📚 Solving Differential Equations with Maclaurin Series | 用麦克劳林级数解微分方程
The Maclaurin series is one of the most powerful tools in the IB Mathematics Analysis and Approaches Higher Level syllabus. While students often encounter it in the context of function approximation, its application to differential equations is a classic exam favourite that tests both conceptual understanding and algebraic fluency.
麦克劳林级数是IB数学分析与方法高级课程中最强大的工具之一。学生通常在学习函数逼近时接触到它，但将其应用于微分方程却是考试中的经典题型，既考查概念理解，也考查代数运算能力。
1. Why Use Maclaurin Series for Differential Equations | 为什么用麦克劳林级数解微分方程
Many differential equations encountered in IB have no closed-form elementary solution. For example, equations of the form y’ = x² + y² or second-order equations with variable coefficients often resist standard techniques. Maclaurin series offer a way to construct an approximate polynomial solution around x = 0, which is often sufficient for practical purposes within a restricted interval.
在IB中遇到的许多微分方程没有初等函数的封闭式解。例如，y’ = x² + y² 形式的方程或变系数二阶方程往往无法用标准方法求解。麦克劳林级数提供了一种在 x = 0 附近构造近似多项式解的方法，这在实际应用中往往足以满足有限区间内的需求。
When the initial condition is given at x = 0, the Maclaurin expansion y(x) = Σₙ₌₀^∞ aₙxⁿ can be used. We simply determine as many coefficients aₙ as required by differentiating the original equation and evaluating at x = 0.
当初值条件在 x = 0 处给出时，可以使用麦克劳林展开式 y(x) = Σₙ₌₀^∞ aₙ xⁿ。我们只需通过对原方程逐次求导并在 x = 0 处取值，来确定所需数量的系数 aₙ。
This method is particularly suited to initial value problems where y(0) and y'(0) are specified. Since the Maclaurin series uses derivatives at zero, these initial values directly become the first coefficients of the series.
这种方法特别适合给定 y(0) 和 y'(0) 的初值问题。由于麦克劳林级数使用零处的各阶导数，这些初值直接成为级数的前几项系数。
2. Key Foundation: The Maclaurin Formula | 关键基础：麦克劳林公式
Recall that any sufficiently differentiable function f(x) can be represented near x = 0 as an infinite series. The coefficients are determined by the derivatives of the function at x = 0.
回顾：任何充分可微的函数 f(x) 在 x = 0 附近都可以表示为一个无穷级数。各项系数由函数在 x = 0 处的各阶导数决定。
f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … + f⁽ⁿ⁾(0)xⁿ/n! + …
The factorial denominators appear because each derivative of xⁿ at zero is n!, and the Taylor coefficient formula involves dividing by n!. In shorthand notation, we write the general term as f⁽ⁿ⁾(0) xⁿ / n!.
阶乘分母的出现是因为 xⁿ 在零处的 n 阶导数是 n!，而泰勒系数公式需要除以 n!。简写记号中，通项写为 f⁽ⁿ⁾(0) xⁿ / n!。
For a differential equation, we denote the series solution as y = a₀ + a₁x + a₂x² + a₃x³ + …, where aₙ = y⁽ⁿ⁾(0)/n!. Thus, computing the series coefficients is equivalent to computing the successive derivatives of y at x = 0.
对于微分方程，我们将级数解记为 y = a₀ + a₁x + a₂x² + a₃x³ + …，其中 aₙ = y⁽ⁿ⁾(0)/n!。因此，计算级数系数等价于计算 y 在 x = 0 处的逐阶导数。
3. The Standard IB Method: Direct Substitution | IB标准方法：直接代入法
For IB, the most tested method begins by assuming the series form of the solution. This approach works well for both first-order and second-order differential equations with polynomial or simple transcendental coefficients.
对于IB考试，最常考的方法从假设解具有级数形式开始。这种方法对一阶和二阶微分方程都非常有效，即使方程的系数是多项式或简单超越函数也不例外。
Consider a typical IB example: find the Maclaurin series up to the x³ term for the solution of y’ = cos(x) + y², given y(0) = 1. The steps involve evaluating the equation at x = 0 to get y'(0), then differentiating both sides and repeating the process.
考虑一个典型的IB例题：设 y’ = cos(x) + y²，y(0) = 1，求解的麦克劳林级数展开到 x³ 项。步骤为：先在 x = 0 处取值得到 y'(0)，然后对等式两边求导并重复该过程。
At x = 0, we have y'(0) = cos(0) + y²(0) = 1 + 1 = 2. Differentiating gives y” = −sin(x) + 2yy’, so y”(0) = 0 + 2(1)(2) = 4. One more derivative yields y”’ = −cos(x) + 2(y’)² + 2yy”, so y”'(0) = −1 + 2(4) + 2(1)(4) = 15.
在 x = 0 处，y'(0) = cos(0) + y²(0) = 1 + 1 = 2。对等式两边求导得到 y” = −sin(x) + 2yy’，因此 y”(0) = 0 + 2(1)(2) = 4。再求一次导数得到 y”’ = −cos(x) + 2(y’)² + 2yy”，所以 y”'(0) = −1 + 2(4) + 2(1)(4) = 15。
Finally, substitute into the series formula: y(x) = 1 + 2x + (4/2!)x² + (15/3!)x³, which simplifies to y(x) = 1 + 2x + 2x² + (5/2)x³.
最后，将这些值代入级数公式：y(x) = 1 + 2x + (4/2!)x² + (15/3!)x³，化简得 y(x) = 1 + 2x + 2x² + (5/2)x³。
4. Step-by-Step Strategy | 分步解题策略
To use Maclaurin series for a differential equation reliably in exams, a consistent methodical approach is essential. Students who skip steps or fail to systematically record values of derivatives often lose credit needlessly.
要在考试中可靠地使用麦克劳林级数求解微分方程，系统化、有条不紊的解题方法是必不可少的。跳步或未能系统记录各阶导数值的学生常常白白丢分。
- Identify the initial values. ✏️ Read the initial conditions from the problem; these give y(0), and if needed, y'(0). | 🎯 确定初值。✏️ 从题目中读取初值条件，它们给出 y(0)，如果题目需要则还有 y'(0)。
- Evaluate the given equation at x = 0. ✏️ Solve for the highest-order derivative appearing in the equation. | 🎯 在 x = 0 处计算给定方程。✏️ 求出方程中出现的最高阶导数在该处的值。
- Differentiate both sides. ✏️ Use implicit differentiation; remember to apply the chain rule to terms involving y. | 🎯 对等式两边求导。✏️ 使用隐函数求导；对含 y 的项应用链式法则。
- Evaluate again at x = 0. ✏️ Substitute all previously found values into the differentiated equation. | 🎯 再次在 x = 0 处取值。✏️ 将所有已知值代入求导后的方程。
- Repeat and record. ✏️ Continue alternating between differentiating and evaluating until the required degree is achieved. | 🎯 重复并记录。✏️ 交替进行求导和取值，直到达到所需阶数。
- Assemble the series. ✏️ Use the formula y(x) = Σ y⁽ⁿ⁾(0) xⁿ / n! with the computed derivative values. | 🎯 组装级数。✏️ 利用公式 y(x) = Σ y⁽ⁿ⁾(0) xⁿ/n! 以及已求得的各阶导数值写出级数。
5. Worked Example 1: First-Order Equation | 例题1：一阶方程
Given y’ = x + 2y, with y(0) = 1, find the series up to the x⁴ term.
设 y’ = x + 2y，y(0) = 1，求展开到 x⁴ 项的麦克劳林级数。
Step 1: At x = 0, the equation gives y'(0) = 0 + 2y(0) = 2. Write this down.
第一步：在 x = 0 处，方程给出 y'(0) = 0 + 2y(0) = 2。记下此值。
Step 2: Differentiate the entire equation with respect to x. The left side becomes y” and the right side becomes 1 + 2y’. Evaluating at x = 0 gives y”(0) = 1 + 2(2) = 5.
第二步：对整个方程关于 x 求导。左边变为 y”，右边变为 1 + 2y’。在 x = 0 处取值得到 y”(0) = 1 + 2(2) = 5。
Step 3: Differentiate again. We obtain y”’ = 2y”. Evaluating at zero yields y”'(0) = 2 × 5 = 10. One more time: y⁽⁴⁾ = 2y”’, so y⁽⁴⁾(0) = 2 × 10 = 20.
第三步：再次求导。得到 y”’ = 2y”。在零处取值得到 y”'(0) = 2 × 5 = 10。再求一次：y⁽⁴⁾ = 2y”’，所以 y⁽⁴⁾(0) = 2 × 10 = 20。
Step 4: Substitute into the Maclaurin formula: y(x) = 1 + 2x + 5x²/2 + 10x³/6 + 20x⁴/24, which simplifies to y(x) = 1 + 2x + (5/2)x² + (5/3)x³ + (5/6)x⁴.
第四步：代入麦克劳林公式：y(x) = 1 + 2x + 5x²/2 + 10x³/6 + 20x⁴/24，化简为 y(x) = 1 + 2x + (5/2)x² + (5/3)x³ + (5/6)x⁴。
A useful check: the equation y’ = x + 2y is actually linear and its exact solution is y = Ce²ˣ − x/2 − 1/4. Using y(0) = 1, we have C = 5/4. If we expand this exact solution, the first few terms match our series, confirming the method works.
一个有用的验证：方程 y’ = x + 2y 实际上是线性的，其精确解为 y = Ce²ˣ − x/2 − 1/4。代入 y(0) = 1 得到 C = 5/4。若展开这个精确解，前几项与我们的级数吻合，确证了方法的正确性。
6. Worked Example 2: Second-Order Equation | 例题2：二阶方程
For y” + x·y = eˣ with y(0) = 1 and y'(0) = 0, find the series to x³.
对于 y” + x·y = eˣ，已知 y(0) = 1、y'(0) = 0，求展开到 x³ 项的级数。
Here the second derivative appears explicitly, so we evaluate at x = 0 immediately. We know e⁰ = 1, so y”(0) + 0 · y(0) = 1, giving y”(0) = 1. Note that the term x·y vanishes at x = 0.
这里二阶导数显式出现，因此直接取 x = 0 进行求值。已知 e⁰ = 1，所以 y”(0) + 0 · y(0) = 1，得到 y”(0) = 1。注意 x·y 项在 x = 0 处为零。
Next, differentiate both sides: y”’ + y + x·y’ = eˣ. Rearranged, y”’ = eˣ − y − x·y’. At x = 0, this becomes y”'(0) = 1 − y(0) − 0 = 1 − 1 = 0.
接下来，对等式两边求导：y”’ + y + x·y’ = eˣ。整理得 y”’ = eˣ − y − x·y’。在 x = 0 处，上式变为 y”'(0) = 1 − y(0) − 0 = 1 − 1 = 0。
The series assembled from derivatives is y(x) = 1 + 0·x + (1/2)x² + (0/6)x³, which is simply y(x) = 1 + x²/2. Thus, the series solution truncated at x³ has no x or x³ terms; they are exactly zero.
由各阶导数组装的级数为 y(x) = 1 + 0·x + (1/2)x² + (0/6)x³，即 y(x) = 1 + x²/2。因此截断到 x³ 的级数解不含 x 项和 x³ 项；它们精确为零。
This example illustrates an important insight: the coefficient of the x³ term is sometimes zero, and students should not force all terms to be present. Any polynomial approximation must be consistent with the differential equation, not with a predetermined pattern.
这个例子说明了一个重要的洞察：x³ 项的系数有时为零，学生不应强行让所有项都出现。任何多项式逼近都必须与微分方程一致，而不是符合预设的规律。
7. Pitfall: Using Standard Series Expansions | 陷阱：直接套用标准展开式
In some IB questions, the differential equation contains functions with known Maclaurin expansions. Students may attempt to substitute the full expansion of functions like eˣ or sin(x) into the equation. This is sometimes valid for finding solutions, but it must be handled with care.
在某些IB题目中，微分方程含有已知麦克劳林展开式的函数。学生可能尝试将 eˣ 或 sin(x) 的完整展开式代入方程。这在某些情况下确实可用于求通解，但必须谨慎对待。
The most common mistake is truncating both the unknown function and the known function at different orders, thus losing consistency. For example, if you expand y to order x³ but use only the first two terms of eˣ, the resulting equation is not accurate to x³.
最常见的错误是将未知函数与已知函数截断到不同阶数，从而失去一致性。例如，若将 y 展开到 x³ 阶但只使用 eˣ 的前两项，最终得到的方程在 x³ 阶上并不准确。
Always maintain a clear order of accuracy. Write down what order the question requires, expand all functions to at least that order, and compare coefficients systematically. When in doubt, use the differentiation method, which avoids these complications entirely.
始终明确所需的精度阶数。将题目要求的阶数写下来，将所有函数展开到至少该阶数，然后系统地比较系数。如果心存疑虑，直接使用逐次求导法，可以完全避免这类难题。
8. Pitfall: Incorrect Chain Rule Application | 陷阱：链式法则误用
Implicit differentiation is central to this method. Whenever a term like y² is differentiated, one must write 2y·y’. Whenever y³ is differentiated, one must write 3y²·y’. The factor y’ is frequently omitted, leading to incorrect derivative values.
隐函数求导是这种方法的核心。每当对 y² 这类项求导时，必须写成 2y·y’。对 y³ 求导时，必须写成 3y²·y’。因子 y’ 经常被遗漏，导致导数值错误。
Similarly, if the equation contains functions of y such as sin(y) or eʸ, their derivatives require the chain rule multiplier y’ as well. For example, d/dx[sin(y)] = cos(y)·y’, not merely cos(y).
同样地，若方程含有 y 的函数如 sin(y) 或 eʸ，其导数同样需要乘以链式因子 y’。例如，d/dx[sin(y)] = cos(y)·y’，而不只是 cos(y)。
A second chain rule subtlety arises when differentiating terms like y·y”. Here one needs the product rule together with the chain rule, producing y’·y” + y·y”’. Students should write out each differentiated term step by step before evaluating at zero.
链式法则的另一个细微之处出现在对 y·y” 这类项求导时。此时需要结合乘积法则与链式法则，得到 y’·y” + y·y”’。学生应在取零值之前逐步写出每项的导数。
Finally, always group terms and evaluate derivatives in a logical order. The safest technique is to write the equation, differentiate it entirely first, and only then substitute x = 0, never substituting partial values before differentiation is complete.
最后，始终按照逻辑顺序分组整理各项并求导值。最安全的技巧是：写出方程 → 完整地对其求导 → 然后再代 x = 0。切勿在求导完成之前代入任何部分值。
9. Recognizing Patterns and Shortcuts | 识别规律与捷径
Once comfortable with the method, students can spot patterns that accelerate computation. If the differential equation is linear with constant coefficients, the terms follow a predictable recurrence relation, and computing each derivative becomes routine.
一旦熟练掌握这种方法，学生就能发现加速计算的规律。若微分方程是常系数线性方程，各阶导数满足可预测的递推关系，计算每个导数就变得非常机械。
For equations of the form y’ = k·y, we know the solution is y = y(0)eᵏˣ. Thus, the Maclaurin series coefficients follow the pattern y⁽ⁿ⁾(0)/n! = kⁿy(0)/n!. Recognizing the exponential pattern saves considerable time in exams.
对于 y’ = k·y 形式的方程，已知解为 y = y(0)eᵏˣ。因此麦克劳林系数满足规律 y⁽ⁿ⁾(0)/n! = kⁿy(0)/n!。在考试中识别出指数模式可以节省大量时间。
For equations where y(0) and y'(0) are given, the Maclaurin series solution can be constructed or verified by direct substitution of a general polynomial form. Students who master both the derivative method and the polynomial comparison method will be best prepared.
对于给定 y(0) 和 y'(0) 的方程，可以通过直接代入一般多项式形式的麦克劳林级数解来构造或验证解。掌握了逐次求导法和多项式比较法这两类方法的学生将有最充分的准备。
However, be careful: pattern recognition must never replace the verification step. In an exam, always show enough working to demonstrate that the computed coefficients truly satisfy the differential equation within the stated order of accuracy.
然而，务必小心：规律识别绝不能取代验证步骤。在考试中，始终要展示足够的工作步骤，以证明计算出的系数在规定的精度阶数内确实满足微分方程。
10. Further Worked Example: Combined Non-linear Equation | 进阶例题：非线性混合方程
Consider y” = y² + eˣ + sin(x), with y(0) = 0 and y'(0) = 1. Obtain terms up to x³.
考虑 y” = y² + eˣ + sin(x)，已知 y(0) = 0 和 y'(0) = 1。求展开到 x³ 项的级数。
Step 1: Evaluate directly at x = 0. We get y”(0) = y²(0) + e⁰ + sin(0) = 0 + 1 + 0 = 1.
第一步：直接在 x = 0 处取值。y”(0) = y²(0) + e⁰ + sin(0) = 0 + 1 + 0 = 1。
Step 2: Differentiate the entire equation: y”’ = 2y·y’ + eˣ + cos(x). Evaluate at x = 0 using y(0) = 0 and y'(0) = 1. So y”'(0) = 2(0)(1) + 1 + 1 = 2.
第二步：对整个方程求导：y”’ = 2y·y’ + eˣ + cos(x)。利用 y(0) = 0 和 y'(0) = 1 在 x = 0 处取值，得 y”'(0) = 2(0)(1) + 1 + 1 = 2。
Step 3: Differentiate once more: y⁽⁴⁾ = 2(y’)² + 2y·y” + eˣ − sin(x). At x = 0 we have y⁽⁴⁾(0) = 2(1)² + 2(0)(1) + 1 − 0 = 3.
第三步：再次求导：y⁽⁴⁾ = 2(y’)² + 2y·y” + eˣ − sin(x)。在 x = 0 处，y⁽⁴⁾(0) = 2(1)² + 2(0)(1) + 1 − 0 = 3。
Step 4: Assemble the series up to degree 3. Given y(0) = 0, y'(0) = 1, y”(0) = 1, and y”'(0) = 2, we obtain y(x) = 0 + x + x²/2 + 2x³/6 = x + x²/2 + x³/3.
第四步：组装到三次项的级数。给定 y(0) = 0、y'(0) = 1、y”(0) = 1、y”'(0) = 2，得到 y(x) = 0 + x + x²/2 + 2x³/6 = x + x²/2 + x³/3。
Note that the x⁴ term requires y⁽⁴⁾(0)/4! = 3/24 = 1/8, which the question does not demand. But knowing how to prompt this extension shows mastery of the deeper mechanics of the method.
注意 x⁴ 项需要 y⁽⁴⁾(0)/4! = 3/24 = 1/8，但题目并未要求。不过，知道如何延伸计算正体现了对该方法深层机制的掌握。
11. An Alternative IB Technique: Series Substitution | IB另一种技巧：级数代入比较系数
In some exam boards-style questions, rather than differentiating the equation step by step, it is preferable to write y = a₀ + a₁x + a₂x² + a₃x³ + … and substitute the entire series into the differential equation.
在某些考试风格的题目中，与其逐次求导，不如写出 y = a₀ + a₁x + a₂x² + a₃x³ + … 并将整个级数代入微分方程，更为可取。
Then one uses the known Maclaurin expansions for any functions appearing on the right-hand side, and compares coefficients of like powers of x on both sides. This can lead to a system of algebraic equations, each one giving the value of aₖ.
然后利用右侧出现任何函数的已知麦克劳林展开式，比较等式两边 x 同次幂的系数。这将得到一个代数方程组，每个方程给出一个 aₖ 的值。
Consider a simple equation: y’ = y with y(0) = 1. Substituting y = Σ aₙxⁿ gives Σ n·aₙxⁿ⁻¹ = Σ aₙxⁿ. Comparing coefficients, we find aₙ = a₀/n!, so y(x) = Σ xⁿ/n! = eˣ. This method verifies the pattern that emerges in derivative-based computations.
考虑简单方程 y’ = y 且 y(0) = 1。代入 y = Σ aₙxⁿ 得到 Σ n·aₙxⁿ⁻¹ = Σ aₙxⁿ。比较系数得 aₙ = a₀/n!，所以 y(x) = Σ xⁿ/n! = eˣ。此方法验证了基于求导的计算所呈现的规律。
For IB exam purposes, we recommend the derivative evaluation method as the primary strategy because it is direct and less prone to error. Leave the substitution method for when it is explicitly required or for verifying results in your working.
就IB考试而言，我们推荐将逐次求导法作为首要策略，因为它更直接、更不易出错。只有在题目明确要求时，或用于验证计算结果时，才使用代入比较系数的方法。
12. Summary and Key Examination Tips | 总结与关键考试要点
The Maclaurin series approach to solving differential equations combines two fundamental calculus concepts: Taylor approximation and differential equation solving. For IBAA HL students, this integration of topics is a hallmark of higher-level questions.
用麦克劳林级数求解微分方程将微积分两大基础概念融为一体：泰勒逼近与求解微分方程。对于IBAA高级水平学生而言，这种主题交叉恰是高水平试题的标志。
Keep in mind that each derivative taken should be recorded in a clear table of values such as y(0), y'(0), y”(0), y”'(0) and so on. This organisation prevents arithmetic mistakes and communicates your working clearly to the examiner.
牢记：每次求导后的值应以清晰表格记录，如 y(0)、y'(0)、y”(0)、y”'(0) 等。这种有组织的做法可以防止算术错误，并让阅卷者清楚看到你的解题过程。
Always round off by writing the final series with factorial denominators simplified, ensuring each coefficient is in its simplest fraction form, exactly as the mark scheme expects.
完成时务必写出最终级数，并将阶乘分母化简，确保每项系数均为最简分数，与官方评分标准的预期完全一致。
Revision practice is vital: attempt at least five past-paper questions of this type before your exam. After each problem, compare your series solution with a numerical estimate of y(0.1) or y(0.2) computed using a Euler or Runge-Kutta method on a calculator or graphing tool.
复习练习至关重要：考试前至少完成五道此类历年真题。每道题完成后，用计算器或绘图工具通过欧拉法或龙格-库塔法计算 y(0.1) 或 y(0.2) 的数值估计，并与你所得级数解进行对比验证。
As a final tip: whenever the series you have found includes all consecutive powers without gaps, check for standard function identities such as eᵏˣ, sin(x), cos(x), or (1+x)ᵖ. The original differential equation sometimes reveals itself as the defining equation of a familiar function.
最后一点提示：若所求级数从头到尾包含连续幂项而没有空缺，检查是否与 eᵏˣ、sin(x)、cos(x)、或 (1+x)ᵖ 等标准函数恒等式对应。原始微分方程有时正是某个常见函数的定义方程。
Master this technique and you will approach every IB paper confidently, knowing that differential equation questions regardless of difficulty can always be cracked with a well-executed Maclaurin expansion.
掌握了这项技术，你就能自信地应对每一份IB试卷，因为你已经明白：无论微分方程题目难度如何，一次执行良好的麦克劳林展开总能为解决它提供可靠途径。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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