In this exercise, the logistic equation models the growth of a population. Use the equation to determine when the population will reach $50 \%$ of its carrying capacity.

$$P(t)=\dfrac{6000}{1+4999e^{-0.8t}}$$

In this exercise, use the slope field to sketch the solution that passes through the given point.

$$\begin{array}{} \text{Differential Equation}\\ \hline y^{\prime}=-x-2 \end{array}\quad \begin{array}{} \text{Point}\\ \hline (-1,1) \end{array}$$

In this exercise, a medical researcher wants to determine the concentration $C$ (in moles per liter) of a tracer drug injected into a moving fluid. Solve this problem by considering a singlecompartment dilution model. Assume that the fluid is continuously mixed and that the volume of the fluid in the compartment is constant.

If the tracer is injected instantaneously at time $t=0$, then the concentration of the fluid in the compartment begins diluting according to the differential equation

$\dfrac{dC}{dt}=\left(-\dfrac{R}{V}\right)C,\quad C=C_0$ when $t=0$.

Solve this differential equation to find the concentration $C$ as a function of time $t$.

In this exercise, sketch the slope field for the differential equation.

$$\begin{array}{} \text{Differential Equation}\\ \hline y^{\prime}=-x-2 \end{array}\quad \begin{array}{} \text{Point}\\ \hline (-1,1) \end{array}$$

In this exercise, the logistic equation models the growth of a population. Use the equation to find the value of $k$.

$$P(t)=\dfrac{6000}{1+4999e^{-0.8t}}$$

In this exercise, verify the particular solution of the differential equation.

$$\begin{array}{} \hline \text{Solution}\\ y=6 e^{-2 x^2} \end{array}\qquad \begin{array}{} \hline \text{Differential Equation and Initial COndition}\\ y^{\prime}=-4xy\\ y(0)=6 \end{array}$$

In this exercise, verify the particular solution of the differential equation.

$$\begin{array}{} \hline \text{Solution}\\ y=6 e^{-2 x^2} \end{array}\qquad \begin{array}{} \hline \text{Differential Equation and Initial COndition}\\ y^{\prime}=-4xy\\ y(0)=6 \end{array}$$

In this exercise, solve the first-order linear differential equation.

$$(x-1)y^{\prime}+y=x^2-1$$

In this exercise, find the general solution of the differential equation.

$$y\ln x-xy^{\prime}=0$$

In this exercise, consider a predator-prey relationship involving foxes (predators) and rabbits (prey). Let $x$ represent the number of rabbits, let $y$ represent the number of foxes, and let $t$ represent the time in months. Assume that the following predator-prey equations model the rates of change of each population.

$$\begin{aligned} \frac{dx}{dt}&=0.8x-0.04xy\quad\text{Rate of change of prey population}\\ \frac{dy}{dt}&=-0.3y+0.006xy\quad\text{Rate of change of predator population} \end{aligned}$$

When $t=0, x=55$ rabbits and $y=10$ foxes.

Use a graphing utility to graph a slope field of the predator-prey equations when $0 \leq x \leq 150$ and $0 \leq y \leq 50$.

In this exercise, write and solve the differential equation that models the verbal statement.

The rate of change of $Q$ with respect to $t$ is inversely proportional to the square of $t$.

In this exercise, the logistic equation models the growth of a population. Use the equation to write a logistic differential equation that has the solution $P(t)$.

$$P(t)=\dfrac{5000}{1+39e^{-0.2t}}$$

In this exercise, the logistic equation models the growth of a population. Use the equation to determine when the population will reach $50 \%$ of its carrying capacity.

$$P(t)=\dfrac{5000}{1+39e^{-0.2t}}$$

Biomass is a measure of the amount of living matter in an ecosystem. Suppose the biomass $s(t)$ in a given ecosystem increases at a rate of about $3.5$ tons per year, and decreases by about $1.9 \%$ per year. This situation can be modeled by the differential equation

$$\frac{ds}{dt}=3.5-0.019s.$$

Explain what happens as $t \rightarrow \infty$.

In this exercise, the logistic equation models the growth of a population. Use the equation to find the initial population.

$$P(t)=\dfrac{5000}{1+39e^{-0.2t}}$$