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theEric
ODE: Finding \(Q(t)\) in \(R\dfrac{\mathrm dQ}{\mathrm dt}+\dfrac{Q}{C}=V\) Would anyone walk me through an ODE problem? Chapter 1 in a book called Elementary Differential Equations, so I don't think it would be so bad for someone who is good at this stuff...
So far all of the example problems solved for \(\dfrac{\mathrm dQ}{\mathrm dT}\) and divided both sided by the other side or it's opposite. Then integrated with respect to \(t\). So then the one side would integrate with respect to \(Q\) and the other side would be like \(\int 1 \mathrm dt\) or \(-\int 1 \mathrm dt\). But this problem isn't that simple. It might be possible to get to it, but I don't see how at the moment.
i can tell you a short method which you can use this this type of differential equation. have you heard of the radioactivity equation?
I have not.. Sounds exciting, for sure. I'm not a quick learner, but I'll be happy to learn. I might of even come across something else in this chapter. Would you like to show me, or should I look that one up? Either works for me!
you can treat the equation like this \[ R\dfrac{\mathrm dQ}{\mathrm dt}+\dfrac{Q}{C}=V \\ R\dfrac{\mathrm dQ}{\mathrm dt}=-(\dfrac{Q}{C}-V)\\ R\dfrac{\mathrm dQ}{\mathrm (\dfrac{Q}{C}-V) } = -dt \]
when you have a differential equation of the form\[\frac{ d N }{ dt }=- \lambda N\] try you solve this equation. Can you?
notice that the derivative of the denominator (Q/C - V) is 1/C dQ so all you need is a 1/C factor up top.
@Abhishek619 \(\int \dfrac{\mathrm dN/\mathrm dt}{N}\mathrm dt=-\int\lambda\mathrm dt\\ =\int N^{-1}\mathrm dN=\ln\left| N\right| + c_1=\lambda t +c_2\\\implies \ln\left| N\right| =\lambda t +c\\\implies N=e^{\lambda t +c}=e^{\lambda t}c \) @phi Why do we take the derivative of \(\dfrac{Q}{C}-V\)? Are we taking the derivative of both sides? Otherwise, I feel like it's something I should know but don't! :P
check once again. there's a '-ve' sign missing somewhere.
\( \int \dfrac{\mathrm dN/\mathrm dt}{N}\mathrm dt=-\int\lambda\mathrm dt\\ =\int N^{-1}\mathrm dN=\ln\left| N\right| + c_1=-\lambda t +c_2\\\implies \ln\left| N\right| =-\lambda t +c\\\implies N=e^{\lambda t +c}=e^{-\lambda t}c \)
a -ve sign in from of lambda in your equation. \[N=e ^{-\lambda t} c\] \[c=N _{t=O}\] the solution to the above DE is \[N=N _{t=O}e ^{- \lambda t}\] Now try to relate it to your actual question.
@Abhishek619 Please, type out the answer to save time. I don't think theEric is a cheating student (me neither). I am waiting for the new.
Haha, this looks good! I did ask for help to get through it :)
@Loser66 Funny you. I believe, those which we can deduce and generalise them in a broader sense, would ease our life in many ways. LOL Anyways, i'll post the solution now.
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I understand your example, @Abhishek619 . Now I need to apply it. I did it similar to how @phi did, at first. \(R\int\left(\dfrac{Q}{C}-V\right)^{-1}\mathrm dQ=-t+c\) ... I got it! I guess I was just afraid to integrate the left side! And I now see why @phi mentioned that derivative.
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Oh, that's a neat technique! Thank you very much! :D