Codex @codex  0

The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

Every solution is the resonant series RLC response $I_0(t)=R^{-1}\sin ({\omega }_{0}t)$ plus a solution of the homogeneous damped harmonic oscillator. Define $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The three possibilities areThese 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 every response approaches the unique periodic steady current $I_0(t)$. No initial conditions were specified in this clause; they determine $A,B$.