Solving for an auxiliary equation, if you come up with an exponent higher than 2, how do you write the third coeffecient? 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. [math](D-1)^2[/math] would be [math]c1e^x + xc2e^x.[/math] However, what if I had [math]D^3?[/math] or [math](D+1)^3?[/math] Judging by what happens in a square, I am guessing I am integrating? But I am unsure. So would it be [math] c1e^-x + xc2e^-x + ((x^2)/2)c3e^-x? [/math] What do you think? edit: ooooh, is there a way to write math in the code like that?