I know that this might be awfully late, but I just started learning about complexification this term and thought I would put up my solution--please excuse any incorrect language I might use as it's the idea I am trying to get across.
First we start off by defining a complex analogue to your function:
eq {1}: $ z''+2z'+2z=2e^xe^{ix} $
where $z=RE(y)+i*IM(y)$
Basically, we can recover the original diffEq by extracting the real part of our complex diffEq. The next step is to use the method of undetermined coefficients to find a guess for what our particular complex solution might be. Guess:
eq {2}: $ z=Ae^{x}e^{ix} $ so that:
$ z'=Ae^{x}e^{ix}+iAe^{x}e^{ix} $ and $ z''=i2Ae^{x}e^{ix} $
Plugging this into {1}:
$ i2Ae^{x}e^{ix} + 2(Ae^{x}e^{ix}+iAe^{x}e^{ix}) + 2(Ae^{x}e^{ix})= 2e^xe^{ix} $
We can simplify by removing the common factor of $ e^{x}e^{ix} $:
$ A(4+4i)=2 => A= \frac{1}{2 + 2i}$
Convert A to complex polar form:
$ A=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}} $
Plugging this into {2}:
$ z=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}}e^{x}e^{ix} $
This can be simplified to eq {3}:
$ z=\frac{\sqrt[]{2}}{4}e^{x}e^{i(x-\frac{\pi}{4})} $
Since our particular solution should be of the form $cos(x)$, we take the real part of {3} and call that our particular x-solution:
$ x = RE(z) = \frac{\sqrt[]{2}}{4}e^{x}cos(x-\frac{\pi}{4}) $
Finally using our difference of cosine identity:
$ \frac{\sqrt[]{2}}{4}e^{x}(cos(x)\frac{\sqrt[]{2}}{2} + sin(x)\frac{\sqrt[]{2}}{2}) $ $ \frac{\sqrt[]{2}}{4}\frac{\sqrt[]{2}}{2}e^{x}(cos(x) + sin(x)) $ $ \frac{2}{8}e^{x}(cos(x) + sin(x)) $ $ \frac{1}{4}e^{x}(cos(x) + sin(x)) $
$QED$