It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = k^2$ :$\map R x = \sin b x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + k^2 y = 0$ From Linear Second Order ODE: $y'' + k^2y = 0$, this has the general solution: :$y_g = C_1 \sin k x + C_2 \cos k x$ We have that: :$\map R x = \sin b x$ There are two cases to address: :$b = k$ :$b \ne k$ First suppose that $b = k$. It is noted that $\sin b x = \sin k x$ is a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A x \sin k x + B x \cos k x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A x \sin k x + B x \cos k x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A k x \cos k x - B k x \sin k x + A \sin k x + B \cos k x       | c = Derivative of Sine Function, Derivative of Cosine Function, Product Rule }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -A k^2 x \sin k x - B k^2 x \cos k x + A k \cos k x - B k \sin k x + A k \cos k x - B k \sin k x       | c = Derivative of Sine Function, Derivative of Cosine Function, Product Rule }} {{eqn | r = -A k^2 x \sin k x - B k^2 x \cos k x + 2 A k \cos k x - 2 B k \sin k x       | c =  }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -A k^2 x \sin k x - B k^2 x \cos k x + 2 A k \cos k x - 2 B k \sin k x + k^2 \paren {A x \sin k x + B x \cos k x}       | r = \sin k x       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {k^2 - k^2} x \sin k x - 2 B k \sin k x       | r = \sin k x       | c = equating coefficients }} {{eqn | l = B \paren {k^2 - k^2} x \cos k x + 2 A k \cos k x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = - 2 B k       | r = 1       | c =  }} {{eqn | l = 2 A k       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = -\frac 1 {2 k}       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k}$ {{qed|lemma}} Now suppose that $b \ne k$. It is noted that $\sin b x$ is not a particular solution of $(2)$. So from the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A \sin b x + B \cos b x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A \sin b x + B \cos b x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A b \cos b x - B b \sin b x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -A b^2 \sin b x - B b^2 \cos b x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = -A b^2 \sin b x - B b^2 \cos b x + k \paren {A \sin b x + B \cos b x}       | r = \sin b x       | c =  }} {{eqn | ll= \leadsto       | l = A \paren {k^2 - b^2} \sin b x       | r = \sin b x       | c = equating coefficients }} {{eqn | l = B \paren {k^2 - b^2} \cos b x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 0       | c =  }} {{eqn | l = B       | r = \frac 1 {k^2 - b^2}       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$ {{qed}}