Deriving second order ODE for an RLC circuit I'm working on deriving a second order DE for an RLC circuit. I'd like to use matrix form to make it easier, but I've come across something I'm not sure how to handle and am having trouble finding a definite answer on.
This is a school project so I'd appreciate the most minimal answers so I can continue working on my own.
$v_{C}, v_{L}$ and $i$ are variable over time, $R, L, C$ and $v_{T}$ are parameters, where $v_{T}$ can be altered.
Using these rules: 
$$v_{T}-Ri=v_{C}+v_{L}$$
$$Cv_{C}'=i$$
$$Li'=v_{L}$$
I've created the following system 
$$i'=\frac{v_{T}}{L}-\frac{Ri}{L}-\frac{v_{C}}{L}$$
$$v_{C}'=\frac{i}{C}$$
When $v_{T}$ is zero, I can easily find a second order eq.
$$Lv_{C}''+Rv_{C}'+\frac{v_{C}}{C}=0$$
But when $v_{T}$ is a non-zero parameter I'm left scratching my head. Specifically I'm confused as to how to approach the $\frac{v_{T}}{L}$ term. This is the first time I've encountered a system with a purely constant term in the system that has no immediately apparent relationship to either $i$ or $v_{C}$ in my original system. 
(We can change the value of $v_{T}$ before we plug it in, but $R,L,C$ are set in stone forever and always, i.e. the final system should have $v_{T}, R, L, C$ as parameters.)
 A: Well, according to 'Faraday's law' in a series RLC-circuit:
$$\text{V}_{\space\text{C}}\left(t\right)+0+\text{V}_{\space\text{R}}\left(t\right)-\text{V}_{\space\text{in}}\left(t\right)=-\text{V}_{\space\text{L}}\left(t\right)\tag1$$
Now, we know a few things:


*

*$$\text{I}_{\space\text{C}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{C}}'\left(t\right)\cdot\text{C}\tag2$$

*$$\text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}=\text{I}_{\space\text{in}}\left(t\right)\cdot\text{R}\tag3$$

*$$\text{V}_{\space\text{L}}\left(t\right)=\text{I}_{\space\text{L}}'\left(t\right)\cdot\text{L}=\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{L}\tag4$$


So, we get:
$$\text{V}_{\space\text{C}}'\left(t\right)+0+\text{V}_{\space\text{R}}'\left(t\right)-\text{V}_{\space\text{in}}'\left(t\right)=-\text{V}_{\space\text{L}}'\left(t\right)\space\Longleftrightarrow\space$$
$$\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+0+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}-\text{V}_{\space\text{in}}'\left(t\right)=-\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$
$$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}\space\Longleftrightarrow\space$$
$$\text{V}_{\space\text{in}}'\left(t\right)=\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\tag5$$
Assuming that the initial conditions are equal to $0$, and using Laplace transform:
$$\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)=\text{s}^2\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{L}+\text{s}\cdot\text{i}_{\space\text{in}}\left(\text{s}\right)\cdot\text{R}+\text{i}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space$$
$$\text{i}_{\space\text{in}}\left(\text{s}\right)=\frac{\text{s}\cdot\text{v}_{\space\text{in}}\left(\text{s}\right)}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag6$$
Using the 'Convolution Theorem' of the Laplace transform:
$$\text{I}_{\space\text{in}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{v}_{\space\text{in}}\left(\text{s}\right)\right]_{\left(\text{y}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}=$$
$$\int_0^t\text{V}_{\space\text{in}}\left(\text{y}\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}\space\text{d}\text{y}\tag7$$
And $\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{L}\cdot\text{s}^2+\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\right]_{\left(t-\text{y}\right)}$, equals:

A: Your equation is
$$ Li' + iR + v_C = v_T $$
Substitute $i = C{v_C}'$ gives
$$ L{v_C}'' + R{v_C}' + \frac{v_C}{C} = \frac{v_T}{C} $$
Which is more or less the same as what you already have, but with another term on the RHS. I'm not really sure what you're confused about. In a mathematical sense, there's nothing more to it. There's nothing strange about this equation other than it's inhomogeneous.
Are you asking what $v_T$ specifically means, and how it affect the other variables? For that, I suggest going to Physics SE for more details. The short answer is, it affects the long-term behavior of the system. If $R$ is sufficiently large (a damped system) then $v_C \to v_T$ as $t\to\infty$
If you're wondering how to solve this inhomogeneous equation, that's a different question entirely,and you should make it clear.
