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: Understanding the Concept:
We use trigonometric identities to separate the variables $x$ and $y$.
Step 2: Key Formula or Approach:
$\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$
The equation is $\frac{\text{d}y}{\text{d}x} = \sin \left( \frac{x-y}{2} \right) - \sin \left( \frac{x+y}{2} \right)$.
Step 3: Detailed Explanation:
Let $C = (x-y)/2$ and $D = (x+y)/2$:
$\frac{C+D}{2} = \frac{x}{2}$ and $\frac{C-D}{2} = \frac{-y}{2}$.
\[ \frac{\text{d}y}{\text{d}x} = 2 \cos \left( \frac{x}{2} \right) \sin \left( \frac{-y}{2} \right) = -2 \cos \frac{x}{2} \sin \frac{y}{2} \] Separating variables:
\[ \frac{\text{d}y}{\sin(y/2)} = -2 \cos \left( \frac{x}{2} \right) \text{d}x \] \[ \csc(y/2) \text{d}y = -2 \cos \left( \frac{x}{2} \right) \text{d}x \] Integrating both sides:
\[ 2 \log \left| \tan \frac{y/2}{2} \right| = -2 \cdot 2 \sin \frac{x}{2} + C' \] \[ 2 \log \left| \tan \frac{y}{4} \right| = -4 \sin \frac{x}{2} + C' \] Divide by 2:
\[ \log \tan \frac{y}{4} = \text{c} - 2 \sin \frac{x}{2} \] Step 4: Final Answer:
The general solution is $\log \tan \left( \frac{y}{4} \right) = \text{c} - 2 \sin \left( \frac{x}{2} \right)$.