4 Introduction to Differential Equations 微分方程引论

本页译自 OpenStax《Calculus Volume 2》第 4 章 Introduction to Differential Equations（微分方程引论）：4.1–4.5 五节（微分方程基础／方向场与数值方法／可分离变量方程／逻辑斯谛方程／一阶线性方程）+ Key Terms/Key Equations/Key Concepts/Review Exercises 全译，段段对照。公式经本地 MathJax 渲染，自定义宏已注入。

4.1 Basics of Differential Equations 4.1 微分方程基础

• 4.1.1 Identify the order of a differential equation.
• 4.1.2 Explain what is meant by a solution to a differential equation.
• 4.1.3 Distinguish between the general solution and a particular solution of a differential equation.
• 4.1.4 Identify an initial-value problem.
• 4.1.5 Identify whether a given function is a solution to a differential equation or an initial-value problem.

• 4.1.1 确定微分方程的阶。
• 4.1.2 解释微分方程的解的含义。
• 4.1.3 区分微分方程的通解与特解。
• 4.1.4 识别初值问题。
• 4.1.5 判断给定函数是否为微分方程或初值问题的解。

Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function $y = f(x)$ and its derivative, known as a *differential equation*. Solving such equations often provides information about how quantities change and frequently provides insight into how and why the changes occur.

Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, or computer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in the course. In this section we study what differential equations are, how to verify their solutions, some methods that are used for solving them, and some examples of common and useful equations.

General Differential Equations 一般的微分方程

Consider the equation $y^{\prime} = 3x^{2},$ which is an example of a differential equation because it includes a derivative. There is a relationship between the variables $x$ and ${y\text{:}}\ y$ is an unknown function of $x.$ Furthermore, the left-hand side of the equation is the derivative of $y.$ Therefore we can interpret this equation as follows: Start with some function $y = f(x)$ and take its derivative. The answer must be equal to $3x^{2}.$ What function has a derivative that is equal to $3x^{2}?$ One such function is $y = x^{3},$ so this function is considered a solution to a differential equation.

A differential equation is an equation involving an unknown function $y = f(x)$ and one or more of its derivatives. A solution to a differential equation is a function $y = f(x)$ that satisfies the differential equation when $f$ and its derivatives are substituted into the equation.

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Some examples of differential equations and their solutions appear in Table 4.1.

Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example, $y = x^{2} + 4$ is also a solution to the first differential equation in Table 4.1. We will return to this idea a little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.

Verifying Solutions of Differential Equations 验证微分方程的解

Verify that the function $y = e^{-3x} + 2x + 3$ is a solution to the differential equation $y^{\prime} + 3y = 6x + 11.$

Solution

To verify the solution, we first calculate $y^{\prime}$ using the chain rule for derivatives. This gives $y^{\prime} = -3e^{-3x} + 2.$ Next we substitute $y$ and $y^{\prime}$ into the left-hand side of the differential equation:

$$(-3e^{-3x} + 2) + 3(e^{-3x} + 2x + 3).$$

The resulting expression can be simplified by first distributing to eliminate the parentheses, giving

$$-3e^{-3x} + 2 + 3e^{-3x} + 6x + 9.$$

Combining like terms leads to the expression $6x + 11,$ which is equal to the right-hand side of the differential equation. This result verifies that $y = e^{-3x} + 2x + 3$ is a solution of the differential equation.

Verify that $y = 2e^{3x} - 2x - 2$ is a solution to the differential equation $y^{\prime} - 3y = 6x + 4.$

It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them. The most basic characteristic of a differential equation is its order.

The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.

Identifying the Order of a Differential Equation 确定微分方程的阶

What is the order of each of the following differential equations?

1. $y^{\prime} - 4y = x^{2} - 3x + 4$

2. $x^{2}y\text{'''} - 3xy^{''} + xy^{\prime} - 3y = \text{sin}\mspace{2mu} x$

3. $\frac{4}{x}y^{(4)} - \frac{6}{x^{2}}y^{''} + \frac{12}{x^{4}}y = x^{3} - 3x^{2} + 4x - 12$

Solution

1. The highest derivative in the equation is $y^{\prime},$ so the order is $1.$

2. The highest derivative in the equation is $y\text{'''}\text{,}$ so the order is $3.$

3. The highest derivative in the equation is $y^{(4)},$ so the order is $4.$

What is the order of the following differential equation?

$$\left( {x^{4} - 3x} \right)y^{(5)} - \left( {3x^{2} + 1} \right)y^{\prime} + 3y = \text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x$$

General and Particular Solutions 通解与特解

We already noted that the differential equation $y^{\prime} = 2x$ has at least two solutions: $y = x^{2}$ and $y = x^{2} + 4.$ The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form $y = x^{2} + C,$ where $C$ represents any constant, is a solution as well. The reason is that the derivative of $x^{2} + C$ is $2x,$ regardless of the value of $C.$ It can be shown that any solution of this differential equation must be of the form $y = x^{2} + C.$ This is an example of a general solution to a differential equation. A graph of some of these solutions is given in Figure 4.2. (*Note*: in this graph we used even integer values for $C$ ranging between $-4$ and $4.$ In fact, there is no restriction on the value of $C;$ it can be an integer or not.)

In this example, we are free to choose any solution we wish; for example, $y = x^{2} - 3$ is a member of the family of solutions to this differential equation. This is called a particular solution to the differential equation. A particular solution can often be uniquely identified if we are given additional information about the problem.

Finding a Particular Solution 求特解

Find the particular solution to the differential equation $y^{\prime} = 2x$ passing through the point $\left( {2,7} \right).$

Solution

Any function of the form $y = x^{2} + C$ is a solution to this differential equation. To determine the value of $C,$ we substitute the values $x = 2$ and $y = 7$ into this equation and solve for $C\text{:}$

$$\begin{array}{l} \\ \\ {y = x^{2} + C} \\ {7 = 2^{2} + C = 4 + C} \\ {C = 3.} \end{array}$$

Therefore the particular solution passing through the point $(2,7)$ is $y = x^{2} + 3.$

Find the particular solution to the differential equation

$$y^{\prime} = 4x + 3$$

passing through the point $\left( {1,7} \right),$ given that $y = 2x^{2} + 3x + C$ is a general solution to the differential equation.

Initial-Value Problems 初值问题

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value, which is an ordered pair that is used to find a particular solution.

A differential equation together with one or more initial values is called an initial-value problem. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation $y^{\prime} = 2x,$ then $y(3) = 7$ is an initial value, and when taken together, these equations form an initial-value problem. The differential equation $y^{''} - 3y^{\prime} + 2y = 4e^{x}$ is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be $y(0) = 2$ and $y^{\prime}(0) = -1.$ These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is $t,$ which represents time. Thus, a value of $t = 0$ represents the beginning of the problem.

Verifying a Solution to an Initial-Value Problem 验证初值问题的解

Verify that the function $y = 2e^{-2t} + e^{t}$ is a solution to the initial-value problem

$$y^{\prime} + 2y = 3e^{t},\quad y(0) = 3.$$

Solution 解答

For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that $y$ satisfies the differential equation, we start by calculating $y^{\prime}.$ This gives $y^{\prime} = -4e^{-2t} + e^{t}.$ Next we substitute both $y$ and $y^{\prime}$ into the left-hand side of the differential equation and simplify:

$$\begin{array}{cl} {y^{\prime} + 2y} & {= \left( {-4e^{-2t} + e^{t}} \right) + 2\left( {2e^{-2t} + e^{t}} \right)} \\ & {= -4e^{-2t} + e^{t} + 4e^{-2t} + 2e^{t}} \\ & {= 3e^{t}.} \end{array}$$

This is equal to the right-hand side of the differential equation, so $y = 2e^{-2t} + e^{t}$ solves the differential equation. Next we calculate $y(0)\text{:}$

$$\begin{array}{cl} {y(0)} & {= 2e^{-2{(0)}} + e^{0}} \\ & {= 2 + 1} \\ & {= 3.} \end{array}$$

This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

Verify that $y = 3e^{2t} + 4\mspace{2mu}\text{sin}\mspace{2mu} t$ is a solution to the initial-value problem

$$y^{\prime} - 2y = 4\mspace{2mu}\text{cos}\mspace{2mu} t - 8\mspace{2mu}\text{sin}\mspace{2mu} t,\quad y(0) = 3.$$

In Example 4.4, the initial-value problem consisted of two parts. The first part was the differential equation $y^{\prime} + 2y = 3e^{t},$ and the second part was the initial value $y(0) = 3.$ These two equations together formed the initial-value problem.

The same is true in general. An initial-value problem consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of $C.$ The family of solutions to the differential equation in Example 4.4 is given by $y = 2e^{-2t} + Ce^{t}.$ This family of solutions is shown in Figure 4.3, with the particular solution $y = 2e^{-2t} + e^{t}$ labeled.

Solving an Initial-value Problem 求解初值问题

Solve the following initial-value problem:

$$y^{\prime} = 3e^{x} + x^{2} - 4,\quad y(0) = 5.$$

Solution 解答

The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

$$\left. \int{y^{\prime}dx} \right. = \left. \int{\left( {3e^{x} + x^{2} - 4} \right)\mspace{2mu} dx} \right.,$$

namely,

$$y + C_{1} = 3e^{x} + \frac{1}{3}x^{3} - 4x + C_{2}.$$ (4.1)

We are able to integrate both sides because the *y* term appears by itself. Notice that there are two integration constants: $C_{1}$ and $C_{2}.$ Solving Equation 4.1 for $y$ gives

$$y = 3e^{x} + \frac{1}{3}x^{3} - 4x + C_{2} - C_{1}.$$

Because $C_{1}$ and $C_{2}$ are both constants, $C_{2} - C_{1}$ is also a constant. We can therefore define $C = C_{2} - C_{1},$ which leads to the equation

$$y = 3e^{x} + \frac{1}{3}x^{3} - 4x + C.$$

Next we determine the value of $C.$ To do this, we substitute $x = 0$ and $y = 5$ into Equation 4.1 and solve for $C\text{:}$

$$\begin{array}{rll} & & \\ 5 & = & {3e^{0} + \frac{1}{3}0^{3} - 4(0) + C} \\ 5 & = & {3 + C} \\ C & = & {2.} \end{array}$$

Now we substitute the value $C = 2$ into Equation 4.1. The solution to the initial-value problem is $y = 3e^{x} + \frac{1}{3}x^{3} - 4x + 2.$

Analysis 分析

The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.

Solve the initial-value problem

$$y^{\prime} = x^{2} - 4x + 3 - 6e^{x},\quad y(0) = 8.$$

In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth's surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton's second law of motion (in equation form $F = ma,$ where $F$ represents force, $m$ represents mass, and $a$ represents acceleration), to derive an equation that can be solved.

In Figure 4.4 we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth's surface, $g,$ is approximately $9.8{\ \text{m/s}}^{2}.$ We introduce a frame of reference, where Earth's surface is at a height of 0 meters. Let $v(t)$ represent the velocity of the object in meters per second. If $v(t) > 0,$ the ball is rising, and if $v(t) < 0,$ the ball is falling (Figure 4.5).

Our goal is to solve for the velocity $v(t)$ at any time $t.$ To do this, we set up an initial-value problem. Suppose the mass of the ball is $m,$ where $m$ is measured in kilograms. We use Newton's second law, which states that the force acting on an object is equal to its mass times its acceleration $(F = ma).$ Acceleration is the derivative of velocity, so $a(t) = v^{\prime}(t).$ Therefore the force acting on the baseball is given by $F = m\ v^{\prime}(t).$ However, this force must be equal to the force of gravity acting on the object, which (again using Newton's second law) is given by $F_{g} = \text{−}mg,$ since this force acts in a downward direction. Therefore we obtain the equation $F = F_{g},$ which becomes $m\ v^{\prime}(t) = \text{−}mg.$ Dividing both sides of the equation by $m$ gives the equation

$$v^{\prime}(t) = \text{−}g.$$

Notice that this differential equation remains the same regardless of the mass of the object.

We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity, or the velocity at time $t = 0.$ This is denoted by $v(0) = v_{0}.$

Velocity of a Moving Baseball 运动棒球的速度

A baseball is thrown upward from a height of $3$ meters above Earth's surface with an initial velocity of $10\ \text{m/s},$ and the only force acting on it is gravity. The ball has a mass of $0.15\ \text{kg}$.

1. Find the velocity $v(t)$ of the baseball at time $t.$

2. What is its velocity after $2$ seconds?

Solution 解答

1. From the preceding discussion, the differential equation that applies in this situation is

$$v^{\prime}(t) = \text{−}g,$$

where $g = 9.8{\ \text{m/s}}^{2}.$ The initial condition is $v(0) = v_{0},$ where $v_{0} = 10\ \text{m/s}\text{.}$ Therefore the initial-value problem is $v^{\prime}(t) = -9.8{\ \text{m/s}}^{2},v(0) = 10\ \text{m/s}\text{.}$

The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives

$$\begin{array}{rll} {\int{v^{\prime}(t)\mspace{2mu} dt}} & = & {\int{-9.8dt}} \\ {v(t)} & = & {-9.8t + C.} \end{array}$$

The next step is to solve for $C.$ To do this, substitute $t = 0$ and $v(0) = 10\text{:}$

$$\begin{array}{rll} {v(t)} & = & {-9.8t + C} \\ {v(0)} & = & {-9.8(0) + C} \\ 10 & = & {C.} \end{array}$$

Therefore $C = 10$ and the velocity function is given by $v(t) = -9.8t + 10.$

2. To find the velocity after $2$ seconds, substitute $t = 2$ into $v(t).$

$$\begin{array}{rll} {v(t)} & = & {-9.8t + 10} \\ {v(2)} & = & {-9.8(2) + 10} \\ {v(2)} & = & -9.6. \end{array}$$

The units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of $9.6\ \text{m/s}\text{.}$

Suppose a rock falls from rest from a height of $100$ meters and the only force acting on it is gravity. Find an equation for the velocity $v(t)$ as a function of time, measured in meters per second.

A natural question to ask after solving this type of problem is how high the object will be above Earth's surface at a given point in time. Let $s(t)$ denote the height above Earth's surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation $s^{\prime}(t) = v(t).$ An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation $s(0) = s_{0}.$ Together these assumptions give the initial-value problem

$$s^{\prime}(t) = v(t),\quad s(0) = s_{0}.$$

If the velocity function is known, then it is possible to solve for the position function as well.

Height of a Moving Baseball 运动棒球的高度

A baseball is thrown upward from a height of $3$ meters above Earth's surface with an initial velocity of $10\ \text{m/s},$ and the only force acting on it is gravity. The ball has a mass of $0.15$ kilogram.

1. Find the position $s(t)$ of the baseball at time $t.$

2. What is its height after $2$ seconds?

Solution 解答

1. We already know the velocity function for this problem is $v(t) = -9.8t + 10.$ The initial height of the baseball is $3$ meters, so $s_{0} = 3.$ Therefore the initial-value problem for this example is

To solve the initial-value problem, we first find the antiderivatives:

$$\begin{array}{rll} {\int{s^{\prime}(t)\mspace{2mu} dt}} & = & {\int{-9.8t + 10dt}} \\ {s(t)} & = & {-4.9t^{2} + 10t + C.} \end{array}$$

Next we substitute $t = 0$ and solve for $C\text{:}$

$$\begin{array}{rll} {s(t)} & = & {-4.9t^{2} + 10t + C} \\ {s(0)} & = & {-4.9(0)^{2} + 10(0) + C} \\ 3 & = & {C.} \end{array}$$

Therefore the position function is $s(t) = -4.9t^{2} + 10t + 3.$

2. The height of the baseball after $2\ \text{s}$ is given by $s(2)\text{:}$

$$\begin{array}{cl} {s(2)} & {= -4.9(2)^{2} + 10(2) + 3} \\ & {= -4.9(4) + 23} \\ & {= 3.4.} \end{array}$$

Therefore the baseball is $3.4$ meters above Earth's surface after $2$ seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.

Section 4.1 Exercises 4.1 节习题

Determine the order of the following differential equations.

1.

$y^{\prime} + y = 3y^{2}$

2\.

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

3.

$y\text{'''} + y^{''}y^{\prime} = 3x^{2}$

4\.

$y^{\prime} = y^{''} + 3t^{2}$

5.

$\frac{dy}{dt} = t$

6\.

$\frac{dy}{dx} + \frac{d^{2}y}{dx^{2}} = 3x^{4}$

7.

$\left( \frac{dy}{dt} \right)^{2} + 8\frac{dy}{dt} + 3y = 4t$

Verify that the following functions are solutions to the given differential equation.

8\.

$y = \frac{x^{3}}{3}$ solves $y^{\prime} = x^{2}$

9\.

$y = 2e^{\text{−}x} + x - 1$ solves $y^{\prime} = x - y$

10\.

$y = e^{3x} - \frac{e^{x}}{2}$ solves $y^{\prime} = 3y + e^{x}$

11\.

$y = \frac{1}{1 - x}$ solves $y^{\prime} = y^{2}$

12\.

$y = e^{x^{2}\text{/}2}$ solves $y^{\prime} = xy$

13\.

$y = 4 + \text{ln}\mspace{2mu} x$ solves $xy^{\prime} = 1$

14\.

$y = 3 - x + x\mspace{2mu}\text{ln}\mspace{2mu} x$ solves $y^{\prime} = \text{ln}\mspace{2mu} x$

15\.

$y = 2e^{x} - x - 1$ solves $y^{\prime} = y + x$

16\.

$y = e^{x} + \frac{\text{sin}\mspace{2mu} x}{2} - \frac{\text{cos}\mspace{2mu} x}{2}$ solves $y^{\prime} = \text{cos}\mspace{2mu} x + y$

17\.

$y = \pi e^{\text{−}\text{cos}\mspace{2mu} x}$ solves $y^{\prime} = y\mspace{2mu}\text{sin}\mspace{2mu} x$

Verify the following general solutions and find the particular solution.

18\.

Find the particular solution to the differential equation $y^{\prime} = 4x^{2}$ that passes through $(-3,-30),$ given that $y = C + \frac{4x^{3}}{3}$ is a general solution.

19.

Find the particular solution to the differential equation $y^{\prime} = 3x^{3}$ that passes through $(1,4.75),$ given that $y = C + \frac{3x^{4}}{4}$ is a general solution.

20\.

Find the particular solution to the differential equation $y^{\prime} = 3x^{2}y$ that passes through $(0,12),$ given that $y = Ce^{x^{3}}$ is a general solution.

21.

Find the particular solution to the differential equation $y^{\prime} = 2xy$ that passes through $\left( {0,\frac{1}{2}} \right),$ given that $y = Ce^{x^{2}}$ is a general solution.

22\.

Find the particular solution to the differential equation $y^{\prime} = \left( {2xy} \right)^{2}$ that passes through $\left( {1, - \frac{1}{2}} \right),$ given that $y = - \frac{3}{C + 4x^{3}}$ is a general solution.

23.

Find the particular solution to the differential equation $y^{\prime}x^{2} = y$ that passes through $\left( {1,\frac{2}{e}} \right),$ given that $y = Ce^{\text{−}{1\text{/}x}}$ is a general solution.

24\.

Find the particular solution to the differential equation $8\frac{dx}{dt} = -2\mspace{2mu}\text{cos}(2t) - \text{cos}(4t)$ that passes through $\left( {\pi,\pi} \right),$ given that $x = C - \frac{1}{8}\text{sin}(2t) - \frac{1}{32}\text{sin}(4t)$ is a general solution.

25.

Find the particular solution to the differential equation $\frac{du}{dt} = \text{tan}\mspace{2mu} u$ that passes through $\left( {1,\frac{\pi}{2}} \right),$ given that $u = \text{sin}^{-1}\left( e^{C + t} \right)$ is a general solution.

26\.

Find the particular solution to the differential equation $\frac{dy}{dt} = e^{(t + y)}$ that passes through $\left( {1,0} \right),$ given that $y = \text{−}\text{ln}(C - e^{t})$ is a general solution.

27.

Find the particular solution to the differential equation $y^{\prime}(1 - x^{2}) = 1 + y$ that passes through $\left( {0,-2} \right),$ given that $y = C\frac{\sqrt{x + 1}}{\sqrt{1 - x}} - 1$ is a general solution.

For the following problems, find the general solution to the differential equation.

28\.

$y^{\prime} = 3x + e^{x}$

29.

$y^{\prime} = \text{ln}\mspace{2mu} x + \text{tan}\mspace{2mu} x$

30\.

$y^{\prime} = \text{sin}\mspace{2mu} xe^{\text{cos}\mspace{2mu} x}$

31.

$y^{\prime} = 4^{x}$

32\.

$y^{\prime} = \text{sin}^{-1}\left( {2x} \right)$

33.

$y^{\prime} = 2t\sqrt{t^{2} + 16}$

34\.

$x^{\prime} = \text{coth}\mspace{2mu} t + \text{ln}\mspace{2mu} t + 3t^{2}$

35.

$x^{\prime} = t\sqrt{4 + t}$

36\.

$y^{\prime} = y$

37.

$y^{\prime} = \frac{y}{x}$

Solve the following initial-value problems starting from $y(0) = 1$ and $y(0) = -1.$ Draw both solutions on the same graph.

38\.

$\frac{dy}{dt} = 2t$

39.

$\frac{dy}{dt} = \text{−}t$

40\.

$\frac{dy}{dt} = 2y$

41.

$\frac{dy}{dt} = \text{−}y$

42\.

$\frac{dy}{dt} = 2$

Solve the following initial-value problems starting from $y_{0} = 10.$ At what time does $y$ increase to $100$ or drop to $1?$

43.

$\frac{dy}{dt} = 4t$

44\.

$\frac{dy}{dt} = 4y$

45.

$\frac{dy}{dt} = -2y$

46\.

$\frac{dy}{dt} = e^{4t}$

47.

$\frac{dy}{dt} = e^{-4t}$

Recall that a family of solutions includes solutions to a differential equation that differ by a constant. For the following problems, use your calculator to graph a family of solutions to the given differential equation. Use initial conditions from $y(t = 0) = -10$ to $y(t = 0) = 10$ increasing by $2.$ Is there some critical point where the behavior of the solution begins to change?

48\.

\[T\] $y^{\prime} = y(x)$

49.

\[T\] $xy^{\prime} = y$

50\.

\[T\] $y^{\prime} = t^{3}$

51.

\[T\] $y^{\prime} = x + y$ (*Hint:* $y = Ce^{x} - x - 1$ is the general solution)

52\.

\[T\] $y^{\prime} = x\mspace{2mu}\text{ln}\mspace{2mu} x + \text{sin}\mspace{2mu} x$

53.

Find the general solution to describe the velocity of a ball of mass $1\ \text{lb}$ that is thrown upward at a rate $a$ ft/sec.

54\.

In the preceding problem, if the initial velocity of the ball thrown into the air is $a = 25$ ft/s, write the particular solution to the velocity of the ball. Solve to find the time when the ball hits the ground.

55.

You throw two objects with differing masses $m_{1}$ and $m_{2}$ upward into the air with the same initial velocity $a$ ft/s. What is the difference in their velocity after $1$ second?

56\.

\[T\] You throw a ball of mass $1$ kilogram upward with a velocity of $a = 25$ m/s on Mars, where the acceleration of gravity is $g = -3.711$ m/s². Use your calculator to approximate how much longer the ball is in the air on Mars than on Earth, where $g = - 9.8m/s^{2}$.

57.

\[T\] For the previous problem, use your calculator to approximate how much higher the ball went on Mars, where $g = - 9.8m/s^{2}$.

58\.

\[T\] A car on the freeway accelerates according to $a = 15\mspace{2mu}\text{cos}(\pi t),$ where $t$ is measured in hours. Set up and solve the differential equation to determine the velocity of the car if it has an initial speed of $50$ mph. After $40$ minutes of driving, what is the driver’s velocity?

59.

\[T\] For the car in the preceding problem, find the expression for the distance the car has traveled in time $t,$ assuming an initial distance of $0.$ How long does it take the car to travel $100$ miles? Round your answer to hours and minutes.

60\.

\[T\] For the previous problem, find the total distance traveled in the first hour.

61.

Substitute $y = Be^{3t}$ into $y^{\prime} - y = 8e^{3t}$ to find a particular solution.

62\.

Substitute $y = a\mspace{2mu}\text{cos}(2t) + b\mspace{2mu}\text{sin}(2t)$ into $y^{\prime} + y = 4\mspace{2mu}\text{sin}(2t)$ to find a particular solution.

63.

Substitute $y = a + bt + ct^{2}$ into $y^{\prime} + y = 1 + t^{2}$ to find a particular solution.

64\.

Substitute $y = ae^{t}\text{cos}\mspace{2mu} t + be^{t}\text{sin}\mspace{2mu} t$ into $y^{\prime} = 2e^{t}\text{cos}\mspace{2mu} t$ to find a particular solution.

65.

Solve $y^{\prime} = e^{kt}$ with the initial condition $y(0) = 0$ and solve $y^{\prime} = 1$ with the same initial condition. As $k$ approaches $0,$ what do you notice?

4.2 Direction Fields and Numerical Methods 4.2 方向场与数值方法

• 4.2.1 Draw the direction field for a given first-order differential equation.
• 4.2.2 Use a direction field to draw a solution curve of a first-order differential equation.
• 4.2.3 Use Euler’s Method to approximate the solution to a first-order differential equation.

• 4.2.1 为给定的一阶微分方程画出方向场。
• 4.2.2 利用方向场画出一阶微分方程的解曲线。
• 4.2.3 使用欧拉方法近似求解一阶微分方程。

For the rest of this chapter we will focus on various methods for solving differential equations and analyzing the behavior of the solutions. In some cases it is possible to predict properties of a solution to a differential equation without knowing the actual solution. We will also study numerical methods for solving differential equations, which can be programmed by using various computer languages or even by using a spreadsheet program, such as Microsoft Excel.

Creating Direction Fields 创建方向场

Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, we consider a first-order differential equation of the form

$$y\prime = f\left( {x,y} \right).$$

An applied example of this type of differential equation appears in Newton’s law of cooling, which we will solve explicitly later in this chapter. First, though, let us create a direction field for the differential equation

$$T^{\prime}(t) = -0.4(T - 72).$$

Here $T(t)$ represents the temperature (in degrees Fahrenheit) of an object at time $t,$ and the ambient temperature is $72\text{°}\text{F}\text{.}$ Figure 4.6 shows the direction field for this equation.

The idea behind a direction field is the fact that the derivative of a function evaluated at a given point is the slope of the tangent line to the graph of that function at the same point. Other examples of differential equations for which we can create a direction field include

$$\begin{array}{l} {y\prime = 3x + 2y - 4} \\ {y\prime = x^{2} - y^{2}} \\ {y\prime = \frac{2x + 4}{y - 2}.} \end{array}$$

To create a direction field, we start with the first equation: $y\prime = 3x + 2y - 4.$ We let $\left( {x_{0},y_{0}} \right)$ be any ordered pair, and we substitute these numbers into the right-hand side of the differential equation. For example, if we choose $x = 1\ \text{and}\ y = 2,$ substituting into the right-hand side of the differential equation yields

$$\begin{array}{cl} y^{\prime} & {= 3x + 2y - 4} \\ & {= 3(1) + 2(2) - 4 = 3.} \end{array}$$

This tells us that if a solution to the differential equation $y\prime = 3x + 2y - 4$ passes through the point $\left( {1,2} \right),$ then the slope of the solution at that point must equal $3.$ To start creating the direction field, we put a short line segment at the point $\left( {1,2} \right)$ having slope $3.$ We can do this for any point in the domain of the function $f\left( {x,y} \right) = 3x + 2y - 4,$ which consists of all ordered pairs $\left( {x,y} \right)$ in $\mathbb{R}^{2}.$ Therefore any point in the Cartesian plane has a slope associated with it, assuming that a solution to the differential equation passes through that point. The direction field for the differential equation $y^{\prime} = 3x + 2y - 4$ is shown in Figure 4.7.

We can generate a direction field of this type for any differential equation of the form $y\prime = f\left( {x,y} \right).$

A direction field (slope field) is a mathematical object used to graphically represent solutions to a first-order differential equation. At each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to the differential equation passing through that point.

Using Direction Fields 使用方向场

We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actual solution. For example, the direction field in Figure 4.7 serves as a guide to the behavior of solutions to the differential equation $y\prime = 3x + 2y - 4.$

To use a direction field, we start by choosing any point in the field. The line segment at that point serves as a signpost telling us what direction to go from there. For example, if a solution to the differential equation passes through the point $\left( {0,1} \right),$ then the slope of the solution passing through that point is given by $y\prime = 3(0) + 2(1) - 4 = -2.$ Now let $x$ increase slightly, say to $x = 0.1.$ Using the method of linear approximations gives a formula for the approximate value of $y$ for $x = 0.1.$ In particular,

$$\begin{array}{cl} {L(x)} & {= y_{0} + f^{\prime}\left( x_{0} \right)\left( {x - x_{0}} \right)} \\ & {= 1 - 2(x - 0)} \\ & {= 1 - 2x.} \end{array}$$

Substituting $x = 0.1$ into $L(x)$ gives an approximate $y$ value of $0.8.$

At this point the slope of the solution changes (again according to the differential equation). We can keep progressing, recalculating the slope of the solution as we take small steps to the right, and watching the behavior of the solution. Figure 4.8 shows a graph of the solution passing through the point $\left( {0,1} \right).$

The curve is the graph of the solution to the initial-value problem

$$y\prime = 3x + 2y - 4,\quad y(0) = 1.$$

This curve is called a solution curve passing through the point $\left( {0,1} \right).$ The exact solution to this initial-value problem is

$$y = - \frac{3}{2}x + \frac{5}{4} - \frac{1}{4}e^{2x},$$

and the graph of this solution is identical to the curve in Figure 4.8.

Create a direction field for the differential equation $y\prime = x^{2} - y^{2}$ and sketch a solution curve passing through the point $\left( {-1,2} \right).$

Go to this website to see more about slope fields.

Now consider the direction field for the differential equation $y\prime = (x - 3)(y^{2} - 4),$ shown in Figure 4.9. This direction field has several interesting properties. First of all, at $y = -2$ and $y = 2,$ horizontal dashes appear all the way across the graph. This means that if $y = -2,$ then $y\prime = 0.$ Substituting this expression into the right-hand side of the differential equation gives

$$\begin{array}{cl} {(x - 3)(y^{2} - 4)} & {= (x - 3)((-2)^{2} - 4)} \\ & {= (x - 3)(0)} \\ & {= 0} \\ & {= y\prime.} \end{array}$$

Therefore $y = -2$ is a solution to the differential equation. Similarly, $y = 2$ is a solution to the differential equation. These are the only constant-valued solutions to the differential equation, as we can see from the following argument. Suppose $y = k$ is a constant solution to the differential equation. Then $y^{\prime} = 0.$ Substituting this expression into the differential equation yields $0 = \left( {x - 3} \right)\left( {k^{2} - 4} \right).$ This equation must be true for all values of $x,$ so the second factor must equal zero. This result yields the equation $k^{2} - 4 = 0.$ The solutions to this equation are $k = -2$ and $k = 2,$ which are the constant solutions already mentioned. These are called the equilibrium solutions to the differential equation.

Consider the differential equation $y\prime = f(x,y).$ An equilibrium solution is any solution to the differential equation of the form $y = c,$ where $c$ is a constant.

To determine the equilibrium solutions to the differential equation $y\prime = f(x,y),$ set the right-hand side equal to zero. An equilibrium solution of the differential equation is any function of the form $y = k$ such that $f\left( {x,k} \right) = 0$ for all values of $x$ in the domain of $f.$

An important characteristic of equilibrium solutions concerns whether or not they approach the line $y = k$ as an asymptote for large values of $x.$

Consider the differential equation $y^{\prime} = f\left( {x,y} \right),$ and assume that all solutions to this differential equation are defined for $x \geq x_{0}.$ Let $y = k$ be an equilibrium solution to the differential equation.

1. $y = k$ is an asymptotically stable solution to the differential equation if there exists $\varepsilon > 0$ such that for any value $c \in \left( {k - \varepsilon,k + \varepsilon} \right)$ the solution to the initial-value problem

$$y^{\prime} = f\left( {x,y} \right),\quad y\left( x_{0} \right) = c$$

approaches $k$ as $x$ approaches infinity.

2. $y = k$ is an asymptotically unstable solution to the differential equation if there exists $\varepsilon > 0$ such that for any value $c \in \left( {k - \varepsilon,k + \varepsilon} \right)$ the solution to the initial-value problem

$$y^{\prime} = f\left( {x,y} \right),\quad y\left( x_{0} \right) = c$$

never approaches $k$ as $x$ approaches infinity.

3. $y = k$ is an asymptotically semi-stable solution to the differential equation if it is neither asymptotically stable nor asymptotically unstable.

Now we return to the differential equation $y\prime = (x - 3)(y^{2} - 4),$ with the initial condition $y(0) = 0.5.$ The direction field for this initial-value problem, along with the corresponding solution, is shown in Figure 4.10.

The values of the solution to this initial-value problem stay between $y = -2$ and $y = 2,$ which are the equilibrium solutions to the differential equation. However, an initial-value problem starting with $- 2 < y < 2$ can never cross the equilibrium solutions $y = 2$ and $y = -2$. Therefore, since $y^{2} - 4 < 0$ and for $x > 3,y' = (x - 3)\left( y^{2} - 4 \right) < 0$, $y$ is decreasing and therefore approaches $y = - 2$. Therefore, $y = - 2$ is an asymptotically stable solution to the differential equation.

What happens when the initial value is below $y = -2?$ This scenario is illustrated in Figure 4.11, with the initial value $y(0) = -3.$

We can see that for initial values $y < - 2$ for $x > 3,y' = (x - 3)\left( y^{2} - 4 \right) > 0$ and $y$ is increasing and therefore approaches $y = - 2$. This reaffirms that $y = - 2$ is an asymptotically stable solution to the differential equation.

Stability of an Equilibrium Solution 平衡解的稳定性

Create a direction field for the differential equation $y\prime = {(y - 3)}^{2}(y^{2} + y - 2)$ and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

Solution 解答

The direction field is shown in Figure 4.12.

The equilibrium solutions are $y = -2,y = 1,$ and $y = 3.$ To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at $y = -2$ the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to $y = -2$ approach $y = -2,$ and the solution is stable. For the solution $y = 1,$ all initial conditions above and below $y = 1$ are repelled (pushed away) from $y = 1,$ so this solution is unstable. The solution $y = 3$ is semi-stable, because for initial conditions slightly greater than $3,$ the solution approaches infinity, and for initial conditions slightly less than $3,$ the solution approaches $y = 3.$

Analysis 分析

It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal to zero and solving for $y.$ This approach gives the same equilibrium solutions as those we saw in the direction field.

Create a direction field for the differential equation $y\prime = (x + 5)(y + 2)(y^{2} - 4y + 4)$ and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

Euler's Method 欧拉方法

Consider the initial-value problem

$$y^{\prime} = 2x - 3,\quad y(0) = 3.$$

Integrating both sides of the differential equation gives $y = x^{2} - 3x + C,$ and solving for $C$ yields the particular solution $y = x^{2} - 3x + 3.$ The solution for this initial-value problem appears as the parabola in Figure 4.13.

The red graph consists of line segments that approximate the solution to the initial-value problem. The graph starts at the same initial value of $\left( {0,3} \right).$ Then the slope of the solution at any point is determined by the right-hand side of the differential equation, and the length of the line segment is determined by increasing the $x$ value by $0.5$ each time (the *step size*). This approach is the basis of Euler's Method.

Before we state Euler's Method as a theorem, let's consider another initial-value problem:

$$y^{\prime} = x^{2} - y^{2},\quad y(-1) = 2.$$

The idea behind direction fields can also be applied to this problem to study the behavior of its solution. For example, at the point $\left( {-1,2} \right),$ the slope of the solution is given by $y\prime = (-1)^{2} - 2^{2} = -3,$ so the slope of the tangent line to the solution at that point is also equal to $-3.$ Now we define $x_{0} = -1$ and $y_{0} = 2.$ Since the slope of the solution at this point is equal to $-3,$ we can use the method of linear approximation to approximate $y$ near $\left( {-1,2} \right).$

$$L(x) = y_{0} + f^{\prime}\left( x_{0} \right)\left( {x - x_{0}} \right).$$

Here $x_{0} = -1,y_{0} = 2,$ and $f^{\prime}\left( x_{0} \right) = -3,$ so the linear approximation becomes

$$\begin{array}{cl} {L(x)} & {= 2 - 3\left( {x - (-1)} \right)} \\ & {= 2 - 3x - 3} \\ & {= -3x - 1.} \end{array}$$

Now we choose a step size. The step size is a small value, typically $0.1$ or less, that serves as an increment for $x;$ it is represented by the variable $h.$ In our example, let $h = 0.1.$ Incrementing $x_{0}$ by $h$ gives our next $x$ value:

$$x_{1} = x_{0} + h = -1 + 0.1 = -0.9.$$

We can substitute $x_{1} = -0.9$ into the linear approximation to calculate $y_{1}.$

$$\begin{array}{cl} y_{1} & {= L\left( x_{1} \right)} \\ & {= -3(-0.9) - 1} \\ & {= 1.7.} \end{array}$$

Therefore the approximate $y$ value for the solution when $x = -0.9$ is $y = 1.7.$ We can then repeat the process, using $x_{1} = -0.9$ and $y_{1} = 1.7$ to calculate $x_{2}$ and $y_{2}.$ The new slope is given by $y\prime = (-0.9)^{2} - (1.7)^{2} = -2.08.$ First, $x_{2} = x_{1} + h = -0.9 + 0.1 = -0.8.$ Using linear approximation gives

$$\begin{array}{cl} {L(x)} & {= y_{1} + f^{\prime}\left( x_{1} \right)\left( {x - x_{1}} \right)} \\ & {= 1.7 - 2.08\left( {x - (-0.9)} \right)} \\ & {= 1.7 - 2.08x - 1.872} \\ & {= -2.08x - 0.172.} \end{array}$$

Finally, we substitute $x_{2} = -0.8$ into the linear approximation to calculate $y_{2}.$

$$\begin{array}{cl} y_{2} & {= L(x_{2})} \\ & {= -2.08x_{2} - 0.172} \\ & {= -2.08(-0.8) - 0.172} \\ & {= 1.492.} \end{array}$$

Therefore the approximate value of the solution to the differential equation is $y = 1.492$ when $x = -0.8.$

What we have just shown is the idea behind Euler's Method. Repeating these steps gives a list of values for the solution. These values are shown in Table 4.2, rounded off to four decimal places.

Table 4.2 Using Euler's Method to Approximate Solutions to a Differential Equation

Euler's Method 欧拉方法

Consider the initial-value problem

$y\prime = f\left( {x,y} \right),\quad y\left( x_{0} \right) = y_{0}.$

To approximate a solution to this problem using Euler's method, define

$$\begin{array}{l} {x_{n} = x_{0} + nh} \\ {y_{n} = y_{n - 1} + hf(x_{n - 1},y_{n - 1}).} \end{array}$$ (4.2)

Here $h > 0$ represents the step size and $n$ is an integer, starting with $1.$ The number of steps taken is counted by the variable $n.$

Typically $h$ is a small value, say $0.1$ or $0.05.$ The smaller the value of $h,$ the more calculations are needed. The higher the value of $h,$ the fewer calculations are needed. However, the tradeoff results in a lower degree of accuracy for larger step size, as illustrated in Figure 4.14.

Using Euler's Method 使用欧拉方法

Consider the initial-value problem

$$y^{\prime} = 3x^{2} - y^{2} + 1,\quad y(0) = 2.$$

Use Euler's method with a step size of $0.1$ to generate a table of values for the solution for values of $x$ between $0$ and $1.$

Solution

We are given $h = 0.1$ and $f\left( {x,y} \right) = 3x^{2} - y^{2} + 1.$ Furthermore, the initial condition $y(0) = 2$ gives $x_{0} = 0$ and $y_{0} = 2.$ Using Equation 4.2 with $n = 0,$ we can generate Table 4.3.

Table 4.3 Using Euler's Method to Approximate Solutions to a Differential Equation

With ten calculations, we are able to approximate the values of the solution to the initial-value problem for values of $x$ between $0$ and $1.$

For more information on Euler's method use this applet.

Consider the initial-value problem

$$y^{\prime} = x^{3} + y^{2},\quad y(1) = -2.$$

Using a step size of $0.1,$ generate a table with approximate values for the solution to the initial-value problem for values of $x$ between $1$ and $2.$

Visit this website for a practical application of the material in this section.

Section 4.2 Exercises 4.2 节习题

For the following problems, use the direction field below from the differential equation $y\prime = -2y.$ Sketch the graph of the solution for the given initial conditions.

66\.

$y(0) = 1$

67.

$y(0) = 0$

68\.

$y(0) = -1$

69.

Are there any equilibria? What are their stabilities?

For the following problems, use the direction field below from the differential equation $y\prime = y^{2} - 2y.$ Sketch the graph of the solution for the given initial conditions.

70\.

$y(0) = 3$

71.

$y(0) = 1$

72\.

$y(0) = -1$

73.

Are there any equilibria? What are their stabilities?

Draw the direction field for the following differential equations, then solve the differential equation.

74\.

$y\prime = t^{3}$

75.

$y\prime = e^{t}$

76\.

$\frac{dy}{dx} = x^{2}\text{cos}\mspace{2mu} x$

77.

$\frac{dy}{dt} = te^{t}$

78\.

$\frac{dx}{dt} = \text{cosh}(t)$

Draw the directional field for the following differential equations. What can you say about the behavior of the solution? Are there equilibria? What stability do these equilibria have?

79.

$y\prime = y^{2} - 1$

80\.

$y\prime = y - x$

81.

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

82\.

$y\prime = t^{2}\text{sin}\mspace{2mu} y$

83.

$y\prime = 3y + xy$

Match the direction field with the given differential equations. Explain your selections.

84\.

$y\prime = -3y$

85.

$y\prime = -3t$

86\.

$y\prime = e^{t}$

87.

$y\prime = \frac{1}{2}y + t$

88\.

$y\prime = \text{−}ty$

Match the direction field with the given differential equations. Explain your selections.

89.

$y\prime = t\mspace{2mu}\text{sin}\mspace{2mu} y$

90\.

$y\prime = \text{−}t\mspace{2mu}\text{cos}\mspace{2mu} y$

91.

$y\prime = t\mspace{2mu}\text{tan}\mspace{2mu} y$

92\.

$y\prime = \text{sin}^{2}y$

93.

$y\prime = y^{2}t^{3}$

Estimate the following solutions using Euler's method with $n = 5$ steps over the interval $t = \lbrack 0,1\rbrack.$ If you are able to solve the initial-value problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler's method. How accurate is Euler's method?

94\.

$y\prime = -3y,\quad y(0) = 1$

95.

${y\prime = t^{2}},~y(0)~ = ~2$

96\.

$y^{\prime} = 3t - y,y(0) = 1.$ Exact solution is $y = 3t + 4e^{\text{−}t} - 3$

97.

$y^{\prime} = y + t^{2},y(0) = 3.$ Exact solution is $y = 5e^{t} - 2 - t^{2} - 2t$

98\.

$y^{\prime} = 2t,y(0) = 0$

99.

\[T\] $y\prime = e^{(x + y)},y(0) = -1.$ Exact solution is $y = \text{−}\text{ln}(e + 1 - e^{x})$

100\.

$y^{\prime} = y^{2}\text{ln}(x + 1),y(0) = 1.$ Exact solution is $y = - \frac{1}{(x + 1)(\text{ln}(x + 1) - 1)}$

101.

$y^{\prime} = 2^{x},y(0) = 0,$ Exact solution is $y = \frac{2^{x} - 1}{\text{ln}(2)}$

102\.

$y^{\prime} = y,y(0) = -1.$ Exact solution is $y = \text{−}e^{x}.$

103.

$y^{\prime} = -5t,y(0) = -2.$ Exact solution is $y = - \frac{5}{2}t^{2} - 2$

Differential equations can be used to model disease epidemics. In the next set of problems, we examine the change of size of two sub-populations of people living in a city: individuals who are infected and individuals who are susceptible to infection. $S$ represents the size of the susceptible population, and $I$ represents the size of the infected population. We assume that if a susceptible person interacts with an infected person, there is a probability $c$ that the susceptible person will become infected. Each infected person recovers from the infection at a rate $r$ and becomes susceptible again. We consider the case of influenza, where we assume that no one dies from the disease, so we assume that the total population size of the two sub-populations is a constant number, $N.$ The differential equations that model these population sizes are

$$\begin{array}{l} {S\prime = rI - cSI\quad\text{and}} \\ {I\prime = cSI - rI.} \end{array}$$

Here $c$ represents the contact rate and $r$ is the recovery rate.

104\.

Show that, by our assumption that the total population size is constant $(S + I = N),$ you can reduce the system to a single differential equation in $I\text{:}\ {I\prime = c(N - I)I - rI.}$

105.

Assuming the parameters are $c = 0.5,N = 5,$ and $r = 0.5,$ draw the resulting directional field.

106\.

\[T\] Use computational software or a calculator to compute an approximate solution to the initial-value problem $y\prime = ty,\quad y(0) = 2$ using Euler's Method with a given step size $h.$ Find the solution at $t = 1$ in terms of $h.$ *Hint*: Here is "pseudo-code" for how to write a computer program to perform Euler's Method for $y\prime = f(t,y),y(0) = y_{0}\text{:}$

• Create function $f(t,y).$
• Define parameters $y(0) = y_{0},\mspace{9mu} t(0) = 0,$ step size $h,$ and total number of steps, $N$
• Write a for loop:

• 创建函数 $f(t,y)$。
• 定义参数 $y(0) = y_{0},\mspace{9mu} t(0) = 0,$ 步长 $h,$ 以及总步数 $N$
• 编写 for 循环：

for $\text{k} = 0\ \textbf{to}\ N$

$\qquad{\text{fn} = \text{f}\left( {\text{t}\left( \text{k} \right),\text{y}\left( \text{k} \right)} \right)}$

$\qquad{\text{y}\left( \text{k+1} \right) = \text{y}\left( \text{k} \right) + \text{h*fn}}$

$\qquad{\text{t}\left( \text{k+1} \right) = \text{t}\left( \text{k} \right) + \text{h}}$

Refer to your answer for Exercise 4.106 to solve the following exercises.

107.

Solve the initial-value problem for the exact solution.

108\.

Draw the directional field

109.

Compute using $h = 1$

110\.

Compute using \[T\] $h = 0.1$

111.

Compute using \[T\] $h = 0.001$

112\.

Compute using \[T\] $h = 0.0001$

113.

\[T\] Evaluate the exact solution at $t = 1.$ Make a table of errors for the relative error between the Euler's method solution and the exact solution. How much does the error change? Can you explain?

Consider the initial-value problem $y\prime = -2y,\quad y(0) = 2.$

114\.

Show that $y = 2e^{-2x}$ solves this initial-value problem.

115.

Draw the directional field of this differential equation.

116\.

\[T\] By hand or by calculator or computer, approximate the solution using Euler's Method at $t = 10$ using $h = 5.$

117.

\[T\] By calculator or computer, approximate the solution using Euler's Method at $t = 10$ using $h = 100.$

118\.

\[T\] Plot exact answer and each Euler approximation (for $h = 5$ and $h = 100)$ at each $h$ on the directional field. What do you notice?

4.3 Separable Equations 4.3 可分离变量的微分方程

• 4.3.1 Use separation of variables to solve a differential equation.
• 4.3.2 Solve applications using separation of variables.

• 4.3.1 使用分离变量法求解微分方程。
• 4.3.2 利用分离变量法求解应用题。

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

Separation of Variables 分离变量法

We start with a definition and some examples.

A separable differential equation is any equation that can be written in the form

$$y\prime = f(x)g{(y).}$$ (4.3)

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of $x$ times a function of $y.$ Examples of separable differential equations include

$$\begin{array}{l} \\ \\ {y\prime = \left( {x^{2} - 4} \right)\left( {3y + 2} \right)} \\ {y\prime = 6x^{2} + 4x} \\ {y\prime = \text{sec}\mspace{2mu} y + \text{tan}\mspace{2mu} y} \\ {y\prime = xy + 3x - 2y - 6.} \end{array}$$

The second equation is separable with $f(x) = 6x^{2} + 4x$ and $g(y) = 1,$ the third equation is separable with $f(x) = 1$ and $g(y) = \text{sec}\mspace{2mu} y + \text{tan}\mspace{2mu} y,$ and the right-hand side of the fourth equation can be factored as $\left( {x - 2} \right)\left( {y + 3} \right),$ so it is separable as well. The third equation is also called an autonomous differential equation because the right-hand side of the equation is a function of $y$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.

Separation of Variables 分离变量法（解题策略）

1. Check for any values of $y$ that make $g(y) = 0.$ These correspond to constant solutions.

2. Rewrite the differential equation in the form $\frac{dy}{g(y)} = f(x)dx.$

3. Integrate both sides of the equation.

4. Solve the resulting equation for $y$ if possible.

5. If an initial condition exists, substitute the appropriate values for $x$ and $y$ into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for $y$ if possible.” It is not always possible to obtain $y$ as an explicit function of $x.$ Quite often we have to be satisfied with finding $y$ as an implicit function of $x.$

Using Separation of Variables 应用分离变量法

Find a general solution to the differential equation $y\prime = \left( {x^{2} - 4} \right)\left( {3y + 2} \right)$ using the method of separation of variables.

Solution

Follow the five-step method of separation of variables.

1. In this example, $f(x) = x^{2} - 4$ and $g(y) = 3y + 2.$ Setting $g(y) = 0$ gives $y = - \frac{2}{3}$ as a constant solution.

2. Rewrite the differential equation in the form

$$\frac{dy}{3y + 2} = (x^{2} - 4)dx.$$

3. Integrate both sides of the equation:

$$\left. \int\frac{dy}{3y + 2} \right. = {\int{\left( {x^{2} - 4} \right)\mspace{2mu} dx}}.$$

Let $u = 3y + 2.$ Then $du = 3\frac{dy}{dx}dx,$ so the equation becomes

$$\begin{array}{rll} {\frac{1}{3}\left. \int{\frac{1}{u}du} \right.} & = & {\frac{1}{3}x^{3} - 4x + C} \\ {\frac{1}{3}\mspace{2mu}\text{ln}|u|} & = & {\frac{1}{3}x^{3} - 4x + C} \\ {\frac{1}{3}\mspace{2mu}\text{ln}\left| {3y + 2} \right|} & = & {\frac{1}{3}x^{3} - 4x + C.} \end{array}$$

4. To solve this equation for $y,$ first multiply both sides of the equation by $3.$

$$\text{ln}\left| {3y + 2} \right| = x^{3} - 12x + 3C$$

Now we use some logic in dealing with the constant $C.$ Since $C$ represents an arbitrary constant, $3C$ also represents an arbitrary constant. If we call the second arbitrary constant $C_{1},$ the equation becomes

$$\text{ln}\left| {3y + 2} \right| = x^{3} - 12x + C_{1}.$$

Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base $e).$

$$\begin{array}{rll} e^{\text{ln}{|{3y + 2}|}} & = & e^{x^{3} - 12x + C_{1}} \\ \left| {3y + 2} \right| & = & {e^{C_{1}}e^{x^{3} - 12x}} \end{array}$$

Again define a new constant $C_{2} = e^{c_{1}}$ (note that $C_{2} > 0)\text{:}$

$$\left| {3y + 2} \right| = C_{2}e^{x^{3} - 12x}.$$

This corresponds to two separate equations: $3y + 2 = C_{2}e^{x^{3} - 12x}$ and $3y + 2 = \text{−}C_{2}e^{x^{3} - 12x}.$

The solution to either equation can be written in the form $y = \frac{-2 \pm C_{2}e^{x^{3} - 12x}}{3}.$

Since $C_{2} > 0,$ it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant $C$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as

$$y = \frac{-2 + Ce^{x^{3} - 12x}}{3}.$$

5. No initial condition is imposed, so we are finished.

Use the method of separation of variables to find a general solution to the differential equation $y\prime = 2xy + 3y - 4x - 6.$

Solving an Initial-Value Problem 求解初值问题

Using the method of separation of variables, solve the initial-value problem

$$y\prime = (2x + 3)(y^{2} - 4),\quad y(0) = -1.$$

Solution

Follow the five-step method of separation of variables.

1. In this example, $f(x) = 2x + 3$ and $g(y) = y^{2} - 4.$ Setting $g(y) = 0$ gives $y = \pm 2$ as constant solutions.

2. Divide both sides of the equation by $y^{2} - 4$ and multiply by $dx.$ This gives the equation

$$\frac{dy}{y^{2} - 4} = \left( {2x + 3} \right)\mspace{2mu} dx.$$

3. Next integrate both sides:

$$\left. \int{\frac{1}{y^{2} - 4}\ dy} \right. = {\int{\left( {2x + 3} \right)\mspace{2mu} dx}}.$$ (4.4)

To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity

$$\frac{1}{y^{2} - 4} = \frac{1}{4}\left( {\frac{1}{y - 2} - \frac{1}{y + 2}} \right).$$

Then Equation 4.4 becomes

$$\begin{array}{rll} {\frac{1}{4}\left. \int{\left( {\frac{1}{y - 2} - \frac{1}{y + 2}} \right)\mspace{2mu} dy} \right.} & = & {\int{\left( {2x + 3} \right)\mspace{2mu} dx}} \\ {\frac{1}{4}\left( {\text{ln}\left| {y - 2} \right| - \text{ln}\left| {y + 2} \right|} \right)} & = & {x^{2} + 3x + C.} \end{array}$$

Multiplying both sides of this equation by $4$ and replacing $4C$ with $C_{1}$ gives

$$\begin{array}{rll} {\text{ln}\left| {y - 2} \right| - \text{ln}\left| {y + 2} \right|} & = & {4x^{2} + 12x + C_{1}} \\ {\text{ln}\left| \frac{y - 2}{y + 2} \right|} & = & {4x^{2} + 12x + C_{1}.} \end{array}$$

4. It is possible to solve this equation for *y*. First exponentiate both sides of the equation and define $C_{2} = e^{C_{1}}\text{:}$

$$\left| \frac{y - 2}{y + 2} \right| = C_{2}e^{4x^{2} + 12x}.$$

Next we can remove the absolute value and let $C_{2}$ be either positive or negative. Then multiply both sides by $y + 2.$

$$\begin{array}{l} \\ \\ {y - 2 = C_{2}\left( {y + 2} \right)e^{4x^{2} + 12x}} \\ {y - 2 = C_{2}ye^{{}^{4x^{2} + 12x}} + 2C_{2}e^{{}^{4x^{2} + 12x}}.} \end{array}$$

Now collect all terms involving *y* on one side of the equation, and solve for $y\text{:}$

$$\begin{array}{rll} {y - C_{2}ye^{4x^{2} + 12x}} & = & {2 + 2C_{2}e^{4x^{2} + 12x}} \\ {y(1 - C_{2}e^{4x^{2} + 12x})} & = & {2 + 2C_{2}e^{4x^{2} + 12x}} \\ y & = & {\frac{2 + 2C_{2}e^{4x^{2} + 12x}}{1 - C_{2}e^{4x^{2} + 12x}}.} \end{array}$$

5. To determine the value of $C_{2},$ substitute $x = 0$ and $y = -1$ into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation $\frac{y - 2}{y + 2} = C_{2}e^{4x^{2} + 12}.$ This is much easier to solve for $C_{2}\text{:}$

$$\begin{array}{rll} \frac{y - 2}{y + 2} & = & {C_{2}e^{4x^{2} + 12x}} \\ \frac{-1 - 2}{-1 + 2} & = & {C_{2}e^{4{(0)}^{2} + 12{(0)}}} \\ C_{2} & = & {-3.} \end{array}$$

Therefore the solution to the initial-value problem is

$$y = \frac{2 - 6e^{4x^{2} + 12x}}{1 + 3e^{4x^{2} + 12x}}.$$

A graph of this solution appears in Figure 4.15.

Find the solution to the initial-value problem

$$6y\prime = \left( {2x + 1} \right)\left( {y^{2} - 2y - 8} \right),\quad y(0) = -3$$

using the method of separation of variables.

Applications of Separation of Variables 分离变量的应用

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations 溶液浓度

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.

Determining Salt Concentration over Time 确定盐浓度随时间的变化

A tank containing $100\ \text{L}$ of a brine solution initially has $4\ \text{kg}$ of salt dissolved in the solution. At time $t = 0,$ another brine solution flows into the tank at a rate of $2\ \text{L/min}\text{.}$ This brine solution contains a concentration of $0.5\ \text{kg/L}$ of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of $2\ \text{L/min},$ so that the level of liquid in the tank remains constant (Figure 4.16). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.

Solution 解答

First we define a function $u(t)$ that represents the amount of salt in kilograms in the tank as a function of time. Then $\frac{du}{dt}$ represents the rate at which the amount of salt in the tank changes as a function of time. Also, $u(0)$ represents the amount of salt in the tank at time $t = 0,$ which is $4$ kilograms.

The general setup for the differential equation we will solve is of the form

$$\frac{du}{dt} = \text{INFLOW RATE} - \text{OUTFLOW RATE}.$$ (4.5)

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of $2$ L/min, and each liter of solution contains $0.5$ kilogram of salt, every minute $2(0.5) = 1\ \text{kilogram}$ of salt enters the tank. Therefore INFLOW RATE = $1.$

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time $t$ is equal to $u(t).$ Thus, the concentration of salt is $\frac{u(t)}{100}$ kg/L, and the solution leaves the tank at a rate of $2$ L/min. Therefore salt leaves the tank at a rate of $\frac{u(t)}{100} \cdot 2 = \frac{u(t)}{50}$ kg/min, and OUTFLOW RATE is equal to $\frac{u(t)}{50}.$ Therefore the differential equation becomes $\frac{du}{dt} = 1 - \frac{u}{50},$ and the initial condition is $u(0) = 4.$ The initial-value problem to be solved is

$$\frac{du}{dt} = 1 - \frac{u}{50},\quad u(0) = 4.$$

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting $1 - \frac{u}{50} = 0$ gives $u = 50$ as a constant solution. Since the initial amount of salt in the tank is $4$ kilograms, this solution does not apply.

Step 2. Rewrite the equation as

$$\frac{du}{dt} = \frac{50 - u}{50}.$$

Then multiply both sides by $dt$ and divide both sides by $50 - u\text{:}$

$$\frac{du}{50 - u} = \frac{dt}{50}.$$

Step 3. Integrate both sides:

$$\begin{array}{rll} \left. \int\frac{du}{50 - u} \right. & = & \left. \int\frac{dt}{50} \right. \\ {- \text{ln}\left| {50 - u} \right|} & = & {\frac{t}{50} + C.} \end{array}$$

Step 4. Solve for $u(t)\text{:}$

$$\begin{array}{rll} {\text{ln}\left| {50 - u} \right|} & = & {- \frac{t}{50} - C} \\ e^{\text{ln}{|{50 - u}|}} & = & e^{\text{−}(t\text{/}50) - C} \\ \left| {50 - u} \right| & = & {C_{1}e^{\text{−}t\text{/}50}.} \end{array}$$

Eliminate the absolute value by allowing the constant to be either positive or negative:

$$50 - u = C_{1}e^{\text{−}t\text{/}50}.$$

Finally, solve for $u(t)\text{:}$

$$u(t) = 50 - C_{1}e^{\text{−}t\text{/}50}.$$

Step 5. Solve for $C_{1}\text{:}$

$$\begin{array}{rll} {u(0)} & = & {50 - C_{1}e^{-0\text{/}50}} \\ 4 & = & {50 - C_{1}} \\ C_{1} & = & {46.} \end{array}$$

The solution to the initial value problem is $u(t) = 50 - 46e^{\text{−}t\text{/}50}.$ To find the limiting amount of salt in the tank, take the limit as $t$ approaches infinity:

$$\begin{array}{cl} {\underset{t\rightarrow\infty}{\text{lim}}u(t)} & {= 50 - 46e^{\text{−}t\text{/}50}} \\ & {= 50 - 46(0)} \\ & {= 50.} \end{array}$$

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is $50$ kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

A tank contains $3$ kilograms of salt dissolved in $75$ liters of water. A salt solution of $0.4\ \text{kg salt/L}$ is pumped into the tank at a rate of $6\ \text{L/min}$ and is drained at the same rate. Solve for the salt concentration at time $t.$ Assume the tank is well mixed at all times.

Newton’s law of cooling 牛顿冷却定律

Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let $T(t)$ represent the temperature of an object as a function of time, then $\frac{dT}{dt}$ represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by $T_{s}.$ Then Newton’s law of cooling can be written in the form

$$\frac{dT}{dt} = k\left( {T(t) - T_{s}} \right)$$

or simply

$$\frac{dT}{dt} = k\left( {T - T_{s}} \right).$$ (4.6)

The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature $T_{0}.$ Therefore the initial-value problem that needs to be solved takes the form

$$\frac{dT}{dt} = k\left( {T - T_{s}} \right),\quad T(0) = T_{0},$$ (4.7)

where $k$ is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example 4.13.

Waiting for a Pizza to Cool 等待披萨冷却

A pizza is removed from the oven after baking thoroughly, and the temperature of the pizza when it comes out of the oven is $200\text{°}\text{F}\text{.}$ The temperature of the kitchen is $75\text{°}\text{F},$ and after $1$ minute the temperature of the pizza is $190\text{°}\text{F}\text{.}$ We would like to wait until the temperature of the pizza reaches $150\text{°}\text{F}$ before cutting and serving it (Figure 4.17). How much longer will we have to wait?

Solution 解答

The ambient temperature (surrounding temperature) is $75\text{°}\text{F},$ so $T_{s} = 75.$ The temperature of the pizza when it comes out of the oven is $200\text{°}\text{F},$ which is the initial temperature (i.e., initial value), so $T_{0} = 200.$ Therefore Equation 4.4 becomes

$$\frac{dT}{dt} = k\left( {T - 75} \right),\quad T(0) = 200.$$

To solve the differential equation, we use the five-step technique for solving separable equations.

1. Setting the right-hand side equal to zero gives $T = 75$ as a constant solution. Since the pizza starts at $200\text{°}\text{F},$ this is not the solution we are seeking.

2. Rewrite the differential equation by multiplying both sides by $dt$ and dividing both sides by $T - 75\text{:}$

$$\frac{dT}{T - 75} = kdt.$$

3. Integrate both sides:

$$\begin{array}{rll} \left. \int\frac{dT}{T - 75} \right. & = & {\int{kdt}} \\ {\text{ln}\left| {T - 75} \right|} & = & {kt + C.} \end{array}$$

4. Solve for $T$ by first exponentiating both sides:

$$\begin{array}{rll} e^{\text{ln}{|{T - 75}|}} & = & e^{kt + C} \\ \left| {T - 75} \right| & = & {C_{1}e^{kt}} \\ {T - 75} & = & {C_{1}e^{kt}} \\ {T(t)} & = & {75 + C_{1}e^{kt}.} \end{array}$$

5. Solve for $C_{1}$ by using the initial condition $T(0) = 200\text{:}$

$$\begin{array}{rll} {T(t)} & = & {75 + C_{1}e^{kt}} \\ {T(0)} & = & {75 + C_{1}e^{k{(0)}}} \\ 200 & = & {75 + C_{1}} \\ C_{1} & = & {125.} \end{array}$$

Therefore the solution to the initial-value problem is

$$T(t) = 75 + 125e^{kt}.$$

To determine the value of $k,$ we need to use the fact that after $1$ minute the temperature of the pizza is $190\text{°}\text{F}\text{.}$ Therefore $T(1) = 190.$ Substituting this information into the solution to the initial-value problem, we have

$$\begin{matrix} {T(t)} & = & {75 + 125e^{kt}} \\ {T(1)} & = & {190 = 75 + 125e^{k}} \\ 115 & = & {125e^{k}} \\ \frac{115}{125} & = & {\frac{23}{25} = e^{k}} \\ {\text{ln}\mspace{2mu} e^{k}} & = & {\text{ln}\mspace{2mu}\left( \frac{23}{25} \right)} \\ k & = & {\mspace{2mu}\text{ln}\mspace{2mu}\left( \frac{23}{25} \right) \approx - .08338} \end{matrix}$$

So now we have $T(t) = 75 + 125e^{- .08338t}.$ When is the temperature $150\text{°}\text{F?}$ Solving for $t,$ we find

$$\begin{matrix} {T(t)} & = & {75 + 125e^{- .08338t}} \\ 150 & = & {75 + 125e^{- .08338t}} \\ 75 & = & {125e^{- .08338t}} \\ \frac{75}{125} & = & {\frac{3}{5} = e^{- .08338t}} \\ {- .08338t} & = & {\text{ln}\ \frac{3}{5}} \\ t & = & {\frac{\text{ln}\left( \frac{3}{5} \right)}{- .08338} \approx 6.12.} \end{matrix}$$

Therefore we need to wait an additional $6.12$ minutes (after the temperature of the pizza reached $200\text{°}\text{F}).$ That should be just enough time to finish this calculation.

A cake is removed from the oven after baking thoroughly, and the temperature of the cake when it comes out of the oven is $450\text{°}\text{F}\text{.}$ The temperature of the kitchen is $70\text{°}\text{F},$ and after $10$ minutes the temperature of the cake is $330\text{°}\text{F}\text{.}$

1. Write the appropriate initial-value problem to describe this situation.

2. Solve the initial-value problem for $T(t).$

3. How long will it take until the temperature of the cake is within $5\text{°}\text{F}$ of room temperature?

Section 4.3 Exercises 4.3 节习题

Solve the following initial-value problems with the initial condition $y_{0} = 0$ and graph the solution.

119.

$\frac{dy}{dt} = y + 1$

120\.

$\frac{dy}{dt} = y - 1$

121.

$\frac{dy}{dt} = {–y} + 1$

122\.

$\frac{dy}{dt} = \text{−}y - 1$

Find the general solution to the differential equation.

123.

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

124\.

$y\prime = \text{tan}(y)x$

125.

$y\prime = 2xy^{2}$

126\.

$\frac{dy}{dt} = y\mspace{2mu}\text{cos}\left( {3t + 2} \right)$

127.

$2x\frac{dy}{dx} = y^{2}$

128\.

$y\prime = e^{y}x^{2}$

129.

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

130\.

$\frac{dx}{dt} = 3t^{2}\left( {x^{2} + 4} \right)$

131.

$t\frac{dy}{dt} = \sqrt{1 - y^{2}}$

132\.

$y\prime = e^{x}e^{y}$

Find the solution to the initial-value problem.

133.

$y\prime = e^{y - x},y(0) = 0$

134\.

$y\prime = y^{2}(x + 1),y(0) = 2$

135.

$\frac{dy}{dx} = y^{3}xe^{x^{2}},y(0) = 1$

136\.

$\frac{dy}{dt} = y^{2}e^{x}\text{sin}(3x),y(0) = 1$

137.

$y\prime = \frac{x}{\text{sech}^{2}y},y(0) = 0$

138\.

$y\prime = 2xy(1 + 2y),y(0) = -1$

139.

$\frac{dx}{dt} = \text{ln}(t)\sqrt{1 - x^{2}},x(1) = 0$

140\.

$y\prime = 3x^{2}(y^{2} + 4),y(0) = 0$

141.

$y\prime = e^{y}5^{x},y(0) = \text{ln}(\text{ln}(5))$

142\.

$y\prime = -2x\mspace{2mu}\text{tan}(y),y(0) = \frac{\pi}{6}$

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?

143.

\[T\] $y\prime = 1 - 2y$

144\.

\[T\] $y\prime = y^{2}x^{3}$

145.

\[T\] $y\prime = y^{3}e^{x}$

146\.

\[T\] $y\prime = e^{y}$

147.

\[T\] $y\prime = y\mspace{2mu}\text{ln}(x)$

148.

Most drugs in the bloodstream decay according to the equation $y\prime = cy,$ where $y$ is the concentration of the drug in the bloodstream. If the half-life of a drug is $2$ hours, what fraction of the initial dose remains after $6$ hours?

149.

A drug is administered intravenously to a patient at a rate $r$ mg/h and is cleared from the body at a rate proportional to the amount of drug still present in the body, $d$. Set up and solve the differential equation, assuming there is no drug initially present in the body.

150\.

\[T\] How often should a drug be taken if its dose is $3$ mg, it is cleared at a rate $c = 0.1$ mg/h, and $1$ mg is required to be in the bloodstream at all times?

151.

A tank contains $1$ kilogram of salt dissolved in $100$ liters of water. A salt solution of $0.1$ kg salt/L is pumped into the tank at a rate of $2$ L/min and is drained at the same rate. Solve for the salt concentration at time $t.$ Assume the tank is well mixed.

152\.

A tank containing $10$ kilograms of salt dissolved in $1000$ liters of water has two salt solutions pumped in. The first solution of $0.2$ kg salt/L is pumped in at a rate of $20$ L/min and the second solution of $0.05$ kg salt/L is pumped in at a rate of $5$ L/min. The tank drains at $25$ L/min. Assume the tank is well mixed. Solve for the salt concentration at time $t.$

153.

\[T\] For the preceding problem, find how much salt is in the tank $1$ hour after the process begins.

154\.

Torricelli’s law states that for a water tank with a hole in the bottom that has a cross-sectional area of $A_{H}$ with a height of water $h$ above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to $\frac{dV}{dt} = \text{−}A_{H}\sqrt{2gh},$ where $g$ is the acceleration due to gravity. Note that $\frac{dV}{dt} = A_{T}\frac{dh}{dt},$ where $A_{T}$ is the cross-sectional area of the tank. Solve the resulting initial-value problem for the height of water, assuming a tank of radius $24\sqrt{2}$ with a circular hole of radius $2$ ft. The initial height of water is $100$ ft.

155.

For the preceding problem, determine how long it takes the tank to drain.

For the following problems, use Newton’s law of cooling.

156\.

The liquid base of an ice cream has an initial temperature of $200\text{°}\text{F}$ before it is placed in a freezer with a constant temperature of $0\text{°}\text{F}\text{.}$ After $1$ hour, the temperature of the ice-cream base has decreased to $140\text{°}\text{F}\text{.}$ Formulate and solve the initial-value problem to determine the temperature of the ice cream.

157.

\[T\] The liquid base of an ice cream has an initial temperature of $210\text{°}\text{F}$ before it is placed in a freezer with a constant temperature of $20\text{°}\text{F}\text{.}$ After $2$ hours, the temperature of the ice-cream base has decreased to $170\text{°}\text{F}\text{.}$ At what time will the ice cream be ready to eat? (Assume $30\text{°}\text{F}$ is the optimal eating temperature.)

158\.

\[T\] You are organizing an ice cream social. The outside temperature is $80\text{°}\text{F}$ and the ice cream is at $10\text{°}\text{F}\text{.}$ After $10$ minutes, the ice cream temperature has risen by $10\text{°}\text{F}\text{.}$ How much longer can you wait before the ice cream melts at $40\text{°}\text{F?}$

For Exercises 159—162, assume a cooling constant of $k = - 0.125$ and assume time $t$ is in minutes.

159.

You have a cup of coffee at temperature $70\text{°}\text{C}$ and the ambient temperature in the room is $20\text{°}\text{C}\text{.}$ Assuming a cooling rate $k\ \text{of}\ 0.125,$ write and solve the differential equation to describe the temperature of the coffee with respect to time.

160\.

\[T\] You have a cup of coffee at temperature $70\text{°}\text{C}$ that you put outside, where the ambient temperature is $0\text{°}\text{C}\text{.}$ After $5$ minutes, how much colder is the coffee?

161.

You have a cup of coffee at temperature $70\text{°}\text{C}$ and you immediately pour in $1$ part milk to $5$ parts coffee. The milk is initially at temperature $1\text{°}\text{C}\text{.}$ Write and solve the differential equation that governs the temperature of this coffee.

162\.

You have a cup of coffee at temperature $70\text{°}\text{C},$ which you let cool $10$ minutes before you pour in the same amount of milk at $1\text{°}\text{C}$ as in the preceding problem. How does the temperature compare to the previous cup after $10$ minutes?

163.

Solve the generic problem $y\prime = ay + b$ with initial condition $y(0) = c.$

164\.

Prove the basic continual compounded interest equation. Assuming an initial deposit of $P_{0}$ and an interest rate of $r,$ set up and solve an equation for continually compounded interest.

165.

Assume an initial nutrient amount of $I$ kilograms in a tank with $L$ liters. Assume a concentration of $c$ kg/L being pumped in at a rate of $r$ L/min. The tank is well mixed and is drained at a rate of $r$ L/min. Find the equation describing the amount of nutrient in the tank.

166\.

Leaves accumulate on the forest floor at a rate of $2$ g/cm²/yr and also decompose at a rate of $90\text{\%}$ per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time $0$ there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?

167.

Leaves accumulate on the forest floor at a rate of $4$ g/cm²/yr. These leaves decompose at a rate of $10\text{\%}$ per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor. Does this amount approach a steady value? What is that value?

4.4 The Logistic Equation 4.4 逻辑斯谛方程

• 4.4.1 Describe the concept of environmental carrying capacity in the logistic model of population growth.
• 4.4.2 Draw a direction field for a logistic equation and interpret the solution curves.
• 4.4.3 Solve a logistic equation and interpret the results.

• 4.4.1 在种群增长的逻辑斯谛模型中描述环境容纳量的概念。
• 4.4.2 为逻辑斯谛方程绘制方向场并解释解曲线。
• 4.4.3 求解逻辑斯谛方程并解释结果。

Differential equations can be used to represent the size of a population as it varies over time. We saw this in an earlier chapter in the section on exponential growth and decay, which is the simplest model. A more realistic model includes other factors that affect the growth of the population. In this section, we study the logistic differential equation and see how it applies to the study of population dynamics in the context of biology.

Population Growth and Carrying Capacity 种群增长与容纳量

To model population growth using a differential equation, we first need to introduce some variables and relevant terms. The variable $t.$ will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem must specify the units used in that particular problem. The variable $P$ will represent population. Since the population varies over time, it is understood to be a function of time. Therefore we use the notation $P(t)$ for the population as a function of time. If $P(t)$ is a differentiable function, then the first derivative $\frac{dP}{dt}$ represents the instantaneous rate of change of the population as a function of time.

In Exponential Growth and Decay, we studied the exponential growth and decay of populations and radioactive substances. An example of an exponential growth function is $P(t) = P_{0}e^{rt}.$ In this function, $P(t)$ represents the population at time $t,P_{0}$ represents the initial population (population at time $t = 0),$ and the constant $r > 0$ is called the growth rate. Figure 4.18 shows a graph of $P(t) = 100e^{0.03t}.$ Here $P_{0} = 100$ and $r = 0.03.$

We can verify that the function $P(t) = P_{0}e^{rt}$ satisfies the initial-value problem

$$\frac{dP}{dt} = rP,\quad P(0) = P_{0}.$$

This differential equation has an interesting interpretation. The left-hand side represents the rate at which the population increases (or decreases). The right-hand side is equal to a positive constant multiplied by the current population. Therefore the differential equation states that the rate at which the population increases is proportional to the population at that point in time. Furthermore, it states that the constant of proportionality never changes.

One problem with this function is its prediction that as time goes on, the population grows without bound. This is unrealistic in a real-world setting. Various factors limit the rate of growth of a particular population, including birth rate, death rate, food supply, predators, and so on. The growth constant $r$ usually takes into consideration the birth and death rates but none of the other factors, and it can be interpreted as a net (birth minus death) percent growth rate per unit time. A natural question to ask is whether the population growth rate stays constant, or whether it changes over time. Biologists have found that in many biological systems, the population grows until a certain steady-state population is reached. This possibility is not taken into account with exponential growth. However, the concept of carrying capacity allows for the possibility that in a given area, only a certain number of a given organism or animal can thrive without running into resource issues.

The carrying capacity of an organism in a given environment is defined to be the maximum population of that organism that the environment can sustain indefinitely.

We use the variable $K$ to denote the carrying capacity. The growth rate is represented by the variable $r.$ Using these variables, we can define the logistic differential equation.

Let $K$ represent the carrying capacity for a particular organism in a given environment, and let $r$ be a real number that represents the growth rate. The function $P(t)$ represents the population of this organism as a function of time $t,$ and the constant $P_{0}$ represents the initial population (population of the organism at time $t = 0).$ Then the logistic differential equation is

$$\frac{dP}{dt} = rP\left( {1 - \frac{P}{K}} \right)$$ (4.8)

See this website for more information on the logistic equation.

The logistic equation was first published by Pierre Verhulst in $1845.$ This differential equation can be coupled with the initial condition $P(0) = P_{0}$ to form an initial-value problem for $P(t).$

Suppose that the initial population is small relative to the carrying capacity. Then $\frac{P}{K}$ is small, possibly close to zero. Thus, the quantity in parentheses on the right-hand side of Equation 4.8 is close to $1,$ and the right-hand side of this equation is close to $rP.$ If $r > 0,$ then the population grows rapidly, resembling exponential growth.

However, as the population grows, the ratio $\frac{P}{K}$ also grows, because $K$ is constant. If the population remains below the carrying capacity, then $\frac{P}{K}$ is less than $1,$ so $1 - \frac{P}{K} > 0.$ Therefore the right-hand side of Equation 4.8 is still positive, but the quantity in parentheses gets smaller, and the growth rate decreases as a result. If $P = K$ then the right-hand side is equal to zero, and the population does not change.

Now suppose that the population starts at a value higher than the carrying capacity. Then $\frac{P}{K} > 1,$ and $1 - \frac{P}{K} < 0.$ Then the right-hand side of Equation 4.8 is negative, and the population decreases. As long as $P > K,$ the population decreases. It never actually reaches $K$ because $\frac{dP}{dt}$ will get smaller and smaller, but the population approaches the carrying capacity as $t$ approaches infinity. This analysis can be represented visually by way of a phase line. A phase line describes the general behavior of a solution to an autonomous differential equation, depending on the initial condition. For the case of a carrying capacity in the logistic equation, the phase line is as shown in Figure 4.19.

This phase line shows that when $P$ is less than zero or greater than $K,$ the population decreases over time. When $P$ is between $0$ and $K,$ the population increases over time.

Chapter Opener: Examining the Carrying Capacity of a Deer Population 章首问题：考察白尾鹿种群的容纳量

Let's consider the population of white-tailed deer (*Odocoileus virginianus*) in the state of Kentucky. The Kentucky Department of Fish and Wildlife Resources (KDFWR) sets guidelines for hunting and fishing in the state. Before the hunting season of $2004,$ it estimated a population of $900,000$ deer. Johnson notes: "A deer population that has plenty to eat and is not hunted by humans or other predators will double every three years." (George Johnson, "The Problem of Exploding Deer Populations Has No Attractive Solutions," January $12,2001,$ accessed April 9, 2015.) This observation corresponds to a rate of increase $r = \frac{\text{ln}(2)}{3} = 0.2311,$ so the approximate growth rate is $23.11\text{\%}$ per year*.* (This assumes that the population grows exponentially, which is reasonable––at least in the short term––with plentiful food supply and no predators.) The KDFWR also reports deer population densities for $32$ counties in Kentucky, the average of which is approximately $27$ deer per square mile. Suppose this is the deer density for the whole state $(39,732$ square miles). The carrying capacity $K$ is $39,732$ square miles times $27$ deer per square mile, or $1,072,764$ deer*.*

1. For this application, we have $P_{0} = 900,000,K = 1,072,764,$ and $r = 0.2311.$ Substitute these values into Equation 4.8 and form the initial-value problem.

2. Solve the initial-value problem from part a.

3. According to this model, what will be the population in $3$ years? Recall that the doubling time predicted by Johnson for the deer population was $3$ years. How do these values compare?

4. Suppose the population managed to reach $1,200,000$ deer. What does the logistic equation predict will happen to the population in this scenario?

Solution 解答

1. The initial value problem is

$\frac{dP}{dt} = 0.2311P\left( {1 - \frac{P}{1,072,764}} \right),\quad P(0) = 900,000.$

2. The logistic equation is an autonomous differential equation, so we can use the method of separation of variables.

Step 1: Setting the right-hand side equal to zero gives $P = 0$ and $P = 1,072,764.$ This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change.

Step 2: Rewrite the differential equation and multiply both sides by:

$$\begin{array}{rll} \frac{dP}{dt} & = & {0.2311P\left( \frac{1,072,764 - P}{1,072,764} \right)} \\ {dP} & = & {0.2311P\left( \frac{1,072,764 - P}{1,072,764} \right)\mspace{2mu} dt.} \end{array}$$

Divide both sides by $P\left( {1,072,764 - P} \right)\text{:}$

$$\frac{dP}{P\left( {1,072,764 - P} \right)} = \frac{0.2311}{1,072,764}dt.$$

Step 3: Integrate both sides of the equation using partial fraction decomposition:

$$\begin{array}{rll} \left. \int\frac{dP}{P\left( {1,072,764 - P} \right)} \right. & = & \left. \int{\frac{0.2311}{1,072,764}dt} \right. \\ {\frac{1}{1,072,764}\left. \int{\left( {\frac{1}{P} + \frac{1}{1,072,764 - P}} \right)\mspace{2mu} dP} \right.} & = & {\frac{0.2311t}{1,072,764} + C} \\ {\frac{1}{1,072,764}\mspace{2mu}\left( {\text{ln}|P| - \text{ln}\left| {1,072,764 - P} \right|} \right)} & = & {\frac{0.2311t}{1,072,764} + C.} \end{array}$$

Step 4: Multiply both sides by $1,072,764$ and use the quotient rule for logarithms:

$$\text{ln}\left| \frac{P}{1,072,764 - P} \right| = 0.2311t + C_{1}.$$

Here $C_{1} = 1,072,764C.$ Next exponentiate both sides and eliminate the absolute value:

$$\begin{array}{rll} e^{\text{ln}{|\frac{P}{1,072,764 - P}|}} & = & e^{0.2311t + C_{1}} \\ \left| \frac{P}{1,072,764 - P} \right| & = & {C_{2}e^{0.2311t}} \\ \frac{P}{1,072,764 - P} & = & {C_{2}e^{0.2311t}.} \end{array}$$

Here $C_{2} = e^{C_{1}}$ but after eliminating the absolute value, it can be negative as well. Now solve for:

$$\begin{array}{rll} P & = & {C_{2}e^{0.2311t}\left( {1,072,764 - P} \right).} \\ P & = & {1,072,764C_{2}e^{0.2311t} - C_{2}Pe^{0.2311t}} \\ {P + C_{2}Pe^{0.2311t}} & = & {1,072,764C_{2}e^{0.2311t}} \\ {P\left( {1 + C_{2}e^{0.2311t}} \right)} & = & {1,072,764C_{2}e^{0.2311t}} \\ {P(t)} & = & {\frac{1,072,764C_{2}e^{0.2311t}}{1 + C_{2}e^{0.2311t}}.} \end{array}$$

Step 5: To determine the value of $C_{2},$ it is actually easier to go back a couple of steps to where $C_{2}$ was defined. In particular, use the equation

$$\frac{P}{1,072,764 - P} = C_{2}e^{0.2311t}.$$

The initial condition is $P(0) = 900,000.$ Replace $P$ with $900,000$ and $t$ with zero:

$$\begin{array}{rll} \frac{P}{1,072,764 - P} & = & {C_{2}e^{0.2311t}} \\ \frac{900,000}{1,072,764 - 900,000} & = & {C_{2}e^{0.2311{(0)}}} \\ \frac{900,000}{172,764} & = & C_{2} \\ C_{2} & = & {\frac{25,000}{4,799} \approx 5.209.} \end{array}$$

Therefore

$$\begin{array}{cl} {P(t)} & {= \frac{1,072,764\left( \frac{25000}{4799} \right)\mspace{2mu} e^{0.2311t}}{1 + \left( \frac{25000}{4799} \right)\mspace{2mu} e^{0.2311t}}} \\ & {= \frac{1,072,764(25000)e^{0.2311t}}{4799 + 25000e^{0.2311t}}.} \end{array}$$

Dividing the numerator and denominator by $25,000$ gives

$$P(t) = \frac{1,072,764e^{0.2311t}}{0.19196 + e^{0.2311t}}.$$

Figure 4.21 is a graph of this equation.

3. Using this model we can predict the population in $3$ years.

$$P(3) = \frac{1,072,764e^{0.2311{(3)}}}{0.19196 + e^{0.2311(3)}} \approx 978,830\ \text{deer}$$

This is far short of twice the initial population of $900,000.$ Remember that the doubling time is based on the assumption that the growth rate never changes, but the logistic model takes this possibility into account.

4. If the population reached $1,200,000$ deer, then the new initial-value problem would be

$$\frac{dP}{dt} = 0.2311P\left( {1 - \frac{P}{1,072,764}} \right),\quad P(0) = 1,200,000.$$

The general solution to the differential equation would remain the same.

$$P(t) = \frac{1,072,764C_{2}e^{0.2311t}}{1 + C_{2}e^{0.2311t}}$$

To determine the value of the constant, return to the equation

$$\frac{P}{1,072,764 - P} = C_{2}e^{0.2311t}.$$

Substituting the values $t = 0$ and $P = 1,200,000,$ you get

$$\begin{array}{rll} {C_{2}e^{0.2311{(0)}}} & = & \frac{1,200,000}{1,072,764 - 1,200,000} \\ C_{2} & = & {- \frac{100,000}{10,603} \approx - 9.431.} \end{array}$$

Therefore

$$\begin{array}{cl} {P(t)} & {= \frac{1,072,764C_{2}e^{0.2311t}}{1 + C_{2}e^{0.2311t}}} \\ & {= \frac{1,072,764\left( {- \frac{100,000}{10,603}} \right)\mspace{2mu} e^{0.2311t}}{1 + \left( {- \frac{100,000}{10,603}} \right)\mspace{2mu} e^{0.2311t}}} \\ & {= - \frac{107,276,400,000e^{0.2311t}}{100,000e^{0.2311t} - 10,603}} \\ & {\approx \frac{10,117,551e^{0.2311t}}{9.43129e^{0.2311t} - 1}.} \end{array}$$

This equation is graphed in Figure 4.22.

Solving the Logistic Differential Equation 求解逻辑斯谛微分方程

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in Example 4.14.

Step 1: Setting the right-hand side equal to zero leads to $P = 0$ and $P = K$ as constant solutions. The first solution indicates that when there are no organisms present, the population will never grow. The second solution indicates that when the population starts at the carrying capacity, it will never change.

Step 2: Rewrite the differential equation in the form

$$\frac{dP}{dt} = \frac{rP\left( {K - P} \right)}{K}.$$

Then multiply both sides by $dt$ and divide both sides by $P\left( {K - P} \right).$ This leads to

$$\frac{dP}{P\left( {K - P} \right)} = \frac{r}{K}dt.$$

Multiply both sides of the equation by $K$ and integrate:

$$\left. \int{\frac{K}{P\left( {K - P} \right)}dP} \right. = \left. \int r \right.dt.$$

The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that

$$\frac{K}{P\left( {K - P} \right)} = \frac{1}{P} + \frac{1}{K - P}.$$

Then the equation becomes

$$\begin{array}{rll} \left. \int{\frac{1}{P} + \frac{1}{K - P}dP} \right. & = & {\int{rdt}} \\ {\text{ln}|P| - \text{ln}\left| {K - P} \right|} & = & {rt + C} \\ {\text{ln}\left| \frac{P}{K - P} \right|} & = & {rt + C.} \end{array}$$

Now exponentiate both sides of the equation to eliminate the natural logarithm:

$$\begin{array}{rll} e^{\text{ln}{|\frac{P}{K - P}|}} & = & e^{rt + C} \\ \left| \frac{P}{K - P} \right| & = & {e^{C}e^{rt}.} \end{array}$$

We define $C_{1} = e^{c}$ and, noting that $K,P > 0$ and $P < K$, so that $\frac{P}{K–P} > 0$ we can drop the absolute value sign, so that the equation becomes

$$\frac{P}{K - P} = C_{1}e^{rt}.$$ (4.9)

To solve this equation for $P(t),$ first multiply both sides by $K - P$ and collect the terms containing $P$ on the left-hand side of the equation:

$$\begin{array}{rll} P & = & {C_{1}e^{rt}\left( {K - P} \right)} \\ P & = & {C_{1}Ke^{rt} - C_{1}Pe^{rt}} \\ {P + C_{1}Pe^{rt}} & = & {C_{1}Ke^{rt}.} \end{array}$$

Next, factor $P$ from the left-hand side and divide both sides by the other factor:

$$\begin{array}{rll} {P\left( {1 + C_{1}e^{rt}} \right)} & = & {C_{1}Ke^{rt}} \\ {P(t)} & = & {\frac{C_{1}Ke^{rt}}{1 + C_{1}e^{rt}}.} \end{array}$$ (4.10)

The last step is to determine the value of $C_{1}.$ The easiest way to do this is to substitute $t = 0$ and $P_{0}$ in place of $P$ in Equation 4.9 and solve for $C_{1}\text{:}$

$$\begin{array}{rll} \frac{P}{K - P} & = & {C_{1}e^{rt}} \\ \frac{P_{0}}{K - P_{0}} & = & {C_{1}e^{r{(0)}}} \\ C_{1} & = & {\frac{P_{0}}{K - P_{0}}.} \end{array}$$

Finally, substitute the expression for $C_{1}$ into Equation 4.10:

$$P(t) = \frac{C_{1}Ke^{rt}}{1 + C_{1}e^{rt}} = \frac{\frac{P_{0}}{K - P_{0}}Ke^{rt}}{1 + \frac{P_{0}}{K - P_{0}}e^{rt}}$$

Now multiply the numerator and denominator of the right-hand side by $\left( {K - P_{0}} \right)$ and simplify:

$$\begin{array}{cl} {P(t)} & {= \frac{\frac{P_{0}}{K - P_{0}}Ke^{rt}}{1 + \frac{P_{0}}{K - P_{0}}e^{rt}}} \\ & {= \frac{\frac{P_{0}}{K - P_{0}}Ke^{rt}}{1 + \frac{P_{0}}{K - P_{0}}e^{rt}} \cdot \frac{K - P_{0}}{K - P_{0}}} \\ & {= \frac{P_{0}Ke^{rt}}{\left( {K - P_{0}} \right) + P_{0}e^{rt}}.} \end{array}$$

We state this result as a theorem.

Solution of the Logistic Differential Equation 逻辑斯谛微分方程的解

Consider the logistic differential equation subject to an initial population of $P_{0}$ with carrying capacity $K$ and growth rate $r.$ The solution to the corresponding initial-value problem is given by

$$P(t) = \frac{P_{0}Ke^{rt}}{\left( {K - P_{0}} \right) + P_{0}e^{rt}}.$$ (4.11)

Now that we have the solution to the initial-value problem, we can choose values for $P_{0},r,$ and $K$ and study the solution curve. For example, in Example 4.14 we used the values $r = 0.2311,K = 1,072,764,$ and an initial population of $900,000$ deer. This leads to the solution

$$\begin{array}{cl} {P(t)} & {= \frac{P_{0}Ke^{rt}}{\left( {K - P_{0}} \right) + P_{0}e^{rt}}} \\ & {= \frac{900,000(1,072,764)e^{0.2311t}}{\left( {1,072,764 - 900,000} \right) + 900,000e^{0.2311t}}} \\ & {= \frac{900,000(1,072,764)e^{0.2311t}}{172,764 + 900,000e^{0.2311t}}.} \end{array}$$

Dividing top and bottom by $900,000$ gives

$$P(t) = \frac{1,072,764e^{0.2311t}}{0.19196 + e^{0.2311t}}.$$

This is the same as the original solution. The graph of this solution is shown again in blue in Figure 4.23, superimposed over the graph of the exponential growth model with initial population $900,000$ and growth rate $0.2311$ (appearing in green). The red dashed line represents the carrying capacity, and is a horizontal asymptote for the solution to the logistic equation.

Working under the assumption that the population grows according to the logistic differential equation, this graph predicts that approximately $20$ years earlier $(1984),$ the growth of the population was very close to exponential. The net growth rate at that time would have been around $23.1\text{\%}$ per year. As time goes on, the two graphs separate. This happens because the population increases, and the logistic differential equation states that the growth rate decreases as the population increases. At the time the population was measured $(2004),$ it was close to carrying capacity, and the population was starting to level off.

The solution to the logistic differential equation has a point of inflection. To find this point, set the second derivative equal to zero:

$$\begin{array}{cll} {P(t)} & = & \frac{P_{0}Ke^{rt}}{\left( {K - P_{0}} \right) + P_{0}e^{rt}} \\ {P^{\prime}(t)} & = & \frac{rP_{0}K\left( {K - P_{0}} \right)e^{rt}}{\left( {\left( {K - P_{0}} \right) + P_{0}e^{rt}} \right)^{2}} \\ {P^{''}(t)} & = & \frac{r^{2}P_{0}K\left( {K - P_{0}} \right)^{2}e^{rt} - r^{2}P_{0}{}^{2}K\left( {K - P_{0}} \right)e^{2rt}}{\left( {\left( {K - P_{0}} \right) + P_{0}e^{rt}} \right)^{3}} \\ & = & {\frac{r^{2}P_{0}K\left( {K - P_{0}} \right)e^{rt}\left( {\left( {K - P_{0}} \right) - P_{0}e^{rt}} \right)}{\left( {\left( {K - P_{0}} \right) + P_{0}e^{rt}} \right)^{3}}.} \end{array}$$

Setting the numerator equal to zero,

$$r^{2}P_{0}K\left( {K - P_{0}} \right)e^{rt}\left( {\left( {K - P_{0}} \right) - P_{0}e^{rt}} \right) = 0.$$

As long as $P_{0} \neq K,$ the entire quantity before and including $e^{rt}$ is nonzero, so we can divide it out:

$$\left( {K - P_{0}} \right) - P_{0}e^{rt} = 0.$$

Solving for $t,$

$$\begin{array}{rll} {P_{0}e^{rt}} & = & {K - P_{0}} \\ e^{rt} & = & \frac{K - P_{0}}{P_{0}} \\ {\text{ln}\mspace{2mu} e^{rt}} & = & {\text{ln}\ \frac{K - P_{0}}{P_{0}}} \\ {rt} & = & {\text{ln}\ \frac{K - P_{0}}{P_{0}}} \\ t & = & {\frac{1}{r}\mspace{2mu}\text{ln}\ \frac{K - P_{0}}{P_{0}}.} \end{array}$$

Notice that if $P_{0} > K,$ then this quantity is undefined, and the graph does not have a point of inflection. In the logistic graph, the point of inflection can be seen as the point where the graph changes from concave up to concave down. This is where the "leveling off" starts to occur, because the net growth rate becomes slower as the population starts to approach the carrying capacity.

A population of rabbits in a meadow is observed to be $200$ rabbits at time $t = 0.$ After a month, the rabbit population is observed to have increased by $4\text{\%}.$ Using an initial population of $200$ and a growth rate of $0.04,$ with a carrying capacity of $750$ rabbits,

1. Write the logistic differential equation and initial condition for this model.

2. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of $200$ rabbits.

3. Solve the initial-value problem for $P(t).$

4. Use the solution to predict the population after $1$ year.

Student Project: Logistic Equation with a Threshold Population 学生项目：带阈值种群的逻辑斯谛方程

An improvement to the logistic model includes a threshold population. The threshold population is defined to be the minimum population that is necessary for the species to survive. We use the variable $T$ to represent the threshold population. A differential equation that incorporates both the threshold population $T$ and carrying capacity $K$ is

$$\frac{dP}{dt} = \text{−}rP\left( {1 - \frac{P}{K}} \right)\left( {1 - \frac{P}{T}} \right)$$ (4.12)

where $r$ represents the growth rate, as before.

1. The threshold population is useful to biologists and can be utilized to determine whether a given species should be placed on the endangered list. A group of Australian researchers say they have determined the threshold population for any species to survive: $5000$ adults. (Catherine Clabby, "A Magic Number," *American Scientist* 98(1): 24, doi:10.1511/2010.82.24. accessed April 9, 2015). Therefore we use $T = 5000$ as the threshold population in this project. Suppose that the environmental carrying capacity in Montana for elk is $25,000.$ Set up Equation 4.12 using the carrying capacity of $25,000$ and threshold population of $5000.$ Assume an annual net growth rate of $18\text{\%}.$

2. Draw the direction field for the differential equation from step $1,$ along with several solutions for different initial populations. What are the constant solutions of the differential equation? What do these solutions correspond to in the original population model (i.e., in a biological context)?

3. What is the limiting population for each initial population you chose in step $2?$ (Hint: use the slope field to see what happens for various initial populations, i.e., look for the horizontal asymptotes of your solutions.)

4. This equation can be solved using the method of separation of variables. However, it is very difficult to get the solution as an explicit function of $t.$ Using an initial population of $18,000$ elk, solve the initial-value problem and express the solution as an implicit function of $t,$ or solve the general initial-value problem, finding a solution in terms of $r,K,T,\text{and}\ P_{0}.$

Section 4.4 Exercises 4.4 节习题

For the following problems, consider the logistic equation in the form $P\prime = CP - P^{2}.$ Draw the directional field and find the stability of the equilibria.

168\.

$C = 3$

169.

$C = 0$

170\.

$C = -3$

171.

Solve the logistic equation for $C = 10$ and an initial condition of $P(0) = 2.$

172\.

Solve the logistic equation for $C = -10$ and an initial condition of $P(0) = 2.$

173.

A population of deer inside a park has a carrying capacity of $200$ and a growth rate of $2\text{\%}.$ If the initial population is $50$ deer, what is the population of deer at any given time?

174\.

A population of frogs in a pond has a growth rate of $5\text{\%}.$ If the initial population is $1000$ frogs and the carrying capacity is $6000,$ what is the population of frogs at any given time?

175.

\[T\] Bacteria grow at a rate of $20\text{\%}$ per hour in a petri dish. If there is initially one bacterium and a carrying capacity of $1$ million cells, how long does it take to reach $500,000$ cells?

176\.

\[T\] Rabbits in a park have an initial population of $10$ and grow at a rate of $4\text{\%}$ per year. If the carrying capacity is $500,$ at what time does the population reach $100$ rabbits?

177.

\[T\] Two monkeys are placed on an island. After $5$ years, there are $8$ monkeys, and the estimated carrying capacity is $25$ monkeys. When does the population of monkeys reach $16$ monkeys?

178\.

\[T\] A butterfly sanctuary is built that can hold $2000$ butterflies, and $400$ butterflies are initially moved in. If after $2$ months there are now $800$ butterflies, when does the population get to $1500$ butterflies?

The following problems consider the logistic equation with an added term for depletion, either through death or emigration.

179.

\[T\] The population of trout in a pond is given by $P\prime = 0.4P\left( {1 - \frac{P}{10000}} \right) - 400,$ where $400$ trout are caught per year. Use your calculator or computer software to draw a directional field and draw a few sample solutions. What do you expect for the behavior?

180\.

In the preceding problem, what are the stabilities of the equilibria $0 < P_{1} < P_{2}?$

181.

\[T\] For the preceding problem, use software to generate a directional field for the value $f = 400.$ What are the stabilities of the equilibria?

182\.

\[T\] For the preceding problems, use software to generate a directional field for the value $f = 600.$ What are the stabilities of the equilibria?

183.

\[T\] For the preceding problems, consider the case where a certain number of fish are added to the pond, or $f = -200.$ What are the nonnegative equilibria and their stabilities?

It is more likely that the amount of fishing is governed by the current number of fish present, so instead of a constant number of fish being caught, the rate is proportional to the current number of fish present, with proportionality constant $k,$ as

$P\prime = 0.4P\left( {1 - \frac{P}{10000}} \right) - kP.$

184\.

\[T\] For the previous fishing problem, draw a directional field assuming $k = 0.1.$ Draw some solutions that exhibit this behavior. What are the equilibria and what are their stabilities?

185.

\[T\] Use software or a calculator to draw directional fields for $k = 0.4.$ What are the nonnegative equilibria and their stabilities?

186\.

\[T\] Use software or a calculator to draw directional fields for $k = 0.6.$ What are the equilibria and their stabilities?

187.

Solve this equation, assuming a value of $k = 0.05$ and an initial condition of $2000$ fish.

188\.

Solve this equation, assuming a value of $k = 0.05$ and an initial condition of $5000$ fish.

The following problems add in a minimal threshold value for the species to survive, $T,$ which changes the differential equation to $P\prime(t) = rP\left( {1 - \frac{P}{K}} \right)\left( {1 - \frac{T}{P}} \right).$

189.

Draw the directional field of the threshold logistic equation, assuming $K = 10,r = 0.1,T = 2.$ When does the population survive? When does it go extinct?

190\.

For the preceding problem, solve the logistic threshold equation, assuming the initial condition $P(0) = P_{0}.$

191.

Bengal tigers in a conservation park have a carrying capacity of $100$ and need a minimum of $10$ to survive. If they grow in population at a rate of $1\text{\%}$ per year, with an initial population of $15$ tigers, solve for the number of tigers present.

192\.

A forest containing ring-tailed lemurs in Madagascar has the potential to support $5000$ individuals, and the lemur population grows at a rate of $5\text{\%}$ per year. A minimum of $500$ individuals is needed for the lemurs to survive. Given an initial population of $600$ lemurs, solve for the population of lemurs.

193.

The population of Florida panthers in the southwestern tip of Florida has an estimated carrying capacity of $250$ and grows at a rate of $0.25\text{\%}$ per year and there must be $25$ for the population to survive. With an initial population of $30$ Florida panthers, how many years will it take to get the Florida panthers off the endangered species list (at least $100)?$

The following questions consider the Gompertz equation, a modification for logistic growth, which is often used for modeling cancer growth, specifically the number of tumor cells.

194\.

The Gompertz equation is given by $P(t)\prime = \alpha\mspace{2mu}\text{ln}\mspace{2mu}\left( \frac{K}{P(t)} \right)P(t).$ Draw the directional fields for this equation assuming all parameters are positive, and given that $K = 1.$

195.

Assume that for a population, $K = 1000$ and $\alpha = 0.05.$ Draw the directional field associated with this differential equation and draw a few solutions. What is the behavior of the population?

196\.

Solve the Gompertz equation for generic $\alpha$ and $K$ and $P(0) = P_{0}.$

197.

\[T\] The Gompertz equation has been used to model tumor growth in the human body. Starting from one tumor cell on day $1$ and assuming $\alpha = 0.1$ and a carrying capacity of $10$ million cells, how long does it take to reach "detection" stage at $5$ million cells?

198\.

\[T\] It is estimated that the world human population reached $3$ billion people in $1959$ and $6$ billion in $1999.$ Assuming a carrying capacity of $16$ billion humans, write and solve the differential equation for logistic growth, and determine what year the population reached $7$ billion.

199.

\[T\] It is estimated that the world human population reached $3$ billion people in $1959$ and $6$ billion in $1999.$ Assuming a carrying capacity of $16$ billion humans, write and solve the differential equation for Gompertz growth, and determine what year the population reached $7$ billion. Was logistic growth or Gompertz growth more accurate, considering world population reached $7$ billion on October $31,2011?$

200\.

Show that the population grows fastest when it reaches half the carrying capacity for the logistic equation $P\prime = rP\left( {1 - \frac{P}{K}} \right).$

201.

When does population increase the fastest in the threshold logistic equation $P\prime(t) = rP\left( {1 - \frac{P}{K}} \right)\left( {1 - \frac{T}{P}} \right)?$

202\.

When does population increase the fastest for the Gompertz equation $P(t)\prime = \alpha\mspace{2mu}\text{ln}\mspace{2mu}\left( \frac{K}{P(t)} \right)P(t)?$

Below is a table of the populations of whooping cranes in the wild from $1940\ \text{to}\ 2000.$ The population rebounded from near extinction after conservation efforts began. The following problems consider applying population models to fit the data. Assume a carrying capacity of $10,000$ cranes. Fit the data assuming years since $1940$ (so your initial population at time $0$ would be $22$ cranes).

203\.

Find the equation and parameter $r$ that best fit the data for the logistic equation.

204\.

Find the equation and parameters $r$ and $T$ that best fit the data for the threshold logistic equation.

205.

Find the equation and parameter $\alpha$ that best fit the data for the Gompertz equation.

206\.

Graph all three solutions and the data on the same graph. Which model appears to be most accurate?

207.

Using the three equations found in the previous problems, estimate the population in $2010$ (year $70$ after conservation). The real population measured at that time was $437.$ Which model is most accurate?

4.5 First-Order Linear Equations 4.5 一阶线性方程

• 4.5.1 Write a first-order linear differential equation in standard form.
• 4.5.2 Find an integrating factor and use it to solve a first-order linear differential equation.
• 4.5.3 Solve applied problems involving first-order linear differential equations.

• 4.5.1 将一阶线性微分方程写成标准形式。
• 4.5.2 求出积分因子并利用它求解一阶线性微分方程。
• 4.5.3 求解涉及一阶线性微分方程的应用问题。

Earlier, we studied an application of a first-order differential equation that involved solving for the velocity of an object. In particular, if a ball is thrown upward with an initial velocity of $v_{0}$ ft/s, then an initial-value problem that describes the velocity of the ball after $t$ seconds is given by

$$\frac{dv}{dt} = -32,\quad v(0) = v_{0}.$$

This model assumes that the only force acting on the ball is gravity. Now we add to the problem by allowing for the possibility of air resistance acting on the ball.

Air resistance always acts in the direction opposite to motion. Therefore if an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 4.24). There is no exact relationship between the velocity of an object and the air resistance acting on it. For very small objects, air resistance is proportional to velocity; that is, the force due to air resistance is numerically equal to some constant $k$ times $v.$ For larger (e.g., baseball-sized) objects, depending on the shape, air resistance can be approximately proportional to the square of the velocity. In fact, air resistance may be proportional to $v^{1.5},$ or $v^{0.9},$ or some other power of $v.$

We will work with the linear approximation for air resistance. If we assume $k > 0,$ then the expression for the force $F_{A}$ due to air resistance is given by $F_{A} = \text{−}kv.$ Therefore the sum of the forces acting on the object is equal to the sum of the gravitational force and the force due to air resistance. This, in turn, is equal to the mass of the object multiplied by its acceleration at time $t$ (Newton’s second law). This gives us the differential equation

$$m\frac{dv}{dt} = \text{−}kv - mg.$$

Finally, we impose an initial condition $v(0) = v_{0},$ where $v_{0}$ is the initial velocity measured in meters per second. This makes $g = 9.8{\ \text{m/s}}^{2}.$ The initial-value problem becomes

$$m\frac{dv}{dt} = \text{−}kv - mg,\quad v(0) = v_{0}.$$ (4.13)

The differential equation in this initial-value problem is an example of a first-order linear differential equation. (Recall that a differential equation is first-order if the highest-order derivative that appears in the equation is $1.)$ In this section, we study first-order linear equations and examine a method for finding a general solution to these types of equations, as well as solving initial-value problems involving them.

A first-order differential equation is linear if it can be written in the form

$$a(x)y^{\prime} + b(x)y = c(x),$$ (4.14)

where $a(x),b(x),$ and $c(x)$ are arbitrary functions of $x.$

Remember that the unknown function $y$ depends on the variable $x;$ that is, $x$ is the independent variable and $y$ is the dependent variable. Some examples of first-order linear differential equations are

$$\begin{array}{rll} {\left( {3x^{2} - 4} \right)y\prime + \left( {x - 3} \right)y} & = & {\text{sin}\mspace{2mu} x} \\ {\left( {\text{sin}\mspace{2mu} x} \right)y\prime - \left( {\text{cos}\mspace{2mu} x} \right)y} & = & {\text{cot}\mspace{2mu} x} \\ {4xy\prime + \left( {3\mspace{2mu}\text{ln}\mspace{2mu} x} \right)y} & = & {x^{3} - 4x.} \end{array}$$

Examples of first-order nonlinear differential equations include

$$\begin{array}{rll} {\left( {y\prime} \right)^{4} - \left( {y\prime} \right)^{3}} & = & {\left( {3x - 2} \right)\left( {y + 4} \right)} \\ {4y\prime + 3y^{3}} & = & {4x - 5} \\ \left( {y\prime} \right)^{2} & = & {\text{sin}\mspace{2mu} y + \text{cos}\mspace{2mu} x.} \end{array}$$

These equations are nonlinear because of terms like $\left( y^{\prime} \right)^{4},y^{3},$ etc. Due to these terms, it is impossible to put these equations into the same form as Equation 4.14.

Standard Form 标准形式

Consider the differential equation

$$\left( {3x^{2} - 4} \right)y^{\prime} + \left( {x - 3} \right)y = \text{sin}\mspace{2mu} x.$$

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of $y^{\prime}$ be equal to $1.$ To make this happen, we divide both sides by $3x^{2} - 4.$

$$y^{\prime} + \left( \frac{x - 3}{3x^{2} - 4} \right)\mspace{2mu} y = \frac{\text{sin}\mspace{2mu} x}{3x^{2} - 4}$$

This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to Equation 4.14, we can divide both sides of the equation by $a(x).$ This leads to the equation

$$y^{\prime} + \frac{b(x)}{a(x)}y = \frac{c(x)}{a(x)}.$$ (4.15)

Now define $p(x) = \frac{b(x)}{a(x)}$ and $q(x) = \frac{c(x)}{a(x)}.$ Then Equation 4.14 becomes

$$y^{\prime} + p(x)y = q(x).$$ (4.16)

We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.

Writing First-Order Linear Equations in Standard Form 把一阶线性方程写成标准形式

Put each of the following first-order linear differential equations into standard form. Identify $p(x)$ and $q(x)$ for each equation.

1. $y\prime = 3x - 4y$

2. $\frac{3xy\prime}{4y - 3} = 2$ (here $x \neq 0)$

3. $y = 3y\prime - 4x^{2} + 5$

Solution 解答

1. Add $4y$ to both sides:

$$y\prime + 4y = 3x.$$

In this equation, $p(x) = 4$ and $q(x) = 3x.$

2. Multiply both sides by $4y - 3,$ then subtract $8y$ from each side:

$$\begin{array}{rll} \frac{3xy\prime}{4y - 3} & = & 2 \\ {3xy\prime} & = & {2\left( {4y - 3} \right)} \\ {3xy\prime} & = & {8y - 6} \\ {3xy\prime - 8y} & = & {-6.} \end{array}$$

Finally, divide both sides by $3x$ to make the coefficient of $y\prime$ equal to $1\text{:}$

$$y\prime - \frac{8}{3x}y = - \frac{2}{x}.$$ (4.17) This is allowable because in the original statement of this problem we assumed that $x \neq 0.$ (If $x = 0$ then the original equation becomes $0 = 2,$ which is clearly a false statement.)

In this equation, $p(x) = - \frac{8}{3x}$ and $q(x) = - \frac{2}{x}.$

3. Subtract $y$ from each side and add $4x^{2} - 5\text{:}$

$$3y\prime - y = 4x^{2} - 5.$$

Next divide both sides by $3\text{:}$

$$y\prime - \frac{1}{3}y = \frac{4}{3}x^{2} - \frac{5}{3}.$$

In this equation, $p(x) = - \frac{1}{3}$ and $q(x) = \frac{4}{3}x^{2} - \frac{5}{3}.$

Put the equation $\frac{\left( {x + 3} \right)y\prime}{2x - 3y - 4} = 5$ into standard form and identify $p(x)$ and $q(x).$

Integrating Factors 积分因子

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

$$y\prime + p(x)y = q{(x).}$$ (4.18)

The first term on the left-hand side of Equation 4.15 is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the product rule from the Differentiation Rules section. If we multiply Equation 4.16 by a yet-to-be-determined function $\mu(x),$ then the equation becomes

$$\mu(x)y^{\prime} + \mu(x)p(x)y = \mu(x)q(x).$$ (4.19)

The left-hand side Equation 4.18 can be matched perfectly to the product rule:

$$\frac{d}{dx}\left\lbrack {f(x)g(x)} \right\rbrack = f^{\prime}(x)g(x) + f(x)g^{\prime}(x).$$

Matching term by term gives $y = f(x),g(x) = \mu(x),$ and $g^{\prime}(x) = \mu(x)p(x).$ Taking the derivative of $g(x) = \mu(x)$ and setting it equal to the right-hand side of $g^{\prime}(x) = \mu(x)p(x)$ leads to

$$\mu^{\prime}(x) = \mu(x)p(x).$$

This is a first-order, separable differential equation for $\mu(x).$ We know $p(x)$ because it appears in the differential equation we are solving. Separating variables and integrating yields

$$\begin{array}{rll} \frac{\mu^{\prime}(x)}{\mu(x)} & = & {p(x)} \\ {\left. \int\frac{\mu^{\prime}(x)}{\mu(x)} \right.dx} & = & {\int{p(x)\mspace{2mu} dx}} \\ {\text{ln}\left| {\mu(x)} \right|} & = & {{\int{p(x)\mspace{2mu} dx}} + C} \\ e^{\text{ln}{|{\mu{(x)}}|}} & = & e^{{\int{p{(x)}\mspace{2mu} dx}} + C} \\ \left| {\mu(x)} \right| & = & {C_{1}e^{\int{p{(x)}\mspace{2mu} dx}}} \\ {\mu(x)} & = & {C_{2}e^{\int{p{(x)}\mspace{2mu} dx}}.} \end{array}$$

Here $C_{2}$ can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of Equation 4.16 by the integrating factor $\mu(x).$ This gives

$$\mu(x)y^{\prime} + \mu(x)p(x)y = \mu(x)q(x).$$ (4.20)

The left-hand side of Equation 4.20 can be rewritten as $\frac{d}{dx}\left( {\mu(x)y} \right).$

$$\frac{d}{dx}\left( {\mu(x)y} \right) = \mu(x)q(x).$$ (4.21)

Next integrate both sides of Equation 4.21 with respect to $x.$

$$\begin{array}{rll} \left. \int{\frac{d}{dx}\left( {\mu(x)y} \right)\mspace{2mu} dx} \right. & = & {\int{\mu(x)q(x)\mspace{2mu} dx}} \\ {\mu(x)y} & = & {{\int{\mu(x)q(x)\mspace{2mu} dx}}.} \end{array}$$ (4.22)

Divide both sides of Equation 4.22 by $\mu(x)\text{:}$

$$y = \frac{1}{\mu(x)}\left\lbrack {\int{\mu(x)q(x)\mspace{2mu} dx + C}} \right\rbrack.$$ (4.23)

Since $\mu(x)$ was previously calculated, we are now finished. An important note about the integrating constant $C\text{:}$ It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving $p(x)$ is necessary in order to find an integrating factor for Equation 4.15. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for $C$ for this integral. We chose $C = 0.$ When calculating the integral inside the brackets in Equation 4.21, it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to Equation 4.15. This integrating factor guarantees just that.

Solving a First-order Linear Differential Equation 求解一阶线性微分方程

1. Put the equation into standard form and identify $p(x)$ and $q(x).$

2. Calculate the integrating factor $\mu(x) = e^{\int{p{(x)}\mspace{2mu} dx}}.$

3. Multiply both sides of the differential equation by $\mu(x).$

4. Integrate both sides of the equation obtained in step $3,$ and divide both sides by $\mu(x).$

5. If there is an initial condition, determine the value of $C.$

Solving a First-order Linear Equation 求解一个一阶线性方程

Find a general solution for the differential equation $xy\prime + 3y = 4x^{2} - 3x.$ Assume $x > 0.$

Solution 解答

1. To put this differential equation into standard form, divide both sides by $x\text{:}$

$$y\prime + \frac{3}{x}y = 4x - 3.$$

Therefore $p(x) = \frac{3}{x}$ and $q(x) = 4x - 3.$

2. The integrating factor is $\mu(x) = e^{\int{{({3\text{/}x})}\mspace{2mu} dx}} = e^{3\mspace{2mu}\text{ln}\mspace{2mu} x} = x^{3}.$

3. Multiplying both sides of the differential equation by $\mu(x)$ gives us

$$\begin{array}{rll} {x^{3}y^{\prime} + x^{3}\left( \frac{3}{x} \right)y} & = & {x^{3}\left( {4x - 3} \right)} \\ {x^{3}y^{\prime} + 3x^{2}y} & = & {4x^{4} - 3x^{3}} \\ {\frac{d}{dx}\left( {x^{3}y} \right)} & = & {4x^{4} - 3x^{3}.} \end{array}$$

4. Integrate both sides of the equation.

$$\begin{array}{rll} {\left. \int{\frac{d}{dx}\left( {x^{3}y} \right)} \right.dx} & = & {{\int{4x^{4} - 3x^{3}}}dx} \\ {x^{3}y} & = & {\frac{4x^{5}}{5} - \frac{3x^{4}}{4} + C} \\ y & = & {\frac{4x^{2}}{5} - \frac{3x}{4} + Cx^{-3}.} \end{array}$$

5. There is no initial value, so the problem is complete.

Analysis 分析

You may have noticed the condition that was imposed on the differential equation; namely, $x > 0.$ For any nonzero value of $C,$ the general solution is not defined at $x = 0.$ Furthermore, when $x < 0,$ the integrating factor changes. The integrating factor is given by Equation 4.19 as $\mu(x) = e^{{\int{p{(x)}}}dx}.$ For this $p(x)$ we get

$$e^{\int{p{(x)}\mspace{2mu} dx =}}e^{\int{(3\text{/}x)dx}} = e^{3\mspace{2mu}\text{ln}\mspace{2mu} |x|} = |x|^{3},$$

since $x < 0.$ The behavior of the general solution changes at $x = 0$ largely due to the fact that $p(x)$ is not defined there.

Find the general solution to the differential equation $\left( {x - 2} \right)y\prime + y = 3x^{2} + 2x.$ Assume $x > 2.$

Now we use the same strategy to find the solution to an initial-value problem.

A First-order Linear Initial-Value Problem 一个一阶线性初值问题

Solve the initial-value problem

$$y^{\prime} + 3y = 2x - 1,\quad y(0) = 3.$$

Solution 解答

1. This differential equation is already in standard form with $p(x) = 3$ and $q(x) = 2x - 1.$

2. The integrating factor is $\mu(x) = e^{\int{3dx}} = e^{3x}.$

3. Multiplying both sides of the differential equation by $\mu(x)$ gives

$$\begin{array}{rll} {e^{3x}y^{\prime} + 3e^{3x}y} & = & {\left( {2x - 1} \right)e^{3x}} \\ {\frac{d}{dx}\left\lbrack {ye^{3x}} \right\rbrack} & = & {\left( {2x - 1} \right)e^{3x}.} \end{array}$$

Integrate both sides of the equation:

$$\begin{array}{rll} \left. \int{\frac{d}{dx}\left\lbrack {ye^{3x}} \right\rbrack dx} \right. & = & {\int{\left( {2x - 1} \right)e^{3x}dx}} \\ {ye^{3x}} & = & {\frac{e^{3x}}{3}\left( {2x - 1} \right) - \left. \int{\frac{2}{3}e^{3x}dx} \right.} \\ {ye^{3x}} & = & {\frac{e^{3x}\left( {2x - 1} \right)}{3} - \frac{2e^{3x}}{9} + C} \\ y & = & {\frac{2x - 1}{3} - \frac{2}{9} + Ce^{-3x}} \\ y & = & {\frac{2x}{3} - \frac{5}{9} + Ce^{-3x}.} \end{array}$$

4. Now substitute $x = 0$ and $y = 3$ into the general solution and solve for $C\text{:}$

$$\begin{array}{rll} y & = & {\frac{2}{3}x - \frac{5}{9} + Ce^{-3x}} \\ 3 & = & {\frac{2}{3}(0) - \frac{5}{9} + Ce^{-3{(0)}}} \\ 3 & = & {- \frac{5}{9} + C} \\ C & = & {\frac{32}{9}.} \end{array}$$

Therefore the solution to the initial-value problem is

$$y = \frac{2}{3}x - \frac{5}{9} + \frac{32}{9}e^{-3x}.$$

Solve the initial-value problem $y\prime - 2y = 4x + 3\quad y(0) = -2.$

Applications of First-order Linear Differential Equations 一阶线性微分方程的应用

We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.

Free fall with air resistance 带空气阻力的自由落体

We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.

A Ball with Air Resistance 带空气阻力的球

A racquetball is hit straight upward with an initial velocity of $2$ m/s. The mass of a racquetball is approximately $0.0427$ kg. Air resistance acts on the ball with a force numerically equal to $0.5v,$ where $v$ represents the velocity of the ball at time $t.$

1. Find the velocity of the ball as a function of time.

2. How long does it take for the ball to reach its maximum height?

3. If the ball is hit from an initial height of $1$ meter, how high will it reach?

Solution 解答

1. The mass $m = 0.0427\ \text{kg},k = 0.5,$ and $g = 9.8{\ \text{m/s}}^{2}.$ The initial velocity is $v_{0} = 2$ m/s. Therefore the initial-value problem is

$$0.0427\frac{dv}{dt} = -0.5v - 0.0427(9.8),\quad v_{0} = 2.$$

Dividing the differential equation by $0.0427$ gives

$$\frac{dv}{dt} = -11.7096v - 9.8,\quad v_{0} = 2.$$

The differential equation is linear. Using the problem-solving strategy for linear differential equations:

Step 1. Rewrite the differential equation as $\frac{dv}{dt} + 11.7096v = -9.8.$ This gives $p(t) = 11.7096$ and $q(t) = -9.8$

Step 2. The integrating factor is $\mu(t) = e^{\int{11.7096dt}} = e^{11.7096t}.$

Step 3. Multiply the differential equation by $\mu(t)\text{:}$

\begin{array}{rll} {e^{11.7096t}\frac{dv}{dt} + 11.7096ve^{11.7096t}} & = & {-9.8e^{11.7096t}} \\ {\frac{d}{dt}\left\lbrack {ve^{11.7096t}} \right\rbrack} & = & {-9.8e^{11.7096t}.} \end{array}

Step 4. Integrate both sides:

\begin{array}{rll} {\left. \int{\frac{d}{dt}\left\lbrack {ve^{11.7096t}} \right\rbrack} \right.\mspace{2mu} dt} & = & {\int{-9.8e^{11.7096t}dt}} \\ {ve^{11.7096t}} & = & {\frac{-9.8}{11.7096}e^{11.7096t} + C} \\ {v(t)} & = & {-0.8369 + Ce^{-11.7096t}.} \end{array}

Step 5. Solve for $C$ using the initial condition $v_{0} = v(0) = 2\text{:}$

\begin{array}{rll} {v(t)} & = & {-0.8369 + Ce^{-11.7096t}} \\ {v(0)} & = & {-0.8369 + Ce^{-11.7096{(0)}}} \\ 2 & = & {-0.8369 + C} \\ C & = & 2.8369. \end{array}

Therefore the solution to the initial-value problem is $v(t) = 2.8369e^{-11.7096t} - 0.8369.$

2. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:

\begin{array}{rll} & & \\ {2.8369e^{-11.7096t} - 0.8369} & = & 0 \\ {2.8369e^{-11.7096t}} & = & 0.8369 \\ e^{-11.7096t} & = & {\frac{0.8369}{2.8369} \approx 0.295} \\ {\text{ln}\mspace{2mu} e^{-11.7096t}} & = & {\text{ln}\mspace{2mu} 0.295 \approx - 1.221} \\ {-11.7096t} & = & -1.221 \\ t & \approx & 0.104. \end{array}

Therefore it takes approximately $0.104$ second to reach maximum height.

3. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if $h(t)$ represents the height at time $t,$ then $h^{\prime}(t) = v(t).$ Because we know $v(t)$ and the initial height, we can form an initial-value problem:

$$h^{\prime}(t) = 2.8369e^{-11.7096t} - 0.8369,\quad h(0) = 1.$$

Integrating both sides of the differential equation with respect to $t$ gives

\begin{array}{rll} {\int{h^{\prime}(t)\mspace{2mu} dt}} & = & {{\int{2.8369e^{-11.7096t} - 0.8369}}dt} \\ {h(t)} & = & {- \frac{2.8369}{11.7096}e^{-11.7096t} - 0.8369t + C} \\ {h(t)} & = & {-0.2423e^{-11.7096t} - 0.8369t + C.} \end{array}

Solve for $C$ by using the initial condition:

\begin{array}{rll} {h(t)} & = & {-0.2423e^{-11.7096t} - 0.8369t + C} \\ {h(0)} & = & {-0.2423e^{-11.7096{(0)}} - 0.8369(0) + C} \\ 1 & = & {-0.2423 + C} \\ C & = & 1.2423. \end{array}

Therefore

$$h(t) = -0.2423e^{-11.7096t} - 0.8369t + 1.2423.$$

After $0.104$ second, the height is given by

$h(0.104) = -0.2423e^{-11.7096t} - 0.8369t + 1.2423 \approx 1.0836$ meter.

The weight of a penny is $2.5$ grams (United States Mint, "Coin Specifications," accessed April 9, 2015), and the upper observation deck of the Empire State Building is $369$ meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to $0.0025v.$ Furthermore, the penny is dropped with no initial velocity imparted to it.

1. Set up an initial-value problem that represents the falling penny.

2. Solve the problem for $v(t).$

3. What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as $t$ approaches infinity)?

Electrical Circuits 电路

A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff's Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:

1. The voltage drop across a resistor is given by

$$E_{R} = Ri,$$

where $R$ is a constant of proportionality called the *resistance,* and $i$ is the current.

2. The voltage drop across an inductor is given by

$$E_{L} = Li^{\prime},$$

where $L$ is a constant of proportionality called the *inductance*, and $i$ again denotes the current.

3. The voltage drop across a capacitor is given by

$$E_{C} = \frac{1}{C}q,$$

where $C$ is a constant of proportionality called the *capacitance*, and $q$ is the instantaneous charge on the capacitor. The relationship between $i$ and $q$ is $i = q^{\prime}.$

We use units of volts $\left( \text{V} \right)$ to measure voltage $E,$ amperes $\left( \text{A} \right)$ to measure current $i,$ coulombs $\left( \text{C} \right)$ to measure charge $q,$ ohms $\left( \text{Ω} \right)$ to measure resistance $R,$ henrys $\left( \text{H} \right)$ to measure inductance $L,$ and farads $\left( \text{F} \right)$ to measure capacitance $C.$ Consider the circuit in Figure 4.25.

Applying Kirchhoff's Loop Rule to this circuit, we let $E$ denote the electromotive force supplied by the voltage generator. Then

$$E_{L} + E_{R} + E_{C} = E.$$

Substituting the expressions for $E_{L},E_{R},$ and $E_{C}$ into this equation, we obtain

$$Li^{\prime} + Ri + \frac{1}{C}q = E.$$ (4.24)

If there is no capacitor in the circuit, then the equation becomes

$$Li^{\prime} + Ri = E.$$ (4.25)

This is a first-order differential equation in $i.$ The circuit is referred to as an $LR$ circuit.

Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so $L = 0,R \neq 0,$ and $C \neq 0.$ Then Equation 4.23 can be rewritten as

$$Rq^{\prime} + \frac{1}{C}q = E,$$ (4.26)

which is a first-order linear differential equation. This is referred to as an *RC* circuit. In either case, we can set up and solve an initial-value problem.

Finding Current in an *RL* Electric Circuit 求 RL 电路中的电流

A circuit has in series an electromotive force given by $E = 50\mspace{2mu}\text{sin}\mspace{2mu} 20t\ \text{V},$ a resistor of $5\text{Ω},$ and an inductor of $0.4\ \text{H}\text{.}$ If the initial current is $0,$ find the current at time $t > 0.$

Solution 解答

We have a resistor and an inductor in the circuit, so we use Equation 4.24. The voltage drop across the resistor is given by $E_{R} = Ri = 5i.$ The voltage drop across the inductor is given by $E_{L} = Li^{\prime} = 0.4i^{\prime}.$ The electromotive force becomes the right-hand side of Equation 4.24. Therefore Equation 4.24 becomes

$$0.4i^{\prime} + 5i = 50\mspace{2mu}\text{sin}\mspace{2mu} 20t.$$

Dividing both sides by $0.4$ gives the equation

$$i^{\prime} + 12.5i = 125\mspace{2mu}\text{sin}\mspace{2mu} 20t.$$

Since the initial current is 0, this result gives an initial condition of $i(0) = 0.$ We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as $i^{\prime} + 12.5i = 125\mspace{2mu}\text{sin}\mspace{2mu} 20t.$ This gives $p(t) = 12.5$ and $q(t) = 125\mspace{2mu}\text{sin}\mspace{2mu} 20t.$

Step 2. The integrating factor is $\mu(t) = e^{\int{12.5dt}} = e^{12.5t}.$

Step 3. Multiply the differential equation by $\mu(t)\text{:}$

$$\begin{array}{rll} {e^{12.5t}i^{\prime} + 12.5e^{12.5t}i} & = & {125e^{12.5t}\text{sin}\mspace{2mu} 20t} \\ {\frac{d}{dt}\mspace{2mu}\left\lbrack {ie^{12.5t}} \right\rbrack} & = & {125e^{12.5t}\text{sin}\mspace{2mu} 20t.} \end{array}$$

Step 4. Integrate both sides:

$$\begin{array}{rll} {\left. \int{\frac{d}{dt}\mspace{2mu}\left\lbrack {ie^{12.5t}} \right\rbrack} \right.\mspace{2mu} dt} & = & {\int{125e^{12.5t}\text{sin}\mspace{2mu} 20t\ dt}} \\ {ie^{12.5t}} & = & {\left( \frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} \right)\mspace{2mu} e^{12.5t} + C} \\ {i(t)} & = & {\frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} + Ce^{-12.5t}.} \end{array}$$

Step 5. Solve for $C$ using the initial condition $v(0) = 2\text{:}$

$$\begin{array}{rll} {i(t)} & = & {\frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} + Ce^{-12.5t}} \\ {i(0)} & = & {\frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} + Ce^{-12.5{(0)}}} \\ 0 & = & {- \frac{400}{89} + C} \\ C & = & {\frac{400}{89}.} \end{array}$$

Therefore the solution to the initial-value problem is $i(t) = \frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t + 400e^{-12.5t}}{89} = \frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} + \frac{400e^{-12.5t}}{89}.$

The first term can be rewritten as a single cosine function. First, multiply and divide by $\sqrt{250^{2} + 400^{2}} = 50\sqrt{89}\text{:}$

$$\begin{array}{cl} \frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{89} & {= \frac{50\sqrt{89}}{89}\left( \frac{250\mspace{2mu}\text{sin}\mspace{2mu} 20t - 400\mspace{2mu}\text{cos}\mspace{2mu} 20t}{50\sqrt{89}} \right)} \\ & {= - \frac{50\sqrt{89}}{89}\left( {\frac{8\mspace{2mu}\text{cos}\mspace{2mu} 20t}{\sqrt{89}} - \frac{5\mspace{2mu}\text{sin}\mspace{2mu} 20t}{\sqrt{89}}} \right).} \end{array}$$

Next, define $\varphi$ to be an acute angle such that $\text{cos}\mspace{2mu}\varphi = \frac{8}{\sqrt{89}}.$ Then $\text{sin}\mspace{2mu}\varphi = \frac{5}{\sqrt{89}}$ and

$$\begin{array}{cl} {- \frac{50\sqrt{89}}{89}\left( {\frac{8\mspace{2mu}\text{cos}\mspace{2mu} 20t}{\sqrt{89}} - \frac{5\mspace{2mu}\text{sin}\mspace{2mu} 20t}{\sqrt{89}}} \right)} & {= - \frac{50\sqrt{89}}{89}\mspace{2mu}\left( {\text{cos}\mspace{2mu}\varphi\mspace{2mu}\text{cos}\mspace{2mu} 20t - \text{sin}\mspace{2mu}\varphi\mspace{2mu}\text{sin}\mspace{2mu} 20t} \right)} \\ & {= - \frac{50\sqrt{89}}{89}\text{cos}\left( {20t + \varphi} \right).} \end{array}$$

Therefore the solution can be written as

$$i(t) = - \frac{50\sqrt{89}}{89}\mspace{2mu}\text{cos}\mspace{2mu}\left( {20t + \varphi} \right) + \frac{400e^{-12.5t}}{89}.$$

The second term is called the *attenuation* term, because it disappears rapidly as *t* grows larger. The phase shift is given by $\varphi,$ and the amplitude of the steady-state current is given by $\frac{50\sqrt{89}}{89}.$ The graph of this solution appears in Figure 4.26:

A circuit has in series an electromotive force given by $E = 20\mspace{2mu}\text{sin}\mspace{2mu} 5t$ V, a capacitor with capacitance $0.02\ \text{F},$ and a resistor of $8\ \text{Ω}.$ If the initial charge is $4\ \text{C},$ find the charge at time $t > 0.$

Section 4.5 Exercises 4.5 节 习题

Are the following differential equations linear? Explain your reasoning.

208\.

$\frac{dy}{dx} = x^{2}y + \text{sin}\mspace{2mu} x$

209.

$\frac{dy}{dt} = ty$

210\.

$\frac{dy}{dt} + y^{2} = x$

211.

$y\prime = x^{3} + e^{x}$

212\.

$y\prime = y + e^{y}$

Write the following first-order differential equations in standard form.

213.

$y\prime = x^{3}y + \text{sin}\mspace{2mu} x$

214\.

$y\prime + 3y - \text{ln}\mspace{2mu} x = 0$

215.

$\text{−}xy\prime = \left( {3x + 2} \right)y + xe^{x}$

216\.

$\frac{dy}{dt} = 4y + ty + \text{tan}\mspace{2mu} t$

217.

$\frac{dy}{dt} = yx\left( {x + 1} \right)$

What are the integrating factors for the following differential equations?

218\.

$y\prime = xy + 3$

219.

$y\prime + e^{x}y = \text{sin}\mspace{2mu} x$

220\.

$y\prime = x\mspace{2mu}\text{ln}(x)y + 3x$

221.

$\frac{dy}{dx} = \text{tanh}(x)y + 1$

222\.

$\frac{dy}{dt} + 3ty = e^{t}y$

Solve the following differential equations by using integrating factors.

223.

$y\prime = 3y + 2$

224\.

$y\prime = 2y - x^{2}$

225.

$xy\prime = 3y - 6x^{2}$

226\.

$\left( {x + 2} \right)y\prime = 3x + y$

227.

$y\prime = 3x + xy$

228\.

$xy\prime = x + y$

229.

$\text{sin}(x)y\prime = y + 2x$

230\.

$y\prime = y + e^{x}$

231.

$xy\prime = 3y + x^{2}$

232\.

$y\prime + \text{ln}\mspace{2mu} x = \frac{y}{x}$

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?

233.

\[T\] $(x + 2)y\prime = 2y - 1$

234\.

\[T\] $y\prime = 3e^{t\text{/}3} - 2y$

235.

\[T\] $xy\prime + \frac{y}{2} = \text{sin}(3t)$

236\.

\[T\] $xy\prime = 2\frac{\text{cos}\mspace{2mu} x}{x} - 3y$

237.

\[T\] $(x + 1)y\prime = 3y + x^{2} + 2x + 1$

238\.

\[T\] $\text{sin}(x)y\prime + \text{cos}(x)y = 2x$

239.

\[T\] $\sqrt{x^{2} + 1}y\prime = y + 2$

240\.

\[T\] $x^{3}y\prime + 2x^{2}y = x + 1$

Solve the following initial-value problems by using integrating factors.

241.

$y\prime + y = x,y(0) = 3$

242\.

$y\prime = y + 2x^{2},y(0) = 0$

243.

$xy\prime = y - 3x^{3},y(1) = 0$

244\.

$x^{2}y\prime = xy - \text{ln}\mspace{2mu} x,y(1) = 1$

245.

$\left( {1 + x^{2}} \right)y\prime = y - 1,y(0) = 0$

246\.

$xy\prime = y + 2x\mspace{2mu}\text{ln}\mspace{2mu} x,y(1) = 5$

247.

$(2 + x)y\prime = y + 2 + x,y(0) = 0$

248\.

$y' = xy + 2xe^{\frac{1}{2}x^{2}},y(0) = 2$

249.

$\sqrt{x}y\prime = y + 2x,y(0) = 1$

250\.

$y\prime = 2y + xe^{x},y(0) = -1$

251.

A falling object of mass $m$ can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant $k.$ Set up the differential equation and solve for the velocity given an initial velocity of $0.$

252\.

Using your expression from the preceding problem, what is the terminal velocity? (*Hint:* Examine the limiting behavior; does the velocity approach a value?)

253.

\[T\] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall $5000$ meters if the mass is $100$ kilograms, the acceleration due to gravity is $9.8$ m/s² and the proportionality constant is $4?$

254\.

A more accurate way to describe terminal velocity is that the drag force is proportional to the square of velocity, with a proportionality constant $k.$ Set up the differential equation and solve for the velocity.

255.

Using your expression from the preceding problem, what is the terminal velocity? (*Hint:* Examine the limiting behavior: Does the velocity approach a value?)

256\.

\[T\] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall $5000$ meters if the mass is $100$ kilograms, the acceleration due to gravity is $9.8{\ \text{m/s}}^{2}$ and the proportionality constant is $4?$ Does it take more or less time than your initial estimate?

For the following problems, determine how parameter $a$ affects the solution.

257.

Solve the generic equation $y\prime = ax + y.$ How does varying $a$ change the behavior?

258\.

Solve the generic equation $y\prime = ay + x.$ How does varying $a$ change the behavior?

259.

Solve the generic equation $y\prime = ax + xy.$ How does varying $a$ change the behavior?

260\.

Solve the generic equation $y\prime = x + axy.$ How does varying $a$ change the behavior?

261.

Solve $y\prime - y = e^{kt}$ with the initial condition $y(0) = 0.$ As $k$ approaches $1,$ what happens to your formula?

Key Terms 术语表

asymptotically semi-stable solution

$y = k$ if it is neither asymptotically stable nor asymptotically unstable

asymptotically stable solution

$y = k$ if there exists $\varepsilon > 0$ such that for any value $c \in \left( {k - \varepsilon,k + \varepsilon} \right)$ the solution to the initial-value problem $y^{\prime} = f\left( {x,y} \right),\quad y\left( x_{0} \right) = c$ approaches $k$ as $x$ approaches infinity

asymptotically unstable solution

$y = k$ if there exists $\varepsilon > 0$ such that for any value $c \in \left( {k - \varepsilon,k + \varepsilon} \right)$ the solution to the initial-value problem $y^{\prime} = f\left( {x,y} \right),\quad y\left( x_{0} \right) = c$ never approaches $k$ as $x$ approaches infinity

autonomous differential equation

an equation in which the right-hand side is a function of $y$ alone

carrying capacity

the maximum population of an organism that the environment can sustain indefinitely

differential equation

an equation involving a function $y = y(x)$ and one or more of its derivatives

direction field (slope field)

a mathematical object used to graphically represent solutions to a first-order differential equation; at each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to the differential equation passing through that point

equilibrium solution

any solution to the differential equation of the form $y = c,$ where $c$ is a constant

Euler’s Method

a numerical technique used to approximate solutions to an initial-value problem

general solution (or family of solutions)

the entire set of solutions to a given differential equation

growth rate

the constant $r > 0$ in the exponential growth function $P(t) = P_{0}e^{rt}$

initial population

the population at time $t = 0$

initial value(s)

a value or set of values that a solution of a differential equation satisfies for a fixed value of the independent variable

initial velocity

the velocity at time $t = 0$

initial-value problem

a differential equation together with an initial value or values

integrating factor

any function $f(x)$ that is multiplied on both sides of a differential equation to make the side involving the unknown function equal to the derivative of a product of two functions

linear

description of a first-order differential equation that can be written in the form $a(x)y^{\prime} + b(x)y = c(x)$

logistic differential equation

a differential equation that incorporates the carrying capacity $K$ and growth rate $r$ into a population model

order of a differential equation

the highest order of any derivative of the unknown function that appears in the equation

particular solution

member of a family of solutions to a differential equation that satisfies a particular initial condition

phase line

a visual representation of the behavior of solutions to an autonomous differential equation subject to various initial conditions

separable differential equation

any equation that can be written in the form $y\prime = f(x)g(y)$

separation of variables

a method used to solve a separable differential equation

solution curve

a curve graphed in a direction field that corresponds to the solution to the initial-value problem passing through a given point in the direction field

solution to a differential equation

a function $y = f(x)$ that satisfies a given differential equation

standard form

the form of a first-order linear differential equation obtained by writing the differential equation in the form $y\prime + p(x)y = q(x)$

step size

the increment $h$ that is added to the $x$ value at each step in Euler’s Method

threshold population

the minimum population that is necessary for a species to survive

Key Equations 关键公式

Euler’s Method $\begin{array}{l} {x_{n} = x_{0} + nh} \\ {y_{n} = y_{n - 1} + hf(x_{n - 1},y_{n - 1}),\text{where}\ h\ \text{is the step size}} \end{array}$

Separable differential equation $y^{\prime} = f(x)g(y)$

Solution concentration $\frac{du}{dt} = \text{INFLOW RATE} - \text{OUTFLOW RATE}$

Newton’s law of cooling $\frac{dT}{dt} = k\left( {T - T_{s}} \right)$

Logistic differential equation and initial-value problem $\frac{dP}{dt} = rP\left( {1 - \frac{P}{K}} \right),\quad P(0) = P_{0}$

Solution to the logistic differential equation/initial-value problem $P(t) = \frac{P_{0}Ke^{rt}}{\left( {K - P_{0}} \right) + P_{0}e^{rt}}$

Threshold population model $\frac{dP}{dt} = \text{−}rP\left( {1 - \frac{P}{K}} \right)\left( {1 - \frac{P}{T}} \right)$

standard form $y\prime + p(x)y = q(x)$

integrating factor $\mu(x) = e^{\int{p{(x)}\mspace{2mu} dx}}$

Key Concepts 关键概念

4.1 Basics of Differential Equations 4.1 微分方程基础

• A differential equation is an equation involving a function $y = f(x)$ and one or more of its derivatives. A solution is a function $y = f(x)$ that satisfies the differential equation when $f$ and its derivatives are substituted into the equation.
• The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.
• A differential equation coupled with an initial value is called an initial-value problem. To solve an initial-value problem, first find the general solution to the differential equation, then determine the value of the constant. Initial-value problems have many applications in science and engineering.

• 微分方程是包含函数 $y = f(x)$ 及其一个或多个导数的方程。所谓解，是指当把 $f$ 及其导数代入方程时能使该微分方程成立的函数 $y = f(x)$。
• 微分方程的阶，是方程中所出现的未知函数任意导数的最高阶数。
• 带有初值的微分方程称为初值问题。求解初值问题时，先求出该微分方程的通解，再确定常数的值。初值问题在科学与工程中有许多应用。

4.2 Direction Fields and Numerical Methods 4.2 方向场与数值方法

• A direction field is a mathematical object used to graphically represent solutions to a first-order differential equation.
• Euler’s Method is a numerical technique that can be used to approximate solutions to a differential equation.

• 方向场是一种数学对象，用于以图形方式表示一阶微分方程的解。
• 欧拉方法是一种数值技术，可用来近似微分方程的若干解。

4.3 Separable Equations 4.3 可分离变量的方程

• A separable differential equation is any equation that can be written in the form $y\prime = f(x)g(y).$
• The method of separation of variables is used to find the general solution to a separable differential equation.

• 可分离变量的微分方程，是指任何可以写成 $y\prime = f(x)g(y)$ 形式的方程。
• 分离变量法用来求可分离变量的微分方程的通解。

4.4 The Logistic Equation 4.4 逻辑斯谛方程

• When studying population functions, different assumptions—such as exponential growth, logistic growth, or threshold population—lead to different rates of growth.
• The logistic differential equation incorporates the concept of a carrying capacity. This value is a limiting value on the population for any given environment.
• The logistic differential equation can be solved for any positive growth rate, initial population, and carrying capacity.

• 在研究种群函数时，不同的假设——如指数增长、逻辑斯谛增长或阈值种群——会导致不同的增长率。
• 逻辑斯谛微分方程融入了容纳量的概念。该值是任一给定环境对种群的限制性取值。
• 对于任意正的增长率、初始种群量和容纳量，逻辑斯谛微分方程都可以求解。

4.5 First-Order Linear Equations 4.5 一阶线性方程

• Any first-order linear differential equation can be written in the form $y\prime + p(x)y = q(x).$
• We can use a five-step problem-solving strategy for solving a first-order linear differential equation that may or may not include an initial value.
• Applications of first-order linear differential equations include determining motion of a rising or falling object with air resistance and finding current in an electrical circuit.

• 任意一阶线性微分方程都可以写成 $y\prime + p(x)y = q(x)$ 的形式。
• 我们可以采用一个五步解题策略来求解一阶线性微分方程，无论其是否含有初值。
• 一阶线性微分方程的应用包括：确定受空气阻力作用下上升或下落物体的运动，以及求电路中的电流。

Review Exercises 章末复习题

*True or False?* Justify your answer with a proof or a counterexample.

262\.

The differential equation $y\prime = 3x^{2}y - \text{cos}(x)y^{''}$ is linear.

263.

The differential equation $y\prime = x - y$ is separable.

264\.

You can explicitly solve all first-order differential equations by separation or by the method of integrating factors.

265.

You can determine the behavior of all first-order differential equations using directional fields or Euler’s method.

For the following problems, find the general solution to the differential equations.

266\.

$y^{\prime} = x^{2} + 3e^{x} - 2x$

267.

$y\prime = 2^{x} + \text{cos}^{-1}x$

268\.

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

269.

$y\prime = e^{\text{−}y}\text{sin}\mspace{2mu} x$

270\.

$y\prime = 3x - 2y$

271.

$y\prime = y\mspace{2mu}\text{ln}\mspace{2mu} y$

For the following problems, find the solution to the initial value problem.

272\.

$y\prime = 8x - \text{ln}\mspace{2mu} x - 3x^{4},y(1) = 5$

273.

$y\prime = 3x - \text{cos}\mspace{2mu} x + 2,y(0) = 4$

274\.

$xy\prime = y\left( {x - 2} \right),y(1) = 3$

275.

$y\prime = 3y^{2}\left( {x + \text{cos}\mspace{2mu} x} \right),y(0) = -2$

276\.

$(x - 1)y\prime = y - 2,y(0) = 0$

277.

$y\prime = 3y - x + 6x^{2},y(0) = -1$

For the following problems, draw the directional field associated with the differential equation, then solve the differential equation. Draw a sample solution on the directional field.

278\.

$y\prime = 2y - y^{2}$

279.

$y\prime = \frac{1}{x} + \text{ln}\mspace{2mu} x - y,$ for $x > 0$

For the following problems, use Euler’s Method with $n = 5$ steps over the interval $t = \lbrack 0,1\rbrack.$ Then solve the initial-value problem exactly. How close is your Euler’s Method estimate?

280\.

$y\prime = -4yx,y(0) = 1$

281.

$y\prime = 3^{x} - 2y,y(0) = 0$

For the following problems, set up and solve the differential equations.

282\.

A car drives along a freeway, accelerating according to $a = 5\mspace{2mu}\text{sin}(\pi t),$ where $t$ represents time in minutes. Find the velocity at any time $t,$ assuming the car starts with an initial speed of $60$ mph.

283.

You throw a ball of mass $2$ kilograms into the air with an upward velocity of $8$ m/s. Find exactly the time the ball will remain in the air, assuming that gravity is given by $g = 9.8{\ \text{m/s}}^{2}.$

284\.

You drop a ball with a mass of $5$ kilograms out an airplane window at a height of $5000$ m. How long does it take for the ball to reach the ground?

285.

You drop the same ball of mass $5$ kilograms out of the same airplane window at the same height, except this time you assume a drag force proportional to the ball’s velocity, using a proportionality constant of $3$ and the ball reaches terminal velocity. Solve for the distance fallen as a function of time. How long does it take the ball to reach the ground?

286\.

A drug is administered to a patient every $24$ hours and is cleared at a rate proportional to the amount of drug left in the body, with proportionality constant $0.2.$ If the patient needs a baseline level of $5$ mg to be in the bloodstream at all times, how large should the dose be?

287.

A $1000$-liter tank contains pure water and a solution of $0.2$ kg salt/L is pumped into the tank at a rate of $1$ L/min and is drained at the same rate. Solve for total amount of salt in the tank at time $t.$

288\.

You boil water to make tea. When you pour the water into your teapot, the temperature is $100\text{°C.}$ After $5$ minutes in your $15\text{°C}$ room, the temperature of the tea is $85\text{°C.}$ Solve the equation to determine the temperatures of the tea at time $t.$ How long must you wait until the tea is at a drinkable temperature $\left( 72\text{°C}\operatorname{)?} \right.$

289.

The human population (in thousands) of Nevada in $1950$ was roughly $160.$ If the carrying capacity is estimated at $10$ million individuals, and assuming a growth rate of $2\text{\%}$ per year, develop a logistic growth model and solve for the population in Nevada at any time (use $1950$ as time = 0). What population does your model predict for $2000?$ How close is your prediction to the true value of $1,998,257?$

290\.

Repeat the previous problem but use Gompertz growth model. Which is more accurate?