If left( frac{2 + sin x}{1 + y} right) frac{d | vedclass

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(C) The given differential equation is $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$. Separating the variables,we get: $\frac{dy}

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$\frac{1}{3}$

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(C) The given differential equation is $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$. Separating the variables,we get: $\frac{dy}{1 + y} = - \frac{\cos x}{2 + \sin x} dx$. Integrating both sides: $\int \frac{dy}{1 + y} = - \int \frac{\cos x}{2 + \sin x} dx$. Let $u = 2 + \sin x$,then $du = \cos x dx$. So,$\ln|1 + y| = - \ln|2 + \sin x| + C$. This can be written as $\ln|1 + y| + \ln|2 + \sin x| = C$,or $\ln|(1 + y)(2 + \sin x)| = C$. Thus,$(1 + y)(2 + \sin x) = K$ (where $K = e^C$). Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $(1 + 1)(2 + \sin 0) = K \implies 2(2 + 0) = K \implies K = 4$. So,$(1 + y)(2 + \sin x) = 4$. To find $y\left( \frac{\pi}{2} \right)$,substitute $x = \frac{\pi}{2}$: $(1 + y)(2 + \sin \frac{\pi}{2}) = 4 (1 + y)(2 + 1) = 4 3(1 + y) = 4 1 + y = \frac{4}{3} y = \frac{4}{3} - 1 = \frac{1}{3}$.

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

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