The solution of the eigenvalue problem $r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 )$ is bounded, $R ^ { \prime } ( 3 ) = 0$ is $\left( J _ { 0 } ^ { t } \left( z _ { n } \right) = 0 \right)$
A) $\lambda = z _ { n } ^ { 2 } , R = J _ { 0 } \left( z _ { n } r \right) , n = 1,2,3 , \ldots$
B) $\lambda = z _ { n } ^ { 2 } / 9 , R = J _ { 0 } \left( z _ { n } r / 3 \right) , n = 1,2,3 , \ldots$
C) $\lambda = z _ { n } , R = J _ { 0 } \left( z _ { n } r \right) , n = 1,2,3 , \ldots$
D) $\lambda = z _ { n } / 3 , R = J _ { 0 } \left( z _ { n } r / 3 \right) , n = 1,2,3 , \ldots$
E) $\lambda = z _ { n } , R = J _ { 0 } ( r ) , n = 1,2,3 , \ldots$

Question

Quizplus · Accepted Answer

The given differential equation is a Bessel's equation, and its solutions are Bessel functions of the first kind, $J_0$. The condition $R'(3) = 0$ implies that the derivative of the solution with respect to $r$ is zero at $r = 3$, which corresponds to the first zero of the derivative of the Bessel function, scaled by a factor of 3. Therefore, the eigenvalues $\lambda$ are given by the squares of the zeros of the Bessel function divided by 9, and the corresponding eigenfunctions are $J_0$ of the scaled argument $z_n r / 3$.

The Solution of the Eigenvalue Problem rR′′+R′+rλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 )rR′′+R′+rλR=0,R(0)

The Solution of the Eigenvalue Problem $r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 )$