Particular solution to y″−3y′+2y=2ex

${\displaystyle y{}^{″}-3y^{\prime }+2y=2e^x}$.
Use operator D:
Let ${\displaystyle D=\frac{d}{dx}}$.
So, ${\displaystyle (D^2-3D+2)y_p=2e^x\Rightarrow (D-1)(D-2)y_p=2e^x\Rightarrow }$
${\displaystyle y_p=\frac{1}{(D-1)(D-2)}\cdot 2e^x}$
${\displaystyle =2e^x\frac{1}{(D+1-1)(D+1-2)}\cdot 1=2e^x\frac{1}{D(D-1)}\cdot 1\Rightarrow }$
${\displaystyle y_p=2e^x\frac{1}{D(0-1)\cdot 1}=-2e^x\frac{1}{D}\cdot 1=-2x{e}^x}$

This is too long for a comment, so I posted it as an answer. First solve for the homogeneous equation ${\displaystyle y{}^{″}-3y^{\prime }+2y=0}$ by setting the right hand side to be zero. The auxiliary equation is ${\displaystyle m^2-3m+2=0}$, which has roots ${\displaystyle m=2,1}$. Therefore the solution for this homogeneous equation is ${\displaystyle e^x\text{and}{e}^{2x}}$. Now we want to find a particular solution ${\displaystyle y{}^{″}-3y^{\prime }+2y=2e^x}$. Normally we set the particular solution to be ${\displaystyle A{e}^x}$. 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 ${\displaystyle Ax{e}^x}$.
Let me do another example: to solve ${\displaystyle y{}^{″}-2y^{\prime }+y=2e^x}$.
First solve the homogenous equation ${\displaystyle y{}^{″}-2y^{\prime }+y=0}$. The auxiliary equation is ${\displaystyle m^2-2m+1=0}$ which has double roots ${\displaystyle m=1}$. Therefore, the solution for this homogeneous equation is ${\displaystyle e^x\text{and}x{e}^x}$. Now if we want to find a particular solution ${\displaystyle y{}^{″}-2y^{\prime }+y=2e^x}$. Normally we set the particular solution to be $A{e}^x$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with x and it becomes ${\displaystyle Ax{e}^x}$, but it still duplicates with ${\displaystyle x{e}^x}$. Therefore, we mupltiply it by ${\displaystyle x^2}$, and the particular solution is given by ${\displaystyle A{x}^2{e}^x}$.

The simplest thing in your case is probably to use this (easily justified) trick: If p(x) is a one variable polynomial with complex coefficients and c a complex number, then the ODE $p(\frac{d}{dx})y=e^{cx}$ has a solution of the form $e^{cx}q(x)$ where q(x) is a polynomial whose degree is the number of roots of p(x) equal to c. In you case, $p(x)=(x-1)(x-2),c=1$, so the trick tells you to look for a solution of the form $(ax+b)e^x$, and you can clearly assume $b=0$.