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The logistic differential equation $$y'=y(b-ay) \, \textrm{with}\, a\neq 0, b\neq 0$$ has the non-trivial solution $$y(t) = \frac{\frac{b}{a}}{1+e^{-bt}}, \quad (1)$$ where $c$ is a constant.

My questions is:

We have that $$\frac{y'}{y}= b-ay$$. But how can I understand this intuitively. What does it give?

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Are you wondering how to find the specific solution or the intuition behind why the graph looks like it does by looking at the differential equation? –  E.O. Nov 12 '12 at 23:48
    
One, you forgot your $c$. Two, what exactly are you looking for? How to solve the equation? –  Mike Nov 13 '12 at 1:38
    
I am thinking what does $\frac{y'}{y}$ tell me. –  Reader Nov 13 '12 at 10:02
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1 Answer 1

up vote 1 down vote accepted

If $y(t)$ represents the size of a population, then $y'$ is the total growth rate for the entire population ("the number of new individuals produced during each unit of time"). Therefore $y'/y$ is the growth rate per capita ("the number of new individuals that each individual produces during each unit of time, on average").

In the logistic model, the growth rate per capita is assumed to decrease linearly with the size of the population (as seen in the right-hand side $b-ay$), for example because of competition about resources.

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Thanks a lot Hans Lundmark. –  Reader Nov 13 '12 at 16:14
    
@ Hans Lundmark If $0<y<\frac{b}{a}$, then we get the s-form curve. If $y>\frac{b}{a}$ then we get the decreasing graph which is above it. In which situations do we get this curve. Can you give an example. We say that y>0, but when do we have that y is also a solution? –  Reader Nov 13 '12 at 23:12
    
@Reader: I'm not sure that I understand the question. The quantity $K=b/a$ is called the carrying capacity of the environment; it's the largest population size that can be supported in the long run. If the initial population is smaller than $K$ it will grow and approach $K$ from below as $t \to \infty$, and if it's larger than $K$ it will decrease and approach $K$ from above. –  Hans Lundmark Nov 14 '12 at 10:10
    
Thanks Hans Lundmark. –  Reader Nov 14 '12 at 16:04
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