Finding two linearly independent solutions (power series)

• April 24th 2013, 03:57 PM
LinearOperator
Finding two linearly independent solutions (power series)
for (x^2+2)y'' +3x-y = 0 about the ordinary point x = 0

my guess was y = sum c_nx^n

I think that I did my sum correctly but if someone can double check it for me, that would be great.

<a href="http://www.codecogs.com/eqnedit.php?latex=\sum_{n=2}^{\infty}n(n-1)c_nx^n @plus; 2 \sum_{ n=0}^{ \infty}(n@plus;2)(n@plus;1)c_{n@plus;2}x^{n}@plus; 3 \sum_{ n=1}^{\ \infty} nc_nx^n-\sum_{ n=0}^{ \infty} c_nx^n=0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\sum_{n=2}^{\infty}n(n-1)c_nx^n + 2 \sum_{ n=0}^{ \infty}(n+2)(n+1)c_{n+2}x^{n}+3 \sum_{ n=1}^{\ \infty} nc_nx^n-\sum_{ n=0}^{ \infty} c_nx^n=0" title="\sum_{n=2}^{\infty}n(n-1)c_nx^n + 2 \sum_{ n=0}^{ \infty}(n+2)(n+1)c_{n+2}x^{n}+3 \sum_{ n=1}^{\ \infty} nc_nx^n-\sum_{ n=0}^{ \infty} c_nx^n=0" /></a>
• April 24th 2013, 04:35 PM
topsquark
Re: Finding two linearly independent solutions (power series)
$\sum_{n=2}^{\infty}n(n-1)c_nx^n @plus; 2 \sum_{ n=0}^{ \infty}(n@plus;2)(n@plus;1)c_{n@plus;2}x^{n}@plus; 3 \sum_{ n=1}^{\ \infty} nc_nx^n-\sum_{ n=0}^{ \infty} c_nx^n=0" target="_blank">$

Wrap it in [tex] tags. Fix it and we can see exactly what you're up to.

-Dan
• April 24th 2013, 06:17 PM
LinearOperator
Re: Finding two linearly independent solutions (power series)
Ok here is my work that I have. The part that I am most unsure about is the part where I have to factor out terms. And also getting the solution(s).

So my guesses were $y=\sum_{n=0}^{\infty}c_nx^n$ and their derivatives which I won't go into doing. Then I got this:
$(x^2+2)\sum_{n=2}^{\infty}n(n-1))c_nx^{n-2}+3x \sum_{n=1}^{\infty} nc_nx^{n-1}-\sum_{n=0}^{\infty}c_nx^n$

I distributed and shifted them down so they could ann be at $x^n$:

$\sum_{n=2}^{\infty}n(n-1)c_nx^{n}+2 \sum_{n=0}^{\infty} (n+2)(n+1)c_{n+2}x^n+ 3\sum_{n=1}^{\infty}nc_nx^n-\sum_{n=0}^{\infty}c_nx^n$

I pulled out n=0 and n=1 from series #2 (starting series number one is on the far left and 4 is the end right) i am not sure if that is correct in doing so. and then i pulled out n = 1 from series 3 and n= 0 and 1 from series #4. Is this correct so far what I am doing?
• April 24th 2013, 06:21 PM
LinearOperator
Re: Finding two linearly independent solutions (power series)
Essenntially, my recusrsive set that I got was: $c_{n+2}=\frac{c_n(1-2n-n^2)}{2(n+2)(n+1)}$
which is where I am also unsure about.