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})
