Series Solution of an ODE: Finding a Non-Recursive Formula

[Mangoes]

Homework Statement

Solve for y' = x^2y

The Attempt at a Solution

There's something that's been really bothering me about this question and similar ones.

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.

 

[clamtrox]

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,

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

 

[vela]

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.