📚 Worked Solutions in IB Mathematics | IB数学解题示范
Mastering mathematics in the IB Diploma requires more than just knowing formulas – it demands clear, logical, and well-structured solutions. In this article, we present fully worked solutions for two typical IB-style questions: a first-order differential equation and a vector problem. Each solution is broken down into small, digestible steps, with explanations in both English and Chinese to support bilingual learners. By studying these examples, you will learn how to organise your reasoning, avoid common pitfalls, and communicate your mathematical thinking effectively.
在IB文凭课程中掌握数学，不仅需要记住公式，更需要清晰、逻辑严谨、结构良好的解题过程。本文展示了两道典型IB风格题目的详细解答：一道是一阶微分方程，另一道是向量问题。每个解答被分解成小而易于理解的步骤，并配有中英双语解释，以帮助双语学习者。通过学习这些示例，你将学会如何组织推理、避免常见错误，并有效传达你的数学思维。
1. Problem Statement (Example 1) | 问题陈述（示例一）
Solve the differential equation dy/dx = y/x + x, with the initial condition y(1) = 2.
求解微分方程 dy/dx = y/x + x，并满足初始条件 y(1) = 2。
2. Identify the Type and Method | 识别方程类型并选择方法
P(x) = −1/x and Q(x) = x. The standard solution method uses an integrating factor μ(x) = e∫P(x)dx.
将方程改写成 dy/dx − (1/x)y = x。这是一个一阶线性微分方程，形式为 dy/dx + P(x)y = Q(x)，其中 P(x) = −1/x，Q(x) = x。标准解法使用积分因子 μ(x) = e∫P(x)dx。
3. Find the Integrating Factor | 求积分因子
Compute ∫P(x)dx = ∫(−1/x) dx = −ln|x|. Therefore, the integrating factor is μ(x) = e−ln|x| = 1/x (taking the positive branch for x > 0).
计算 ∫P(x)dx = ∫(−1/x) dx = −ln|x|。因此积分因子为 μ(x) = e−ln|x| = 1/x（取正分支，x > 0）。
4. Multiply Through by the Integrating Factor and Integrate | 乘以积分因子并积分
Multiply both sides of the standard form by 1/x:
(1/x) × (dy/dx) − (1/x²) × y = 1
This can be written as the derivative of a product:
d/dx (y/x) = 1
Integrate both sides with respect to x: ∫ d/dx (y/x) dx = ∫ 1 dx ⟹ y/x = x + C, where C is the constant of integration.
将标准形式两边乘以 1/x：
(1/x) × (dy/dx) − (1/x²) × y = 1
这可以写成某个乘积的导数：
d/dx (y/x) = 1
对 x 积分两边：∫ d/dx (y/x) dx = ∫ 1 dx ⟹ y/x = x + C，其中 C 为积分常数。
5. Solve for y and Apply the Initial Condition | 解出 y 并代入初始条件
Multiply through by x to obtain the general solution: y = x² + Cx.
两边乘以 x 得到通解：y = x² + Cx。
Now use y(1) = 2: 2 = (1)² + C(1) ⟹ 2 = 1 + C ⟹ C = 1.
现在代入 y(1) = 2：2 = (1)² + C(1) ⟹ 2 = 1 + C ⟹ C = 1。
6. Final Solution for Example 1 | 示例一的最终解
The particular solution satisfying the initial condition is y = x² + x.
满足初始条件的特解为 y = x² + x。
7. Problem Statement (Example 2) | 问题陈述（示例二）
and b = i − 2j + 2k, find the angle between a and b and the vector projection of a onto b.
给定向量 a = 2i + 3j − k 和 b = i − 2j + 2k，求 a 与 b 之间的夹角以及 a 在 b 上的向量投影。
8. Compute the Dot Product and Magnitudes | 计算点积与模长
The dot product a·b = (2)(1) + (3)(−2) + (−1)(2) = 2 − 6 − 2 = −6.
点积 a·b = (2)(1) + (3)(−2) + (−1)(2) = 2 − 6 − 2 = −6。
Magnitude of a: |a| = √(2² + 3² + (−1)²) = √4 + 9 + 1 = √14.
a 的模：|a| = √(2² + 3² + (−1)²) = √4 + 9 + 1 = √14。
Magnitude of b: |b| = √(1² + (−2)² + 2²) = √1 + 4 + 4 = √9 = 3.
b 的模：|b| = √(1² + (−2)² + 2²) = √1 + 4 + 4 = √9 = 3。
9. Calculate the Angle Between the Vectors | 计算向量间夹角
Using the formula cos θ = (a·b) / (|a| |b|):
cos θ = (−6) / (√14 × 3) = −2 / √14
Therefore, θ = arccos(−2/√14). This is an obtuse angle since the cosine is negative.
使用公式 cos θ = (a·b) / (|a| |b|)：
cos θ = (−6) / (√14 × 3) = −2 / √14
因此，θ = arccos(−2/√14)。由于余弦值为负，该角度为钝角。
10. Find the Scalar Projection of a onto b | 求 a 在 b 上的标量投影
The scalar projection (also called the component) of a onto b is given by (a·b) / |b|.
a 在 b 上的标量投影（也称为分量）为 (a·b) / |b|。
Scalar projection = −6 / 3 = −2
标量投影 = −6 / 3 = −2
11. Determine the Vector Projection of a onto b | 确定 a 在 b 上的向量投影
The vector projection is found by multiplying the scalar projection by the unit vector in the direction of b:
projb a = ((a·b) / |b|²) × b
向量投影通过将标量投影乘以 b 方向上的单位向量得到：
projb a = ((a·b) / |b|²) × b
Since |b| and a·b = −6, we have ((−6) / 9) × b = (−2/3) × (i − 2j + 2k) = (−2/3)i + (4/3)j − (4/3)k.
因为 |b|² = 9 且 a·b = −6，所以 ((−6) / 9) × b = (−2/3) × (i − 2j + 2k) = (−2/3)i + (4/3)j − (4/3)k。
12. Summary of Key Takeaways | 主要收获总结
Worked solutions reveal the logical flow behind each problem. For differential equations, always start by rewriting into standard form and identifying the integrating factor. For vector problems, systematic computation of dot products and magnitudes leads directly to angle and projection results. Practice writing out every step – even the elementary algebraic simplifications – to build accuracy and confidence for the IB exams.
详细的解题过程揭示了每个问题背后的逻辑流程。对于微分方程，始终要先改写成标准形式并找出积分因子。对于向量问题，系统化地计算点积与模长，便可直接得出夹角和投影结果。在练习中写出每一个步骤——即使是基本的代数化简——也有助于提高准确性并增强应对IB考试的信心。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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