First order ordinary differential equation about fish populations in a lake.

Question

I have this question which has me stumped, here's the problem.

A lake is stocked with a fish population of size $ N(t)$ at time t. Initially the population is $N(0) = N_0$ the evolution of he population is given by this ODE:$$ \frac{dN}{dt}=\frac{aN}{b}(b-N)$$ Where a and b are both positive constants. a) find N(t) b) what is the behavior of N(t) and t tend to infinity

So the way the question is set up makes me think its somehow separable here's what i've tried Rearranging this gets me: $$\int \frac{1}{N(b-N)}dN=\int \frac{a}{b}dt$$ Partial fractions on LHS give: $$\int \frac{1}{b(b-N)}+\frac{1}{bN} dN=\int \frac{a}{b}dt$$ evaluating the integrals and simplifying: $$\frac{1}{b}ln(\frac{N}{b-N})=\frac{a}{b}t+C$$ some more fiddling: $$\frac{N}{b-N}=Ae^{at}$$ where A is a postive constant. So i can mess around further getting this sort of thing $N(t)=\large{\frac{Abe^{at}}{1-Ae^{at}}}$ (thanks to martin for picking up the sign error that should be $N(t)=\large{\frac{Abe^{at}}{1+Ae^{at}}}$)thing i'm missing. I've tried subbing in the initial conditions but it really doesn't help i get something no nice looking for my constant A as a result. hopefully you guys can see it. thanks for the help.

Answer 1

What you did is almost right (sign mistake when solving for $N$). You can "prettify" your $N$ a bit by writing $$ N(t)=\frac{Ab}{e^{-at}+A} $$ Zill-Wright (7th Edition, at least) call your equation the logistic equation.

When you introduce the initial condition, you get $$ N_0=\frac{Ab}{1+A}, $$ so $$ A=\frac{N_0}{b-N_0}. $$ Now $N$ looks like $$ N(t)=\frac{N_0b(b-N_0)}{e^{-at}+N_0/(b-N_0)} =\frac{N_0b}{(b-N_0)e^{-at}+N_0}. $$ Or even $$ N(t)=\frac{b}{(b/N_0-1)e^{-at}+1}. $$

Answer 2

It seems to me you made a sign mistake.. $$\int \frac{1}{N(b-N)}dN=\int \frac{a}{b}dt$$ $$\frac 1 b(\int \frac {dN}{N}+\int\frac {dN}{b-N})=\frac ab t+k$$ $$\frac 1 b(\int \frac {dN}{N}-\int\frac {dN}{N-b})=\frac ab t+k$$ $$ \ln|\frac {N}{N-b}|=a t+K$$ $$ \frac {N}{N-b}=Ke^{a t}$$ $$ {N}(1-Ke^{at})=-bKe^{a t}$$ $$ {N}=\frac {-bKe^{a t}} {(1-Ke^{at})}=\frac {b} {1-Ke^{-at}}$$ $$N(0)=\frac b {1-K} \to K=1-\frac b {N_0} \to N=\frac {b} {1+(\frac b {N_0} -1)e^{-at}}$$