Linear Second Order ODE/y'' - 3 y' + 2 y = 14 sine 2 x - 18 cosine 2 x

Theorem
The second order ODE:
 * $(1): \quad y'' - 3 y' + 2 y = 14 \sin 2 x - 18 \cos 2 x$

has the general solution:
 * $y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$

Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
 * $y'' + p y' + q y = R \left({x}\right)$

where:
 * $p = -3$
 * $q = 2$
 * $R \left({x}\right) = 14 \sin 2 x - 18 \cos 2 x$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $y'' - 3 y' + 2 y = 0$

From Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution:
 * $y_g = C_1 e^x + C_2 e^{2 x}$

We have that:
 * $R \left({x}\right) = 14 \sin 2 x - 18 \cos 2 x$

and it is noted that $14 \sin 2 x - 18 \cos 2 x$ is not itself a particular solution of $(2)$.

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

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:


 * $y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + 2 \sin 2 x + 3 \cos 2 x$