So to define a new population growth model, this is the (Cauchy problem for a) differential equation of growth I am considering: $$ \begin{cases} \dfrac{\mathrm{d}N}{\mathrm{d}t} = rN(t) - r\dfrac{N^2(t)}{kt + b} \\ N(0) = N_0, \end{cases} $$ where $k<0$, $b>0$ are two constant parameters. From this, I need to find: $$ \lim_{{t \to -\frac{b}{k}}}{N'(t)} $$ Until now, I have tried to use L'Hopital's Rule and also explored ways to break down the limit into different compositions.
Can I get some help?
Maybe it's not the most elegant method, but one can compute the limit by first solving the ODE. To simplify it, let's define the variables $$ n=-\frac{r}{k}N, \qquad s=\frac{r}{k}(kt+b), \tag{1} $$ in terms of which one can rewrite the ODE as $$ \frac{dn}{ds}=n+\frac{n^2}{s}. \tag{2} $$ That's a Bernoulli equation; one can transform it into a linear ODE with the substitution $u=\frac{1}{n}$: $$ n'=-\frac{u'}{u^2}=\frac{1}{u}+\frac{1}{su^2} \implies u'+u=-\frac{1}{s}. \tag{3} $$ Multiplying $(3)$ by $e^{s}$ and integrating yields $$ e^{s}(u'+u)=(e^su)'=-\frac{e^s}{s}\implies u=e^{-s}(C-\text{Ei}(s)), \tag{4} $$ where $\text{Ei}(s)=\int_{-\infty}^s\frac{e^x}{x}dx$ is the exponential integral. Finally, $$ n(s)=\frac{1}{u}=\frac{e^s}{C-\text{Ei(s)}}. \tag{5} $$ Now we are ready to compute $\lim_{t\to-b/k}N'(t)$. From $(1)$ and $(2)$ it follows that $$ \lim_{t\to-b/k}N'(t)=-k\lim_{s\to 0^-}\frac{dn}{ds}=-k\lim_{s\to 0^-} \left(n(s)+\frac{n^2(s)}{s}\right). \tag{6} $$ Since $\text{Ei}(s)= \gamma+\ln(-s)+O(s)$ for $0<-s\ll 1$, it follows from $(5)$ that $\lim_{s\to 0^-}n(s)=0$ and $$ \lim_{s\to 0^-}\frac{n^2(s)}{s}=\lim_{s\to 0^-}\frac{1}{s(\ln(-s))^2}=-\infty. \tag{7} $$ Since $k<0$, it follows from $(6)$ and $(7)$ that $$ \lim_{t\to-b/k}N'(t)=-\infty. \tag{8} $$