In the circuit shown, the switch S has been closed for long time and then opened at $t=0$. Current through the inductor as a function of time after the switch is opened__

 

At $t=\infty$  circuit will be in steady state. Then circuit becomes as slow

At $t=0$ current through indicator is

Applying K V L to loop ABCDA
$6I+L\frac{di}{dt}=12$
$6I-12=-\frac{di}{dt}(\therefore L=1H)$
$\frac{di}{6i-12}=-dt\Rightarrow {\int }_6^i\frac{dI}{6i-12}=-{\int }_0^{t}dt$
$=\frac{1}{6}{\left[\ln \right(i-2\left)\right]}_6^i=-[t-0]$

$\ln \left(\frac{i-2}{4}\right)=-6t\Rightarrow \frac{i-2}{4}=e^{-6t}\Rightarrow i=2+4e^{-6t}$