# Differential equations, I can't understand a textbook example.

1. Jul 15, 2008

### itunescape

1. The problem statement, all variables and given/known data
Find the recurrence relation and the general term for the solution:
y'' - xy' - y = 0 xo=1

2. Relevant equations

y= sum (n=0 to infinity) an (x-1)^n

3. The attempt at a solution
i get:

y= sum (n=0 to infinity) an x^n
y'= sum (n=0 to infinity) (n+1)an+1 (x-1)^n
y'' = sum (n=0 to infinity) (n+2)(n+1)an+2(x-1)^n

here all of the sums start at zero and the powers of (x-1) all are n, but within the orginal equation there is an +and this is where the textbook doesn't make sense:
the book sets x= 1+ (x-1). Why do you do this? i don't understand.

I thought that: xy' = sum (n=0 to infinity) (n+1) an+1 (x-1)^n+1
but i'm not sure what to do from here...bc the indexes and powers need to be the same before calculating the recurrence relation right?

2. Jul 16, 2008

### Mute

Why are you using $y = \sum a_n (x-1)^n$ as opposed to $y = \sum a_n x^n$? If you just use x^n instead of (x-1)^n, you get

$$\sum_{n=0}^{\infty} n(n+1)a_n x^{n-2} - x\sum_{n=0}^{\infty}a_n n x^{n-1} - \sum_{n=0}^{\infty} a_n x^n = 0$$

In the second term you multiply the x through to get x^n in the sum. You then have two sums with x^n and one with x^{n-2}. So, you change dummy indices in the first summation to get it to be x^n as well. Do you know how to do that?

There may be a reason for using (x-1) instead, such as if for some reason the series doesn't converge well for x < 0 or something. In that case,

$$\sum_{n=0}^{\infty} n(n+1)a_n (x-1)^{n-2} - x\sum_{n=0}^{\infty}a_n n (x-1)^{n-1} - \sum_{n=0}^{\infty} a_n (x-1)^n = 0$$

In this case, you want the (x-1)^(n-1) in the second term to absorb the factor of x out front of it, but you can't do this directly as you have x(x-1)^(n-1), whereas in the above example I wrote you have x x^(n-1) = x^n. So, instead you write x = 1 + (x-1)so that you can multiply that into the second term to get one term that still looks like (x-1)^(n-1) and another that looks like (x-1)^n:

$$\sum_{n=0}^{\infty} n(n+1)a_n (x-1)^{n-2} - \sum_{n=0}^{\infty}a_n n (x-1)^{n-1} - \sum_{n=0}^{\infty}a_n n (x-1)^{n} - \sum_{n=0}^{\infty} a_n (x-1)^n = 0$$

Now you can do the usual business of shifting indices to get your series in terms of powers of (x-1).

Last edited: Jul 16, 2008
3. Jul 16, 2008

### HallsofIvy

Staff Emeritus
Do you mean y(0)= 1 or are you specifically asked to find a series solution expanded around x0= 1?

If you want an expansion about x0= 1, then you mean (x-1)n.

You do understand that x does equal 1+ (x-1) don't you? And they do this because in the xy you need to to have (x-1) to multiply into the sum.

Why would you think that?
$$x y'= x\sum_{n=0}^\infty (n+1)a_n (x-1)^n= \sum_{n=0}^\infty (n+1)a_n x(x-1)^n$$
of course- you can't multiply x times (x-1)n and get (x-1)n+1!
Change the "dummy indices" in the different sums so that you have the same powers.

4. Jul 16, 2008

### itunescape

Thank you guys for helping me clarify what I've been doing wrong.
The series needs to be expanded around xo=1 so it becomes (x-1) from (x-xo).
I've tried doing the work over again and this is what i obtained:

y=sum (n=0 to infinity) an (x-1)^n
y= ao+a1(x-1)+a2(x-1)^2+a3(x-1)^3...an(x-1)^n

y'= sum(n=1 to infinity) n*an (x-1)^n-1
y'=a1+ 2a2(x-1)+3a3(x-1)^2...n*an(x-1)^n-1

(x-1)y'= sum(n=1 to infinity) n*an (x-1)^n
(x-1)y'= a1(x-1)+ 2a2(x-1)^2+ 3a3(x-1)^3...n*an (x-1)^n

y''= sum(n=2 to infinity) (n-1)(n)*an (x-1)^n-2
y''= 2a2+3a3(x-1)+...(n-1)(n)*an (x-1)^n-2
y''= sum(n=0 to infinity) (n+1)(n+2) an+2 (x-1)^n

I need to add up the sums to try and get a recurrence, but i have to change the indexes to n=1 so:

y= ao + sum(n=1 to infinity) an (x-1)^n
y''= 2a2 + sum(n=1 to infinity) (n+1)(n+2)an+2(x-1)^n

2a2 + ao=0
recurrence:
{ (n+1)(n+2)an+2 -(n)an -an=0}

Is all this right so far? the text book has the recurrence as:
{(n+2)an+2 -an+1 -an=0}
is it suppose to be written this way?