Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x + 6 cosine x + 8 x^2 - 4 x

Theorem
The second order ODE:
 * $(1): \quad y'' + 4 y = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$

has the general solution:
 * $y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$

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:
 * $p = 0$
 * $q = 4$
 * $\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$

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

We have that:
 * $\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$

Let:
 * $\map R x = \map {R_1} x + \map {R_2} x + \map {R_3} x$

where:
 * $\map {R_1} x = 4 \cos 2 x$
 * $\map {R_2} x = 6 \cos x$
 * $\map {R_3} x = 8 x^2 - 4 x$

Consider in turn the solutions to:
 * $y'' + p y' + q y = \map {R_1} x$

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


 * $y'' + p y' + q y = \map {R_2} x$

From Second Order ODE: $y'' + 4 y = 6 \cos x$, this has the general solution:
 * $y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$


 * $y'' + p y' + q y = \map {R_3} x$

From Second Order ODE: $y'' + 4 y = 8 x^2 - 4 x$, this has the general solution:
 * $y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$

It follows from Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part that the general solution to $(1)$ is:
 * $y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$