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I've got this circuit. It contains a switch that will open at T = 0:

I'm tasked with finding the voltage on the capacitor. I calculate the initial conditions (assuming that the capacitor and the inductor are fully charged). The initial C voltage is 16 V and the initial current through the inductor is 0 A.

Then I proceed to simplify the remaining components and I get to this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Then I want to apply Kirchoff's Voltage Law. I get this equation:

And solving for the ODE (without the constants):

However, according to the solution manual, this equation describes the voltage in this circuit. How can that be? I can't understand the gap between having the ODE and its correlation to the voltage. Here is the solution:

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Whatever you did is correct. You have got an expression for current. Solution manual has used a different approach and they have obtained expression for voltage.

Finally what you want is voltage. Let's start from where you have stopped.

$$i(t)=Ae^{-9t}+Be^{-t}$$

so voltage across capacitor: $$\begin{align*} v(t)&=\frac{1}{C}\int i(t) dt\\ &=-27\left[\frac{A}{9}e^{-9t}+Be^{-t}\right] \end{align*}$$

Now we have to evaluate the constants \$A\$ and \$B\$. From the initial conditions, you know that \$i(0^-)=0\$ and current through an inductor can not vary instantly. Which gives: $$0=A+B\tag1$$

Now we need one more condition. We know that voltage across capacitor can not vary instantly, and \$v(0^-)=16\$.

$$16=-3A-27B\tag2$$

From (1) and (2) you can solve for \$A\$ and \$B\$. You will get the same answer as given in solution manual once you submit these values to voltage expression.

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