The general solution of frac{dy}{dx} + sin le | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ Rearranging the terms: $\frac

Vedclass · Accepted Answer

$\log 	an \left(\frac{y}{4}ight) = C - 2 \sin \left(\frac{x}{2}ight)$

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(B) Given the differential equation: $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}ight) = \sin \left(\frac{x-y}{2}ight)$ Rearranging the terms: $\frac{dy}{dx} = \sin \left(\frac{x-y}{2}ight) - \sin \left(\frac{x+y}{2}ight)$ Using the trigonometric identity $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)$,we get: $\frac{dy}{dx} = 2 \cos \left(\frac{x}{2}ight) \sin \left(-\frac{y}{2}ight) = -2 \sin \left(\frac{y}{2}ight) \cos \left(\frac{x}{2}ight)$ Separating the variables: $\int \operatorname{cosec} \left(\frac{y}{2}ight) dy = -\int 2 \cos \left(\frac{x}{2}ight) dx$ Integrating both sides: $2 \log 	an \left(\frac{y}{4}ight) = -4 \sin \left(\frac{x}{2}ight) + c_1$ Dividing by $2$: $\log 	an \left(\frac{y}{4}ight) = -2 \sin \left(\frac{x}{2}ight) + C$,where $C = \frac{c_1}{2}$.

The general solution of $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ is

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