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Near zero, the Taylor expansions areThe constant, linear and quadratic terms cancel in the numerator. ConsequentlyThe square root and logarithm are both real in a sufficiently small neighborhood of zero, so the two-sided real limit is well defined.

Let $a=R/(2L)$ and ${\omega }_d=\sqrt{1/(LC)-a^2}>0$. The roots of the characteristic polynomial are $-a\pm i{\omega }_d$, so the real underdamped solution is $Q=e^{-at}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$. The initial charge gives $A=Q_0$ and the zero initial current gives $B=a{Q}_0/{\omega }_d$. HenceThis solves the homogeneous RLC circuit equation supplied in the question. A nonzero applied voltage would appear as a forcing term and require a particular solution; it is not present in that equation.

Set $v=\dot{z}$. The second-order position equation becomes the first-order linear ordinary differential equation $\dot{v}+\gamma v=-g$. For $\gamma >0$, multiply by the integrating factor $e^{\gamma t}$ and impose $v(0)=v_0$:The limiting terminal velocity is $-g/\gamma$, consistent with upward-positive position and downward gravity. The continuous zero-drag limit is $v_0-gt$; one must take that limit of the whole expression rather than set $\gamma =0$ in its two singular-looking terms separately.

Using the complex-angle identity,Since $\cosh y>0$, the real part can vanish only when $x=\pi /2+k\pi$. Then the imaginary part requires $\sinh y=(-1{)}^{k+1}$. Therefore all roots areSubstitution gives $\cos z=i$ for every integer $k$, and the positivity of $\cosh y$ shows that no other real parts were missed. The imaginary offsets are the two values $\pm \mathrm{arsinh}1$.

On an interval with $x\ne 0$, the zero function is one solution. For a nonzero solution, set $u=y^{-2}$. Dividing the equation by two and differentiating $u$ gives the linear ordinary differential equationIts integrating factor is $x^{-2}$, so $(u/x^2{)}^{\prime }=-x^{-5}$ and $u=1/(4x^2)+C{x}^2$. Absorbing the factor four into the integration constant givesFor real solutions, choose a connected interval avoiding $x=0$ on which $1+K{x}^4>0$, with a fixed sign. The special straight solutions $y=\pm 2x$ are already included at $K=0$. The separate zero solution would be lost by the division used in the substitution.