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Differential Equation Initial Value Problem

  1. Feb 14, 2017 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    A Solve the following initial value problem:

    ##\frac{dx}{dt}=-x(1-x)##
    ##x(0)=\frac{3}{2}##

    B. At what finite time does ##x→∞##

    2. Relevant equations


    3. The attempt at a solution

    ##\frac{dx}{dt}=x(x-1)##
    ##\frac{dx}{x(x-1)}=dt##
    Partial fractions: ##dx(\frac{-1}{x}-\frac{1}{x-1})=dt##
    Integrating both sides: ##ln|\frac{1}{x}|-ln|x-1|=t+c##
    ##ln|\frac{1}{x(x-1)}|=t+c##
    e to the power of both sides and taking the constant ##e^c## as A: ##\frac{1}{x(x-1)}=Ae^t##
    Plugging in the initial value gives me ##A=\frac{4}{3}##

    My final equation is: ##\frac{1}{x(x-1)} = \frac{4}{3}e^t##

    What I don't understand is how ##x## is related to ##t## and how to figure out at what time x goes to infinity. Offhand I don't see any way for X to increase to infinity since t is an exponent of e, unless t goes to negative infinity.
     
  2. jcsd
  3. Feb 14, 2017 #2

    LCKurtz

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    Those fractions don't add up to what you started with.

    You can always solve for ##x## in terms of ##t##.
     
  4. Feb 14, 2017 #3

    cnh1995

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    Fix the partial fractions. A sign error can turn division into multiplication when you take logarithm.
     
  5. Feb 14, 2017 #4

    Drakkith

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    Gah! That's no good! Looks like I dropped a negative sign.

    New fractions: ##\frac{1}{x-1}-\frac{1}{x}##
    That means the problem becomes: ##\frac{x-1}{x}=Ae^t##
    ##1-\frac{1}{x}=Ae^t##
    ##1-\frac{1}{\frac{3}{2}}=A##
    ##A=\frac{1}{3}##

    That puts the final equation as: ##x=\frac{3}{3-e^t}##
    That means that x goes to infinity at t=3.

    Indeed. It's just more difficult when your equation isn't correct in the first place. :rolleyes:
     
  6. Feb 14, 2017 #5

    LCKurtz

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    Hopefully, that ##t=3## is just a typo.
     
  7. Feb 15, 2017 #6

    Drakkith

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    I take it you mean that it should be ##t→3##?
     
  8. Feb 15, 2017 #7

    LCKurtz

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    No. ##t=3## is not the correct answer.
     
  9. Feb 15, 2017 #8

    cnh1995

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    The entire denominator should tend to zero.
     
  10. Feb 15, 2017 #9

    Drakkith

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    Christ, please tell me it's when ##e^t→3##...
     
  11. Feb 15, 2017 #10

    Drakkith

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    Indeed. I promise I'm not as stupid as I appear... maybe. :rolleyes:
     
  12. Feb 15, 2017 #11

    cnh1995

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    Right.
     
  13. Feb 15, 2017 #12

    LCKurtz

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    What value of ##t## gives ##e^t = 3##?
     
  14. Feb 15, 2017 #13

    Drakkith

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    I believe it's ##ln(3)##. But my basic math has been so bad for this question that I don't know if I'd double down on that bet.
     
  15. Feb 15, 2017 #14

    LCKurtz

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    ##\ln 3## is correct.
     
  16. Feb 15, 2017 #15

    cnh1995

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    Relax! That's just a silly mistake..No big deal!
    Yes.
     
  17. Feb 15, 2017 #16

    Drakkith

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    Relax? I don't see that equation on my formula sheets... :-p
     
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