Second Order Differential Equations | 二阶微分方程

📚 Second Order Differential Equations | 二阶微分方程

Second order differential equations are equations that involve a function and its second derivative. They appear throughout physics, engineering, and economics, describing phenomena such as oscillations, electric circuits, and population dynamics.

二阶微分方程是包含函数及其二阶导数的方程。它们广泛出现在物理、工程和经济领域，用于描述振动、电路以及人口动态等自然现象。

1. Standard Form and Classification | 标准形式与分类

The general linear second order differential equation with constant coefficients is written as:

常系数线性二阶微分方程的一般形式为：

a y″ + b y′ + c y = f(x)

where a, b, c are constants with a ≠ 0, and f(x) is a given function. If f(x) = 0, the equation is called homogeneous; otherwise it is non-homogeneous.

其中 a、b、c 为常数，且 a ≠ 0，f(x) 为已知函数。若 f(x) = 0，方程称为齐次的；否则称为非齐次的。

Equations are classified according to the form of f(x) and the nature of the roots of the auxiliary equation, which we define in the next section.

方程根据 f(x) 的形式以及辅助方程根的性质进行分类，我们将在下一节中定义辅助方程。

2. The Auxiliary Equation | 辅助方程

For the homogeneous equation a y″ + b y′ + c y = 0, we seek solutions of the form y = eᵐˣ. Substituting y = eᵐˣ, y′ = m eᵐˣ, and y″ = m² eᵐˣ gives:

对于齐次方程 a y″ + b y′ + c y = 0，我们尝试形如 y = eᵐˣ 的解。代入 y = eᵐˣ、y′ = m eᵐˣ、y″ = m² eᵐˣ 后得到：

a m² + b m + c = 0

This quadratic is called the auxiliary equation (or characteristic equation). Solving it gives two roots m₁ and m₂, which determine the general solution of the homogeneous equation.

该二次方程称为辅助方程（或特征方程）。解出两个根 m₁ 和 m₂，即可确定齐次方程的通解。

3. Real Distinct Roots | 相异实根

If the auxiliary equation has two real distinct roots m₁ ≠ m₂, the general solution of the homogeneous equation is:

若辅助方程有两个相异实根 m₁ ≠ m₂，则齐次方程的通解为：

y = A eᵐ¹ˣ + B eᵐ²ˣ

where A and B are arbitrary constants. This form arises from the principle of superposition: any linear combination of two independent solutions is also a solution.

其中 A 和 B 为任意常数。此形式源于叠加原理：两个独立解的任意线性组合仍然是方程的解。

Example: Solve y″ − 5y′ + 6y = 0. The auxiliary equation is m² − 5m + 6 = 0 = (m − 2)(m − 3), giving m = 2 and m = 3. Hence the general solution is y = A e²ˣ + B e³ˣ.

示例：求 y″ − 5y′ + 6y = 0 的通解。辅助方程为 m² − 5m + 6 = 0 = (m − 2)(m − 3)，得 m = 2 和 m = 3。因此通解为 y = A e²ˣ + B e³ˣ。

4. Repeated Roots | 重根

If the auxiliary equation has a repeated root m, then eᵐˣ is only one solution. To find a second linearly independent solution, we multiply by x:

若辅助方程有重根 m，则 eᵐˣ 仅为一个解。为了找到第二个线性无关的解，我们将原解乘以 x：

y = (A + Bx) eᵐˣ

This occurs when the discriminant b² − 4ac = 0. The factor x ensures that the two functions eᵐˣ and x eᵐˣ are independent, spanning the full solution space.

这种情况发生在判别式 b² − 4ac = 0 时。因子 x 保证了 eᵐˣ 和 x eᵐˣ 两个函数线性无关，从而张成完整的解空间。

Example: Solve y″ − 6y′ + 9y = 0. Here m² − 6m + 9 = 0 = (m − 3)², so m = 3 (repeated). The solution is y = (A + Bx)e³ˣ.

示例：求 y″ − 6y′ + 9y = 0 的通解。此处 m² − 6m + 9 = 0 = (m − 3)²，故 m = 3（重根）。解为 y = (A + Bx)e³ˣ。

5. Complex Roots | 复根

If the auxiliary equation has complex roots m = p ± qi, where i = √(−1), the general solution can be written using Euler’s formula as:

若辅助方程有复根 m = p ± qi，其中 i = √(−1)，则利用欧拉公式可将通解写为：

y = eᵖˣ (A cos(qx) + B sin(qx))

where p is the real part and q is the imaginary part of the roots. This form is equivalent to C eᵖˣ cos(qx + φ), but the A, B version is usually easier to work with in exams.

其中 p 为根的实部，q 为虚部。该形式等价于 C eᵖˣ cos(qx + φ)，但含 A、B 的形式在考试中通常更容易处理。

Example: Solve y″ + 2y′ + 5y = 0. The auxiliary equation m² + 2m + 5 = 0 gives m = −1 ± 2i. Thus y = e⁻ˣ (A cos 2x + B sin 2x).

示例：求 y″ + 2y′ + 5y = 0 的通解。辅助方程 m² + 2m + 5 = 0 给出 m = −1 ± 2i。因此 y = e⁻ˣ (A cos 2x + B sin 2x)。

6. Non-Homogeneous Equations and Particular Integrals | 非齐次方程与特解

For the non-homogeneous equation a y″ + b y′ + c y = f(x), the general solution is:

对于非齐次方程 a y″ + b y′ + c y = f(x)，其通解为：

y = y_c + y_p

where y_c is the complementary function (the general solution of the corresponding homogeneous equation) and y_p is a particular integral — any single solution of the full equation.

其中 y_c 为补函数（对应齐次方程的通解），y_p 为特解——即完整方程的任意一个解。

The method of undetermined coefficients is used to find y_p by assuming a form that matches f(x), substituting into the equation and solving for the unknown constants.

待定系数法通过假设一个与 f(x) 形式匹配的特定解，代入方程并求解未知常数来确定 y_p。

7. Choosing the Form of the Particular Integral | 选择特解的形式

The table below shows the standard trial function y_p for common types of f(x).

下表展示了常见 f(x) 类型对应的标准试探解 y_p。

f(x)	Trial y_p	中文试探解
Polynomial of degree n	aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀	n 次多项式
k eˣ	A eˣ (unless eˣ appears in y_c, then Ax eˣ)	A eˣ（若与 y_c 重合，则用 Ax eˣ）
k cos(nx) or k sin(nx)	A cos(nx) + B sin(nx)	A cos(nx) + B sin(nx)
k xⁿ eˣ	(aₙxⁿ + … + a₀)eˣ	多项式与 eˣ 的乘积

If the trial function duplicates any term in the complementary function, multiply the entire trial by x (or x² if necessary) to ensure linear independence.

若试探解与补函数中的任何项重复，则需将整个试探解乘以 x（必要时乘以 x²）以保证线性无关。

8. Worked Example: Polynomial Forcing | 实例：多项式非齐次项

Problem: Solve y″ + y′ − 2y = 3x.

题目：求 y″ + y′ − 2y = 3x 的通解。

Step 1: Complementary function. The auxiliary equation m² + m − 2 = 0 = (m + 2)(m − 1) gives m = −2, 1. Thus y_c = A e⁻²ˣ + B eˣ.

步骤 1：求补函数。辅助方程 m² + m − 2 = 0 = (m + 2)(m − 1) 得 m = −2, 1。因此 y_c = A e⁻²ˣ + B eˣ。

Step 2: Try y_p = Cx + D. Then y_p′ = C, y_p″ = 0.

步骤 2：设 y_p = Cx + D。则 y_p′ = C，y_p″ = 0。

Step 3: Substitute: 0 + C − 2(Cx + D) = 3x. Equate coefficients: −2C = 3 and C − 2D = 0, giving C = −3/2 and D = −3/4.

步骤 3：代入：0 + C − 2(Cx + D) = 3x。比较系数：−2C = 3 且 C − 2D = 0，解得 C = −3/2，D = −3/4。

Step 4: General solution:

步骤 4：通解为：

y = A e⁻²ˣ + B eˣ − (3/2)x − 3/4

9. Worked Example: Exponential Forcing | 实例：指数非齐次项

Problem: Solve y″ − 3y′ + 2y = 4e³ˣ.

题目：求 y″ − 3y′ + 2y = 4e³ˣ 的通解。

Step 1: y_c: m² − 3m + 2 = 0 = (m − 1)(m − 2), so y_c = A eˣ + B e²ˣ.

步骤 1：求 y_c：m² − 3m + 2 = 0 = (m − 1)(m − 2)，故 y_c = A eˣ + B e²ˣ。

Step 2: Since e³ˣ does not appear in y_c, try y_p = K e³ˣ. Then y_p′ = 3K e³ˣ, y_p″ = 9K e³ˣ.

步骤 2：因为 e³ˣ 未出现在 y_c 中，设 y_p = K e³ˣ。则 y_p′ = 3K e³ˣ，y_p″ = 9K e³ˣ。

Step 3: Substitute: 9K e³ˣ − 9K e³ˣ + 2K e³ˣ = 4e³ˣ → 2K = 4 → K = 2.

步骤 3：代入：9K e³ˣ − 9K e³ˣ + 2K e³ˣ = 4e³ˣ → 2K = 4 → K = 2。

Step 4: General solution: y = A eˣ + B e²ˣ + 2e³ˣ.

步骤 4：通解：y = A eˣ + B e²ˣ + 2e³ˣ。

If f(x) were 4e²ˣ, the initial trial K e²ˣ would clash with B e²ˣ in y_c, so we would use K x e²ˣ instead.

若 f(x) 为 4e²ˣ，则初始试探解 K e²ˣ 会与 y_c 中的 B e²ˣ 重复，此时应改用 K x e²ˣ。

10. Worked Example: Trigonometric Forcing | 实例：三角非齐次项

Problem: Solve y″ + 4y = 5 sin x.

题目：求 y″ + 4y = 5 sin x 的通解。

Step 1: y_c: m² + 4 = 0 → m = ±2i, so y_c = A cos 2x + B sin 2x.

步骤 1：求 y_c：m² + 4 = 0 → m = ±2i，故 y_c = A cos 2x + B sin 2x。

Step 2: Try y_p = C cos x + D sin x. Then y_p″ = −C cos x − D sin x.

步骤 2：设 y_p = C cos x + D sin x。则 y_p″ = −C cos x − D sin x。

Step 3: Substitute: (−C cos x − D sin x) + 4(C cos x + D sin x) = 5 sin x → 3C cos x + 3D sin x = 5 sin x. Equating coefficients: C = 0, D = 5/3.

步骤 3：代入：(−C cos x − D sin x) + 4(C cos x + D sin x) = 5 sin x → 3C cos x + 3D sin x = 5 sin x。比较系数：C = 0，D = 5/3。

Step 4: General solution: y = A cos 2x + B sin 2x + (5/3) sin x.

步骤 4：通解：y = A cos 2x + B sin 2x + (5/3) sin x。

Note that forcing a sine term sometimes requires both sine and cosine in the trial, because the derivative of a sine is a cosine.

注意：当非齐次项为正弦函数时，试探解通常需要同时包含正弦和余弦，因为正弦的导数为余弦。

11. Boundary Conditions and Particular Solutions | 边界条件与特解

To determine the arbitrary constants in the general solution, we need additional conditions. Two common types are:

为了确定通解中的任意常数，我们需要附加条件。两种常见类型为：

Initial conditions: y(x₀) = y₀ and y′(x₀) = y₁, often given at x = 0.

初始条件：y(x₀) = y₀ 和 y′(x₀) = y₁，通常给定在 x = 0 处。
Boundary conditions: y(x₀) = y₀ and y(x₁) = y₁, given at two different points.

边界条件：y(x₀) = y₀ 和 y(x₁) = y₁，给定在两个不同的点。

Example: For y″ + y = 0, with y(0) = 2 and y′(0) = 3. The general solution is y = A cos x + B sin x. Applying y(0) = 2 gives A = 2. Differentiating y′ = −A sin x + B cos x, so y′(0) = B = 3. Hence the particular solution is y = 2 cos x + 3 sin x.

示例：对于 y″ + y = 0，且有 y(0) = 2，y′(0) = 3。通解为 y = A cos x + B sin x。代入 y(0) = 2 得 A = 2。求导得到 y′ = −A sin x + B cos x，故 y′(0) = B = 3。因此特解为 y = 2 cos x + 3 sin x。

12. Real-World Applications | 实际应用

Second order differential equations model many physical systems. The standard mass-spring-damper equation is:

二阶微分方程可模拟许多物理系统。标准的质量-弹簧-阻尼方程为：

m x″ + c x′ + k x = F(t)

where m is mass, c is damping coefficient, k is spring stiffness, and F(t) is an external force. The complementary function describes free vibration (transient response), while the particular integral describes the forced response.

其中 m 为质量，c 为阻尼系数，k 为弹簧刚度，F(t) 为外力。补函数描述自由振动（瞬态响应），特解描述受迫响应。

In electrical circuits, the RLC loop equation is:

在电路中，RLC 回路方程为：

L Q″ + R Q′ + (1/C) Q = E(t)

where L is inductance, R is resistance, C is capacitance, Q is charge, and E(t) is the voltage source. The same mathematical structure applies to many different systems, which is why mastering this topic is essential.

其中 L 为电感，R 为电阻，C 为电容，Q 为电荷，E(t) 为电压源。相同的数学结构适用于许多不同系统，因此掌握该主题至关重要。

In AQA A-Level Mathematics, you are expected to solve such equations with constant coefficients, apply initial or boundary conditions, and interpret the physical meaning of the solution.

在 AQA A-Level 数学中，你需要能够求解常系数二阶微分方程，应用初始条件或边界条件，并解释解的物理意义。

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Second Order Differential Equations | 二阶微分方程