= Solution
Let $Y=\widehat y$. Transform the three terms of the order-zero <Bessel differential equation> using part (a):
$$
\widehat{ty''}=i(-\omega^2Y)'=-i(2\omega Y+\omega^2Y'),\qquad\widehat{y'}=i\omega Y,\qquad\widehat{ty}=iY'.
$$
Adding and dividing by $i$ gives
$$
\boxed{(1-\omega^2)Y'-\omega Y=0.}
$$
On $-1<\omega<1$, <separation of variables> gives $Y'/Y=\omega/(1-\omega^2)$, hence $Y=C/\sqrt{1-\omega^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extension
$$
Y(\omega)=\frac{C\,\mathbf1_{(-1,1)}(\omega)}{\sqrt{1-\omega^2}}.
$$
This is a genuine <distributional weak solution>: the differential equation can be written $[(1-\omega^2)Y]'+\omega Y=0$, and $(1-\omega^2)Y=C\sqrt{1-\omega^2}\,\mathbf1_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint <Dirac delta> terms.
<Fourier inversion> now gives
$$
y(t)=\frac{C}{2\pi}\int_{-1}^1\frac{e^{i\omega t}}{\sqrt{1-\omega^2}}\,d\omega.
$$
This integral is smooth at the origin and solves the original equation. Its value there is $C/2$. Normalizing $y(0)=1$ fixes $C=2$, and identifies the <Bessel function of the first kind> of order zero:
$$
\boxed{J_0(t)=\frac1\pi\int_{-1}^1\frac{\cos(tu)}{\sqrt{1-u^2}}\,du=\frac1\pi\int_0^\pi\cos(t\cos\theta)\,d\theta.}
$$
The last equality uses $u=\cos\theta$. Without normalization, the regular solution is an arbitrary constant times this integral; this is the <Fourier transform of the Bessel function J0> pair.
\b[Regularity qualification.] The source specifies the differential equation but no condition at its singular point $t=0$. The transformed equation separately has solutions $C_\pm/\sqrt{\omega^2-1}$ on the two exterior intervals. Thus it does not, by itself, determine the support or a single global multiplicative constant. The compactly supported choice above constructs the regular Bessel branch. The other independent branch, the <Bessel function of the second kind>, is singular at the origin. Also $J_0$ is not absolutely integrable on the real line, so its transform is properly taken as a <tempered distribution>, rather than relying on the stronger classical integrability assumptions of part (a).
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