Question

[Key Concept: dN/dt = α − λN: the population relaxes exponentially from N₀ to the equilibrium value α/λ.]

a) $\frac{dN}{dt} = + \alpha - \lambda N$

$$\begin{matrix} & \ \int_{}^{}\ \frac{dN}{\alpha - \lambda N} = \int_{}^{}\ dt \\ & \left. \ \frac{\mathcal{l}n(\alpha - \lambda N)}{- \lambda} \right|_{N_{0}}^{N} = t \\ & \alpha - \lambda N = \left( a - \lambda N_{0} \right)e^{- \lambda t} \\ & \text{ }N = \frac{\alpha}{\lambda}\left( 1 - e^{- \lambda t} \right) + N_{0}e^{- \lambda t} \end{matrix}$$

b) $\alpha = \lambda N_{0}$

$$\begin{matrix} & N = N_{0}\left( 1 - e^{- \lambda t} \right) + N_{0}e^{- \lambda t} \\ & \text{ }N = N_{0} = \text{~constant~} \end{matrix}$$

c)$\ \alpha = 2\lambda{\text{ }N}_{0}\ t \rightarrow \infty \Rightarrow e^{- \lambda t} = 0$

$$N = 2{\text{ }N}_{0}$$

d) $\alpha = 2\lambda{\text{ }N}_{0}\ t = \frac{\mathcal{l}n2}{\lambda} \Rightarrow e^{- \lambda t} = \frac{1}{2}$

$$N = 2{\text{ }N}_{0}\left( 1 - \frac{1}{2} \right) + N_{0}\frac{1}{2} = \frac{3{\text{ }N}_{0}}{2}$$