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Ok I have the equation:

$$ R\frac{dq(t)}{dt} + \frac{q(t)}{C} - V = 0 $$

and have been asked to find the general solution using separation of variables. I am unsure if I am rearranging the the equation correctly. Here is my attempt:

$$ CR \ dq(t) = (-q(t) + VC) \ dt $$

Which then I integrate both sides giving me:

$$ CR \ q(t)=-q(t)t+VCt $$

Which I rearrange in terms of q(t):

$$ q(t) = \frac{VCt}{CR+t} $$

I just wanted to confirm I performed the steps correctly. Alot of the online examples are in terms of $x$ and $y$, seeing the $dq(t)$ has really confused my perspective of the solution.

Thanks

colormegone
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user142973
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  • You could always use find and replace on a text edit to perform $q\to y$. – Git Gud May 02 '14 at 01:40
  • Yeah that's true, but then would $\frac{dq(t)}{dt}$ equal $y'$? If so then there's only one 'function' in the equation? – user142973 May 02 '14 at 01:42
  • Yes, the notation $\dfrac{dq(t)}{dt}$ is short for $\dfrac{dq}{dt}(t)$ which is the derivative of $q$ evaluated at $t$, just like $q'(t)$. – Git Gud May 02 '14 at 01:44
  • The thing that bothers me is that when I integrate both sides, I'm integrating a function $q(t)$ and treating it like a variable, whereas it should have some sort of function with respect to $t$ within it. Am I correct in assuming it's a variable and just integrating it? – user142973 May 02 '14 at 01:49
  • Very nice observation. Your question is answered here. TL.DR: you can treat as a variable, it works, even though it doesn't make sense. – Git Gud May 02 '14 at 01:51
  • In differential equations, functions are among the "variables"... – colormegone May 02 '14 at 01:52
  • @RecklessReckoner But they are variables of a different order-type than what the OP is talking about. – Git Gud May 02 '14 at 01:53
  • I am using the term in the sense of what the method is called -- it is a little strange to say "separation of variables", as much as doing so in the method "variation of parameters". These, I believe, are survivals of terminology from the early history of differential equation study. – colormegone May 02 '14 at 01:57
  • I wonder if this has a cleaner interpretation with nonstandard analysis / hyperreals. – joeA May 02 '14 at 02:22

1 Answers1

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The "variables", which are $ \ q \ $ and $ \ t \ $ , have not actually been separated in your expression: you need to divide $ \ CR \ dq = (-q(t) + VC) \ dt \ $ through to write

$$ \frac{dq}{q \ - \ VC} \ = \ -\frac{1}{CR} \ dt \ \ , $$

before proceeding to integrate both sides. This can now be accomplished with a $ \ u-$ substitution.

colormegone
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  • Thankyou for that, I just had an epiphany when I replaced the $q(t)$ with $y$ and realised that there is no dependant $t$ variable as with most $x$ and $y$ examples. That really helps, thanks again. – user142973 May 02 '14 at 01:56
  • Epiphanies are always good -- I'm glad this was helpful. – colormegone May 02 '14 at 01:59