When $b^{2} = 4ac$ the characteristic equation has a single repeated root $r = -\frac{b}{2a}$. The exponential $e^{rt}$ gives one solution, but a second-order equation needs two independent ones.
Key idea
A repeated root $r$ gives the solutions $e^{rt}$ and $te^{rt}$, so $$y = (c_1 + c_2t)e^{rt}.$$
Why it works
Write the equation as $a(D - r)^{2}y = 0$, where $D$ means $\frac{d}{dt}$. Apply $D - r$ to $te^{rt}$: $(te^{rt})' - rte^{rt} = e^{rt}$. Apply it again: $(D - r)e^{rt} = 0$. So $te^{rt}$ is a solution. Another way to see it: for two close roots $r_1, r_2$, the combination $\frac{e^{r_2t} - e^{r_1t}}{r_2 - r_1}$ is a solution, and as $r_2 \to r_1$ it becomes the derivative of $e^{rt}$ with respect to $r$, which is $te^{rt}$. The Wronskian $W(e^{rt}, te^{rt}) = e^{2rt}$ is never 0.
Worked example: Critically damped
Solve $y'' + 4y' + 4y = 0$, $y(0) = 3$, $y'(0) = -1$.
- $r^{2} + 4r + 4 = (r + 2)^{2}$: $r = -2$ twice.
- $y = (c_1 + c_2t)e^{-2t}$, so $y(0) = c_1 = 3$.
- $y' = c_2e^{-2t} - 2(c_1 + c_2t)e^{-2t}$, so $y'(0) = c_2 - 2c_1 = -1$ and $c_2 = 5$.
Answer: $y = (3 + 5t)e^{-2t}$
Worked example: A fractional root
Find the general solution of $9y'' - 6y' + y = 0$.
- $9r^{2} - 6r + 1 = (3r - 1)^{2}$: $r = \frac{1}{3}$ twice.
Answer: $y = (c_1 + c_2t)e^{t/3}$
Worked example: Starting at zero
Solve $y'' - 2y' + y = 0$, $y(0) = 0$, $y'(0) = 2$.
- $(r - 1)^{2} = 0$, so $y = (c_1 + c_2t)e^{t}$.
- $y(0) = c_1 = 0$, and $y'(0) = c_2 + c_1 = 2$.
Answer: $y = 2te^{t}$
Common mistake
Writing $y = c_1e^{rt} + c_2e^{rt}$. That is just $(c_1 + c_2)e^{rt}$, one constant in disguise, and it cannot match both $y(0)$ and $y'(0)$ in general (try the first example). The second solution needs the factor $t$; check independence with the Wronskian.
Physically, a repeated root is the borderline case. For a spring $my'' + cy' + ky = 0$ it happens when $c^{2} = 4mk$, called critical damping: a little less damping and the spring oscillates, a little more and it creeps back slowly. A critically damped solution $(c_1 + c_2t)e^{rt}$ crosses zero at most once, since $c_1 + c_2t$ has at most one root, and when $r < 0$ it still decays to 0, because the exponential eventually beats the factor $t$. That is why door closers and car suspensions are tuned close to critical damping.