# Series solutions to ODE

1. Mar 24, 2012

### physicsjock

Hey,

I've been trying to solve this ODE using the power series method,

y'' + x^2y = 0,

I end up with (the first sum can start from 0 or 2, i just left it as starting from n=0)

\begin{align} & \sum\limits_{n=0\,}^{\infty }{n(n-1){{a}_{n}}{{x}^{n-2}}+}\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n+2}}}=0 \\ & \sum\limits_{n=0\,}^{\infty }{n(n-1){{a}_{n}}{{x}^{n}}+}\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n+4}}}=0 \\ & \sum\limits_{n=0\,}^{\infty }{n(n-1){{a}_{n}}{{x}^{n}}+}\sum\limits_{n=4}^{\infty }{{{a}_{n-4}}{{x}^{n}}}=0 \\ & \sum\limits_{n=0\,}^{\infty }{n(n-1){{a}_{n}}{{x}^{n}}+}\sum\limits_{n=0}^{\infty }{{{a}_{n-4}}{{x}^{n}}}-{{a}_{-4}}{{x}^{0}}-{{a}_{-3}}x-{{a}_{-2}}{{x}^{2}}-{{a}_{-1}}{{x}^{3}}=0 \\ & By\,\,Thrm\,\,of\,\,vanishing\,\,coefficients: \\ & {{a}_{-4}}{{x}^{0}}={{a}_{-3}}x={{a}_{-2}}{{x}^{2}}={{a}_{-1}}{{x}^{3}}=0 \\ & -n(n-1){{a}_{n}}={{a}_{n-4}}\,\,\,\,\,n=4,5,6,... \\ \end{align}

but I'm having trouble getting an expression for a2n and a2n+1 since n starts at 4 there's no a2 to write a6 in terms of a0 like im used to doing,

\begin{align} & n(n-1){{a}_{n}}={{a}_{n-4}} \\ & {{a}_{4}}=\frac{{{a}_{0}}}{4\cdot 3},{{a}_{3}}=\frac{{{a}_{6}}}{6\cdot 5},.... \\ \end{align}

Is there something I'm missing?

2. Mar 24, 2012

### dikmikkel

y'' + x^2y = 0
So you use Frobenius method expanding around x0=0, because there are no singularities in the finite interval:
The guess is: $\sum\limits_{n=0}^{\infty}a_nx^{n+j},\hspace{10pt}a_0\neq 0$
So insert like you did: $\sum\limits_{n=0}^{\infty}a_n(n+j)(n+j-1)x^{n+j-2}+a_nx^{n+j+2}$
You would like to sum from the same indicies, try taking out the first 4 terms in the 1 term:
$a_0j(j-1)x^{j-2}+a_1(1+j)jx^{j-1}+a_2(2+j)(1+j)x^{j}+a_3(3+j)(2+j)x^{j+1}+\sum_{n=4}^{\infty}a_n(n+j)(n+j-1)x^{n+j-2}+\sum_{n=0}^{\infty}a_jx^{n+j+2}$
So switch the index in the first sequence:
$a_0j(j-1)x^{j-2}+a_1(1+j)jx^{j-1}+a_2(2+j)(1+j)x^{j}+a_3(3+j)(2+j)x^{j+1}+\sum_{n=0}^{\infty}a_{n+4}(n+4+j)(n+j+3)x^{n+j+2}+\sum_{n=0}^{\infty}a_jx^{n+j+2}$
Now you can use linear independence of polynomials like you did before and analyse the inicidial equation.

Last edited: Mar 24, 2012
3. Mar 24, 2012

### HallsofIvy

Staff Emeritus
No, you don't need the "n+ j" precisely because x= 0 is NOT a singular point.

4. Mar 24, 2012

### dikmikkel

Yeah, guess i did the robot here. Sorry. But isn't it right that fuchs theorem would lead to the conclusion that j=0 from the inicidial equation?

5. Mar 24, 2012

### physicsjock

Thanks for your replies guys,

I worked on it a little more and for my solutions I get:

\begin{align} & {{y}_{1}}(x)={{a}_{o}}-\frac{{{a}_{o}}}{3\cdot 4}{{x}^{4}}+\frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8}{{x}^{8}}-\frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8\cdot 11\cdot 12}{{x}^{12}}+... \\ & {{y}_{2}}(x)={{a}_{1}}x-\frac{{{a}_{1}}}{4\cdot 5}{{x}^{5}}+\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9}{{x}^{9}}-\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9\cdot 12\cdot 13}{{x}^{13}}+... \\ \end{align}

Do they look correct?
I've been trying to find a way to write each of them as sums rather then individual terms but I'm finding it a little hard to write an a2n and an a2n+1 which satisfies the two above solutions.

For example in y2 the factorials go like

3!/5!, 3!7!/5!9!, 3!7!11!/5!9!13!

Is there a way that can be written in terms of n?

Last edited: Mar 25, 2012