>> Welcome to the Cypress College math review
on power series solutions.
Verify that X equals zero is an ordinary point
for the differential equation.
And express the general solution in terms
of power series about this point.
A solution Y equals the sum N equals zero
to infinity A sub N X to the Nth
that contains two arbitrary constants A sub zero
and A sub 1.
Converges inside a circle centered at X equals zero.
And extends out to the singular point
or points nearest X equals zero.
First let's find the singular points.
So we take the differential equation.
We divide by the coefficient of the second derivative.
1 minus X squared and we see
that the singular points are plus or minus 1.
Which tells us that X equals zero is an ordinary point.
Since this series is centered at X equals zero,
and the singular points are plus or minus 1.
That tells us that this series will converge
for absolute value of X less than one.
Now let's go back
to the original version of our equation.
Before we actually get started
on this problem let's take a look
at power series solutions in general.
They look like this the general form.
So if we write them out.
And if we take the derivative term by term.
This would be A sub 1 because we lose the constant.
And how about the 2nd derivative.
So 2 A sub 2 plus 3 times 2 A sub 3 X plus 4 times 3 A sub 4 X
squared and so on and so forth.
So we're losing a term each time aren't I?
So here we're starting at 1.
Here we're starting at 2.
Yes. We still have A sub N each time.
And notice -- well this is 1 times it.
This is 2 times it, so we have N times it.
And then we have X to the 1 smaller power
than it was before.
This is 2 now it's 1.
Power rule yes.
And we have N here.
And then we have N minus 1, we have 3 times 2, 4 times 3,
and then we have N minus 2 good cool.
Distributing we get the 2nd derivative.
Minus X squared times the 2nd derivative.
Minus 5X times the 1st derivative.
Minus 3Y is equal to zero.
Here are the series that we just came up with.
And I'm going to use the following shorthand notation
for this video.
Series starts at 2 so instead of writing N equal 2 to infinity,
I'm simply going to write 2.
So N, N minus 1, A sub N, X to the N minus 2,
and then we have minus X squared times the next series.
Starting at 2.
And then we have minus 5X times the next series.
That series starts at 1.
And then we have minus 3 times the next series starting
at zero.
Equals zero.
Now we notice that in front of 2
of these series we have variables.
So we have X squared in front of this series and we have X
in front of this series.
We need to distribute those inside the series.
So that's our next step.
So we distribute the X squared inside
which will change the exponent.
Instead of N minus 2 it will now be N. Here instead
of the exponent being N minus 1 it will be N.
Next step.
We notice that all of the exponents are N except
for the one on the left.
Over here we have an exponent of N minus 2.
We need to make them all match.
So we're going to make a substitution.
So let's let K equal N minus 2.
So here we're going to have X to the K
so this would be A sub what?
Well if K is N minus 2 then N is equal to K plus 2?
So that's going to be K plus 2 N minus 1 Then would be K plus 1.
N is going to be K plus 2.
Our starting value was N equaled 2.
Well if N starts out at 2 then K is going to start out at zero.
Everything else on this line is just being copied down.
That was the only thing we did on that step.
We now wish to combine the series into one series.
We notice that we have different starting values.
So the latest starting value is the one that we're going to use.
And any series that starts earlier then that, we will pull
out those earlier terms.
So this first series starts out at zero.
So we'll pull out the term with N equals zero and N equal 1.
So when N is equal to zero or K whatever.
So when we substitute in zero we get 2 times 1 A sub 2 X
to the zero.
We substitute in 1 we get 3 times 2 X sub 3 X to the first.
The next series starts out at 2 so we have no extra terms.
The next series starts out at 1.
So we have to pull out the term with N equal 1
so we have minus 5 times 1 A sub 1 X to the first.
The next series starts out at zero so we pull out the term
where N is equal to zero.
So we have minus 3 A sub zero X to the zero and when N is equal
to 1 we have minus 3 A sub 1 X to the first.
Now we have our series starting off at 2.
So then we have N plus 2 N plus 1 A sub N plus 2 minus N N minus
1 A sub N minus 5 N A sub N minus 3 A sub N and all
of that is being multiplied times X to the Nth power
and that's equal to zero.
Now we equate coefficients.
Alright the constant on the left hand side is 2 A sub 2 minus 3 A
sub zero that's equal to the constant on the right.
Which tells us that A sub 2 must be 3 halves A sub zero.
The coefficient of X on the left hand side is 6 A sub 3 minus 8 A
sub 1.
The coefficient of X on the right hand side is zero.
Which tell us that A sub 3 must be 4 thirds A sub 1.
The coefficient of all other powers of X
on the left hand side is what's inside the brackets.
That's equal to zero and then do the algebra.
And eventually you'll get that A sub N plus 2 is equal
to N plus 3 over N plus 2 times A sub N and this was
for N greater than or equal to 2.
Initially we had that A sub 2 was 3 halves times A sub zero.
And that A sub 3 was 4 thirds times A sub 1.
That would be for N equals zero and N equal 1.
Now our recurrence relation gives us other values.
If N is equal to 2 we substitute it into this recurrence relation
and we get A sub 4 is 5 fourths times A sub 2.
But we know that A sub 2 is 3 halves A sub zero.
If N is 3 A sub 5 is 6 fifths A sub 3.
But A sub 3 is 4 thirds A sub 1.
N equal 4 gives us the A sub 6 is 7 over 6 times A sub 4.
But A sub 4 was 5 times 3 over the product
of 4 times 2 times A sub zero.
N equal 5 if A sub 7 8 sevenths A sub 5
which gives us 8 sevenths.
And we have 6 times 4
over 5 times 3 times A sub 1 N equals 6 we have A sub 8
which is 9 over 8 times A sub 6 and we can replace A sub 6
with 7 times 5 times 3
over 6 times 4 times 2 times A sub zero.
And you can continue on as long as you want to with this.
So in general our solution looks like this.
Our series.
Substitute in the values we know.
Here we go.
So replacing A sub 2 with 3 halves A sub zero.
Replacing A sub 3 with 4 thirds A sub 1.
Replace A sub 4.
Replace A sub 5.
And continue on.
Substituting in the values that we just determined.
Next let's factor out our constants.
So we have 1 plus 3 halves X squared.
And then we have 5 times 3 over 4 times 2 X to the 4th.
For the next term we have 7 times 5 times 3
over 6 times 4 times 2 X to the 6th.
And so on and so forth.
Then we factor out A sub 1 from the other terms.
We have X plus 4 thirds X cubed.
Then we have 6 times 4 over 5 times 3 X to the 5th.
Then we have 8 times 6 times 4 over 7 times 5 times 3 X
to the 7th and so on and so forth.
They'll be times when they'll be no obvious pattern.
And you'll be allowed to just simplify
and write your answer like this.
For this problem we can definitely determine
the pattern.
And we're going to do so.
We're going to go ahead and start at 1.
Alright so this term is N is equal to 1.
This is N equal 2.
This is N equal 3.
So let's write down the term that has the most information
for us, which is the last term.
So 7 times 5 times 3 over 6 times 4 times 2 X to the 6th.
And that's when N is equal to 3.
So for the numerator we have 7 when N is equal to 3.
So possibilities for odd representations
of numbers are 2N plus 1, 2N minus 1, 2N plus 3, 2N minus 5,
there's all kinds of different representations of odd numbers.
Play around with it until you find the one that works.
And yes it's 2N plus 1.
So we need a representation of an odd integer where we get 7
as our result when N is equal to 3.
Subtract 2 to go down to the next odd integer.
And so on and so forth.
Down to 5 times 3 for our numerator.
Denominators a little bit easier in my opinion.
We're going to pull out a 2 from each of these factors.
We have 2 times 3.
Here we have 2 times 2.
Here we have 2 times 1.
So I have 2 to the third times 3 factorial,
ut that was when N was equal to 3.
So my denominator is 2 to the Nth times N factorial.
And what's my exponent for X?
Well it's 2N yes.
Alright. So that's my first summation.
Second one.
So the term with the most information for us.
So we have 8.
That's 8 times 6 times 4 over 7 times 5 times 3 X to the 7th.
And that's when N is equal to 3 once again.
So if you look at it the denominator
of this fraction is the same as the numerator
for the last fraction.
So we don't have to reinvent the wheel.
And the exponent it's the same representation
because we have 7 here and when N is equal to 3 we have 7
so it's 2N plus 1 again.
Now the numerator.
So we're going to pull out a 2 from each of those factors.
So we have 2 times 4, 2 times 3, and 2 times 2 yeah.
Let's clean that up a little bit.
Looking just at that.
So I have 2 to the 3rd times 4 times 3 times 2 basically
times 1.
So I have 4 factorial when N is equal to 3.
So I have 2 to the 3rd times 4 factorial when N is equal to 3.
So I have 2 to the Nth but I don't have N factorial,
I have N plus 1 factorial.
And now I have my solution written in terms
of summations with sigma notation.
By the ratio test it can be shown that these series converge
for the absolute value of X less than 1.
Which is the result that we originally expected.