The solution of cos (x + y) , dy = dx is | vedclass

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(A) Given the differential equation: $\cos (x + y) \, dy = dx$ ... $(i)$ Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx

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

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(A) Given the differential equation: $\cos (x + y) \, dy = dx$ ... $(i)$ Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting these into $(i)$,we have $\cos v \left( \frac{dv}{dx} - 1 ight) = 1$. This simplifies to $\cos v \frac{dv}{dx} = 1 + \cos v$,or $\frac{\cos v}{1 + \cos v} \, dv = dx$. Using the identity $\cos v = 2 \cos^2(v/2) - 1$,we get $\frac{2 \cos^2(v/2) - 1}{2 \cos^2(v/2)} \, dv = dx$,which simplifies to $\left( 1 - \frac{1}{2} \sec^2(v/2) ight) \, dv = dx$. Integrating both sides,we get $v - 	an(v/2) = x + c$. Substituting $v = x + y$ back,we get $(x + y) - 	an \left( \frac{x + y}{2} ight) = x + c$,which simplifies to $y = 	an \left( \frac{x + y}{2} ight) + c$.

The solution of $\cos (x + y) \, dy = dx$ is

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