If (2+sin x) frac{dy}{dx}+(y+1) cos x=0 and y | vedclass

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(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$. Rearranging the terms to separate the variables: $\frac{1}{y+1} dy

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$\frac{1}{3}$

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(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$. Rearranging the terms to separate the variables: $\frac{1}{y+1} dy = -\frac{\cos x}{2+\sin x} dx$. Integrating both sides: $\int \frac{1}{y+1} dy = -\int \frac{\cos x}{2+\sin x} dx$. This gives: $\ln(y+1) = -\ln(2+\sin x) + C$. Using the initial condition $y(0)=1$: $\ln(1+1) = -\ln(2+\sin 0) + C \implies \ln 2 = -\ln 2 + C \implies C = 2\ln 2 = \ln 4$. Substituting $C$ back into the equation: $\ln(y+1) = -\ln(2+\sin x) + \ln 4 = \ln\left(\frac{4}{2+\sin x}\right)$. Taking the exponential of both sides: $y+1 = \frac{4}{2+\sin x} \implies y = \frac{4}{2+\sin x} - 1$. Now,evaluating at $x = \frac{\pi}{2}$: $y\left(\frac{\pi}{2}\right) = \frac{4}{2+\sin(\pi/2)} - 1 = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

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

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