What is the physical meaning of the voltage across an inductor in the Laplace domain?
The Laplace transform of a derivative is

$$\mathcal{L}\left\{ \frac{dy}{dt} \right\} = s\mathcal{L}\left\{f\right\} - y(0^+)$$

Ie. \$y(0^+)\$ is the initial condition, for time approaching zero from the right side.

Since the "law" of inductors is

$$v_L = L\cdot \frac{di_L}{dt}$$

Its Laplace transform is given by

$$\begin{align}
V_L(s) &amp;= L\left( sI_L(s) - i_L(0^+) \right) \\
&amp;= Ls\cdot I_L(s) - L\cdot i_L(0^+)
&amp;\Downarrow \\
I_L(s) &amp;= \frac{V_L(s)}{Ls} + \frac{i_L(0^+)}{s}
\end{align}$$

So in this formula, \$i_L(0^+)\$ takes the meaning of the initial current through the inductor. From the last equation, you also can assign a physical meaning to it. This equation coincides to having a parallel constant current source with the inductor!

In other words, you can replace an inductor with







simulate this circuit &ndash; Schematic created using CircuitLab

For a capacitor, the same can be done. It will result in putting a constant voltage source in series with the capacitor.