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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: $\frac{dv}{dx} - 1 = \sin v + \cos v$.$\frac{dv}{dx} = 1 + \sin v + \cos v$.$\frac{dv}{1 + \sin v + \cos v} = dx$.Using half-angle formulas $\sin v = \frac{2 	an(v/2)}{1 + 	an^2(v/2)}$ and $\cos v = \frac{1 - 	an^2(v/2)}{1 + 	an^2(v/2)}$:$1 + \sin v + \cos v = 1 + \frac{2 	an(v/2) + 1 - 	an^2(v/2)}{1 + 	an^2(v/2)} = \frac{1 + 	an^2(v/2) + 2 	an(v/2) + 1 - 	an^2(v/2)}{1 + 	an^2(v/2)} = \frac{2 + 2 	an(v/2)}{1 + 	an^2(v/2)} = \frac{2(1 + 	an(v/2))}{\sec^2(v/2)}$.Thus, $\int \frac{\sec^2(v/2) dv}{2(1 + 	an(v/2))} = \int dx$.Let $u = 1 + 	an(v/2)$, then $du = \frac{1}{2} \sec^2(v/2) dv$.The integral becomes $\int \frac{du}{u} = \int dx$, which gives $\log |u| = x + c$.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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