Let $ y=y(x) $ be the Solutions of the differential equation, $ x y' -y=x^{2}(x \cos x+\sin x), x0 $. If $ y(\pi)=\pi $, then. $ y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right) $ is equal to : [JEE Main 2020, 4 Sep (Shift 1)]

Question

Let $ y=y(x) $ be the Solutions of the differential equation, $ x y' -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 : [JEE Main 2020, 4 Sep (Shift 1)]

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Let \( y=y(x) \) be the Solutions of the differential equation, \( x y' -y=x^{2}(x \cos x+\sin x), x>0 \). If \( y(\p…

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Let \( y=y(x) \) be the Solutions of the differential equation, \( x y' -y=x^{2}(x \cos x+\sin x), x&gt;0 \). If \( y(\p…

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Let \( y=y(x) \) be the Solutions of the differential equation, \( x y' -y=x^{2}(x \cos x+\sin x), x>0 \). If \( y(\p…