📚 Transformations of Non-Linear Differential Equations to Linear Form | 非线性微分方程变换为线性形式
Many differential equations encountered in A-Level Mathematics are non-linear. Although their solutions may seem difficult to obtain directly, a clever substitution can often transform them into a linear differential equation, which is far easier to solve. This article explores the essential transformations required by AQA and shows you how to apply them step by step.
在A-Level数学中，我们遇到的许多微分方程都是非线性的。虽然直接求解它们可能显得困难，但通过巧妙的代换，往往可以将它们转化为线性微分方程，从而大大简化求解过程。本文将系统介绍AQA要求掌握的变换方法，并逐步演示如何应用这些方法。
1. Recognising Non-Linear Differential Equations | 识别非线性微分方程
A first-order differential equation is linear if it can be written in the form dy/dx + P(x)y = Q(x), where P and Q are functions of x only. Any term involving y², y³, e^y, sin(y), or products like y(dy/dx) makes the equation non-linear. Recognising this distinction is the first step toward choosing the correct substitution.
一阶微分方程若能写成 dy/dx + P(x)y = Q(x) 的形式（其中 P 和 Q 仅为 x 的函数），则该方程是线性的。任何包含 y²、y³、e^y、sin(y) 或 y(dy/dx) 等乘积项的项都会使方程成为非线性。识别这种区别是选择正确代换的第一步。
For example, dy/dx = x² + y is linear, but dy/dx = x² + y² is non-linear. Similarly, dy/dx = xy is linear, whereas dy/dx = xy² is non-linear because of the y² term.
例如，dy/dx = x² + y 是线性的，但 dy/dx = x² + y² 是非线性的。同样，dy/dx = xy 是线性，而 dy/dx = xy² 因为含有 y² 项而是非线性的。
2. The Substitution u = y¹⁻ⁿ for Bernoulli Equations | 伯努利方程的代换 u = y¹⁻ⁿ
A Bernoulli equation has the form dy/dx + P(x)y = Q(x)yⁿ, where n ≠ 0 and n ≠ 1. The standard transformation is to set u = y¹⁻ⁿ. Differentiating u with respect to x gives du/dx = (1 – n)y⁻ⁿ dy/dx. Substituting this into the original equation eliminates the yⁿ factor and leaves a linear equation in u.
伯努利方程具有形式 dy/dx + P(x)y = Q(x)yⁿ，其中 n ≠ 0 且 n ≠ 1。标准变换是令 u = y¹⁻ⁿ。对 x 求导得 du/dx = (1 – n)y⁻ⁿ dy/dx。将其代入原方程可消去 yⁿ 因子，得到关于 u 的线性方程。
After solving for u, you must substitute back u = y¹⁻ⁿ to obtain y in terms of x. This method is particularly useful for equations such as dy/dx + (1/x)y = x²y³.
在解出 u 之后，必须回代 u = y¹⁻ⁿ 才能得到用 x 表示的 y。这种方法对 dy/dx + (1/x)y = x²y³ 这类方程尤为有效。
3. Worked Example: Solving a Bernoulli Equation | 示例：解伯努利方程
Consider the equation dy/dx + (1/x)y = x²y³. Here n = 3, so set u = y¹⁻³ = y⁻². Then du/dx = -2y⁻³ dy/dx, which gives dy/dx = -(y³/2) du/dx. Substituting into the original equation:
考虑方程 dy/dx + (1/x)y = x²y³。这里 n = 3，故令 u = y¹⁻³ = y⁻²。则 du/dx = -2y⁻³ dy/dx，从而 dy/dx = -(y³/2) du/dx。代入原方程得：
-(y³/2) du/dx + (1/x)y = x²y³
Divide through by y³ (which is non-zero for typical solutions) to get -(1/2) du/dx + (1/x)y⁻² = x². Since u = y⁻², this becomes:
除以 y³（在通常解中非零），得到 -(1/2) du/dx + (1/x)y⁻² = x²。由于 u = y⁻²，上式化为：
du/dx – (2/x)u = -2x²
This is a linear equation in u. Using the integrating factor IF = e^(∫(-2/x)dx) = x⁻², the solution is u = -x³ + Cx². Since u = y⁻², we finally obtain y⁻² = -x³ + Cx², or y² = 1/(Cx² – x³).
这是关于 u 的线性方程。积分因子为 IF = e^(∫(-2/x)dx) = x⁻²，解得 u = -x³ + Cx²。由于 u = y⁻²，最终得到 y⁻² = -x³ + Cx²，即 y² = 1/(Cx² – x³)。
4. Homogeneous Equations and the Substitution y = vx | 齐次方程与代换 y = vx
A differential equation of the form dy/dx = f(y/x) is called homogeneous. To linearize it, set y = vx, where v is a new dependent variable. Then dy/dx = v + x(dv/dx). Substituting y = vx transforms the right-hand side into a function of v only, giving:
形如 dy/dx = f(y/x) 的微分方程称为齐次方程。为了将其线性化，令 y = vx，其中 v 是新的因变量。则 dy/dx = v + x(dv/dx)。代入 y = vx 后，右端变为仅含 v 的函数，得到：
v + x(dv/dx) = f(v)
This new equation is separable: x(dv/dx) = f(v) – v. It may not be linear in v, but it is often easier to solve by separating variables. The original substitution then gives the general solution in x and y.
这个新方程是可分离变量的：x(dv/dx) = f(v) – v。它虽然不一定关于 v 线性，但通常更容易通过分离变量求解。最后回代原变量即可得到关于 x 和 y 的通解。
5. Worked Example: Homogeneous Differential Equation | 示例：齐次微分方程
Solve dy/dx = (x² + y²)/(xy). First rewrite the right-hand side as (x/y) + (y/x) = (y/x)⁻¹ + (y/x). Since this is a function of y/x, the equation is homogeneous. Let y = vx, so dy/dx = v + x(dv/dx). Substituting gives:
求解 dy/dx = (x² + y²)/(xy)。先将右端改写为 (x/y) + (y/x) = (y/x)⁻¹ + (y/x)。由于它是 y/x 的函数，所以该方程是齐次的。令 y = vx，则 dy/dx = v + x(dv/dx)。代入得：
v + x(dv/dx) = v⁻¹ + v
Hence x(dv/dx) = v⁻¹, so v dv = dx/x. Integrating gives v²/2 = ln|x| + C. Replacing v by y/x yields (y/x)²/2 = ln|x| + C, or y² = 2x²(ln|x| + C).
因此 x(dv/dx) = v⁻¹，即 v dv = dx/x。积分得 v²/2 = ln|x| + C。将 v = y/x 代回，得到 (y/x)²/2 = ln|x| + C，即 y² = 2x²(ln|x| + C)。
6. Linearising by Reversing the Dependent and Independent Variables | 交换因变量与自变量进行线性化
Sometimes an equation that is non-linear in y is linear in x. If the equation can be written as dx/dy + P(y)x = Q(y), then we can solve for x as a function of y, then invert the result. This is especially useful for equations involving powers or reciprocals of y that are awkward to handle directly.
有时一个关于 y 非线性的方程，关于 x 却是线性的。如果方程能写成 dx/dy + P(y)x = Q(y) 的形式，那么我们可以把 x 解为 y 的函数，再反演得到 y 关于 x 的表达式。这种方法对包含 y 的高次幂或倒数的方程特别有用。
For example, dy/dx = 1/(x + y) can be rewritten as dx/dy = x + y, which is linear in x with P(y) = -1 and Q(y) = y. Solving gives x = Ce^y – y – 1.
例如，dy/dx = 1/(x + y) 可以改写为 dx/dy = x + y，这是关于 x 的线性方程，其中 P(y) = -1，Q(y) = y。解得 x = Ce^y – y – 1。
7. Using the Substitution z = ax + by + c | 使用代换 z = ax + by + c
When a differential equation contains a linear combination of x and y, such as dy/dx = f(ax + by + c), it is often convenient to set z = ax + by + c. Then dz/dx = a + b(dy/dx), and the equation becomes dz/dx = a + b f(z), which is separable.
当微分方程包含 x 与 y 的线性组合，例如 dy/dx = f(ax + by + c) 时，通常令 z = ax + by + c。则 dz/dx = a + b(dy/dx)，原方程变为 dz/dx = a + b f(z)，这是可分离变量的。
This transformation is not a linearization in the strict sense, but it reduces many non-linear first-order equations to a simpler form. For instance, dy/dx = (x + y)² can be handled by setting z = x + y.
这种变换严格来说不是线性化，但它能将许多非线性一阶方程化为更简单的形式。例如，dy/dx = (x + y)² 可以通过令 z = x + y 来处理。
8. Euler-Cauchy Equations and the Substitution x = e^t | 欧拉-柯西方程与代换 x = e^t
The second-order Euler-Cauchy equation x²(d²y/dx²) + ax(dy/dx) + by = 0 is not linear in the variables x and y in the usual sense, but it can be transformed into a linear equation with constant coefficients by substituting x = e^t. Then t = ln x, and using the chain rule we obtain:
二阶欧拉-柯西方程 x²(d²y/dx²) + ax(dy/dx) + by = 0 在通常意义下关于变量 x 和 y 并不是线性的，但通过代换 x = e^t 可以将其转化为常系数线性方程。令 t = ln x，利用链式法则可得：
x dy/dx = dy/dt, x² d²y/dx² = d²y/dt² – dy/dt
Substituting these into the original equation eliminates all powers of x and gives a linear differential equation with constant coefficients in t. This is a powerful technique for equations with terms xⁿ dⁿy/dxⁿ.
将这些表达式代入原方程，可以消去所有 x 的幂，得到关于 t 的常系数线性微分方程。这是处理含有 xⁿ dⁿy/dxⁿ 项方程的强有力技巧。
9. Linearising via Small-Angle Approximations | 利用小角度近似进行线性化
In many physical applications, a non-linear term like sin(θ) is approximated by θ when θ is small. The pendulum equation d²θ/dt² + (g/L)sin(θ) = 0 becomes d²θ/dt² + (g/L)θ = 0, which is linear with constant coefficients. While this is an approximation rather than an exact transformation, it is a vital technique in modelling real phenomena.
在许多物理应用中，当 θ 很小时，sin(θ) 等非线性项可以近似为 θ。单摆方程 d²θ/dt² + (g/L)sin(θ) = 0 就变为 d²θ/dt² + (g/L)θ = 0，这是一个常系数线性方程。虽然这是一种近似而非精确变换，但它是实际建模中的重要技术。
This idea extends to Taylor series: keeping only the first-order term of a non-linear function about an equilibrium point yields a linear approximation. The AQA syllabus often tests this in mechanics contexts, especially for small oscillations.
这一思想可推广到泰勒展开：在平衡点附近只保留非线性函数的一阶项，便得到线性近似。AQA考纲常在力学背景下考查这一内容，尤其是小振动问题。
10. Choosing the Right Transformation | 选择合适的变换
There is no single rule for linearizing every non-linear equation, but a logical sequence of checks helps. First, check if it is Bernoulli (yⁿ term). If not, try homogeneity (f(y/x)). Next, look for combinations like ax + by + c. Finally, consider whether swapping x and y would make it linear. Practice with past papers is the best way to build this intuition.
并不存在一种能线性化所有非线性方程的通用规则，但按逻辑顺序进行检查会很有帮助。首先，判断是否为伯努利方程（含 yⁿ 项）。如果不是，尝试是否齐次（f(y/x)）。接着，寻找像 ax + by + c 这样的组合。最后，考虑交换 x 与 y 是否能使其线性化。通过练习真题来建立这种直觉是最有效的方法。
When applying any substitution, always state clearly what the new variable is and compute the derivative correctly. A common mistake is forgetting to substitute for both y and dy/dx in the original equation.
在应用任何代换时，务必清楚地写出新变量，并正确计算导数。常见的错误是忘记在原始方程中同时替换 y 和 dy/dx。
11. Worked Example: Multiple-Step Transformation | 示例：多步变换
Solve dy/dx = (x + y + 1)². This is not Bernoulli or homogeneous, but it contains the combination x + y + 1. Let z = x + y + 1. Then dz/dx = 1 + dy/dx, so dy/dx = dz/dx – 1. The equation becomes dz/dx – 1 = z², or dz/dx = z² + 1. Separating variables gives ∫ dz/(z² + 1) = ∫ dx, so arctan(z) = x + C. Hence z = tan(x + C), and y = tan(x + C) – x – 1.
求解 dy/dx = (x + y + 1)²。这不是伯努利方程也不是齐次的，但它含有组合 x + y + 1。令 z = x + y + 1，则 dz/dx = 1 + dy/dx，所以 dy/dx = dz/dx – 1。原方程变为 dz/dx – 1 = z²，即 dz/dx = z² + 1。分离变量得 ∫ dz/(z² + 1) = ∫ dx，所以 arctan(z) = x + C。因此 z = tan(x + C)，于是 y = tan(x + C) – x – 1。
This example shows how a single substitution can reduce a seemingly intractable equation to a simple integral.
这个例子表明，一次代换就能将看似难以处理的方程化为简单的积分。
12. Summary of Key Transformations | 关键变换总结
The table below summarises the common non-linear forms and the substitutions that lead to a linear or separable equation.
下表总结了常见的非线性形式以及能化为线性或可分离方程的代换。
| Form / 形式 | Substitution / 代换 | Result / 结果 |
| dy/dx + P(x)y = Q(x)yⁿ | u = y¹⁻ⁿ | Linear in u |
| dy/dx = f(y/x) | y = vx | Separable in v |
| dy/dx = f(ax + by + c) | z = ax + by + c | Separable in z |
| x² y” + ax y’ + b y = 0 | x = e^t | Constant coefficient linear |
| Non-linear in y, linear in x | Swap variables | Linear in x |
Always check that your substitution is valid, that the domain is appropriate, and that you reverse the substitution at the end. For AQA examinations, clear working and correct use of the integrating factor are essential for full marks.
始终注意检查代换是否有效、定义域是否合适，并在最后将代换回代。对于AQA考试，清晰的步骤和正确使用积分因子是取得满分的关键。
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