Second order constant coefficients (I)

Problem:

$y'' + 4y' + 4y = 0$

Solution:

Note that we do not have the independent variable $x$, in principle we could have reduce to order to 1, but for this one let's try the solution $y = e^{rx}$. Putting the solution in the problem, we have

$r^2e^{rx} + 4re^{rx} + 4e^{rx} = 0$

Solving, get $r = -2$ (repeated).

The repeated solution worried me because then we have only one degree of freedom.

Let's try $y = xe^{rx}$, $y' = e^{rx} + rxe^{rx}$, $y'' = re^{rx} + re^{rx} + r^2xe^{rx}$

$y'' + 4y' + 4y = (re^{rx} + re^{rx} + r^2xe^{rx}) + 4(e^{rx} + rxe^{rx}) + 4(xe^{rx})$
$= e^{rx}((r + r + r^2x) + 4(1 + rx) + 4(x))$
$= e^{rx}(2r + 4+ (r^2 + 4r + 4))$

So if we put $r = -2$, the equation still work out, so the solution is

$Ae^{-2x} + Bxe^{-2x}$.

To be honest, I kind of know $xe^{-2x}$ is an answer when I see the double root. I know it because of my experience (sort of my teacher told me so thing). But wouldn't it be more satisfying if we figured out the general rule and why the general rule work? Let's investigate in the next post.