Question

The general solution of the differential equation $\frac{dy}{dx}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$ is

• A. ${\log }_e\left|\tan \frac{y}{2}\right|=-2\cos \frac{x}{2}+C$
• B. ${\log }_e\left|\tan \frac{y}{2}\right|=2\cos \frac{x}{2}+C$
• C. ${\log }_e\left|\tan \frac{y}{2}\right|=2\sin \frac{x}{2}+C$
• D. ${\log }_e\left|\tan \frac{y}{2}\right|=-2\sin \frac{x}{2}+C$

Answer 1

Answer: (D)

${\log }_e\left|\tan \frac{y}{2}\right|=-2\sin \frac{x}{2}+C$

We have
$\frac{dy}{dx}+\sin \frac{x+y}{2}=\sin \frac{x-y}{2}$
$\Rightarrow \frac{dy}{dx}=\sin \frac{x-y}{2}-\sin \frac{x+y}{2}$
$\Rightarrow \frac{dy}{dx}=2\cos \frac{(\frac{x-y}{2}+\frac{x+y}{2})}{2}\sin \frac{(\frac{x-y}{2}-\frac{x+y}{2})}{2}$ ..... $[\because \sin C-\sin D=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2}]$
$\Rightarrow \frac{dy}{dx}=2\cos \frac{x}{2}.\sin \frac{-y}{2}$

$\Rightarrow \frac{dy}{dx}=-2\cos \left(\frac{x}{2}\right).\sin \left(\frac{y}{2}\right)$
$\Rightarrow \frac{dy}{\sin \left(\frac{y}{2}\right)}=-2\cos \left(\frac{x}{2}\right)$
On integrating both sides, we get
$\Rightarrow \int \text{cosec}\left(\frac{y}{2}\right)dy=-2{\int }^{}\cos \left(\frac{x}{2}\right)dx$
$\Rightarrow 2{\log }_e\left|\tan \frac{y}{2}\right|=-4\sin \left(\frac{x}{2}\right)+C$

$\Rightarrow {\log }_e\left|\tan \frac{y}{2}\right|=-2\sin \left(\frac{x}{2}\right)+C$