Question

If \( \sin\left(\frac{y}{x}\right) = \log_e |x| + \frac{\alpha}{2} \) is the solution of the differential equation \[x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x\]and \( y(1) = \frac{\pi}{3} \), then \( \alpha^2 \) is equal to

• A. 3
• B. 12
• C. 4
• D. 9

Answer 1

The Correct Option is A

Starting with the differential equation:  
\(x \cos \left( \frac{y}{x} \right) \frac{dy}{dx} = y \cos \left( \frac{y}{x} \right) + x\)

Step 1. Divide both sides by \( x^2 \cos \left( \frac{y}{x} \right) \):
 \(\cos \left( \frac{y}{x} \right) \left( \frac{y}{x} \frac{dy}{dx} - \frac{y}{x^2} \right) = \frac{1}{x}\)

Step 2. Let \( \frac{y}{x} = t \), then \( y = tx \) and \( \frac{dy}{dx} = t + x \frac{dt}{dx} \), substituting into the equation:
\(\cos t \left( \frac{dt}{dx} \right) = \frac{1}{x}\)

Step 3. Integrate both sides:
  \(\sin t = \ln |x| + c\)  
  \(\sin \frac{y}{x} = \ln |x| + c\)
Step 4. Using the initial condition \( y(1) = \frac{\sqrt{3}}{2} \), we find \( c = \frac{\sqrt{3}}{2} \).
  Thus,  \( \alpha = \sqrt{3} \implies \alpha^2 = 3\)
The Correct Answer is: 3