Differential Eq- Power Series Solution

[sami23]

Find a power series sol'n: (x²-1)y'' + 3xy' + xy = 0

Homework Equations

let y = \Sigma (from \infty to n=0) C_nxⁿ
let y' = \Sigma (from \infty to n=1) nC_nx^n-1
let y'' = \Sigma (from \infty to n=2) n(n-1)C_nx^n-2

The Attempt at a Solution

I wrote the differential eq as: x²y''-y''+3xy'+xy=0

Substituting back into the differential eq and multiplying the x² gives:
\Sigma (from \infty to n=2) n(n-1)C_nxⁿ - \Sigma (from \infty to n=2) n(n-1)C_nx^n-2 + 3\Sigma (from \infty to n=1) nC_nxⁿ + \Sigma (from \infty to n=0) C_nxⁿ⁺¹ = 0

For the first 2 terms I let k=n-2 , n=k+2 which would give:
\Sigma (from \infty to k=0) (k+2)(k+1)C_k+2x^k+2 - \Sigma (from \infty to k=0) (k+2)(k+1)C_k+2x^k

For the 3rd and 4th term I let k=n which would give:
3\Sigma (from \infty to k=1) kC_kx^k + \Sigma (from \infty to k=0) C_kx^k+1

The new series would be:
\Sigma (from \infty to k=0) (k+2)(k+1)C_k+2x^k+2 - \Sigma (from \infty to k=0) (k+2)(k+1)C_k+2x^k + 3\Sigma (from \infty to k=1) kC_kx^k + \Sigma (from \infty to k=0) C_kx^k+1

I don't know how to make the starting value of each index the same (from \infty to k=0) and how would I get to the general solution?

 

[Cyosis]

First off a tip. Write your summations like this: \sum_{n=a}^\infty. This will look like \sum_{n=a}^\infty, which is a lot neater. Secondly put the tex brackets around your entire equation, not just the summation symbols.

This will make your equation look like:
\sum_{n=0}^\infty c_n x^n

Let's start with a simple example. Take \sum_{n=1}^\infty c_{n-1}x^{n-1}. We want to write this series as \sum_{n=0}^\infty c_k x^k. Let's take a look at the first series.

<br /> \sum_{n=1}^\infty c_{n-1}x^{n-1}=c_0 x^0+c_1 x^1+c_2 x^2+...<br />

We obviously want \sum_{n=0}^\infty c_k x^k to be equal to this. So how would you express k in terms of n?

\sum_{n=0}^\infty c_k x^k=c_0 x^0+c_1 x^1+c_2 x^2+...?