Linear Second Order ODE/y'' + 4 y = 3 sin 2 x

Theorem
The second order ODE:
 * $(1): \quad y'' + 4 y = 3 \sin 2 x$

has the general solution:
 * $y = C_1 \sin k x + C_2 \cos k x - \dfrac 3 4 x \cos 2 x$

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:
 * $p = 0$
 * $q = 4$
 * $\map R x = 3 \sin 2 x$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' + 4 y = 0$

From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution:
 * $y_g = C_1 \sin 2 x + C_2 \cos 2 x$

It is noted that $3 \sin 2 x$ is a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for Sine and Cosine:
 * $y_p = A x \sin 2 x + B x \cos 2 x$

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:


 * $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac 3 4 x \cos 2 x$

is the general solution to $(1)$.