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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
The point x=0 is an ordinary point, there is no need to use Frobenius Method here.
Look for a solution of the form,
$\displaystyle y=\sum_{n=0}^{\infty}a_nx^n$
I was told to use the method to get two independant linear sets.
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 ?
Say that you are going to guess a solution of the form:Quote:
Originally Posted by neven87
$\displaystyle y = \sum_{n = 0}^{\infty}a_nx^n$
Then
$\displaystyle y^{\prime} = \sum_{n = 1}^{\infty}na_nx^{n-1}$
and
$\displaystyle y^{\prime \prime} = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$
Put this into your differential equation:
$\displaystyle (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0$
The first term becomes:
$\displaystyle (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} = x \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - 1 \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$
$\displaystyle = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$
The second term becomes:
$\displaystyle x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} = \sum_{n = 1}^{\infty}na_nx^n$
So your differential equation becomes:
$\displaystyle \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - \sum_{n = 1}^{\infty}na_nx^n + \sum_{n = 0}^{\infty}a_nx^n = 0$
-Dan
