Where does the $t$ come from in the solution to a critically damped differential equation?

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 $2ar+b=0$.

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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:

$$\lim_{h\to0}\frac{e^{(r+h)t}-e^{rt}}h=e^{rt}\lim_{h\to0}\frac{e^{ht}-1}h=e^{rt}t.$$

The general theory guarantees a two-dimensional solution space, whatever the given coefficients $a\ne0$, $b$, $c$. If the characteristic equation has a double root we obtain just one solution of the form $t\mapsto e^{\lambda t}$. There have to be other solutions, and it so happens that the $t$ thrown in works.

But you want to know why $t$, and not $\log t$, or somethig even more complicated. In this regard consider the following example: $$y'' -\lambda y=0\ .$$ If $\lambda\ne0$ we have two different characteristic values, namely $\lambda$ and $0$, and the general solution is $$y(t)=c_1 e^{\lambda t}+c_2\ .$$ The solution space ${\cal L}$ contains in particular the function $$y_{\rm special}(t):={e^{\lambda t}-1\over\lambda}\ .$$ Now, if the underlying dynamical system is continuously modified (by turning a knob) such that $\lambda\to0$ then this special solution will converge to $y_*(t)=t$. It is then plausible that this $y_*(\cdot)$ will in fact be a solution in the limiting case.

We lose nothing by starting with the equation $$ y'' + 2by' + cy = 0, $$ since the coefficient of $y''$ must be nonzero. We can proceed in a couple of different ways:

Reduction of Order

The simplest way is to suppose that $\alpha$ is a root of the characteristic equation $k^2+2bk+c=0$, and substitute $ y=ue^{\alpha x} $. This gives $$ 0 = y'' + 2by' + cy = e^{\alpha x} \left( (\alpha^2+2b\alpha+c)u + u''+ 2(\alpha+b) u' \right). $$ The first term in the bracket vanishes since $\alpha$ is a root, so we end up having to solve $$ u'' + 2(\alpha+b)u' = 0, $$ a first-order equation. Now, suppose that $\beta$ is the other root of the characteristic equation. We have $$ k^2+2bk+c = (k-\alpha)(k-\beta) = k^2 - (\alpha+\beta)k+\alpha\beta, $$ so $2b = -\alpha-\beta$. Hence the new differential equation can be written as $$ u'' + (\alpha-\beta)u' = 0. $$ There are now two possibilities:

• if $\alpha=\beta$, this is $u''=0$, which has solution $At+B$
• on the other hand, if $\alpha \neq \beta$, this equation can be solved by using the integrating factor $e^{(\alpha-\beta)x}$, so $$ (e^{(\alpha-\beta)x}u')' = 0 \implies u' = Ae^{(\beta-\alpha)x}, $$ and hence $$ u = A\frac{e^{(\beta-\alpha)x}}{\beta-\alpha} +B, $$ which obviously only works if $\alpha-\beta \neq 0$.

Limit of the Initial Value Problem

A differential equation normally comes with boundary conditions of some kind; the type that are easiest to use here for understanding are initial conditions, say $y(0) = q$, $y'(0) = p $. We can write down a general solution with unequal roots to this, $$ y = Ae^{-bx}\cos{\omega t} + B e^{-bx} \frac{\sin{\omega t}}{\omega}, $$ where $\omega = \sqrt{c-b^2}$ is the square root of the discriminant. (This solution works for both real and imaginary $\omega$, and is always real, because $(\sin{\omega t})/\omega$ is real in both cases.) Note that this works because the quadratic formula gives the roots as $-b \pm i\omega$.

Now, the solution with the right initial conditions can be found to be $$ y = q e^{-bx}\cos{\omega t} + (p+bq) e^{-bx} \frac{\sin{\omega t}}{\omega} $$ by solving the equations simultaneously, and this is a good reason for writing the solution in the form we did in the first place. Now, we can rewrite this in terms of the roots: $$ b = -\frac{\alpha+\beta}{2}, i\omega = \frac{\alpha-\beta}{2}, $$ where we chose which is $\alpha$ and which $\beta$ so that the sign makes sense. (Since $\cos{\omega t}$ and $(\sin{\omega t})/\omega$ are both even, it actually makes no difference.)

The upshot of all this is that taking $\beta \to \alpha$ is the same as taking $\omega$ to $0$. We have the limits $$ \cos{\omega t} \to 1, \quad \frac{\sin{\omega t}}{\omega} \to t $$ as $\omega \to 0$, so the solution assumes the form we wanted.

The moral of the latter story is that $(\sin{\omega t})/\omega$ is the important function to consider in initial value problems, and has a limit that gives the answer we expect. (This function is also important in Fourier theory, and geometry of surfaces with constant curvature, as it happens.)