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(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substitut

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$	an (\frac{x + y}{2}) = x + c$

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(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \cos v$.$\frac{dv}{dx} = 1 + \cos v = 2 \cos^2 (\frac{v}{2})$.Separating variables: $\frac{dv}{2 \cos^2 (\frac{v}{2})} = dx$.$\frac{1}{2} \sec^2 (\frac{v}{2}) dv = dx$.Integrating both sides: $\int \frac{1}{2} \sec^2 (\frac{v}{2}) dv = \int dx$.$	an (\frac{v}{2}) = x + c$.Substituting $v = x + y$ back: $	an (\frac{x + y}{2}) = x + c$.

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

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