TLDR: When going from series to parallel (parallel to series): switch V with I, L with C, R with 1/R, and Q with \$\Phi\$, the magnetic flux of the inductor.
Each term of the first two ODEs is a voltage. In the second ODE the standard defining relationships make this easy to see.
So in your parallel case it is the currents that add, so this is the standard starting point.
Rewriting the second equation:
$$L \frac{dI}{dt}+RI+\frac{Q}{C}=0\tag{Equ 1}$$
reveals that the middle term is Ohm's law in a voltage form.
So replace current with voltage and write the middle term for the parallel case as Ohm's law in a current form:
$$C\frac{dV}{dt}+\frac{V}{R}+\frac{X}{L}=0$$
What could X be?
Notice that changing L to C reveals the capacitor current in the first term on the left. So then it makes sense to change C to L as well. But the X-factor need to be clarified.
Comparing to your first equation we can write
$$
C \frac{d^2X}{dt^2}+\frac{1}{R}\frac{dX}{dt}+\frac{X}{L}=0
$$
only if \$V=\frac{dX}{dt}\$.
Using Faraday's law of Induction:$$V=-\frac{d\Phi}{dt}$$
where \$\Phi\$ is the magnetic flux produced by the inductor current.
The last term still needs verification, but by substituting \$-\Phi\$ for X, the equation now is in the same form as your first one.
$$
C \frac{d^2\Phi}{dt^2}+\frac{1}{R}\frac{d\Phi}{dt}+\frac{\Phi}{L}=0
$$
To verify the last term, look to Faraday's Law again and recognize that inductance is:$$L=\frac{-d\Phi}{dI_L}$$
Then integrating and solving yields: $$I_L=\frac{-\Phi}{L}$$
Of course this can be done with Kirchhoff, but he did not include charge or magnetic flux.
So now we have the equation in the form that your desire:
$$C\frac{dV}{dt}+\frac{V}{R}+\frac{\Phi}{L}=0$$
So, "Which we can happily solves using matrices etc."
