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

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$(1, 1)$

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(D) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{dy}{dx} = -\cos x$. Separating the variables,we get: $\frac{dy}{y+1} = \frac{-\cos x}{2+\sin x} dx$. Integrating both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$. This gives: $\ln(y+1) = -\ln(2+\sin x) + C$. Using the initial condition $y(0) = 1$: $\ln(1+1) = -\ln(2+\sin 0) + C \Rightarrow \ln 2 = -\ln 2 + C \Rightarrow C = 2\ln 2 = \ln 4$. Thus,$\ln(y+1) = \ln\left(\frac{4}{2+\sin x}\right)$,which implies $y+1 = \frac{4}{2+\sin x}$,or $y(x) = \frac{4}{2+\sin x} - 1$. For $x = \pi$,$a = y(\pi) = \frac{4}{2+\sin \pi} - 1 = \frac{4}{2} - 1 = 1$. Now,find $b = \frac{dy}{dx}$ at $x = \pi$: $\frac{dy}{dx} = \frac{-\cos x}{2+\sin x} (y+1) = \frac{-\cos x}{2+\sin x} \left(\frac{4}{2+\sin x}\right) = \frac{-4\cos x}{(2+\sin x)^2}$. At $x = \pi$,$b = \frac{-4\cos \pi}{(2+\sin \pi)^2} = \frac{-4(-1)}{(2+0)^2} = \frac{4}{4} = 1$. Therefore,the ordered pair $(a, b) = (1, 1)$.

Let $y=y(x)$ be the solution of the differential equation $\frac{2+\sin x}{y+1} \cdot \frac{dy}{dx} = -\cos x$,where $y > 0$ and $y(0) = 1$. If $y(\pi) = a$ and $\frac{dy}{dx}$ at $x = \pi$ is $b$,then the ordered pair $(a, b)$ is equal to:

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