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. 
 A: 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.
A: 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 $y_p$.
A: 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.
A: Note that the "lowest" derivative in the differential equation is $y'$. If your solution was to be chosen then the $Ax+B$ will become $A$. Thus the highest degree polynomial expression you can get with your solution is a constant. Essentially this means you will be lacking an $x$ term on the LHS. This is what the $x^2$ term in $y_p = Ax^2 + Bx + Cxe^{-2x}$ does. It provides the $x$ term and is why
$$y_p = Ax^2 + Bx + Cxe^{-2x}$$
and not
$$y_p = Ax + B + Cxe^{-2x}$$
