Let $ y=y(x) $ be the solution of the differential equation, $ \frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y0, y(0)=1 $. If $ y(\pi)=a $ and $ \frac{d y}{d x} $ at $ x=\pi $ is $ b $, then the ordered pair $ (a, b) $ is equal to: [JEE Main 2020, 2 Sep (Shift 1)]

Question

Let $ y=y(x) $ be the solution of the differential equation, $ \frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1 $. If $ y(\pi)=a $ and $ \frac{d y}{d x} $ at $ x=\pi $ is $ b $, then the ordered pair $ (a, b) $ is equal to: [JEE Main 2020, 2 Sep (Shift 1)]

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Let \( y=y(x) \) be the solution of the differential equation, \( \frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y&…

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