📚 Edexcel A-Level Maths Problem pdfjoiner_(4)-021 | Edexcel A-Level 数学题 pdfjoiner_(4)-021
ves solving a first-order linear differential equation using an integrating factor and applying an initial condition to find the particular solution. The detailed steps, common pitfalls, and conceptual explanations are included to strengthen your exam technique.
本文为一道典型的 Edexcel A-Level 纯数题目（编号 pdfjoiner_(4)-021）提供详尽解析。该题要求使用积分因子求解一阶线性微分方程，并结合初始条件确定特解。文中包含详细步骤、常见错误以及概念讲解，助你提升应试技巧。
1. Problem Statement | 题目叙述
Consider the differential equation dy/dx + 2y = e⁻ˣ, with the initial condition y(0) = 1. Find the particular solution for y in terms of x.
考虑微分方程 dy/dx + 2y = e⁻ˣ，初始条件为 y(0) = 1。求 y 关于 x 的特解。
2. Recognising the Equation Type | 识别方程类型
inear differential equation of the form dy/dx + P(x)y = Q(x). Here, P(x) = 2 and Q(x) = e⁻ˣ. Such equations can be solved by determining an integrating factor.
给定方程属于一阶线性微分方程，标准形式为 dy/dx + P(x)y = Q(x)。此处 P(x) = 2，Q(x) = e⁻ˣ。此类方程可通过确定积分因子求解。
3. Integrating Factor Formula | 积分因子公式
) dx). For a constant P(x) = 2, the integral is straightforward: ∫ 2 dx = 2x, so μ(x) = e²ˣ.
积分因子记作 μ(x)，计算公式为 μ(x) = e^(∫ P(x) dx)。由于 P(x) = 2 是常数，积分极为简单：∫ 2 dx = 2x，因此 μ(x) = e²ˣ。
4. Multiplying Through by the Integrating Factor | 方程两边同乘积分因子
Multiply every term of the differential equation by μ(x) = e²ˣ. The left-hand side becomes e²ˣ dy/dx + 2e²ˣ y, which is exactly the derivative of (e²ˣ y) with respect to x. The right-hand side becomes e²ˣ · e⁻ˣ = eˣ.
将微分方程的各项同时乘以积分因子 μ(x) = e²ˣ。左边化为 e²ˣ dy/dx + 2e²ˣ y，这恰好是 (e²ˣ y) 对 x 的导数。右边变为 e²ˣ · e⁻ˣ = eˣ。
d/dx (e²ˣ y) = eˣ
5. Integrating Both Sides | 两边积分
Now integrate both sides with respect to x. The left side gives e²ˣ y directly, while the integral of eˣ is simply eˣ. Do not forget the constant of integration, C.
现在对 x 积分两边。左边直接得到 e²ˣ y，而 eˣ 的积分就是 eˣ。切勿遗漏积分常数 C。
e²ˣ y = eˣ + C
6. Solving for y | 解出 y
Divide the entire equation by e²ˣ to isolate y. This yields y = e⁻ˣ + C e⁻²ˣ. This expression represents the general solution of the differential equation.
将整个方程除以 e²ˣ 以分离 y。得到 y = e⁻ˣ + C e⁻²ˣ。该表达式即微分方程的通解。
7. Applying the Initial Condition | 代入初始条件
en condition y(0) = 1. Substitute x = 0 and y = 1 into the general solution: 1 = e⁰ + C e⁰ ⟹ 1 = 1 + C. Hence, C = 0.
利用给定条件 y(0) = 1。将 x = 0 和 y = 1 代入通解：1 = e⁰ + C e⁰ ⟹ 1 = 1 + C。因此 C = 0。
8. Particular Solution | 特解
With C = 0, the particular solution simplifies remarkably to y = e⁻ˣ ify that this satisfies both the original differential equation and the initial condition.
由 C = 0 可知，特解显著简化为 y = e⁻ˣ。良好的习惯是验证它同时满足原微分方程和初始条件。
- y(0) = e⁰ = 1 ✓
- dy/dx = -e⁻ˣ, so LHS = -e⁻ˣ + 2e⁻ˣ = e⁻ˣ = RHS ✓
9. Step-by-Step Summary | 分步总结
The systematic approach for such problems is summarised below. Following these steps minimises errors, especially under timed conditions.
现将此类题目的系统解法总结如下。遵循下列步骤可最大程度减少失误，尤其在限时作答时。
| Step | Action | Mathematical Operation |
|---|---|---|
| 1 | Write in standard form | dy/dx + P(x)y = Q(x) |
| 2 | Find integrating factor | μ = e∫ P dx |
| 3 | Multiply through by μ | μ dy/dx + μ P y = μ Q |
| 4 | Recognise LHS as derivative | d/dx (μ y) = μ Q |
| 5 | Integrate both sides | μ y = ∫ μ Q dx + C |
| 6 | Apply initial condition | Solve for C |
10. Common Mistakes and Examiner Tips | 常见错误与考官建议
Many students forget to compute the integrating factor correctly or omit the constant of integration. Another frequent error is misapplying the product rule when differentiating μ y. Examiners look for clear layout and explicit checking against the initial condition. Always present your working logically, and double-check that the exponent in the integrating factor does not have a sign error.
许多学生未能正确计算积分因子，或遗漏积分常数。另一常见错误是在微分 μ y 时误用乘法法则。考官看重清晰的排版以及对初始条件的明确验证。务必条理分明地展示推导过程，并仔细检查积分因子中的指数符号是否正确。
11. Extension: Using Alternative Methods | 拓展：其他方法的对比
Although the integrating factor method is optimal here, the equation dy/dx + 2y = e⁻ˣ h, treating it as a linear ODE with constant coefficients. The complementary function is yc = A e⁻²ˣ, and a trial particular integral yp = k e⁻ˣ to k = 1, giving the same general solution. This demonstrates consistency across methods.
虽然积分因子法是本题最优解法，但方程 dy/dx + 2y = e⁻ˣ 也可通过“余函数 + 特积分”方法处理，将其视为常系数线性常微分方程。余函数为 yc = A e⁻²ˣ，尝试特积分 yp = k e⁻ˣ 解得 k = 1，得到相同的通解。这展示了不同解法间的一致性。
12. Practice Recommendation | 练习建议
To master these questions, practice with varying P(x) and Q(x), including cases where P is not constant and Q is trigonometric or polynomial. Always attempt Edexcel past paper questions that combine integration by parts with the integrating factor technique.
要精通此类题目，应针对不同的 P(x) 和 Q(x) 进行练习，包括 P 不为常数以及 Q 为三角函数或多项式的情形。务必尝试将分部积分与积分因子技巧相结合的 Edexcel 历年真题。
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