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

To solve the given differential equation: 

\[ x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \]

We note that it is in a format that can be addressed by assuming a substitution:

Let \( z = \frac{y}{x} \). This gives \( y = zx \) and differentiating with respect to \( x \), we have:

\(\frac{dy}{dx} = z + x \frac{dz}{dx}\)

Substitute in the differential equation:

\[ x \cos(z) (z + x \frac{dz}{dx}) = zx \cos(z) + x \]

On simplifying, we get:

\[ xz \cos(z) + x^2 \cos(z) \frac{dz}{dx} = zx \cos(z) + x \]

Cancel terms and simplify:

\[ x^2 \cos(z) \frac{dz}{dx} = x(1 - z \cos(z)) \]

Which simplifies to:

\[ \frac{dz}{dx} = \frac{1 - z \cos(z)}{x \cos(z)} \]

Let's separate the variables and integrate:

\[ \int \frac{\cos(z)}{1 - z \cos(z)} \, dz = \int \frac{1}{x} \, dx \]

On the right side, integrate:

\[ \int \frac{1}{x} \, dx = \log_e |x| + C \]

Let's consider the solution format provided in the question:

\[ \sin\left(\frac{y}{x}\right) = \log_e |x| + \frac{\alpha}{2} \]

At point \( x = 1, y = \frac{\pi}{3} \), substituting these values gives:

\[ \sin\left(\frac{\pi}{3}\right) = \log_e |1| + \frac{\alpha}{2} \]

Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\log_e |1| = 0\), we have:

\[ \frac{\sqrt{3}}{2} = 0 + \frac{\alpha}{2} \]

This implies:

\[ \alpha = \sqrt{3} \]

Thus, \(\alpha^2 = (\sqrt{3})^2 = 3\).

Therefore, the correct answer is 3.

Answer 2

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