Yes, your complimentary solution is right (via your characteristic equation) except you want x's and not t's. Let me explain a little bit more about the construction of this particular solution. There are primary two methods to do this (1) the method of undetermined coefficients and (2) the variation of parameters. The first method is usually easier to use but involves some educated guessing. The method basic seeks the form of the particular solution by looking at the rhs of the ODE. If it has power's guess power's, exponential's guess exponential's, sin's and cos's guess sin's and cos's. In your example

it has powers

and exponential's

Therefore one would try a guess of

Taking derivatives

and substituting gives

Expanding gives

(yes, the B's cancel but why?) The reason is that

is part of the complimentary solution so

identically! So in this case, what do we do for a guess. Let us onsider the following first order example

The complimentary equation is

the characteristic equations is

the complimentary solution is

However, since eqn. (1) is linear we solve giving

In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use

substituting gives

and when we expand we get

from which we obtain

and the particular solution