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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

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

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

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

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

when you have a differential equation of the form$\frac{ d N }{ dt }=- \lambda N$ try you solve this equation. Can you?

6. phi

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

@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

check once again. there's a '-ve' sign missing somewhere.

9. theEric

Alright, thanks!

10. theEric

$$\int \dfrac{\mathrm dN/\mathrm dt}{N}\mathrm dt=-\int\lambda\mathrm dt\\ =\int N^{-1}\mathrm dN=\ln\left| N\right| + c_1=-\lambda t +c_2\\\implies \ln\left| N\right| =-\lambda t +c\\\implies N=e^{\lambda t +c}=e^{-\lambda t}c$$

11. anonymous

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

@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

Haha, this looks good! I did ask for help to get through it :)

14. theEric

Thank you, though!

15. anonymous

@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

|dw:1388965390357:dw|

17. theEric

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

|dw:1388965549025:dw|

19. theEric

Oh, that's a neat technique! Thank you very much! :D

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