The following content is
provided under a Creative
Commons license.
Your support will help
MIT OpenCourseWare
continue to offer high-quality
educational resources for free.
To make a donation or
view additional materials
from hundreds of MIT courses,
visit MIT OpenCourseWare
at ocw.mit.edu.
HERBERT GROSS: Hi,
today, we're going
to emphasize the
role of power series
in the solution of linear
differential equations.
And let me just
say at the outset
that we have paid no attention
to nonlinear differential
equations in this course.
And we won't pay any attention
to the nonlinear equation
simply because the nonlinear
equation is very, very
difficult to handle.
We usually tackle it
only by special cases.
And, even of more
importance, most
of the problems that
we have to tackle
are at least reasonably
well approximated
by linear differential
equations.
Now what I'd like to do before
we launch into power series
today is to pull
together everything
that we've said so far about
linear differential equations.
And, rather than to drone
on here saying that,
I thought that I would write
the summary on the board,
and we could go through this
fairly rapidly together.
At any rate, the
lecture for today
is power series solutions, but
the summary to date is this.
Given the general linear
differential equation L of y
equals f of x with
non-constant coefficients,
this equation always
has a general solution,
provided p, q, and
f are continuous.
Not only does the
solution exist,
but the solution is always
given by y sub h plus y sub p
where y sub h is the general
solution of L of y equals 0,
and y sub p is any
solution, in other words,
a particular solution,
of L of y equals f of x.
And then you see the
rest of our study
has been how do you find y sub
h and how do you find y sub p.
The point is it's very
easy to find y sub h
if L sub y has
constant coefficients.
That was the first
case that we tackled.
If L of y has
constant coefficient,
then y sub h has one of these
three forms, depending on what?
This is the case that
holds if the roots
of the characteristic
equation are real and unequal.
This is the form that holds if
the roots are real but equal.
And this is the equation that
holds if the reals are non--
if the roots are
non-real where alpha
is the real part of the root,
and beta is the imaginary part
of the root, OK?
As far as y sub p
was concerned, recall
that the method of
undetermined coefficients
yields y sub p when L of y
has constant coefficients
and when f of x has the very
special form e to the mx or x
to the n or cosine
mx or sine mx.
We then turned our attention
to variation of parameters,
and the key point about
variation of parameters
was that, by use of
variation of parameters,
we could always obtain a y
sub p when y sub h was known.
By the way, in
particular, notice
that, if we are dealing with
constant coefficients, then
we do know y sub h.
That's what we were just
talking about earlier.
So, in particular, the
variation of parameters method
will work whenever we
have constant coefficients
because we know y
sub h in that case.
But quite in general,
you see, even
if the coefficients
aren't constant,
variation of parameters works.
And, in fact, once
we know y sub h--
say y sub h is c1
u1 plus c2 u2, then
y sub p is g1 u1 plus g2 u2
where g1 prime and g2 prime
satisfy this pair of equations.
The other important thing
about variation of parameters
was that it reduces
the order of L of y
equals 0 once one solution
of the equation is known.
And that, you see,
completes our summary.
What we want to tackle next
is what's still lacking.
And that is that the success of
the method known as variation
of parameters hinges
on the fact that we
have the general solution
of the homogeneous equation.
In other words,
our next problem is
to find the general
solution of L of y
equals 0 when the
coefficients are not constant.
In other words, if I can find
this general solution of L of y
equals 0 when coefficients are
not constant, then, you see,
I can use the method of
variation of parameters, which
always yields y sub p
once y sub h is known,
to find the particular solution.
Then y sub p plus y sub h
will be my general solution.
At any rate, it's
into this environment
that we introduce the
concept of power series.
And the key theorem in
using power series is this.
Let's suppose that p of x
and q of x are analytic.
And, by the way, I've
use that word for complex
valued functions.
Analytic meant that all of
the derivatives existed.
It meant that the function
could be expanded or represented
by a convergent power series.
That's the meaning of analytic,
even in the real case.
When I say that p
and q are analytic
for absolute value
of x minus x0 less
than some value
R, what I mean is
that p and q can be
represented by convergent power
series for all x, which are
within R of some fixed point
x0.
But, at any rate,
all I'm saying is,
if p and q are analytic
in this interval,
then every solution of this
equation, every solution which
exists, which is defined at x
equals x0, is itself analytic,
at least in the same interval
that p and q were analytic.
In other words,
what this means is
that, in the first
part of our review,
we said, lookit, as long as
p, q, and f are continuous,
there will always be
a general solution.
Now we're going one step further
or maybe several steps further.
We're saying, lookit, as
long as p and q are not only
just continuous, but they
also happen to be analytic,
we can not only guarantee that
there'll be a general solution,
but we can guarantee,
more to the point,
that whatever
solution there is will
be found in the form of a
convergent power series.
And that gives us the step
that we need to solve problems
by this technique.
Now what I'm going
to do is this.
I am going to start
with an example
that we already knew
how to solve before.
First of all, notice
that the equation y
double prime plus y
equals 0, first of all,
has constant coefficients.
And notice that the lead
in to today's lesson
was we we're going to tackle
situations where we didn't
have constant coefficients.
This does have
constant coefficients.
And, even more to the
point, solution is known.
In fact, what is
the solution here?
It's y equals an arbitrary
constant times sine x
plus an arbitrary
constant times cosine x.
I picked one with
constant coefficients
where the solution
was known simply
so that we could illustrate
what the method is.
By the way, notice that,
in this particular problem,
p of x, which is the
coefficient of y prime, is 0.
q of x, the
coefficient of y, is 1.
And, certainly, 0 and 1
are analytic functions.
In fact, their power series--
the power series for 0 is 0.
And the power series for 1 is 1.
So, certainly, our coefficients
meet the requirement.
And what we do is we
start off saying, OK,
if any solution exists, it
must look like a power series.
Let's go find it.
Notice this is a
glorified version
of undetermined coefficients.
We say, lookit, let's assume
that our solution has the form
of a power series, in
other words, summation,
n goes from 0 to infinity, a sub
n x to the n where all I have
to know now are the
values of the a sub n's.
And, if you don't like the sigma
notation here, all we're saying
is we're trying for a
solution in the form y
equals a0 plus a1 x
plus a2 x squared,
et cetera, where this series
is a convergent power series.
And why is convergent
power series important?
Well, the answer, recall
from part one of our course
that, in the interval
of convergence,
the power series behaves
precisely like a polynomial.
We can add power
series term by term.
We can subtract
them term by term.
We can multiply them the
way we do polynomials.
We can integrate term by term.
We can differentiate
term by term.
We can rearrange the terms
the way we wish, et cetera.
In other words, it has
all of the niceties
of a finite
polynomial expression.
At any rate, what I
then do is, because I'm
assuming this is a
convergent power series,
I differentiate this
thing term by term.
And, if I differentiate
this, notice what I'm doing.
I differentiate each term
here, but, because I can--
because I can
differentiate term by term,
a very nice generic
way of doing this
is notice that this
is the n-th term.
The derivative of this
term would be what?
n an x to the n minus 1,
n an x to the n minus 1.
Notice that, every
time you differentiate,
a term drops out.
Like, when I differentiate
this, the a0 term drops out.
When I differentiate this,
the a1 one term drops out.
So what I do is I differentiate
the general expression
inside the sigma sign and then
start the index of summation
one up further.
In other words, instead of
going from 0 to infinity,
I now go from 1 to infinity.
In a similar way,
y double prime,
I differentiate
this term by term.
That's n times n minus 1 times a
sub n times x to the n minus 2.
And now my sum goes
from 2 to infinity.
That's what I've
written over here.
Now I'm trying to find out
what the a's are equal to.
I could actually
call these y sub p's.
See, I'm looking for a
particular solution perhaps,
not particular in the sense of
a solution, but this is a trial.
Maybe it should have
been a T over here,
y sub T. I'm looking
for a trial solution
over here, a trial solution.
I take yT double prime,
yT, plug it in here,
and see what this means.
And, simply replacing y double
prime by this and y by this,
I see that this is
the equation that
must be satisfied identically.
Now the point is the only way a
power series can be identically
equal to 0 is if
coefficient by coefficient
the power series is 0.
Now what has me bugged
a little bit here
is that the exponents
don't match up.
You see, notice that the general
term here has an exponent n,
and the general term here
has an exponent n minus 2.
A very nice trick
that I can use here
is I say, you know, why
don't I jack up n by 2
every place I see it?
In other words, let me
replace n by n plus 2.
I'll jack it up by 2.
And, to compensate
for that, I'll
lower the summation of index--
the index summation by
2 every place I see it.
Now let me show you what
that means so that we
don't get too confused on that.
Let's suppose I had written
down here summation n goes
from 0 to infinity a sub n.
This means what?
a0 plus a1 plus a2, et cetera.
Suppose, for some
reason, I would
like to either lower or
raise the index here.
Let's suppose I want
to lower the index.
In other words, I want each--
I want the subscript here to be
k less than what it was before.
In other words, I'm
going to replace a sub
n by a sub n minus k.
I claim that all I have to do
is start my counting process
at n equals k now,
instead of n equals 0.
In fact, look what this says.
When n is k, n minus k is 0.
So the first term here is a0.
The next term would be k plus
1. k plus 1 minus k is 1.
So the next term would
be a1, et cetera.
And this is just another
way of saying this.
So the idea is, if
I want to get this
to be an n, that means I
have to raise everything
by 2 inside the summation sign.
So I will lower everything by
2 outside the summation sign.
Now, to do that, notice
what that's going to do.
My sum here will now
go from 0 to infinity.
This will become n plus 2.
This will become n plus 1.
You see, I'm raising it by 2.
Every place I'm seeing an n,
I'm replacing it by n plus 2.
This becomes a sub n plus
2, and this will become n.
In summary, then this expression
is the same as summation,
n goes from 0 to infinity, n
plus 2 times n plus 1 a sub n
plus 2 x to the n plus
summation, n goes from 0
to infinity, an x to the n.
And that must be
identically zero.
Now the point is that the
exponents now line up.
I'm starting the sum in both
series at the same place.
And, consequently,
because these are
assumed to be
convergent power series,
I can add them term by term.
If I add them term
by term, that means
I come inside the
summation sign here.
x to the n is a common factor.
I factor that out.
And what's left is what?
n plus 2 times n plus
1 times a sub n plus 2
plus a sub n, that's
this expression here.
And this is what must
be identically zero.
Now the point is that the only
way that a convergent power
series can be identically zero
is if each coefficient is 0.
And that's how I now handle
the undetermined coefficients.
I come to the conclusion
that, because this
is to be identically zero,
every one of these terms
must be itself 0.
Setting this equal to zero
and solving for a sub n plus 2
in terms of a sub n, I
find that a sub n plus 2
must be minus an over n
plus 2 times n plus 1.
And what this tells me is that
I now know what a term looks
like as soon as I know the
one that comes two before it.
You see, notice that this tells
me how to find a sub n plus 2
once I happen to
know what a sub n is.
Now, lookit, before
I can get to two
away from a
particular subscript,
I must be at least at
the second subscript.
In other words, it seems
that this recipe will tell me
how to find a sub 2
in terms of a sub 0,
how to find a sub 4
in terms of a sub 2,
how to find a sub 3
in terms of a sub 1,
but it gives me no hold on what
a sub 0 and a sub 1 must be.
So I say, OK, what
I will do is I
will use this
recipe, recognizing
that I am free to pick a0 and
a1 completely arbitrarily.
Now look at what happens here.
If I pick a0 and a1 arbitrarily,
what does the recipe tell me?
If I pick n to be 0,
the recipe says what?
a sub n plus 2, a sub
2, is minus a0 over n
plus 2 times n plus 1.
That's just 2 times 1.
a sub 2 is just minus
a0 over 2 factorial.
In a similar way,
the recipe tells me
that a sub 4 is minus
a sub 2 over 4 times 3.
You see, in this case,
I take n to be 2.
So n plus 2 is 4.
And, if I now say a sub n
plus 2 is minus a sub n over n
plus 2 times n plus 1, this just
says that a4 is minus a sub 2
over 4 times 3.
I know what a sub
2 is from here.
So I can now express a
sub 2 in terms of a sub 0.
In fact, it now turns
out that a sub 4
is a sub 0 over 4 factorially--
4 factorial.
Similarly, a sub 6 is minus
a sub 4 over 6 times 5.
a sub 4 is a0 over 4 factorial.
So, substituting that
value of a4 in here,
I get that a6 is minus
a0 over 6 factorial.
And, without beating
this thing to death,
notice that I can find every
even subscript of a, a2, a4,
a6, a8, et cetera, in terms
of a0 times something.
In fact, I think
you can see what's
starting to develop over here.
In a similar way, picking n
to be 1 so that n plus 2 is 3,
the recipe a sub n plus 2
equals minus a sub n over n
plus 2 times n plus 1 says that
a3 is minus a1 over 3 times 2.
That's minus a1
over 3 factorial.
Similarly, a5 is minus
a3 over 5 times 4.
Noticing that a3 is minus
a1 over 3 factorial,
I see that a5 is a1
over 5 factorial.
And, without going
on further here,
notice that a3, a5,
a7, a9, a11, et cetera,
will all be expressible
in terms of a1.
And, in fact, I think you
see what's happening here,
that the terms will
alternate in sign,
and the coefficients
will be things like what?
1, 1 over 3 factorial, 1 over 5
factorial, 1 over 7 factorial,
et cetera.
But I'll leave
those details to you
because I think that's the
easy part to meditate upon.
But now what we're
saying is I now
know what all these
coefficients look
like in terms of a0 and a1.
Remember what I'm looking for.
I'm looking for a
power series solution
that's y equals a 0 plus a1 x
plus a2 x squared, et cetera.
We're assuming that this is
a convergent power series.
That means I can group the
terms any way that I want.
In particular, since all
of the even subscripts
seem to be expressible
all in terms of a0,
and all the odd subscripts
seem to be expressible in terms
of a1, let me group the
terms with even powers
together and the terms
with odd powers together.
And now, remembering what
a2 and a4 are-- remember,
a2 was minus x squared--
was minus a0 over 2 factorial.
a4 was a0 over 4
factorial, et cetera.
I now factor out a0 from here.
And, substituting in the
values of a2, a4, a6, et
cetera, that we
found from before,
this tells me that the
coefficient that multiplies a0
is 1 minus x squared
over 2 factorial plus x
to the fourth over 4
factorial, et cetera.
And, similarly, since a3 is
minus a1 over 3 factorial,
and a5 was a1 over 5
factorial, et cetera,
I can now express all
of these in terms of a1.
I factor out a1.
And what's left this x minus x
cubed over 3 factorial plus x
to the fifth over 5
factorial, et cetera.
And now it appears that my
solution is this plus this.
Keep in mind that a0 and
a1 are arbitrary constants.
By the ratio test,
even if I didn't
know what these
series represented,
I can show that these
series converge absolutely
for all finite values of x, but
that's not too crucial here.
The important point is that,
even though it may be implicit,
this does name some function of
x, whatever it happens to be.
This names some function of
x, whatever it happens to be.
I can find particular
solutions by choosing a0 and a1
to be certain fixed constants
that I want them to be.
These two power series
are not constant multiples
of one another, as you
can see by a glance.
Therefore, these two solutions
are linearly independent.
And, consequently, I now
have a general solution here,
even though this may
be awkward for me.
Well, the reason that I picked
constant coefficients and y
double prime plus y
equals 0 was that I
wanted our first
experience to be pleasant,
at least in the sense that we
would get a feeling for what
the answer meant.
Keep in mind that,
if I had never
heard of sine x and cosine
x, this solution here
would make sense.
But, if I happen to recall
that the power series
expansion for cosine
x is precisely this,
it just happens, you
see, as an afterthought
that this is cosine x.
This also happens to be sine x.
Notice now what this tells me is
that the solution to I've found
is a0 cosine x plus a1 sine x.
And that just happens
to agree with the result
that I already knew.
Now, rather than have
you feel that I've
wasted your time by
giving you something
that you've already known,
let me go one step further
and give you the same
kind of a problem.
Only now, I'm going
to give you one
that you couldn't solve
before and, in fact,
I don't think you
could solve now
if you don't use power series.
And the example I have in mind
is almost the same as the one
we just did.
Instead of y double
prime plus y equals 0,
I pick y double prime
plus xy equals 0.
And now you see I'm back--
not back, but now I
have arrived at the case
of non-constant coefficients.
You see the coefficient of y is
x, and that's not a constant.
So you see now I'm at a bona
fide problem where I couldn't
find y sub h in the
trivial way that I
could if the coefficients
all happened to be constant.
Now the way I tackle
this is precisely
the same way as before.
What I do is I again try
for a solution in the form y
equals summation, n goes
from 0 to infinity, a sub n
x to the n.
And what I must do
now is to determine
what the a sub n's are.
And, again, as before, I find
y prime the same mechanical way
as before.
I differentiate term by
term, meaning, generically, I
differentiate this to get this.
And, similarly, to
get y double prime,
I differentiate this
term to get this one, OK?
And now what I do, in terms of
my so-called trial solution--
see, remember, these
are trial solutions--
I'm going to try to feed
these back into here
and see if that will tell me
what the coefficients a0, a1,
a2, a3, et cetera
have to look like.
Now, if I do that,
you see, I get what?
y double prime is
just this term here.
And now I want x times y.
That means I have to take
this y and multiply it by x.
Multiplying by x on the
outside, since we're
assuming that this is
a convergent series,
that means I can
multiply term by term.
Therefore, I can
bring the x inside,
and that makes this x to the n
plus 1, rather than x to the n.
And so I wind up with
the fact that I now
want to find the constants
a0, a1, et cetera,
from this particular
relationship.
You see what's happened here?
This is just y double prime.
This is y multiplied by x.
That's why that n
plus 1 is in here.
And that must be
identically zero.
Now what I'm going to do is
the same trick as before.
What I'm going to do is I want
to add these term by term.
And I notice that their
exponents are different.
In fact, this exponent
is three larger
for each value of
n than this one,
right? n plus 1 minus the
quantity n minus 2 is 3.
So what I want to do is I
want to lower this subscript--
this exponent by 3.
So I'm going to
inside the integral--
I'm sorry, inside
the summation sign,
I am going to replace
n by n minus 3.
And, in line with our previous
note, to compensate for this,
instead of my summation going
from 0 to infinity here,
it will now go
from 3 to infinity.
In other words, I
am now simply going
to replace this by
the equivalent sum,
n equals 3 to infinity, a sub
n minus 3 x to the n minus 2,
all right?
And you see, if I do that, what
I will end up with is simply
this expression here.
And now a new wrinkle comes
up that didn't happen before,
but I just wanted to make sure
a few funny things happened here
so I can tell you
directly what's going on.
And then we can drill
with the remaining fine
points in the exercises.
But the key point
here is, lookit,
I've got the exponents
lined up now.
But look at the
indices of summation.
One begins at 2, and
the other begins at 3.
And it seems that
they're out of phase.
All I want you to observe is
that, to make this start at 3,
I could split off separately
the term that corresponds to n
equals 2.
In other words,
notice that, when n
equals 2, what I have is what?
2 times 1 times a sub
2 times x to the 0.
In other words, all I have
is 2 a sub 2, 2 a sub 2
when n is equal to 2.
So what I can do is
I'll split that n
equals 2 term off separately.
That's 2 a sub 2.
Then what's left is the sum
as n goes from 3 to infinity.
You see, splitting off these
terms isn't hard at all.
All you do is, if these don't
match up, take the smaller
index, and split off the
number of terms necessary so
that the resulting sum
will begin at a higher
index matching this one.
Since they only
differed by 1 here,
I only have to split
off the n equals 2 term.
Now what I have is what?
I have 2 a2 plus this summation,
now going from 3 to infinity,
plus this summation, which
also goes from 3 to infinity.
And now, since both the
exponents and the indices
match up, now I can add term by
term for these convergent power
series.
I bring these inside
one integral sign.
See, the sigma of a sum
is the sum of the sigmas.
I add these up term by term.
So that leaves me what?
I have my 2 a2
still outside here.
And then, inside the sigma
notation, from 3 to infinity,
I have what?
n times n minus 1 times a
sub n plus a sub n minus 3,
that whole quantity,
times x to the n minus 2.
And that must be
identically zero.
Now the only way that this
can be identically zero
is if each coefficient is 0.
By the way, notice that,
since the first term that
appears here corresponds to
n equals 3, when n equals 3,
the exponent here is 1.
So notice that this is
my only constant term.
All the terms in here begin with
at least a factor of x in them.
Since my power series,
to be identically zero,
must have all of its
coefficients identically zero,
it means that not only must each
of these be 0, but 2 a2, which
is my constant term on the
left-hand side, that must also
be 0.
So, in other words,
to summarize this up,
what we're saying so far
is that, for this to be
identically zero, a2 must be 0.
And, once n is at
least as big as 3,
this condition here
must be obeyed.
What condition?
That the expression in
brackets must be 0 for each n
once n is at least as big as 3.
In other words, then,
in summary, a2 is 0.
And, for n at least as big as
3, a sub n is minus a sub n
minus 3 over n times n minus 1.
And keep in mind the way I
got that quite trivially was
I set this equal to 0.
I transposed the
a sub n minus 3,
divided through by
n times n minus 1,
and solved for a sub n.
What this tells me
now is that, once I
know a particular a, look
at what this tells me.
This says, to find a sub n,
once I know what the a was three
before this one, I'm home free.
In other words, to find a sub 4,
all I have to know is a sub 1.
To find a sub 5, all I
have to know is a sub 2.
You see?
To find a sub 6, all I have
to know is a sub 3, et cetera.
So the idea is I can
now pick a0 and a1
at random, just as before.
And now, using this
recipe with n equal to 3,
I have that a sub 3 is minus
a sub 0 over 3 times 2.
a sub 6 is minus a
sub 3 over 6 times 5.
And, remembering what
a3 is in terms of a0,
this tells me that a6 is a0
over 5 times 6 times 3 times 2.
And this is not a misprint
that the four is missing here.
This is not a factorial.
You see, I have a
way of generating
what these
coefficients look like,
but now I've picked
a problem where
I may not remember or recognize
what power series I'm getting.
That's irrelevant again.
All I'm saying is
I can determine
a3, a6, a9, et cetera,
just knowing what a0 is.
And, just knowing what a1
is, I can determine a4.
Just by using this
recipe with n equal to 4,
a4 is minus a1 over 4 times 3.
a7 is minus a4 over 7 times
6, using that same recipe.
Putting in the value of
a4 from here into here,
I have that a7 is a1 over
7 times 6 times 4 times 3.
And this keeps on going.
I can now find a10
in terms of a7.
That will give me
a10 in terms of a1,
et cetera, et cetera, et cetera.
Finally, to find
a5 or a8, notice
that, using the same recipe,
a5 is minus a2 over 5 times 4.
But, since a2 is
0, a5 will be 0.
And, similarly,
because a5 is 0, a8
will be 0 and so will
a11 and a14, et cetera.
Now the idea is the series
that I was looking for,
being a convergent
series, can be rearranged.
You see, the idea
in this problem
was that I would like to
group the exponents according
to whether they are divisible
by 3, leave a remainder of 1
when I divide by 3, or
leave a remainder of 2
when I divide by 3.
So I group the terms
this way-- see,
a0 plus a sub 3 x cubed
plus a sub 6 x to the sixth,
et cetera, plus a1 x plus a4
x to the fourth, et cetera,
plus a2 x squared plus
a5 x fifth, et cetera--
the idea being that
I know how to express
a3, a6, a9 in terms of a0.
I know how to express a4, a7,
a10, et cetera, in terms of a1.
And I know how to express a5,
a8, et cetera, in terms of a2.
In fact, in this simple case,
which turned out nicely,
they all turned out to
be 0, these coefficients.
So what's left is what?
Replacing a4, a7,
et cetera, by what
they look like in terms
of a1 and a3, a6, et
cetera, by what they
look like in terms of a0,
I wind up with y
equals a0 times 1
minus x cubed over 3 times
2 plus x to the sixth
over 6 times 5 times
3 times 2, et cetera,
plus a1 times this thing here.
Now the only difference
between this problem
and the previous one was
that, in the previous one,
I happened to recognize
what well-known function had
this as a power series.
In this case, I don't know
that, but this is no less
a bona fide solution than
we had in the previous case.
This is a convergent
power series.
We give it a name--
h of x, k of x.
If this were the solution to
an important enough problem,
we would say, lookit, instead
of calling it h of x and k of x,
let's give it a special
name because it's going
to come up over and over again.
I'm not going to go into this.
If I do go into it all, it'll be
very lightly in the exercises.
But things like that
you may have heard,
in terms of name
dropping, Bessel functions
and the like,
Legendre polynomials,
these were all special
power series solutions
to very special
differential equations
where the equation
itself was so important
an application that
the power series that
represented the equation
was given a special name.
So you see, in terms of
power series solutions,
one extremely powerful
use of power series
is that it's used
to help us find
solutions of the homogeneous
equation L of y equals 0.
And, once we have the general
solution of the equation L of y
equals 0, we can then use
variation of parameters
to find a solution of
L of y equals f of x.
Then we add these two
solutions together
to get the general solution.
And that ends our theory on
linear differential equations,
except for one more topic
called the Laplace transform
that I would like to
introduce for you to see.
And so I will have
one more lecture
in the guise of linear
differential equations in order
that I can bring up
the Laplace transform,
but that won't be
until next time.
Until next time then, goodbye.
Funding for the
publication of this video
was provided by the Gabriella
and Paul Rosenbaum Foundation.
Help OCW continue to provide
free and open access to MIT
courses by making a donation
at ocw.mit.edu/donate.
