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Doomsday equation help

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data
    P: Let c be a positive number. A differential equation of the form: dy/dt = ky^(1+c)

    where k is a positive constant, is called a doomsday equation because the equation in the expression ky^(1+c) is larger than that for natural growth (that is, ky).

    (a) Determine the solution that satisfies the initial condition y(0)=y(subzero)

    (b) Show that there is a finite time t = ta (doomsday) such that lim(t->T-) wy(t) = infinity

    (c) An especially prolific breed of rabbits has the growth term ky^(1.01). If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is doomsday?

    2. Relevant equations

    integration

    3. The attempt at a solution

    First i wanted to find solution to the dy/dx (meaning i integrated it)

    I got y^c = -c(kx+T)

    but i could not define it as function y because of the negative sign in front of C

    What should i do?
     
  2. jcsd
  3. Jan 30, 2010 #2

    Mark44

    Staff: Mentor

    I'm not sure what you mean by "solution to the dy/dx." There is no x in this problem, so it doesn't make sense to talk about dy/dx. And maybe you are using "dy/dx" as shorthand for differential equation, which is needlessly confusing.
    Your solution above is incorrect, and not only because you have x instead of t.

    The differential equation dy/dt = ky1+c is separable. Your next equation should be dy/(y1 + c) = k dt. Integrate both sides. What do you get?
     
  4. Jan 30, 2010 #3
    sorry what i meant by dx or x were dt and t
    i apologize for that confusion.
    so starting from dy/(y1 + c) = k dt
    If i integrate that, 1/(1-1-c)y^(-1-c+1) =kt + T (upper case T is the constant)
    -c^(-1)*y^(-c)=kt+T
    I got y^(-c)=-c(kt+T)

    what am i doing wrong?
     
  5. Jan 30, 2010 #4
    Not knowing how to isolate Y by getting rid of that the exponent, -c, i decided to plug in the initial condition to define the constant T.

    at t=0, y^(-c)=-c*T
    A(initial)^(-c) =-cT
    Therefore, T= ((A(initial)^(-c))/(-c)
     
  6. Jan 30, 2010 #5

    Mark44

    Staff: Mentor

    Looks fine so far. You can simplify it some more, though.
    [tex]y^{-c} = -c(kt + T)[/tex]
    [tex]\Rightarrow \frac{1}{y^c}= -(ckt + cT)[/tex]
    [tex]\Rightarrow y^c= \frac{-1}{ckt + cT}[/tex]
    [tex]\Rightarrow y = y(t) = \left(\frac{-1}{ckt + cT}\right)^{1/c}[/tex]

    Now use your initial condition that y(0) = y0 and continue from there to parts b and c of your problem.
     
  7. Jan 30, 2010 #6
    so at t=0, ((-1)/(cT))^(1/c) =A initial

    -1/(cT) = A (initial)^c
    T = -1 /(cA^c)

    after substituting that for the constant T, i get, y(t) = (-A^c)/((A^c)*(kt)-1)

    you can see that at t=1/k, as t approaches T from the negative side, the y value approahces infinity
    does this answer part b?
     
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