Question

Consider the following differential equation:

\[ 2x(x-2)^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + (x-2) y = 0 \]

Which of the following statements is/are TRUE?

• A. The point \(x = 2\) is not a singular point.
• B. The point \(x = 2\) is a singular point.
• C. The point \(x = 2\) is a regular singular point.
• D. The point \(x = 2\) is not a regular singular point.

Hint:

Write the equation as y'' + P(x)y' + Q(x)y = 0 and check if (x-2)P(x) stays finite at x=2.

Answer 1

The Correct Option is B, D

Instead of computing limits of $P(x)$ and $Q(x)$ directly, we can classify the point $x=2$ by counting the order of the zero in each coefficient of the equation $a(x)y'' + b(x)y' + c(x)y = 0$, where $a(x) = 2x(x-2)^2$, $b(x)=3x$, and $c(x)=x-2$.

1. Order of zero in $a(x)$ at $x=2$: $a(x)=2x(x-2)^2$ has the factor $(x-2)^2$, so $a(x)$ vanishes to order 2 at $x=2$. This confirms $x=2$ is a singular point, since the coefficient of the highest derivative vanishes there. So statement (A) is false and statement (B) is true.
2. Order of zero in $b(x)$ at $x=2$: $b(x)=3x$ gives $b(2)=6$, which is nonzero, so $b(x)$ has order 0 (no zero) at $x=2$.
3. Order of zero in $c(x)$ at $x=2$: $c(x)=x-2$ vanishes to order 1 at $x=2$.
4. Regular singular point test: A singular point is regular only if the order of zero of $a(x)$ exceeds that of $b(x)$ by at most 1, and exceeds that of $c(x)$ by at most 2. Here the order of $a(x)$ exceeds the order of $b(x)$ by $2-0=2$, which is more than the allowed 1, so this test fails, meaning $x=2$ is an irregular singular point.

Since the excess order condition on $b(x)$ is violated, $x=2$ cannot be a regular singular point, confirming statement (D) is true and statement (C) is false.

Let's summarize:

• $x=2$ makes the leading coefficient $a(x)$ vanish, so it is a singular point.
• The order of vanishing of $a(x)$ is too high compared to $b(x)$, so the singular point is irregular, not regular.

So the true statements are (B) and (D).

$$\boxed{(B) \text{ and } (D)}$$