Question Recap
I'd like to start with a quick recap taken directly from my question:
$$\begin{align*}
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A}
\\\\
\frac{\text{d}}{\text{d}t}\left[\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t\right.&=\bigg.0\:\text{A}\bigg]
\\\\
\frac1{R_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+C_1\frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac{v_{_\text{C}}}{L_1}&=0\:\text{A}
\\\\
\frac{\text{d}^2}{\text{d}t^2}v_{_\text{C}}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac{1}{L_1\,C_1}v_{_\text{C}}&=0\:\text{A}
\\\\
\left[\frac{\text{d}^2}{\text{d}t^2}+\frac1{R_1\,C_1}\frac{\text{d}}{\text{d}t}+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A}
\\\\
\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}&=0\:\text{A}
\end{align*}$$
I had snuck in that \$D=\frac{\text{d}}{\text{d}t}\$ in the last line above. It's just short-hand to save on scribbling.
To zero out the left side above there are two choices. One is to set \$v_{_\text{C}}=0\$. But given that the problem allows the specification of \$V_0\ne 0\$ and \$I_0\ne 0\$, that option is readily disposed of.
The Quadratic Solution
There's only one other choice. \$D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\$ must somehow always be zero.
Given the quadratic form of \$ax^2+bx+c=0\$ the roots are found at \$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\$.
But we have something more like \$x^2+\textsf{a}\,x+\textsf{b}=0\$, with roots at: \$x=-\frac12\textsf{b}\pm\frac12\sqrt{\textsf{a}^2-4\textsf{b}}\$.
By setting \$\beta=\sqrt{\textsf{b}}\$ and \$\alpha=\frac{\textsf{a}}{2\sqrt{\textsf{b}}}\$ (or setting \$\beta=\sqrt{\frac{c}{a}}\$ and \$\alpha=\frac{b}{2\sqrt{a\,c}}\$) then the roots (of \$x^2+2\alpha\beta\,x+\beta^2=0\$) are at \$x=\beta\left(-\alpha\pm\sqrt{\left(-\alpha\right)^2-1}\right)\$ and where \$\alpha\ge 0\$. This separates things out so that \$-\alpha\$ is the relative axis of symmetry and \$\beta\$ is a scaling factor. Now, a simple inspection informs us immediately whether or not roots are complex conjugates. If \$\alpha\lt 1\$ then they are complex conjugates. Otherwise, they are real and different or else real and the same.
In electronic transfer functions where the homogeneous response is 2nd order, \$\omega_{_0}\$ (or \$\omega_{_\text{c}}\$ for crossover or \$\omega_{_\text{p}}\$ for pole) is a special angular frequency used instead of \$\beta\$ and \$\zeta=\frac1{2\,Q}\$ is the damping factor and is used instead of \$\alpha\$.
Annihilators
If you knew that \$\frac{\text{d}}{\text{d}t}\,f\left(t\right)=0\$, then you'd be pretty sure that \$f\left(t\right)\$ wasn't a function of time, but instead was constant. So you'd guess that the general solution is \$f\left(t\right)=A_1\$, where \$A_1\$ was some constant determined by some initial condition. You could get to the specific form if you knew the initial condition.
In similar fashion, it follows that if we knew that \$\left[\frac{\text{d}}{\text{d}t}+\alpha\right]f\left(t\right)=0\$, we could guess that the general solution is \$f\left(t\right)=A_1\,e^{^{-\alpha\,t}}\$. We could test this:
$$\begin{align*}
\left[\frac{\text{d}}{\text{d}t}+\alpha\right]f\left(t\right)&=0
\\\\
\left[\frac{\text{d}}{\text{d}t}+\alpha\right]A_1\,e^{^{-\alpha\,t}}&= 0
\\\\
\frac{\text{d}}{\text{d}t}A_1\,e^{^{-\alpha\,t}}+\alpha\,A_1\,e^{^{-\alpha\,t}}&= 0
\\\\
-\alpha\,A_1\,e^{^{-\alpha\,t}}+\alpha\,A_1\,e^{^{-\alpha\,t}}&= 0
\\\\
0&=0
\end{align*}$$
So the operator \$\left[D+\alpha\right]\$ (just snuck \$D\$ back in) annihilates functions of the general form of \$f\left(t\right)=A_1\,e^{^{-\alpha\,t}}\$. (It's also conversely true that the operator \$\left[D-\alpha\right]\$ annihilates functions of the general form of \$f\left(t\right)=A_1\,e^{^{\alpha\,t}}\$.)
What does the operator \$\left[D+\alpha\right]^2\$ annihilate? In general, any operator of the form \$\left[D+\alpha\right]^m\$ annihilates any solution of the general form \$f\left(t\right)=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{-\alpha\,t}}\$. When \$m=2\$, this means it annihilates any function of the general form, \$f\left(t\right)=A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\$. Let's see:
$$\begin{align*}
\left[D+\alpha\right]^2 f\left(t\right)&=0
\\\\
\left[D+\alpha\right]\left[D+\alpha\right]\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]&= 0
\\\\
\left[D+\alpha\right]\left\{D\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\left[ A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\right]\right\}&= 0
\\\\
\left[D+\alpha\right]\left\{ -\alpha\,A_0\,e^{^{-\alpha\,t}}+D\left[A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\,A_0\,e^{^{-\alpha\,t}}+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&= 0
\\\\
\left[D+\alpha\right]\left\{D\left[A_1\,t\,e^{^{-\alpha\,t}}\right]+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&=0
\\\\
\left[D+\alpha\right]\left\{A_1\,e^{^{-\alpha\,t}}-\alpha\,A_1\,t\,e^{^{-\alpha\,t}}+\alpha\,A_1\,t\,e^{^{-\alpha\,t}}\right\}&=0
\\\\
\left[D+\alpha\right]A_1\,e^{^{-\alpha\,t}}&=0
\end{align*}$$
And we already know where that leads to. So \$\left[D+\alpha\right]^2\$ annihilates \$f\left(t\right)=A_0\,e^{^{-\alpha\,t}}+A_1\,t\,e^{^{-\alpha\,t}}\$!
The following are perhaps the two more important general annihilator forms (in electronics) to have in an annihilator toolbox:
\$\left[D-\alpha\right]^m\$ annihilates any solution of the general form:
\$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{\alpha\,t}}\$.
\$\left[\left(D-\alpha\right)^2+\beta^2\right]^m\$ annihilates any solution of the general form:
\$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_{2k}\,t^k\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right) + A_{2k+1}\,t^k\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right)\$
(When I need to know how to annihilate some function, \$f\left(t\right)\$, then I look at the denominator of the Laplace transform, \$F\left(s\right)=\mathcal{L}\, f\left(t\right)\$. Let's call that denominator \$X\left(s\right)\$. If \$X\left(s\right)\$ is a polynomial then \$X\left(D\right)\$ will be an annihilator of \$f\left(t\right)\$. I don't have the time before I leave to find or create a proof, or a counter-factual. But that's been my experience.)
Annihilators are rigorous and powerful tools.
They can be used in cases where we do not have an homogeneous equation -- the left side isn't zero and instead includes a non-homogeneity. All we need to do is to annihilate the non-homogeneity on the right side by applying the appropriate annihilator to both sides, equally.
Doing so causes the non-homogeneity to become zero (annihilated), turning a non-homogeneous equation into an homogeneous one (read: easier to solve.)
It's a way of zeroing out the driving function. Had this circuit been one that included an input source through a resistor for example, we'd then have had a non-homogeneous equation to mess around with. But by annihilating the input source, we get back a simpler problem (even if some more zeroes to worry over.) Nice.
When you squash the problem down until there's only zeroes left, it's just easier!
In real-estate it's all about "location, location, & location." Perhaps in electronics math it's all about "annihilation, annihilation, & annihilation."
Critically Damped
Returning to the problem at hand, \$\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}=0\$, the critically damped case is where \$\zeta=1\$. This happens when \$L_1=4\, C_1 R_1^2\$. (Seeing a \$\text{Farad}\cdot\text{Ohm}^2\$ don't we all just think, "Oh, that's a \$\text{Henry}\$!")
Substituting, we find \$\left[D^2+2\frac1{2\,R_1\,C_1}D+\left(\frac1{2\,R_1\,C_1}\right)^2\right]v_{_\text{C}}=0\$. If we set \$\alpha=-\frac1{2\,R_1\,C_1}\$ then this is:
$$\left[D^2-2\alpha D+\alpha^2\right]v_{_\text{C}}=\left[D-\alpha\right]^2\:v_{_\text{C}}=0$$
We now easily recognize the \$\left[D-\alpha\right]^{m=2}\$ annihilator! And clearly, the roots happen when \$D=\alpha\$.
As \$m=2\$, the general solution is:
$$\begin{align*}
v_{_\text{C}}&=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{\alpha\,t}}
\\\\
&= A_{_0}\,e^{^{\alpha\,t}}+A_{_1}\,t\,e^{^{\alpha\,t}}
\\\\
&=e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right)
\end{align*}$$
At \$t=0\$ we know that \$v_{_\text{C}}=V_{_0}\$. Therefore, \$A_{_0}=V_{_0}\$.
The other initial condition is the current in the inductor. Let's recall the KCL again, replace that initial current into the equation, and follow through:
$$\begin{align*}
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+I_{_0}&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}\left[e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right)\right]+I_{_0}&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\left[\alpha\,e^{^{\alpha\,t}}\left(A_{_0}+A_{_1}\,t\right)+e^{^{\alpha\,t}}\left(A_{_1}\right)\right]+I_{_0}&=0\:\text{A}
\end{align*}$$
At \$t=0\$, which is when we care about finding \$A_{_1}\$, this collapses down to:
$$\begin{align*}
\frac{v_{_\text{C}}}{R_1}+C_1\bigg[\alpha\,A_{_0}+A_{_1}\bigg]+I_{_0}&=0\:\text{A}
\end{align*}$$
From there, and substituting in for \$\alpha=-\frac1{2\,R_1\,C_1}\$, \$v_{_\text{C}}=V_{_0}\$ at \$t=0\$ and \$A_{_0}=V_{_0}\$, we find that \$A_{_1}=\frac{-1}{C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$.
And now we can finally lay out the specific solution for the critically damped case:
$$v_{_\text{C}}=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[1-\frac1{C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\cdot t\right]$$
This incorporates the initial conditions.
Now, keep in mind something. If one of the initial conditions is \$V_{_0}=0\:\text{V}\$, then propagate \$V_{_0}\$ through and into the right side, first, before setting it to zero. Then the above solution reduces to:
$$v_{_\text{C}}= -\frac{I_{_0}}{C_1}\cdot t\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}$$
Obviously, if both \$V_{_0}=0\:\text{V}\$ and \$I_{_0}=0\:\text{A}\$ then the whole thing collapses, as it should, to \$v_{_\text{C}}=0\:\text{V}\$.
Note that \$L_1\$ is nowhere to be seen in the specific solution. The reason is simple. If we know \$R_1\$ and \$C_1\$ then we know \$L_1\$ as the situation is critically damped and there's only one possible value for this unique situation. We do have to know two of the three values, but it doesn't have to be the same two I picked above.
Under-Damped
In the under-damped case (\$\zeta\lt 1\$), we have \$L_1\lt 4\, C_1 R_1^2\$. We are still solving for \$\left[D^2+\frac1{R_1\,C_1}D+\frac{1}{L_1\,C_1}\right]v_{_\text{C}}=0\$ and we will still set \$\alpha=-\frac1{2\,R_1\,C_1}\$.
Expanding \$\left[D-\alpha\right]^2\$ gives \$D^2+\frac1{R_1\,C_1}D+\frac{1}{4\,R_1^2\,C_1^2}\$. Everything is good except for the last term.
To get to the \$\left(D-\alpha\right)^2+\beta^2\$ form, we now also set \$\beta=\sqrt{\frac1{L_1\,C_1}-\frac1{4\, R_1^2\,C_1^2}}\$.
The general solution is:
$$\begin{align*}
v_{_\text{C}}&=\sum_{k=0}^{m-1} A_{2k}\,t^k\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right) + A_{2k+1}\,t^k\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right)
\\\\
&= A_{_0}\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right)+A_{_1}\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right)
\\\\
&=e^{^{\alpha\,t}}\left[A_{_0}\cos\left(\beta\,t\right)+A_{_1}\sin\left(\beta\,t\right)\right]
\end{align*}$$
Again, at \$t=0\$ we know that \$v_{_\text{C}}=V_{_0}\$. And just as before, \$A_{_0}=V_{_0}\$.
Like before, the other initial condition is still the current in the inductor. So let's recall the KCL again, replace that initial current into the equation, and follow through:
$$\begin{align*}
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+\frac1{L_1}\int v_{_\text{C}}\:\text{d}t&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}v_{_\text{C}}+I_{_0}&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\frac{\text{d}}{\text{d}t}\left[e^{^{\alpha\,t}}\left[A_{_0}\cos\left(\beta\,t\right)+A_{_1}\sin\left(\beta\,t\right)\right]\right]+I_{_0}&=0\:\text{A}
\\\\
\frac{v_{_\text{C}}}{R_1}+C_1\,e^{^{\alpha\,t}}\left[\left(\alpha\,A_{_0}+\beta\,A_{_1}\right)\cos\left(\beta\,t\right)+\left(\alpha\,A_{_1}-\beta\,A_{_0}\right)\sin\left(\beta\,t\right)\vphantom{e^{^{\alpha\,t}}}\right]+I_{_0}&=0\:\text{A}
\end{align*}$$
At \$t=0\$, which is when we care about finding \$A_{_1}\$, this collapses down to:
$$\begin{align*}
\frac{v_{_\text{C}}}{R_1}+C_1\left(\alpha\,A_{_0}+\beta\,A_{_1}\right)+I_{_0}&=0\:\text{A}
\end{align*}$$
From there, and substituting in for \$\alpha=-\frac1{2\,R_1\,C_1}\$, \$v_{_\text{C}}=V_{_0}\$ at \$t=0\$ and \$A_{_0}=V_{_0}\$, we find that \$A_{_1}=\frac{-1}{\beta\,C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$. (Yeah, I didn't want to expand \$\beta\$ into that.)
And now we can finally lay out the specific solution for the critically damped case:
$$\begin{align*}
v_{_\text{C}}&=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cos\left(\beta\,t\right)-\frac1{\beta\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sin\left(\beta\,t\right)\right]
\end{align*}$$
(You can incorporate the sine and cosine terms, using a phase computed from the constants. If you want to see where I do that, see near the bottom of my answer to Andy at the link given in the first sentence of the question.)
Over-Damped
In the over-damped case (\$\zeta\gt 1\$), we have \$L_1\gt 4\, C_1 R_1^2\$.
In this case, we also set \$\beta=\sqrt{\frac1{L_1\,C_1}-\frac1{4\, R_1^2\,C_1^2}}\$.
The general solution follows the same process as for the under-damped case and that at \$t=0\$ then \$A_{_0}=V_{_0}\$ and \$A_{_1}=\frac{-1}{\beta\,C_1}\left(\frac{V_{_0}}{2\,R_1}+I_{_0}\right)\$. Same as before.
But note here that \$\beta\$ will be imaginary!
So let's set \$\beta^{'}=\sqrt{\frac1{4\, R_1^2\,C_1^2}-\frac1{L_1\,C_1}}\$, which is real, and let \$\beta=j\,\beta^{'}\$.
And now we can finally lay out the specific solution for the critically damped case:
$$\begin{align*}
v_{_\text{C}}&=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cos\left(\beta\,t\right)-\frac1{\beta\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sin\left(\beta\,t\right)\right]
\\\\
&=V_{_0}\cdot e^{^{\frac{-t}{2\,R_1\,C_1}}}\cdot \left[\cosh\left(\beta^{'}\,t\right)-\frac{1}{\beta^{'}\,C_1}\cdot \left(\frac1{2\,R_1}+\frac{I_{_0}}{V_{_0}}\right)\sinh\left(\beta^{'}\,t\right)\right]
\end{align*}$$
All real values, now. The \$j\$ is all gone.
Demonstration
Let's use \$R_1=1\:\text{k}\Omega\$ and \$C_1=10\:\mu\text{F}\$ and supply \$\zeta\$ to vary \$L_1\$. This means using the following equation: \$L_1=\zeta^2\left( 4\,C_1\,R_1^{\,2}\right)\$.
I'll do a run with enough lines to make up a reasonable demonstration (16 lines) going from \$\zeta=0.25\$ to \$\zeta=2\$ (three octaves) with 5 lines per octave.
Good match!
