📚 A-Level Mathematics: Methods & Steps for Solving Differential Equations | A-Level 数学:解微分方程的方法与步骤
Differential equations are a cornerstone of A-Level Mathematics and Further Mathematics. They describe how a quantity changes with respect to another variable, and mastering the standard solution techniques is essential for both pure mathematics questions and real-world modelling problems in exams.
微分方程是 A-Level 数学与进阶数学的核心内容之一。它描述了一个量相对于另一个量的变化规律。无论是纯数学题还是实际建模题,熟练掌握标准的求解方法都是取得高分的关键。
1. What is a Differential Equation? | 什么是微分方程?
A differential equation (DE) is an equation that involves an unknown function and one or more of its derivatives. For example, dy/dx = 3x² + 2 is a first-order DE, while d²y/dx² – 5 dy/dx + 6y = 0 is a second-order DE. The order of a DE is the highest derivative it contains.
微分方程是含有未知函数及其一阶或高阶导数的方程。例如,dy/dx = 3x² + 2 是一阶微分方程,而 d²y/dx² – 5 dy/dx + 6y = 0 是二阶微分方程。方程的阶数等于其中出现的最高阶导数的阶数。
In A-Level Mathematics, you are mainly required to solve first-order differential equations — those containing only dy/dx — using two core techniques: separation of variables and the integrating factor method. Some boards also ask for homogeneous first-order DEs solved by substitution.
在 A-Level 数学中,主要要求求解一阶微分方程(只含 dy/dx 的方程),核心方法有变量分离法与积分因子法。部分考试局还要求掌握用换元法求解齐次一阶微分方程。
2. Separation of Variables | 变量分离法
If a first-order DE can be written in the form dy/dx = g(x) · f(y), we can separate the variables. The key step is to rearrange so that every y-term is with dy and every x-term is with dx:
若一阶微分方程可以写成 dy/dx = g(x) · f(y) 的形式,我们就可以分离变量。关键步骤是重新整理,使所有含 y 的项与 dy 放在一起,所有含 x 的项与 dx 放在一起:
∫ 1/f(y) dy = ∫ g(x) dx + C
- Rearrange the equation to separate y and x. | 重新整理方程,将 y 与 x 分离到等号两侧。
- Integrate both sides: the left with respect to y, the right with respect to x. | 两边分别积分:左边对 y,右边对 x。
- Add the constant of integration C on one side only. | 只在其中一侧加上积分常数 C。
- Solve for y explicitly if possible; otherwise leave the answer as an implicit equation. | 如可能,尽量解出 y;否则保留为隐式方程。
Worked example: solve dy/dx = 2xy.
典型例题:求解 dy/dx = 2xy。
1/y dy = 2x dx → ∫ 1/y dy = ∫ 2x dx → ln|y| = x² + C
Since C is arbitrary, we may write ln|y| = x² + C and then exponentiate both sides:
由于 C 为任意常数,可将 ln|y| = x² + C 两边取指数:
|y| = e^(x² + C) → y = A e^(x²), where A = ±e^C
The absolute value disappears because the constant A can absorb the positive or negative sign.
绝对值符号可以去掉,因为常数 A 已经包含了正负号。
3. Integrating Factor Method | 积分因子法
For a first-order linear DE written in the standard form dy/dx + P(x)y = Q(x), we multiply both sides by the integrating factor (IF):
对于标准形式的一阶线性微分方程 dy/dx + P(x)y = Q(x),在等式两边同时乘以积分因子(IF):
IF = e^(∫ P(x) dx)
This transforms the left-hand side into the derivative of a product, so the equation becomes:
这样做可以将左边化为一个乘积的导数,方程变为:
d/dx [ y · e^(∫ P(x) dx) ] = Q(x) · e^(∫ P(x) dx)
Then integrate both sides and solve for y. Do not forget the constant of integration.
然后两边积分并解出 y。不要忘记积分常数。
Worked example: solve dy/dx + 2y = e^x, given y(0) = 1.
典型例题:求解 dy/dx + 2y = e^x,已知 y(0) = 1。
Here P(x) = 2, so IF = e^(∫ 2 dx) = e^(2x).
这里 P(x) = 2,所以 IF = e^(∫ 2 dx) = e^(2x)。
d/dx ( y e^(2x) ) = e^x · e^(2x) = e^(3x)
y e^(2x) = ∫ e^(3x) dx = ⅓ e^(3x) + C
y = ⅓ e^x + C e^(-2x)
Using y(0) = 1: 1 = ⅓ + C, so C = ⅔. Therefore the particular solution is y = ⅓ e^x + ⅔ e^(-2x).
代入初值 y(0) = 1:1 = ⅓ + C,得 C = ⅔。因此特解为 y = ⅓ e^x + ⅔ e^(-2x)。
4. Homogeneous Differential Equations | 齐次微分方程
A first-order DE is homogeneous if it can be expressed as dy/dx = F(y/x). A typical example is dy/dx = (x² + y²)/(xy).
如果一阶微分方程可以写成 dy/dx = F(y/x) 的形式,则称其为齐次微分方程。典型例子如 dy/dx = (x² + y²)/(xy)。
Method: use the substitution y = vx, where v is a new function of x. By the product rule:
解法:令 y = vx,其中 v 是 x 的新函数。由乘法法则可得:
dy/dx = v + x dv/dx
Substituting into dy/dx = F(y/x) gives v + x dv/dx = F(v), which is a separable equation in v and x. Solve it, then replace v with y/x.
代回 dy/dx = F(y/x) 得到 v + x dv/dx = F(v),这是关于 v 与 x 的可分离变量方程。求出 v 后,再代回 v = y/x。
Worked example: solve dy/dx = (x² + y²)/(xy).
典型例题:求解 dy/dx = (x² + y²)/(xy)。
First rewrite the right-hand side:
首先改写右边:
dy/dx = x/y + y/x = 1/v + v
With y = vx, we get:
令 y = vx,得到:
v + x dv/dx = v + 1/v → x dv/dx = 1/v → v dv = dx/x
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