Gompertz growth

**brogers** · October 24th 2009, 10:29 PM

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!

**CaptainBlack** · October 24th 2009, 11:34 PM

Originally Posted by brogers

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

Thread: Gompertz growth

LinkBack

Thread Tools

Display

Gompertz growth

Similar Math Help Forum Discussions

Gompertz differential equation proof

Gompertz equation

Multi-part Gompertz growth problem

Population growth with non-constant growth factor

Growth Periods - Continuous Growth

Search Tags