Applying Kirchhoff Rule to an LC circuit In a simple LC circuit (just one capacitor and one inductor), using the Kirchhoff rule has $$\frac{Q}{C} + L\frac{dI}{dt} = 0.$$ But isn't the voltage drop across an inductor $-L\frac{dI}{dt}$? What happened to the negative sign?
 A: According to the convention in AC circuits, voltage = potential drop on a two-terminal device  (sometimes misleadingly called "voltage drop" on the device) is defined as positive, when moving from one terminal to the other in the same direction as positive current does makes the potential to decrease. On ideal inductor, this potential drop is always
$$
potential~drop = L\frac{dI}{dt},
$$
that is, there is no minus sign.
When using Kirchhoff's voltage law (KVL) one writes sum of all potential drops in a closed path equals zero. The above is why there is no minus sign in front of $L$ in the resulting equation.
The minus sign appears when induced electromotive force (emf) as opposed to potential drop is discussed. Self-induced EMF on inductor, ideal or not, is always
$$
emf = -L\frac{dI}{dt}.
$$
As you can see, induced EMF has opposite sign to the potential drop. This means that the induced emf acts against the Coulomb field of charges on the inductor surface and its terminals (whose effect on the mobile charges as they move from one terminal to the other can be quantified by means of the above potential drop), so the net impressed force on mobile charge carriers in the ideal inductor is zero. This force has to be zero since there is no ohmic resistance in an ideal inductor; it takes negligible macroscopic force to accelerate the charge carriers there, and we simplify the model by assuming it is zero.
In real inductors with ohmic resistance, this is not true; while the self-induced EMF is still given by the same formula, potential drop is no longer equal in magnitude to this EMF.
This effect of the resistance can be derived from the generalized Ohm's law and simple model of the real inductor. The generalized Ohm's law means that integral of net impressed force acting on mobile charge, per unit charge, along the conductive path, equals $R_cI$ where $R_c$ is resistance of that conductive path.
In the usual case where all emfs are either due to localized voltages sources or induced emfs on inductors, there are only two impressed forces acting on current in the real inductor, the conservative electric force characterized by the potential drop, and the self-induction electric force, characterized by the induced EMF, so we end up with potential drop on real inductor obeying the equation
$$
P.D. + emf = R_cI,
$$
where $R_c$ is resistance of the inductor coiled path and $I$ is current through the inductor (here we simplify and assume that the current is the same everywhere). At the same time, induced EMF is still given by the formula
$$
emf = -L\frac{dI}{dt},
$$
so we obtain approximate expression for the potential drop
$$
P.D. = L\frac{dI}{dt} + R_cI.
$$
A: According to the ubiquitous passive sign convention, the voltage drop across an inductor is 
$$ V = L\frac{dI}{dt}. $$
A: You have to remember that, when a capacitor is discharging and the current on the inductor is increasing, then:
$$q=q_o-it$$
Therefore:
$$\frac{dq}{dt}=-i
\quad\Rightarrow\quad\frac{d^2q}{dt^2}=-\frac{di}{dt}$$
Upon doing the loop rule, you get:
$$-L\frac{di}{dt}+\frac{q}{C}=0
\quad\Rightarrow\quad
L\frac{d^2q}{dt^2}+\frac{q}{C}=0$$
That last equation is the equation we were looking for.
