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theEric

  • 2 years ago

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...

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  1. theEric
    • 2 years ago
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    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.

  2. anonymous
    • 2 years ago
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    i can tell you a short method which you can use this this type of differential equation. have you heard of the radioactivity equation?

  3. theEric
    • 2 years ago
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    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!

  4. phi
    • 2 years ago
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    you can treat the equation like this \[ R\dfrac{\mathrm dQ}{\mathrm dt}+\dfrac{Q}{C}=V \\ R\dfrac{\mathrm dQ}{\mathrm dt}=-(\dfrac{Q}{C}-V)\\ R\dfrac{\mathrm dQ}{\mathrm (\dfrac{Q}{C}-V) } = -dt \]

  5. anonymous
    • 2 years ago
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    when you have a differential equation of the form\[\frac{ d N }{ dt }=- \lambda N\] try you solve this equation. Can you?

  6. phi
    • 2 years ago
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    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.

  7. theEric
    • 2 years ago
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    @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

  8. anonymous
    • 2 years ago
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    check once again. there's a '-ve' sign missing somewhere.

  9. theEric
    • 2 years ago
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    Alright, thanks!

  10. theEric
    • 2 years ago
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    \( \int \dfrac{\mathrm dN/\mathrm dt}{N}\mathrm dt=-\int\lambda\mathrm dt\\ =\int N^{-1}\mathrm dN=\ln\left| N\right| + c_1=-\lambda t +c_2\\\implies \ln\left| N\right| =-\lambda t +c\\\implies N=e^{\lambda t +c}=e^{-\lambda t}c \)

  11. anonymous
    • 2 years ago
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    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.

  12. Loser66
    • 2 years ago
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    @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.

  13. theEric
    • 2 years ago
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    Haha, this looks good! I did ask for help to get through it :)

  14. theEric
    • 2 years ago
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    Thank you, though!

  15. anonymous
    • 2 years ago
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    @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.

  16. anonymous
    • 2 years ago
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    |dw:1388965390357:dw|

  17. theEric
    • 2 years ago
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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.

  18. anonymous
    • 2 years ago
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    |dw:1388965549025:dw|

  19. theEric
    • 2 years ago
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    Oh, that's a neat technique! Thank you very much! :D

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