LR circuit and work?

Question

An LR circuit is hooked up to a battery , with the switch initially open. The resistance in the circuit is R=140 Ω, the inductance is L=2.40 H, and the battery maintains a voltage of E=48.0 V. At time t=0 the switch is closed

How much work has the battery done from the time the switch was closed until t=2.54E-2 s?

I know that you must integrate the power (P=VI) over the time interval to get the total work done by the battery, but I can't seem to get the right answer.

Answer 1

V = L dI/dt + IR

The key is when you flip the switch a step function occurs in V.

It's been a while so I can't remember the solution to the linear system.