The general solution of the differential equa | vedclass

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(C) Given differential equation: $\cos (x+y) dy = dx$ Rearranging, we get: $\frac{dx}{dy} = \cos (x+y)$ Let $x+y = v$. Then, differentiating with resp

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

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(C) Given differential equation: $\cos (x+y) dy = dx$ Rearranging, we get: $\frac{dx}{dy} = \cos (x+y)$ Let $x+y = v$. Then, differentiating with respect to $y$, we get: $\frac{dx}{dy} + 1 = \frac{dv}{dy} \implies \frac{dx}{dy} = \frac{dv}{dy} - 1$ Substituting this into the equation: $\frac{dv}{dy} - 1 = \cos v$ $\frac{dv}{dy} = 1 + \cos v$ Separating variables: $\int \frac{dv}{1 + \cos v} = \int dy$ Using the identity $1 + \cos v = 2 \cos^2 \frac{v}{2}$: $\int \frac{dv}{2 \cos^2 \frac{v}{2}} = \int dy$ $\frac{1}{2} \int \sec^2 \frac{v}{2} dv = \int dy$ Integrating both sides: $	an \frac{v}{2} = y + c$ Substituting $v = x+y$ back: $	an \frac{x+y}{2} = y + c$ Thus, $y = 	an \frac{x+y}{2} + c$.

The general solution of the differential equation $\cos (x+y) dy = dx$ is

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