If (2 + sin x) frac{dy}{dx} + (y + 1) cos x = | vedclass

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(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] =

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$\frac{1}{3}$

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(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] = 0$.Integrating both sides with respect to $x$,we get: $(2 + \sin x)(y + 1) = C$,where $C$ is a constant.Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$:$(2 + \sin 0)(1 + 1) = C \Rightarrow (2 + 0)(2) = C \Rightarrow C = 4$.Thus,the equation becomes: $(2 + \sin x)(y + 1) = 4$.Solving for $y$: $y + 1 = \frac{4}{2 + \sin x} \Rightarrow y = \frac{4}{2 + \sin x} - 1$.Now,we find $y(\frac{\pi}{2})$:$y(\frac{\pi}{2}) = \frac{4}{2 + \sin(\frac{\pi}{2})} - 1 = \frac{4}{2 + 1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

If $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$ and $y(0) = 1$,then $y(\frac{\pi}{2}) = \dots$

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