The solution of the initial-value problem $x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0$ is called a Bessel function of order 0. Solve the initial - value problem to find a power series expansion for the Bessel function.
A) $y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { n x ^ { 2 n - 1 } } { 2 ^ { 2 n - 1 } ( n ! ) ^ { 2 } }$
B) $y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { n } } { 2 ^ { n } ( 2 n ! ) }$
C) $y ( x ) = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }$
D) $y ( x ) = 2 ^ { x } + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { ( n ! ) ^ { 2 } }$
E) $y ( x ) = x + \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 2 n } } { 2 ^ { 2 n } ( n ! ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

The Bessel function of order 0 is defined as $J_0(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}\left(\frac{x}{2}\right)^{2n}$. 
To derive this, we can assume that $y(x)$ can be expressed as the power series $\sum_{n=0}^{\infty}a_nx^n$, where $a_0=y(0)$ and $a_1=y'(0)$. Then, we can substitute this into the differential equation and solve for the coefficients $a_n$ using the recursive formula 
$$a_n = \frac{(-1)^n}{2^{2n}(n!)^2}a_0.$$
Plugging in the initial conditions, we get $a_0=1$ and $a_1=0$, and so the coefficient formula simplifies to $a_n= \frac{(-1)^n}{2^{2n}(n!)^2}$. Thus, we have $y(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n}(n!)^2}x^{2n}$, which matches choice (C).

The Solution of the Initial-Value Problem x2y′′+xy′+x2y=0,y(0)=1,y′(0)=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0x2y′′+xy′+x2y=0,y(0)=1,y′(0)=0

The Solution of the Initial-Value Problem $x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0$