# L-R-C Series Circuits - Help With Differential Equation

1. Jan 20, 2015

### BOAS

Hello,

this is a maths problem that is related to a physics problem, but I think it's best posted here due to what i'm asking about.

1. The problem statement, all variables and given/known data

$\frac{d^{2}q}{dt^{2}} + \frac{R}{L} \frac{dq}{dt} + \frac{1}{LC}q = 0$ is a differential equation describing how charge and current change with time in an LRC series circuit (found via Kirchoff's loop rule).

It is stated that when $R^{2} < \frac{4L}{C}$ the solution has the form $$q = Ae^{-(\frac{R}{2L})t}cos(\sqrt{(\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t)} + \phi)$$

2. Relevant equations

3. The attempt at a solution

What I would like to do is verify this, by finding the first and second derivative and substituting them in to the original equation.

$q = f(x)g(x)$ where $f(x) = Ae^{-(\frac{R}{2L})t}$ and $g(x) = cos(\sqrt{(\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t)} + \phi)$

$\frac{dq}{dt} = f(x)g'(x) + g(x)f'(x)$

$f'(x) = - \frac{R}{2L}Ae^{-\frac{R}{2L}t}$

$g'(x) = - \frac{R^{2} \sin(\sqrt{(\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t} + \phi))}{8L^{2} \sqrt{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t}}$

$\frac{dq}{dt} = Ae^{-(\frac{R}{2L})t} (-\frac{R^{2} \sin{\sqrt{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t}+ \phi}}{8L^{2} \sqrt{\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t}})+ cos(\sqrt{(\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t)} + \phi) (- \frac{R}{2L}Ae^{-\frac{R}{2L}t})$

I would then need to use the quotient rule and the product rule again, meaning i'm going to collect even more terms. Am I going about doing this in the correct manner? I assume everything is meant to cancel at the end, but this seems somewhat absurd...

I don't expect anyone to check my work, but I would greatly appreciate someone confirming or denying my method.

Thanks!

Last edited: Jan 20, 2015
2. Jan 20, 2015

### LCKurtz

I'm glad you don't expect anyone to check your work. For one thing, probably everywhere you have an $x$ it should be a $t$. What I would do if I were you is call $a = \frac R L$ and $b = \frac 1 {LC}$ and just solve the DE $q'' + aq' +bq = 0$. It's just a constant coefficient equation and you can take the case where the roots are complex conjugate. That's probably the reason for $R^2<\frac{4L} C$. Put the constants in at the end. It should be easy.

3. Jan 20, 2015

### epenguin

In your first equation q = ... and elsewhere, shouldn't the t be outsid the ) bracket?

4. Jan 21, 2015

Right!