Solving a circuit containing a resistor and inductor in parallel
For t &lt; 0:

Since it was left open for "a long time", then its safe to assume that the inductor has been completely drained of energy. So \$V_L = 0\$.

For 0 &lt; t &lt; 1ms:

Taking the Laplace transform of the components and simplifying (as a voltage divider):

$$V_L(s) = \dfrac{120sLR_2}{R_1R_2 + sLR_1 + sLR_2}$$
$$= \dfrac{24s}{s + 500} = 24\dfrac{s}{s + 500} = 24(1 - \dfrac{500}{s + 500})$$

Inverse Laplace gives:

$$V_L(t) = 24(\delta(t) - 500e^{-500t})$$

Since t > 0, this results in:

$$V_L(t) = - 12000e^{-500t}$$

For t > 1ms:

The inductor drains all its stored energy through the resistor using the standard equation.