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anonymous

  • 5 years ago

can anyone help me get started on this? dh/dt= 0.016(2.5-h) Solve the differential equation to find h in terms of t.

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  1. anonymous
    • 5 years ago
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    dh/(2.5-h)=0.016 dt now integrating, -log(2.5-h)=0.016 t+c

  2. anonymous
    • 5 years ago
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    Is this enough, ersp?

  3. anonymous
    • 5 years ago
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    i think so my brains kinda died doing this work XD thanks :)

  4. anonymous
    • 5 years ago
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    ok :)

  5. anonymous
    • 5 years ago
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    actually i dont get where the minus log comes from when integrating

  6. anonymous
    • 5 years ago
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    Comes from the -h...

  7. anonymous
    • 5 years ago
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    \[\int\limits_{}{}\frac{dh}{2.5-h}=-\int\limits_{}{}\frac{(2.5-h)'dh}{2.5-h}=-\log (2.5-h)+c\]

  8. anonymous
    • 5 years ago
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    I got bumped. Is this explanation okay?

  9. anonymous
    • 5 years ago
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    i think so cheers

  10. anonymous
    • 5 years ago
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    so what method is this?

  11. anonymous
    • 5 years ago
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    Method? To solve the differential equation?

  12. anonymous
    • 5 years ago
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    The d.e. is separable.

  13. anonymous
    • 5 years ago
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    Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.

  14. anonymous
    • 5 years ago
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    thanks makes a bit more sense to me now, my mind is really failing me today!

  15. anonymous
    • 5 years ago
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    np ;)

  16. anonymous
    • 5 years ago
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    shouldnt both sides be logged?

  17. anonymous
    • 5 years ago
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    No. It goes like this:\[\frac{dh}{dt}=0.016(2.5-h) \rightarrow \frac{dh}{2.5-h}=0.016dt \rightarrow \int\limits_{}{}\frac{dh}{2.5-h}=\int\limits_{}{}0.016dt\]\[\rightarrow -\log (2.5-h) = 0.016t + c \rightarrow \log (2.5-h)=-0.016t+c_1 \rightarrow 2.5-h=c_2e^{-0.016t}\]\[\rightarrow h=2.5-ce^{-0.016t}\]

  18. anonymous
    • 5 years ago
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    I added subscripts as I went along, but we just lump constants into constants, so normally, the subscripts are left off...which is what I did in the end.

  19. anonymous
    • 5 years ago
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    ah that makes sense thanks very much, this work ive been throwing my head in to a wall for the last few days!

  20. anonymous
    • 5 years ago
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    lol, it happens :p

  21. anonymous
    • 5 years ago
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    Thanks for fanning me, ersp :D

  22. anonymous
    • 5 years ago
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    after that much help it seems logical! XD

  23. anonymous
    • 5 years ago
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    ah it doesnt seem to fit for the next part of the question, at max h is 2.5, how long does it take to get to this point, so i have to rearrange it to make t the subject but the values dont seem to fit at all!

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