$$y''-5y'+6y=e^{2x}$$ The characteristic polynomial is: $$r^2-5r+6 =0 \implies (r-2)(r-3)=0$$ $$ \implies S_r= \{2,3\}$$ The solution of the homogeneous equation is : $$y(x)=c_1e^{2x}+c_2e^{3x}$$ So the particular solution should be $$y_p(x)=Axe^{2x}$$ Normally the guess should be $Ae^{2x}$. But since $e^{2x}$ is already solution of the homogeneous equation, you need to multiply by $x$ the guess.

EDIT

A good exercice is to solve the following equation : $$y''-4y'+4y=e^{2x}$$ Try it ..

The intrinsic structure of this phenomenon can be seen via the operator viewpoint:

If ${D_x}^2(y) - 5·D_x(y) + 6 y = e^{2x}$ then:

  $( {D_x}^2 - 5·D_x + 6 )(y) = e^{2x}$.

  $(D_x-2)( {D_x}^2 - 5·D_x + 6 )(y) = (D_x-2)(e^{2x})$.

  $(D_x-2)(D_x-2)(D_x-3)(y) = 0$.

It should be clear that the extra polynomial factor arises from the multiplicity of $(D_x-2)$.

Of course, there are other details to fill in, but the intrinsic structure is here. An exactly analogous phenomenon happens with the forward difference operator for the same reason.