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

  • Thread starter BOAS
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  • #1
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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. Homework Statement


[itex]\frac{d^{2}q}{dt^{2}} + \frac{R}{L} \frac{dq}{dt} + \frac{1}{LC}q = 0[/itex] 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 [itex]R^{2} < \frac{4L}{C}[/itex] the solution has the form [tex]q = Ae^{-(\frac{R}{2L})t}cos(\sqrt{(\frac{1}{LC} - \frac{R^{2}}{4L^{2}}t)} + \phi)[/tex]

Homework Equations




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.

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

[itex]\frac{dq}{dt} = f(x)g'(x) + g(x)f'(x)[/itex]

[itex]f'(x) = - \frac{R}{2L}Ae^{-\frac{R}{2L}t}[/itex]

[itex]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}}[/itex]

[itex]\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})[/itex]

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:

Answers and Replies

  • #2
LCKurtz
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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
epenguin
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In your first equation q = ... and elsewhere, shouldn't the t be outsid the ) bracket?
 
  • #4
rude man
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In your first equation q = ... and elsewhere, shouldn't the t be outsid the ) bracket?
Right!
 

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