Question

Solve:

$\frac{dy}{dx}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$

• A. $\ln \left|\tan \frac{y}{2}\right|=c_1-2\cos \frac{x}{4}$
• B. $\ln \left|\tan \frac{y}{4}\right|=c_1-2\sin \frac{x}{2}$
• C. $\ln \left|\cot \frac{y}{4}\right|=c_1-2\cos \frac{x}{2}$
• D. $\ln \left|\cot \frac{y}{2}\right|=c_1-2\sin \frac{x}{4}$

Answer 1

The correct option is B $\ln \left|\tan \frac{y}{4}\right|=c_1-2\sin \frac{x}{2}$

$\frac{dy}{dx}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$

we know $\sin (A-B)-\sin (A+B)=-\cos A\sin B$

$\Rightarrow \frac{dy}{dx}=-2\cos \frac{x}{2}\sin \frac{y}{2}$

$\Rightarrow \text{cosec}\frac{y}{2}dy=-2\cos \frac{x}{2}dx$

On integration we get

$\Rightarrow 2\left(\ln \left|\tan \frac{y}{4}\right|\right)=2(-2\sin \frac{x}{2})+c$

$\Rightarrow \ln \left|\tan \frac{y}{4}\right|=c_1-2\sin \frac{x}{2}$