The general solution of the differential equa | vedclass

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

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

Vedclass · Answer

(B) Given differential equation is $\frac{dy}{dx} + \sin \left( \frac{x + y}{2} ight) = \sin \left( \frac{x - y}{2} ight)$.Rearranging the terms,we get $\frac{dy}{dx} = \sin \left( \frac{x - y}{2} ight) - \sin \left( \frac{x + y}{2} ight)$.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 $\sin \left( \frac{x - y}{2} ight) - \sin \left( \frac{x + y}{2} ight) = 2 \cos \left( \frac{x}{2} ight) \sin \left( -\frac{y}{2} ight) = -2 \cos \left( \frac{x}{2} ight) \sin \left( \frac{y}{2} ight)$.Thus,$\frac{dy}{dx} = -2 \cos \left( \frac{x}{2} ight) \sin \left( \frac{y}{2} ight)$.Separating the variables,we get $	ext{cosec} \left( \frac{y}{2} ight) dy = -2 \cos \left( \frac{x}{2} ight) dx$.Integrating both sides,$\int 	ext{cosec} \left( \frac{y}{2} ight) dy = -2 \int \cos \left( \frac{x}{2} ight) dx + c$.Using $\int 	ext{cosec} (ax) dx = \frac{1}{a} \log 	an \left( \frac{ax}{2} ight)$,we get $\frac{\log 	an (y/4)}{1/2} = -2 \frac{\sin (x/2)}{1/2} + c$.This simplifies to $2 \log 	an (y/4) = -4 \sin (x/2) + c$,or $\log 	an (y/4) = c - 2 \sin (x/2)$.

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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