Define the Bessel function $J_{0}$ by $$J_{0}(x) = \frac{1}{\pi}\int_{-1}^{1}\frac{\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t.$$ Prove that $J_{0}'' + (\frac{1}{x})J_{0}' + J_{0} = 0$ for all $x > 0$.
The hint of my professor was to use the Leibniz Rule (permutation of partial derivatives), but I don't see how that can help. Can someone help me? I appreciate it!
Simply use Leibniz rule
$$J''_{0}(x)+ J_{0}(x)= \frac{1}{\pi}\frac {d^2}{dx^2}\int_{-1}^{1}\frac{\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t.+\frac{1}{\pi}\int_{-1}^{1}\frac{\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t$$ $$J''_{0}(x)+ J_{0}(x)= \frac{1}{\pi}\int_{-1}^{1}\partial^2_x(\frac{\cos xt}{\sqrt{1-t^{2}}})\mathrm{d}t.+\frac{1}{\pi}\int_{-1}^{1}\frac{\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t$$ $$J''_{0}(x)+ J_{0}(x)= \frac{1}{\pi}\int_{-1}^{1}\frac{-t^2\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t.+\frac{1}{\pi}\int_{-1}^{1}\frac{\cos xt}{\sqrt{1-t^{2}}}\mathrm{d}t$$ $$J''_{0}(x)+ J_{0}(x)= \frac{1}{\pi}\int_{-1}^{1}{\sqrt{1-t^{2}}}\cos(xt)\mathrm{d}t.$$ Integrate by part $$J''_{0}(x)+ J_{0}(x)= \frac{1}{\pi}\left |{\frac {\sqrt{1-t^{2}}}x}\sin(xt)\right |_{-1}^{1}+\frac{1}{x\pi}\int_{-1}^{1}\frac {t\sin(xt)} {{\sqrt{1-t^{2}}}}\mathrm{d}t.$$ $$J''_{0}(x)+ J_{0}(x)= \frac{1}{x\pi}\int_{-1}^{1}\frac {t\sin(xt)} {{\sqrt{1-t^{2}}}}\mathrm{d}t.$$ $$J''_{0}(x)+ J_{0}(x)=-\frac 1 xJ_{0}'(x) $$