Nodal analysis with Laplace of simple circuit

Using nodal analysis with Laplace transform find “v” for “t > 0”

Founded initial conditions:

i(0-)=i(0+)=0A

v(0-)=v(0+)=0V

i(infinite)=3A

v(infinite)=4*3A=12V

My nodal analysis with Laplace transform equations are:

v1 – node where current source flows into circuit

The Laplace of “v” is in the second row, but it seems to be an error in my equations – the answer in book isn’t equal to reverse transform of second row! May anyone help to properly write that equations?

Answer

I see two errors. #1; a switch in series will NOT stop a current source. You must use a switch in parallel with the source and open it at t=0 to enable. #2; the Laplace transform of the suddenly applied 3A source is 3/S. This results in Vc=60/[(s)(S+5)] from which Vc(t)=12(1-e^(-5t)). Note that the final value of Vc is 12volts as expected.