Question

Fluctuating Population The differential equation$\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\left(}\mathbf{k}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{t}\mathbf{\right)}\mathit{P}$, where k is a positive constant,$\mathit{P}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ is a mathematical model for a population P(t) that undergoes yearly seasonal fluctuations. Solve the equation subject to$\mathit{P}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{P}}_{\mathbf{0}}$.

Use a graphing utility to graph the solution for different choices

of ${\mathit{P}}_{\mathbf{0}}$.

Short answer

$P=P_0{e}^{\sin t}$

Step-by-step solution

Given Information

The given value is:

$\frac{dp}{dt}=\left(kcost\right)P$

Determining the population at time t

From the differential equation, we have a population P(t) for a country that experiences yearly seasonal changes.

$\frac{dp}{dt}=\left(kcost\right)P$

with the predicament

$P\left(0\right)=P_0$

and we must solve the problem as follows:

We can solve this differential equation as follows because it is a first order and separable D.E.

$\begin{array}{rcl}\frac{dP}{P}& =& kcostdt\int \frac{1}{P}dP=k\int costdt\\ ln\left(P\right)& =& ksint+c_1\\ e^{ln\left(P\right)}& =& e^{ksint+c_1}\end{array}$

Then there's

$\begin{array}{rcl}P& =& e^{c_1}{e}^{k\sin t}\\ & =& c{e}^{k\sin t}\end{array}$

After that, we must use the point of condition$(P,t)=(P_0,0)$ into equation (1) to obtain the value of constant c.

$$\begin{gathered} P_0=c{e}^{sin0} \\ {P}_0=c{e}^0 \end{gathered}$$

Then there's

$c=P_0$

then, in equation (1), substitute the value of constant c, and we have

$P=P_0{e}^{sint}$

is the population at time t.

Graphing the population at time t

This is a graph illustrating the population's solution at time t.