Question

Population ModelAnother population model is given by$\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\mathit{P}\mathbf{-}\mathit{h}$where h and k are positive constants. For what initial values P(0) = P₀ does this model predict that the population will go extinct?

Short answer

$0<P_0<\frac{h}{k}$

Step-by-step solution

Step 1:Definition of differential equation

Differential Equation is defined as the equation that consists of the derivatives of one or more dependent functions in respect to one or more independent functions.

Linear differential equation

The given differential equation, can be written as

$\frac{dp}{dt}-kP=-h\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Linear differential equation of the first order,

$P\text{'}+Q\left(t\right)P=f\left(t\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Based on (1) and (2), we obtain

$Q\left(t\right)=-k\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}f\left(t\right)=-h$

Now substitute the value of $Q\left(t\right)$and $f\left(t\right)$in the formula,

$P\left(t\right)=e^{-\int Q\left(t\right)dt}[c+\int f\left(t\right)e^{\int Q\left(t\right)dt}dt]$

To get,

$\begin{array}{l}P\left(t\right)=e^{\int kdt}[c-\int h{e}^{\int -kdt}dt]\\ P\left(t\right)=e^{kt}[c-h\int e^{-kt}dt]\\ P\left(t\right)=e^{kt}[c+h\frac{e^{-kt}}{k}]\\ P\left(t\right)=e^{kt}c+\frac{h}{k}\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Use initial conditions

Use the initial condition$P\left(0\right)=P_0$ to get c,

$\begin{array}{l}P\left(0\right)=e^{0}c+\frac{h}{k}\\ P_0=c+\frac{h}{k}\\ c=P_0-\frac{h}{k}\end{array}$

Substitute the value of$c=P_0-\frac{h}{k}$ in the equation (3).

$P\left(t\right)=e^{kt}(P_0-\frac{h}{k})+\frac{h}{k}\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)$

Predict the model that the population will go extinct

The population will extinct,if.$P\left(t\right)=0$

$P\left(t\right)\to -\infty$as$t\to \infty$gives same result as$P\left(t\right)=0$, because population is always positive.

Using equation (4), $P\left(t\right)\to -\infty$as $t\to \infty$if.$P_0-\frac{h}{k}<0\Rightarrow P_0<\frac{h}{k}$ The population is positive, therefore, initial condition is positive.$(P_0>0)$

So, the condition for the population to extinct is.$0<P_0<\frac{h}{k}$