📚 Separable First-Order Differential Equations in IB Mathematics | IB数学：可分离变量型非线性方程解法
Separable differential equations are among the most important techniques in the IB Mathematics Analysis and Approaches (AA) higher level curriculum. They appear frequently in Paper 2 and Paper 3, often embedded in contexts from physics, biology, or economics. In this article, we will explore the full method of solving first-order nonlinear differential equations of the form dy/dx = f(x)g(y), including handling of absolute values, constant combinations, and initial value problems.
可分离变量微分方程是IB数学分析与方法（AA）高级课程中最重要的技巧之一。它们频繁出现在Paper 2和Paper 3中，通常嵌入在物理、生物或经济学的实际情境里。本文将深入探讨形如 dy/dx = f(x)g(y) 的一阶非线性微分方程的完整解法，包括绝对值处理、常数合并以及初值问题的求解。
1. What Is a Separable Differential Equation? | 什么是可分离变量微分方程？
A first-order differential equation is said to be separable if it can be written in the form dy/dx = f(x) · g(y), where f is a function of x only and g is a function of y only. The word “separable” means that the two variables x and y can be completely separated to opposite sides of the equation.
如果一个一阶微分方程可以写成 dy/dx = f(x) · g(y) 的形式，其中 f 只是 x 的函数，g 只是 y 的函数，那么这个方程称为可分离变量的。所谓”可分离”意味着变量 x 和 y 可以被完全分到等式的两侧。
For example, the equation dy/dx = x²y is separable because we can write it as dy/dx = (x²)(y). However, the equation dy/dx = x + y is not separable because the right-hand side cannot be factored into a product of a function of x and a function of y.
例如，方程 dy/dx = x²y 是可分离的，因为它可以写成 dy/dx = (x²)(y)。但方程 dy/dx = x + y 是不可分离的，因为右边无法分解为 x 的函数与 y 的函数的乘积。
2. The Core Method: Separate, Integrate, Rearrange | 核心方法：分离、积分、整理
The standard procedure for solving a separable differential equation involves three steps. Step one: rearrange the equation so that all terms involving y are on one side with dy, and all terms involving x are on the other side with dx. Step two: integrate both sides with respect to the appropriate variable. Step three: solve for y explicitly if possible, or otherwise leave the solution in implicit form.
解可分离微分方程的标准流程包含三个步骤。第一步：重新整理方程，使所有含 y 的项与 dy 在一边，所有含 x 的项与 dx 在另一边。第二步：对两边分别关于对应变量积分。第三步：如果可能则显式解出 y，否则将解保留为隐式形式。
Consider the equation dy/dx = 3x²y. To separate the variables, we divide both sides by y (assuming y ≠ 0) and multiply both sides by dx:
考虑方程 dy/dx = 3x²y。为了分离变量，我们在两边同时除以 y（假设 y ≠ 0）并乘以 dx：
(1/y) dy = 3x² dx
Now integrate both sides. The left side gives ln|y| and the right side gives x³ + C:
现在对两边积分。左边得到 ln|y|，右边得到 x³ + C：
ln|y| = x³ + C
Exponentiating both sides, we obtain |y| = e^(x³+C) = e^C · e^(x³). Since e^C is a positive constant, we can absorb the absolute value and write the general solution as y = A·e^(x³), where A is an arbitrary real constant.
两边取指数，我们得到 |y| = e^(x³+C) = e^C · e^(x³)。由于 e^C 是正数常数，我们可以去掉绝对值符号，将通解写成 y = A·e^(x³)，其中 A 是任意实数常数。
3. Handling Nonlinearity in g(y) | 处理 g(y) 中的非线性项
When g(y) is not simply y but a nonlinear expression such as y², sin y, or √(1 – y²), the integration step becomes more involved. The key is to recognise which standard integral applies on the y-side.
当 g(y) 不仅仅是 y，而是一个非线性表达式如 y²、sin y 或 √(1-y²) 时，积分步骤会变得更加复杂。关键在于识别 y 侧对应的是哪个标准积分。
For instance, consider dy/dx = x·cos y. Separating variables gives:
例如，考虑 dy/dx = x·cos y。分离变量得到：
sec y dy = x dx
The left side integrates to ln|sec y + tan y|, a standard result. The right side integrates to x²/2 + C. Therefore the implicit solution is:
左边的积分结果为 ln|sec y + tan y|，这是一个标准结果。右边积分得到 x²/2 + C。因此隐式解为：
ln|sec y + tan y| = x²/2 + C
Notice that solving this equation for y explicitly is nontrivial and often unnecessary. The IB exam accepts answers in implicit form as long as they are obtained from a correct separation and integration.
注意，从这个方程显式解出 y 并不容易，而且往往没有必要。IB考试只要分离和积分过程正确，就接受隐式形式的解。
4. The Role of Initial Conditions | 初值条件的作用
A first-order differential equation has infinitely many solutions distinguished by the constant of integration. An initial condition, typically given as y(x₀) = y₀, selects exactly one particular solution from the family of general solutions.
一阶微分方程有无限多个解，它们由积分常数区分。一个初值条件，通常给为 y(x₀) = y₀，从通解族中确定唯一的特解。
In IB problems, it is often easier to substitute the initial condition into the general solution after integration rather than before. However, some students find it helpful to set the limits of integration directly using the initial condition. For example, from dy/dx = 2xy with y(0) = 1, we write:
在IB题目中，通常更容易在积分之后将初值条件代入通解，而不是在积分之前。然而，有些学生发现直接用初值条件设定积分限更方便。例如，对 dy/dx = 2xy 且 y(0) = 1，我们写成：
∫₁^y (1/t) dt = ∫₀^x 2s ds
Evaluating gives ln y – ln 1 = x² – 0, so ln y = x² and hence y = e^(x²). Note that since the initial condition gives y(0) = 1 > 0, we do not need the absolute value in this case.
求值得到 ln y – ln 1 = x² – 0，因此 ln y = x²，从而 y = e^(x²)。注意，因为初值条件给出 y(0) = 1 > 0，这种情况下不需要绝对值符号。
5. When dy/dx Is Given as a Ratio | 当 dy/dx 以比值形式给出时
Some IB exam questions do not present the differential equation in the explicit form dy/dx = f(x)g(y). Instead, they may write an equation involving differentials, such as x dy = y dx, or a relation like (1 + y²)dx = x dy.
有些IB考题并不以显式形式 dy/dx = f(x)g(y) 给出微分方程，而是写出含微分的方程，如 x dy = y dx，或关系式 (1 + y²)dx = x dy。
To solve such equations, we first rewrite them in the form dy/dx by dividing both sides by dx. For example, (1 + y²)dx = x dy can be rewritten as dy/dx = (1 + y²)/x. Then we separate:
要解此类方程，我们首先通过两边除以 dx 改写为 dy/dx 的形式。例如，(1 + y²)dx = x dy 可以改写为 dy/dx = (1 + y²)/x。然后分离变量：
dy/(1 + y²) = (1/x) dx
Integrating the left side yields arctan y, and integrating the right side yields ln|x| + C. Hence the implicit solution is:
左边积分得到 arctan y，右边积分得到 ln|x| + C。因此隐式解为：
arctan y = ln|x| + C
Taking the tangent of both sides gives y = tan(ln|x| + C), provided the domain is appropriate.
两边取正切得到 y = tan(ln|x| + C)，前提是定义域合适。
6. Common IB Exam Trap: Losing the Singular Solution | 常见IB考试陷阱：丢失奇异解
When we divide by g(y) to separate variables, we implicitly assume that g(y) ≠ 0. If g(y) = 0 for some constant value y = k, then y = k is itself a solution to the differential equation, because dy/dx = 0 and f(x)·g(k) = f(x)·0 = 0.
当我们除以 g(y) 来分离变量时，隐含地假设了 g(y) ≠ 0。如果 g(y) = 0 对某个常数值 y = k 成立，那么 y = k 本身是微分方程的一个解，因为此时 dy/dx = 0 且 f(x)·g(k) = f(x)·0 = 0。
Consider dy/dx = y² – 1. Separating gives dy/(y²-1) = dx, which requires y² ≠ 1. The general solution is found by partial fractions:
考虑 dy/dx = y² – 1。分离得到 dy/(y²-1) = dx，这要求 y² ≠ 1。通解通过部分分式求得：
(1/2)ln|(y-1)/(y+1)| = x + C
However, the constant functions y = 1 and y = -1 are also valid solutions. These singular solutions are sometimes missed if we blindly divide. In an initial value problem, check whether the given initial value y₀ makes g(y₀) = 0; if so, the singular solution is the answer.
然而，常数函数 y = 1 和 y = -1 也是有效解。如果我们盲目地做除法，这些奇异解有时会被忽略。在初值问题中，检查给定的初值 y₀ 是否使 g(y₀) = 0；如果是，那么奇异解就是答案。
7. Worked Example: Logistic-Type Growth | 例题精解：逻辑斯谛型增长
A classic IB context involves population growth modelled by dP/dt = kP(1 – P/M), where P(t) is the population, M is the carrying capacity, and k is a positive constant. This is a separable nonlinear equation because the right-hand side is a product of P and (1 – P/M).
一个经典的IB情境是用 dP/dt = kP(1 – P/M) 来模拟种群增长，其中 P(t) 是种群数量，M 是环境承载力，k 是正常数。这是一个可分离的非线性方程，因为右边是 P 与 (1 – P/M) 的乘积。
Suppose k = 0.1, M = 1000, and P(0) = 100. We separate the variables:
假设 k = 0.1，M = 1000，P(0) = 100。我们分离变量：
dP/[P(1 – P/1000)] = 0.1 dt
Using partial fractions, 1/[P(1 – P/1000)] = 1/P + 1/(1000 – P). Therefore:
使用部分分式，1/[P(1 – P/1000)] = 1/P + 1/(1000 – P)。因此：
∫ [1/P + 1/(1000 – P)] dP = ∫ 0.1 dt
This gives ln P – ln(1000 – P) = 0.1t + C, that is, ln[P/(1000 – P)] = 0.1t + C. Using P(0) = 100 gives C = ln(100/900) = ln(1/9). Hence:
得到 ln P – ln(1000 – P) = 0.1t + C，即 ln[P/(1000 – P)] = 0.1t + C。利用 P(0) = 100 得到 C = ln(100/900) = ln(1/9)。因此：
P/(1000 – P) = (1/9)e^(0.1t)
Solving for P explicitly, we obtain P(t) = 1000e^(0.1t)/(9 + e^(0.1t)). Notice that as t→∞, P→1000, consistent with the carrying capacity.
显式解出 P，我们得到 P(t) = 1000e^(0.1t)/(9 + e^(0.1t))。注意当 t→∞ 时 P→1000，与环境承载力一致。
8. Using Substitutions to Reach Separable Form | 运用换元化为可分离形式
Occasionally, an IB question presents an equation that is not immediately separable but becomes separable after a clever substitution. One common type is dy/dx = f(ax + by + c), using the substitution u = ax + by + c.
有时，IB题目给出的方程并非直接可分离，但通过巧妙的变量替换可以化为可分离形式。一种常见类型是 dy/dx = f(ax + by + c)，使用换元 u = ax + by + c。
For example, consider dy/dx = (x + y)². Let u = x + y. Then du/dx = 1 + dy/dx, so dy/dx = du/dx – 1. The equation becomes:
例如，考虑 dy/dx = (x + y)²。令 u = x + y。则 du/dx = 1 + dy/dx，所以 dy/dx = du/dx – 1。方程变为：
du/dx – 1 = u², that is, du/dx = u² + 1
This is now separable: du/(u² + 1) = dx. Integrating gives arctan u = x + C, so u = tan(x + C). Reverting to original variables, x + y = tan(x + C), or y = tan(x + C) – x.
现在方程可分离：du/(u² + 1) = dx。积分得到 arctan u = x + C，所以 u = tan(x + C)。代回原变量，x + y = tan(x + C)，即 y = tan(x + C) – x。
9. Summary of Techniques for Nonlinear Cases | 非线性情形的技巧总结
The following table summarises the strategies for handling common nonlinear expressions that appear in separable equations:
下表总结了处理可分离方程中常见非线性表达式的策略：
| Form of g(y) | g(y) 的形式 | Integral Needed | 所需积分 | Result | 结果 |
| 1 / y | ∫ (1/y) dy | ln|y| + C |
| 1 / (y² + a²) | ∫ dy / (y² + a²) | (1/a) arctan(y/a) + C |
| 1 / √(a² – y²) | ∫ dy / √(a² – y²) | arcsin(y/a) + C |
| sec y | ∫ sec y dy | ln|sec y + tan y| + C |
| 1 / [y(1 – y/M)] | ∫ dy / [y(1 – y/M)] | ln y – ln(1 – y/M) + C |
In addition to memorising these integrals, students should watch for the special case where g(y) contains terms that factor, making partial fractions necessary. Always verify that solutions obtained from integration are consistent with the domain of the original equation.
除了记住这些积分之外，学生还需要注意 g(y) 含有可因式分解项的情况，此时需要使用部分分式。永远要验证通过积分得到的解与原始方程的定义域是否一致。
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