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

[itunescape]

Homework Statement

Find the recurrence relation and the general term for the solution:
y'' - xy' - y = 0 xo=1

Homework Equations

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

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?

 

[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).

 

[HallsofIvy]

Homework Statement

Find the recurrence relation and the general term for the solution:
y'' - xy' - y = 0 xo=1

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

Homework Equations

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

The Attempt at a Solution

i get:

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

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

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.

I thought that: xy' = sum (n=0 to infinity) (n+1) an+1 (x-1)^n+1

Why would you think that?
x y&#039;= 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!

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?

Change the "dummy indices" in the different sums so that you have the same powers.

 

[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 textbook has the recurrence as:
{(n+2)an+2 -an+1 -an=0}
is it suppose to be written this way?