Laplace Transform (Voltage Transformation)
To be solved by Laplace transforms, the entire circuit must be in the frequency domain.  Therefore I2 must be represented by 2/(S+1).  The 4 for V1 states that Il=2A at t=0+.  When the topological graph is drawn for the circuit (opening current sources and shorting voltage sources), there is only one loop around which to write a mesh equation which is: (2/s)Ic+(2S)Il+Ir2+3Ir1=4-4/S; where Ir2 is the downward current in R2. Along with ancillary equations relating the various currents, we finally get:  Il=(2S^2+5S+2)/(S+1)^3.  Using partial fraction expansion and taking the inverse Laplace transform we get:
i(t)=[(-1/2)t^2 + t + 2)]e^-t).  As a check, evaluating at t=0+, we get il(0+)=2 which is expected.