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?
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?
Show 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.
Updated On: Jul 21, 2026
- The point \(x = 2\) is not a singular point.
- The point \(x = 2\) is a singular point.
- The point \(x = 2\) is a regular singular point.
- The point \(x = 2\) is not a regular singular point.
Show Solution
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The Correct Option is B, D
Solution and Explanation
Step 1: Write the equation in standard form.
The equation is \(2x(x-2)^2 y'' + 3x y' + (x-2) y = 0\).
Divide every term by the coefficient of \(y''\), which is \(2x(x-2)^2\), to get the standard form \(y'' + P(x) y' + Q(x) y = 0\).
\[ P(x) = \frac{3x}{2x(x-2)^2} = \frac{3}{2(x-2)^2} \]
\[ Q(x) = \frac{x-2}{2x(x-2)^2} = \frac{1}{2x(x-2)} \]
Step 2: Check whether \(x=2\) is a singular point.
A point \(x_0\) is an ordinary point only if \(P(x)\) and \(Q(x)\) are both analytic (finite and well behaved) there.
Here \(P(x)\) has \((x-2)^2\) in the denominator and \(Q(x)\) has \((x-2)\) in the denominator, so both blow up at \(x=2\).
This means \(x=2\) is not an ordinary point, so it is a singular point. So option (A) is false and option (B) is true.
Step 3: Check whether it is a regular singular point.
A singular point \(x_0\) is a regular singular point only if \((x-x_0)P(x)\) and \((x-x_0)^2 Q(x)\) are both analytic (finite) at \(x_0\).
Compute \((x-2)P(x)\):
\[ (x-2) P(x) = (x-2) \cdot \frac{3}{2(x-2)^2} = \frac{3}{2(x-2)} \]
As \(x \to 2\), this expression still blows up (goes to infinity), so it is not analytic at \(x=2\).
Step 4: Conclude the type of singular point.
Since \((x-2)P(x)\) fails to be analytic at \(x=2\), the point \(x=2\) does not satisfy the condition for a regular singular point.
So \(x=2\) is an irregular singular point, meaning option (C) is false and option (D) is true.
Final Answer:
The point \(x=2\) is a singular point but not a regular one, so statements (B) and (D) are true. \[ \boxed{\text{(B) and (D)}} \]
The equation is \(2x(x-2)^2 y'' + 3x y' + (x-2) y = 0\).
Divide every term by the coefficient of \(y''\), which is \(2x(x-2)^2\), to get the standard form \(y'' + P(x) y' + Q(x) y = 0\).
\[ P(x) = \frac{3x}{2x(x-2)^2} = \frac{3}{2(x-2)^2} \]
\[ Q(x) = \frac{x-2}{2x(x-2)^2} = \frac{1}{2x(x-2)} \]
Step 2: Check whether \(x=2\) is a singular point.
A point \(x_0\) is an ordinary point only if \(P(x)\) and \(Q(x)\) are both analytic (finite and well behaved) there.
Here \(P(x)\) has \((x-2)^2\) in the denominator and \(Q(x)\) has \((x-2)\) in the denominator, so both blow up at \(x=2\).
This means \(x=2\) is not an ordinary point, so it is a singular point. So option (A) is false and option (B) is true.
Step 3: Check whether it is a regular singular point.
A singular point \(x_0\) is a regular singular point only if \((x-x_0)P(x)\) and \((x-x_0)^2 Q(x)\) are both analytic (finite) at \(x_0\).
Compute \((x-2)P(x)\):
\[ (x-2) P(x) = (x-2) \cdot \frac{3}{2(x-2)^2} = \frac{3}{2(x-2)} \]
As \(x \to 2\), this expression still blows up (goes to infinity), so it is not analytic at \(x=2\).
Step 4: Conclude the type of singular point.
Since \((x-2)P(x)\) fails to be analytic at \(x=2\), the point \(x=2\) does not satisfy the condition for a regular singular point.
So \(x=2\) is an irregular singular point, meaning option (C) is false and option (D) is true.
Final Answer:
The point \(x=2\) is a singular point but not a regular one, so statements (B) and (D) are true. \[ \boxed{\text{(B) and (D)}} \]
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