The solution of frac{dy}{dx} = sin(x + y) + c | vedclass

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

Vedclass · Accepted Answer

$\log \left[ 1 + 	an \left( \frac{x + y}{2} ight) ight] = x + c$

Vedclass · Answer

(B) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting this into the given equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.$\frac{dv}{dx} = 1 + \cos v + \sin v$.Using trigonometric identities: $1 + \cos v = 2\cos^2(v/2)$ and $\sin v = 2\sin(v/2)\cos(v/2)$.$\frac{dv}{dx} = 2\cos^2(v/2) + 2\sin(v/2)\cos(v/2) = 2\cos^2(v/2) [1 + 	an(v/2)]$.Separating variables: $\frac{\sec^2(v/2) dv}{2[1 + 	an(v/2)]} = dx$.Let $u = 1 + 	an(v/2)$,then $du = \frac{1}{2}\sec^2(v/2) dv$.The integral becomes $\int \frac{du}{u} = \int dx$.$\log|u| = x + c$.Substituting back $u = 1 + 	an(v/2)$ and $v = x + y$: $\log \left[ 1 + 	an \left( \frac{x + y}{2} ight) ight] = x + c$.

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

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