Calculating Differential equation RLC Circuit

Question

Hi,
I tried to find the voltage on the capacitor as a function of time, Vc(t).
In the initial condition (t<0) the switch was closed and the current source was open circuit, so I found the initial conditions for the capacitor and the inductor ( VC(0)=65v, and IL(0)=1A).
At t=0, the switch will open and the current source connect to circuit like the picture here.
I'd tried to write the differential equation of the circuit and got something weird:

I know that the equation of RLC circuit 2d must be positive, otherwise, I will get one of the roots positive which is impossible.
Please, I need your help, what is important to me is to proof the equation so I can show the lecturer that he was wrong, but I'm not sure if what I did was right.
Here is the solution of the lecturer:
(A1 and A2 easy to find)

the decition of current in the inductor is true:

Answer 1

Since everyone seems to have been variously confused at different times about your question, let me re-draw and re-phrase it:

^{simulate this circuit – Schematic created using CircuitLab}

(The above reflects the case after the switch opens and after the current source has been added, both at \$t=0\$.)

On the left is your new circuit (without initial conditions showing, as I think it is unnecessary in order to answer your question.) On the right is the same circuit through a simple, obvious Thevenin transformation about which I'm sure you agree. I prefer the right side because it makes it a little easier to answer your question about the lecturer's answer.

Let's use KCL for the node marked as \$V_t\$:

$$C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac1{L}\int V_{\left(t\right)} \:\text{d}t=\frac1{L}\int V_{s\left(t\right)} \:\text{d}t$$

(I've placed out-flowing currents on the left and in-flowing currents on the right.)

Some development results in:

$$\begin{align*} \frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac1{L\,C}\int V_{\left(t\right)} \:\text{d}t&=\frac1{L\,C}\int V_{s\left(t\right)} \:\text{d}t\\\\ \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{V_{s\left(t\right)}}{L\,C}\\\\ \end{align*}$$

But we also know that the current, \$I_{\left(t\right)}\$, is everywhere the same throughout the right-hand circuit. So it follows that:

$$V_{s\left(t\right)}=39\:\text{V}-R\,I_{\left(t\right)} = 39\:\text{V}-R\,C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}$$

Substituting in:

$$\begin{align*} \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{1}{L\,C}\left(39\:\text{V}-R\,C\,\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}\right)\\\\ \frac{\text{d}^2\,V_{\left(t\right)}}{\text{d}t^2}+\frac{R}{L}\:\frac{\text{d}\,V_{\left(t\right)}}{\text{d}t}+\frac{V_{\left(t\right)}}{L\,C}&=\frac{39\:\text{V}}{L\,C} \end{align*}$$

Which is just a 2nd order diff-eq I'm sure you can solve.

But just looking at the characteristic equation, (\$s^2+650\,s+1\times 10^5=0\$), you would easily find \$s=-400\$ and \$s=-250\$. Which is a huge clue for setting up the general solution.

The lecturer's general solution is correct.

Answer 2

One way to get the initial condition is through the Thevenin equivalent circuit of the current source and the resistor in parallel of it. Then, the initial DC voltage of the cap and DC current of the inductor can be solved easily, considering L1 as a short, and C1 as an open circuit.