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Doomsday differential equation

  1. Feb 23, 2008 #1
    1. Let c be a positive number. a differential equation of the form
    dy/dx=ky^(1+c)
    Determine the solution that satisfies the initial condition y(0)=y(subzero)

    The solution is y(subzero)=(1)/((ckT)^(1/c))

    but I can't understand how to find it. ok when i integrate

    dy/y^(1+c)=kdt, I get y^(-c)/(-c)=kt+C and I don't know what do from there.
     
  2. jcsd
  3. Feb 23, 2008 #2

    Defennder

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    I googled the term "doomsday differential equation" and guess what? The top 2 entries were this thread itself and an older one which concerns the same DE:

    https://www.physicsforums.com/showthread.php?t=73150

    EDIT: Does anyone know why it's called the doomsday equation?
     
  4. Feb 23, 2008 #3
    that doesn't help me.
     
  5. Feb 23, 2008 #4
    Try leaving it in the form

    [tex]y^c = f(x)[/tex]

    and then solve for C (the integration constant, not the one in the form I just mentioned).
     
  6. Feb 23, 2008 #5

    arildno

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    Dearly Missed

    1. It is called the Doomsday equation because for any c>0, there exists a finite time T when y blows up (i.e, graphing y(t) with t as the argument axis, there would be a vertical asymptote to the graph at t=T, which the graph of y(t) will follow into infinity as t approaches T)

    2. Now, for the solution:
    We have that [tex]-\frac{1}{cy(t)^{c}}=kt+C in (')[/tex] where our initial condition y(0)=y_{0} now can be used to determine C:
    [tex]-\frac{1}{cy_{0}^{c}}=C[/tex]
    Inserting in (*) and simplifying, we have:
    [tex]\frac{1}{y(t)^{c}}=\frac{1}{y_{0}^{c}}-ckt=\frac{1-y_{0}^{c}ckt}{y_{0}^{c}}[/tex]
    Or, reciprocating, and taking the c-root on both sides:
    [tex]y(t)=\frac{y_{0}}{(1-y_{0}^{c}ckt)^{c}}[/tex]

    Now, the blow-up instant T must fulfill the equation:
    [tex]1-y_{0}^{c}ckT=0\to{T}=\frac{1}{cky_{0}^{c}}[/tex]
    Using the blow-up time T as our independent parameter, rather than y_0, we get from the same equation: [tex]y_{0}=\frac{1}{(ckT)^{\frac{1}{c}}}[/tex]

    The first result is simplest to use when you DO know the value of y_0, and wants to know WHEN the solution will blow up.

    The second result is most convenient when you know when the solution is going to blow up, and wants to know what was the initial value of y, i.e, y_0.
     
  7. Feb 23, 2008 #6
    thank you.
     
  8. Feb 23, 2008 #7

    HallsofIvy

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    I have no idea what you are saying. "The solution is y(subzero)=(1)/((ckT)^(1/c))" can't possibly be right- there is no y(t) in it so it can't be a solution to the differenital equation. Also, is T supposed to be t are is it a specific value of t?
     
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