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## Homework Statement

http://img396.imageshack.us/img396/2781/chapter8problem55oy6.jpg [Broken]

For the circuit above, find [itex]v(t)[/itex] for [itex]t\,>\,0[/itex].

Assume that [itex]v\left(0^+\right)\,=\,4\,V[/itex] and [itex]i\left(0^+\right)\,=\,2\,A[/itex].

## Homework Equations

[tex]i_c\,=\,C\,\frac{d\,v_c}{dt}[/tex]

## The Attempt at a Solution

I made a new diagram:

http://img507.imageshack.us/img507/4142/chapter8problem55part2zm4.jpg [Broken]

[tex]i_1\,=\,C_1\,\frac{d\,V_{C_1}}{dt}[/tex]

[tex]i_2\,=\,\frac{V_1\,-\,V_2}{2\,\Omega}[/tex]

[tex]i\,=\,C_2\,\frac{d\,V_{C_2}}{dt}[/tex]

KCL @ [itex]V_1[/itex]) [tex]i_1\,+\,\frac{i}{4}\,-\,i_2\,=\,0[/tex]

[tex]C_1\,\frac{d\,V_{C_1}}{dt}\,+\,\frac{C_2}{4}\,\frac{d\,V_{C_2}}{dt}\,-\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]

[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,0.25\frac{d\,V_{C_2}}{dt}\,-\,V_1\,+\,V_2\,=\,0[/tex]

KCL @ [itex]V_2[/itex]) [tex]i\,+\,i_2\,=\,0[/tex]

[tex]C_2\,\frac{d\,V_{C_2}}{dt}\,+\,\frac{V_1\,-\,V_2}{2\,\Omega}\,=\,0[/tex]

[tex]\frac{d\,V_{C_2}}{dt}\,+\,V_1\,-\,V_2\,=\,0[/tex]

[tex]V_2\,=\,\frac{d\,V_{C_2}}{dt}\,+\,V_1[/tex]

Now substituting the KCL @ [itex]V_2[/itex] equation into the other equation:

[tex]0.2\,\frac{d\,V_{C_1}}{dt}\,+\,1.25\,\frac{d\,V_{C_2}}{dt}\,=\,0[/tex]

Here I am stuck. I don't know how to proceed, any hints? I know that I am supposed to get a second order differential equation for the circuit, but where from?

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