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

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

The correct answer is option (4) : $2+\frac{\pi }{2}$

We have,

$x\frac{dy}{x}-y = x^2 (x cos x + sin x)$

$⇒\frac{dy}{dx}-\frac{1}{x}y=x(x\, cos x+sinx)$ ..............(i)

This is a linear differential equation with

$I.F. =e^{-∫1/xdx}=e^{-logx}=1/x.$ Multiplying both sides of (i) by 1/x and integrating, we obtain

$y×\frac{1}{x}=∫(xcosx+sinx) dx$

$⇒\frac{y}{x}=xsin x + C$

$⇒y=x^2 sinx +cx$ ...................(ii)

It is given that $ y = \pi $ when $x= \pi $. Putting $x = \pi , y = \pi $ in(ii), we  obtain

$\pi = C\pi ⇒ C= 1.$

Putting $C=1$ in (ii), we obtain

$y = x^2 sin x+ x$

$⇒y'= 2xsinx +x^2 cos x+ 1$

and , $y''=4xcos x -x^2 sin x + 2sin x $

$∴y''\left(\frac{\pi }{2} \right) =-\frac{\pi^2}{4} + 2 $ and $y \left(\frac{\pi }{2} \right) = \frac{\pi^2}{4}+\frac{\pi }{2}$

Hence, $y''\left(\frac{\pi }{2} \right) + y \left(\frac{\pi }{2}\right) =\frac{\pi }{2} + 2$