# Math Help - Power series question

1. ## Power series question

Determine the recurrence relation for the coefficients of the power series solutions, and find explicitly the …first four non-zero terms in each of the two linear independent series solutions to: $y^{''} +xy^{'} +2y = 0$ about Xo=0

I found the recurrence relation $(n+1)(n+2)An+2 + (n+2)An) = 0$

I dont understand what it means "find explicitly" the first four non-zero terms.

2. Originally Posted by Oblivionwarrior
Determine the recurrence relation for the coefficients of the power series solutions, and find explicitly the …first four non-zero terms in each of the two linear independent series solutions to: $y^{''} +xy^{'} +2y = 0$ about Xo=0
Let $y = \sum_{n=0}^{\infty}a_n x^n$ then it means,
$\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} + 2\sum_{n=0}^{\infty} a_n x_n = 0$
This becomes,
$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=1}^{\infty} na_nx^n + 2\sum_{n=0}^{\infty}a_nx^n = 0$
Change index,
$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n + \sum_{n=1}^{\infty}na_nx^n + 2\sum_{n=0}^{\infty}a_nx^n = 0$
Evaluate the first and third sum at $n=0$ and combine,
$2a_2 + 2a_0 + \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2} + (n+2)a_n]x^n = 0$
But this is the same as writing,
$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n+2)a_n]x^n = 0$

The recurrence relation is therefore, $(n+2)(n+1)a_{n+2} + (n+2)a_n = 0 \implies a_{n+2} = -\frac{a_n}{n+1} \text{ for }n\geq 0$

3. I have got that part, but I am not sure how they are getting these answers from the recurrence formula. These are the answers.

$Y1 = 1-x^{2} +\frac{1}{3}x^4 -\frac{1}{15}x^6+...$
$Y2 = x -\frac{1}{2}x^3 +\frac{1}{8}x^5 - \frac{1}{48}x^7$

4. I'll do the even ones, $y_1(x)$. Try and do the odd ones:

$a_2=-\frac{a_0}{1}$

$a_4=-\frac{a_2}{3}$

$a_6=-\frac{a_4}{5}$

$\vdots$

$a_{2k}=-\frac{a_{2k-1}}{2k-1}$

Now multiply all corresponding equations on each side:

$a_2 a_4 a_6\cdots a_{2k}=\frac{(-1)^k}{1\cdot 3\cdot 5\cdot\cdots (2k-1)} a_0 a_2 a_4\cdots a_{2k-2}$

Simplifying:

$a_{2k}=\frac{(-1)^k}{1\cdot 3\cdot 5\cdots (2k-1)} a_0$

Let $a_0=1$ and we can then apply the arbitrary constant to $y_1(x)$. We then have:

$y_1(x)=1+\sum_{k=1}^{\infty}\frac{(-1)^k}{1\cdot 3\cdot 5\cdots (2k-1)} x^{2k}$