Kirchhoff’s laws yield $I_{R}=I_{L}=I_{C}$ and $V_{R}+V_{L}+V_{C}=V$. Using the properties of our three elements we arrive at the second-order linear differential equation
$$
L \vec{I}(t)+R \dot{I}(t)+\frac{1}{C} I(t)=\dot{V}(t)
$$
for the current $I$. Let us try to solve this equation for an external sinusoidal voltage
$$
V(t)=V_{0} \cos (\omega t)
$$
It turns out convenient to use the complex voltage $V(t)=V_{0} \mathrm{e}^{\mathrm{i} u t}$ :
$$
\vec{I}+\frac{R}{L} \dot{I}+\frac{1}{L C} I=i \frac{\omega V_{0}}{L} \mathrm{e}^{\mathrm{i} t} \text {. }
$$
We get the solutions for $V(t)=V_{0} \cos (\omega t)$ and $V(t)=V_{0} \sin (\omega t)$ by taking real and imaginary part of the complex solution, respectively.
The eigenvalues are
$$
\alpha_{1,2}=-\eta \pm \sqrt{\eta^{2}-\omega_{0}^{2}}
$$
where we have introduced the convenient abbreviations
$$
\eta=\frac{R}{2 L} \quad \text { and } \quad \omega_{0}=\frac{1}{\sqrt{L C}} .
$$
If $\eta>\omega_{0}$ (over damping), both eigenvalues are negative and the solution of the homogenous equation is given by
$$
I_{h}(t)=k_{1} e^{\alpha_{1} t}+k_{2} e^{\alpha_{2} t} \text {. }
$$
If $\eta=\omega_{0}$ (critical damping), both eigenvalues are equal and the solution of the homogenous equation is given by
$$
I_{h}(t)=\left(k_{1}+k_{2} t\right) \mathrm{e}^{-i t}
$$
Finally, for $\eta<\omega_{0}$ (under damping) we have complex conjugate eigenvalues and the solution of the homogenous equation is given by $$ I_{h}(t)=k_{1} \mathrm{e}^{-\eta t} \cos (\beta t)+k_{2} \mathrm{e}^{-\eta t} \sin (\beta t), \quad \beta=\sqrt{\omega_{0}^{2}-\eta^{2}}>0
$$
MATH4063 COURSE NOTES ：
Look at the second-order equation
$$
\vec{x}+c_{1} \dot{x}+c_{0} x=g
$$
and let $\alpha_{1}, \alpha_{2}$ be the corresponding eigenvalues (not necessarily distinet). Show that the equation an be factorized as
$$
\vec{x}+c_{1} \dot{x}+c_{0} x=\left(\frac{d}{d t}-\alpha_{2}\right)\left(\frac{d}{d t}-\alpha_{1}\right) x .
$$
Hence the equation can be reducel to solving two first order equations
$$
\left(\frac{d}{d t}-\alpha_{2}\right) y=g, \quad\left(\frac{d}{d t}-\alpha_{1}\right) x=y .
$$
Use this to prove in the case $n=2$. What can you say about the structure of the solution if $g(t)=p(t) e^{\beta t}$, where $p(t)$ is a polynomial. Can you also do the general case $n \in \mathbb{N}$ ? (Hint: The solution for the first order case is given Moreover, $\int p(t) \mathrm{e}^{3 t} d t=q(t) \mathrm{e}^{3 t}$, where $q(t)$ is a polynomial of degree $\operatorname{deg}(q)=\operatorname{deg}(p)$ if $\beta \neq 0$ and $\operatorname{deg}(q)=\operatorname{deg}(p)+1$ if $\beta=0$. For the general case use induction.)