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Question

(A) Given the differential equation: $\frac{dy}{dx} 	an y = \sin(x + y) + \sin(x - y)$.Using the trigonometric identity $\sin(A + B) + \sin(A - B) =

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$\sec y + 2 \cos x = c$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} 	an y = \sin(x + y) + \sin(x - y)$.Using the trigonometric identity $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$,we get:$\frac{dy}{dx} 	an y = 2 \sin x \cos y$.Rewrite $	an y$ as $\frac{\sin y}{\cos y}$:$\frac{dy}{dx} \cdot \frac{\sin y}{\cos y} = 2 \sin x \cos y$.Separate the variables:$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$.Integrate both sides:$\int \frac{\sin y}{\cos^2 y} dy = 2 \int \sin x dx$.Let $u = \cos y$,then $du = -\sin y dy$,so $\int -u^{-2} du = u^{-1} = \frac{1}{\cos y} = \sec y$.Thus,$\sec y = -2 \cos x + c$,which simplifies to $\sec y + 2 \cos x = c$.

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

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