Question

Find the solution of $\cos (x+y)dy=dx$

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

Answer 1

The correct option is A $y-c=\tan \left(\frac{x+y}{2}\right)$

$\cos (x+y)dy=dx..\left(1\right)$
Put, $x+y=v\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$
Thus (1) becomes,
$\cos v(dv-dx)=dx$
$\therefore dx=\frac{\cos v}{1+\cos v}dv=(1-\frac{1}{1+\cos v})dv$
or $dx=(1-\frac{1}{2}{\sec }^2\frac{v}{2})dv$
Integrating we get,
$x+c=(v-\tan \frac{v}{2})$
Now replace $v$ by $(x+y)$
$\therefore y-c=\tan \left(\frac{x+y}{2}\right)$