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(A) The given differential equation is $x \frac{dy}{dx}-y=x^{2}(x \cos x+\sin x)$.Dividing by $x$,we get $\frac{dy}{dx}-\frac{y}{x}=x(x \cos x+\sin x)

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$2+\frac{\pi}{2}$

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(A) The given differential equation is $x \frac{dy}{dx}-y=x^{2}(x \cos x+\sin x)$.Dividing by $x$,we get $\frac{dy}{dx}-\frac{y}{x}=x(x \cos x+\sin x)$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=-\frac{1}{x}$ and $Q=x^{2} \cos x+x \sin x$.The integrating factor $I.F. = e^{\int P dx} = e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.The solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$\frac{y}{x} = \int \frac{1}{x} \cdot x(x \cos x+\sin x) dx = \int (x \cos x+\sin x) dx$.Using integration by parts for $\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x$.So,$\frac{y}{x} = x \sin x + \cos x - \cos x + C = x \sin x + C$.Given $y(\pi)=\pi$,we have $\frac{\pi}{\pi} = \pi \sin(\pi) + C \Rightarrow 1 = 0 + C \Rightarrow C=1$.Thus,$y = x^{2} \sin x + x$.Now,$y\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^{2} \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} = \frac{\pi^{2}}{4} + \frac{\pi}{2}$.Next,$\frac{dy}{dx} = x^{2} \cos x + 2x \sin x + 1$.Then,$\frac{d^{2}y}{dx^{2}} = -x^{2} \sin x + 2x \cos x + 2x \cos x + 2 \sin x = -x^{2} \sin x + 4x \cos x + 2 \sin x$.Evaluating at $x=\frac{\pi}{2}$,$\frac{d^{2}y}{dx^{2}}\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right)^{2} \sin\left(\frac{\pi}{2}\right) + 4\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + 2 \sin\left(\frac{\pi}{2}\right) = -\frac{\pi^{2}}{4} + 0 + 2 = 2 - \frac{\pi^{2}}{4}$.Finally,$y^{\prime \prime}\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right) = \left(2 - \frac{\pi^{2}}{4}\right) + \left(\frac{\pi^{2}}{4} + \frac{\pi}{2}\right) = 2 + \frac{\pi}{2}$.

Let $y=y(x)$ be the solution of the differential equation,$x y^{\prime}-y=x^{2}(x \cos x+\sin x), x>0$. If $y(\pi)=\pi$,then $y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right)$ is equal to

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