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Help with First Order Differential

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    AWzLt5n.jpg


    3. The attempt at a solution

    The first part is fairly simple I think. It's just rate of accumulation = rate of generation - rate of output(losses)

    I'm not too sure how to solve this differential equation. I divide the whole equation through by mc and rearrange but I keep getting an extra constant and I have no idea what to do with it.

    I get someone in the form dtheta/dt + P(t)theta = Q(t) + constant and I have no idea how to solve that. Any help will be appreciated, thank you.
     
  2. jcsd
  3. Mar 8, 2015 #2

    wabbit

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    The term that is messing things up is ## mc_vkt ##. So, first solve the differential equation without that term. You'll get a solution of the form Constant*Something. This gives you an idea of how the solution might look like : try the form C(t)*Something (same Something, now C(t) is what you solve for.)
     
  4. Mar 8, 2015 #3
    Hey, thanks for replying. Surely if I just remove a term from the differential I won't be solving what I'm supposed to or am I missing something else? :S
     
  5. Mar 8, 2015 #4

    wabbit

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    Of course you won't, but it's like a warm up lap. The solution to the simplified equation serves you as a basis from which you build the solution to the full equation - this is a very common and useful strategy in diff equations (and not only there)

    I told you how in first post
     
  6. Mar 8, 2015 #5

    LCKurtz

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    Divide your equation through by ##mc_v## and put it in the form$$
    \frac{d\theta}{dt}+\alpha \theta = \alpha \theta_w + kt$$That is a constant coefficient linear DE. You can solve it using constant coefficient methods for homogeneous and non-homogeneous equations, or use an integrating factor ##e^{\alpha t}##.
     
  7. Mar 10, 2015 #6
    Hey thanks for answering, I thought about that but surely that's not in the correct form to use an integrating factor? From my notes on the left it should just be Q(t) but we have a constant term in there as well.
     
  8. Mar 10, 2015 #7

    SammyS

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    What do you mean by "it" in ... on the left it should just be Q(t) ?

    That is the correct integrating factor.

    What is the derivative of ## \theta\,e^{\alpha\, t}\ ##?
     
  9. Mar 10, 2015 #8
    No I get that, it's just that I thought in order to use the integrating factor differentials had to be in the form dy/dx + p(x)y = Q(x)

    and LCKurtz reply mentioned dividing through by mc to get dy/dx + p(x)y = Q(x) + constant
    I'm just wondering is it still okay to use the IF here?
     
  10. Mar 10, 2015 #9

    LCKurtz

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    Why don't you just try it and see what happens?
     
  11. Mar 10, 2015 #10
    Fine, I'll give it a go now.
     
  12. Mar 10, 2015 #11

    Ah, it worked out, thank you for your help.
     
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