2nd Order diffn eqn using Euler's eqn

Need help to find the particular integral $\displaystyle y_p$ of a 2nd order diffn eqn using Euler's eqn $\displaystyle only$;

I know that there are other methods of solution. Just want to understand the principle (Euler's) that i'm trying to apply to my type of solution.

$\displaystyle \frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y = 2sin4x$

$\displaystyle D=\frac{d}{dx}, y_p = $ particular integral,

$\displaystyle (D^2-5D+6)y_p = 2sin4x$

$\displaystyle y_p \times [(D-2)(D-3)] = 2sin4x$ ,

$\displaystyle \therefore y_p = \frac{2sin4x}{(D-2)(D-3)}$

By Euler's eqn $\displaystyle e^{4ix} = cos4x + isin4x$

This is where i'm uncertain,

$\displaystyle \therefore y_p = Real[ \frac{e^{4ix}}{(D-2)(D-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$

$\displaystyle 2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$ because question has $\displaystyle 2sin4x$ on RHS

Question:

1) Should the Real part be included seeing that the question has the imaginary part (sin4x) only?

next, Remember $\displaystyle D = \frac{d}{dx}$

$\displaystyle \therefore y_p = Real \frac{e^{4ix}}{(4i-2)(4i-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $\displaystyle e^{4ix}$

My final answer wasn't correct. Seek help using my method and not another approach.

Thanks.