Proof: homogeneous second order differential equation This is an homogeneous second order differential equation
$Ay'' + By' + Cy = 0$
Its solution can be retrieved assuming $y = e^{rx}$: in this way, we will have
$y = e^{rx}$
$y' = re^{rx}$
$y'' = r^{2}e^{rx}$
replace:
$Ar^{2}e^{rx} + Bre^{rx} + Ce^{rx} = 0$
$(e^{rx}) (Ar^{2} + Br + C) = 0$
The last is a second degree equation. One can manifest a case in which the roots of this equation are real and coincident. If $r_1 = r_2$, a solution will be the function $y_1 = e^{rx}$ surely. At this point, we must build a linearly independent second solution: we take, for example, $y_2 = xe^{rx}$.
I should prove that $y_2 = xe^{rx}$ is a solution as well: how can I do?
Thank you in advance
 A: The_lost is right in his comment, we show $xe^{rx}$ solves the equation
$Ay''(x) + By'(x) + Cy(x) = 0, \; A \ne 0, \tag 0$
by simple substitution (here I assume $A \ne 0$ to ensure (0) is a bona-fide second order equation); however, it helps to know something about $r$ first, to wit:
According to the quadratic formula, the roots of
$Ar^2 + Br + C = 0, \; A \ne 0, \tag 1$
are
$r_1, r_2 = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}; \tag 2$
we see that there is one root
$r = r_1 = r_2 = -\dfrac{B}{2A} \tag 3$
precisely when
$B^2 - 4AC = 0, \tag 4$
or, since $A \ne 0$,
$C = \dfrac{B^2}{4A}, \tag 5$
or
$\dfrac{C}{A} = \dfrac{B^2}{4A^2} = \left ( \dfrac{B}{2A} \right )^2; \tag 6$
for then,
$Ar^2 + Br + C = A \left (r^2 + \dfrac{B}{A} r + \dfrac{C}{A}  \right )$
$= A \left ( r^2 + \dfrac{B}{A} +  \left ( \dfrac{B}{2A} \right )^2 \right ) = A \left ( r + \dfrac{B}{2A} \right )^2; \tag 7$
now if
$y(x) = x e^{rx}, \tag 8$
we calculate:
$y'(x) = e^{rx} + rxe^{rx}; \tag 9$
$y''(x) = re^{rx} + re^{rx} + r^2xe^{rx} = r^2xe^{rx} + 2re^{rx}, \tag{10}$
whence, via (3), 
$y''(x) + \dfrac{B}{A}y'(x) + \dfrac{C}{A} y(x) =  r^2xe^{rx} + 2re^{rx} + \dfrac{B}{A}(rxe^{rx} + e^{rx}) + \dfrac{C}{A} x e^{rx}$
$= \left (r^2  + \dfrac{B}{A}r  + \dfrac{C}{A} \right ) xe^{rx} + \left (2r + \dfrac{B}{A} \right )e^{rx}$
$= \left ( r + \dfrac{B}{2A} \right )^2 xe^{rx} + 2 \left (r + \dfrac{B}{2A} \right ) e^{rx} = 0; \tag{11}$
it then follows that
$Ay''(x) + By'(x) + Cy(x) = A \left (y''(x) + \dfrac{B}{A}y'(x) + \dfrac{C}{A} y(x) \right ) = 0 \tag{12}$
as well.
A: $$y_2 = x\exp(rx)$$
$$y_2' = (rx+1)\exp(rx)$$
$$y_2^" = (r^2x+2r)\exp(rx)$$
Substitute these into the original equation.
$$\exp(rx) (Ar^2x+2Ar+Brx+B+Cx)=0$$
$$\exp(rx) ((Ar^2+Br+C)x+2Ar+B)=0$$
For the $r$ that you found, we already know that $Ar^2+Br+c=0$. 
Since this is a repeated root, the derivative also vanishes, that is $2Ar+B=0$.
