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.
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 $$ \begin{align} N(t)&=\frac{-1.5\mathrm{e}^t-2-\sqrt{4+2.25\mathrm{e}^{2t}}}{-3\mathrm{e}^t}\\ &=\frac{1.5\mathrm{e}^t+2+\sqrt{4+2.25\mathrm{e}^{2t}}}{3\mathrm{e}^t}\\ \end{align} $$