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(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} -

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$\log |1 + 	an (\frac{x + y}{2})| = x + c$, where $c$ is the constant of integration

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(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.$\frac{dv}{dx} = 1 + \sin v + \cos v$.Using half-angle formulas: $1 + \sin v + \cos v = 2\cos^2(\frac{v}{2}) + 2\sin(\frac{v}{2})\cos(\frac{v}{2}) = 2\cos^2(\frac{v}{2}) [1 + 	an(\frac{v}{2})]$.So, $\int \frac{dv}{2\cos^2(\frac{v}{2}) [1 + 	an(\frac{v}{2})]} = \int dx$.Let $u = 1 + 	an(\frac{v}{2})$, then $du = \frac{1}{2} \sec^2(\frac{v}{2}) dv = \frac{1}{2\cos^2(\frac{v}{2})} dv$.Thus, $\int \frac{du}{u} = \int dx \implies \log |u| = x + c$.Substituting back: $\log |1 + 	an(\frac{x + y}{2})| = x + c$.

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

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