Getting the differential equation back from its solutions. 
Find a linear differential equation with constant coefficients satisfied by all the given
  functions.$u_1(x) = x^2$, $u_2(x) = e^x$, $u_3(x) = xe^x$.

How do I proceed with this? I know that 1 is a twice repeated root of characteristic equation, but what can I say about $x^2$? If $1$ and $x$ are also given as solutions, I could have claimed 0 is a thrice repeated root of the characteristic equation. Can I still do this or is there a third order DE that satisfies all the functions?
 A: 
''I know that $1$ is a twice repeated root of characteristic equation''

So you know that $k=1$ solves the characteristic equation $ak^2 + bk + c =0$. Then, you could have, for example $a = 1, b = -2, c = 1$. Since the characteristic equation corresponds directly to the homogeneous linear SODE, we have
\begin{equation}
y'' - 2y' + y = 0
\end{equation}
This has the complementary solution:
\begin{equation}
y(x) = C_1e^x + C_2xe^x
\end{equation}
To match your given solution $u_2(x) = e^x$ and $u_3(x) = xe^x$, you need to thrown in an initial condition like $y(0) = 3$ and $y'(0) = \pi$ or something so that $C_1 = C_2 = 1$. This is easy, I'll leave you to do it.
The tricky bit is how do you get the particular solution to be $u_3(x) = x^2$? 
If you want $u_3(x) = x^2$ to be a particular solution, it has to solve
\begin{equation}
y'' - 2y' + y = \text{something}
\end{equation}
Well, let's forget about the something part and just focus on the LHS. Suppose $y = x^2$. Then, $y' = 2x$ and $y'' = 2$. Plugging this into the LHS we have:
\begin{equation}
y'' - 2y' + y = x^2 - 4x + 2
\end{equation}
Turns out $x^2 - 4x + 2$ is the something part that we needed. So in fact, the last equation above is a linear SODE which $u_1, u_2$ and $u_3$ satisfies. Of course, you still need the initial condition bit.
