Showing a Result of the Logisitic Equation

Question

The logisitc equation is given by $$\frac{d N}{dt}=rN\left(1-\frac{N}{K}\right)\tag{1},$$ where $K$ is the carrying capacity and $r$ is the intrinsic growth rate. I am trying to show for the logisitc equation that $$r=-\frac{d}{dt}\left(\ln\left(\frac{K-N}{N}\right)\right), \tag{2}$$ so that the closeness of the population size to the carrying capacity determines its rate of approach to $K$. I solved equation $(1)$ in an attempt to show the required result, which is given below \begin{align} \frac{dN}{dt}&=rN\left(1-\frac{N}{K}\right) \\ \frac{dN}{N(K-N)}&=\frac{r}{K} \ dt \\ \int\left(\frac{1/K}{N}+\frac{1/K}{K-N}\right) \ dN&=\int \frac{r}{K} \ dt \\ \frac{1}{K}\left(\ln(N)-\ln(K-N)\right)&=\frac{r}{K}t+C, \ \ C\in\mathbb{R} \\ \ln\left(\frac{K-N}{N}\right)&=-rt-KC. \end{align} I am unsure how to proceed from this point.

Answer 1

You're really close to finishing off.
Differentiating both sides gives us: $$\frac{\mathrm{d}}{\mathrm{d}t}\ln\left(\frac{K-N}{N}\right)=-r$$