📚 Introduction to Differential Equations | 微分方程导论
A differential equation is an equation that involves a function and its derivatives. Instead of asking for a number, it asks for a function that satisfies a relationship between the function and its rate of change. Differential equations are central to AQA A-level Mathematics because they connect calculus to real-world change.
微分方程是包含函数及其导数的方程。它要求得到的不是一个数，而是一个满足函数与其变化率之间关系的函数。微分方程是AQA A-level数学的核心内容，因为它将微积分与现实世界的变化联系起来。
1. What is a Differential Equation? | 什么是微分方程？
s equal to 2x. Another example is d²y/dx² + y = 0, which describes simple harmonic motion.
从本质上讲，微分方程包含一个未知函数及其一个或多个导数。例如，dy/dx = 2x 是一个一阶微分方程：它表示 y 关于 x 的梯度等于 2x。另一个例子是 d²y/dx² + y = 0，它描述简谐运动。
These equations appear in physics, engineering, biology and economics because many natural laws are expressed as rates of change. Learning to formulate, solve and interpret them is a key skill at A-level.
这类方程出现在物理、工程、生物和经济中，因为许多自然规律都是以变化率的形式表达的。学会建立、求解并解释它们是A-level阶段的关键技能。
2. The Order of a Differential Equation | 微分方程的阶
The order of a differential equation is the highest derivative that appears in it. A first-order equation contains only dy/dx. A second-order equation contains d²y/dx², and may also contain dy/dx and y.
微分方程的阶是指方程中出现的最高阶导数。一阶方程只含有 dy/dx。二阶方程含有 d²y/dx²，也可能含有 dy/dx 和 y。
| Equation | Highest derivative | Order |
| dy/dx = 3x² | dy/dx | 1 |
| d²y/dx² - 5 dy/dx + 6y = 0 | d²y/dx² | 2 |
In this introduction we focus on first-order differential equations, which form the main AQA requirement.
在这篇导论中，我们主要关注一阶微分方程，这也是AQA考试的主要要求。
3. General and Particular Solutions | 通解与特解
ctions that satisfy it. For dy/dx = 2x, integrating both sides gives the general solution:
求解微分方程意味着找出所有满足该方程的函数。对于 dy/dx = 2x，两边积分得到通解：
y = ∫ 2x dx = x² + C
given, such as y(0) = 3, then we can find C and obtain a particular solution: y = x² + 3.
这里 C 是任意常数。它被称为通解，因为它代表一族曲线。如果给出了额外条件，例如 y(0) = 3，那么我们就可以求出 C，得到特解：y = x² + 3。
4. Verifying a Solution | 验证解
A common exam question asks you to show that a given function satisfies a differential equation. To do this, differentiate the proposed function and substitute it into the original equation.
常见考试题会要求你证明某个给定函数满足一个微分方程。做法是先对所给函数求导，再代入原方程验证。
Example: Show that y = e2x satisfies dy/dx - 2y = 0.
示例：证明 y = e2x 满足 dy/dx - 2y = 0。
dy/dx = 2e2x, so dy/dx - 2y = 2e2x - 2e2x = 0.
Since the left-hand side becomes zero, the solution is valid. Notice that verification does not require solving an equation; it only requires careful differentiation and substitution.
因为左边等于零，所以该解成立。注意，验证解并不需要求解方程，只需要仔细求导并代入。
5. Initial Conditions and Boundary Conditions | 初始条件与边界条件
An initial condition is given at a single starting point, for example y(0) = 1 or y'(0) = -2. These are common when the variable is time, such as the population at t = 0.
初始条件是在一个起始点给出的条件，例如 y(0) = 1 或 y'(0) = -2。当变量是时间时，这类条件很常见，例如 t = 0 时的种群数量。
th types of condition allow us to determine the arbitrary constants in a general solution.
边界条件用于涉及两个不同端点的问题，例如 y(0) = 0 和 y(1) = 2。这两类条件都能帮助我们确定通解中的任意常数。
For a first-order equation you normally need one condition. For a second-order equation you normally need two.
对于一阶方程，通常需要一个条件；对于二阶方程，通常需要两个条件。
6. Slope Fields (Direction Fields) | 斜率场（方向场）
t (x, y), you draw a short line segment whose gradient is f(x, y). Together, these segments show how solution curves behave without solving the equation algebraically.
斜率场是 dy/dx = f(x, y) 形式微分方程的直观表示。在每个网格点 (x, y) 处，画一条梯度为 f(x, y) 的短线段。这些线段共同显示出解曲线的大致走向，而不需要代数求解。
pe is negative for x < 0 and positive for x > 0; the direction field clearly suggests a family of parabolas. AQA questions may ask you to match a direction field to a differential equation or to sketch a solution curve through a given point.
例如，若 dy/dx = x，则当 x < 0 时斜率为负，当 x > 0 时斜率为正；方向场清楚地暗示一族抛物线。AQA题目可能要求你将方向场与微分方程匹配，或绘制通过某一点的大致解曲线。
7. Separation of Variables | 分离变量法
The main algebraic method required at A-level is separation of variables. It applies to first-order equations that can be written as:
A-level阶段要求掌握的主要代数方法是分离变量法。它适用于可以写成以下形式的一阶方程：
dy/dx = g(x) h(y)
Rearrange by dividing by h(y) and multiplying by dx:
通过除以 h(y) 并乘以 dx 进行变形：
∫ 1/h(y) dy = ∫ g(x) dx
Worked example: Solve dy/dx = x/y, given y(0) = 2.
示例：求解 dy/dx = x/y，已知 y(0) = 2。
∫ y dy = ∫ x dx → ½y² = ½x² + C
by 2 and rename 2C as A: y² = x² + A. Using y(0) = 2 gives 4 = 0 + A, so A = 4. Therefore:
两边乘以2，并将 2C 改写为 A：y² = x² + A。利用 y(0) = 2 可得 4 = 0 + A，即 A = 4。因此：
y² = x² + 4 → y = √x² + 4
We take the positive root because the initial value is positive. Always check whether a positive or negative branch is required.
因为初值为正，所以我们取正根。一定要判断需要正分支还是负分支。
8. Exponential Growth and Decay | 指数增长与衰减
The simplest model of growth and decay is the differential equation:
最简单的增长与衰减模型是微分方程：
dP/dt = kP
where P is the quantity and k is a constant. Separating variables gives:
其中 P 是数量，
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