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#1

$Old$ October 24th, 2009, 11:29 PM

brogers $brogers is offline$

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$Default$ Gompertz growth

I am given $\frac{dN}{dt}=\gamma N$ where $\frac{d\gamma}{dt}=-\alpha \gamma$

How do I get $\frac{dN}{dt}=\gamma_{0}e^{-\alpha t}N$ if I know $\frac{1}{N}\frac{dN}{dt}=\frac{d}{dt}(ln (N))$ ?
Thanks!

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$Old$ October 25th, 2009, 12:34 AM

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Originally Posted by brogers $View Post$

I am given $\frac{dN}{dt}=\gamma N$ where $\frac{d\gamma}{dt}=-\alpha \gamma$

How do I get $\frac{dN}{dt}=\gamma_{0}e^{-\alpha t}N$ if I know $\frac{1}{N}\frac{dN}{dt}=\frac{d}{dt}(ln (N))$ ?
Thanks!

Just solve:

$\frac{d\gamma}{dt}=-\alpha \gamma$

for $\gamma$ and substitute back into the original equation.

CB

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