Question

Solution of the differential equation $\frac{dy}{dx}$ $=\sin (x+y)+\cos (x+y)$ is

• A. $\log |1+\tan \left(\frac{x+y}{2}\right)|=x+c$
• B. $\log |2+\sec \left(\frac{x+y}{2}\right)|=x+c$
• C. $\log |1+\tan (x+y\left)\right|=y+c$
• D. None of these

Answer 1

The correct option is A $\log |1+\tan \left(\frac{x+y}{2}\right)|=x+c$

$\frac{dy}{dx}=sin(x+y)+cos(x+y)$

let $x+y=u$

then

$\frac{dy}{dx}=\frac{du}{dx}-1$

$\frac{du}{dx}=2sinu+cosu+1$

$\frac{du}{dx}=2sin\frac{u}{2}cos\frac{u}{2}+2{cos}^2\frac{u}{2}$

$\frac{du}{dx}=2{cos}^2\frac{u}{2}(1+tan\frac{u}{2})$

$\left[\frac{\frac{1}{2}{sec}^2\frac{u}{2}}{1+tan\frac{u}{2}}\right]du=dx$

On integrating, we get

$log[1+tan\frac{x+y}{2}]=x+c$

Hence option (A) is correct.