If y=y(x) satisfies left(frac{2+sin x}{1+y}ri | vedclass

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

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(D) Given the differential equation: $\left(\frac{2+\sin x}{1+y}\right) \frac{dy}{dx} = -\cos x$. Separating the variables,we get: $\frac{dy}{1+y} = -\frac{\cos x}{2+\sin x} dx$. Integrating both sides: $\int \frac{1}{1+y} dy = -\int \frac{\cos x}{2+\sin x} dx$. This gives: $\ln|1+y| = -\ln|2+\sin x| + C$. Using the initial condition $y(0)=2$: $\ln|1+2| = -\ln|2+\sin 0| + C \implies \ln 3 = -\ln 2 + C \implies C = \ln 3 + \ln 2 = \ln 6$. So,$\ln(1+y) = -\ln(2+\sin x) + \ln 6 = \ln\left(\frac{6}{2+\sin x}\right)$. Therefore,$1+y = \frac{6}{2+\sin x} \implies y = \frac{6}{2+\sin x} - 1$. Now,find $y\left(\frac{\pi}{2}\right)$: $y\left(\frac{\pi}{2}\right) = \frac{6}{2+\sin(\pi/2)} - 1 = \frac{6}{2+1} - 1 = \frac{6}{3} - 1 = 2 - 1 = 1$.

If $y=y(x)$ satisfies $\left(\frac{2+\sin x}{1+y}\right) \frac{dy}{dx} = -\cos x$ such that $y(0)=2$,then $y\left(\frac{\pi}{2}\right)$ is equal to

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