# Series solution of an ODE

1. Mar 9, 2013

### Mangoes

1. The problem statement, all variables and given/known data

Solve for $$y' = x^2y$$

3. The attempt at a solution

We assume that the solution to the ODE will take the form

$$y = \sum_{n=0}{a_nx^n}$$

After finding y', plugging in the expressions into the ODE, and distributing x2,

$$\sum_{n=1}na_nx^{n-1} - \sum_{n=0}a_nx^{n+2} = 0$$

I want to eventually combine the summations, so first I make the exponents match.

$$\sum_{n=0}(n+1)a_{n+1}x^n - \sum_{n=2}a_{n-2}x^n = 0$$

I need to make the indices match without manipulating exponents, so I just peel off the first two terms (n = 0 and n = 1) from the first summation. After doing so and combining the summations:

$$\sum_{n=2}((n+1)(a_{n+1} - a_{n-2}))x^n + a_1 + 2a_3x = 0$$

Now here's where I start becoming uncertain...

The summation begins at n = 2, so the smallest power of x that can come out is x2. Since the RHS is 0,

$$a_1 = 0$$
$$2a_3x = 0$$

This means that the coefficients a_1 and a_3 must be equal to 0. Also, since the coefficients for all powers of x on the RHS are zero,

$$(n+1)a_{n+1} - a_{n-2} = 0$$

Rearranging gives me the recursion formula for n ≥ 2:

$$a_{n+1} = \frac{c_{n-2}}{n+1}$$

My end goal in here is to write the sum in a way that doesn't involve any past terms so that there's no recursion. I start plugging in numbers and hope that I see some pattern...

n = 2, $a_3 = \frac{a_0}{3}$

n = 3, $a_4 = \frac{a_1}{5} = 0$

n = 4, $a_5 = \frac{a_2}{6}$

n = 5, $a_6 = \frac{a_3}{7} = 0$

n = 6, $a_7 = \frac{a_4}{8} = \frac{a_1}{5*8}$

At this point it's clear that the odd n will give a zero and that I only have to worry about even n.

What bothers me is the term I get when n = 4. I'm not given any information as far as I know about a2 and since the ODE is first-order, there's only one arbitrary constant which I assume is a0 (if I don't assume it's arbitrary then I have no idea what to say about a0 either). So I'm stumped. How do I write a non-recursive formula for the sum with the mystery number a2? What am I missing in here?

Apologies for the lengthy post

EDIT: Disregard the thing I said about odd and even n. n = 7 doesn't equal 0.

Last edited: Mar 9, 2013
2. Mar 10, 2013

### clamtrox

I guess there's a typo there. You should have a1 = a2 = 0.

3. Mar 10, 2013

### vela

Staff Emeritus
Because of the error clamtrox pointed out, you had a3=0, and the n=2 equation then tells you a0=0. So a2 would be the arbitrary constant.

Since you're really supposed to have $a_1=a_2=0$, the non-vanishing coefficients are $a_{3k}$ (k=0, 1, 2, …), and a0 plays the role of the arbitrary constant.