Finding Bessel function $J_{1/2}$ using power series method. The Bessel function $J_{1/2}$ is a solution of the differential equation:
$$x^2 y '' + xy' +(x^2 - (1/2)^2)y = 0 $$
I am looking to find $J_{1/2}$ using a power series method, but letting $y = \sum_{n=0}^\infty a_n x^n$ gives $a_0 = a_1 = 0$ and $a_n = \frac{a_{n-2}}{n^2 - 1/4}$ which gives all coefficients equal to $0$, which is not desirable.
How do I find the correct power series expansion of $J_{1/2}?$

If you instead want to use the differential equation, let
$$
u(x)=\sqrt{x}y(x).
$$
If I did the calculations correctly (you confirm), the differential equation transforms to
$$
x^{3/2}(u''(x)+u(x))=0.
$$
Please try to take it from here.
