When solving homogeneous second order constant coefficient linear ODEs ($ay''+by'+cy=0$), there are three 'cases' that solutions fall into, based on the roots $r_1$, $r_2$ of the characteristic equation $ar^2+br+c=0$:
- $r_1, r_2 \in \Bbb R,\; r_1 \ne r_2$, aka "overdamped", for which the general solution is of the form $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$.
- $r_1, r_2 \in \Bbb C,\; r_1=\overline{r_2}$, aka "underdamped", for which the general solution is $y(t) = c_1 e^{\Re(r)\,t}
\cos\Im(r)\,t + c_2 e^{\Re(r)\,t} \sin\Im(r)\,t$. - $r_1, r_2 \in \Bbb R,\; r_1 = r_2$, aka "critically damped", for which the general solution is $y(t) = c_1 e^{rt} + c_2 t e^{rt}$.
I understand the underdamped solution as essentially the same thing as the overdamped solution, equivalent to picking complex $c_1, c_2$ such that $y(t)$ is always real. But the critically damped case doesn't seem to fit neatly as a generalization of the other cases – that extra 't' makes it at least appear like it's something else entirely.
Where does the "$t$" actually come from in the critically damped case?
Why don't the other cases require or allow this?
Best Answer
For a critically damped system, you have a double root so that the usual exponentials $e^{r_1t}$ and $e^{r_2t}$ merge in $e^{rt}$ and no longer form two linearly independent functions. Hence they cannot generate all solutions.
Now if you plug $te^{rt}$ in the equation, you get
$$a(2r+r^2t)e^{rt}+b(1+rt)e^{rt}+cte^{rt}=2ar+b+(ar^2+br+c)t.$$ The $t$ term obviously cancels out (whatever the damping), and so does the constant one because the double root verifies $2ar+b=0$.
A heuristic reasoning can justify the choice of such a term.
Let the damping converge to the critical one, with roots $r$ and $r+h$ getting closer and closer. We consider two independent solutions $e^{rt}$ and $e^{(r+h)t}-e^{rt}$, and amplify the second to avoid cancellation:
$$\lim_{h\to0}\frac{e^{(r+h)t}-e^{rt}}h=e^{rt}\lim_{h\to0}\frac{e^{ht}-1}h=e^{rt}t.$$