Show Bessel integral function is solution of $x^2y''+xy'+(x^2-n^2)y=0$. 
Show Bessel integral function $$J_n(t)=\frac{1}{2\pi} \int_{0}^{2\pi} e^{i t \sin \theta} e^{-in \theta}  d \theta$$ is a solution of $$x^2y''+xy'+(x^2-n^2)y=0.$$

The suggest is  that $n^2J_n(x)=-\frac{1}{2\pi} \int_{0}^{2\pi} e^{i x \sin \theta} \frac{d^2}{d \theta^2}(e^{-i n \theta})  d \theta$
I want prove $y(x)=J_n(x)$ is solution of $x^2y''+xy'+(x^2-n^2)y=0$
$J_n(x)=\frac{1}{2\pi} \int_{0}^{2\pi} e^{i (x \sin \theta -n \theta)}  d \theta$
$e^{i (x \sin \theta -n \theta)}=\cos(x \sin \theta -n \theta)+i \sin(x \sin \theta -n \theta)$
$\sin(x \sin \theta -n \theta)$ is odd on $[0,2\pi]$ and $\cos(x \sin \theta -n \theta)$ is even on $[0,2\pi]$ so  
$J_n(x)=\frac{1}{\pi} \int_{0}^{\pi} \cos{(x \sin \theta -n \theta)}  d \theta$
I don't know how show the rest of the proof using the observation in the second line
 A: This integral representation of the Bessel functions holds only when $n\in\mathbb{Z}$, as kindly remarked by Maxim.  Therefore, $n$ is assumed to be an integer throughout.  
Recall that
$$J_n(x)=\frac1{2\pi}\,\int_0^{2\pi} \,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta\,.$$
So
$$J'_n(x)=\frac1{2\pi}\,\int_0^{2\pi}\, \frac{\partial}{\partial x}\,\left(\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\right)\,\text{d}\theta=\frac1{2\pi}\,\int_0^{2\pi}\, \text{i}\sin(\theta)\, \text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta$$
and
$$J''_n(x)=\frac1{2\pi}\,\int_0^{2\pi}\, \frac{\partial}{\partial x}\,\left(\text{i}\sin(\theta)\, \text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\right)\,\text{d}\theta=\frac{1}{2\pi}\,\int_0^{2\pi}\big(\text{i}\sin(\theta)\big)^2\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta\,.$$
Because $\sin^2(\theta)+\cos^2(\theta)=1$ and $\big(\text{i}\sin(\theta)\big)^2=-\sin^2(\theta)$, we get
$$J''_n(x)=-J_n(x)+\frac{1}{2\pi}\,\int_0^{2\pi}\,\cos^2(\theta)\, \text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta.$$
Therefore,
$$\begin{align}x^2\,J''_n(x)+x^2\,J_n(x)&=-\frac{1}{2\pi}\,\int_0^{2\pi}\,\big(\text{i}x\cos(\theta)\big)^2\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta\\
&=-\frac{1}{2\pi}\,\int_0^{2\pi}\,\Big(\big(\text{i}x\cos(\theta)\big)\,\text{e}^{-\text{i}n\theta}\Big)\,\left(\big(\text{i}x\cos(\theta)\big)\text{e}^{\text{i}x\sin(\theta)}\right)\,\text{d}\theta\,.\end{align}$$
Using integration by parts twice, we have
$$\begin{align}x^2\,J''_n(x)+x^2\,J_n(x)&=\frac{1}{2\pi}\,\int_0^{2\pi}\,\big(-\text{i}n-\tan(\theta)\big)\,\big(\text{i}x\cos(\theta)\big)\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta
\\&=\frac{1}{2\pi}\,\int_0^{2\pi}\,\Big(\big(-\text{i}n-\tan(\theta)\big)\,\text{e}^{-\text{i}n\theta}\Big)\,\left(\big(\text{i}x\cos(\theta)\big)\,\text{e}^{\text{i}x\sin(\theta)}\right)\,\text{d}\theta
\\&=\frac{1}{2\pi}\,\int_0^{2\pi}\,\left(n^2-\text{i}n\tan(\theta)+\sec^2(\theta)\right)\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta
\\&=n^2\,J_n(x)-\frac{1}{2\pi}\,\int_0^{2\pi}\,\big(\text{i}n\tan(\theta)-\sec^2(\theta)\big)\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta\,.
\end{align}$$
Also with integration by parts,
$$\begin{align}J'_n(x)&=\frac{1}{2\pi\,x}\,\int_0^{2\pi}\,\Big(\tan(\theta)\,\text{e}^{-\text{i}n\theta}\Big)\,\Big(\big(\text{i}x\cos(\theta)\big)\,\text{e}^{\text{i}x\sin(\theta)}\Big)\,\text{d}\theta\\
&=\frac{1}{2\pi\, x}\int_0^{2\pi}\, \left(\text{i}n\tan(\theta)-\sec^2(\theta)\right)\,\text{e}^{\text{i}x\sin(\theta)}\,\text{e}^{-\text{i}n\theta}\,\text{d}\theta\,.\end{align}$$
This shows that $$x^2\,J_n''(x)+x^2\,J_n(x)=n^2\,J_n(x)-x\,J'_n(x)\,,$$
and so
$$x^2\,J_n''(x)+x\,J'_n(x)+(x^2-n^2)\,J_n(x)=0$$
as required.
