uniqueness of solution of diff equation second order In the case of a diff equation where $r_1=r_2$ the solution to $y''+py'+qy=0$ is
$y=Ce^{r_1 x}+Dxe^{r_1 x}$  (I)
In a proof I have for uniqueness of solution they start by defining $y=u e^{r_1 x}$ if $u''=0$ $ u=Cx+D$ and then $y=(Cx+D)e^{r_1 x}=ue^{r_1 x}$ But why does this show that only solution to y is(I). They also argue like this:
$y''+py'+qy=u''e^{r_1 x}+2r_1 u'e^{r_1 x}+ur_1^2 e^{r_1 x}+p(u'e^{r_1 x}+ur_1 e^{r_1 x}) +qe^{r_1 x}=[u''+(2r_1+p)u']e^{r_1 x}=0  (II)$
$r^2+pr+q=(r-r_1)^2=r^2-2r_1r+r_1^2$  $p=-2r_1$ 
which shows that   (II) gives $u''=0$ which is used above but that does not answer my question why using $y=ue^{r_1 x}$ shows that the only solution is (I)? 
 A: On one hand the existence and uniqueness theorem for second order linear differential equation states:
If $p(t)$, $q(t)$, and $r(t)$ are continuous functions on an interval $[a,b]$, then the initial value problem
$$y'' + py' + qy  =  r \quad\quad y(t_0)  =  y_0,\quad y'(t_0)  =  y'_0$$
has a unique solution defined for all $t$ in $[a,b]$.
On the other hand, a second order linear ODE has two fundamental solutions (it can be put in the form of a $2 \times 2$ linear system). You have found the two fundamental solution, namely, $e^x$ and $xe^x$. Since the ODE is homogeneous, the general solution is a linear combination of your fundamental solutions.
A: The idea in the proof is the following: Suppose you have any solution $y$. Then you can write the solution as $y(x) = u(x)e^{r_1x}$ for some differentiable function $u$ (why?). Through the calculations above, you obtain the condition $u'' = 0$. The solution to this differential equation must have the form $u(x) = C+Dx$ for some $C,D\in\mathbb{R}$, which means that your original solution $y$ must have the form $y(x) = (C+Dx)e^{r_1x}$ for some $C,D\in\mathbb{R}$.
