Let y=y(x) be the solution of the differential equation, xy'-y=x2(xcosx+sinx),x>0 . If y(π)=π , then y''π2+yπ2 is equal to

y'-1xy=x(xcos+sinx)I.F. =e∫-1xdx=e-logx=x-1y·1x=∫1xx(xcosx+sinx)dx+Cyx=xsinx+Cput x=π. y(π)=πππ=πsinπ+C⇒C=1yx=xsinx+1y=x2sinx+xy(π2)=π24sinπ2+π2=π24+π2y'=x2cosx+sinx 2x+1 y"=x2(-sinx)+cosx 2x+2(sinx+xcosx) y"(π2)=-π24sinπ2+2π2cosπ2+2(sinπ2+π2cosπ2)           =-π24 +0+2+0 y(π2)+y"(π2)=π2+2