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

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

has the general solution:
 * $y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$

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

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:
 * $y_g = C_1 \cos 2 x + C_2 \sin 2 x$

We have that:
 * $\map R x = 3 \sin x$

and it is noted that $\sin x$ is not itself a particular solution of $(2)$.

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

for some $A$ and $B$ to be determined.

Hence:

Substituting into $(1)$:

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


 * $y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$