00:01
Okay, so for this program, we first need to rewrite the lde in the following way.
00:10
W double prime plus px, w prime, plus w equals w equals qx times w equals to 0.
00:20
So in this case, px equals to minus 1 over x, and qx equals to minus 1.
00:27
So x equals to 0 is a singular value.
00:32
We can see that the following to limit x times px, when x goes to 0, we have minus 1.
00:41
The limit x goes to 0, x squared times qx is 0.
00:48
That means x equals 0 is a regular singular value with initial equation, r times r minus r minus r equals 0.
00:59
This quadratic equation has two solutions, r equals to 0 and i.
01:04
Equals to 2.
01:05
We choose the larger one to be the exponent.
01:09
So the solution to this ode has an expansion.
01:14
We denote this expansion by w2x.
01:18
So it's a summation k from 0 to infinity 8k, x to k plus 2.
01:23
And we take the derivative for w and the second order derivative.
01:45
So we plug in this expansion to the left side of the ode.
01:50
That gives us the left and side equals to summation k from 0 to infinity, k plus 2 times k plus 1, a sub k x 3 k plus 1, minus summation k from 0 to infinity, k plus 2 times a sub k x to the k plus 1, minus summation k from 0 to infinity, a sub k, x to k plus 3.
02:20
We need to shift the index to merge all these submissions into a single one.
02:27
So first we can merge the first two terms.
02:31
So we have summation k from zero to infinity, k plus two times k times 8k, x to the k plus 1, minus summation k from zero to infinity, 8k, x to the k plus 3.
02:48
So if we consider the first term, if we plug in k equals 0 into the first term, we have 0.
02:54
That means, in fact, this summation should be a start with k from 1.
03:01
Now we can shift the index...