Power series and Frobenius' method

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August 1st 2007, 06:30 AM
neven87

Power series and Frobenius' method

I dont work with power seris that much and i have a problem:

(x-1)y''-xy'+y=0

I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)

Anywhos it...i could also use help with a power series in which i have to find two linealy indepentdent series about x=0 (using Frobenius' method) on the porblem 2xy''-y'+2y=0

I really appreciate any help, thanks. :D
August 1st 2007, 07:59 AM
ThePerfectHacker

The point x=0 is an ordinary point, there is no need to use Frobenius Method here.

Look for a solution of the form,
$y=\sum_{n=0}^{\infty}a_nx^n$
August 1st 2007, 08:09 AM
neven87

I was told to use the method to get two independant linear sets.
August 1st 2007, 08:22 AM
neven87

never mind, figured it out, but i was wondering if anyone could tel me about the x+1...

im guessing it would basically become (X^(n+1)) +1 ?
August 1st 2007, 01:19 PM
topsquark

Quote:

Originally Posted by neven87

I dont work with power seris that much and i have a problem:

(x-1)y''-xy'+y=0

I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)

Say that you are going to guess a solution of the form:
$y = \sum_{n = 0}^{\infty}a_nx^n$

Then
$y^{\prime} = \sum_{n = 1}^{\infty}na_nx^{n-1}$
and
$y^{\prime \prime} = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

Put this into your differential equation:
$(x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0$

The first term becomes:
$(x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} = x \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - 1 \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

$= \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

The second term becomes:
$x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} = \sum_{n = 1}^{\infty}na_nx^n$

So your differential equation becomes:
$\sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - \sum_{n = 1}^{\infty}na_nx^n + \sum_{n = 0}^{\infty}a_nx^n = 0$

-Dan