Transformations of Non-Linear Differential Equations to Linear Form | 非线性微分方程到线性形式的变换

📚 Transformations of Non-Linear Differential Equations to Linear Form | 非线性微分方程到线性形式的变换

Non-linear differential equations are often difficult to solve directly. By applying a suitable substitution or transformation, many of them can be rewritten as linear differential equations, for which standard solution methods exist. This article explains the key transformation techniques, with worked examples and examination-style guidance.

非线性微分方程通常难以直接求解。通过适当的代换或变换，许多非线性微分方程可以改写为线性微分方程，从而使用已有的标准解法。本文将解释关键的变换技巧，并给出例题和考试风格的指导。

1. Why Linearise? | 为什么要线性化？

Linear differential equations have well-defined methods: separation of variables, integrating factors, and characteristic equations. Non-linear equations do not obey superposition, and only a few special types can be solved exactly. Transforming a non-linear equation into a linear one allows us to apply these powerful tools.

线性微分方程有成熟的解法：分离变量法、积分因子法、特征方程法。非线性方程不满足叠加原理，只有少数特殊类型能精确求解。将非线性方程变换为线性方程，就可以应用这些强大工具。

The main idea is to choose a new dependent variable, say z, that is a function of the original variable y, so that the derivative terms become linear in z. The success of the method depends on identifying the nonlinearity that can be absorbed by the substitution.

主要思想是选择一个新的因变量，比如 z，它是原变量 y 的函数，使得导数项在 z 中变为线性。方法能否成功，取决于识别出可以通过代换吸收的非线性项。

2. Recognising Linear and Non-Linear Equations | 识别线性与非线性方程

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 depend only on x. Any term involving y², eʸ, sin(y), y·(dy/dx), or (dy/dx)² makes the equation non-linear.

一阶微分方程如果能写成形式 dy/dx + P(x)y = Q(x)，则是线性的，其中 P 和 Q 仅依赖于 x。任何包含 y²、eʸ、sin(y)、y·(dy/dx) 或 (dy/dx)² 的项都会使方程变为非线性。

For example, dy/dx + 3y = sin x is linear, while dy/dx + y² = x is non-linear because of the y² term.

例如，dy/dx + 3y = sin x 是线性的，而 dy/dx + y² = x 因含有 y² 项而是非线性的。

3. Bernoulli Equations | 伯努利方程

A Bernoulli equation has the form:

dy/dx + P(x)y = Q(x)yⁿ

where n is a real number, and n ≠ 0, n ≠ 1. When n = 0 or n = 1, the equation is already linear.

伯努利方程具有形式 dy/dx + P(x)y = Q(x)yⁿ，其中 n 为实数，且 n ≠ 0，n ≠ 1。当 n = 0 或 n = 1 时，方程已经是线性的。

The standard transformation is z = y^(1-n). Differentiating with respect to x gives:

dz/dx = (1-n)y^(-n) · dy/dx

Multiplying the original equation by (1-n)y^(-n) and substituting leads to a linear equation in z:

dz/dx + (1-n)P(x)z = (1-n)Q(x)

This linear equation can then be solved using an integrating factor.

标准变换是 z = y^(1-n)。对 x 求导得到 dz/dx = (1-n)y^(-n) · dy/dx。将原方程乘以 (1-n)y^(-n) 并代换，得到关于 z 的线性方程 dz/dx + (1-n)P(x)z = (1-n)Q(x)。然后用积分因子法求解。

4. Worked Example: Bernoulli Equation | 伯努利方程例题

Solve the differential equation:

dy/dx + 2x y = x y³

Here P(x)=2x, Q(x)=x, n=3. Since n=3, we set z = y^(1-3) = y^(-2). Then dz/dx = -2 y^(-3) dy/dx.

这里 P(x)=2x，Q(x)=x，n=3。因为 n=3，令 z = y^(1-3) = y^(-2)。则 dz/dx = -2 y^(-3) dy/dx。

Multiply the original equation by -2 y^(-3):

-2 y^(-3) dy/dx – 4x y^(-2) = -2x

Since -2 y^(-3) dy/dx = dz/dx and y^(-2)=z, we obtain:

dz/dx – 4x z = -2x

This is linear in z. The integrating factor is I = exp(∫ -4x dx) = e^(-2x²). Solving gives:

z · e^(-2x²) = ∫ -2x e^(-2x²) dx = ½ e^(-2x²) + C

Thus z = ½ + C e^(2x²). Recalling z = y^(-2), the general solution is:

y^(-2) = ½ + C e^(2x²)

5. Transforming Equations of the Form dy/dx = f(ax + by + c) | 形如 dy/dx = f(ax + by + c) 的变换

Consider equations where the right-hand side is a function of a linear combination of x and y, such as:

dy/dx = (ax + by + c)²

These are non-linear, but they become separable after the substitution u = ax + by + c. Then:

du/dx = a + b dy/dx

so dy/dx = (du/dx − a)/b. Substitution yields a separable equation in u and x.

考虑右侧为 x 和 y 的线性组合函数的情况，比如 dy/dx = (ax + by + c)²。这类方程是非线性的，但通过代换 u = ax + by + c 后变为可分离变量。由于 du/dx = a + b dy/dx，所以 dy/dx = (du/dx − a)/b，代换后得到关于 u 和 x 的可分离方程。

6. Example: Linear Combination Substitution | 线性组合代换例题

Solve dy/dx = (x + y)². Let u = x + y. Then du/dx = 1 + dy/dx, so dy/dx = du/dx − 1. The equation becomes:

du/dx − 1 = u²

Hence du/dx = u² + 1. Separating variables:

∫ du/(1 + u²) = ∫ dx

Therefore arctan(u) = x + C. Replacing u = x + y gives tan(x + C) = x + y, or y = tan(x + C) − x.

解 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，即 du/dx = u² + 1。分离变量得 ∫ du/(1 + u²) = ∫ dx，所以 arctan(u) = x + C。代回 u = x + y，得到 tan(x + C) = x + y，即 y = tan(x + C) − x。

7. Homogeneous Equations and y = vx | 齐次方程与 y = vx

A first-order equation is homogeneous if it can be written as dy/dx = F(y/x). The substitution y = v x transforms it into a separable equation in v and x. Although the result is not always linear, the substitution reduces the order or simplifies the structure. In some special cases, further transformation can yield a linear equation.

一阶方程如果能写成 dy/dx = F(y/x) 的形式，则为齐次方程。代换 y = v x 可将其转化为关于 v 和 x 的可分离方程。虽然结果不总是线性，但这种代换能降低阶数或简化结构。在某些特殊情况下，进一步变换可得到线性方程。

For example, an equation of the form dy/dx = (y/x) + (y/x)² becomes, after y = vx, a separable equation. Sometimes the v-equation itself is Bernoulli and can be linearised further.

例如，形如 dy/dx = (y/x) + (y/x)² 的方程，经过 y = vx 后变为可分离方程。有时关于 v 的方程本身是伯努利方程，可以进一步线性化。

8. Riccati Equations | 里卡蒂方程

The Riccati equation is a non-linear equation of the form:

dy/dx = p(x) + q(x)y + r(x)y²

If one particular solution, say y₁(x), is known, the substitution y = y₁ + 1/u transforms the Riccati equation into a linear equation in u:

du/dx + (q(x) + 2r(x)y₁)u = −r(x)

This is a valuable technique when a single solution can be guessed or spotted.

里卡蒂方程是非线性方程，形式为 dy/dx = p(x) + q(x)y + r(x)y²。如果已知一个特解 y₁(x)，则代换 y = y₁ + 1/u 可将里卡蒂方程转化为关于 u 的线性方程：du/dx + (q(x) + 2r(x)y₁)u = −r(x)。当能猜出一个解时，这是非常有用的技巧。

9. Second-Order Non-Linear Equations | 二阶非线性方程

Some second-order non-linear equations can also be linearised. For instance, equations where the dependent variable y is missing can be reduced to first order by setting p = dy/dx. If the independent variable x is missing, we set p = dy/dx and use d²y/dx² = p · dp/dy. These substitutions often produce a first-order equation which may then be linear or Bernoulli.

某些二阶非线性方程也可以线性化。例如，当方程中缺少因变量 y 时，可以令 p = dy/dx 降为一阶。如果缺少自变量 x，则令 p = dy/dx 并利用 d²y/dx² = p · dp/dy。这些代换通常会产生一阶方程，该方程可能是线性的或伯努利型的。

A classic example is the nonlinear pendulum equation d²θ/dt² + (g/L) sin θ = 0. For small angles, sin θ ≈ θ, giving a linear equation. This linearisation is an approximation, not an exact transformation.

一个经典例子是非线性单摆方程 d²θ/dt² + (g/L) sin θ = 0。在小角度近似下，sin θ ≈ θ，得到线性方程。这是近似线性化，并非精确变换。

10. Practical Strategy for Examinations | 考试实用策略

In an exam, when asked to ‘transform a non-linear differential equation to linear form’, follow these steps:

在考试中，当要求“将非线性微分方程变换为线性形式”时，遵循以下步骤：

Identify the type: Bernoulli, Riccati, or a linear combination argument.
Write down the required substitution explicitly.
Differentiate the substitution correctly to express dy/dx in terms of the new variable.
Substitute fully and simplify, cancelling all terms that make the equation non-linear.
If the resulting equation is linear, solve using an integrating factor; if separable, solve by integration.

识别类型：伯努利、里卡蒂还是线性组合形式。
明确写出所需的代换。
正确求导，用新变量表示 dy/dx。
完全代换并化简，消去所有导致非线性的项。
如果结果是线性方程，用积分因子求解；如果是可分离的，则用积分求解。

Always check the domain of the solution. If a substitution involves y^(1-n), then y = 0 may need special attention. Also remember that the initial condition, if any, must be transformed into the new variable.

始终检查解的定义域。如果代换涉及 y^(1-n)，则 y = 0 可能需要特别处理。还要记住，初始条件（如果有）必须变换到新变量中。

11. Summary | 总结

Linearisation by transformation is a powerful technique for solving certain non-linear differential equations. The most examinable types are Bernoulli equations, equations of the form dy/dx = f(ax + by + c), and Riccati equations with a known particular solution. Mastering these substitutions not only saves time but also deepens insight into the structure of differential equations.

通过变换实现线性化是求解某些非线性微分方程的强大技巧。最常考的类型包括伯努利方程、形如 dy/dx = f(ax + by + c) 的方程，以及已知特解的里卡蒂方程。掌握这些代换不仅能节省时间，还能加深对微分方程结构的理解。

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Transformations of Non-Linear Differential Equations to Linear Form | 非线性微分方程到线性形式的变换