Answer
$$\eqalign{
& \left( {\bf{a}} \right)x{e^{ - x}} - {x^2}{e^{ - x}} = x{e^{ - x}} - {x^2}{e^{ - x}} \cr
& \left( {\bf{b}} \right)x{e^{ - {x^2}/2}} - {x^3}{e^{ - {x^2}/2}} = x{e^{ - {x^2}/2}} - {x^3}{e^{ - {x^2}/2}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)y = x{e^{ - x}} \cr
& {\text{Calculate }}y' \cr
& y' = \frac{d}{{dx}}\left[ {x{e^{ - x}}} \right] \cr
& y' = x\left( { - {e^{ - x}}} \right) + {e^{ - x}} \cr
& y' = {e^{ - x}} - x{e^{ - x}} \cr
& {\text{Substituting }}y{\text{ and }}y'{\text{ into the equation }}xy' = \left( {1 - x} \right)y \cr
& x\left( {{e^{ - x}} - x{e^{ - x}}} \right) = \left( {1 - x} \right)x{e^{ - x}} \cr
& x{e^{ - x}} - {x^2}{e^{ - x}} = x{e^{ - x}} - {x^2}{e^{ - x}} \cr
& {\text{Therefore, }}y = x{e^{ - x}}{\text{ satisfies the equation}} \cr
& \cr
& \left( {\bf{b}} \right)y = x{e^{ - {x^2}/2}} \cr
& {\text{Calculate }}y' \cr
& y' = \frac{d}{{dx}}\left[ {x{e^{ - {x^2}/2}}} \right] \cr
& y' = x\left( { - x{e^{ - {x^2}/2}}} \right) + {e^{ - {x^2}/2}} \cr
& y' = {e^{ - {x^2}/2}} - {x^2}{e^{ - {x^2}/2}} \cr
& {\text{Substituting }}y{\text{ and }}y'{\text{ into the equation }}xy' = \left( {1 - {x^2}} \right)y \cr
& x\left( {{e^{ - {x^2}/2}} - {x^2}{e^{ - {x^2}/2}}} \right) = \left( {1 - {x^2}} \right)x{e^{ - {x^2}/2}} \cr
& x{e^{ - {x^2}/2}} - {x^3}{e^{ - {x^2}/2}} = x{e^{ - {x^2}/2}} - {x^3}{e^{ - {x^2}/2}} \cr
& {\text{Therefore, }}y = x{e^{ - {x^2}/2}}{\text{ satisfies the equation}} \cr} $$
