The inductance term must be with negative sign if one is to treat it as EMF of self-inductance. Also, if we use passive sign convention, the KVL (or Faraday's law to be precise) equation is $V_{C_0} + V_C + V_L = \frac{q_{c_0}}{C0}+\frac{q_c}{C}+Lq′′ = 0$, which gives the correct expression for frequency, actually. And I don't understand how the part $L \frac{dI}{dt} = V_0 cos(\omega t)$ is obtained. The only way is to find current and it is the derivative of charge. The only way to obtain charge is to solve the differential equation I got. But for this I need the innitial conditions

And where am I wrong in the part that we have a wire with potential difference between the capacitors and thus a short circuit?

I am not sure how much more hints I can give before it is a direct solution to this homework. What you are asking are all the important points, and if you solve them, the solution would present itself. What I can tell you is that the book is correct, with only that one mistake. The equations are not that difficult to solve, even all the way to obtaining $\omega$. Maybe if you solve the equations and get the solution, you might be able to backwards deduce why they are correct.

What "equations" are you talking about? I told you why I can't solve this: $$\frac{q_{c_0}}{C_0} + \frac{q_{c}}{C} + L q'' = 0$$ due to not having the innitial charge condition

That equation is just wrong (as in not applicable to this question), and you actually do have an initial charge condition; you just do not realise that you do.

Solve the problem with the magical two equations that the book wrote down. (It is actually 3 equations, if not 4, but no matter what, it is already sufficient to solve.)

I literally do not have the innitial condition for charge as the charge is by the very specification of the problem different in each capacitor. So there is no "circuit charge function"

No, there is. It is hidden as another initial condition. i.e. You know the initial charge of both capacitors.

it seems like you think I was assigned this problem by a teacher and all I want is an answer. And this is likely the reason why my question was closed despite me showing my work and asking about the physical concepts, not the mere answer. I want to understand the behaviour of the system. I want to know how to rigorously obtain the answer from the very postulates. I have NO interest in doing anything with some "magical equations" untill I completely understand where they come from

I do know the charge of each capacitor, but that charge is different. Thus, the charge does not depend only on time, but also on what part of the circuit we are considering

I can tell you that your misapplication of KVL led to you getting the wrong equation, instead of the left half of the equation that the book had. If you did it correctly, you will not see it as coming out of nowhere. With some insight you would also be able to get the right half. Charge conservation is just charge conservation, and it should be obvious from hindsight, even if it is not obvious that you needed it beforehand. Once you have all 3 ingredients, couplied with the initial charges, you could solve the whole thing.

You should only have the charges on the capacitors to contend with. No other place to store charges. This should be an assumption that is blatantly supported by physical evidence and not too egregious to assume.

I know where the charge conservation comes from. As I stated, the equation out of nowhere was this: $$V_{C_0} - V_C = L \frac{dI}{dt} = V_0 cos(\omega t)$$ I don't understand where this comes from. And I don't understand why this: $$\frac{q_{c_0}}{C_0} + \frac{q_{c}}{C} + L q'' = 0$$ is wrong, as it is the mere application of KVL with passive sign convention, which accounts for magnetic flux stuff in the inductor. And also my equation reduces to $q'' = - \frac{1}{LC_{equiv}} q$, which is the expected result

You have to have a consistent set of sign conventions between the charge conservation equation, the Faraday's Law, and taking some simplified limits to make sure that your choices are not violating basic mathematical sensibilities. I am not even sure that you applied passive sign convention correctly.

Passive sign convention implies choosing an arbitraty current direction, then assuming that current always enters the positive terminal. This way, $U_R = IR$, $U_C = \frac{q}{C}$ and $U_L = L \frac{dI}{dt}$. For charge conservation I'm not sure how to determine the sign and I may be wrong there, since I can't know for sure which terminal of a capacitor is positively charged. That's why I need help with understanding where to get these equation

Sigh, I'll give the answer to the sign problem. The charge convention had $V_0 C_0 = V_{C_0} C_0 + V_C C$. Intentionally consider the short-circuited case, which means that the two RHS voltages are the same $V_\infty$, and the sign is dictated by $V_0$, which WLoG we can pick to be positive. Let us take the switch K as the reference point, essentially grounded as V=0 for that line. It is just convenient. Then, as you go from the left branch downwards, you pass by the left capacitor. WLoG you can pick this to be positively charged, integrating parallel to E field, getting $V_{C_0}$.

Then you have to integrate the E field in the capacitor on the right, getting $\pm V_C$, but it is not clear yet which sign it should be. Now, again consider the intentionally short-circuited case, then you would have $V_\infty$ on the both capacitors. The only way for you to integrate a whole loop and get zero, is the case for you to have $V_{C_0} - V_C = 0 = \pm L \dot I$ for some choice of the sign of the inductance term. So, the choice of the signs on the charge conservation equation, dictates the sign that is in this Faraday's Law. You are not free to choose the voltages' sign.

1) As I understand, you defined capacitor voltages as $\varphi_\text{positively charged plate} - \varphi_\text{negatively charged plate}$, right? 2) What do you mean by "short-circuited case" and what is $V_\infty$ 3) In your convention, voltage having positive sign means that while traversing the chosen loop, we encounter the positive terminal first? 4) How to determine the sign of $L\dot{I}$ and that it equals $V_0 cos(\omega t)$? 5) Why can't I just assume the direction of current and use the passive sign convention?

1) yes. 2) the inductor prevented it from being short-circuited; simply change the inductor to a simple non-coiling wire. Then the conservation of charge says $V_0 C_0 = V_{C_0} C_0 + V_C C = V_\infty ( C_0 + C )$ because the two capacitors will have common voltage $V_\infty$, where if you wait for $\infty$ time it should equilibrate to that. 3) No, negative terminal first. The stuff connected to switch K on both sides are negatively charged. 4) sign of $L \dot I$ requires choice of sign of $I$. That it equals $V_0 \cos \omega t$ is another physical insight.

5) proper application of Faraday's Law: going against E field raises voltage, going along E field drops voltage.

1) "That it equals $V_0 cosωt$ is another physical insight" - explain this please. 2) The very reason of implementing the passive sign convention is that we cannot know in advance the direction of the E field, so we just assume the current and voltage direction and then see if the negative sign shows up. In fact, the E field can oscillate and change direction. 3) I still don't understand those $V_\infty$ and short circuit arguments. How did you obtain this from some Maxwell's/Kirchhoff's or other fundamental equations and how exactly it influences the capacitot voltage at any given time?

4) And you state that both capasitor plates connected to K are negatively charged. Do you mean at any time? Why is that so?