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Math Help - Growth model word problem w/differential equations

  1. #1
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    Growth model word problem w/differential equations

    Instructions:

    Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?

    Gompertz growth model:

    \frac{dy}{dt}=kyln\frac{60}{y}

    Here's what I did.

    dy=kyln\frac{60}{y}(dt)

    \frac{dy}{yln\frac{60}{y}}=kydt

    \int \frac{1}{y(ln60-lny)}dy=\int kdt

    u=ln60-lny
    du=-\frac{1}{y}dy
    -du=\frac{1}{y}dy

    - \int \frac {1}{u}du = \int kdt

    -ln|u| = kt + C

    -|ln60-lny| = e^{kt+c}

    -ln\frac{60}{y} = Ce^{kt}

    \frac{60}{y}=e^{-Ce^{kt}}

    60=e^{-Ce^{kt}}y

    My values for the variables are:

    y=0, t=0<br />
y=15, t=3<br />
y=?, t=10

    However, the solutions manual states the answer for the general soln is

     y=60e^{-Ce^{kt}} which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by emttim84 View Post
    Instructions:

    Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?

    Gompertz growth model:

    \frac{dy}{dt}=kyln\frac{60}{y}

    Here's what I did.

    dy=kyln\frac{60}{y}(dt)

    \frac{dy}{yln\frac{60}{y}}=kydt

    \int \frac{1}{y(ln60-lny)}dy=\int kdt

    u=ln60-lny
    du=-\frac{1}{y}dy
    -du=\frac{1}{y}dy

    - \int \frac {1}{u}du = \int kdt

    -ln|u| = kt + C

    -|ln60-lny| = e^{kt+c}

    -ln\frac{60}{y} = Ce^{kt}

    \frac{60}{y}=e^{-Ce^{kt}}

    60=e^{-Ce^{kt}}y

    My values for the variables are:

    y=0, t=0<br />
y=15, t=3<br />
y=?, t=10

    However, the solutions manual states the answer for the general soln is

     y=60e^{-Ce^{kt}} which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please?
    Your absolute value dissapeared
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  3. #3
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    Isn't -|ln\frac{60}{y}| = whatever the same as - ln\frac{60}{y}=whatever since the negative sign is on the outside of the absolute value anyway?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by emttim84 View Post
    Isn't -|ln\frac{60}{y}| = whatever the same as - ln\frac{60}{y}=whatever since the negative sign is on the outside of the absolute value anyway?
    no...no it doesnt...because say you have \pm{x}\cdot{-1}

    all this will do is change "when" the negative occurs..not the existence of it...it is still +/-
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    no...no it doesnt...because say you have \pm{x}\cdot{-1}

    all this will do is change "when" the negative occurs..not the existence of it...it is still +/-
    Ahh ok...so because of the +/-, since we're talking about population growth, and you can't have a - if it's growth, then it's assumed that the + value of the absolute value should be used?
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