Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Exponential Form

Proof
From Second Order ODE: $y'' - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is:
 * $y_g = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$

We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.

From the Method of Undetermined Coefficients for Sine and Cosine:
 * $y_p = A \cos 3 x + B \sin 3 x$

where $A$ and $B$ are to be determined.

The of $(1)$ is the real part of $\paren {2 + i} e^{3 i x}$.

Thus, to find a particular solution of $(1)$ when the is $\paren {2 + i} e^{3 i x}$, we substitute $y = A e^{3 i x}$.

Hence:

{{eqn | r = -\dfrac {\paren {2 + i} \paren {14 - 6 i} {14^2 + 6^2} | c = }}

Taking the real part of the corresponding solution $A e^{3 i x}$, we get:

Hence the result:
 * $y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$