Complementary Functions and Particular Integrals | 补函数与特积分

📚 Complementary Functions and Particular Integrals | 补函数与特积分

In A-Level AQA Mathematics, solving second-order linear differential equations with constant coefficients is a key skill. The general solution of a non-homogeneous equation is formed by adding the complementary function (CF) and a particular integral (PI). This article explains both components clearly, with worked examples and exam-style guidance.

在A-Level AQA数学中，求解常系数二阶线性微分方程是一项核心技能。非齐次方程的通解由补函数（CF）和特积分（PI）相加而成。本文将清晰讲解这两个部分，并配以例题和考试风格指导。

1. Standard Form | 标准形式

A second-order linear differential equation with constant coefficients has the general form:

常系数二阶线性微分方程的一般形式为：

a(d²y/dx²) + b(dy/dx) + cy = f(x)

where a, b, c are constants and a ≠ 0. When f(x) = 0, the equation is homogeneous; when f(x) ≠ 0, it is non-homogeneous.

其中a、b、c为常数且a ≠ 0。当f(x) = 0时，方程为齐次的；当f(x) ≠ 0时，方程为非齐次的。

For the non-homogeneous case, the general solution is:

对于非齐次情形，通解为：

y = CF + PI

The complementary function solves the associated homogeneous equation; the particular integral provides one specific solution to the full equation.

补函数求解对应的齐次方程；特积分给出原方程的一个特解。

2. The Auxiliary Equation and Complementary Function | 辅助方程与补函数

To find the CF, set f(x) = 0 and replace d²y/dx² by m², dy/dx by m, and y by 1. This gives the auxiliary equation:

为求CF，令f(x) = 0，并将d²y/dx²换成m²，dy/dx换成m，y换成1。得到辅助方程：

am² + bm + c = 0

Solve this quadratic for m. The form of the CF depends on the nature of the roots:

解此二次方程求m。CF的形式取决于根的性质：

Real and distinct roots m₁ ≠ m₂: y = Ae^(m₁x) + Be^(m₂x)

两个不同实根 m₁ ≠ m₂： y = Ae^(m₁x) + Be^(m₂x)
Real repeated root m: y = (A + Bx)e^(mx)

重实根 m： y = (A + Bx)e^(mx)
Complex roots m = p ± qi: y = e^(px)(A cos(qx) + B sin(qx))

共轭复根 m = p ± qi： y = e^(px)(A cos(qx) + B sin(qx))

Here A and B are arbitrary constants, determined later using initial or boundary conditions.

这里A和B为任意常数，稍后由初值或边界条件确定。

3. Finding a Particular Integral | 求特积分

The PI is a specific function that satisfies the original non-homogeneous equation. The method of undetermined coefficients (also called the method of trial functions) assumes a form for the PI based on f(x), with unknown constants to be found.

特积分是满足原非齐次方程的一个特定函数。待定系数法（也称试函数法）根据f(x)的形式假设PI的形式，其中含待定常数。

General rules for choosing the trial form:

选择试函数形式的一般规则：

If f(x) is a polynomial of degree n, try a polynomial of degree n (including all lower powers).

若f(x)是n次多项式，则试设一个n次多项式（包括所有较低次幂）。
If f(x) is a constant multiple of e^(kx), try Ce^(kx).

若f(x)是e^(kx)的常数倍，则试设Ce^(kx)。
If f(x) involves sin(kx) or cos(kx), try D sin(kx) + E cos(kx).

若f(x)涉及sin(kx)或cos(kx)，则试设D sin(kx) + E cos(kx)。

Substitute the trial PI into the differential equation and solve for the unknown coefficients.

将试设的PI代入微分方程，解出未知系数。

4. Table of Trial Particular Integrals | 特积分试设表

The table below shows common f(x) forms and the corresponding trial PI. Use this as your first guess.

下表展示了常见的f(x)形式及对应的试设PI，可作首选猜测。

f(x)	Trial PI
polynomial degree n	aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
Ae^(kx)	Ce^(kx)
A sin(kx) or A cos(kx)	D sin(kx) + E cos(kx)
Ae^(kx) sin(mx) or Ae^(kx) cos(mx)	e^(kx)[D sin(mx) + E cos(mx)]
sum of different types	sum of corresponding trial forms

If the trial PI already appears in the CF, multiply it by x (or x² for a repeated root) to make it linearly independent.

如果试设的PI已出现在CF中，则将其乘以x（重根时乘以x²）使其线性无关。

5. Worked Example 1: Polynomial f(x) | 例题1：多项式型f(x)

Solve d²y/dx² – 3(dy/dx) + 2y = 2x².

求解 d²y/dx² – 3(dy/dx) + 2y = 2x²。

Step 1: Find the CF. The auxiliary equation is m² – 3m + 2 = 0, so (m – 1)(m – 2) = 0, giving m = 1, 2. Hence:

第一步：求CF。辅助方程为 m² – 3m + 2 = 0，即 (m – 1)(m – 2) = 0，得 m = 1, 2。因此：

CF = Ae^x + Be^(2x)

Step 2: Find the PI. Since f(x) = 2x² is quadratic, try y = ax² + bx + c.

第二步：求PI。由于f(x) = 2x²是二次多项式，试设 y = ax² + bx + c。

Then dy/dx = 2ax + b, d²y/dx² = 2a. Substitute:

则 dy/dx = 2ax + b，d²y/dx² = 2a。代入得：

2a – 3(2ax + b) + 2(ax² + bx + c) = 2x²

Simplify: 2ax² + (2b – 6a)x + (2a – 3b + 2c) = 2x².

化简：2ax² + (2b – 6a)x + (2a – 3b + 2c) = 2x²。

Compare coefficients:

比较系数：

x²: 2a = 2 → a = 1
x: 2b – 6a = 0 → 2b – 6 = 0 → b = 3
constant: 2a – 3b + 2c = 0 → 2 – 9 + 2c = 0 → c = 7/2

So PI = x² + 3x + 7/2.

因此 PI = x² + 3x + 7/2。

Step 3: General solution. y = Ae^x + Be^(2x) + x² + 3x + 7/2.

第三步：通解。 y = Ae^x + Be^(2x) + x² + 3x + 7/2。

6. Worked Example 2: Exponentials and Resonance | 例题2：指数型与共振

Solve d²y/dx² + 4y = 3 sin(2x).

求解 d²y/dx² + 4y = 3 sin(2x)。

Step 1: CF. Auxiliary equation m² + 4 = 0 gives m = ±2i. Thus CF = A cos(2x) + B sin(2x).

第一步：CF。辅助方程 m² + 4 = 0 得 m = ±2i。所以 CF = A cos(2x) + B sin(2x)。

Step 2: Trial PI. Since f(x) = 3 sin(2x), the natural trial is P cos(2x) + Q sin(2x). But this is the same as the CF! Therefore multiply by x:

第二步：试设PI。由于f(x) = 3 sin(2x)，自然试设 P cos(2x) + Q sin(2x)。但这与CF相同！因此乘以x：

y = x(P cos(2x) + Q sin(2x))

Differentiate using the product rule:

用乘积法则求导：

dy/dx = P cos(2x) + Q sin(2x) + x(-2P sin(2x) + 2Q cos(2x)).

dy/dx = P cos(2x) + Q sin(2x) + x(-2P sin(2x) + 2Q cos(2x))。

d²y/dx² = -2P sin(2x) + 2Q cos(2x) + [-2P sin(2x) + 2Q cos(2x)] + x(-4P cos(2x) – 4Q sin(2x)).

d²y/dx² = -2P sin(2x) + 2Q cos(2x) + [-2P sin(2x) + 2Q cos(2x)] + x(-4P cos(2x) – 4Q sin(2x))。

Substitute into d²y/dx² + 4y = 3 sin(2x). The x terms cancel:

代入 d²y/dx² + 4y = 3 sin(2x)。x项相互抵消：

-4P sin(2x) + 4Q cos(2x) = 3 sin(2x)

Comparing coefficients: -4P = 3 → P = -3/4; 4Q = 0 → Q = 0. So PI = -3x cos(2x)/4.

比较系数：-4P = 3 → P = -3/4；4Q = 0 → Q = 0。所以 PI = -3x cos(2x)/4。

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

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

7. Superposition of Forcing Terms | 逼迫项的叠加

When f(x) is a sum of several terms, find a particular integral for each term separately and add them. Because the differential operator is linear, the sum of particular integrals is itself a particular integral for the combined equation.

当f(x)是多项之和时，分别对每一项求特积分然后相加。因为微分算子是线性的，特积分之和就是整个方程的特积分。

For example, if f(x) = 2x + 3e^(5x), try a PI of the form ax + b + ce^(5x). Substitute directly, or use the table.

例如，若f(x) = 2x + 3e^(5x)，试设 PI 的形式为 ax + b + ce^(5x)，直接代入，或查表处理。

Be careful: if one of the trial components duplicates part of the CF, apply the multiplication rule only to that component, not to the whole trial function.

注意：如果某个试设分量与CF的一部分重复，只需对该分量应用乘以x的规则，而不是对整个试设函数。

8. Special Cases: When the Trial Form Fails | 特殊情况：试设失败时

If the trial PI is not independent of the CF, the standard form will lead to an identity 0 = f(x) after substitution, giving no solution. Fix this by multiplying the failing part by x. If multiplying by x still overlaps (because the CF has a repeated root), multiply by x².

如果试设PI与CF不独立，代入后会出现0 = f(x)的恒等式，无法求解。解决方法是将失败部分乘以x。如果乘以x后仍然重叠（因为CF有重根），则乘以x²。

Example: for d²y/dx² – 2(dy/dx) + y = e^x, the CF is (A + Bx)e^x. The trial Ce^x fails (it is contained in CF). Trial Cx²e^x works because it is linearly independent.

例如：对于 d²y/dx² – 2(dy/dx) + y = e^x，CF为 (A + Bx)e^x。试设 Ce^x 失败（它包含在CF中）。试设 Cx²e^x 成功，因为它是线性无关的。

To verify the correct power of x, check the multiplicity of the root in the auxiliary equation:

要验证x的正确幂次，检查辅助方程中根的重数：

If e^(kx) is not a solution of the homogeneous equation, try Ce^(kx).
如果e^(kx)不是齐次方程的解，试设Ce^(kx)。
If e^(kx) is a simple root solution, try Cxe^(kx).
如果e^(kx)是单根解，试设Cxe^(kx)。
If e^(kx) is a repeated root solution, try Cx²e^(kx).
如果e^(kx)是重根解，试设Cx²e^(kx)。

Similar logic applies to sine/cosine terms when the auxiliary equation has purely imaginary roots.

当辅助方程有纯虚根时，对正弦/余弦项也适用类似逻辑。

9. Applying Initial and Boundary Conditions | 应用初值与边界条件

The general solution contains two arbitrary constants from the CF. To find a particular solution, apply two conditions — typically y(0) and dy/dx at 0, or two boundary values.

通解包含来自CF的两个任意常数。为求特解，需应用两个条件——通常是y(0)和0处的dy/dx，或两个边界值。

Steps:

步骤：

Write the full general solution: y = CF + PI.

写出完整通解：y = CF + PI。
Differentiate it to get dy/dx in terms of A, B and x.

对其求导，得到含A、B、x的dy/dx表达式。
Use the given conditions to set up two equations in A and B.

用给定条件建立含A、B的两个方程。
Solve for A and B and substitute back.

解出A和B并代回。

Remember that the PI itself does not contain arbitrary constants, so differentiating the general solution mixes CF derivative and PI derivative.

请记住PI本身不含有任意常数，所以对通解求导时要分别对CF和PI求导并相加。

10. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Here are frequent errors students make, and how to avoid them.

以下是学生常犯的错误以及如何避免它们。

Forgetting to include both sine and cosine terms when f(x) contains only one of them. Always try D sin(kx) + E cos(kx).

忘记同时包含正弦和余弦项，当f(x)只含其中一种时。始终试设 D sin(kx) + E cos(kx)。
Not checking resonance: if the trial form duplicates the CF, multiply by x immediately.

未检查共振：如果试设形式与CF重复，立即乘以x。
Algebraic errors when substituting: differentiate carefully, especially when the trial function includes x times sine/cosine or exponential terms.

代入时代数错误：求导要仔细，尤其是当试设函数含x乘以正弦/余弦或指数项时。
Ignoring constants of integration — the CF’s A and B are not part of the PI.

忽略积分常数——CF中的A和B不属于PI。
Using degrees instead of radians in trigonometric functions; A-Level maths always uses radians.

在三角函数中使用度而非弧度；A-Level数学中始终使用弧度。

11. Worked Example 3: Complete Solution with Conditions | 例题3：带条件的完整求解

Solve d²y/dx² + dy/dx – 6y = 10 sin(x), given y(0) = 1 and dy/dx(0) = 2.

求解 d²y/dx² + dy/dx – 6y = 10 sin(x)，已知 y(0) = 1，dy/dx(0) = 2。

Step 1: CF. m² + m – 6 = 0 → (m + 3)(m – 2) = 0 → m = -3, 2. Hence CF = Ae^(-3x) + Be^(2x).

第一步：CF。 m² + m – 6 = 0 → (m + 3)(m – 2) = 0 → m = -3, 2。所以 CF = Ae^(-3x) + Be^(2x)。

Step 2: PI. Try y = D sin(x) + E cos(x). Then dy/dx = D cos(x) – E sin(x), d²y/dx² = -D sin(x) – E cos(x).

第二步：PI。试设 y = D sin(x) + E cos(x)。则 dy/dx = D cos(x) – E sin(x)，d²y/dx² = -D sin(x) – E cos(x)。

Substitute:

代入：

[-D sin(x) – E cos(x)] + [D cos(x) – E sin(x)] – 6[D sin(x) + E cos(x)] = 10 sin(x)

Group terms:

合并同类项：

(-D – E – 6D) sin(x) + (-E + D – 6E) cos(x) = 10 sin(x)

Simplify:

化简：

(-7D – E) sin(x) + (D – 7E) cos(x) = 10 sin(x)

Thus D – 7E = 0 and -7D – E = 10. Solving gives D = -7E and -7(-7E) – E = 10 → 49E – E = 10 → 48E = 10 → E = 5/24, D = -35/24.

因此D – 7E = 0且-7D – E = 10。解得D = -7E，且 -7(-7E) – E = 10 → 49E – E = 10 → 48E = 10 → E = 5/24，D = -35/24。

PI = (-35/24) sin(x) + (5/24) cos(x).

PI = (-35/24) sin(x) + (5/24) cos(x)。

Step 3: General solution.

第三步：通解。

y = Ae^(-3x) + Be^(2x) – (35/24) sin(x) + (5/24) cos(x)

Step 4: Apply conditions. y(0) = 1 gives A + B + 5/24 = 1, so A + B = 19/24.

第四步：应用条件。 y(0) = 1 得 A + B + 5/24 = 1，即 A + B = 19/24。

dy/dx = -3Ae^(-3x) + 2Be^(2x) – (35/24) cos(x) – (5/24) sin(x).

dy/dx = -3Ae^(-3x) + 2Be^(2x) – (35/24) cos(x) – (5/24) sin(x)。

dy/dx(0) = 2 gives -3A + 2B – 35/24 = 2, so -3A + 2B = 83/24.

dy/dx(0) = 2 得 -3A + 2B – 35/24 = 2，即 -3A + 2B = 83/24。

Solve the pair:

解方程组：

A + B = 19/24, -3A + 2B = 83/24

Multiply the first by 3: 3A + 3B = 57/24. Add to the second: 5B = 140/24 = 35/6, so B = 7/6. Then A = 19/24 – 7/6 = 19/24 – 28/24 = -9/24 = -3/8.

第一式乘以3：3A + 3B = 57/24。加到第二式：5B = 140/24 = 35/6，所以B = 7/6。则A = 19/24 – 7/6 = 19/24 – 28/24 = -9/24 = -3/8。

Final answer:

最终答案：

y = -(3/8)e^(-3x) + (7/6)e^(2x) – (35/24) sin(x) + (5/24) cos(x)

12. Summary and Formula Sheet | 总结与公式速查

To solve any second-order linear differential equation with constant coefficients:

解任何常系数二阶线性微分方程的步骤：

Find the CF via the auxiliary equation.

通过辅助方程求CF。
Choose a trial PI from the table.

从表中选择试设PI。
Modify by x if necessary to avoid duplication with CF.

必要时乘以x以避免与CF重叠。
Substitute to determine the coefficients.

代入求解系数。
Add CF and PI; apply conditions to find A and B.

将CF与PI相加；应用条件求A和B。

Key formulas:

关键公式：

am² + bm + c = 0 → CF depends on roots

Distinct real: Ae^(m₁x) + Be^(m₂x)

Repeated real: (A + Bx)e^(mx)

Complex p ± qi: e^(px)(A cos(qx) + B sin(qx))

Always check your final solution by substituting it back into the original differential equation — a quick and powerful verification tool.

始终通过将最终解代回原微分方程进行检验——这是一个快速而强大的验证工具。

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Complementary Functions and Particular Integrals | 补函数与特积分