I'm trying to solve this question:
We can prove that $u$ is a solution of Bessel's Equation $z^2 u'' + zu + (z^2 -\nu^2) = 0 $ if and only if $w(z) = z^{\frac{1}{2}}u(z)$ is a solution of $$ z^2 w'' + (z^2 + \frac{1}{4} - \nu^2)w=0. $$ Using this result, express $J_{\pm \frac{1}{2}}(z)$ and $Y_{\pm \frac{1}{2}}(z)$, the Bessel function of first kind and second kind respectively, in terms of $\sin(z)$ and $\cos(z)$.
This is not an assignment question, it is a question that we had in class a few weeks ago. While catching up I've tried to do this but I haven't been able to finish the question. This is what I've tried:
Attempt. If $\nu = \pm \frac{1}{2}$ then the ODE with $w$ simply becomes $$ z^2 ( w'' + w) = 0. $$ Since we are looking for a solution $\forall z$, we require $w'' + w = 0 $ which leads to a general solution $$ w(z) = A\cos(z) + B\sin(z).$$
Now if we know that $ w(z) = z^{\frac{1}{2}} u(z)$ then the general solution to Bessel's equation for $\nu =\pm \frac{1}{2}$ $$C J_{\pm 1/2}(z) + DY_{\pm 1/2} = \frac{1}{\sqrt{z}}(A\cos(z) + B\sin(z)). $$
And at this point I'm stuck how to move forward. The answers provided are:
- $J_{1/2}(z) = \sqrt{\frac{2}{\pi z}} \sin(z)$
- $J_{-1/2}(z) = \sqrt{\frac{2}{\pi z}} \cos(z) $
- $Y_{1/2}(z) = -\sqrt{\frac{2}{\pi z}} \cos(z)$
- $Y_{-1/2}(z) = \sqrt{\frac{2}{\pi z}} \sin(z) $
But I have no idea how to continue with the question from here. I've tried subbing in values for $z$ but with not much success. Can someone please show how to finish the question from here?