Let $y_p(x)$ be a nontrivial solution of Bessel's DE of order $p$. Let $x_1$ and $x_2$ be successive zeroes. By using the normal form of the DE and comparing this to the well-known solutions of $$y''+y'=0,$$ show that if $0 \le p< 1/2$, then $x_2 - x_1 < \pi$ and $$\lim_{x_1\to \infty}(x_2-x_1) = \pi.$$
So I know that the solution to $y''+y'=0$ is $$y=c_0\cos(x)+c_1\sin(x).$$ I can understand why $x_2-x_1$ would be less than $\pi$. But why does the order $p$ matter when it comes to the Bessel DE? Also, how would I prove the last expression?