📚 AQA A-Level Further Maths Paper 3 (FP2) Example Responses | AQA A-Level 高等数学 Paper 3 (FP2) 高分范例
This article demonstrates how to structure full-mark solutions for AQA A-Level Further Maths Paper 3 (Unit FP2). Each example is based on a question style commonly seen in the exam, with model answers, examiner commentary, and marks-focused written solutions. The goal is to show you the level of rigour and clarity expected in your own work.
本文以 AQA A-Level 高等数学 Paper 3（FP2 单元）常见题型为例，提供完整的范例作答、考官点评与得分要点。我们希望通过这些规范的书写示范，帮助你在考试中写出清晰、严谨、拿满分的答案。
1. Integrating Factor for First-Order Differential Equations | 一阶微分方程的积分因子法
Typical question: Solve, for x > 0, the differential equation dy/dx + (2/x)y = x³, given that y = 3 when x = 1.
典型例题：解微分方程 dy/dx + (2/x)y = x³，其中 x > 0，且当 x = 1 时 y = 3。
Model response:
范例作答：
- Identify the integrating factor (IF): IF = e^(∫ 2/x dx) = e^(2 ln x) = x².
- 确定积分因子：IF = e^(∫ 2/x dx) = e^(2 ln x) = x²。
Integrating factor = x²
积分因子 = x²
- Multiply both sides by x²: x² dy/dx + 2xy = x⁵.
- 两边同时乘以 x²：x² dy/dx + 2xy = x⁵。
The left-hand side now equals d/dx (x² y), so we can integrate directly:
此时左边等于 d/dx (x² y)，因此可直接积分：
d/dx (x² y) = x⁵ ⇒ x² y = x⁶/6 + C
- Use the initial condition y = 3 when x = 1: 1² × 3 = 1⁶/6 + C ⇒ C = 17/6.
- 代入初值条件 x = 1，y = 3：1² × 3 = 1⁶/6 + C ⇒ C = 17/6。
y = x⁴/6 + 17/(6x²)
Commentary: Always state the integrating factor explicitly. Show the product-rule step before integrating. Use the initial condition even if it appears trivial. These three habits guarantee method marks.
点评：务必明确写出积分因子；积分前展示乘积法则的转化步骤；即使初值条件看似简单也一定要代入。这三点习惯能确保你拿到方法分。
2. Second-Order Differential Equations with a Particular Integral | 二阶微分方程的特解
Typical question: Solve d²y/dx² + 3 dy/dx + 2y = e^(2x), given that y = 0 and dy/dx = 1 when x = 0.
典型例题：解微分方程 d²y/dx² + 3 dy/dx + 2y = e^(2x)，其中 x = 0 时 y = 0，dy/dx = 1。
Model response:
范例作答：
Write the auxiliary equation and solve it:
先写出辅助方程并求解：
m² + 3m + 2 = 0 ⇒ (m + 1)(m + 2) = 0 ⇒ m = -1, -2
Thus the complementary function is:
因此补函数为：
y_c = Ae^(-x) + Be^(-2x)
For the particular integral, try y = λe^(2x). Since y’ = 2λe^(2x) and y” = 4λe^(2x), substitution gives:
试特解 y = λe^(2x)。由 y’ = 2λe^(2x)，y” = 4λe^(2x)，代入得：
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