 Solution

Let $Y=\hat{y}$. Transform the three terms of the order-zero Bessel differential equation using part (a):Adding and dividing by $i$ givesOn $-1<\omega <1$, separation of variables gives $Y^{\prime }/Y=\omega /(1-{\omega }^2)$, hence $Y=C/\sqrt{1-{\omega }^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extensionThis is a genuine distributional weak solution: the differential equation can be written $[(1-{\omega }^2)Y{]}^{\prime }+\omega Y=0$, and $(1-{\omega }^2)Y=C\sqrt{1-{\omega }^2}{1}_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint Dirac delta terms.

Fourier inversion now givesThis 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: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.

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).