Solving Bessel's equation by Laplace transform I am learning Bessel function the solution of Bessel equation by book 'Advanced Engineering Mathematics' by Peter V.O'Neil and here i found its derivation by Laplace transform. In this derivation of order $n$ the substitution $y(t)=t^{-n}w(t)$ is  taken and then by Laplace transform $$W(s) = (1+s^2)^{-\frac{2n+1}{2}}$$ then by binomial expansion and then    term by term inverting we have Bessel function of order n by  substituting  $y(t)=t^{-n}w(t)$. But the standard generalized part marked by me as 2 is not equal to solution obtained marked as 1. It looks nothing wrong in derivation but still i can not point out why this is. Please  help me about this. 

 A: The geneal term in (1) is $ (-1)^ka_kt^{n+2k} $ with
$$
a_k=\frac{\frac{1}{2n-1}\prod_{m=0}^k(2n+2m-1)}{2^kk!\,\big(2(n+k)\big)!}=\frac{(2n+1)(2n+3)\cdots (2n+2k-1)}{2^kk!\,\big(2(n+k)\big)!}=\frac{\big(2(n+k)-1\big)!!}{2^kk!\,\big(2(n+k)\big)!}
$$
and the geneal term in (2) is $ (-1)^kb_kt^{n+2k} $ with
$$
b_k=\frac{1}{2^{2k+n}k!\,(n+k)!}
$$
We have
$$
a_k=\frac{\big(2(n+k)-1\big)!!}{2^kk!\,\big(2(n+k)\big)!}=\frac{(2(n+k))!}{2^{n+k}(n+k)!}\cdot\frac{1}{2^kk!\,\big(2(n+k)\big)!}=\frac{1}{2^{2k+n}k!\,(n+k)!}=b_k
$$
using the fact that $(2m-1)!!={(2m)!\over2^m m!}$
A: Their $W(s)$ is supposed to be the Laplace transform of $t^n J_n(t)$. The solution to the ODE on $W$ is $W(s) = C (1 + s^2)^{-n - 1/2}$, and they set $C = 1$.  However, the correct normalization constant is $(2n - 1)!!$ (related to the fact that $\int_0^\infty t^n J_n(t) \, dt = (2 n - 1)!!$ for negative integers). For $C = 1$, the inverse transform is
$$\mathcal L^{-1}[s \mapsto (1 + s^2)^{-n - 1/2}](t) =
\frac {t^n J_n(t)} {(2 n - 1)!!}.$$
(1) is the correct expansion for $J_n(t)/(2 n - 1)!!$ and is certainly not equal to (2).
