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

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(A) The given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$.Rearranging the terms to separate the variables, w

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$\frac{1}{3}$

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(A) The given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$.Rearranging the terms to separate the variables, we get:$\frac{d y}{1+y} + \frac{\cos x}{2+\sin x} dx = 0$.Integrating both sides:$\int \frac{d y}{1+y} + \int \frac{\cos x}{2+\sin x} dx = C_0$.Let $t = 2+\sin x$, then $dt = \cos x dx$. The integral becomes:$\ln|1+y| + \ln|2+\sin x| = C_1$.This simplifies to $(1+y)(2+\sin x) = C$.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 particular solution is $(1+y)(2+\sin x) = 4$.Now, find $y\left(\frac{\pi}{2}\right)$ by substituting $x=\frac{\pi}{2}$:$(1+y(\frac{\pi}{2}))(2+\sin(\frac{\pi}{2})) = 4$.$(1+y(\frac{\pi}{2}))(2+1) = 4$.$3(1+y(\frac{\pi}{2})) = 4$.$1+y(\frac{\pi}{2}) = \frac{4}{3}$.$y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

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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