Let y = y(x) be the solution of the differential equation,
xy'- y = x²(xcosx + sinx), x > 0. if y ($\pi$) = $\pi$ then
$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$ is equal to :

$$xy' - y = {x^2}(x{\mathop{\rm cosx}\nolimits} + sinx),\,x &gt; 0,\,y(\pi ) = \pi $$<br><br>$y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\}$<br><br>$I.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}$<br><br>$\therefore$ $y.{1 \over x} = \int {{1 \over x}.x(x\,cos\,x + sin\,x)<br>dx}$<br><br>${{y \over x}}$ = $\int ( x\,\cos \,x + \sin \,x)dx$<br><br>$= x\,\sin \,x + C$<br><br>${{y \over x}}$ = $x\,\sin \,x + C$<br><br>$\Rightarrow y = {x^2} = \sin \,x + cx$<br><br>x = $\pi$, y = $\pi$<br><br>$\pi$ = $\pi$c $\Rightarrow$ C = 1<br><br>$$y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$<br><br>$y' = 2x\sin x + {x^2}\cos x + 1$<br><br>y" = $2\sin x + 2x\cos x + 2x\cos x - {x^2}\sin x$<br><br>y" $\left( {{\pi \over 2}} \right) = 2 - {{{\pi ^2}} \over 4}$ <br><br>$\Rightarrow y\left( {{\pi \over 2}} \right)$ + y" $\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}$