Solution of the differential equation (dy)/(dx) = (x + y) + (x + y)...

We have,

$\dfrac{dy}{dx} = \sin \left(x + y\right) + \cos \left(x + y\right)$

Put $x + y = v ⇒ 1 + \dfrac{dy}{dx} = \dfrac{dv}{dx}$

$\dfrac{dv}{dx} - 1 = \sin v + \cos v$

$⇒ \dfrac{dv}{dx} = \sin v + \cos v + 1$

$⇒ \dfrac{dv}{\sin v + \cos v + 1} = dx$

$⇒ ∫ \left(\dfrac{1}{\sin v + \cos v + 1}\right)dv = ∫ dx$

$⇒ ∫ \dfrac{1}{\left\{\dfrac{2\tan \left(\dfrac{v}{2}\right)}{1 + \tan^{2}\left(\dfrac{v}{2}\right)}\right\} + \left\{\dfrac{1 - \tan^{2}\left(\dfrac{v}{2}\right)}{1 + \tan^{2}\left(\dfrac{v}{2}\right)}\right\} + 1}dv = ∫ dx$

$⇒ ∫ \left[\dfrac{\left\{1 + \tan^{2}\left(\dfrac{v}{2}\right)\right\}}{2\tan \left(\dfrac{v}{2}\right) + 1 - \tan^{2}\left(\dfrac{v}{2}\right) + 1 + \tan^{2}\left(\dfrac{v}{2}\right)}\right]dv = x + C$

$⇒ \dfrac{1}{2} ∫ \left[\dfrac{\sec^{2}\left(\dfrac{v}{2}\right)}{\tan \left(\dfrac{v}{2}\right) + 1}\right]dv = x + C$

$⇒ \log \left(\tan \left(\dfrac{v}{2}\right) + 1\right) = x + C$

$⇒ \log \left(\tan \left(\dfrac{x + y}{2}\right) + 1\right) = x + C$