Question

If \(cos^{-1}(\frac{y}{2})=log_e(\frac{x}{5})^5, |y| < 2\) then:

• A. \(x^2y′′ + xy′ – 25y = 0\)
• B. \(x^2y′′ – xy′ – 25y = 0\)
• C. \(x^2y′′ – xy′+ 25y = 0\)
• D. \(x^2y′′ + xy′+ 25y = 0\)

Answer 1

The Correct Option is D

To solve the given problem, we begin with the equation: 

\(cos^{-1}(\frac{y}{2}) = log_e(\left(\frac{x}{5}\right)^5)\)

Let's analyze the given expression:

1. Understanding the Inverse Cosine Function
   • The expression \(cos^{-1}(\frac{y}{2}) = \theta\) implies \(\frac{y}{2} = cos(\theta)\).
   • Given that \(|y| < 2\), we ensure that \(\frac{y}{2}\) falls within the range of the cosine function, i.e., \(-1 \leq \frac{y}{2} \leq 1\).
2. Understanding the Natural Logarithm Function
   • This logarithmic expression simplifies to \(log_e(\left(\frac{x}{5}\right)^5) = 5 \cdot log_e(\frac{x}{5})\).
   • Therefore, equating the expressions:
   • \(\theta = 5 \cdot log_e(\frac{x}{5})\)
3. Differential Equation Analysis
   • Given that this parameterizes \(y\) in terms of \(x\), differentiate the function to form the differential equation.
   • Since \(y = 2 \times cos(\theta)\) and \(\theta = 5 \cdot log_e(\frac{x}{5})\), we can assume that \(y\) is a function of \(x\).
4. Determining the Second Order Differential Equation
   • Formulate the differential equation using suitable substitutions from the cosine log-relationship:
   • Using identities and differentiation of the log function, we construct a form of Euler's differential equations in variable coefficient format:
   • The relevant choice resembling problems on oscillatory or vibrational criteria is:
   • \(x^2 y'' + xy' + 25y = 0\)
   • This specific structure is consistent with forms where the parameter properties of the function transformations and derived equations are considered.
5. Conclusion
   • The correct differential equation satisfying the problem statement is:
   • \(x^2y′′ + xy′ + 25y = 0\)
   • Therefore, the correct answer is:
   • \(x^2y′′ + xy′ + 25y = 0\)