For the equation $\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0$ , the point $x = 1$ is
A) an ordinary point
B) a regular singular point
C) an irregular singular point
D) a special point
E) none of the above

Question

Quizplus · Accepted Answer

The point $x = 1$ is a regular singular point because the equation can be written in the standard form $y'' + p(x)y' + q(x)y = 0$, where $p(x)$ and $q(x)$ may have singularities at $x = 1$. Specifically, the coefficient of $y''$ becomes zero at $x = 1$ and $x = 4$, indicating that these points could be singular. However, for a point to be considered a regular singular point, the limits $(x - 1)p(x)$ and $(x - 1)^2q(x)$ must exist and be finite as $x$ approaches the singular point. In this case, the structure of the given differential equation suggests that after simplification, these conditions are met for $x = 1$, making it a regular singular point.

For the Equation (x2−16)3(x−1)y′′−2xy′+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0(x2−16)3(x−1)y′′−2xy′+y=0

For the Equation $\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0$