Mastering Differential Equations for Edexcel A-Level Pure 4 | 精通爱德思A-Level纯数4：微分方程

📚 Mastering Differential Equations for Edexcel A-Level Pure 4 | 精通爱德思A-Level纯数4：微分方程

Differential equations connect an unknown function with its own derivatives, allowing us to model continuous change in population, temperature, finance, and many other fields. In Edexcel A-Level Pure Mathematics 4, the focus is on forming, solving, and interpreting first-order separable differential equations, especially in applied contexts.

微分方程将未知函数与其自身的导数联系起来，使我们能够对人口、温度、金融及许多其他领域中的连续变化进行建模。在爱德思A-Level纯数4中，重点在于建立、求解和解释一阶可分离变量微分方程，尤其是在实际应用情境中。

1. What Is a Differential Equation? | 什么是微分方程？

A differential equation is an equation that involves an unknown function and one or more of its derivatives. In A-Level Pure 4, the unknown function is often written as y = f(x), and the equation contains dy/dx or occasionally a second derivative.

微分方程是包含未知函数及其一个或多个导数的方程。在A-Level纯数4中，未知函数通常写作 y = f(x)，方程中含有 dy/dx，偶尔也会出现二阶导数。

The order of a differential equation is the highest derivative that appears. For example, dy/dx = 3x² is a first-order equation, while d²y/dx² + 5 dy/dx = 0 is a second-order equation. In Pure 4, only first-order separable equations are examined in depth.

微分方程的阶是指方程中出现的最高导数的阶数。例如，dy/dx = 3x² 是一阶方程，而 d²y/dx² + 5 dy/dx = 0 是二阶方程。在纯数4中，只深入考查一阶可分离变量方程。

dy/dx = 3x² – 2y

2. First-Order Separable Equations | 一阶可分离变量方程

A first-order differential equation is said to be separable if it can be rearranged into the form dy/dx = g(x)h(y), where g(x) is a function of x only and h(y) is a function of y only. This structure means the variables x and y can be separated onto opposite sides of the equation before integration.

如果一阶微分方程可以整理为 dy/dx = g(x)h(y) 的形式，其中 g(x) 只含有 x，h(y) 只含有 y，就称该方程是可分离变量的。这种结构意味着在积分之前可以把变量 x 和 y 分离到方程的两边。

For example, dy/dx = x²y is separable because g(x) = x² and h(y) = y. However, dy/dx = x + y is not separable because the right-hand side cannot be written as a product of a function of x alone and a function of y alone.

例如，dy/dx = x²y 是可分离变量的，因为 g(x) = x²，h(y) = y。而 dy/dx = x + y 不是可分离变量的，因为右边无法写成仅含 x 的函数与仅含 y 的函数的乘积。

dy/dx = g(x)h(y)

3. General Method of Separation of Variables | 分离变量法的一般步骤

The first step is to rewrite the derivative using differential notation: dy/dx can be treated as a ratio of dy and dx. Next, divide both sides by h(y) and multiply both sides by dx so that all y terms are on the left and all x terms are on the right.

第一步是用微分记号重写导数：dy/dx 可以看作 dy 与 dx 的比值。接下来，两边同时除以 h(y)，再同时乘以 dx，使所有含 y 的项都在左边，所有含 x 的项都在右边。

After separation, the equation becomes (1/h(y)) dy = g(x) dx. Then integrate both sides separately. Do not forget to include the constant of integration on one side after integration.

分离后，方程变为 (1/h(y)) dy = g(x) dx。然后两边分别积分。注意积分后要在其中一侧加上积分常数。

∫ 1/h(y) dy = ∫ g(x) dx

This method works because the left side depends only on y and the right side only on x, so both sides must equal the same constant for all x and y values.

这种方法之所以有效，是因为左边只依赖于 y，右边只依赖于 x，因此对所有 x 和 y 值两边必须等于同一个常数。

4. Finding the Arbitrary Constant | 求任意常数

When both sides are integrated, the left integral gives a constant and the right integral gives another constant. These two constants are usually combined into a single arbitrary constant C written on one side of the equation only.

当两边都积分后，左边积分产生一个常数，右边积分也产生一个常数。通常将这两个常数合并成一个任意常数 C，只写在方程的一侧。

For example, solving dy/dx = 2x gives ∫ 1 dy = ∫ 2x dx, so y = x² + C. The constant C represents infinitely many solution curves, all differing by a vertical translation.

例如，求解 dy/dx = 2x 得到 ∫ 1 dy = ∫ 2x dx，因此 y = x² + C。常数 C 代表无穷多条解曲线，它们之间只相差一个垂直平移。

y = x² + C

5. Particular Solutions from Initial Conditions | 由初始条件求特解

A general solution contains an arbitrary constant C. If a question gives an initial condition, such as y(x₀) = y₀, substitute the given x and y values into the general solution to find C. This gives the particular solution that passes through the specified point.

通解含有一个任意常数 C。如果题目给出了初始条件，例如 y(x₀) = y₀，就将给定的 x 和 y 值代入通解求出 C。这样就得到经过指定点的特解。

For example, if dy/dx = 3x² and the condition is y(1) = 4, integrating gives y = x³ + C. Substituting x = 1 and y = 4 gives 4 = 1 + C, so C = 3. The particular solution is y =

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