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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

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  • $\begingroup$ You could always use find and replace on a text edit to perform $q\to y$. $\endgroup$ – Git Gud May 2 '14 at 1:40
  • $\begingroup$ Yeah that's true, but then would $\frac{dq(t)}{dt}$ equal $y'$? If so then there's only one 'function' in the equation? $\endgroup$ – user142973 May 2 '14 at 1:42
  • $\begingroup$ 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 $q'(t)$. $\endgroup$ – Git Gud May 2 '14 at 1:44
  • $\begingroup$ 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? $\endgroup$ – user142973 May 2 '14 at 1:49
  • $\begingroup$ 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. $\endgroup$ – Git Gud May 2 '14 at 1:51
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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.

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  • $\begingroup$ 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. $\endgroup$ – user142973 May 2 '14 at 1:56
  • $\begingroup$ Epiphanies are always good -- I'm glad this was helpful. $\endgroup$ – colormegone May 2 '14 at 1:59

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