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Have the following population model, $$ \frac {dN}{dt}= cN(N-k)(1-N)$$

The first stage of the question is to investigate the steady states, however im a little stuck on finding the solution to the differential equation given our initial condition $N(0)=2$ and assuming that $c=1$ and $k=0.5$,

$$\int \frac {1}{N(N-0.5)(1-N)} dN = \int \left(\frac {-2}{N}+\frac{4}{N-0.5}+\frac{2}{1-N}\right) dN = \int dt$$

$$\displaylines{\Rightarrow -2\log(N)+4\log(N-0.5)-2\log(1-N)=t+{\rm constant} \Rightarrow \log \frac{2(N-0.5)}{N(1-n)}= t+c\cr \Rightarrow \frac{2(N-0.5)}{N(1-N)}=Ae^{t}\cr}$$

Now there is a tip at the bottom of the question say consider a substitution of $k=N-0.5$.

$\Rightarrow$ $\frac{2(k)}{k+0.5(0.5-k)}$=$Ae^{t}$ $\Rightarrow$ $8k+4k^{2}Ae^{t}=Ae^{t}$

Therefore I must have gone wrong somewhere, any help would be much appreciated, many thanks.

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If the first stage is just to investigate steady-states, you do not need to solve the equation. You only need to see which $N$-values cause $\frac{dN}{dt}=0$. With an initial condition of $N=2$, $\frac{dN}{dt}$ is negative, so the population will decline toward the next steady state below $2$. Assuming that you have solved the integral correctly, you can solve for $N$ explicitly by using the quadratic formula and choosing the $\pm$ based on which one provides an initial value of $2$. –  alex.jordan Apr 1 '12 at 19:12
    
Just a bit confused to what i can use the quadratic formula on –  user24930 Apr 1 '12 at 19:34
    
After some algebraic rearranging, $\frac{2(N-0.5)}{N(1-N)}=A\,\mathrm{e}^t$ is a quadratic equation in $N$: $(A\,\mathrm{e}^t)N^2+(2-A\,\mathrm{e}^t)N-1=0$ –  alex.jordan Apr 1 '12 at 23:30
    
Attempted to improve formatting. Appreciate any efforts to do better, correct any mistakes I've introduced, etc. –  Gerry Myerson Apr 1 '12 at 23:56
    
Are you sure that both the parameter from the equation and the variable that the hint suggests you substitute to are both called $k$? That would be a poor choice on the part of the author. –  alex.jordan Apr 2 '12 at 1:50

1 Answer 1

up vote 1 down vote accepted

Steady-states happen when $\frac{dN}{dt}=0$. This happens for $N=0$, $N=k=0.5$, and $N=1$.

For $N=2$, $\frac{dN}{dt}=-3$. In fact for all $N>1$, $\frac{dN}{dt}$ is negative. $N(t)$ will approach $1$ as $t\to\infty$.

To find an explicit solution, as you have worked out $\frac{2(N-0.5)}{N(1-N)}=A\,\mathrm{e}^t$. This is equivalent to $$(A\,\mathrm{e}^t)N^2+(2-A\,\mathrm{e}^t)N-1=0$$ From this we can solve for $N$: $$ \begin{align} N&=\frac{(A\,\mathrm{e}^t-2)\pm\sqrt{(2-A\,\mathrm{e}^t)^2+4(A\,\mathrm{e}^t)}}{2(A\,\mathrm{e}^t)}\\ &=\frac{(A\,\mathrm{e}^t-2)\pm\sqrt{4+A^2\mathrm{e}^{2t}}}{2(A\,\mathrm{e}^t)}\\ \end{align}$$ The earlier equation $\frac{2(N-0.5)}{N(1-N)}=A\,\mathrm{e}^t$ let's us solve for $A$ when we consider time $t=0$: $A=-1.5$. $$ \begin{align} N&=\frac{-1.5\,\mathrm{e}^t-2\pm\sqrt{4+2.25\mathrm{e}^{2t}}}{-3\mathrm{e}^t}\\ \end{align}$$ The solution where $+$ is used instead of $-$ yields a solution where $N(0)\neq2$, so $$ \begin{align} N(t)&=\frac{-1.5\mathrm{e}^t-2-\sqrt{4+2.25\mathrm{e}^{2t}}}{-3\mathrm{e}^t}\\ &=\frac{1.5\mathrm{e}^t+2+\sqrt{4+2.25\mathrm{e}^{2t}}}{3\mathrm{e}^t}\\ \end{align} $$

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