📚 IB Math: Variable Substitution in Differential Equations | IB数学：微分方程中的变量代换法
Variable substitution is a powerful technique for solving differential equations that do not fall directly into standard separable or linear forms. By introducing a new variable that simplifies the structure of the equation, we can transform a seemingly intractable problem into one we already know how to solve. This article explores the main substitution strategies in the IB Mathematics curriculum.
变量代换法是求解微分方程的一种强大工具，适用于那些不能直接归为可分离变量型或线性型的方程。通过引入一个新变量来简化方程的结构，我们可以将看似棘手的问题转化为已经掌握求解方法的形式。本文将探讨IB数学课程中主要的变量代换策略。
1. Why Use Variable Substitution? | 为何使用变量代换？
A differential equation often presents itself in a form that is not immediately recognisable as separable or linear. However, a clever change of variable can reveal an underlying structure. The goal is to choose a new variable v such that the derivative dy/dx becomes a simpler expression in terms of v and x (or v and y), thereby converting the original equation into a solvable type.
微分方程通常呈现出的形式，使我们无法立即辨认其是否属于可分离变量型或线性型。然而，巧妙的变量代换可以揭示其内在结构。我们的目标是选择一个新的变量 v，使得导数 dy/dx 变成关于 v 与 x（或 v 与 y）的更简单表达式，从而将原方程转化为可求解的类型。
2. Homogeneous First-Order Differential Equations | 一阶齐次微分方程
A first-order differential equation dy/dx = f(x, y) is called homogeneous if f(x, y) can be written as a function of the ratio y/x alone, i.e. dy/dx = g(y/x). Such equations are not separable in x and y directly, but they become separable after the substitution y = vx.
如果一个一阶微分方程 dy/dx = f(x, y) 中的 f(x, y) 可以仅表示为 y/x 的函数，即 dy/dx = g(y/x)，则称该方程为齐次微分方程。此类方程不能直接对 x 和 y 分离变量，但在代换 y = vx 之后即可化为可分离变量的形式。
3. The Substitution y = vx | 代换 y = vx 的基本步骤
Let y = vx, where v is a function of x. Differentiating with respect to x using the product rule gives: dy/dx = v + x(dv/dx). Substituting both y and dy/dx into the original equation dy/dx = g(y/x) = g(v) yields an equation in v and x, which is separable:
设 y = vx，其中 v 是 x 的函数。运用乘积法则对 x 求导得：dy/dx = v + x(dv/dx)。将 y 与 dy/dx 同时代入原方程 dy/dx = g(y/x) = g(v)，得到关于 v 和 x 的方程，该方程是可分离变量的：
v + x(dv/dx) = g(v) → x(dv/dx) = g(v) − v → ∫ dv/(g(v) − v) = ∫ dx/x
4. Worked Example: Solving a Homogeneous Equation | 实例：求解齐次方程
Solve the differential equation dy/dx = (x² + y²)/(xy), for x > 0, y > 0.
求解微分方程 dy/dx = (x² + y²)/(xy)，其中 x > 0，y > 0。
First, rewrite the right-hand side: (x² + y²)/(xy) = x²/(xy) + y²/(xy) = x/y + y/x = 1/(y/x) + y/x. Let v = y/x, so y/x = v and x/y = 1/v. Thus dy/dx = v + 1/v. This agrees with the homogeneous form g(v).
首先，重写等号右端：(x² + y²)/(xy) = x²/(xy) + y²/(xy) = x/y + y/x = 1/(y/x) + y/x。设 v = y/x，则 y/x = v 且 x/y = 1/v。于是 dy/dx = v + 1/v，这符合齐次形式 g(v)。
Since y = vx, dy/dx = v + x(dv/dx). Equating: v + x(dv/dx) = v + 1/v, so x(dv/dx) = 1/v. Separate variables: v dv = dx/x. Integrate both sides: ∫ v dv = ∫ dx/x, giving v²/2 = ln|x| + C.
由于 y = vx，所以 dy/dx = v + x(dv/dx)。令两边相等：v + x(dv/dx) = v + 1/v，故 x(dv/dx) = 1/v。分离变量：v dv = dx/x。两边同时积分得：∫ v dv = ∫ dx/x，即 v²/2 = ln|x| + C。
Substitute v = y/x back: (y/x)²/2 = ln|x| + C. Multiply by 2: y²/x² = 2ln|x| + 2C. Using the initial condition y(1) = 1 gives C = ½. Thus y² = x²(2ln|x| + 1).
将 v = y/x 代回： (y/x)²/2 = ln|x| + C，两边乘2得 y²/x² = 2ln|x| + 2C。利用初值条件 y(1) = 1 得到 C = ½。因此 y² = x²(2ln|x| + 1)。
5. Recognising Equations Reducing to Homogeneous Form | 可化为齐次方程的方程识别
Some equations that appear non-homogeneous can be transformed into homogeneous ones via a translation of variables. Consider dy/dx = (ax + by + c)/(dx + ey + f), where the determinant ae − bd ≠ 0. The substitution X = x − h, Y = y − k, with suitable h and k chosen to eliminate the constants c and f, reduces the equation to a homogeneous form in X and Y, solvable by Y = VX.
某些看似非齐次的方程可以通过变量平移化为齐次方程。考虑 dy/dx = (ax + by + c)/(dx + ey + f)，其中行列式 ae − bd ≠ 0。选取适当的 h 和 k，令 X = x − h，Y = y − k 以消去常数 c 与 f，即可将方程化为关于 X 和 Y 的齐次形式，再用 Y = VX 求解。
6. Linear Equations and the Need for an Integrating Factor (Revision) | 线性方程与积分因子（回顾）
A first-order linear differential equation has the form dy/dx + P(x)y = Q(x). This is solved using an integrating factor I = e^(∫P(x)dx). Multiplying through by I converts the left side into d/dx(I·y), which integrates directly. The substitution method extends this idea to nonlinear equations that can be reduced to linear form.
一阶线性微分方程具有形式 dy/dx + P(x)y = Q(x)，其解法依赖于积分因子 I = e^(∫P(x)dx)。等式两边乘以 I 后，左端可化为 d/dx(I·y)，从而直接积分求解。变量代换法将这一思想推广到可以化为线性形式的非线性方程中。
7. Bernoulli Differential Equations | 伯努利微分方程
A Bernoulli equation is of the form dy/dx + P(x)y = Q(x)yⁿ, where n is a real number (n ≠ 0, n ≠ 1). When n = 0 the equation is linear; when n = 1 it is separable. For other values of n, the appropriate substitution is u = y^(1−n).
伯努利方程具有形式 dy/dx + P(x)y = Q(x)yⁿ，其中 n 为实数（n ≠ 0，n ≠ 1）。当 n = 0 时方程为线性；当 n = 1 时可分离变量。对于 n 的其他取值，适当代换为 u = y^(1−n)。
8. Deriving the Bernoulli Substitution | 伯努利代换的推导
Given dy/dx + P(x)y = Q(x)yⁿ, divide the entire equation by yⁿ to obtain y⁻ⁿ(dy/dx) + P(x)y^(1−n) = Q(x). Let u = y^(1−n). Then du/dx = (1−n)y⁻ⁿ(dy/dx), which means y⁻ⁿ(dy/dx) = (1/(1−n)) du/dx. Substituting yields a linear equation in u:
对于 dy/dx + P(x)y = Q(x)yⁿ，先将方程各项除以 yⁿ 得 y⁻ⁿ(dy/dx) + P(x)y^(1−n) = Q(x)。设 u = y^(1−n)，则 du/dx = (1−n)y⁻ⁿ(dy/dx)，所以 y⁻ⁿ(dy/dx) = (1/(1−n)) · du/dx。代入后得到关于 u 的线性方程：
du/dx + (1−n)P(x)u = (1−n)Q(x)
This linear form in u can then be solved using an integrating factor.
这个关于 u 的线性方程随后即可用积分因子法求解。
9. Worked Example: Solving a Bernoulli Equation | 实例：求解伯努利方程
Solve: dy/dx − y = xy².
求解：dy/dx − y = xy²。
Here P(x) = −1, Q(x) = x, n = 2. Let u = y^(1−2) = y⁻¹. Divide the original equation by y²: y⁻²(dy/dx) − y⁻¹ = x. Since u = y⁻¹, du/dx = −y⁻²(dy/dx), so y⁻²(dy/dx) = −du/dx. Substitution gives:
此处 P(x) = −1，Q(x) = x，n = 2。令 u = y^(1−2) = y⁻¹。将原方程除以 y²：y⁻²(dy/dx) − y⁻¹ = x。由于 u = y⁻¹，du/dx = −y⁻²(dy/dx)，故 y⁻²(dy/dx) = −du/dx。代入后得到：
−du/dx − u = x → du/dx + u = −x
This is linear in u with P = 1. Integrating factor I = e^(∫1 dx) = eˣ. Multiply: eˣ(du/dx) + eˣu = −xeˣ, i.e. d/dx(eˣ·u) = −xeˣ. Integrate: eˣ·u = −∫xeˣ dx = −(x−1)eˣ + C. Thus u = 1 − x + Ce⁻ˣ. Finally, back-substitute u = 1/y:
这是关于 u 的线性方程，其中 P = 1。积分因子 I = e^(∫1 dx) = eˣ。两边乘以 I：eˣ(du/dx) + eˣu = −xeˣ，即 d/dx(eˣ·u) = −xeˣ。积分得：eˣ·u = −∫xeˣ dx = −(x−1)eˣ + C。因此 u = 1 − x + Ce⁻ˣ。最后代回 u = 1/y：
1/y = 1 − x + Ce⁻ˣ → y = 1/(1 − x + Ce⁻ˣ)
10. Substitutions for Special Non-Linear Terms | 针对特殊非线性项的代换
Sometimes a differential equation contains a repeated expression such as ax + by or a trigonometric combination. The substitution z = ax + by reduces dy/dx to dz/dx = a + b(dy/dx), eliminating y entirely and often producing a separable equation in z and x. For example, dy/dx = (x + y)² is not separable directly, but with z = x + y, we get dz/dx = 1 + dy/dx = 1 + z² — a separable equation.
有时微分方程含有重复出现的表达式，如 ax + by 或某种三角组合。令 z = ax + by，则 dz/dx = a + b(dy/dx)，可完全消去 y，通常得到关于 z 和 x 的可分离方程。例如，dy/dx = (x + y)² 不能直接分离，但设 z = x + y 后，dz/dx = 1 + dy/dx = 1 + z²，即化为可分离方程。
11. Common Pitfalls and Exam Tips | 常见误区与考试提示
Students often confuse the homogeneous equation (type dy/dx = f(y/x)) with a homogeneous linear equation (dy/dx + P(x)y = 0), although both share the word ‘homogeneous’. In this substitution context, the key test is whether f(x, y) depends only on y/x.
学生经常混淆齐次方程（dy/dx = f(y/x) 型）与齐次线性方程（dy/dx + P(x)y = 0），尽管两者共享“齐次”一词。在变量代换的语境下，关键判断标准是 f(x, y) 是否仅取决于 y/x。
Check conditions carefully: for a Bernoulli equation, n must not equal 0 or 1. Also be careful with signs when computing du/dx for u = y^(1−n) — the factor (1−n) appears in both the derivative and the transformed equation. In examinations, always verify your substitution by differentiating it with respect to x before substituting into the original equation.
仔细检查条件：对于伯努利方程，n 不能等于 0 或 1。同时，在计算 u = y^(1−n) 的导数 du/dx 时注意符号——因子 (1−n) 会同时出现在导数和变换后的方程中。考试中，务必先对代换式关于 x 求导，确认后再代入原方程。
Finally, do not forget to back-substitute to the original variables. The final answer must be expressed in terms of x and y unless the question explicitly requests otherwise.
最后，不要忘记代回原变量。除非题目另有明确要求，最终答案必须用 x 和 y 表示。
12. Practice Questions | 练习题目
- Solve dy/dx = (x + 2y)/(2x + y) using the substitution y = vx. (Hint: divide numerator and denominator by x first)
- Solve dy/dx + (2/x)y = x³y², for x > 0, using the Bernoulli method.
- Solve dy/dx = cos²(x − y) by letting u = x − y.
- 用代换 y = vx 求解 dy/dx = (x + 2y)/(2x + y)。（提示：先将分子分母同除以 x）
- 用伯努利方法求解 dy/dx + (2/x)y = x³y²，其中 x > 0。
- 令 u = x − y 来求解 dy/dx = cos²(x − y)。
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