For the differential equation $\left( x ^ { 2 } - 4 \right) ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0$ , the point $x = 0$ 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 differential equation is in the form $P(x)y'' + Q(x)y' + R(x)y = 0$. At $x = 0$, $P(x) = (x^2 - 4)^2$ is not zero, so $x = 0$ is an ordinary point.

For the Differential Equation (x2−4)2y′′−2xy′+y=0\left( x ^ { 2 } - 4 \right) ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0(x2−4)2y′′−2xy′+y=0

For the Differential Equation $\left( x ^ { 2 } - 4 \right) ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0$