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:

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

This equation can be solved using the substitution \( z = \frac{y}{x} \).

This implies \( y = zx \). Differentiating with respect to \( x \), we get:

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

Substituting these into the differential equation yields:

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

Simplifying the equation leads to:

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

After canceling terms, we obtain:

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

This further simplifies to:

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

Separating variables and integrating both sides:

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

The integral on the right side is:

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

Considering the provided solution format:

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

Using the point \( x = 1, y = \frac{\pi}{3} \), we substitute these values:

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

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

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

Solving for \( \alpha \):

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

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

Therefore, the final answer is 3.