This is too long for a comment, so I posted it as an answer. First solve for the homogeneous equation y'' - 3y' + 2y =0 by setting the right hand side to be zero. The auxiliary equation is $m^2-3m+2=0$, which has roots $m=2,1$. Therefore the solution for this homogeneous equation is $e^x$ and $e^{2x}$. Now we want to find a particular solution y'' - 3y' + 2y = 2e^x. Normally we set the particular solution to be $Ae^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with $x$ until no duplication occurs. Therefore, the particular solution is given by $Axe^{x}$.
Let me do another example: to solve y'' - 2y' + y = 2e^x. First solve the homogenous equation y'' - 2y' + y = 0. The auxiliary equation is $m^2-2m+1=0$ which has double roots $m=1$. Therefore, the solution for this homogeneous equation is $e^x$ and $xe^{x}$. Now if we want to find a particular solution y'' - 2y' + y = 2e^x. Normally we set the particular solution to be $Ae^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with $x$ and it becomes $Axe^x$, but it still duplicates with $xe^{x}$. Therefore, we mupltiply it by $x^2$, and the particular solution is given by $Ax^2e^{x}$.