What is the particular solution for $ y''+2y'=2x+5-e^{-2x}$? How is the particular solution for $y''+2y'=2x+5-e^{-2x}$ be the following?
$$y_p = Ax^2 + Bx + Cxe^{-2x}$$
Shouldn't it be $y_p = Ax + B + Cxe^{-2x}$? 
Anything of degree one should be in the form $Ax + B$, and $2x+5$ is in degree one and not squared... I just don't get it. 

Using The Annihilator Method, we have:
$$P(D)y=D^3(D+2)^2y=0$$ so the probable suitable general solution is: $$y(x)=C_1+C_2x+C_3x^2+C_4xe^{-2x}+C_5e^{-2x}$$ Now pick the $y_p$.
