Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I stumbled upon this question in an old exam (I'm preparing for an exam of a course about ODEs). I didn't have much difficulty solving the Legendre and Hermite equations using power series, but this one is different.

First of all, can you say this is a Bessel equation? Generally that would be $$x^2u''+xu'+(x^2-w^2)u = 0$$ which is quite similiar to my equation $$x^2u''+xu'-(x^2+\frac{1}{4})u=0$$ except for the minus-sign for the $x^2$. My usual approach is just expanding $u(x)$ as a power series, i.e. $$u(x) = \sum_{n=0}^{\infty} a_n x^n$$ and then substituting $u$, $u'$ and $u''$ into the original equation, in this case resulting in

$$\sum_{n=0}^\infty n(n-1)a_n x^n + \sum_{n=0}^\infty n \, a_n x^n - \frac{1}{4} \sum_{n=0}^\infty a_n x^n - \sum_{n=0}^\infty a_n x^{n+2}$$

It's also possible to start the first summation at $2$ and the second one at $1$.

The actual question is: Determine two independent solutions in the form of a power series about $x=0$. Provide the general expression for both power series and determine the functions they represent.

Any hints on how to continue?

[Edit]: My approach would be to combine the four summations into one, and try to rewrite all powers of $x$ to $x^n$. Therefore, the summation with $x^{n+2}$ should be rewritten to start at $2$, resulting in $x^n$ with coefficients $a_{n-2}$. But when this summation starts at $2$, the others should too. As I already mentioned, the first one can be chosen to start at $2$, but the second one starts at either $0$ or $1$ and the third summation starts at $0$.

share|cite|improve this question
Why did you stop there? – Qiaochu Yuan Jun 15 '12 at 14:30
@QiaochuYuan, my usual approach is to combine all summations into one -- leading to an expression for $a_n$, like $a_{n+2} = (\text{expression}) a_n$. But I don't see how to do that here. – Ailurus Jun 15 '12 at 14:36
What happens when you try to do that? – Qiaochu Yuan Jun 15 '12 at 14:39
@QiaochuYuan Just a sec, I'll update my question :) – Ailurus Jun 15 '12 at 14:47
Just write the extra terms in the summation separately, and start all of them at the same index. – Mariano Suárez-Alvarez Jun 15 '12 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.