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 initial-value problem given is a Bessel differential equation of order 0. The power series solution for this equation is of the form:

$$y(x) = \sum_{n=0}^\infty a_nx^{n+r}$$

where $a_n$ are constants and $r$ is the singular point. To find the power series solution, we first need to determine the indicial equation, which is obtained by substituting $y(x) = x^r\sum_{n=0}^\infty a_nx^n$ into the differential equation and equating the coefficients of like powers of $x$.

For the given differential equation, we have:

$$r(r-1) a_0 x^r + r(r+1)a_1x^{r+1} + \sum_{n=2}^\infty (n+r)(n+r-1)a_n x^{n+r} + \sum_{n=2}^\infty (n+r)a_n x^{n+r} + \sum_{n=2}^\infty a_n x^{n+r+1} = 0$$

Equating the coefficient of $x^r$, we get:

$$r(r-1)a_0 = 0 \implies r = 0 	ext{ or } r = 1$$

Equating the coefficient of $x^{r+1}$, we get:

$$2a_1 = 0 \implies a_1 = 0$$

Substituting $r=0$ in the original differential equation and equating coefficients, we get:

$$a_0 = 1 \quad 	ext{and} \quad a_n = \frac{(-1)^n}{2^{2n}(n!)^2} \quad 	ext{for } n\geq 1$$

Hence, the power series solution for the given initial-value problem is:

$$y(x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{2^{2n}(n!)^2}$$

which is exactly the option (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 $x ^ { 2 } y ^ { tt } + x y ^ { t } + x ^ { 2 } y = 0 , y ( 0 ) = 1 , y ^ { t } ( 0 ) = 0$