Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

population differential eq problem?

I'm given the eqn dP/dt= cln(K/P)P where c= .15, K= 1000 and P(0)=100 and I'm asked to solve the diff eq for P(t) with the given conditions and to find at what P does P grow fastest?

Any help would be great thanks!

Update:

c is a constant and K is the carrying capacity

1 Answer

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  • 1 decade ago
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    dP/dt= c ln(K/P)P

    dP/{P ln (k/P)} = c.dt.......................(i)

    put ln (k/P) = y

    (k/P) = e^y

    P= k e^-y

    differentiate

    dP = -k(e^-y) dy......................(ii)

    integrate[ -k dy/y]

    -k ln y = ct + C

    -k ln{ln (k/P)} = ct +C...................(iii)

    ln{ ln(k/P)} = -ct/k -C/ k

    ln (k/P) = e^(-ct-C)

    ln k - ln P = e^(-ct-C)

    ln P = ln k + e^(-ct -C).......................(iv)

    Put K= 1000, c = 0.15 when t= 0 P(0) =100

    ln P(0)= ln 1000 + e^(-0-C)

    ln P(0) = 3 + e^-C

    ln P(o) -3 = e^-c...................................(v)

    ln P = ln k + e^(-ct){ln P(o) - 3}.......................(iv)..................Ans

    since dP/dt= c {ln(K/P)}P > 0 because P=100, c=0.15 and ln(k/P) also positive,

    therefore P is increasing...................................ans

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