I'd multiply by e^(-2t) : e^(-2t) y" - 4e^(-2t) y' + 4e^(-2t) y = t but e^(-2t) y" - 4e^(-2t) y' + 4e^(-2t) y = (e^(-2t) y" - 2e^(-2t) y') + (- 2e^(-2t) y' + 4e^(-2t) y) = (e^(-2t) y ' ) ' - 2 (e^(-2t) y ) ' = (e^(-2t) y '- 2 e^(-2t) y) ' = (e^(-2t) y ) " So the differential equation becomes: (e^(-2t) y ) " = t and so: (e^(-2t) y) ' = (1/2) t^2 + a (a is constant) e^(-2t) y = (1/6) t^3 + at + b (a and b constants)
so the general solution is: y = (1/6) t^3 e^(2t) + a t e^(2t)+ be^(2t)
the last two terms are your homogenous solution, the first term is the particular one.
Note that even if the right side was different from te^(2t), I'd still multiply by e^(-2t), since this is what makes the left side become (e^(-2t) y ) " , so that we just have to anti-differentiate two times.
So if we had y"-4y'+4y=t^5 instead, then the problem would become: (e^(-2t) y ) " = t^5 e^(-2t) from which we'd find the solution by finding the primitive of the right side twice in a row, and then multiplying by e^(2t) to get the general solution (again, the constants of integration will give the homogenous part).