Parallel RLC Circuit: how to write differential equation?
I would like to thank @Jonk for the hint. I have discovered that I was very close to the solution, but missed only one step.

Let's start. Here is a parallel RLC-network and we have to find the equation for the waveform (voltage and current). As I have mentioned in the question, whether it is a voltage waveform or a current one, damping factor (alpha) and resonant frequency (omega) always be the same and they always lead to 3 different cases:

alpha &gt; omega; overdamped response
alpha = omega; critically damped response
alpha &lt; omega; underdamped response


Let's calculate these values:

alpha = 1/2RC = 3.5 Hz;
omega = 1/(LC)^0.5 = (6)^0.5.


Here we can see that it is a overdamped response case. Let's calculate natural frequencies:

s = -alpha +/- (alpha^2 - omega^2)^0.5 = -1; -6


In this case the general solution has the following form:

x(t) = A1*exp(-1t) + A2*exp(-6t).


Now we have to decide what is X - voltage or current? Let it be voltage:

v(t) = A1*exp(-1t) + A2*exp(-6t); general solution for the voltage waveform


Let's define A1 and A2. We have to compose a system of two independent equations for the volatge and the voltage derivative at t=0. Initial conditions can be obtained from the circuit: when t&lt;0, the capacitor is shunted by the inductor, so v(0-) = v(0+) = 0. At t>0, there is no voltage source in the circuit, thus final voltage = 0. 

v(0) = A1*exp(-1*0) + A2*exp(-6t*0) = A1 + A2;


Let's find out the voltage derivative. Lazy way: just open the textbook and pick the formula:

dv/dt = - (v(0) + Ri(0))/RC = -168;


Not lazy way (it will give the same result). There are 3 currents in the parallel source-free RLC network from the top node downwards to the ground:

0 A = CdV/dt + iL(0) + v(0+)/R = CdV/dt + 4 + 0/R;
CdV/dt = -4;
dV/dt = -4 * 42 = -168.


Let's differentiate the general solution equation:

dv/dt = d(A1*exp(-1t) + A2*exp(-6t))/dt = -A1*exp(-1t) - 6*A2*exp(-6t));
dv(0)/dt = -A1 - 6*A2;


Now it is time to compose a system of two equations:

0 = A1 + A2;
-168 = -A1 - 6*A2;
--------------------&gt; A2 = 33.6, A1 = -33.6.


Also, the general solution for the voltage waveform is:

v(t) = 33.6*exp(-6t) - 33.6*exp(-t)).


Now let's remember the advice of @Jonk. This is the parallel RLC-circuit and the volatge is the same across all the branches. Thus, inductor has the same voltage. The current across the inductor is the integral of the voltage:

i(t) = (1/L)* ∫ v(t)dt = (33.6/7)*((-1/6)exp(-6t) + exp(-1t)) = 4.8*exp(-t) - 0.8*exp(-6t);


Bingo!

Now lets' go back to the general equation and assume that X is the current waveform:

x(t) = A1*exp(-1t) + A2*exp(-6t);
i(t) = A1*exp(-1t) + A2*exp(-6t).


Initial conditions: iL(0-) = iL(0+) = 4A. Current derivative:

vL = LdI/dt; voltage across the inductor
dI(0+)/dt = vL(0+)/L = 0.


The system of equations:

4 = A1 + A2;
0 = -A1 - 6*A2;
------------------&gt; A2 = -0.8, A1 = 4.8.


Current waveform equation is

i(t) = 4.8*exp(-t) - 0.8*exp(-6t);


Here we are.

PS! I sincerely hope that this is not a coincidence and this reasoning can be applied to all circuits.