If y=y(x) is the solution of frac{dy}{dx}=fra | vedclass

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(A) Given the differential equation $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$.Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\cos x}{1+\s

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$\frac{5 \pi^2}{8}$

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(A) Given the differential equation $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$.Rearranging the terms,we get $\frac{dy}{dx} + \left(\frac{\cos x}{1+\sin x}\right)y = \frac{x}{1+\sin x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\cos x}{1+\sin x}$ and $Q(x) = \frac{x}{1+\sin x}$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{\cos x}{1+\sin x} dx} = e^{\ln(1+\sin x)} = 1+\sin x$.The general solution is $y(I.F.) = \int Q(x)(I.F.) dx + C$.$y(1+\sin x) = \int \frac{x}{1+\sin x} (1+\sin x) dx + C = \int x dx + C = \frac{x^2}{2} + C$.Using the condition $y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}$:$\frac{\pi^2}{8}(1+\sin(\frac{\pi}{2})) = \frac{(\pi/2)^2}{2} + C \Rightarrow \frac{\pi^2}{8}(2) = \frac{\pi^2}{8} + C \Rightarrow \frac{\pi^2}{4} - \frac{\pi^2}{8} = C \Rightarrow C = \frac{\pi^2}{8}$.Thus,$y(1+\sin x) = \frac{x^2}{2} + \frac{\pi^2}{8}$.For $x = \pi$,$y(1+\sin \pi) = \frac{\pi^2}{2} + \frac{\pi^2}{8}$.Since $\sin \pi = 0$,$y(1) = \frac{4\pi^2 + \pi^2}{8} = \frac{5\pi^2}{8}$.

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

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