I've never seen a complete derivation of Bessel's function. I put this one together and had a question.
$$0 = t^2 y''(t) + t y'(t) + (t^2 + \alpha) y(t) $$
Let $\alpha = 0$ and $t \not= 0$
$$0 = t y''(t) + y'(t) + t y(t)$$
Laplace Transform
$$ L[ty''(t)] = (-1)^1 \frac{d}{ds}L[y''(t)] = -\frac{d}{ds}(-y'_0 -s y_0 + s^2Ly) = y_0 -2s Ly - s^2 \frac{d}{ds}Ly $$
$$L[y'(t)]= -y_0 +sLy$$
$$L[t y(t)]= -\frac{d}{ds}Ly$$
Simplify $$(s^2+1)\frac{d}{ds}Ly + sLy =0 $$
$$ \frac{dLy}{Ly} = -\frac{sds}{s^2+1}$$
Finessing this next expression was the key. I have a question about it.
Solve (Let $z=\frac{1}{s^2}$) $$\tag{*} Ly = \frac{1}{\sqrt{s^2+1}} = \frac{1}{s}\frac{1}{\sqrt{1+\frac{1}{s^2}}} = \frac{1}{s}\frac{1}{\sqrt{1+z}}$$
Taylor's Theorem $$ Ly = \frac{1}{s} [\frac{1}{0!}z^0 +\frac{\frac{-1}{2}}{1!}z^2 + \frac{\frac{1}{2}\frac{3}{2}}{2!}s^4+\frac{\frac{-1}{2}\frac{3}{2}\frac{5}{2}}{3!}z^6+...]$$ $$ = \frac{1}{s} [\frac{1}{0!}z^0 +\frac{-1}{1!}\frac{1}{2}z^1 + \frac{1}{2!}\frac{1}{2}\frac{3}{2}z^2+\frac{1}{3!}\frac{-1}{2}\frac{3}{2}\frac{5}{2}z^3+...] $$ $$\tag{**} = \frac{1}{s}\sum_{n=0}^\infty \frac{(-1)^n}{n!}\frac{(2n)!}{n!2^{2n}} z^{n} $$
Substitute and rearrange $$= \sum_{n=0}^\infty \frac{(-1)^n}{n!}\frac{1}{n!2^{2n}} \frac{(2n)!}{s^{2n+1}}$$
Inverse Laplace Transform $$ y = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\frac{1}{n!2^{2n}} L^{-1}[\frac{(2n)!}{s^{2n+1}}] = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\frac{1}{n!2^{2n}} t^{2n} $$
Which is Bessel's function of 0th order. The elephant in the room is the fudge in step (*). If I didn't do that, then I would have ended up at step (**) with $s^{2n}$ instead of $z^n$. I wouldn't have been able to apply the inverse laplace transform to that expression to obtain the correct formula. How is it that the two expressions are equal?