I'm trying to solve a problem from the textbook mathematics for physicist by Susan Lea and I have a few questions about it.
First of all, I have to find 2 solutions as power of $x$ for the Bessel series $$x^2 \frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 -\frac{9}{4})y = 0$$ and then I have to verify that my solutions agree with the standard form $\sqrt{\frac{2}{\pi x}} (\frac{\sin x}{x} - \cos x) $ and $-\sqrt{\frac{2}{\pi x}} (\frac{\cos x}{x} + \sin x)$ which I'm not sure what that mean.
Here my attempt
Standard form $$ \frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} + (x^2 -\frac{9}{4})\frac{y}{x^2} = 0$$
Plugging $$(x-x_0)^p \sum_{n=0}^{\infty}a_n(x-x_0)^n,$$ where $x_0 = 0$ in the first equation I get $$\sum[(n+p)(n+p-1)a_n x^{n+p}] \\ + \sum[(n+p)a_n x^{n+p}] + \sum[a_n x^{n+p+2}] - 9/4 \sum[a_n x^{n+p}] = 0$$
The smallest power is $x^p$ for $n = 0$, thus I find $p = \pm \frac{3}{2}$
For $x^{m+p}$ I have $$(m+p)(m+p-1)a_m + (m+p)a_m + a_{m-2} - 9/4 a_m = 0$$ $$a_{m-2} = [(m+p(m+p-1) + (m+p) -9/4)] a_m$$
Plugging $p = \frac{3}{2}$ I have my recurrence relation $$a_{m-2} = -m(m+3)a_m$$
From there arise some questions.
From the recurrence relation I'm trying to find a series and those series should be a series of sine or cosine to agree with the standard form.
Thus, $$a_m = -\frac{a_{m-2}}{m(m+3)} = - \frac{1}{m(m+3)} \cdot - \frac{a_{m-4}}{(m-2)(m+1)}$$
For $m$ even $$a_m = \frac{-1^{m/2}}{m!!(m+3)!!}a_0$$ $m = 2k$ $$a_{2k} = \frac{-1^{k}}{k!!(k+3)!!}a_0$$
As far as I can tell, I can't see a cosine or sine series yet. I'm wondering if I have to find a series for $m$ odd then find a way to put it together with $m$ even, or can I find a solution with $m$ even and one with $m$ odd even if I have 2 $p$ where I have to find a solution for each one.
I hope my question is clear even if it's not perfectly clear in my head.