Solving a modified logistic equation

Question

How can we solve a modified logistic equation?

$$\frac{dP}{dt}=rP \left(1-\frac{P}{k}\right)-c$$

Where $k$, $r$ and $c$ are constants.

Answer 1

Hint: Observe \begin{align} rP-\frac{r}{k}P^2-c = -\frac{rk}{4}+rP-\frac{r}{k}P^2-c+\frac{rk}{4} = -\frac{r}{k}\left(P-\frac{k}{2}\right)^2+\frac{rk}{4}-c \end{align} which means you have \begin{align} \frac{dP}{\left(P-\frac{k}{2}\right)^2+\frac{ck}{r}-\frac{k^2}{4}} = -\frac{r}{k}dt. \end{align} You will need to consider three cases i) $\frac{ck}{r}=\frac{k^2}{4}$, ii) $\frac{ck}{r}>\frac{k^2}{4}$ iii) $\frac{ck}{r}<\frac{k^2}{4}$.

Answer 2

$$\frac{dP}{dt}=rP\left(1-\frac{P}{k}\right)-c=-\frac{r}{k}P^2+rP-c$$

$$\implies \int\frac{dP}{-\frac{r}{k}P^2+rP-c}=\int dt$$

Can you solve integrals such as : $$\int \frac{1}{ax^2+bx+c}dx$$

If yes, you are done. You need to solve :$$ \int\frac{dP}{rP^2-krP+kc}=-\frac{t}{k}$$

I leave that part for you to solve.