Using KVL for a mesh with a switch and inductor in parallel with capactior

Question

I'm currently studying for second order circuits, and I'm stuck understanding this example in my book pictured at the bottom.

1. I got the same KCL equation, but I don't understand why the KVL equation on the right-hand mesh isn't $$v = \frac{1}{C}\int i(t)dt +6i+{L}\frac{di}{dt}$$ The book just has: $$v = 6i+{L}\frac{di}{dt}$$ and I don't understand why you don't account for the voltage across the capacitor. Even if it doesn't initially have voltage across it, it would still be changing, right? Because I think that we're concerned about the rate of change, not just the initial value or the infinity value.
2. I am also a little confused why you cannot just take the derivative of the KCL equation like this: $$\frac{v}{4}+i_{L}+i_{C}= \frac{v_{s}}{4}$$ $$\frac{d}{dt}(\frac{v}{4}+\frac{1}{L}\int v(t)dt +C\frac{dv}{dt}=\frac{v_{s}}{4})$$

$$\frac{d^{2}v}{dt^{2}}+\frac{dv}{dt}+4 = \frac{dv_{s}}{dt}$$ which leads to the characteristic equation for the problem $$\lambda ^{2}+\lambda +4 =0$$

Thank you in advance for any help!

Answer 1

I don't understand why you don't account for the voltage across the capacitor.

\$v\$ is the voltage across the capacitor. They've accounted for it by putting it in the left hand side of the equation. They just haven't written it in terms of the mesh current and constitutive relation of the capacitor yet.

I am also a little confused why you cannot just take the derivative of the KCL equation like this ...

Your version has

$$i_L = \frac{1}{L}\int v(t)dt$$

This isn't correct because it ignores the effect of the 6 ohm resistor.