The "characteristic equation" for this differential equation is \(\displaystyle r^2- 5r+ 6= (r 3)(r- 2)= 0\) so the "characteristic numbers" are 2 and 3. That means that the general solution to the "associated homogeneous equation, y''- 5y'+ 6y= 0, is \(\displaystyle Ae^{2x}+ Be^{3x}\).
Normally, with the "non-homogeneous" part, \(\displaystyle 3e^{3x}\), involving \(\displaystyle e^{3x}\) we would try a solution to the entire equation of the form \(\displaystyle y= Ae^{3x}\). But since that is already a solution to the associated homogeneous equation we try, instead, \(\displaystyle y= Axe^{3x}\).
Then \(\displaystyle y'= Ae^{3x}+ 3Axe^{3x}\) and \(\displaystyle y''= 6Ae^{3x}+ 9Axe^{3x}\).
\(\displaystyle y''- 5y'+ 6y= 6Ae^{3x}+ 9Axe^{3x}- 5Ae^{3x}- 15Axe^{3x}+ Axe^{3x}= Ae^{3x}= 3e^{3x}\) so A= 3.
The general solution to the differential equation is \(\displaystyle y(x)= Ae^{2x}+ Be^{3x}+ 3xe^{3x}\).
