GATE PH 2022 • Q27
MSQ • 1 Mark
Section 1: Mathematical Physics
Differential equations
Subtopic: First and Second Order ODEs
Question Statement:
The ordinary differential equation $(1 - x^2) y'' - x y' + 9y = 0$ has a regular singularity at
Options:
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(A) $-1$
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(B) $0$
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(C) $+1$
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(D) no finite value of $x$
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.27)**
**1. Singular Points of Ordinary Differential Equations**:
- Standard form of ODE:
$y'' + P(x) y' + Q(x) y = 0$
- Given ODE: $(1 - x^2)y'' - xy' + 9y = 0$.
- Divide by $(1 - x^2)$:
$y'' - \frac{x}{1 - x^2} y' + \frac{9}{1 - x^2} y = 0$
$P(x) = \frac{-x}{1 - x^2} = \frac{x}{(x-1)(x+1)}, \quad Q(x) = \frac{9}{1 - x^2} = \frac{-9}{(x-1)(x+1)}$
**2. Step-by-Step Singularity Analysis**:
- Singular points occur where $1 - x^2 = 0 \implies x = +1$ and $x = -1$.
- **At $x = 1$**:
$\lim_{x \to 1} (x - 1) P(x) = \lim_{x \to 1} \frac{x}{x + 1} = \frac{1}{2} < \infty$
$\lim_{x \to 1} (x - 1)^2 Q(x) = \lim_{x \to 1} \frac{-9(x - 1)}{x + 1} = 0 < \infty$
Since both limits are finite and analytic, **$x = +1$ is a regular singular point**.
- **At $x = -1$**:
$\lim_{x \to -1} (x + 1) P(x) = \lim_{x \to -1} \frac{x}{x - 1} = \frac{-1}{-2} = \frac{1}{2} < \infty$
$\lim_{x \to -1} (x + 1)^2 Q(x) = \lim_{x \to -1} \frac{-9(x + 1)}{x - 1} = 0 < \infty$
Both limits are finite and analytic, so **$x = -1$ is a regular singular point**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, C)** (`-1` and `+1`).
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