[Math] Solving Bessel’s equation by Laplace transform

Question

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.

Answer 1

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 $(2m-1)!!={(2m)!\over2^m m!}$