Question

If \(y^{1/m} + y^{-1/m} = 2x\), then \((x^2 - 1)y'' + xy'\) is equal to

• A. \(m^2 y\)
• B. \(-m^2 y\)
• C. \(\pm m^2 y\)
• D. None of these

Hint:

The expression \(x \pm \sqrt{x^2 - 1}\) appears in the solution of \((x^2 - 1)y'' + xy' = m^2 y\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let \(y^{1/m} = t\), then \(t + \frac{1}{t} = 2x\).

Step 2: Detailed Explanation:
\(t + \frac{1}{t} = 2x \implies t^2 - 2xt + 1 = 0 \implies t = x \pm \sqrt{x^2 - 1}\).
So \(y^{1/m} = x + \sqrt{x^2 - 1}\) or its reciprocal.
Then \(y = (x + \sqrt{x^2 - 1})^m\).
This is a standard form that satisfies the differential equation \((x^2 - 1)y'' + xy' = m^2 y\).

Step 3: Final Answer:
Option (A) \(m^2 y\).