The Bessel Function of the First Kind $J_0(x)$ is given as the unique function that solves Bessel's Differential Equation
$$x^2y'' + xy' + x^2y = 0$$
subject to $y(0) = 1$.
It also has the following integral representation:
$$J_0(x) = \frac{1}{2\pi}\int_0^{2\pi}e^{ix\cos t}\text{dt}$$
I have been trying to show that $J_0(x)$, defined using the integral representation, solves Bessel's Differential Equation, but I haven't had any luck so far.
I've tried to plug the integral representation in for $y(x)$, which gave me the following expression $I(x)$, which should be identically zero everywhere.
$$I(x) = \frac{x^2}{2\pi}\int_0^{2\pi}\frac{\partial^2}{\partial x^2}e^{ix\cos t}\text{dt} + \frac{x}{2\pi}\int_0^{2\pi}\frac{\partial}{\partial x}e^{ix\cos t}\text{dt} + \frac{x^2}{2\pi}\int_0^{2\pi}e^{ix\cos t}\text{dt}$$
Multiplying across by $2\pi$ and calculating the derivatives, we obtain:
\begin{align*} 2\pi I(x)&= -x^2\int_0^{2\pi}\cos^2(t)e^{ix\cos t}\text{dt} + ix\int_0^{2\pi}\cos(t)e^{ix\cos t}\text{dt} + x^2\int_0^{2\pi}e^{ix\cos t}\text{dt} \\ &= x^2\int_0^{2\pi}\sin^2(t)e^{ix\cos t}\text{dt} + ix\int_0^{2\pi}\cos(t)e^{ix\cos t}\text{dt} \\ &= -\frac{x^2}{4}\int_{0}^{2\pi}(e^{it}-e^{-it})^2 e^{ix\cos t}\text{dt} + \frac{ix}{2}\int_0^{2\pi}(e^{it}+e^{-it})e^{ix\cos t}\text{dt}\\ &= -\frac{x^2}{4}\int_0^{2\pi}e^{i(2t + x\cos t)} + e^{i(-2t + x\cos t)} - 2e^{ix\cos t}\text{dt} + \frac{ix}{2}\int_0^{2\pi}e^{i(t+x\cos t)} + e^{i(-t+x\cos t)}\text{dt}\\ &=x^2\left[\pi J_0(x) -\frac{1}{4}\int_0^{2\pi}e^{i(2t + x\cos t)} + e^{i(-2t + x\cos t)}\text{dt}\right]+\frac{ix}{2}\int_0^{2\pi}e^{i(t+x\cos t)} + e^{i(-t+x\cos t)}\text{dt} \end{align*}
Now for any integer $n$, the following holds:
$$J_n(x) = \frac{1}{2\pi}\int_0^{2\pi}e^{i(-nt+ix\cos t)}\text{dt}$$
Thus we have: $$2\pi I(x)=\pi x^2\left[J_0(x) - \frac{1}{2}J_{-2}(x) - \frac{1}{2}J_2(x)\right] + i\pi x\left[J_{-1}(x) + J_1(x)\right]$$
Because $J_{-2}(x) = J_2(x)$ and $J_{-1}(x) = -J_1(x)$, this simplifies to:
$$2\pi I(x)=\pi x^2\left[J_0(x) - J_2(x)\right]$$
Rearranging, we obtain:
$$I(x) = \frac{x^2}{2}\left[J_0(x) - J_2(x)\right]$$
But this expression is not identically zero - so I must have made a mistake somewhere, and I don't have the faintest clue where. Can anybody point out my mistake to me?