Here is the equation:

$\displaystyle y''+3y'+2y=2cos(2x)-6sin(2x)$

I am supposed to solve this differential equation using 2 different methods. Method 1, I understand fine, but method 2 is what I need help understanding.

METHOD 1: Find complete solution using real trial functions. By using $\displaystyle Asin(2x)+Bcos(2x)$, differentiating it and putting it back into the origional equation I got the answer of $\displaystyle y=C_{1} e^{-x}+C_{2} e^{-2}+\frac{3}{5}sin(2x)+\frac{4}{5}cos(2x)$ which IS correct. I got this method down.

METHOD 2: Here are my given directions for "using a complex valued approach"

a) Solve the equation $\displaystyle y''+3y'+2y=2cos(2x)-6sin(2x)=e^{2ix}$ for a patricular solution $\displaystyle z_{p}=c e^{2ix}$, where $\displaystyle c$ is a complex constant.

b) Expanding $\displaystyle z_{p}=c e^{2ix}=z_{1}(x)+i z_{2}(x)$;

c)Noting that: $\displaystyle z_{1}(x)=Re[ce^{2ix}] and z_{2}(x)=Im[ce^{2ix}]$ are solutions to both solutions to $\displaystyle cos(2x)$ and $\displaystyle sin(2x)$ respectively.

d)Now by taking the right combination of $\displaystyle z_{1}(x), z_{2}(x)$ to obtain a particular solution to the original differential equation.

Hopefully I made what I need to do in method 2 clear, unfortuneatly, I am not clear on how to do it. If anyone would be willing to run throught the solution procedure using this "complex valued approach" I would be thankful.