these two,
420
00:28:16 --> 00:28:22
an arbitrary constant.
I am going to change the name
421
00:28:19 --> 00:28:25
since I don't want to confuse it
with the c1 I used before,
422
00:28:22 --> 00:28:28
times the first solution which
is (2, 1) e to the negative t
423
00:28:26 --> 00:28:32
plus c2, another arbitrary
constant, times 1 negative 2 e
424
00:28:29 --> 00:28:35
to the minus 6t.
425
00:28:32 --> 00:28:38
Now you notice that is exactly
426
00:28:37 --> 00:28:43
the same solution I got before.
The only difference is that I
427
00:28:43 --> 00:28:49
have renamed the arbitrary
constants.
428
00:28:47 --> 00:28:53
The relationship between them,
c1 over 2,
429
00:28:53 --> 00:28:59
I am now calling c1 tilda,
and c2 I am calling c2 tilda.
430
00:29:00 --> 00:29:06
If you have an arbitrary
constant, it doesn't matter
431
00:29:04 --> 00:29:10
whether you divide it by two.
It is still just an arbitrary a
432
00:29:09 --> 00:29:15
constant.
It covers all values,
433
00:29:12 --> 00:29:18
in other words.
Well, I think you will agree
434
00:29:16 --> 00:29:22
that is a different procedure,
yet it has only one
435
00:29:20 --> 00:29:26
coincidence.
It is like elimination goes
436
00:29:24 --> 00:29:30
this way and comes to the
answer.
437
00:29:28 --> 00:29:34
And this method goes a
completely different route and
438
00:29:31 --> 00:29:37
comes to the answer,
except it is not quite like
439
00:29:34 --> 00:29:40
that.
They walk like this and then
440
00:29:36 --> 00:29:42
they come within viewing
distance of each other to check
441
00:29:40 --> 00:29:46
that both are using the same
characteristic equation,
442
00:29:44 --> 00:29:50
and then they again go their
separate ways and end up with
443
00:29:48 --> 00:29:54
the same answer.
444
00:29:50 --> 00:29:56
445
00:29:58 --> 00:30:04
There is something special of
these values.
446
00:30:00 --> 00:30:06
You cannot get away from those
two values of lambda.
447
00:30:03 --> 00:30:09
Somehow they are really
intrinsically connected.
448
00:30:06 --> 00:30:12
Occurs the exponential
coefficient, and they are
449
00:30:09 --> 00:30:15
intrinsically connected with the
problem of the egg that we
450
00:30:12 --> 00:30:18
started with.
Now what I would like to do is
451
00:30:15 --> 00:30:21
very quickly sketch how this
method looks when I remove all
452
00:30:18 --> 00:30:24
the numbers from it.
In some sense,
453
00:30:20 --> 00:30:26
it becomes a little clearer
what is going on.
454
00:30:23 --> 00:30:29
And that will give me a chance
to introduce the terminology
455
00:30:26 --> 00:30:32
that you need when you talk
about it.
456
00:30:30 --> 00:30:36
457
00:30:55 --> 00:31:01
Well, you have notes.
Let me try to write it down in
458
00:31:03 --> 00:31:09
general.
459
00:31:05 --> 00:31:11
460
00:31:10 --> 00:31:16
I will first write it out
two-by-two.
461
00:31:13 --> 00:31:19
I am just going to sketch.
The system looks like (x,
462
00:31:18 --> 00:31:24
y) equals, I will still put it
up in colors.
463
00:31:23 --> 00:31:29
464
00:31:30 --> 00:31:36
Except now, instead of using
twos and fives,
465
00:31:33 --> 00:31:39
I will use (a,
b; c, d).
466
00:31:35 --> 00:31:41
467
00:31:40 --> 00:31:46
The trial solution will look
how?
468
00:31:44 --> 00:31:50
The trial is going to be (a1,
a2).
469
00:31:48 --> 00:31:54
That I don't have to change the
name of.
470
00:31:53 --> 00:31:59
I am going to substitute in,
and what the result of
471
00:31:59 --> 00:32:05
substitution is going to be
lambda (a1, a2).
472
00:32:06 --> 00:32:12
I am going to skip a step and
pretend that the e to the lambda
473
00:32:11 --> 00:32:17
t's have already
been canceled out.
474
00:32:16 --> 00:32:22
Is equal to (a,
b; c, d) times (a1,
475
00:32:19 --> 00:32:25
a2).
What does that correspond to?
476
00:32:22 --> 00:32:28
That corresponds to the system
as I wrote it here.
477
00:32:28 --> 00:32:34
And then we wrote it out in
terms of two equations.
478
00:32:31 --> 00:32:37
And what was the resulting
thing that we ended up with?
479
00:32:35 --> 00:32:41
Well, you write it out,
you move the lambda to the
480
00:32:38 --> 00:32:44
other side.
And then the homogeneous system
481
00:32:41 --> 00:32:47
is we will look in general how?
Well, we could write it out.
482
00:32:45 --> 00:32:51
It is going to look like a
minus lambda,
483
00:32:48 --> 00:32:54
b, c, d minus lambda.
484
00:32:50 --> 00:32:56
That is just how it looks there
485
00:32:53 --> 00:32:59
and the general calculation is
the same.
486
00:32:56 --> 00:33:02
Times (a1, a2) is equal to
zero.
487
00:33:00 --> 00:33:06
488
00:33:05 --> 00:33:11
This is solvable nontrivially.
In other words,
489
00:33:13 --> 00:33:19
it has a nontrivial solution if
an only if the determinant of
490
00:33:25 --> 00:33:31
coefficients is zero.
491
00:33:30 --> 00:33:36
492
00:33:35 --> 00:33:41
Let's now write that out,
calculate out once and for all
493
00:33:39 --> 00:33:45
what that determinant is.
I will write it out here.
494
00:33:43 --> 00:33:49
It is a minus lambda times d
minus lambda,
495
00:33:46 --> 00:33:52
the product of the diagonal
elements, minus the
496
00:33:50 --> 00:33:56
anti-diagonal minus bc is equal
to zero.
497
00:33:55 --> 00:34:01
And let's calculate that out.
498
00:34:00 --> 00:34:06
It is lambda squared minus a
lambda minus d lambda plus ad,
499
00:34:07 --> 00:34:13
the constant term from here,
negative bc from there,
500
00:34:14 --> 00:34:20
plus ad minus bc,
501
00:34:21 --> 00:34:27
where have I seen that before?
This equation is the general
502
00:34:29 --> 00:34:35
form using letters of what we
calculated using the specific
503
00:34:36 --> 00:34:42
numbers before.
Again, I will code it the same
504
00:34:45 --> 00:34:51
way with that color salmon.
Now, most of the calculations
505
00:34:54 --> 00:35:00
will be for two-by-two systems.
I advise you,
506
00:35:00 --> 00:35:06
in the strongest possible
terms, to remember this
507
00:35:03 --> 00:35:09
equation.
You could write down this
508
00:35:06 --> 00:35:12
equation immediately for the
matrix.
509
00:35:08 --> 00:35:14
You don't have to go through
all this stuff.
510
00:35:11 --> 00:35:17
For God's sakes,
don't say let the trial
511
00:35:13 --> 00:35:19
solution be blah,
blah, blah.
512
00:35:15 --> 00:35:21
You don't want to do that.
I don't want you to repeat the
513
00:35:19 --> 00:35:25
derivation of this every time
you go through a particular
514
00:35:23 --> 00:35:29
problem.
It is just like in solving
515
00:35:25 --> 00:35:31
second order equations.
You have a second order
516
00:35:29 --> 00:35:35
equation.
You immediately write down its
517
00:35:32 --> 00:35:38
characteristic equation,
then you factor it,
518
00:35:35 --> 00:35:41
you find its roots and you
construct the solution.
519
00:35:39 --> 00:35:45
It takes a minute.
The same thing,
520
00:35:42 --> 00:35:48
this takes a minute,
too.
521
00:35:43 --> 00:35:49
What is the constant term?
Ad minus bc,
522
00:35:47 --> 00:35:53
what is that?
Matrix is (a,
523
00:35:49 --> 00:35:55
b; c, d).
Ad minus bc is its determinant.
524
00:35:52 --> 00:35:58
This is the determinant of that
matrix.
525
00:35:55 --> 00:36:01
I didn't give the matrix a
name, did I?
526
00:36:00 --> 00:36:06
I will now give the matrix a
name A.
527
00:36:03 --> 00:36:09
What is this?
Well, you are not supposed to
528
00:36:07 --> 00:36:13
know that until now.
I will tell you.
529
00:36:10 --> 00:36:16
This is called the trace of A.
Put that down in your little
530
00:36:16 --> 00:36:22
books.
The abbreviation is trace A,
531
00:36:19 --> 00:36:25
and the word is trace.
The trace of a square matrix is
532
00:36:25 --> 00:36:31
the sum of the d elements down
its main diagonal.
533
00:36:31 --> 00:36:37
If it were a three-by-three
there would be three terms in
534
00:36:35 --> 00:36:41
whatever you are up to.
Here it is a plus b,
535
00:36:40 --> 00:36:46
the sum of the diagonal
elements.
536
00:36:43 --> 00:36:49
You can immediately write down
this characteristic equation.
537
00:36:48 --> 00:36:54
Let's give it a name.
This is a characteristic
538
00:36:52 --> 00:36:58
equation of what?
Of the matrix,
539
00:36:55 --> 00:37:01
now.
Not of the system,
540
00:36:57 --> 00:37:03
of the matrix.
You have a two-by-two matrix.
541
00:37:02 --> 00:37:08
You could immediately write
down its characteristic
542
00:37:06 --> 00:37:12
equation.
Watch out for this sign,
543
00:37:08 --> 00:37:14
minus.
That is a very common error to
544
00:37:11 --> 00:37:17
leave out the minus sign because
that is the way the formula
545
00:37:15 --> 00:37:21
comes out.
Its roots.
546
00:37:17 --> 00:37:23
If it is a quadratic equation
it will have roots;
547
00:37:20 --> 00:37:26
lambda1, lambda2 for the moment
let's assume are real and
548
00:37:25 --> 00:37:31
distinct.
549
00:37:27 --> 00:37:33
550
00:37:37 --> 00:37:43
For the enrichment of your
vocabulary, those are called the
551
00:37:39 --> 00:37:45
eigenvalues.
552
00:37:40 --> 00:37:46
553
00:37:50 --> 00:37:56
They are something which
belonged to the matrix A.
554
00:37:53 --> 00:37:59
They are two secret numbers.
You can calculate from the
555
00:37:56 --> 00:38:02
coefficients a,
b, and c, and d,
556
00:37:58 --> 00:38:04
but they are not in the
coefficients.
557
00:38:00 --> 00:38:06
You cannot look at a matrix and
see what its eigenvalues are.
558
00:38:05 --> 00:38:11
You have to calculate
something.
559
00:38:07 --> 00:38:13
But they are the most important
numbers in the matrix.
560
00:38:10 --> 00:38:16
They are hidden,
but they are the things that
561
00:38:13 --> 00:38:19
control how this system behaves.
Those are called the
562
00:38:17 --> 00:38:23
eigenvalues.
Now, there are various purists,
563
00:38:20 --> 00:38:26
there are a fair number of them
in the world who do not like
564
00:38:24 --> 00:38:30
this word because it begins
German and ends English.
565
00:38:29 --> 00:38:35
Eigenvalues were first
introduced by a German
566
00:38:32 --> 00:38:38
mathematician,
you know, around the time
567
00:38:35 --> 00:38:41
matrices came into being in
or so.
568
00:38:38 --> 00:38:44
A little while after
eigenvalues came into being,
569
00:38:41 --> 00:38:47
too.
And since all this happened in
570
00:38:44 --> 00:38:50
Germany they were named
eigenvalues in German,
571
00:38:47 --> 00:38:53
which begins eigen and ends
value.
572
00:38:50 --> 00:38:56
But people who do not like that
call them the characteristic
573
00:38:54 --> 00:39:00
values.
Unfortunately,
574
00:38:56 --> 00:39:02
it is two words and takes a lot
more space to write out.
575
00:39:02 --> 00:39:08
An older generation even calls
them something different,
576
00:39:06 --> 00:39:12
which you are not so likely to
see nowadays,
577
00:39:10 --> 00:39:16
but you will in slightly older
books.
578
00:39:13 --> 00:39:19
You can also call them the
proper values.
579
00:39:17 --> 00:39:23
Characteristic is not a
translation of eigen,
580
00:39:21 --> 00:39:27
but proper is,
but it means it in a funny
581
00:39:25 --> 00:39:31
sense which has almost
disappeared nowadays.
582
00:39:30 --> 00:39:36
It means proper in the sense of
belong to.
583
00:39:33 --> 00:39:39
The only example I can think of
is the word property.
584
00:39:37 --> 00:39:43
Property is something that
belongs to you.
585
00:39:40 --> 00:39:46
That is the use of the word
proper.
586
00:39:43 --> 00:39:49
It is something that belongs to
the matrix.
587
00:39:46 --> 00:39:52
The matrix has its proper
values.
588
00:39:49 --> 00:39:55
It does not mean proper in the
sense of fitting and proper or I
589
00:39:54 --> 00:40:00
hope you will behave properly
when we go to Aunt Agatha's or
590
00:39:59 --> 00:40:05
something like that.
But, as I say,
591
00:40:03 --> 00:40:09
by far the most popular thing,
slowly the word eigenvalue is
592
00:40:07 --> 00:40:13
pretty much taking over the
literature.
593
00:40:11 --> 00:40:17
Just because it's just one
word, that is a tremendous
594
00:40:15 --> 00:40:21
advantage.
Okay.
595
00:40:16 --> 00:40:22
What now is still to be done?
Well, there are those vectors
596
00:40:21 --> 00:40:27
to be found.
So the very last step would be
597
00:40:24 --> 00:40:30
to solve the system to find the
vectors a1 and a2.
598
00:40:30 --> 00:40:36
599
00:40:35 --> 00:40:41
For each (lambda)i,
find the associated vector.
600
00:40:40 --> 00:40:46
The vector, we will call it
(alpha)i.
601
00:40:44 --> 00:40:50
That is the a1 and a2.
Of course it's going to be
602
00:40:50 --> 00:40:56
indexed.
You have to put another
603
00:40:53 --> 00:40:59
subscript on it because there
are two of them.
604
00:41:00 --> 00:41:06
And a1 and a2 is stretched a
little too far.
605
00:41:04 --> 00:41:10
By solving the system,
and the system will be the
606
00:41:09 --> 00:41:15
system which I will write this
way, (a minus lambda,
607
00:41:15 --> 00:41:21
b, c, d minus lambda).
608
00:41:19 --> 00:41:25
It is just that system that was
609
00:41:24 --> 00:41:30
over there, but I will recopy
it, (a1, a2) equals zero,
610
00:41:30 --> 00:41:36
zero.
And these are called the
611
00:41:34 --> 00:41:40
eigenvectors.
Each of these is called the
612
00:41:39 --> 00:41:45
eigenvector associated with or
belonging to,
613
00:41:44 --> 00:41:50
again, in that sense of
property.
614
00:41:48 --> 00:41:54
Eigenvector,
let's say belonging to,
615
00:41:52 --> 00:41:58
I see that a little more
frequently, belonging to lambda
616
00:41:58 --> 00:42:04
i.
So we have the eigenvalues,
617
00:42:01 --> 00:42:07
the eigenvectors and,
of course, the people who call
618
00:42:04 --> 00:42:10
them characteristic values also
call these guys characteristic
619
00:42:08 --> 00:42:14
vectors.
I don't think I have ever seen
620
00:42:11 --> 00:42:17
proper vectors,
but that is because I am not
621
00:42:13 --> 00:42:19
old enough.
I think that is what they used
622
00:42:16 --> 00:42:22
to be called a long time ago,
but not anymore.
623
00:42:20 --> 00:42:26
And then, finally,
the general solution will be,
624
00:42:24 --> 00:42:30
by the superposition principle,
(x, y) equals the arbitrary
625
00:42:30 --> 00:42:36
constant times the first
eigenvector times the eigenvalue
626
00:42:35 --> 00:42:41
times the e to the corresponding
eigenvalue.
627
00:42:40 --> 00:42:46
And then the same thing for the
second one, (a1,
628
00:42:44 --> 00:42:50
a2), but now the second index
will be 2 to indicate that it
629
00:42:50 --> 00:42:56
goes with the eigenvalue e to
the lambda 2t.
630
00:42:56 --> 00:43:02
I have done that twice.
And now in the remaining five
631
00:43:02 --> 00:43:08
minutes I will do it a third
time because it is possible to
632
00:43:06 --> 00:43:12
write this in still a more
condensed form.
633
00:43:09 --> 00:43:15
And the advantage of the more
condensed form is A,
634
00:43:13 --> 00:43:19
it takes only that much space
to write, and B,
635
00:43:16 --> 00:43:22
it applies to systems,
not just the two-by-two
636
00:43:20 --> 00:43:26
systems, but to end-by-end
systems.
637
00:43:23 --> 00:43:29
The method is exactly the same.
Let's write it out as it would
638
00:43:27 --> 00:43:33
apply to end-by-end systems.
The vector I started with is
639
00:43:34 --> 00:43:40
(x, y) and so on,
but I will simply abbreviate
640
00:43:39 --> 00:43:45
this, as is done in 18.02,
by x with an arrow over it.
641
00:43:45 --> 00:43:51
The matrix A I will abbreviate
with A, as I did before with
642
00:43:51 --> 00:43:57
capital A.
And then the system looks like
643
00:43:56 --> 00:44:02
x prime is equal to --
644
00:44:00 --> 00:44:06
645
00:44:05 --> 00:44:11
x prime is what?
Ax.
646
00:44:06 --> 00:44:12
That is all there is to it.
647
00:44:10 --> 00:44:16
There is our green system.
Now notice in this form I did
648
00:44:15 --> 00:44:21
not even tell you whether this a
two-by-two matrix or an
649
00:44:20 --> 00:44:26
end-by-end.
And in this condensed form it
650
00:44:23 --> 00:44:29
will look the same no matter how
many equations you have.
651
00:44:30 --> 00:44:36
Your book deals from the
beginning with end-by-end
652
00:44:33 --> 00:44:39
systems.
That is, in my view,
653
00:44:36 --> 00:44:42
one of its weaknesses because I
don't think most students start
654
00:44:40 --> 00:44:46
with two-by-two.
Fortunately,
655
00:44:43 --> 00:44:49
the book double-talks.
The theory is end-by-end,
656
00:44:46 --> 00:44:52
but all the examples are
two-by-two.
657
00:44:49 --> 00:44:55
So just read the examples.
Read the notes instead,
658
00:44:53 --> 00:44:59
which just do two-by-two to
start out with.
659
00:44:58 --> 00:45:04
The trial solution is x equals
what?
660
00:45:01 --> 00:45:07
An unknown vector alpha times e
to the lambda t.
661
00:45:07 --> 00:45:13
Alpha is what we called a1 and
662
00:45:11 --> 00:45:17
a2 before.
Plug this into there and cancel
663
00:45:15 --> 00:45:21
the e to the lambda t's.
664
00:45:19 --> 00:45:25
What do you get?
Well, this is lambda alpha e to
665
00:45:24 --> 00:45:30
the lambda t equals A alpha e to
the lambda t.
666
00:45:29 --> 00:45:35
667
00:45:36 --> 00:45:42
These two cancel.
And the system to be solved,
668
00:45:40 --> 00:45:46
A alpha equals lambda alpha.
669
00:45:46 --> 00:45:52
And now the question is how do
you solve that system?
670
00:45:51 --> 00:45:57
Well, you can tell if a book is
written by a scoundrel or not by
671
00:45:57 --> 00:46:03
how they go --
A book, which is in my opinion
672
00:46:02 --> 00:46:08
completely scoundrel,
simply says you subtract one
673
00:46:07 --> 00:46:13
from the other,
and without further ado writes
674
00:46:12 --> 00:46:18
A minus lambda,
and they tuck a little I in
675
00:46:16 --> 00:46:22
there and write alpha equals
zero.
676
00:46:19 --> 00:46:25
Why is the I put in there?
Well, this is what you would
677
00:46:25 --> 00:46:31
like to write.
What is wrong with this
678
00:46:28 --> 00:46:34
equation?
This is not a valid matrix
679
00:46:32 --> 00:46:38
equation because that is a
square end-by-end matrix,
680
00:46:36 --> 00:46:42
a square two-by-two matrix if
you like.
681
00:46:39 --> 00:46:45
This is a scalar.
You cannot subtract the scalar
682
00:46:42 --> 00:46:48
from a matrix.
It is not an operation.
683
00:46:45 --> 00:46:51
To subtract matrices they have
to be the same size,
684
00:46:49 --> 00:46:55
the same shape.
What is done is you make this a
685
00:46:52 --> 00:46:58
two-by-two matrix.
This is a two-by-two matrix
686
00:46:56 --> 00:47:02
with lambdas down the main
diagonal and I elsewhere.
687
00:47:01 --> 00:47:07
And the justification is that
lambda alpha is the same thing
688
00:47:05 --> 00:47:11
as the lambda I times alpha
because I is an identity matrix.
689
00:47:10 --> 00:47:16
Now, in fact,
jumping from here to here is
690
00:47:13 --> 00:47:19
not something that would occur
to anybody.
691
00:47:16 --> 00:47:22
The way it should occur to you
to do this is you do this,
692
00:47:21 --> 00:47:27
you write that,
you realize it doesn't work,
693
00:47:24 --> 00:47:30
and then you say to yourself I
don't understand what these
694
00:47:29 --> 00:47:35
matrices are all about.
I think I'd better write it all
695
00:47:33 --> 00:47:39
out.
And then you would write it all
696
00:47:36 --> 00:47:42
out and you would write that
equation on the left-hand board
697
00:47:39 --> 00:47:45
there.
Oh, now I see what it should
698
00:47:41 --> 00:47:47
look like.
I should subtract lambda from
699
00:47:44 --> 00:47:50
the main diagonal.
That is the way it will come
700
00:47:47 --> 00:47:53
out.
And then say,
701
00:47:48 --> 00:47:54
hey, the way to save lambda
from the main diagonal is put it
702
00:47:51 --> 00:47:57
in an identity matrix.
That will do it for me.
703
00:47:54 --> 00:48:00
In other words,
there is a little detour that
704
00:47:57 --> 00:48:03
goes from here to here.
And one of the ways I judge
705
00:48:01 --> 00:48:07
books is by how well they
explain the passage from this to
706
00:48:05 --> 00:48:11
that.
If they don't explain it at all
707
00:48:08 --> 00:48:14
and just write it down,
they have never talked to
708
00:48:11 --> 00:48:17
students.
They have just written books.
709
00:48:14 --> 00:48:20
Where did we get finally here?
The characteristic equation
710
00:48:17 --> 00:48:23
from that, I had forgotten what
color.
711
00:48:20 --> 00:48:26
That is in salmon.
The characteristic equation,
712
00:48:23 --> 00:48:29
then, is going to be the thing
which says that the determinant
713
00:48:27 --> 00:48:33
of that is zero.
That is the circumstances under
714
00:48:32 --> 00:48:38
which it is solvable.
In general, this is the way the
715
00:48:35 --> 00:48:41
characteristic equation looks.
And its roots,
716
00:48:38 --> 00:48:44
once again, are the
eigenvalues.
717
00:48:40 --> 00:48:46
And from then you calculate the
corresponding eigenvectors.
718
00:48:44 --> 00:48:50
Okay.
Go home and practice.
719
00:48:46 --> 00:48:52
In recitation you will practice
on both two-by-two and
720
00:48:50 --> 00:48:56
three-by-three cases,
and we will talk more next
721
00:48:53 --> 00:48:59
time.