A-Level Further Maths Unit 4 Mark Scheme Jan 20 Key Points | A-Level 进阶数学第四单元（2020年1月评分方案）知识点精讲

📚 A-Level Further Maths Unit 4 Mark Scheme Jan 20 Key Points | A-Level 进阶数学第四单元（2020年1月评分方案）知识点精讲

This comprehensive guide breaks down the core topics tested in the A-Level Further Mathematics Unit 4 examination, with a particular focus on the insights provided by the January 2020 mark scheme. Whether you are preparing for Edexcel IAL Further Pure 2 or a similar specification, understanding how marks are awarded for method, accuracy, and final answers is just as critical as knowing the mathematical content itself. The January 2020 paper demanded a strong command of complex numbers, polar coordinates, hyperbolic functions, advanced integration, and differential equations. By studying the mark scheme, we can identify exactly where candidates commonly lost marks due to insufficient justification, missing limits, or algebraic slips.

本精讲全面解析A-Level进阶数学第四单元考试涉及的核心专题，并特别聚焦2020年1月评分方案所揭示的得分要点。无论你备考的是Edexcel IAL Further Pure 2还是类似大纲的考试，理解评分标准中对方法、准确性和最终答案的给分方式，与掌握数学内容本身同样重要。2020年1月试卷重点考察了复数、极坐标、双曲函数、高级积分技巧及微分方程等知识点。通过分析评分方案，我们可以准确找出考生因理由阐述不充分、遗漏积分限或代数疏漏而失分的高频原因。

1. Overview of Unit 4 (Further Pure 2) and Mark Scheme Insights | 第四单元概览与评分方案洞察

Unit 4 in the International A-Level Further Mathematics sequence typically corresponds to Further Pure Mathematics 2, a paper that extends pure mathematical techniques well beyond the standard A-Level. The January 2020 paper featured questions on solving inequalities involving moduli, using de Moivre’s theorem to sum trigonometric series, converting between polar and Cartesian coordinates to calculate areas, differentiating and integrating hyperbolic functions, and solving second-order differential equations. The mark scheme reveals that examiners explicitly reward correct manipulation of trigonometric identities in complex number proofs and award accuracy marks only when final expressions are fully simplified.

国际A-Level进阶数学序列中的第四单元通常对应Further Pure Mathematics 2，该试卷将纯数学技巧拓展到远超标准A-Level的程度。2020年1月试卷包含的题型有：求解含绝对值的不等式、利用棣莫弗定理化简三角级数的和、在极坐标与直角坐标之间转换以计算面积、对双曲函数进行微分和积分，以及求解二阶微分方程。评分方案显示，阅卷人会明确奖励复数证明过程中三角恒等式的正确运用，并且只有当最终表达式完全化简后才给予精确答案分。

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2. Inequalities with Modulus and Polynomials | 含绝对值和多项式的不等式

A common question type requires solving inequalities such as |2x + 1| > 3|x − 2|. The safest approach is to square both sides, because both sides are non‑negative, to obtain (2x + 1)² > 9(x − 2)². Next, expand and simplify to a quadratic inequality: 4x² + 4x + 1 > 9x² − 36x + 36, which leads to 0 > 5x² − 40 x + 35, or equivalently x² − 8x + 7 < 0. Factorising gives (x − 1)(x − 7) < 0, so the solution is 1 < x < 7. The mark scheme clearly awards a method mark for squaring and an accuracy mark for the correct interval; omitting the strict inequality sign or writing the interval in an incorrect format causes a loss of an answer mark.

一种常见题型是求解类似于 |2x + 1| > 3|x − 2| 的不等式。最稳妥的方法是两边同时平方，因为两边均为非负数，得到 (2x + 1)² > 9(x − 2)²。接着展开并化简为一个二次不等式：4x² + 4x + 1 > 9x² − 36x + 36，进而推出 0 > 5x² − 40x + 35，即 x² − 8x + 7 < 0。因式分解得 (x − 1)(x − 7) < 0，因此解集为 1 < x < 7。评分方案明确对平方步骤给予方法分，对正确的区间给予准确分；如果遗漏严格不等号或以错误格式书写区间，会导致丢失答案分。

For rational inequalities like (x + 2)/(x − 3) ≤ 0, candidates often forget to consider where the denominator is zero. The critical values are x = −2 and x = 3. A sign diagram shows the expression is negative between −2 and 3; however, x = 3 makes the denominator zero, so it must be excluded, giving the final solution −2 ≤ x < 3. The mark scheme insists on strictly open intervals at any value that makes a denominator zero – a single inclusion error loses the final accuracy mark.

对于类似 (x + 2)/(x − 3) ≤ 0 的有理不等式，考生常忽略分母为零的情况。关键值为 x = −2 和 x = 3。符号表显示该表达式在 −2 和 3 之间取负值；但 x = 3 会使分母为零，因此必须排除，最终解为 −2 ≤ x < 3。评分方案要求在使分母为零的点处必须使用开区间 —— 一次包含错误就会丢失最后的准确性分数。

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3. Complex Numbers: De Moivre’s Theorem and Roots of Unity | 复数：棣莫弗定理与单位根

De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. The January 2020 paper frequently tested the application of this theorem to express cos 3θ in terms of cos θ, or to find sums such as Σ cos kθ. One typical solution expands (cos θ + i sin θ)³ using the binomial theorem: cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ. Equating the real part to cos 3θ and using sin²θ = 1 − cos²θ yields cos 3θ = 4 cos³θ − 3 cos θ. The mark scheme assigns marks for both the expansion and the use of the identity, but a significant number of candidates lost marks by miswriting the imaginary unit or failing to replace sin²θ correctly.

棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。2020年1月试卷多次考查了应用该定理将 cos 3θ 用 cos θ 表示，或求如 Σ cos kθ 的和。典型解法是利用二项式定理展开 (cos θ + i sin θ)³：cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ。将实部与 cos 3θ 对应，并使用 sin²θ = 1 − cos²θ，即可得出 cos 3θ = 4 cos³θ − 3 cos θ。评分方案对展开步骤和恒等式应用各赋分值，但不少考生因误写虚数单位或未正确替换 sin²θ 而失分。

Roots of unity appear when solving zⁿ = 1. The n distinct roots are given by z = cis(2kπ/n) for k = 0, 1, …, n−1. When asked to find the roots of z⁵ = 1 and plot them on an Argand diagram, candidates must label the points clearly and note that they lie on the unit circle. The mark scheme penalises diagrams that lack an indication of the unit radius or that misplace the roots symmetrically. Moreover, questions frequently ask for the sum of the roots, which, by Vieta’s formulas, is zero, and the product, which is (−1)ⁿ⁻¹. However, simple geometric reasoning alone can verify that the vectors sum to zero.

求解 zⁿ = 1 时即出现单位根。n 个互异的根由 z = cis(2kπ/n) (k = 0, 1, …, n−1) 给出。当题目要求找出 z⁵ = 1 的根并在阿根图上绘制时，考生必须清晰标记各点并指出它们位于单位圆上。评分方案会扣罚未标注单位半径或根的对称位置错误的图。此外，常见题目是求根的和，根据韦达定理和为零；求根的积，为 (−1)ⁿ⁻¹。不过，仅凭简单的几何推理也能验证这些向量之和为零。

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4. Polar Coordinates: Area and Arc Length | 极坐标：面积与弧长

Polar curves defined by r = f(θ) often require the calculation of the area enclosed by a curve between two rays. The formula is A = ½ ∫ r² dθ, with the limits taken from the least to the greatest θ that trace out the region exactly once. In a typical Jan 2020 problem, candidates had to find the area of one loop of r = a sin 2θ. The loop is traced when 2θ runs from 0 to π, so θ ranges from 0 to π/2. The area becomes ½ ∫₀^{π/2} a² sin² 2θ dθ. Using cos 4θ = 1 − 2 sin² 2θ gives sin² 2θ = (1 − cos 4θ)/2, and integration yields A = ½ a² [θ/2 − (sin 4θ)/8]₀^{π/2} = (π a²)/8. The mark scheme strictly requires candidates to state the correct limits – a wrong lower limit from confusion about where the pole occurs leads to no marks for the integration.

由 r = f(θ) 定义的极坐标曲线常需计算介于两射线间的曲线所围面积。公式为 A = ½ ∫ r² dθ，积分限取恰好遍历该区域一次的最小 θ 至最大 θ。在2020年1月一道典型题目中，考生需求出 r = a sin 2θ 一瓣的面积。当 2θ 从 0 变到 π 时遍历一瓣，因此 θ 的范围为 0 到 π/2。面积变为 ½ ∫₀^{π/2} a² sin² 2θ dθ。利用 cos 4θ = 1 − 2 sin² 2θ 得到 sin² 2θ = (1 − cos 4θ)/2，积分后得 A = ½ a² [θ/2 − (sin 4θ)/8]₀^{π/2} = (π a²)/8。评分方案严格规定考生必须写出正确的积分限 —— 因混淆极点位置而写错下限，将导致整个积分部分不得分。

For arc length, the formula is s = ∫ √(r² + (dr/dθ)²) dθ. While the Jan 2020 paper focused more on area, the mark scheme indicated that a small number of candidates mistakenly attempted to use arc length for area problems, mixing up the two formulae. A solid tip is to always write down the appropriate formula before substituting any values; the mark scheme awards a method mark for quoting the correct formula even if subsequent algebra contains errors.

对于弧长，公式为 s = ∫ √(r² + (dr/dθ)²) dθ。虽然2020年1月试卷更侧重面积题，但评分方案显示，少数考生错误地在面积题中套用了弧长公式，混淆了两个公式。一个扎实的技巧是，在代入任何数值之前先写出相应的公式；即使后续代数运算有错，评分方案仍会奖励正确引用公式的方法分。

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5. Hyperbolic Functions: Identities and Equations | 双曲函数：恒等式与方程

Hyperbolic functions are defined as sinh x = (eˣ − e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2, and tanh x = sinh x / cosh x. Their identities closely mirror trigonometric ones but with key sign differences, such as cosh² x − sinh² x = 1 and sinh 2x = 2 sinh x cosh x. A common Jan 2020 question required solving an equation like 5 sinh x − 3 cosh x = 4. Expressing sinh and cosh in exponential form gives 5((eˣ − e⁻ˣ)/2) − 3((eˣ + e⁻ˣ)/2) = 4, which simplifies to (2eˣ − 8e⁻ˣ)/2 = 4, leading to eˣ − 4e⁻ˣ = 4. Multiplying by eˣ yields the quadratic in eˣ: e²ˣ − 4eˣ − 4 = 0, so eˣ = 2 ± √8. Since eˣ > 0, we take eˣ = 2 + 2√2, giving x = ln(2 + 2√2). The mark scheme gives credit for the correct substitution, but many students lose marks by poor handling of the negative root or by failing to simplify the logarithm.

双曲函数定义为 sinh x = (eˣ − e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2，tanh x = sinh x / cosh x。它们的恒等式与三角函数很相似，但存在关键符号差别，比如 cosh² x − sinh² x = 1 和 sinh 2x = 2 sinh x cosh x。2020年1月一道常见考题要求解方程 5 sinh x − 3 cosh x = 4。将 sinh 和 cosh 写成指数形式得到 5((eˣ − e⁻ˣ)/2) − 3((eˣ + e⁻ˣ)/2) = 4，化简后为 (2eˣ − 8e⁻ˣ)/2 = 4，进而推出 eˣ − 4e⁻ˣ = 4。两边乘以 eˣ 得到关于 eˣ 的二次方程：e²ˣ − 4eˣ − 4 = 0，所以 eˣ = 2 ± √8。由于 eˣ > 0，我们取 eˣ = 2 + 2√2，从而 x = ln(2 + 2√2)。评分方案对正确代换给分，但许多学生因处理负根不当或未简化对数而失分。

A second frequent task is to prove identities such as tanh² x + sech² x = 1. Starting from the definitions, cosh² x − sinh² x = 1, divide through by cosh² x to obtain 1 − tanh² x = sech² x, which rearranges to the required form. The mark scheme explicitly states that merely quoting the standard formula without showing division by cosh² x does not earn the proof mark; the intermediate step must be shown.

另一常见任务是证明如 tanh² x + sech² x = 1 的恒等式。从定义出发，cosh² x − sinh² x = 1，两边同除以 cosh² x 即得 1 − tanh² x = sech² x，整理后即为所求形式。评分方案明确说明，仅引用标准公式而不展示除以 cosh² x 的步骤，不能获得证明分；必须呈现中间推导步骤。

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6. Differentiation: Inverse Trigonometric and Hyperbolic Functions | 微分：反三角函数与双曲函数

Unit 4 expects fluency in differentiating inverse functions. For example, the derivative of y = arsinh x (also written sinh⁻¹ x) is dy/dx = 1/√(1 + x²). Similarly, d/dx (arcosh x) = 1/√(x² − 1) for x > 1, and d/dx (artanh x) = 1/(1 − x²) for |x| < 1. In the January 2020 paper, candidates were asked to differentiate y = x² arsinh(2x). Using the product rule, dy/dx = 2x arsinh(2x) + x² ⋅ [1/√(1+(2x)²)] ⋅ 2. The mark scheme rewards the correct application of the chain rule inside the arsinh derivative; forgetting the factor of 2 from the argument 2x was a pervasive error that lost the accuracy mark.

第四单元要求熟练微分反函数。例如，y = arsinh x（也写作 sinh⁻¹ x）的导数为 dy/dx = 1/√(1 + x²)。同理，对于 x > 1，d/dx (arcosh x) = 1/√(x² − 1)；对于 |x| < 1，d/dx (artanh x) = 1/(1 − x²)。在2020年1月试卷中，考生被要求微分 y = x² arsinh(2x)。利用乘法法则，dy/dx = 2x arsinh(2x) + x² ⋅ [1/√(1+(2x)²)] ⋅ 2。评分方案对 arsinh 导数中正确运用链式法则得分奖励；而忘记由自变量 2x 产生的因子 2 则是一个普遍错误，导致失去准确分。

For inverse trigonometric functions, the derivatives are d/dx (arcsin x) = 1/√(1 − x²), d/dx (arccos x) = −1/√(1 − x²), and d/dx (arctan x) = 1/(1 + x²). The mark scheme places heavy emphasis on the domain and the sign; for instance, the derivative of arccos x is negative, and using the positive version can invalidate a following integration problem. A helpful exam technique is to always write the derivative with its valid domain next to it, ensuring you don’t apply a formula outside its region of validity.

对于反三角函数，其导数为 d/dx (arcsin x) = 1/√(1 − x²)，d/dx (arccos x) = −1/√(1 − x²)，以及 d/dx (arctan x) = 1/(1 + x²)。评分方案非常注重定义域和符号；例如，arccos x 的导数为负，使用正号版本会使后续的积分问题失效。一个有用的应试技巧是，写导数时总在旁边注明其有效定义域，确保没有在有效区间之外套用公式。

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7. Integration Techniques: Reduction Formulae and Substitution | 积分技巧：递推公式与换元法

Reduction formulae feature prominently in Further Pure 2. A classic example is Iₙ = ∫₀^{π/2} sinⁿ x dx. By writing sinⁿ x = sinⁿ⁻¹ x sin x and integrating by parts, one obtains the recurrence Iₙ = ((n − 1)/n) Iₙ₋₂. The mark scheme for a proof question gives marks for setting u = sinⁿ⁻¹ x and dv = sin x dx, correctly applying integration by parts, and then using cos² x = 1 − sin² x to reintroduce Iₙ and Iₙ₋₂. Many candidates lose marks by failing to explain the limits properly or by mishandling the boundary term [−cos x sinⁿ⁻¹ x]₀^{π/2}, which actually evaluates to zero for n > 1.

递推公式在Further Pure 2中占据显著地位。一个经典例子是 Iₙ = ∫₀^{π/2} sinⁿ x dx。写出 sinⁿ x = sinⁿ⁻¹ x sin x 并利用分部积分，可得到递推关系 Iₙ = ((n − 1)/n) Iₙ₋₂。证明题的评分方案会对令 u = sinⁿ⁻¹ x、dv = sin x dx，正确应用分部积分，然后利用 cos² x = 1 − sin² x 重新引入 Iₙ 和 Iₙ₋₂ 的步骤分别给分。许多考生因未恰当解释积分限，或错误处理边界项 [−cos x sinⁿ⁻¹ x]₀^{π/2} （对于 n > 1 该项实际为零）而失分。

Hyperbolic substitution is another key tool, particularly for integrals involving √(x² + a²) or √(x² − a²). For instance, to integrate √(x² + 4) dx, one can substitute x = 2 sinh u. Then dx = 2 cosh u du, and the integral becomes ∫ 2 cosh u ⋅ 2 cosh u du = 4 ∫ cosh² u du. Using the double-angle identity, 4 ∫ (1 + cosh 2u)/2 du = 2u + sinh 2u + C. Back-substitution through u = arsinh(x/2) and simplifying the sinh 2u term to (x/2)√(1 + (x/2)²) times 2 gives the final answer in terms of x. The mark scheme rewards careful handling of the constant and the explicit back-substitution steps; simply writing the answer without showing u → x conversions is heavily penalised.

双曲换元是另一关键工具，尤其适用于包含 √(x² + a²) 或 √(x² − a²) 型的积分。例如，计算 ∫ √(x² + 4) dx，可令 x = 2 sinh u。则 dx = 2 cosh u du，积分变为 ∫ 2 cosh u ⋅ 2 cosh u du = 4 ∫ cosh² u du。利用倍角恒等式，4 ∫ (1 + cosh 2u)/2 du = 2u + sinh 2u + C。通过 u = arsinh(x/2) 进行反代换，并把 sinh 2u 项化简为 (x/2)√(1 + (x/2)²) 乘以 2，得到以 x 表示的最终答案。评分方案奖励对常数项的细致处理以及明确展示反代换步骤；只写出答案而缺乏 u 到 x 的转换过程会被严重扣分。

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8. Maclaurin Series and Series Expansions | 麦克劳林级数与级数展开

The Maclaurin series expansion of a function f(x) is f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + …. In the Jan 2020 paper, a question required the series expansion of ln(1 + sin x) up to the term in x³. The derivative approach is demanding; a more elegant method uses the known expansions sin x = x − x³/6 + … and ln(1 + u) = u − u²/2 + u³/3 − …, substituting u = sin x, and retaining terms up to x³. This gives (x − x³/6) − ½(x − x³/6)² + ⅓(x³) = x − x³/6 − ½(x² − x⁴/3 + x⁶/36) + ⅓x³. Discarding terms higher than x³ yields x − ½x² + (−1/6 + ⅓)x³ = x − ½x² + x³/6. The mark scheme explicitly requires the justification of discarding higher-order terms; a statement like “terms in x⁴ and above are ignored” is adequate and earns a method mark.

函数 f(x) 的麦克劳林级数展开式为 f(0) + f ‘(0)x + f ”(0)x²/2! + f ”'(0)x³/3! + …。在2020年1月试卷中，一道题要求展开 ln(1 + sin x) 至 x³ 项。直接求导的方法较为繁琐；更优雅的方法是使用已知展开式 sin x = x − x³/6 + … 和 ln(1 + u) = u − u²/2 + u³/3 − …，代入 u = sin x，并保留至 x³ 项。这样得到 (x − x³/6) − ½(x − x³/6)² + ⅓(x³) = x − x³/6 − ½(x² − x⁴/3 + x⁶/36) + ⅓x³。舍去高于 x³ 的项后得到 x − ½x² + (−1/6 + ⅓)x³ = x − ½x² + x³/6。评分方案明确要求对舍弃高阶项给出说明；像 “x⁴ 及以上项忽略” 这样的陈述即足以得到方法分。

Another Maclaurin-based question involved approximating a definite integral using a series. For example, evaluating ∫₀^{0.5} (eˣ² − 1)/x dx can be done by expanding eˣ² = 1 + x² + x⁴/2! + … , so (eˣ² − 1)/x = x + x³/2 + …. Integrating term by term gives ∫₀^{0.5} (x + x³/2) dx = [x²/2 + x⁴/8]₀^{0.5} = 0.125 + 0.0078125 = 0.1328125. The mark scheme awards accuracy marks for correct term-by-term integration but insists that the approximate value be expressed to an appropriate number of significant figures as requested in the question; failing to round correctly loses a mark.

另一道基于麦克劳林级数的题目涉及用级数近似计算定积分。例如，求 ∫₀^{0.5} (eˣ² − 1)/x dx，可通过展开 eˣ² = 1 + x² + x⁴/2! + …，从而 (eˣ² − 1)/x = x + x³/2 + …。逐项积分得 ∫₀^{0.5} (x + x³/2) dx = [x²/2 + x⁴/8]₀^{0.5} = 0.125 + 0.0078125 = 0.1328125。评分方案对正确逐项积分给予准确分，但坚持要求近似值按题目要求的有效数字位数表示；未正确四舍五入会丢分。

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9. First-Order Differential Equations: Integrating Factor | 一阶微分方程：积分因子法

First-order linear differential equations of the form dy/dx + P(x)y = Q(x) are solved using an integrating factor μ = e^{∫ P dx}. A typical Jan 2020 example asked to solve x dy/dx + 2y = sin x / x. First, divide through by x to get the standard form dy/dx + (2/x)y = (sin x)/x². The integrating factor is μ = e^{∫ (2/x) dx} = e^{2 ln x} = x². Multiplying through by x² gives x² dy/dx + 2xy = sin x. The left-hand side is the derivative of (x²y), so d/dx (x² y) = sin x. Integrating both sides yields x² y = −cos x + C, so y = (−cos x + C)/x². The mark scheme assigns a method mark for correctly finding the integrating factor and another for recognising the left side as an exact derivative; however, a frequent mistake is forgetting to divide by x initially, which prevents the equation from being standard and loses all subsequent marks.

形如 dy/dx + P(x)y = Q(x) 的一阶线性微分方程可用积分因子 μ = e^{∫ P dx} 求解。2020年1月一道典型题要求解 x dy/dx + 2y = sin x / x。首先，两边除以 x 得到标准形式 dy/dx + (2/x)y = (sin x)/x²。积分因子为 μ = e^{∫ (2/x) dx} = e^{2 ln x} = x²。两边乘以 x² 得 x² dy/dx + 2xy = sin x。左边正是 (x²y) 的导数，因此 d/dx (x² y) = sin x。两边积分得 x² y = −cos x + C，故 y = (−cos x + C)/x²。评分方案对正确求出积分因子给予方法分，并对识别左边为恰当导数再给一分；然而，一个常见错误是忘记初始除以 x，导致方程不标准，从而失去全部后续分数。

When an initial condition is given, say y = 1 at x = π/2, it must be substituted promptly to find C. In the above, when x = π/2, we have (π/2)² ⋅ 1 = −cos(π/2) + C ⇒ π²/4 = 0 + C, so C = π²/4. The final answer with the particular solution is expected. The mark scheme reveals that many candidates attempted to rearrange for C after writing the general solution, but made errors in evaluating the trigonometric function at the given x; using a calculator in radian mode is essential.

当给定初始条件，比如 x = π/2 时 y = 1，则必须立即代入以求出常数 C。在上例中，当 x = π/2 时，有 (π/2)² ⋅ 1 = −cos(π/2) + C ⇒ π²/4 = 0 + C，因此 C = π²/4。此时期望给出带特解的最终答案。评分方案揭示，许多考生在写出通解后尝试求解 C，但在计算给定 x 处的三角函数值时出错；务必使用计算器的弧度模式。

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10. Second-Order Differential Equations with Constant Coefficients | 常系数二阶微分方程

The general solution to a d²y/dx² + b dy/dx + c y = f(x) consists of the complementary function (CF) and a particular integral (PI). For the homogeneous case, the auxiliary equation am² + bm + c = 0 yields roots m₁ and m₂. If real and distinct, the CF is y = A e^{m₁ x} + B e^{m₂ x}; if repeated, y = (A + Bx)e^{mx}; if complex (p ± iq), y = e^{px} (C cos qx + D sin qx). In the Jan 2020 paper, a problem with f(x) = eᵏˣ required finding the PI by trying y = λ eᵏˣ, provided k is not a root of the auxiliary equation; when it is a root, the trial function must be multiplied by x. Many marks were lost because candidates failed to adjust the trial function correctly, leading to an inconsistent equation and no marks for the PI.

方程 a d²y/dx² + b dy/dx + c y = f(x) 的通解由余函数 (CF) 和特积分 (PI) 组成。对于齐次情形，辅助方程 am² + bm + c = 0 给出根 m₁ 和 m₂。若为相异实根，则 CF 为 y = A e^{m₁ x} + B e^{m₂ x}；若为重根，为 y = (A + Bx)e^{mx}；若为共轭复根 (p ± iq)，则为 y = e^{px} (C cos qx + D sin qx)。在2020年1月试卷中，有一道 f(x) = eᵏˣ 的题目，当 k 不是辅助方程的根时，尝试特积分 y = λ eᵏˣ 可求出 PI；而当它是根时，试函数必须乘以 x。许多考生因未能正确调整试函数而失分，导致恒等式矛盾，整道 PI 题不得分。

For f(x) = poly(x) sin ωx or cos ωx, the PI trial form needs both sine and cosine terms of the same polynomial degree. The Jan 2020 mark scheme indicates that the examiners award method marks for writing the correct trial form and substituting it into the left-hand side, then equating coefficients. A slip in the derivative of the trial PI is the most common algebraic mistake; double-checking the differentiation, especially with product rule terms involving x, is strongly recommended.

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