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anonymous

  • one year ago

Write a DE in the form dy/dt=ay+b ;y=2/3 t->0 1) dy/dt=a(y+b/a) 2)dy/(y+b/a)=a dt I understand why it integrates as but do not understand why the there is a -ln(c) and not +c on both sids. 3)ln(y+b/a)-ln(c)=a*t

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  1. Loser66
    • one year ago
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    Can you take a snapshot of the original problem?

  2. freckles
    • one year ago
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    minus ln(c) is still subtract a constant why can't it take this form?

  3. phi
    • one year ago
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    first , if you have +c1 and +c2 on both sides, you can combine them into , for example, +c2-c1 and rename that as C (an arbitrary constant) so people usually put the +C on only one side

  4. anonymous
    • one year ago
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    Why is it ln(c). It does not seem very intuitive.

  5. freckles
    • one year ago
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    I think they wrote the constant as ln(c) so they can write the left hand side as one term

  6. freckles
    • one year ago
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    \[\ln(y+\frac{b}{a})-\ln(c) = \ln(\frac{y+\frac{b}{a}}{c})\]

  7. phi
    • one year ago
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    second, ln(C) is also arbitrary. they are doing that because usually what we do is ln y = x + C make each side the exponent of e y = e^(x+C) y= e^C * e^x and e^C is renamed an arbitrary constant , A (for example) y = A e^x

  8. anonymous
    • one year ago
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    Oh ok!!!

  9. phi
    • one year ago
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    the idea is if you use ln(C) rather than c you can do some algebra, and simplify the answer

  10. anonymous
    • one year ago
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    I thought there was some hidden rule

  11. anonymous
    • one year ago
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    That is just something that you have to train yourself to see?!

  12. anonymous
    • one year ago
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    is ln(c) more of a substitute for C so that you can simplify easier?

  13. freckles
    • one year ago
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    yes the C took on the form ln(c) just to write things prettier like

  14. anonymous
    • one year ago
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    THANK YOU EVERYONE!!!

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