Question

The general solution of the differential equation $\frac{\text{d}y}{\text{d}x} + \sin \left( \frac{x+y}{2} \right) = \sin \left( \frac{x-y}{2} \right)$ is

• A. $\log \tan \left( \frac{y}{2} \right) = \text{c} - 2 \sin \frac{x}{2}$
• B. $\log \tan \left( \frac{y}{4} \right) = \text{c} - 2 \sin \left( \frac{x}{2} \right)$
• C. $\log [\tan \left( \frac{y}{2} + \frac{\pi}{4} \right)] = \text{c} - 2 \sin x$
• D. $\log [\tan \left( \frac{y}{4} + \frac{\pi}{4} \right)] = \text{c} - 2 \sin \frac{x}{2}$

Hint:

$\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$.

Answer 1

The Correct Option is B

Step 1: Simplify Trigonometry
$\frac{dy}{dx} = \sin(\frac{x-y}{2}) - \sin(\frac{x+y}{2}) = 2 \cos(\frac{x}{2}) \sin(\frac{-y}{2}) = -2 \cos \frac{x}{2} \sin \frac{y}{2}$.
Step 2: Variable Separable
$\frac{dy}{\sin(y/2)} = -2 \cos(\frac{x}{2}) dx \implies \csc(\frac{y}{2}) dy = -2 \cos(\frac{x}{2}) dx$.
Step 3: Integration
$2 \log \tan(\frac{y}{4}) = -2 (2 \sin \frac{x}{2}) + C' \implies \log \tan(\frac{y}{4}) = -2 \sin \frac{x}{2} + C$.
Final Answer: (B)