I believe I get a wrong result...
$$ \frac{d²i}{dt²} + \frac{R}{L} \frac{di}{dt} - \frac{1}{LC} = 0$$
I think besides being wrong or right, a result should also be useful. If you arrived at a differential equation for $i$, but have boundary value for $U_C(t)$, that is $U_C(0) = U_{C,0}$, then I wonder if this differential equation is the result you are looking for.
I am lead to believe that my convention for $U_C$ is the problem
The convention is arbitrary as long as you stick to it.
Let's add some currents to the schematic:
---------------
| | | |
| V Ic R | Ur
| | | V
Uc | C Irc V -
| | | |
| | L | Ul
V | | V
---------------
The indices in the schematic are lower case for readability while in the formulas they are upper case.
As you said
$$U_C = U_R + U_L$$
additionally
$$I_C = C\dot U_C$$
$$U_L = L\dot I_{RL}$$
$$U_R = RI_{RL}$$
from the circuit it looks like
$$I_C = -I_{RL}$$
which gives
$$U_L = L\dot I_{RL}= L\dot I_C$$
$$U_R = RI_C$$
and with $I_C = C\dot U_C$ and $\dot I_C = C\ddot U_C$, respectively
$$U_L = L\dot I_C = LC\ddot U_C$$
$$U_R = RI_C = RC\dot U_C $$
you end up with a differential equation for $U_C$ which is what you are looking for (I guess)
$$0= LC\ddot U_C + RC\dot U_C - U_C$$
clumsy me taking over, trying to solve this
$$0= LC\lambda^2 + RC\lambda - 1$$
$$0= \lambda^2 + \frac{R}{L}\lambda - \frac{1}{LC}$$
$$\lambda_{1,2}=\frac{-RC\pm\sqrt{(RC)^2-4}}{2LC}$$
with the general solution for $U_C$
$$U_C(t) = a_1e^{\lambda_1 t} + a_2e^{\lambda_2 t}$$
interpreting the different kinds of possible $\lambda_{1,2}$
- $(RC)^2 = 4$, yielding only one solution from the root
$$\lambda_{1,2}=\lambda=\frac{-RC\pm\sqrt{(RC)^2-4}}{2LC}=\frac{-2\pm\sqrt{0}}{2LC}=-\frac{1}{2LC}$$
R,C and L are all > 0, which is why I can conclude that $RC=2$, not $\pm2$. I also say that $\lambda <0$ for the same reason. The general solution
$$U_C(t) = ae^{\lambda t}$$
with the boundary value yields
$$U_C(0) = U_{C,0}= ae^{\lambda 0} = a$$
hence
$$U_C(t) = U_{C,0}e^{\lambda t}$$
With $\lambda <0$ the function decays over time or is damped over time as you put it. So it actually does what you expect it to do. Good?
- if $(RC)^2>4$ then $\lambda_{1,2} \in \mathbb{R}$, which is the same as the general solution
$$U_C(t) = a_1e^{\lambda_1 t} + a_2e^{\lambda_2 t}$$
With two variables $a_1$ and $a_2$, the single boundary value is not enough to determine the two. But I say that $\lambda_{1,2} <0$, because $RC>\sqrt{(RC)^2-4}$. The same reasoning as in the first case applies: $U_C(t)$ is damped over time. Ok?
- the last case is $(RC)^2<4$ so that $\lambda_{1,2} \in \mathbb{C}$
$$\lambda_{1,2}=\alpha \pm\beta i$$
with the solution for $U_C(t)$
$$U_C(t)=e^{\alpha t}\sin (\beta t) + e^{\alpha t}\cos (\beta t)$$
Only the real part of $\lambda_{1,2}$ influences the damping of the function. If split into real and imaginary part $\lambda_{1,2}$ looks like this
$$\lambda_{1,2}=\underbrace{{\frac{-RC}{2LC}}}_{\alpha}\pm\underbrace{\frac{\sqrt{(RC)^2-4}}{2LC}}_{\beta i}$$
Given that $\alpha <0$, the function $U_C(t)$ is also damped in this case. It does some fancy wobbling, but it's still damped.
The conclusion is that no matter how the values are, $U_C(t)$ is always damped.
Best Answer
You can use current divider law and we have $i_{R_3}=\frac{i\omega L_2}{\sqrt{(R_3)^2+(\omega L_2)^2}}$ or $i_{L_2}=\frac{iR_3}{\sqrt{(R_3)^2+(\omega L_2)^2}}$ where $\omega$ is angular frequency of the circuit. Plugging any of the two into your equation will yield a differential equation in $i,t$ which I think should be solvable.Note that $i_{L_2}$ will be complex part of the current $i$ . I have ignored that as calculations can be done with real part too.