y prime and y double prime.
So, q of x times y 
equal zero.
The linearity of the equation,
that is, the form in which it
appears is going to be the key
idea today.
Today is going to be
theoretical, but some of the
ideas in it are the most
important in the course.
So, I don't have to apologize
for the theory.
Remember, the solution method
was to find two independent y
one, y two
independent solutions.
And now, I'll formally write
out what independent means.
There are different ways to say
it.
But for you,
I think the simplest and most
intelligible will be to say that
y2 is not to be a constant
multiple of y1.
And, unfortunately,
it's necessary to add,
nor is y1 to be a constant
multiple.
I have to call it by different
constants.
So, let's call this one c prime
of y2.
Well, I mean,
the most obvious question is,
well, look.
If this is not a constant times
that, this can't be there
because I would just use one
over c if it was.
Unfortunately,
the reason I have to write it
this way is to take account of
the possibility that y1 might be
zero.
If y1 is zero,
so, the bad case that must be
excluded is that y1 equals zero,
y2 nonzero.
I don't want to call those
independent.
But nonetheless,
it is true that y2 is not a
constant multiple of y1.
However, y1 is a constant
multiple of y2,
namely, the multiple zero.
It's just to exclude that case
that I have to say both of those
things.
And, one would not be sufficed.
That's a fine point that I'm
not going to fuss over.
But I just have,
of course.
Now, why do you do that?
That's because,
then, all solutions,
and this is what concerns us
today, are what?
The linear combination with
constant coefficients of these
two, and the fundamental
question we have to answer today
is, why?
Now, there are really two
statements involved in that.
On the one hand,
I'm claiming there is an easier
statement, which is that they
are all solutions.
So, that's question one,
or statement one.
Why are all of these guys
solutions?
That, I could trust you to
answer yourself.
I could not trust you to answer
it elegantly.
And, it's the elegance that's
the most important thing today
because you have to answer it
elegantly.
Otherwise, you can't go on and
do more complicated things.
If you answer it in an ad hoc
basis just by hacking out a
computation, you don't really
see what's going on.
And you can't do more difficult
things later.
So, we have to answer this,
and answer it nicely.
The second question is,
so, if that answers why there
are solutions at all,
why are they all the solutions?
Why all the solutions?
In other words,
to say it as badly as possible,
why are all solutions,
why all the solutions-- Never
mind.
Why are all the solutions.
This is a harder question to
answer, but that should make you
happy because that means it
depends upon a theorem which I'm
not going to prove.
I'll just quote to you.
Let's attack there for problem
one first.
q1 is answered by what's called
the superposition.
The superposition principle
says exactly that.
It says exactly that,
that if y1 and y2 are solutions
to a linear equation,
to a linear homogeneous ODE,
in fact it can be of higher
order, too, although I won't
stress that.
In other words,
you don't have to stop with the
second derivative.
You could add a third
derivative and a fourth
derivative.
As long as the former makes the
same, but that implies
automatically that c1 y1 plus c2
y2 is a solution.
Now, the way to do that nicely
is to take a little detour and
talk a little bit about linear
operators.
And, since we are going to be
using these for the rest of the
term, this is the natural place
for you to learn a little bit
about what they are.
So, I'm going to do it.
Ultimately, I am aimed at a
proof of this statement,
but there are going to be
certain side excursions I have
to make.
The first side side excursion
is to write the differential
equation in a different way.
So, I'm going to just write its
transformations.
The first, I'll simply recopy
what you know it to be,
q y equals zero.
That's the first form.
The second form,
I'm going to replace this by
the differentiation operator.
So, I'm going to write this as
D squared y.
That means differentiate it
twice.
D it, and then D it again.
This one I only have to
differentiate once,
so I'll write that as p D(y),
p times the derivative of Y.
The last one isn't
differentiated at all.
I just recopy it.
Now, I'm going to formally
factor out the y.
So this, I'm going to turn into
D squared plus pD plus q.
Now, everybody reads this as
times y equals zero.
But, what it means is this guy,
it means this is shorthand for
that.
I'm not multiplying.
I'm multiplying q times y.
But, I'm not multiplying D
times y.
I'm applying D to y.
Nonetheless,
the notation suggests this is
very suggestive of that.
And this, in turn,
implies that.
I'm just transforming it.
And now, I'll take the final
step.
I'm going to view this thing in
parentheses as a guy all by
itself, a linear operator.
This is a linear operator,
called a linear operator.
And, I'm going to simply
abbreviate it by the letter L.
And so, the final version of
this equation has been reduced
to nothing but Ly equals zero.
Now, what's L?
You can think of L as,
well, formally,
L you would write as D squared
plus pD plus q.
But, you can think of L,
the way to think of it is as a
black box, a function of what
goes into the black box,
well, if this were a function
box, what would go it would be a
number, and what would come out
with the number.
But it's not that kind of a
black box.
It's an operator box,
and therefore,
what goes in is a function of
x.
And, what comes out is another
function of x,
the result of applying this
operator to that.
So, from this point of view,
differential equations,
trying to solve the
differential equation means,
what should come out you want
to come out zero,
and the question is,
what should you put in?
That's what it means solving
differential equations in an
inverse problem.
The easy thing is to put it a
function, and see what comes
out.
You just calculate.
The hard thing is to ask,
you say, I want such and such a
thing to come out,
for example,
zero; what should I put in?
That's a difficult question,
and that's what we're spending
the term answering.
Now, the key thing about this
is that this is a linear
operator.
And, what that means is that it
behaves in a certain way with
respect to functions.
The easiest way to say it is,
I like to make two laws of it,
that L of u1,
if you have two functions,
I'm not going to put up the
parentheses, x,
because that just makes things
look longer and not any clearer,
actually.
What does L do to the sum of
two functions?
If that's a linear operator,
if you put in the sum of two
functions, what you must get out
is the corresponding L's,
the sum of the corresponding
L's of each.
So, that's a law.
And, the other law,
linearity law,
and this goes for anything in
mathematics and its
applications,
which is called linear,
basically anything is linear if
it does the following thing to
functions or numbers or
whatever.
The other one is of a constant
times any function,
I don't have to give it a
number now because I'm only
using one of them,
should be equal to c times L of
u.
So, here, c is a constant,
and here, of course,
the u is a function,
functions of x.
These are the two laws of
linearity.
An operator is linear if it
satisfies these two laws.
Now, for example,
the differentiation operator is
such an operator.
D is linear,
why?
Well, because of the very first
things you verify after you
learn what the derivative is
because the derivative of,
well, I will write it in the D
form.
I'll write it in the form in
which you know it.
It would be D apply to u1 plus
u2.
How does one write that in
ordinary calculus?
Well, like that.
Or, maybe you write d by dx out
front.
Let's write it this way,
is equal to u1 prime plus u2
prime.
That's a law.
You prove it when you first
study what a derivative is.
It's a property.
From our point of view,
it's a property of the
differentiation operator.
It has this property.
The D of u1 plus u2 is D of u1
plus D of u2.
And, it also has the property
that c u prime,
you can pull out the constant.
That's not affected by the
differentiation.
So, these two familiar laws
from the beginning of calculus
say, in our language,
that D is a linear operator.
What about the multiplication
of law?
That's even more important,
that u1 times u2 prime,
I have nothing whatever to say
about that in here.
In this context,
it's an important law,
but it's not important with
respect to the study of
linearity.
So, there's an example.
Here's a more complicated one
that I'm claiming is the linear
operator.
And, since I don't want to have
to work in this lecture,
the work is left to you.
So, the proof,
prove that L is linear,
is this particular operator.
L is linear.
That's in your part one
homework to verify that.
And you will do some simple
exercises in recitation tomorrow
to sort of warm you up for that
if you haven't done it already.
Well, you shouldn't have
because this only goes with this
lecture, actually.
It's forbidden to work ahead in
this class.
All right, where are we?
Well, all that was a prelude to
proving this simple theorem,
superposition principle.
So, finally,
what's the proof?
The proof of the superposition
principle: if you believe that
the operator is linear,
then L of c1,
in other words,
the ODE is L.
L is D squared plus pD plus q.
So, the ODE is Ly equals zero.
And, what am I being asked to
prove?
I'm being asked to prove that
if y1 and y2 are solutions,
then so is that thing.
By the way, that's called a
linear combination.
Put that in your notes.
Maybe I better write it even on
the board because it's something
people say all the time without
realizing they haven't defined
it.
This is called a linear
combination.
This expression is called a
linear combination of y1 and y2.
It means that particular sum
with constant coefficients.
Okay, so, the ODE is Ly equals
zero.
And, I'm trying to prove that
fact about it,
that if y1 and y2 are
solutions, so is a linear
combination of them.
So, the proof,
then, I start with apply L to
c1 y1 plus c2 y2.
Now, because this operator is
linear, it takes the sum of two
functions into the corresponding
sum up what the operator would
be.
So, it would be L of c1 y1 plus
L of c2 y2.
That's because L is a linear
operator.
But, I don't have to stop
there.
Because L is a linear operator,
I can pull the c out front.
So, it's c1 L of y1 plus c2 L
of y2. Now, where am I?
Trying to prove that this is
zero.
Well, what is L of y1?
At this point,
I use the fact that y1 is a
solution.
Because it's a solution,
this is zero.
That's what it means to solve
that differential equation.
It means, when you apply the
linear operator,
L, to the function,
you get zero.
In the same way,
y2 is a solution.
So, that's zero.
And, the sum of c1 times zero
plus c2 times zero is zero.
That's the argument.
Now, you could make the same
argument just by plugging c1 y1,
plugging it into the equation
and calculating,
and calculating,
and calculating,
grouping the terms and so on
and so forth.
But, that's just calculation.
It doesn't show you why it's
so.
Why it's so is because the
operator, this differential
equation is expressed as a
linear operator applied to y is
zero.
And, the only properties that
are really been used as the fact
that this operator is linear.
That's the key point.
L is linear.
It's a linear operator.
Well, that's all there is to
the superposition principle.
As a prelude to answering the
more difficult question,
why are these all the
solutions?
Why are there no other
solutions?
We need a few definitions,
and a few more ideas.
And, they are going to occur in
connection with,
so I'm now addressing,
ultimately, question two.
But, it's not going to be
addressed directly for quite
awhile.
Instead, I'm going to phrase it
in terms of solving the initial
value problem.
So far, we've only talked about
the general solution with those
two arbitrary constants.
But, how do you solve the
initial value problem,
in other words,
fit initial conditions,
find the solution with given
initial values for the function
and its derivatives.
Now, --
-- the theorem is that this
collection of functions with
these arbitrary constants,
these are all the solutions we
have so far.
In fact, they are all the
solutions there are,
but we don't know that yet.
However, if we just focus on
the big class of solutions,
there might be others lurking
out there somewhere lurking out
there.
I don't know.
But let's use what we have,
that just from this family is
enough to satisfy any initial
condition, to satisfy any
initial values.
In other words,
if you give me any initial
values, I will be able to find
the c1 and c2 which work.
Now, why is that?
Well, I'd have to do a song and
dance at this point,
if you hadn't been softened up
by actually calculating for
specific differential equations.
You've had exercises and
actually how to calculate the
values of c1 and c2.
So, I'm going to do it now in
general what you have done so
far for particular equations
using particular values of the
initial conditions.
So, I'm relying on that
experience that you've had in
doing the homework to make
intelligible what I'm going to
do now in the abstract using
just letters.
So, why is this so?
Why is that so?
Well, we are going to need it,
by the way, here,
too.
I'll have to,
again, open up parentheses.
But let's go as far as we can.
Well, you just try to do it.
Suppose the initial conditions
are, how will we write them?
So, they're going to be at some
initial point,
x0.
You can take it to be zero if
you want, but I'd like to be,
just for a little while,
a little more general.
So, let's say the initial
conditions, the initial values
are being given at the point x0,
all right, that's going to be
some number.
Let's just call it a.
And, the initial value also has
to specify the velocity or the
value of the derivative there.
Let's say these are the initial
values.
So, the problem is to find the
c which work.
Now, how do you do that?
Well, you know from
calculation.
You write y equals c1 y1 plus
c2 y2.
And you write y prime,
and you take the derivative
underneath that,
which is easy to do.
And now, you plug in x equals
zero.
And, what happens?
Well, these now turn into a set
of equations.
What will they look like?
Well, y of x0 is a,
and this is b.
So, what I get is let me flop
it over onto the other side
because that's the way you're
used to looking at systems of
equations.
So, what we get is c1 times y1
of x0 plus c2 times y2 of x0.
What's that supposed to be
equal to?
Well, that's supposed to be
equal to y of x0.
It's supposed to be equal to
the given number,
a.
And, in the same way,
c1 y1 prime of x0 plus c2 y2
prime of x0, that's supposed to
turn out to 
be the number, b.
In the calculations you've done
up to this point,
y1 and y2 were always specific
functions like e to the x
or cosine of 22x,
stuff like that.
Now I'm doing it in the
abstract, just calling them y1
and y2, so as to include all
those possible cases.
Now, what am I supposed to do?
I'm supposed to find c1 and c2.
What kind of things are they?
This is what you studied in
high school, right?
The letters are around us,
but it's a pair of simultaneous
linear equations.
What are the variables?
What are the variables?
What are the variables?
Somebody raise their hand.
If you have a pair of
simultaneous linear equations,
you've got variables and you've
got constants,
right?
And you are trying to find the
answer.
What are the variables?
Yeah?
c1 and c2.
Very good.
I mean, it's extremely
confusing because in the first
place, how can they be the
variables if they occur on the
wrong side?
They're on the wrong side;
they are constants.
How can constants be variables?
Everything about this is wrong.
Nonetheless,
the c1 and the c2 are the
unknowns, if you like the high
school terminology.
c1 and c2 are the unknowns.
These messes are just numbers.
After you've plugged in x0,
this is some number.
You've got four numbers here.
So, c1 and c2 are the
variables.
The two find,
in other words,
to find the values of.
All right, now you know general
theorems from 18.02 about when
can you solve such a system of
equations.
I'm claiming that you can
always find c1 and c2 that work.
But, you know that's not always
the case that a pair of
simultaneous linear equations
can be solved.
There's a condition.
There's a condition which
guarantees their solution,
which is what?
What has to be true about the
coefficients?
These are the coefficients.
What has to be true?
The matrix of coefficients must
be invertible.
The determinant of coefficients
must be nonzero.
So, they are solvable if,
for the c1 and c2,
if this thing,
I'm going to write it.
Since all of these are
evaluated at x0,
I'm going to write it in this
way.
y1, the determinant,
whose entries are y1,
y2, y1 prime,
and y2 prime,
evaluated at zero,
x0, that means that I evaluate
each of the functions in the
determinant at x0.
I'll write it this way.
That should be not zero.
So, in other words,
the key thing which makes this
possible, makes it possible for
us to solve the initial value
problem, is that this funny
determinant should not be zero
at the point at which we are
interested.
Now, this determinant is
important in 18.03.
It has a name,
and this is when you're going
to learn it, if you don't know
it already.
That determinant is called the
Wronskian.
The Wronskian of what?
If you want to be pompous,
you say this with a V sound
instead of a W.
But, nobody does except people
trying to be pompous.
The Wronskian,
we'll write a W.
Now, notice,
you can only calculate it when
you know what the two functions
are.
So, the Wronskian of the two
functions, y1 and y2,
what's the variable?
It's not a function of two
variables, y1 and y2.
These are just the names of
functions of x.
So, when you do it,
put it in, calculate out that
determinant.
This is a function of x,
a function of the independent
variable after you've done the
calculation.
Anyway, let's write its
definition, y1,
y2, y1 prime,
y2 prime.
Now, in order to do this,
the point is we must know that
that Wronskian is not zero,
that the Wronskian of these two
functions is not zero at the
point x0.
Now, enter a theorem which
you're going to prove for
homework, but this is harder.
So, it's part two homework.
It's not part one homework.
In other words,
I didn't give you the answer.
You've got to find it yourself,
alone or in the company of good
friends.
So, anyway, here's the
Wronskian.
Now, what can we say for sure?
Note, suppose y1 and y,
just to get you a feeling for
it a little bit,
suppose they were not
independent.
The word for not independent is
dependent.
Suppose they were dependent.
In other words,
suppose that y2 were a constant
multiple of y1.
We know that's not the case
because our functions are
supposed to be independent.
But suppose they weren't.
What would the value of the
Wronskian be?
If y2 is a constant times y1,
then y2 prime is that same
constant times y1 prime.
What's the value of the
determinant?
Zero.
For what values of x is it zero
for all values of x?
And now, that's the theorem
that you're going to prove,
that if y1 and y2 are solutions
to the ODE, I won't keep,
say, it's the ODE we've been
talking about,
y double prime plus py.
But the linear homogeneous with
not constant coefficients,
just linear homogeneous second
order.
Our solutions,
as there are only two
possibilities,
either-or.
Either the Wronskian of y,
there are only two
possibilities.
Either the Wronskian of y1 and
y2 is always zero,
identically zero,
zero for all values of x.
This is redundant.
When I say identically,
I mean for all values of x.
But, I am just making assurance
doubly sure.
Or, or the Wronskian is never
zero.
Now, there is no notation as
for that.
I'd better just write it out,
is never zero,
i.e.
for no x is it,
i.e.
for all x.
There's no way to say that.
I mean, for all values of x,
it's not zero.
That means, there is not a
single point for which it's
zero.
In particular,
it's not zero here.
So, this is your homework:
problem five,
part two.
I'll give you a method of
proving it, which was discovered
by the famous Norwegian
mathematician,
Abel, who is,
I guess, the centenary of his
birth, I guess,
was just celebrated last year.
He has one of the truly tragic
stories in mathematics,
which I think you can read.
It must be a Simmons book,
if you have that.
Simmons is very good on
biographies.
Look up Abel.
He'll have a biography of Abel,
and you can weep if you're
feeling sad.
He died at the age of 26 of
tuberculosis,
having done a number of
sensational things,
none of which was recognized in
his lifetime because people
buried his papers under big
piles of papers.
So, he died unknown,
uncelebrated,
and now he's Norway's greatest
culture hero.
In the middle of a park in
Oslo, there's a huge statue.
And, since nobody knows what
Abel looked like,
the statue is way up high so
you can't see very well.
But, the inscription on the
bottom says Niels Henrik Abel,
1801-1826 or something like
that.
Now, --
-- the choice,
I'm still, believe it or not,
aiming at question two,
but I have another big
parentheses to open.
And, when I closed it,
the answer to question two will
be simple.
But, I think it's very
desirable that you get this
second big parentheses.
It will help you to understand
something important.
It will help you on your
problem set tomorrow night.
I don't have to apologize.
I'm just going to do it.
So, the question is,
the thing you have to
understand is that when I write
this combination,
I'm claiming that these are all
the solutions.
I haven't proved that yet.
But, they are going to be all
the solutions
The point is,
there's nothing sacrosanct
about the y1 and y2.
This is exactly the same
collection as a collection which
I would write using other
constants.
Let's call them u1 and u2.
They are exactly the same,
where u1 and u2 are any other
pair of linearly independent
solutions.
Any other pair of independent
solutions, they must be
independent, either a constant
multiple of each other.
In other words,
u1 is some combination,
now I'm really stuck because I
don't know how to,
c1 bar, let's say,
that means a special value of
c1, and a special value of c2,
and u2 is some other special
value, oh my God,
c1 double bar,
how's that?
The notation is getting worse
and worse.
I apologize for it.
In other words,
I could pick y1 and y2 and make
up all of these.
And, I'd get a bunch of
solutions.
But, I could also pick some
other family,
some other two guys in this
family, and just as well express
the solutions in terms of u1 and
u2.
Now, well, why is he telling us
that?
Well, the point is that the y1
and the y2 are typically the
ones you get easily from solving
the equations,
like e to the x and e
to the 2x.
That's what you've gotten,
or cosine x and sine x,
something like that.
But, for certain ways,
they might not be the best way
of writing the solutions.
There is another way of writing
those that you should learn,
and that's called finding
normalized, the normalized.
They are okay,
but they are not normalized.
For some things,
the normalized solutions are
the best.
I'll explain to you what they
are, and I'll explain to you
what they're good for.
You'll see immediately what
they're good for.
Normalized solutions,
now, you have to specify the
point at which you're
normalizing.
In general, it would be x
nought,
but let's, at this point,
since I don't have an infinity
of time, to simplify things,
let's say zero.
It could be x nought,
any point would do just as
well.
But, zero is the most common
choice.
What are the normalized
solutions?
Well, first of all,
I have to give them names.
I want to still call them y.
So, I'll call them capital Y1
and Y2.
And, what they are,
are the solutions which satisfy
certain, special,
very special,
initial conditions.
And, what are those?
So, they're the ones which
satisfy, the initial conditions
for Y1 are, of course there are
going to be guys that look like
this.
The only thing that's going to
make them distinctive is the
initial conditions they satisfy.
Y1 has to satisfy at zero.
Its value should be one,
and the value of its derivative
should be zero.
For Y2, it's just the opposite.
Here, the value of the function
should be zero at zero.
But, the value of its
derivative, now,
I want to be one.
Let me give you a trivial
example of this,
and then one,
which is a little less trivial,
so you'll have some feeling for
what I'm asking for.
Suppose the equation,
for example,
is y double prime plus,
well, let's really make it
simple.
Okay, you know the standard
solutions are y1 is cosine x,
and y2 is sine x.
These are functions,
which, when you take the second
derivative, they turn into their
negative.
You know, you could go the
complex roots are i and minus i,
and blah, blah,
blah, blah, blah,
blah.
If you do it that way,
fine.
But at some point in the course
you have to be able to write
down and, right away,
oh, yeah, cosine x,
sine x.
Okay, what are the normalized
things?
Well, what's the value of this
at zero?
It is one.
What's the value of its
derivative at zero?
Zero.
This is Y1.
This is the only case in which
you locked on immediately to the
normalized solutions.
In the same way,
this guy is Y2 because its
value at zero is zero.
It's value of its derivative at
zero is one.
So, this is Y2.
Okay, now let's look at a case
where you don't immediately lock
on to the normalized solutions.
Very simple:
all I have to do is change the
sign.
Here, you know,
think through r squared minus
one equals zero.
The characteristic roots are
plus one and minus one,
right?
And therefore,
the solution is e to the x,
and e to the negative x.
So, the solutions you find by
the usual way of solving it is
y1 equals e to the x,
and y2 equals e to the
negative x.
Those are the standard
solution.
So, the general solution is of
the form.
So, the general solution is of
the form c1 e to the x plus c2 e
to the negative x.
Now, what I want to find out is
what is Y1 and Y2?
How do I find out what Y1 is?
Well, I have to satisfy initial
conditions.
So, if this is y,
let's write down here,
if you can still see that,
y prime is c1 e to the x minus
c2 e to the negative x .
So, if I plug in,
I want y of zero to be one,
I want this guy at the point
zero to be one.
What equation does that give
me?
That gives me c1 plus c2,
c1 plus c2,
plugging in x equals zero,
equals the value of this thing
at zero.
So, that's supposed to be one.
How about the other guy?
The value of its derivative is
supposed to come out to be zero.
And, what is its derivative?
Well, plug into this
expression.
It's c1 minus c2.
Okay, what's the solution to
those pair of equations?
c2 has to be equal to c1.
The sum of the two of them has
to be one.
Each one, therefore,
is equal to one half.
And so, what's the value of Y1?
Y1, therefore,
is the function where c1 and c2
are one half.
It's the function e to the x
plus e to the negative x divided
by two.
In the same way,
I won't repeat the calculation.
You can do yourself.
Same calculation shows that Y2,
so, put in the initial
conditions.
The answer will be that Y2 is
equal to e to the x minus e to
the minus x divided by two.
These are the special
functions.
For this equation,
these are the normalized
solutions.
They are better than the
original solutions because their
initial values are nicer.
Just check it.
The initial value,
when x is equal to zero,
the initial value,
this has is zero.
Here, when x is equal to zero,
the value of the function is
zero.
But, the value of its
derivative, these cancel,
is one.
So, these are the good guys.
Okay, there's no colored chalk
this period.
Okay, there was colored chalk.
There's one.
So, for this equation,
these are the good guys.
These are our best solutions.
e to the x and e to the
minus x are good solutions.
But, these are our better
solutions.
And, this one,
of course, is the function
which is called hyperbolic sine
of x, and this is the one which
is called hyperbolic cosine of
x.
This is one of the most
important ways in which they
enter into mathematics.
And, this is why the engineers
want them.
Now, why do the engineers want
normalized solutions?
Well, I didn't explain that.
So, what's so good about
normalized solutions?
Very simple:
if Y1 and Y2 are normalized at
zero, let's say,
then the solution to the IVP,
in other words,
the ODE plus the initial
values, y of zero equals,
let's say, a and y
prime of zero equals b.
So, the ODE I'm not repeating.
It's the one we've been talking
about all term since the
beginning of the period.
It's the one with the p of x
and q of x.
And, here are the initial
values.
I'm going to call them a and b.
You can also call them,
if you like,
maybe that's better to call
them y0, as they are individual
in the homework.
They are called,
I'm using the,
let's use those.
What is the solution?
I say the solution is,
if you use y1 and y2,
the solution is y0,
in other words,
the a times Y1,
plus y0 prime,
in other words,
b times Y2.
In other words,
you can write down instantly
the solution to the initial
value problem,
if instead of using the
functions, you started out with
the little Y1 and Y2,
you use these better functions.
The thing that's better about
them is that they instantly
solve for you the initial value
problem.
All you do is use this number,
initial condition as the
coefficient of Y1,
and use this number as the
coefficient of Y2.
Now, just check that by looking
at it.
Why is that so?
Well, for example,
let's check.
What is its value of this
function at zero?
Well, the value of this guy at
zero is one.
So, the answer is y0 times one,
and the value of this guy at
zero is zero.
So, this term disappears.
And, it's exactly the same with
the derivative.
What's the value of the
derivative at zero?
The value of the derivative of
this thing is zero.
So, this term disappears.
The value of this derivative at
zero is one. And so, 
the answer is y0 prime.
So, check, check,
this works.
So, these better solutions have
the property,
what's good about them,
and why scientists and
engineers like them,
is that they enable you
immediately to write down the
answer to the initial value
problem without having to go
through this business,
which I buried down here,
of solving simultaneous linear
equations.
Okay, now, believe it or not,
that's all the work.
We are ready to answer question
number two: why are these all
the solutions?
Of course, I have to invoke a
big theorem.
A big theorem:
where shall I invoke a big
theorem?
Let's see if we can do it here.
The big theorem says,
it's called the existence and
uniqueness theorem.
It's the last thing that's
proved at the end of an analysis
course, at which real analysis
courses, over which students
sweat for one whole semester,
and their reward at the end is,
if they are very lucky,
and if they have been very good
students, they get to see the
proof of the existence and
uniqueness theorem for
differential equations.
But, I can at least say what it
is for the linear equation
because it's so simple.
It says, so,
the equation we are talking
about is the usual one,
homogeneous equation,
and I'm going to assume,
you have to have assumptions
that p and q are continuous for
all x.
So, they're good-looking
functions.
Coefficients aren't allowed to
blow up anywhere.
They've got to look nice.
Then, the theorem says there is
one and only one solution,
one and only solution
satisfying, given initial values
such that y of zero,
let's say y of zero is equal to
some given number, A,
and y, let's make y0,
and y prime of zero equals B.
The initial value problem has
one and only one solution.
The existence is,
it has a solution.
The uniqueness is,
it only has one solution.
If you specify the initial
conditions, there's only one
function which satisfies them
and at the same time satisfies
that differential equation.
Now, this answers our question.
This answers our question,
because, look,
what I want is all solutions.
What we want are all solutions
to the ODE.
And now, here's what I say:
a claim that this collection of
functions, c1 Y1 plus c2 Y2
are all
solutions.
Of course, I began a period by
saying I'd show you that c1
little y1 c little y2 are all
the solutions.
But, it's the case that these
two families are the same.
So, the family that I started
with would be exactly the same
as the family c1 prime Y1
because,
after all, these are two
special guys from that
collection.
So, it doesn't matter whether I
talk about the original ones,
or these.
The theorem is still the same.
The final step,
therefore, if you give me one
more minute, I think that will
be quite enough.
Why are these all the
solutions?
Well, I have to take an
arbitrary solution and show you
that it's one of these.
So, the proof is,
given a solution,
u(x), what are its values?
Well, u of x zero is u zero,
and u prime of x zero, zero,
let's say, is equal to u zero,
is equal to some other
number.
Now, what's the solution?
Write down what's the solution
of these using the Y1's?
Then, I know I've just shown
you that u zero times Y1 plus u
zero prime Y2
satisfies the same initial
conditions, satisfies these
initial conditions,
initial values.
In other words,
I started with my little
solution.
u of x walks up to and
says, hi there.
Hi there, and the differential
equation looks at it and says,
who are you?
You say, oh,
I satisfy you and my initial,
and then it says what are your
initial values?
It says, my initial values are
u0 and u0 prime.
And, it said,
sorry, but we've got one of
ours who satisfies the same
initial conditions.
We don't need you because the
existence and uniqueness theorem
says that there can only be one
function which does that.
And therefore,
you must be equal to this guy
by the uniqueness theorem.
Okay, we'll talk more about
stuff next time,
linear equation next time.
