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anonymous

  • 5 years ago

looking to integrate a differential equation on both sides... d^2C/dz^2 = -zA* dC/dz how would I go about doing this?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    thanks! i'll be sure to give credit where credit is due. just to let you know, I simplified the constants in the equation into "A"

  3. amistre64
    • 5 years ago
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    you replied to an automated message that really does nothing but post that you have aquestion in the chatroom box :)

  4. amistre64
    • 5 years ago
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    there you are :)

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  5. anonymous
    • 5 years ago
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    thanks I noticed that when i clicked around. need an answer though!

  6. amistre64
    • 5 years ago
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    d^2C/dz^2 = -zA* dC/dz hmm.... havent done to many of these so I dont know alot of the techniques, except for swapping variables when doing a first derivative..

  7. anonymous
    • 5 years ago
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    its a secondary order differential equation.

  8. amistre64
    • 5 years ago
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    C'' = -zA C' ...

  9. amistre64
    • 5 years ago
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    Id say divide both sides by C'... but thats just a guess :)

  10. amistre64
    • 5 years ago
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    its outta my league.... good luck with it :)

  11. anonymous
    • 5 years ago
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    d^2C/dz^2 = -zA* dC/dz (D^2+Az)C=0 where D=d/dz

  12. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=dC%28dC%2Fdz%29%2Fdz+%3D+-zA+%28dC%2Fdz%29

  13. anonymous
    • 5 years ago
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    thanks anyway. its for transport phenomenon in biological systems :P

  14. anonymous
    • 5 years ago
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    solution says that form is in a form of a natural exponential. can anyone explain?

  15. anonymous
    • 5 years ago
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    i tried to recall the solution of higher order DE, but could not,away from it for a long time,i just remember the formation of auxilliary eq

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