Let y(x) be a solution of frac{(2 + sin x) dy | vedclass

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(C) The given differential equation is $\frac{(2 + \sin x)}{(1 + y)} \frac{dy}{dx} = \cos x.$Separating the variables,we get $\frac{dy}{1 + y} = \frac

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$\frac{7}{2}$

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(C) The given differential equation is $\frac{(2 + \sin x)}{(1 + y)} \frac{dy}{dx} = \cos x.$Separating the variables,we get $\frac{dy}{1 + y} = \frac{\cos x}{2 + \sin x} dx.$Integrating both sides,we have $\int \frac{dy}{1 + y} = \int \frac{\cos x}{2 + \sin x} dx.$This yields $\ln(1 + y) = \ln(2 + \sin x) + \ln C.$Exponentiating both sides,we get $1 + y = C(2 + \sin x).$Given $y(0) = 2,$ we substitute $x = 0$ and $y = 2$ to find $C$: $1 + 2 = C(2 + \sin 0) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}.$Now,we find $y\left( \frac{\pi}{2} \right)$ by substituting $x = \frac{\pi}{2}$ and $C = \frac{3}{2}$ into the equation $1 + y = \frac{3}{2}(2 + \sin x).$$1 + y\left( \frac{\pi}{2} \right) = \frac{3}{2}(2 + \sin \frac{\pi}{2}) = \frac{3}{2}(2 + 1) = \frac{3}{2}(3) = \frac{9}{2}.$Therefore,$y\left( \frac{\pi}{2} \right) = \frac{9}{2} - 1 = \frac{7}{2}.$

Let $y(x)$ be a solution of $\frac{(2 + \sin x) dy}{(1 + y) dx} = \cos x.$ If $y(0) = 2,$ then $y\left( \frac{\pi}{2} \right)$ equals

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