📚 Example 6.11.4: Separation of Variables | 例题 6.11.4：分离变量法
In this article, we will work through Example 6.11.4 from the AQA A-Level Mathematics syllabus, which focuses on solving a first-order differential equation using the separation of variables method. This example is a classic exam-style question that tests your ability to rearrange, integrate, and apply an initial condition correctly. By the end, you will have a clear, step-by-step template for tackling similar problems.
在本文中，我们将一起完成 AQA A-Level 数学大纲中的例题 6.11.4，该题重点考查使用分离变量法求解一阶微分方程。这是一道经典的考试风格题目，考查你正确变形、积分和应用初始条件的能力。学完本例题后，你将拥有一个清晰、分步的解题模板，可用来应对类似的题目。
1. Problem Statement | 问题陈述
al condition y = 1 when x = 1. We need to find an explicit expression for y in terms of x.
问题如下：求解微分方程 dy/dx = (y² + 1) / x，其中 x > 0，并满足初始条件 y = 1 当 x = 1 时。我们需要求出 y 关于 x 的显式表达式。
2. Identifying the Method | 方法的识别
ion, the right-hand side is a product of a function of y, (y² + 1), and a function of x, (1/x). This structure makes it ideal for the separation of variables technique.
一阶微分方程如果能写成 g(y) dy/dx = h(x) 的形式，就可以用分离变量法求解。在我们的方程中，右边是 y 的函数 (y² + 1) 与 x 的函数 (1/x) 的乘积。这种结构非常适合使用分离变量法。
3. Step 1: Separate the Variables | 第一步：分离变量
acing all y terms on one side and all x terms on the other is the entire goal of this first step.
我们将 dy/dx 看作一个比例，两边同时乘以 dx，再除以 (y² + 1)，得到 dy/(y² + 1) = (1/x) dx。把所有含 y 的项放在一边，含 x 的项放在另一边，这就是这一步的全部目标。
4. Step 2: Integrate Both Sides | 第二步：两边积分
the standard integral for arctan; on the right, we have a standard natural logarithm. Remember that since x > 0, we can write ln x rather than ln|x|.
接下来，我们对分离后的方程两边分别求积分：∫ dy/(y² + 1) = ∫ (1/x) dx。左边我们识别出反正切函数的标准积分；右边则是标准自然对数。记住，因为 x > 0，我们可以直接写 ln x 而不是 ln|x|。
5. Step 3: Evaluate the Integrals | 第三步：计算积分
ning the two constants into a single arbitrary constant C, we obtain arctan(y) = ln x + C.
左边的积分为 arctan(y) + C₁，因为 d/dy[arctan(y)] = 1/(1 + y²)。右边的积分为 ln x + C₂。将两个常数合并为一个任意常数 C，我们得到 arctan(y) = ln x + C。
6. Step 4: Apply the Initial Condition | 第四步：应用初始条件
ln(1) = 0, we find C = π/4. Therefore the general solution becomes arctan(y) = ln x + π/4.
我们使用条件 x = 1 时 y = 1。代入这些值得到 arctan(1) = ln(1) + C。由于 arctan(1) = π/4，ln(1) = 0，我们求出 C = π/4。因此通解变为 arctan(y) = ln x + π/4。
7. Step 5: Write the Particular Solution | 第五步：写出特解
s the particular solution. Since tangent is a periodic function, we must be aware that the solution is valid on an interval where the argument does not cross an asymptote, i.e. ln x + π/4 ≠ π/2 + kπ, where k is an integer.
对两边取正切函数，得到 y = tan(ln x + π/4)。这就是特解。由于正切函数是周期函数，我们必须注意解有效的区间，即自变量不能穿过渐近线，即 ln x + π/4 ≠ π/2 + kπ（k为整数）。
8. Step 6: Check the Solution | 第六步：验证解
(1 + y²)/x, which matches the original equation. The initial condition is also satisfied.
我们可以通过求导来快速验证解。若 y = tan(ln x + π/4)，则 dy/dx = sec²(ln x + π/4) × (1/x)。利用恒等式 sec²θ = 1 + tan²θ，得到 dy/dx = [1 + tan²(ln x + π/4)] / x = (1 + y²)/x，与原方程一致。初始条件也满足。
9. Common Mistakes and AQA Exam Tips | 常见错误与 AQA 考试建议
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Forgetting the constant of integration: Always include the constant before applying initial conditions.
忘记积分常数：应用初始条件之前一定要保留常数。
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Mixing up variables during separation: Make sure all y terms are with dy and all x terms with dx.
分离过程中混淆变量：确保所有含 y 的项和 dy 在一起，含 x 的项和 dx 在一起。
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Neglecting the absolute value in logarithms when x can be negative: In this question, x > 0, so we ignore | |, but for full generality use ln|x|.
当 x 可能为负数时，忽略对数中的绝对值：本题 x > 0，所以不用管 | |，但为了完整性，应写成 ln|x|。
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for y explicitly if requested: Read the question carefully; sometimes y = ... is required.
若题目要求显式解却忘了解出 y：仔细读题；有时需要 y = ... 的形式。
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Incorrectly applying the tangent to both sides: Remember that 'cancelling' arctan is not correct; you must take tan of both sides.
对两边取正切时出错：记住不能直接'消去'arctan；必须对两边取 tan。
tandard integral ∫ 1/(a² + x²) dx = (1/a) arctan(x/a) is extremely useful in this context.
为了在 AQA 考试中取得好成绩，务必写出解的成立区间。很多分数都来自对定义域限制的说明。另外，请善用公式册：标准积分 ∫ 1/(a² + x²) dx = (1/a) arctan(x/a) 在此类问题中非常有用。
10. Related Practice Example | 相关练习例题
ence y = 1/cos x = sec x, with x ≠ π/2 + kπ. This demonstrates the same four-step process.
为巩固这一方法，请尝试求解：dy/dx = y² sin x，且 y(0) = 1。首先分离得到 dy/y² = sin x dx。积分得到 -1/y = -cos x + C。利用 y(0)=1 得 -1 = -1 + C，所以 C = 0。因此 y = 1/cos x = sec x，其中 x ≠ π/2 + kπ。这演示了相同的四步过程。
11. Summary | 总结
In this article, we have solved Example 6.11.4 by separating variables, integrating both sides, applying the initial condition, and writing the particular solution. We also verified the solution and discussed common pitfalls. With practice, solving such differential equations becomes a quick and reliable part of your exam technique. For more worked examples and revision resources, visit aleveler.com.
在本文中，我们通过分离变量、两边积分、应用初始条件并写出特解，成功解决了例题 6.11.4。我们还验证了解，并讨论了常见错误。通过练习，求解此类微分方程会变得迅速而可靠。更多例题与复习资源，请访问 aleveler.com。
Published by TutorHao | Maths Revision Series | aleveler.com
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