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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 \left( \frac{x + y}{2} ight) = 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: $\frac{dv}{dx} - 1 = \cos(v)$.$\frac{dv}{dx} = 1 + \cos(v) = 2 \cos^2 \left( \frac{v}{2} ight)$.Separating variables: $\frac{dv}{2 \cos^2 \left( \frac{v}{2} ight)} = dx$.$\frac{1}{2} \sec^2 \left( \frac{v}{2} ight) dv = dx$.Integrating both sides: $\int \frac{1}{2} \sec^2 \left( \frac{v}{2} ight) dv = \int dx$.$	an \left( \frac{v}{2} ight) = x + c$.Substituting $v = x + y$ back: $	an \left( \frac{x + y}{2} ight) = x + c$.

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

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