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