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

1+ \frac{\\pi}{2}+ {\\pi^{2}}/{4}

0 {{dy}}/{ {dx}}-{{y}}/{ {x}}= {x}({x} \cos{x}+\sin {x}) \Rightarrow{{dy}}/{ {dx}}+ {Py}= {Q} so, I.F. = {e}^{\int- {1}/{ {x}} {dx}}= {1}/{| {x}|} = {1}/{ {x}}( {x}>0) Thus, {{y}}/{ {x}}=\int{1}/{ {x}}( {x}( {x} \cos {x}+\sin {x})) {d} {x} \Rightarrow{{y}}/{ {x}}= {x} \sin {x}+ {C} ∵{y}(\\pi)=\\pi\Rightarrow{C}=1 so, {y}= {x}^{2} \sin {x}+{x} \Rightarrow({y})_{\\pi\/ 2}= {\\pi^{2}}/{4}+ \frac{\\pi}{2} Also, {{dy}}/{ {dx}}= {x}^{2} \cos {x} +2 {x} \sin {x}+1 \Rightarrow{{d}^{2} {y}}/{ {dx}^{2}}=- {x}^{2} \sin {x} +4 {x} \cos {x}+2 \sin {x} \Rightarrow{{{d}^{2} {y}}/{ {dx}^{2}} |}_{\frac{\\pi}{2}}=- {\\pi^{2}}/{4}+2 Thus, {y}_{(\frac{\\pi}{2} )}+{{d}^{2} {y}}/{ {dx}^{2}(\frac{\\pi}{2} )}= \frac{\\pi}{2}+2" >