Every solution is the <resonant series RLC response> $I_0(t)=R^{-1}\sin(\omega_0t)$ plus a solution of the homogeneous <damped harmonic oscillator>. Define $\gamma=R/(2L)$ and $\omega_0^2=1/(LC)$. The three possibilities are
$$
\boxed{I(t)=\frac{\sin(\omega_0t)}R+
\begin{cases}
e^{-\gamma t}\bigl(A\cos(\Omega t)+B\sin(\Omega t)\bigr),
&\gamma<\omega_0,\quad\Omega=\sqrt{\omega_0^2-\gamma^2},\\
e^{-\gamma t}(A+Bt),&\gamma=\omega_0,\\
Ae^{(-\gamma+\kappa)t}+Be^{(-\gamma-\kappa)t},
&\gamma>\omega_0,\quad\kappa=\sqrt{\gamma^2-\omega_0^2}.
\end{cases}}
$$
These are respectively the <underdamped RLC response>, <critically damped RLC response> and <overdamped RLC response>. Both exponents in the last case are negative because $0<\kappa<\gamma$. All three homogeneous contributions tend to zero as $t\to\infty$, so \b[every response approaches the unique periodic steady current $I_0(t)$.] No initial conditions were specified in this clause; they determine $A,B$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 5C / c / ii / Solution

When $R^2<4L/C$, the homogeneous <series RLC circuit> response oscillates with angular frequency $\sqrt{1/(LC)-R^2/(4L^2)}$ and envelope $e^{-Rt/(2L)}$. Its <characteristic roots> form a complex-conjugate pair with negative <real part> when $R>0$.

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 Incoming links: Underdamped RLC response