The solution of the initial-value problem $x ^ { 2 } y ^ { tt } + x y ^ {t } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 0$ 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 given differential equation is a Bessel's equation of order 0, which has a solution in terms of Bessel functions of the first kind denoted as J₀(x). Since the initial conditions of the problem correspond to J₀(x), the solution must be of the form of the Bessel function of the first kind of order 0. 
The power series representation of J₀(x) is given by:
$$J_{0}(x)=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(k!)^{2}}$$
which matches option C. Thus, the correct answer is C.

The Solution of the Initial-Value Problem x2ytt+xyt+x2y=0,y(0)=1,yt(0)=0x ^ { 2 } y ^ { tt } + x y ^ {t } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 0x2ytt+xyt+x2y=0,y(0)=1,yt(0)=0

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