And you would have to split the particular solution up because the non homogeneous part is the sum of two terms?
In other words find Yp for then find Yp for and the two of them together make up the particular solution for the original DE?
From the homogenous solution to the differential you find that solutions are:
.
However, when you go to solve for the two particular solutions to the differential, you notice that you already have an exponential to the 2x as a solution of your homogenous. So you need to multiple your original "guess" by , where k is the smallest exponent needed to get a unique guess that is not a solution to our homogenous differential.
Therefore I would solve my particular as:
Now substituting into your original d.e.:
Terms cancel out leaving:
Therefore our particular solution for our exponential is