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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

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  • $\begingroup$ Can't you do it by a simple derivation? $\endgroup$ – Obvious Aug 18 '18 at 17:29
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$$y_2 = x\exp(rx)$$

$$y_2' = (rx+1)\exp(rx)$$

$$y_2^" = (r^2x+2r)\exp(rx)$$

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 $2Ar+B=0$.

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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.

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