solve DE: ${y}''-6{y}'+9y=0$ how to factor when y=x$e^{rx}$

Question

solve DE: ${y}''-6{y}'+9y = 0$

Guess => $y=e^{rx}$

${y}'=re^{rx}$

${y}''=r^{2}e^{rx}$

backsub into y, y', & y'' into DE.

$e^{rx}(r^{2} - 6r + 9) = 0$

$e^{rx}(r-3)^{2} = 0$

$r = 3$

Now the textbook says that I need to find another root since its a double root.
So we are are to guess:

$y = xe^{rx}$

${y}' = e^{rx} + rxe^{rx}$

${y}'' = 2re^{rx} + r^{2}xe^{rx}$

${y}''-6{y}'+9y = 0$

back sub y, y', y'' into DE again:

$2re^{rx} + r^{2}xe^{rx} + -6 (e^{rx} + rxe^{rx}) + 9xe^{rx} = 0$

$e^{rx} (2r + r^{2}x + -6 - rx + 9x) = 0$

Ok here's where I get stuck, how to factor this beast?
The textbook is mute on the subject.

Answer 1

Let / guess $y_2(x) = xe^{rx}$, where $r = 3$ as previously found, then

$$\begin{align*} y_2'(x) &= e^{rx} + rxe^{rx}\\ y_2''(x) &= re^{rx} + re^{rx} + r^2xe^{rx}\\ &= 2re^{3x} + r^2xe^{rx}\\ y_2'' -6y_2' + 9y_2 &= 2re^{rx} + r^2xe^{rx} -6\left( e^{rx} + rxe^{rx}\right) + 9xe^{rx}\\ &= (2r-6)e^{rx} + (r^2-6r+9)xe^{rx}\\ &= 0 \end{align*}$$

So $y_2(x)$ and hence $c_2y_2(x)$ are solutions to the differential equation.

This is not a coincidence, because it is known that $r=3$ is a root of $r^2-6r+9 = 0$ and, since $r=3$ is a double root, the first derivative of $r^2-6r+9$:

$$\frac d{dr} (r^2-6r+9) = 2r-6$$

is zero at $r=3$.

Answer 2

There is no "beast",

$$2r + r^{2}x-6 - rx + 9x=0$$ when both

$$2r-6=0$$ and $$r^2-r+9=0,$$

which is not possible.

But looking closely, we find a typo and the second equation should be

$$r^2-6r+9=0.$$

You can conclude.