A) \[\log \tan \left( \frac{y}{2} \right)=c-2\sin x\]
B) \[\log \tan \,\left( \frac{y}{4} \right)=c-2\sin \left( \frac{x}{2} \right)\]
C) \[\log \tan \,\left( \frac{y}{2}+\frac{\pi }{4} \right)=c-2\sin x\]
D) \[\log \tan \left( \frac{y}{4}+\frac{\pi }{4} \right)=c-2\sin \left( \frac{x}{2} \right)\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}+\sin \left( \frac{x+y}{2} \right)=\sin \left( \frac{x-y}{2} \right)\] Þ \[\frac{dy}{dx}=\sin \left( \frac{x-y}{2} \right)-\sin \left( \frac{x+y}{2} \right)\] Þ \[\frac{dy}{dx}=-2\sin \,\left( \frac{y}{2} \right)\,.\cos \,\left( \frac{x}{2} \right)\] Þ \[\text{cos}\text{ec}\left( \frac{y}{2} \right).dy=-2\cos \left( \frac{x}{2} \right)\,dx\] Integrating both sides, \[\int{\text{cosec}\left( \frac{y}{\text{2}} \right)dy=-\int{2\cos \left( \frac{x}{2} \right)dx+c}}\]. Þ \[\frac{\log \,\tan \frac{y}{4}}{1/2}=-\frac{2\sin \left( x/2 \right)}{1/2}+c\] Þ \[\log (\tan \frac{y}{4})=c-2\sin (x/2)\].
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