How to design a differential equation to match a given general solution? I am in a first year differential equations course, and in class on Friday, the teacher did a problem from the book that I wasn't quite sure how to solve (yet I'm sure has a possibility of showing up on a test!). 
The question I have written in my notes is: "create a differential equation that has $ y = C_1e^{-2x} + C_2e^{3x} + C_3xe^{3x} + e^x + x^2 + x $ as its general solution".
How would I go about doing this?
I see that the general solution has $ C_1e^{-2x} + C_2e^{3x} + C_3xe^{3x} $, meaning the characteristic equation for the homogeneous equation ($Y_h$) should have roots $-2, 3, 3$. I guess for that I should make a polynomial which yields these roots?
I also notice that the particular solution ($Y_p$) should be in the form $Ae^x + Bx^2 + Cx$. If we assume the DE I make is in the form $y'' + p(x)y' + q(x)y = f(x)$ - sorry if this isn't standard convention!), then $f(x)$ should contain something like $e^x + x^2 + x$ ? 
Perhaps I'm way off.

Yes, you're on the right track.
Presuming you're looking for a linear d.e. with that general solution, note that since the solution space has three parameters, the d.e. must be third order (also, since $0$ is evidently not a solution, we know immediately that the d.e. cannot be homogeneous).
First, we find the (homogeneous) linear, constant-coefficient d.e. whose general solution is
$$C_1 e^{-2x} + C_2 e^{3x} + C_3 x e^{3x}.$$
Like you say, this form tells us that the characteristic polynomial of the d.e. has roots $-2, 3, 3$ and so it is
$$(r + 2) (r - 3)^2 = r^3 - 4 r^2 - 3 r + 18,$$
and the corresponding homogeneous d.e. is
$$y''' - 4 y'' - 3 y' + 18 y = 0.$$
Now, to account for the inhomogeneous part $e^x + x^2 + x$, we can simply substitute this expression in the l.h.s: We get
$$(e^x) - 4(e^x + 2) - 3(e^x + 2x + 1) + 18(e^x + x^2 + x) = 12 e^x + 18 x^2 + 12 x - 11.$$
In particular $e^x + x^2 + x$ is a solution of the inhomogeneous d.e.
$$\phantom{(\ast)} \qquad y''' - 4 y'' - 3 y' + 18 y = 12 e^x + 18 x^2 + 12 x - 11, \qquad (\ast)$$
and the difference of any two solutions of this equation is a solution $C_1 e^{-2x} + C_2 e^{3x} + C_3 x e^{3x}$ to the corresponding homogeneous equation; thus the general solution to $(\ast)$ is $$C_1 e^{-2x} + C_2 e^{3x} + C_3 x e^{3x} + e^x + x^2 + x$$ as desired.
