Solution of Bessel equation

Question

Prove that for a Bessel function in its normal form that is: $$u'' + \left(1 + \frac{1-(4*p^2)}{4x^2}\right)u=0$$ if $p > \frac12$ then every interval of length $\pi$ contains at most one zero of a non trivial solution of Bessel's equation. I was able to solve for $p < \frac12$ and p between $-\frac12$ and $\frac12$ by Sturm comparison theorem but this for $p > \frac12$ not able to solve.

Answer 1

Anyway if it's about $$u''+\left(1+\frac{1-4p^2}{x^2}\right)u=0$$ here's my solution:

Compare it with $$u''+u=0$$. Observe that $$u(x) = A \sin(x-a)$$ is a solution of $u''+u=0$, & has zeros at $a$ and $a+\pi$.

If $p>\frac{1}{2}$, $1-4p^2<0$, so $1+\frac{1-4p^2}{x^2}<1$ for $x\in[a,a+\pi)$.

Now, by the Sturm comparison theorem, a solution to the proposed ODE cannot have more than one zero in $[a,a+\pi)$, as $u(x) = A \sin(x-a)$ does not have a zero in $(a,a+\pi)$.