AQA A-Level Further Maths Paper 3 (FP2) Example Responses | AQA A-Level 高等数学 Paper 3 (FP2) 高分范例

📚 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。

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1. Integrating Factor for First-Order Differential Equations | 一阶微分方程的积分因子法

Typical question: ve, : 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。

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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：

Published by TutorHao | Mathematics Revision Series | aleveler.com

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