At #t<0# the system is in steady state. So, the inductor behaves like a short and hence no current flows through the #4\ Omega# resistance. So, the initial condition is

#i = 0# at #t = 0#

For #t>0# the current through the #8\ Omega# resistance is #9\ "A"-i#, and so the voltage across it is

#8\ Omega times (9\ "A"-i) = 72\ "V" - 8\ Omega times i#

If the current through the inductance is #i^'# then that through the #6\ Omega# resistance is #i-i^'# and so, we have

# 72\ "V" - 8\ Omega times i = 4\ Omega times i +6\Omega times (i-i^') + 6\ "V"#

and hence

#18i-6i^' = 66\ "A" implies i^' = 3i-11\ "A" implies {di^'}/{dt} = 3{di}/{dt}#

Again, by using the fact that the voltage drop across the inductor is given by #2\ "H" times {di^'}/dt# we get

# 72\ "V" - 8\ Omega times i = 4\ Omega times i +2\ "H" times {di^'}/dt#

So, the differential equation for #i(t)# finally takes the form

#{di}/dt + lambda i= \Lambda#

where #\lambda = 2\ "s"^-1# and #\Lambda = 12\ "A"\ "s"^-1#

This differential equation can finally be cast in the form

#d/dt (Lambda/lambda-i) = -lambda(Lambda/lambda-i)#

which is easy to integrate to give

#ln (i_oo-i) = -lambda t+C#

where #i_oo = Lambda/lambda = 6\ "A"#. From the initial condition, we get

#C = ln (i_oo)#

so that

#1-i/i_oo = e^{-lambda t}# and thus finally

#i(t) = i_oo(1-e^{-lambda t})#