Your problem can be solved analytically.
The switch between capacitor and 2? opened at t=0, it means that the capacitor is charged to 12V*2/3=8V at t=0.
The problem reduced to RLC circuit with initial conditions:
t=0;
I=0, Vc=8V, V[sub]R[/sub]=0V, V[sub]L[/sub]=2V (V[sub]L[/sub]=10-Vc-V[sub]R[/sub] )
The differential equation need to be solved is $10=L \frac{dI}{dt}+IR+\frac{1}{C}\int I dt$
or $L\frac{d^Q}{dt^2}+I\frac{dQ}{dt}+\frac{Q}{c}=0$

The solution is similar to a spring with constant k, attached with mass m and damping constant b. $m\frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$

The analytical solution can be found in standard textbook.

You are also welcomed to check out [url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=30.0] RLC circuit simulation (DC Voltage source)[/url] to find out the simulated solution.
[/quote]

Oh, ok, thanks, this problem already solved.

I must put in the initial condition for capacitor, 8V.

So my circuit will become like this:

[img]http://img2.pict.com/e7/67/54/1738217/0/1.jpg[/img]
I simplified it. IC=Initial condition.

Then, settings:

[img]http://img2.pict.com/d9/91/de/1738218/0/2.jpg[/img]
Tick "Skip initial operating..."

Finally, become like this:

[img]http://img2.pict.com/65/49/43/1738219/0/3.jpg[/img]

Then I just run the simulation for the graph. Must put the equation also for V(t) and Vr(t)...