Question

Solution of the differential equation \(x \sin \frac{y}{x} \, dy = \left( y \sin \frac{y}{x} - x \right) dx\) is:

• A. \(\log \frac{y}{x} - \cos x + c = 0\)
• B. \(\log x - \sin \frac{y}{x} + c = 0\)
• C. \(\log \frac{1}{x} - \cos y + c = 0\)
• D. \(\log x - \cos \frac{y}{x} + c = 0\)

Hint:

Exam Tip:
For homogeneous differential equations:

• Use \(y = vx\).
• Separate variables after substitution.
• Substitute back to get the solution.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given equation is homogeneous in \(x\) and \(y\). We can use the substitution \(y = vx\) to solve it.

Step 2: Key Formula or Approach:
The given equation is: \[ x \sin \frac{y}{x} \, dy = \left( y \sin \frac{y}{x} - x \right) dx \] Rewrite as: \[ \frac{dy}{dx} = \frac{y \sin(y/x) - x}{x \sin(y/x)} \] Let \(y = vx\), so \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).
Also, \(\sin \frac{y}{x} = \sin v\).

Step 3: Detailed Explanation:
Substitute: \[ v + x \frac{dv}{dx} = \frac{vx \sin v - x}{x \sin v} = \frac{v \sin v - 1}{\sin v} = v - \frac{1}{\sin v} \] So, \[ v + x \frac{dv}{dx} = v - \csc v \Rightarrow x \frac{dv}{dx} = - \csc v \] Separate variables: \[ \sin v \, dv = -\frac{dx}{x} \] Integrate: \[ \int \sin v \, dv = -\int \frac{dx}{x} \Rightarrow -\cos v = -\ln x + c \] \[ \cos v = \ln x - c \Rightarrow \ln x - \cos v + c = 0 \] Substitute \(v = y/x\): \[ \ln x - \cos \frac{y}{x} + c = 0 \] This matches option (D).

Step 4: Final Answer:
Therefore, option (D) is correct.