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: :$p = 2$ :$q = 5$ :$\map R x = x \sin x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 2 y' + 5 y = 0$ From Linear Second Order ODE: $y'' + 2 y' + 5 y = 0$, this has the general solution: :$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$ 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 Function of Sine and Cosine, we assume a solution: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {A_1 \sin x + A_2 \cos x} \paren {B_1 x + B_2}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = \paren {A_1 \cos x - A_2 \sin x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \sin x + A_2 \cos x}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \cos x - A_2 \sin x} + B_1 \paren {A_1 \cos x - A_2 \sin x}       | c =  }} {{eqn | r = \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + 2 B_1 \paren {A_1 \cos x - A_2 \sin x}       | c =  }} {{end-eqn}} Substituting in $(1)$: {{begin-eqn}} {{eqn | o =        | r = \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + 2 B_1 \paren {A_1 \cos x - A_2 \sin x}       | c =  }} {{eqn | o =        | ro= +       | r = 2 \paren {\paren {A_1 \cos x - A_2 \sin x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \sin x + A_2 \cos x} }       | c =  }} {{eqn | o =        | ro= +       | r = 5 \paren {\paren {A_1 \sin x + A_2 \cos x} \paren {B_1 x + B_2} }       | c =  }} {{eqn | r = x \sin x       | c =  }} {{eqn | ll= \leadsto       | o =        | r = x \sin x \paren {-A_1 B_1 - 2 A_2 B_1 + 5 A_1 B_1}       | c =  }} {{eqn | o =        | ro= +       | r = x \cos x \paren {-A_2 B_1 + 2 A_1 B_1 + 5 A_2 B_1}       | c =  }} {{eqn | o =        | ro= +       | r = \sin x \paren {-A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1 + 5 A_1 B_2}       | c =  }} {{eqn | o =        | ro= +       | r = \cos x \paren {-A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1 + 5 A_2 B_2}       | c =  }} {{eqn | r = x \sin x       | c =  }} {{eqn | n = 2       | ll= \leadsto       | o =        | r = x \sin x \paren {4 A_1 B_1 - 2 A_2 B_1}       | c =  }} {{eqn | o =        | ro= +       | r = x \cos x \paren {4 A_2 B_1 + 2 A_1 B_1}       | c =  }} {{eqn | o =        | ro= +       | r = \sin x \paren {4 A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1}       | c =  }} {{eqn | o =        | ro= +       | r = \cos x \paren {4 A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1}       | c =  }} {{eqn | r = x \sin x       | c =  }} {{end-eqn}} Hence by equating coefficients: {{begin-eqn}} {{eqn | l = 4 A_1 B_1 - 2 A_2 B_1       | r = 1       | c =  }} {{eqn | l = 4 A_2 B_1 + 2 A_1 B_1       | r = 0       | c =  }} {{eqn | l = 4 A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1       | r = 0       | c =  }} {{eqn | l = 4 A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1       | r = 0       | c =  }} {{end-eqn}} {{finish|possibly the least fun you can have while doing mathematics}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$