the repeated-root case
Sometimes the characteristic equation has a double root — its two solutions coincide at a single value r. The exponential guess then hands you only ONE solution, e^(r x), but a second-order equation needs two independent ones to fill its solution space. So where does the missing second solution come from? The answer is one of the prettiest small surprises in the subject: it is x times the first.
When a r^2 + b r + c = 0 has the repeated root r = -b/(2a) (the case b^2 - 4ac = 0), the two members of the fundamental set are e^(r x) and x e^(r x). You can verify the second works by direct substitution, or derive it by reduction of order: assume a second solution of the form v(x) e^(r x), plug in, and the resulting equation for v collapses (because r is a double root) to v'' = 0, whose simplest non-constant answer is v = x. Hence the extra factor of x. The general solution is therefore y = (c1 + c2 x) e^(r x): an exponential multiplied by a linear polynomial.
This is exactly the borderline between the over- and under-damped worlds — physically the critically damped case, the fastest possible return to equilibrium without oscillation, as in a well-tuned door closer or a car suspension at the verge of bouncing. The same x-multiplier pattern generalises: a root of multiplicity m in a higher-order equation contributes e^(r x), x e^(r x), ..., x^(m-1) e^(r x). The lesson is that a repeated root forces you to reach beyond pure exponentials to recover full dimension.
y'' - 6 y' + 9 y = 0 has r^2 - 6 r + 9 = (r - 3)^2 = 0, a double root r = 3. So {e^(3x), x e^(3x)} is the fundamental set and y = (c1 + c2 x) e^(3x). Substituting x e^(3x) gives 0, confirming the x-factor really works.
A double root supplies only one exponential; multiplying by x produces the needed independent second solution.
The second solution is x e^(r x), not e^(2 r x) or x^2 e^(r x). A common slip is to invent a second exponent; the truth is the SAME exponent r, multiplied by x, and the x-factor is forced by reduction of order.