If y=y(x) is the solution of the differential | vedclass

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(A) Given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$. Separating the variables,we get $\frac{d y}{y+1} = -\

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$\frac{1}{3}$

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(A) Given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$. Separating the variables,we get $\frac{d y}{y+1} = -\frac{\cos x}{2+\sin x} d x$. Integrating both sides,we have $\int \frac{d y}{y+1} = -\int \frac{\cos x}{2+\sin x} d x$. Let $2+\sin x = t$,then $\cos x dx = dt$. Thus,$\ln|y+1| = -\ln|2+\sin x| + C$. This simplifies to $\ln|y+1| + \ln|2+\sin x| = C$,or $(y+1)(2+\sin x) = C'$. Given $y(0)=1$,we substitute $x=0$ and $y=1$: $(1+1)(2+\sin 0) = C' \Rightarrow 2(2) = C' \Rightarrow C'=4$. So,the equation is $(y+1)(2+\sin x) = 4$. At $x=\frac{\pi}{2}$,we have $(y(\frac{\pi}{2})+1)(2+\sin \frac{\pi}{2}) = 4$. $(y(\frac{\pi}{2})+1)(2+1) = 4$. $3(y(\frac{\pi}{2})+1) = 4 \Rightarrow y(\frac{\pi}{2})+1 = \frac{4}{3}$. $y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

If $y=y(x)$ is the solution of the differential equation $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$ with $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

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