so this is a question I have been given and I have no idea where to even start, my bessel functions is not all that great but if someone could just help me get a handle on this thing, i.e. just help me start it, point me in the right direction...that'd be great!
Hint
Define $$y=u(x) \sqrt x $$ $$y'=\sqrt{x} u'(x)+\frac{u(x)}{2 \sqrt{x}}$$ $$y''=\sqrt{x} u''(x)+\frac{u'(x)}{\sqrt{x}}-\frac{u(x)}{4 x^{3/2}}$$ So $$y''+k^2x^2y=\frac{\left(4 k^2 x^4-1\right) u(x)+4 x \left(x u''(x)+u'(x)\right)}{4 x^{3/2}}$$ and, since the expression is equal to zero, we then need to consider $$4 x^2 u''(x)+4 x u'(x)+\left(4 k^2 x^4-1\right) u(x)=0$$ Now remember that Bessel functions are solutions of $$x^2z''(x)+x z'(x)+(x^2-\alpha^2)z(x)=0$$
I am sure that you can take from here.