Example 6.11.2: Solving a First-Order Linear Differential Equation | 例题6.11.2：解一阶线性微分方程

📚 Example 6.11.2: Solving a First-Order Linear Differential Equation | 例题6.11.2：解一阶线性微分方程

In this article, we work through Example 6.11.2, a standard A-Level mathematics problem from the AQA specification. The problem asks us to solve a first-order linear differential equation using an integrating factor, and then to apply an initial condition to find the particular solution. We will explain every stage of the solution, verify the result, and highlight common mistakes that students often make.

在本文中，我们将详细讲解 AQA 大纲中一个典型的 A-Level 数学例题 6.11.2。该题要求我们使用积分因子求解一阶线性微分方程，并利用初始条件求出特解。我们会逐步解释整个解题过程，验证结果，并指出学生常犯的错误。

1. Problem Statement | 问题陈述

Solve the differential equation

dy/dx + 2y = e^−x

subject to the initial condition y(0) = 1. This is Example 6.11.2 from the AQA A-level mathematics textbook, and it tests both algebraic manipulation and integration techniques.

求解微分方程

dy/dx + 2y = e^−x

并满足初始条件 y(0) = 1。这是 AQA A-Level 数学教材中的例题 6.11.2，它同时考查代数变换和积分技巧。

2. Recognising the Type of Equation | 识别方程类型

Before solving, we must identify the structure of the differential equation. The equation has the form

dy/dx + P(x)y = Q(x)

where P(x) = 2 and Q(x) = e^−x. Because the right-hand side is not zero, the equation is non-homogeneous. However, it is a linear first-order differential equation because y and dy/dx appear only to the first power and are not multiplied together.

求解之前，我们必须识别微分方程的结构。该方程具有如下标准形式

dy/dx + P(x)y = Q(x)

其中 P(x) = 2，Q(x) = e^−x。由于右边不为零，方程是非齐次的。但它是线性一阶微分方程，因为 y 和 dy/dx 都只出现一次幂，并且没有相乘。

3. Computing the Integrating Factor | 计算积分因子

For a linear first-order ordinary differential equation in the standard form shown above, the integrating factor is defined by

μ(x) = e^{∫ P(x) dx}

Here P(x) = 2, so we integrate with respect to x: ∫2 dx = 2x. Therefore the integrating factor is

μ(x) = e^2x

It is important not to add a constant of integration when computing the integrating factor; the extra constant would cancel later and is unnecessary.

对于上述标准形式的一阶线性常微分方程，积分因子定义为

μ(x) = e^{∫ P(x) dx}

这里 P(x) = 2，所以对 x 积分得 ∫2 dx = 2x。因此积分因子为

μ(x) = e^2x

注意：在计算积分因子时不要添加积分常数；这个常数之后会被抵消，所以没有必要写。

4. Multiplying Both Sides | 两边同时乘以积分因子

Multiply every term of the original differential equation by e^2x:

e^2x · dy/dx + 2e^2x · y = e^2x · e^−x

Simplify the right-hand side using the index law e^ae^b = e^a+b:

e^2x · dy/dx + 2e^2x · y = e^x

The left-hand side is now exactly the derivative of the product y · e^2x. Indeed, using the product rule:

d/dx ( y e^2x ) = dy/dx · e^2x + y · 2e^2x

which matches the left-hand side perfectly.

将原微分方程的每一项都乘以 e^2x：

e^2x · dy/dx + 2e^2x · y = e^2x · e^−x

利用指数法则 e^ae^b = e^a+b 化简右边：

e^2x · dy/dx + 2e^2x · y = e^x

此时左边恰好是乘积 y · e^2x 的导数。由乘法法则：

d/dx ( y e^2x ) = dy/dx · e^2x + y · 2e^2x

这与左边完全一致。

5. Integrating Both Sides | 两边积分

Now we write the equation in compact form:

d/dx ( y e^2x ) = e^x

Integrate both sides with respect to x:

∫ d/dx ( y e^2x ) dx = ∫ e^x dx

The left-hand side is simply y e^2x (by the fundamental theorem of calculus), and the right-hand side integrates to e^x + C, where C is an arbitrary constant. Thus

y e^2x = e^x + C

It is essential to include the constant of integration here; it will be determined later using the initial condition.

现在将方程写成紧凑形式：

d/dx ( y e^2x ) = e^x

两边对 x 积分：

∫ d/dx ( y e^2x ) dx = ∫ e^x dx

由微积分基本定理，左边等于 y e^2x；右边积分得 e^x + C，其中 C 为任意常数。因此

y e^2x = e^x + C

这里务必写上积分常数 C，稍后将由初始条件确定。

6. Rearranging for y | 解出 y

To obtain an explicit expression for y, divide both sides of the equation by e^2x:

y = ( e^x + C ) / e^2x

Using the law of exponents, this can be split into two terms:

y = e^−x + C e^−2x

This is the general solution of the differential equation. It contains the arbitrary constant C, which represents a family of solutions corresponding to different initial values.

为了得到 y 的显式表达式，将方程两边同除以 e^2x：

y = ( e^x + C ) / e^2x

利用指数法则，可拆成两项：

y = e^−x + C e^−2x

这就是微分方程的通解。它包含任意常数 C，代表一组对应于不同初始条件的解族。

7. Applying the Initial Condition | 应用初始条件

We are given that y(0) = 1. Substitute x = 0 and y = 1 into the general solution:

1 = e⁰ + C e⁰

Since e⁰ = 1, this simplifies to 1 = 1 + C, which gives C = 0.

Therefore the particular solution is

y = e^−x

Notice that the exponential term e^−2x disappears because its coefficient C is zero. This happens because the initial condition exactly matches the particular integral part of the solution.

已知条件为 y(0) = 1。将 x = 0，y = 1 代入通解：

1 = e⁰ + C e⁰

因为 e⁰ = 1，所以 1 = 1 + C，解得 C = 0。

因此特解为

y = e^−x

注意，e^−2x 这一项消失是因为其系数 C 为零。这是因为初始条件恰好与解的“特解部分”完全匹配。

8. Verification | 验证解

It is always good practice to check that the obtained function really satisfies the original differential equation and the initial condition.

Given y = e^−x, we differentiate with respect to x:

dy/dx = −e^−x

Substitute into the left-hand side of the original equation:

dy/dx + 2y = −e^−x + 2e^−x = e^−x

This equals the right-hand side, so the differential equation is satisfied. Also y(0) = e⁰ = 1, so the initial condition holds.

好的做法是检验所得函数是否确实满足原微分方程和初始条件。

已知 y = e^−x，对 x 求导：

dy/dx = −e^−x

代入原方程左边：

dy/dx + 2y = −e^−x + 2e^−x = e^−x

这等于右边，所以微分方程成立。同时 y(0) = e⁰ = 1，初始条件也成立。

9. Numerical Check | 数值检验

We can also test a few numerical values using the solution y = e^−x. The table below shows the exact values at selected points:

x	y = e^−x
0	1
0.2	0.8187
0.5	0.6065
1.0	0.3679
2.0	0.1353

These values show a clear exponential decay, which matches the physical interpretation of the equation y’ + 2y = e^−x: the forcing term e^−x decays, and the solution also decays, but the rate is controlled by the homogeneous part.

我们也可以用解 y = e^−x 来检验一些数值。下表给出了若干点处的精确值：

x	y = e^−x
0	1
0.2	0.8187
0.5	0.6065
1.0	0.3679
2.0	0.1353

这些数值显示出明显的指数衰减，这与方程 y’ + 2y = e^−x 的物理含义一致：驱动项 e^−x 在衰减，解也在衰减，但衰减速率受到齐次部分的影响。

10. Common Mistakes and Tips | 常见错误与提示

Students frequently make errors in problems like Example 6.11.2. The most common ones are listed below:

Forgetting to identify the standard form before finding the integrating factor. Always rewrite the equation as dy/dx + P(x)y = Q(x) first.
Neglecting the constant of integration when integrating both sides. The constant C is crucial for applying the initial condition.
Making sign errors when differentiating or integrating exponentials. Remember that the derivative of e^−x is −e^−x, and the integral of e^x is e^x + C.
Not simplifying the general solution before substitution. Splitting e^2x into the denominator correctly avoids algebra mistakes.

学生在处理类似例题 6.11.2 时经常出错。最常见的错误如下：

忘记先将方程写成标准形式 dy/dx + P(x)y = Q(x) 再求积分因子。请务必先整理成标准形式。
两边积分时忽略了积分常数。常数 C 对应用初始条件至关重要。
在指数函数求导或积分时出现符号错误。记住 e^−x 的导数是 −e^−x，e^x 的积分是 e^x + C。
没有化简通解就代入初始条件。正确处理分母中的 e^2x 可以避免代数错误。

11. Generalisation | 方法推广

This method works for any first-order linear differential equation of the form dy/dx + P(x)y = Q(x), where P and Q are continuous functions. The integrating factor is always e^{∫P dx}, and after multiplying through, the left-hand side becomes the derivative of y times the integrating factor.

In our example, the homogeneous equation dy/dx + 2y = 0 has solution y = C e^−2x. The particular solution e^−x was found because the forcing term Q(x) = e^−x is also an exponential. In general, students should solve the homogeneous equation first to understand the structure, then find a particular solution.

该解法适用于任何形如 dy/dx + P(x)y = Q(x) 的一阶线性微分方程，其中 P 和 Q 都是连续函数。积分因子永远是 e^{∫P dx}，两边乘以积分因子后，左边就变成 y 乘以积分因子后的导数。

在本题中，齐次方程 dy/dx + 2y = 0 的解为 y = C e^−2x。特解 e^−x 之所以出现，是因为驱动项 Q(x) = e^−x 也是一个指数函数。一般来说，学生应先求解齐次方程以理解结构，再求特解。

12. Conclusion | 总结

Example 6.11.2 demonstrates a complete procedure for solving a first-order linear differential equation: identify the type, compute the integrating factor, multiply through, integrate, apply the initial condition, and verify the solution. The final answer is simply y =

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Example 6.11.2: Solving a First-Order Linear Differential Equation | 例题6.11.2：解一阶线性微分方程