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. 

Hint:
The particular solution is of the form:
$$y_p = a x + b x^2 + c x e^{-2x}$$
We have to take $a  + b x$ and multiply by $x$ and multiply $e^{-2x}$ by $x$ because we already have a constant in homogeneous and also have $e^{-2x}$ in homogeneous.
