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

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

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.

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

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

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

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 $t \rightarrow \infty$.

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

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

Use a graphing utility to graph the slope field for the differential equation. What do you notice?

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

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

Solve the differential equation.

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

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

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

$$\begin{array}{} \hline \text{Solution}\\ y=\frac{1}{2} x^2-4 \cos x+2 \end{array}\qquad \begin{array}{} \hline \text{Differential Equation and Initial Condition}\\ y^{\prime}=x+4 \sin x\\y(0)=-2 \end{array}$$

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 the functions $x$ and $y$ when $0 \leq t \leq 36$. Describe the behavior of each solution as $t$ increases.

In this exercise, solve the differential equation.

$$xy+y^{\prime}=100x$$

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

$$(y-1)\sin xdx-dy=0$$

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

$$\sqrt{x^2-9}y^{\prime}=5x$$

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

$$\dfrac{dy}{dx}=\dfrac{x}{y}$$