Question

The general solution of the differential equation $\frac{dy}{dx}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$ is (where $c$ is a constant of integration)

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

Answer 1

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

We have
$\frac{dy}{dx}=\sin \frac{x-y}{2}-\sin \frac{x+y}{2}$
$=-2\cos \frac{x}{2}\sin \frac{y}{2}$
$(\because \sin A-\sin B=2\cdot \cos \frac{A+B}{2}\cdot \sin \frac{A-B}{2})$
$\Rightarrow \frac{dy}{2\sin \frac{y}{2}}=-\cos \frac{x}{2}dx$
On integrating bith sides,we have
$\Rightarrow \frac{\frac{1}{2}\cdot \ln \left|\tan \frac{y}{4}\right|}{\frac{1}{2}}=-\frac{\sin \frac{x}{2}}{\frac{1}{2}}+c$
$\Rightarrow \ln \left|\tan \left(\frac{y}{4}\right)\right|=c-2\sin \left(\frac{x}{2}\right)$