📚 Solving Differential Equations with the y = vx Substitution | 利用 y = vx 换元求解微分方程
First-order differential equations appear throughout the IB Mathematics Analysis & Approaches HL syllabus. Many of them are not directly separable, but can be transformed into a separable form using a clever substitution. One of the most important techniques is the substitution y = vx, used for solving homogeneous differential equations.
在 IB 数学分析与方法 HL 课程中，一阶微分方程频繁出现。许多方程虽然不能直接分离变量，但借助巧妙的换元可以化为可分离变量的形式。其中最重要的技巧之一就是利用 y = vx 换元，用于求解齐次微分方程。
1. What is a Homogeneous Differential Equation? | 什么是齐次微分方程？
A first-order differential equation is called homogeneous of the same degree if the right-hand side can be written as a function of y/x only. In other words, it has the form:
如果一个一阶微分方程的右端可以写成仅关于 y/x 的函数，那么这个方程就称为齐次微分方程。也就是说，它具有如下形式：
dy/dx = f(y/x)
For example, dy/dx = (x² + 3y²)/(2xy) can be rewritten by dividing the numerator and denominator by x², giving f(v) with v = y/x. This is a homogeneous equation.
例如，dy/dx = (x² + 3y²)/(2xy) 可以通过分子分母同除以 x² 来改写，得到以 v = y/x 为变量的 f(v)。这就是齐次方程。
2. How to Recognise a Homogeneous Equation | 如何识别齐次方程
There are two quick tests. First, try to rewrite the expression so that y always appears together with x as the ratio y/x. Second, replace x by tx and y by ty. If every term gains the same factor tⁿ, then the equation is homogeneous of degree n.
这里有两种快速检验方法。第一种，尝试改写表达式，使 y 总是与 x 一起出现，即 y/x 的比例形式。第二种，用 tx 替换 x、用 ty 替换 y。如果每一项都乘上相同的因子 tⁿ，则该方程是 n 次齐次的。
Example: dy/dx = (y² − x²)/(xy). Dividing by x² gives ((y/x)² − 1)/(y/x) = (v² − 1)/v, so it is homogeneous.
例：dy/dx = (y² − x²)/(xy)，同除以 x² 得到 ((y/x)² − 1)/(y/x) = (v² − 1)/v，因此它是齐次的。
Example: dy/dx = (x + y + 1)/(x − y). The constant term 1 breaks the homogeneity; here y = vx does not work directly.
例：dy/dx = (x + y + 1)/(x − y)。常数项 1 破坏了齐次性；此时直接使用 y = vx 是行不通的。
3. Deriving the Key Substitution | 推导核心换元公式
Let v = y/x, so that y = vx. Differentiate y = vx with respect to x using the product rule:
设 v = y/x，即 y = vx。对 y = vx 关于 x 求导，使用乘积法则：
dy/dx = v + x·dv/dx
This single formula is the engine of the whole method. It replaces both y and dy/dx in the original equation and reduces it to a differential equation in v and x, which is often separable.
这一个公式就是整个方法的核心。它同时替换了原方程中的 y 和 dy/dx，将方程化为关于 v 和 x 的微分方程，而后者通常可以分离变量。
4. Step-by-Step Procedure | 分步解题步骤
Step 1: Confirm the equation can be written as dy/dx = f(y/x). If not, try another method.
步骤 1：确认方程可以写成 dy/dx = f(y/x) 的形式。如果不能，则尝试其他方法。
Step 2: Let y = vx, and write dy/dx = v + x·dv/dx.
步骤 2：令 y = vx，并写出 dy/dx = v + x·dv/dx。
Step 3: Substitute into the differential equation. Simplify to obtain x·dv/dx = g(v).
步骤 3：代入原微分方程，化简得到 x·dv/dx = g(v)。
Step 4: Separate variables and integrate both sides: ∫ 1/g(v) dv = ∫ 1/x dx.
步骤 4：分离变量并两边积分：∫ 1/g(v) dv = ∫ 1/x dx。
Step 5: After integration, replace v by y/x to return to x and y.
步骤 5：积分完成后，用 v = y/x 代回，回到 x 和 y 的表达形式。
Step 6: Apply any initial condition, if given, to find the particular solution.
步骤 6：如果题目给出初始条件，则代入求出特解。
5. Worked Example 1 | 例题精讲 1
Solve the differential equation:
求解微分方程：
dy/dx = (x² + 3y²)/(2xy)
Divide the numerator and denominator by x²:
分子分母同除以 x²：
dy/dx = (1 + 3v²)/(2v), where v = y/x
Using dy/dx = v + x·dv/dx:
利用 dy/dx = v + x·dv/dx：
v + x·dv/dx = (1 + 3v²)/(2v)
Subtract v from both sides:
两边同时减去 v：
x·dv/dx = (1 + 3v² − 2v²)/(2v) = (1 + v²)/(2v)
Separate variables:
分离变量：
∫ 2v/(1 + v²) dv = ∫ 1/x dx
The left-hand side integrates to ln(1 + v²). Hence:
左边积分得到 ln(1 + v²)。因此：
ln(1 + v²) = ln|Cx|
So 1 + v² = Cx, and v² = Cx − 1. Replacing v by y/x:
所以 1 + v² = Cx，即 v² = Cx − 1。用 v = y/x 代回：
(y/x)² = Cx − 1
y² = Cx³ − x²
This is the general solution. You can verify by differentiating implicitly that it satisfies the original equation.
这就是通解。你可以通过隐式求导验证它满足原方程。
6. Worked Example 2 | 例题精讲 2
Solve the differential equation:
求解微分方程：
dy/dx = (x − y)/(x + y)
Write the right-hand side in terms of v = y/x. Dividing the numerator and denominator by x gives:
用 v = y/x 表示右端。分子分母同除以 x 得：
v + x·dv/dx = (1 − v)/(1 + v)
Hence:
因此：
x·dv/dx = (1 − v)/(1 + v) − v = (1 − 2v − v²)/(1 + v)
Separate variables:
分离变量：
∫ (1 + v)/(1 − 2v − v²) dv = ∫ 1/x dx
Let u = 1 − 2v − v². Then du/dv = −2(1 + v), so the integrand becomes −du/(2u):
令 u = 1 − 2v − v²，则 du/dv = −2(1 + v)，因此被积函数变为 −du/(2u)：
−½·ln|1 − 2v − v²| = ln|x| + C
Multiplying by −2 and taking exponentials:
两边乘以 −2 并取指数：
1 − 2v − v² = K/x²
Now replace v by y/x:
现在用 v = y/x 代回：
1 − 2y/x − y²/x² = K/x²
Multiplying through by x² gives the neat implicit solution:
两边同乘以 x² 得到简洁的隐式解：
x² − 2xy − y² = K
7. Worked Example 3: Initial Value Problem | 例题精讲 3：初值问题
Solve the initial value problem:
求解初值问题：
dy/dx = y/x + x/y, y(1) = 2
Let y = vx. Then dy/dx = v + x·dv/dx, and y/x + x/y = v + 1/v. So:
令 y = vx，则 dy/dx = v + x·dv/dx，且 y/x + x/y = v + 1/v。因此：
v + x·dv/dx = v + 1/v ⇒ x·dv/dx = 1/v
Separate variables:
分离变量：
∫ v dv = ∫ 1/x dx ⇒ v²/2 = ln|x| + C
Thus v² = 2·ln|x| + C′. Replacing v by y/x:
所以 v² = 2·ln|x| + C′。用 v = y/x 代回：
y²/x² = 2·ln|x| + C′ ⇒ y² = 2x²·ln|x| + C′x²
Apply y(1) = 2:
代入 y(1) = 2：
4 = 0 + C′ ⇒ C′ = 4
Therefore the particular solution is:
因此特解为：
y² = 2x²·ln|x| + 4x²
Since y(1) = 2 is positive, we may take y = x·√2·ln|x| + 4 on the interval containing x = 1.
由于 y(1) = 2 为正，在包含 x = 1 的区间上可以取 y = x·√2·ln|x| + 4。
8. Common Mistakes | 常见错误警示
Mistake 1: Forgetting the term x·dv/dx. The derivative of y = vx is not v alone, but v + x·dv/dx.
错误 1：漏掉 x·dv/dx 这一项。y = vx 的导数不是 v 本身，而是 v + x·dv/dx。
Mistake 2: Trying to separate variables too early. Only after substituting and simplifying should you attempt to separate.
错误 2：过早尝试分离变量。只有在完成换元和化简之后，才应尝试分离变量。
Mistake 3: Forgetting to substitute v = y/x back at the end. The final answer must be in terms of x and y only.
错误 3：最后忘记用 v = y/x 代回。最终的答案必须只包含 x 和 y。
Mistake 4: Using the initial condition too early, before converting back to x and y.
错误 4：在代回 x 和 y 之前就过早地使用初始条件。
Mistake 5: Dropping absolute values when integrating 1/x, then missing sign cases for the constant.
错误 5：积分 1/x 时丢掉绝对值，然后遗漏常数的符号情形。
9. Exam Tips for IB | IB 考试技巧
Tip 1: The examiner rewards clearly structured work. Write "Let y = vx" and "dy/dx = v + x·dv/dx" explicitly before substituting.
技巧 1：阅卷老师青睐结构清晰的解答。在代入前要明确写出 "令 y = vx" 以及 "dy/dx = v + x·dv/dx"。
Tip 2: You do not need to solve for y explicitly. An implicit solution accepted if it is logically derived and simplified.
技巧 2：你不必一定要把 y 解成显式。只要推导逻辑严谨并化简得当，隐式解也是可以接受的。
Tip 3: In the exam, integration of rational functions often requires a u-substitution, so keep your substitution steps visible.
技巧 3：考试中，有理函数的积分常常需要再进行一次换元，请保留你的换元步骤以便得分。
Tip 4: Always check whether dy/dx = f(y/x) before committing to this method. Some equations that look similar are not homogeneous.
技巧 4：在决定使用本方法之前，务必检查方程是否为 dy/dx = f(y/x) 的形式。有些看似相似的方程其实并不是齐次的。
10. Practice Problems | 练习与检验
Try these two exam-style problems on your own:
请独立完成以下两道考试风格题目：
| Problem 1: Solve dy/dx = (y² − x²)/(xy) | 题 1：求解 dy/dx = (y² − x²)/(xy) |
| Problem 2: Solve dy/dx = y/(x + y) with y(1) = 1 | 题 2：求解 dy/dx = y/(x + y)，且 y(1) = 1 |
Answers: 1. v = y/x gives ∫ v/(1 − v²) dv, resulting in x² − y² = Cx. 2. v + x·dv/dx = v/(1 + v), giving y = x·(v), with solution x + y = 2y·ln|x| + 2y after applying the condition; more standard form: y = x·(W(...)) — check by implicit differentiation.
参考答案：1. 令 v = y/x，得 ∫ v/(1 − v²) dv，通解为 x² − y² = Cx。2. 令 v + x·dv/dx = v/(1 + v)，代入初值得 x + y = 2y·ln|x| + 2y；可通过隐式求导验证。
11. Conclusion | 总结
The substitution y = vx turns a homogeneous first-order differential equation into a separable equation involving v and x. With the derivative formula dy/dx = v + x·dv/dx, you can systematically solve any equation of the form dy/dx = f(y/x). Master this technique, and you will gain a reliable method for an entire class of IB HL differential equation problems.
换元 y = vx 可以将齐次一阶微分方程转化为关于 v 与 x 的可分离变量方程。借助导数公式 dy/dx = v + x·dv/dx，你可以系统地求解一切形如 dy/dx = f(y/x) 的方程。掌握这一技巧，你便掌握了一整类 IB HL 微分方程题目的可靠解法。
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