Originally Posted by

**Ulysses** Hi there. Well, I'm stuck with this problem, which says:

When p=0 the Bessel equation is: $\displaystyle x^2y''+xy'+x^2y=0$

Show that its indicial equation only has one root and find the Frobenius solution correspondingly. (Answer: $\displaystyle y=\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }x^{2n} $

Well, this is what I did:

At first I normalized the equation:

$\displaystyle y''+\frac{y'}{x}+y=0$

Then

$\displaystyle P(x)=\frac{1}{ x} \rightarrow xP(x)=1$

$\displaystyle Q(x)=1 \rightarrow x^2Q(x)=x^2$

So x=0 is regular singular point.

Then the indicial equation is: $\displaystyle \alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0$

$\displaystyle y=\sum_{n = 0}^\infty a_n x^n \rightarrow y'=\sum_{n = 1}^\infty a_n n x^{n-1} \rightarrow y''=\sum_{n = 2}^\infty a_n n(n-1) x^{n-2}$

Then $\displaystyle x^2y''+xy'+x^2y=\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_n x^{n+2} = $

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

So from here I took

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