The general solution of the differential equa | vedclass

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(B) Given the differential equation $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$.Using the trigonometric identi

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$\log 	an \left(\frac{y}{4}ight) = c - 2 \sin \left(\frac{x}{2}ight)$,where $c$ is the constant of integration

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(B) Given the differential equation $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}ight) = \sin \left(\frac{x-y}{2}ight)$.Using the trigonometric identity $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)$,we have:$\sin \left(\frac{x+y}{2}ight) - \sin \left(\frac{x-y}{2}ight) = 2 \cos \left(\frac{x}{2}ight) \sin \left(\frac{y}{2}ight)$.Thus,the equation becomes $\frac{dy}{dx} = -2 \cos \left(\frac{x}{2}ight) \sin \left(\frac{y}{2}ight)$.Separating the variables,we get $\frac{dy}{\sin(y/2)} = -2 \cos(x/2) dx$.Integrating both sides: $\int \csc(y/2) dy = -2 \int \cos(x/2) dx$.$2 \log |	an(y/4)| = -2(2 \sin(x/2)) + c_1$.Dividing by $2$: $\log |	an(y/4)| = -2 \sin(x/2) + c$,where $c = c_1/2$.Therefore,the general solution is $\log 	an \left(\frac{y}{4}ight) = c - 2 \sin \left(\frac{x}{2}ight)$.

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

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