It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 2$ :$q = 1$ :$\map R x = x e^{-x}$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 2 y' + y = 0$ From Linear Second Order ODE: $y'' + 2 y' + y = 0$, this has the general solution: :$y_g = \paren {C_1 + C_2 x} e^{-x}$ It remains to find a particular solution $y_p$ to $(1)$. We have that: :$\map R x = x e^{-x}$ and so from the Method of Undetermined Coefficients for Product of Polynomial and Exponential, we assume a solution: {{begin-eqn}} {{eqn | l = y_p       | r = e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = -e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4} + e^{-x} \paren {3 A_1 x^2 + 2 A_2 x + A_3}       | c =  }} {{eqn | r = e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4} + e^{-x} \paren {-3 A_1 x^2 + 2 \paren {3 A_1 - A_2} x + \paren {2 A_2 - A_3} }       | c =  }} {{eqn | r = e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + 2 A_2 - 2 A_3 + A_4}       | c =  }} {{end-eqn}} Substituting in $(1)$: {{begin-eqn}} {{eqn | o =        | r = e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + 2 A_2 - 2 A_3 + A_4}       | c =  }} {{eqn | o =        | ro= +       | r = 2 e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4}       | c =  }} {{eqn | o =        | ro= +       | r = e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}       | c =  }} {{eqn | r = x e^{-x}       | c =  }} {{eqn | ll= \leadsto       | n = 2       | o =        | r = e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + \paren {2 A_2 - 2 A_3 + A_4} }       | c =  }} {{eqn | o =        | ro= +       | r = e^{-x} \paren {-2 A_1 x^3 + \paren {6 A_1 - 2 A_2} x^2 + \paren {4 A_2 - 2 A_3} x + \paren {2 A_3 - 2 A_4} }       | c =  }} {{eqn | o =        | ro= +       | r = e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}       | c =  }} {{eqn | r = x e^{-x}       | c =  }} {{end-eqn}} Hence by equating coefficients: {{begin-eqn}} {{eqn | l = A_1 - 2 A_1 + A_1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = 0       | c =  }} {{eqn | l = \paren {-6 A_1 + A_2} + \paren {6 A_1 - 2 A_2} + A_2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = 0       | c =  }} {{eqn | l = \paren {6 A_1 - 4 A_2 + A_3} + \paren {4 A_2 - 2 A_3} + A_3       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = 6 A_1       | r = 1       | c =  }} {{eqn | l = \paren {2 A_2 - 2 A_3 + A_4} + \paren {2 A_3 - 2 A_4} + A_4       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A_2       | r = 0       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$ {{qed}}