Separation of Variables: Finding the Original Function | 分离变量法求原函数

📚 Separation of Variables: Finding the Original Function | 分离变量法求原函数

In A-level Mathematics, many exam questions ask you to solve a first-order differential equation by separating the variables and then integrating both sides to recover the original function y = f(x). This article explains the method step by step with worked examples, common pitfalls, and exam tips.

在A-level数学考试中，许多题目要求你通过分离变量法求解一阶微分方程，并在两边积分后还原原函数 y = f(x)。本文将分步骤讲解这一方法，搭配例题、常见错误与考试技巧。

1. What Is Separation of Variables | 什么是分离变量法

Separation of variables is a technique for solving a differential equation of the form dy/dx = f(x)g(y), where the right-hand side can be written as a product of a function of x and a function of y.

分离变量法是求解形如 dy/dx = f(x)g(y) 的微分方程的一种技巧，其中等号右边可以写成 x 的函数与 y 的函数的乘积。

The key idea is to rearrange the equation so that every term involving y (and dy) is on one side, and every term involving x (and dx) is on the other side. Then we integrate both sides separately.

核心思路是重新整理方程，使所有含 y（及 dy）的项位于一边，所有含 x（及 dx）的项位于另一边，然后分别对两边积分。

It is essential that the expression can indeed be factorised as f(x)g(y). If the equation contains sums or differences that cannot be separated, this method does not apply directly.

必须注意表达式确实能因式分解为 f(x)g(y)。如果方程含有无法分离的加和或差项，则不能直接使用该方法。

2. The General Procedure | 一般求解步骤

Step 1: Write the derivative in the form dy/dx = f(x)g(y).

步骤1：将导数写成 dy/dx = f(x)g(y) 的形式。

Step 2: Divide both sides by g(y) and multiply by dx, giving (1/g(y)) dy = f(x) dx.

步骤2：两边除以 g(y) 并乘以 dx，得到 (1/g(y)) dy = f(x) dx。

Step 3: Place an integral sign on both sides: ∫ (1/g(y)) dy = ∫ f(x) dx.

步骤3：两边加上积分号：∫ (1/g(y)) dy = ∫ f(x) dx。

Step 4: Integrate the left side with respect to y and the right side with respect to x. Do not forget the constant of integration on one side.

步骤4：左边对 y 积分，右边对 x 积分。不要忘记在其中一边写上积分常数。

Step 5: Solve the resulting equation for y in terms of x if possible, to obtain the original function.

步骤5：如果可能，解出 y 关于 x 的表达式，从而得到原函数。

3. Worked Example 1 – Basic Case | 例1：基础情形

Solve the differential equation dy/dx = x²y, given that y > 0.

求解微分方程 dy/dx = x²y，已知 y > 0。

Step 1: The right-hand side is already a product: f(x) = x², g(y) = y. Separate the variables:

步骤1：右边已经是乘积形式：f(x) = x²，g(y) = y。分离变量：

(1/y) dy = x² dx

Step 2: Integrate both sides:

步骤2：两边积分：

∫ (1/y) dy = ∫ x² dx

This gives ln|y| = x³/3 + C.

得到 ln|y| = x³/3 + C。

Step 3: Since y > 0, we can drop the absolute value. Write y = e^(x³/3 + C) = A e^(x³/3), where A = e^C is a positive constant.

步骤3：由于 y > 0，可去掉绝对值。写成 y = e^(x³/3 + C) = A e^(x³/3)，其中 A = e^C 为正的常数。

So the original function is y = A e^(x³/3).

因此原函数为 y = A e^(x³/3)。

4. Worked Example 2 – With Initial Conditions | 例2：含初始条件

Solve dy/dx = 3y, given that y = 2 when x = 0.

求解 dy/dx = 3y，已知 x = 0 时 y = 2。

Separating variables:

分离变量：

(1/y) dy = 3 dx

Integrate:

积分：

ln|y| = 3x + C

Rewrite using the exponential function: y = e^(3x + C) = A e^(3x).

利用指数函数改写：y = e^(3x + C) = A e^(3x)。

Apply the initial condition y(0) = 2: 2 = A e⁰ = A, so A = 2.

代入初始条件 y(0) = 2：2 = A e⁰ = A，因此 A = 2。

Hence the particular solution is y = 2 e^(3x).

因此特解为 y = 2 e^(3x)。

5. Worked Example 3 – Using Partial Fractions | 例3：使用部分分式

Solve dy/dx = (1 – y)(y + 2), with y = 0 when x = 0.

求解 dy/dx = (1 – y)(y + 2)，当 x = 0 时 y = 0。

Separate the variables:

分离变量：

1/((1 – y)(y + 2)) dy = dx

The left side must be split by partial fractions. We seek constants P and Q such that:

左边需要用部分分式展开。我们寻求常数 P 和 Q，使得：

1/((1 – y)(y + 2)) = P/(1 – y) + Q/(y + 2)

Multiplying through gives 1 = P(y + 2) + Q(1 – y). Setting y = 1 gives 1 = 3P, so P = 1/3. Setting y = -2 gives 1 = 3Q, so Q = 1/3.

两边乘以后得到 1 = P(y + 2) + Q(1 – y)。令 y = 1 得 1 = 3P，故 P = 1/3；令 y = -2 得 1 = 3Q，故 Q = 1/3。

Therefore:

因此：

∫ [1/(3(1 – y)) + 1/(3(y + 2))] dy = ∫ dx

The integrals yield -(1/3) ln|1 – y| + (1/3) ln|y + 2| = x + C.

积分得到 -(1/3) ln|1 – y| + (1/3) ln|y + 2| = x + C。

Multiply by 3 and combine the logarithms: ln|(y + 2)/(1 – y)| = 3x + C’. Take exponentials: (y + 2)/(1 – y) = A e^(3x).

乘以3并合并对数：ln|(y + 2)/(1 – y)| = 3x + C’。取指数：(y + 2)/(1 – y) = A e^(3x)。

Using y(0) = 0: (0 + 2)/(1 – 0) = A, so A = 2. Solving for y gives y + 2 = 2 e^(3x)(1 – y), and therefore y(1 + 2 e^(3x)) = 2 e^(3x) – 2.

利用 y(0) = 0：(0 + 2)/(1 – 0) = A，故 A = 2。解出 y：y + 2 = 2 e^(3x)(1 – y)，因此 y(1 + 2 e^(3x)) = 2 e^(3x) – 2。

y = (2 e^(3x) – 2)/(1 + 2 e^(3x))

6. Worked Example 4 – Exponential Growth Model | 例4：指数增长模型

A population P grows at a rate proportional to its size. This is modelled by dP/dt = kP, where k is a positive constant. Find P as a function of t if P(0) = P₀.

某个种群数量的增长速率与其规模成正比。该模型为 dP/dt = kP，其中 k 为正的常数。若 P(0) = P

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Separation of Variables: Finding the Original Function | 分离变量法求原函数