For the equation $\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0$ , the point $x = 4$ 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 Answer is C

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