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

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(B) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{d y}{d x} = -\cos x$ Separate the variables: $\int \frac{d y}{y+1} = \int \frac{-\cos

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$\frac{1}{3}$

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(B) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{d y}{d x} = -\cos x$ Separate the variables: $\int \frac{d y}{y+1} = \int \frac{-\cos x}{2+\sin x} d x$ Integrate both sides: $\ln|y+1| = -\ln|2+\sin x| + C_1$ This simplifies to: $\ln|y+1| + \ln|2+\sin x| = C_1$ Using logarithmic properties: $\ln|(y+1)(2+\sin x)| = C_1$ Exponentiating both sides: $(y+1)(2+\sin x) = C$ Given $y(0)=1$,substitute $x=0$ and $y=1$: $(1+1)(2+\sin 0) = C \Rightarrow 2(2+0) = C \Rightarrow C=4$ So,the equation is: $(y+1)(2+\sin x) = 4$ To find $y\left(\frac{\pi}{2}\right)$,substitute $x=\frac{\pi}{2}$: $(y+1)(2+\sin\frac{\pi}{2}) = 4$ $(y+1)(2+1) = 4 \Rightarrow 3(y+1) = 4 \Rightarrow y+1 = \frac{4}{3}$ $y = \frac{4}{3} - 1 = \frac{1}{3}$

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

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