Linear Second Order ODE/y'' + k^2 y = sine b x

Theorem
The second order ODE:
 * $(1): \quad y'' + k^2 y = \sin b x$

has the general solution:
 * $y = \begin{cases} C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2} & : b \ne k \\

C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k} & : b = k \end{cases}$

Proof
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:

Substituting into $(1)$:

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:


 * $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k}$

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:

Substituting into $(1)$:

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:


 * $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$