First, we will transform the left side of the circuit.
We transform the voltage source of 20 V and the 4Ω resistor into a
current source.
The equivalent source will have a current of I = U/R = 20/4 = 5A, the
resistor still being of 4Ω.
This source sums with the 9A one, thus we get an equivalent current
source of It = 9 + 5= 14 A, and the two resistors in parallel give a Re =
4*12/16 = 3Ω.
Before the switch opens we have this 3Ω resistor in parallel with another
3Ω resistor.
This gives a 1.5 Ω resistor, which is in parallel with the 9Ω
one. This is thus an equivalent resistor Re of 9/7 Ω.
We needed this equivalent circuit to find the V(t-) and i(t) in the
right side of the circuit before the switch is opened.
We have V = Re*It = 9/7*14 = 18 V and now we can find i = 18/9 = 2A because
the voltage V is for the 9Ω resistor.
Once the switch is opened, we will only have a closed circuit on the
right hand side, formed by the inductance, the 3Ω and the 9Ω
resistor, which is a 12Ω resistor (series).
The equation is 12*i(t) = - L*di(t)/dt
L = 0.5H so we get di(t)/dt = - 24*i(t)
Thus i(t) = A*exp(- 24*t)
To find A, we take i(0) = A = 2A from the first part of the problem.
So i(t) = 2*exp( -24*t) and finally V(t) = 9*i(t) = 18*exp( -24*t)