Question

The solution of the differential equation $\frac{dy}{dx}+\sin \left(\frac{y+x}{2}\right)+\sin \left(\frac{y-x}{2}\right)=0$ is

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

Answer 1

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

Given differential equation can be rewritten as
$\frac{dy}{dx}=\sin \left(\frac{x-y}{2}\right)-\sin \left(\frac{x+y}{2}\right)$
$=2\cos \left(\frac{x}{2}\right)\sin \left(\frac{-y}{2}\right)$
$=-2\cos \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)$
$\Rightarrow \int \frac{dy}{2\sin \left(\frac{y}{2}\right)}=-\int \cos \left(\frac{x}{2}\right)dx$
$\Rightarrow \frac{1}{2}\int \csc \left(\frac{y}{2}\right)dy=-\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{1}{2}\right)}+C$
$\Rightarrow \frac{1}{2}\frac{\log (\csc \left(\frac{y}{2}\right)-\cot \left(\frac{y}{2}\right))}{\left(\frac{1}{2}\right)}=-\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{1}{2}\right)}+C$
$\Rightarrow \log \tan \left(\frac{y}{4}\right)=-2\sin \left(\frac{x}{2}\right)+C$