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frank0520

  • one year ago

Use separation of variables to solve the differential equation: K dN/dt = -r(N-K)(N-A) , K,r,A are constants.

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  1. frank0520
    • one year ago
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    here is a picture

  2. freckles
    • one year ago
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    \[K \frac{dN}{dt}=-r(N-K) (N-A) \\ K dN=-r(N-K)(N-A) dt \text{ divide both sides by } (N-K)(N-A) \\ \frac{K}{(N-K)(N-A)} dN=-r dt \text{ now integrate both sides }\]

  3. frank0520
    • one year ago
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    that's the part I'm stuck on. the left hand side because the right is easy

  4. freckles
    • one year ago
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    \[\frac{K}{(N-K)(N-A)}=\frac{a}{N-K}+\frac{b}{N-A} \\ \frac{K}{(N-K)(N-A)}=\frac{a(N-A)+b(N-K)}{(N-K)(N-A)} \\ \frac{K}{(N-K)(N-A)}=\frac{(a+b)N+(-aA-bK)}{(N-K)(N-A)} \\ \text{ so we have teh system } \\ a+b=0 \\ -aA-bK=K \\ \text{ so the first equation gives } a=-b \\ \text{ and using the first \in the second we have } \\ bA-bK=K \\ \text{ solving for } b \\ b=\frac{K}{A-K} \\ \text{ and so } a=-b=\frac{-K}{A-K}\]

  5. freckles
    • one year ago
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    what I'm trying to suggest from the above is to use partial fractions

  6. freckles
    • one year ago
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    are you there? still having trouble?

  7. freckles
    • one year ago
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    the left hand side you should be evaluating.. \[\frac{-K}{A-K} \int\limits \frac{1}{N-K} dN+ \frac{K}{A-K} \int\limits \frac{1}{N-A} dN\]

  8. frank0520
    • one year ago
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    so would it be: \[\frac{ -K }{ A-K }\ln|N-K| + \frac{ K }{ A-K }\ln|N-A| = -r t + C\]

  9. freckles
    • one year ago
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    yep

  10. frank0520
    • one year ago
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    Ok, Thank for the Help!

  11. freckles
    • one year ago
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    did you understand what I did above for the partial fraction part?

  12. frank0520
    • one year ago
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    Yes, when I did myself I put the -r with the partial fractions, which made it harder like: \[\frac{ K }{ -r(N-K)(N-A) } dN\] but now it more clearer with your way.

  13. freckles
    • one year ago
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    cool stuff

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