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微分方程

AP 微积分 BC · 第 7 主题

训练
讲义 词汇表
7.1

用微分方程为情境建模

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.A
Interpret verbal statements of problems as differential equations involving a derivative expression.

  • FUN-7.A.1 Differential equations relate a function of an independent variable and the function's derivatives.

来源:美国大学理事会 AP 课程与考试说明

一个微分方程(differential equation)把一个函数与它的导数关联。许多真实的情况由一个率描述:"种群以与它的大小成比例的一个率增长"变成 $\dfrac{dP}{dt}=kP$。从一个文字描述设置方程——辨别什么变化以及它与什么成比例——是第一个技能。

词汇表 训练
英文 中文 拼音
differential equation 微分方程 wēi fēn fāng chéng
7.2

验证微分方程的解

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.B
Verify solutions to differential equations.

  • FUN-7.B.1 Derivatives can be used to verify that a function is a solution to a given differential equation.
  • FUN-7.B.2 There may be infinitely many general solutions to a differential equation.

来源:美国大学理事会 AP 课程与考试说明

一个(solution)是满足方程的一个函数。要验证一个提出的解,求导它并代入方程,检查两侧一致。一个通解(general solution)包含一个常数 $C$;一个特解(particular solution)从一个条件确定 $C$

7.3

绘制斜率场

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.C
Estimate solutions to differential equations.

  • FUN-7.C.1 A slope field is a graphical representation of a differential equation on a finite set of points in the plane.
  • FUN-7.C.2 Slope fields provide information about the behavior of solutions to first-order differential equations.

来源:美国大学理事会 AP 课程与考试说明

斜率场与解曲线

一个斜率场(slope field)在许多点画一个短线段,每个带方程在那里给出的斜率 $\dfrac{dy}{dx}$。它描绘解曲线的族而不求解。要草绘一个,在每个网格点求右侧的值并画一个那个斜率的线段。

一个斜率场处处显示斜率;解曲线跟随它
一个斜率场处处显示斜率;解曲线跟随它
探索

Read a differential equation as a slope field

A slope field draws the slope $dy/dx$ at each point. A solution curve threads through, always tangent to the little segments — you can sketch it by following the flow.

词汇表 训练
英文 中文 拼音
slope field 斜率场 xié lǜ chǎng
7.4

利用斜率场进行推理

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.C
Estimate solutions to differential equations.

  • FUN-7.C.3 Solutions to differential equations are functions or families of functions.

来源:美国大学理事会 AP 课程与考试说明

一条解曲线跟随这些线段,像一条船跟随一股水流。从一个斜率场你能草绘通过一个给定点的特解、描述长期行为,并定位解在哪里趋平(斜率接近零)——纯粹从图片推理解。

7.5

用欧拉法近似求解

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.C
Estimate solutions to differential equations.

  • FUN-7.C.4 Euler's method provides a procedure for approximating a solution to a differential equation or a point on a solution curve. BC ONLY

来源:美国大学理事会 AP 课程与考试说明

欧拉法

欧拉方法(Euler's method)通过沿着斜率场步进来数值地近似一个解。从一个已知点开始,取一个小步长 $\Delta x$ 并更新:

$$y_{\text{new}}=y_{\text{old}}+\frac{dy}{dx}\cdot\Delta x.$$
对每一步重复。更小的步长给出更好的近似。(这是一个 BC 独有的技法。)

考试技能: 能够从一张表手工执行两三个欧拉步,并知道欧拉方法根据解的凹凸性低估或高估

Worked example. 用步长 $\Delta x=0.5$$\dfrac{dy}{dx}=x+y$$y(0)=1$ 近似 $y(1)$。第 1 步:$(0,1)$ 处的斜率是 $0+1=1$,所以 $y(0.5)\approx 1+1(0.5)=1.5$。第 2 步:$(0.5,1.5)$ 处的斜率是 $0.5+1.5=2$,所以 $y(1)\approx 1.5+2(0.5)=2.5$

词汇表 训练
英文 中文 拼音
Euler's method 欧拉方法 ōu lā fāng fǎ
7.6

用分离变量法求通解

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.D
Determine general solutions to differential equations.

  • FUN-7.D.1 Some differential equations can be solved by separation of variables.
  • FUN-7.D.2 Antidifferentiation can be used to find general solutions to differential equations.

来源:美国大学理事会 AP 课程与考试说明

一个可分离(separable)微分方程能被写成所有 $y$ 在一侧而所有 $x$ 在另一侧,然后积分:

$$\frac{dy}{dx}=g(x)h(y)\ \Rightarrow\ \int\frac{dy}{h(y)}=\int g(x)\,dx.$$
这产生通解(带一个 $+C$),课程里解微分方程的主要解析方法。

Worked example.$y(0)=2$$\dfrac{dy}{dx}=xy$。分离,$\int\frac{dy}{y}=\int x\,dx$ 给出 $\ln|y|=\frac{x^2}{2}+C$,所以 $y=Ae^{x^2/2}$。条件 $y(0)=2$ 给出 $A=2$,所以 $y=2e^{x^2/2}$

词汇表 训练
英文 中文 拼音
separable 可分离 kě fēn lí
7.7

用初始条件与分离变量法求特解

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.E
Determine particular solutions to differential equations.

  • FUN-7.E.1 A general solution may describe infinitely many solutions to a differential equation. There is only one particular solution passing through a given point.
  • FUN-7.E.2 The function $F$ defined by $F(x) = y_0 + \int_a^x f(t)\,dt$ is a particular solution to the differential equation $\dfrac{dy}{dx} = f(x)$, satisfying $F(a) = y_0$.
  • FUN-7.E.3 Solutions to differential equations may be subject to domain restrictions.

来源:美国大学理事会 AP 课程与考试说明

一个初始条件(initial condition)(一个已知点,例如 $y(0)=5$)确定常数 $C$。求通解、代入条件以求 $C$,然后写特解。注意定义域——一个特解只在包含初始点的区间上有效。

常数给出一族曲线;一个初始条件选出一条
常数给出一族曲线;一个初始条件选出一条
词汇表 训练
英文 中文 拼音
initial condition 初始条件 chū shǐ tiáo jiàn
7.8

微分方程的指数模型

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.F
Interpret the meaning of a differential equation and its variables in context.

  • FUN-7.F.1 Specific applications of finding general and particular solutions to differential equations include motion along a line and exponential growth and decay.
  • FUN-7.F.2 The model for exponential growth and decay that arises from the statement "The rate of change of a quantity is proportional to the size of the quantity" is $\dfrac{dy}{dt} = ky$.

FUN-7.G
Determine general and particular solutions for problems involving differential equations in context.

  • FUN-7.G.1 The exponential growth and decay model, $\dfrac{dy}{dt} = ky$, with initial condition $y = y_0$ when $t = 0$, has solutions of the form $y = y_0 e^{kt}$.

来源:美国大学理事会 AP 课程与考试说明

指数增长与逻辑斯蒂增长

方程 $\dfrac{dy}{dt}=ky$ 说变化率与数量成比例——给出指数增长或衰减(exponential growth or decay)。分离变量产生

$$y=y_0 e^{kt},$$
$k>0$ 为增长而 $k<0$ 为衰减。这为不受限的种群增长、放射性衰变和连续复利建模。

A cooling cup of coffee: Newton's law of cooling is a classic differential equation model
A cooling cup of coffee: Newton's law of cooling is a classic differential equation model
探索

An exponential growth/decay model

y = a·e^(bx) + c

The equation $dy/dt=ky$ has exponential solutions: quantity changes at a rate proportional to itself, giving unbounded growth ($k>0$) or decay to zero ($k<0$).

词汇表 训练
英文 中文 拼音
exponential growth or decay 指数增长 zhǐ shù zēng zhǎng
7.9

微分方程的逻辑斯谛模型

大纲
Enduring UnderstandingLearning ObjectiveEssential Knowledge

FUN-7
Solving differential equations allows us to determine functions and develop models.

FUN-7.H
Interpret the meaning of the logistic growth model in context. BC ONLY

  • FUN-7.H.1 The model for logistic growth that arises from the statement "The rate of change of a quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity" is $\dfrac{dy}{dt} = ky(a - y)$. BC ONLY
  • FUN-7.H.2 The logistic differential equation and initial conditions can be interpreted without solving the differential equation. BC ONLY
  • FUN-7.H.3 The limiting value (carrying capacity) of a logistic differential equation as the independent variable approaches infinity can be determined using the logistic growth model and initial conditions. BC ONLY
  • FUN-7.H.4 The value of the dependent variable in a logistic differential equation at the point when it is changing fastest can be determined using the logistic growth model and initial conditions. BC ONLY

来源:美国大学理事会 AP 课程与考试说明

真实的增长受资源限制,所以逻辑斯蒂模型(logistic model)加上一个环境容纳量(carrying capacity)$L$:

$$\frac{dP}{dt}=kP\!\left(1-\frac{P}{L}\right).$$
$P$ 小时增长几乎是指数的、随着 $P$ 接近 $L$ 减慢,并在 $P=L$ 停止。关键的 BC 事实:种群$L$ 处趋平($\lim_{t\to\infty}P=L$),而它在 $P=\tfrac{L}{2}$增长最快(S 形曲线的拐点)。你被期望直接从方程读出 $L$ 和最快增长值。

Worked example. 对于 $\dfrac{dP}{dt}=0.05\,P\!\left(1-\dfrac{P}{2000}\right)$,环境容纳量是 $L=2000$(种群在那里趋平),而增长在 $P=\dfrac{L}{2}=1000$ 时最快——两者都直接从方程读出,不需要求解。

逻辑斯蒂模型在 P=L/2 增长最快并在环境容纳量 L 处趋平
逻辑斯蒂模型在 P=L/2 增长最快并在环境容纳量 L 处趋平
词汇表 训练
英文 中文 拼音
logistic model 逻辑斯蒂模型 luó jí sī dì mó xíng
carrying capacity 环境容纳量 huán jìng róng nà liàng
7.9

考试技巧

  • 通过把所有 $y$ 移到一侧而所有 $x$ 移到另一侧、然后积分两侧(加一次 $+C$)来解一个可分离方程。
  • 用初始条件求 $C$(一个特解)。
  • 草绘或读一个斜率场:小线段在每个点显示 $\tfrac{dy}{dx}$,而一条解曲线跟随它们。
  • 辨认指数模型 $\tfrac{dy}{dt}=ky\Rightarrow y=Ce^{kt}$(增长/衰减)。
  • 一个微分方程给出斜率——你必须积分以恢复函数。

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