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Math Help - doomsday equation..

  1. #1
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    doomsday equation..

    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?


    so....
    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
    does it always have to be function y of c ? can we just leave it at that?

    What should i do?
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  2. #2
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    Quote Originally Posted by hangainlover View Post
    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?


    so....
    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
    does it always have to be function y of c ? can we just leave it at that?

    What should i do?

    dy\slash dx=ky^{1+c}\Longleftrightarrow \int\frac{dy}{y^{1+c}}=\int k\,dx\Longrightarrow y^{-c}=kx+T\,,\,\,T= a constant \Longrightarrow y=(kx+T)^{-1\slash c} \Longrightarrow y_0=y(0)=T^{-1\slash c} \Longrightarrow T=y_0^{-c}\Longrightarrow y=\left(kx+y_0^{-c}\right)^{-1\slash c}

    Tonio
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