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. 

Suppose you wanted to find the particular solution to $y'=x$. The answer is obviously $x^2$: notice that the degree of $x^2$ is one greater than $x$. You can't assume the particular solution will have degree $\le$ the degree of the inhomogeneous part on the right-hand side of the original ODE.
As for why the constant is left out of the particular solution form, it is redundant. Since there is no term with just "$y$" on the left-hand side, only derivatives of $y$, and constant part will be killed off and so it doesn't need to be considered.
