= Solution
Differentiating the integral representation gives
$$
J_0'(t)=-\frac2\pi\int_0^{\pi/2}\cos x\sin(t\cos x)\,dx,
\qquad
J_0''(t)+J_0(t)=\frac2\pi\int_0^{\pi/2}\sin^2x\cos(t\cos x)\,dx.
$$
Now integrate the derivative
$$
\frac d{dx}\big(\sin x\sin(t\cos x)\big)
=\cos x\sin(t\cos x)-t\sin^2x\cos(t\cos x).
$$
The boundary terms vanish at both endpoints. Consequently $J_0'(t)=-t(J_0''(t)+J_0(t))$, proving
$$
\boxed{tJ_0''(t)+J_0'(t)+tJ_0(t)=0.}
$$
The identity includes $t=0$, where it reads $J_0'(0)=0$.
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