Annihilator Operators and higher than second order results for aux equation

Hello,

I am solving a DE by the Annihilator method.

I know that if you have a solution for the auxillary repeated then you must add an (x) before the next constant.

i.e. $\displaystyle (D-1)^2$ would be $\displaystyle c1e^x + xc2e^x.$

However, what if I had $\displaystyle D^3?$ or $\displaystyle (D+1)^3?$ Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

So would it be $\displaystyle c1e^-x + xc2e^-x + ((x^2)/2)c3e^-x? $

What do you think?