A-Level Maths 微分方程 分离变量 特解 建模

A-Level Maths 微分方程 分离变量 特解 建模

微分方程在 A-Level 数学里是一个小而固定的得分点，位置通常在纯数学第二年积分的尾巴上。很多学生能背出分离变量的步骤，真正动笔时却在常数项和最后一行上丢分。这篇按解题动作的顺序拆开讲，从识别题型到写出特解，再到判断解在现实中还有没有意义。

文中的考纲口径来自 Pearson Edexcel 9MA0、AQA 7357 与 OCR H240 的现行 specification。三个考试局对同一块内容的措辞并不一样，这一点会在后面单独用一节对照。

Differential equations sit at the end of the second year of pure mathematics, and they reward a fixed routine rather than talent. Students can usually recite the separation method, yet most lost marks come from the constant of integration and the final rearranged line. This piece follows the order in which you actually write a solution, then asks what the answer means in context.

微分方程解出来的是函数不是数

辨别题型的第一件事是看方程里除了 x 和 y 还有没有导数，出现 dy/dx 就是微分方程，最高只出现一阶导数就是一阶方程，这也是 A-Level 单科数学的全部要求。解出来的是一个函数关系，把它和它的导数同时代回原方程，两边要恒等。通解里带一个任意常数，题目给了初始条件才需要把它定下来。很多学生以为求出积分就完成了，卷面上最后留下的却是含 C 的一堆式子，而题目问的是 y 的表达式。判断自己的答案对不对，最快的办法是求导一次，看能不能回到原方程。判分时通解与特解通常分开给分，中间步骤写清楚比一步到底更安全。

The first job is recognition: if the equation contains a derivative as well as x and y, it is a differential equation, and single A-Level Mathematics only asks for first order cases. A solution is a relationship that makes the original equation true when both y and its derivative are substituted back. The general solution always carries an arbitrary constant, and only an initial condition forces it to a single value. Marks are usually split between the general and the particular solution, so a clear intermediate line is worth writing.

分离变量的三步与那一行写全的积分式

分离变量的动作只有三个，把含 y 的部分连同 dy 归到左边，含 x 的部分连同 dx 归到右边，然后两边各自积分。写成完整的一行很重要，左边应该是 ∫(1/g(y)) dy，右边是 ∫f(x) dx，而不是省略了积分号的草稿式。常数 C 只在一侧出现，两侧都加 C 等于两个常数合成一个，白写一行还容易被判记号不清。常见的错误是把乘积当成可以直接积分的对象，例如把 dy/dx = xy 拆开后写成 ∫xy dx。分离之后两边的积分变量各自独立，左边的结果里只该出现 y，右边的只该出现 x。积分变量要跟分母上的微分对齐，这一点在 x 与 y 混合出现时最容易被忽略。

Separation of variables is three moves: put everything containing y and dy on one side, everything containing x and dx on the other, then integrate each side. Write the integrals out in full, with one over g of y on the left and f of x on the right, rather than a shorthand that drops the integral signs. A single constant on one side is enough, because two of them would only combine into one. The recurring error is carrying a product across as if it could be integrated on its own, which is what happens when a mixed term such as x times y is split without being placed on the correct side.

先提取公因式再分离的题型

有些方程不能直接分离，因为右边是两项相加，看起来像 x 和 y 搅在一起。AQA 7357 的考纲 H7 专门提到分离变量前可能需要提取公因式，例如 dy/dx = xy + 2x 要先写成 x(y + 2)，才能把 y + 2 挪到左边。判断标准很直接，右边的每一项都含有同一个含 x 的因子时，就一定能提出去。题目常会在括号里给一点提示，看到 dy/dx = x(y + 2) 这样的括号就要想到分离。提不出来才是真的不可分离，那类方程超出了单科数学的范围。这一步不做，后面的积分会变成一个无法处理的整体，写多少行都算不上方法分。

Some equations look inseparable because the right hand side is a sum, mixing x and y in a way that seems to block the method. The AQA specification makes this explicit in H7, noting that separation may need factorisation involving a common factor, so dy/dx = xy + 2x has to become x times (y + 2) first. The test is simple: if every term on the right shares a factor in x, that factor can be taken out. Equations that survive that attempt are not separable at all and sit outside single Mathematics.

常数 C 的处理与绝对值脱掉时的符号

得到 ln|y| = sin x + c 之后，下一步是把它化成显函数，常数真正出问题就在这一步。两边取指数得到 |y| = e^c 乘 e^(sin x)，e^c 是一个新的正数，再乘上正负号就合成一个可以取负值的任意常数，习惯写成 A。常见的错误是把 e^c 直接写成 c，看似省事，实际上把常数的取值范围写窄了，严谨的阅卷会认为不完整。题里有初始条件时，宁可先保留 c 到代入之后再化简，能少一次改名。取对数之前先确认 y 的符号是否已知，初值往往能直接把符号定下来。对数里带绝对值也是同样的道理，脱掉绝对值时要让常数承担符号。

After reaching ln of the modulus of y equals sin x plus c, turning it into an explicit function is where the constant is mishandled. Exponentiating gives the modulus of y as e to the c multiplied by e to the sin x, and that first factor is a positive number which, once the sign is absorbed, becomes an arbitrary constant usually written as A. Writing e to the c as plain c looks tidy but narrows the constant to positive values only. Keeping the raw constant until the initial condition is substituted saves one rename and protects the sign.

特解 初始条件代入的时机

特解不是新题型，只是把给定的 x 与 y 代进通解，解出一个具体的常数值。代入有两种写法都能得分，先在含 C 的形式里定出 C 再化简，或者先化到 y = A 乘 e^(f(x)) 再定 A。第二种更省事，因为指数上的常数已经并进去了。先解出常数再回头化简，比边化简边代入少出错。很多学生会漏掉初值的含义，例如 t = 0 时数量是 2000，结果把 2000 代成 y 却把 t 留成未知。最后一行要明确写 y = ...，只写一个常数的值不算完成任务。

A particular solution is not a new technique, only the general solution with a given pair of values substituted to pin the constant down. Both orders of working earn marks: fix the constant in the implicit form first, or rewrite as y equals A times an exponential and then fix A. The second is quicker because the constant on the exponent has already been absorbed. The usual slip is misreading the condition, for instance substituting a quantity for the wrong variable, or stopping at a value of the constant instead of a finished expression for y.

指数增长与衰减 dy/dt = ky

dy/dt = ky 是这类题里出现频率最高的形式，k 取正就是增长，取负就是衰减，解都是 y = A e^(kt)。A 是 t = 0 时的初值，k 的符号决定曲线方向，这两个结论从解的结构直接读出来，不需要额外记忆。考题常把 k 藏起来，给出每 3 小时数量翻倍这样的条件，此时先写 y = A e^(kt)，再用 y(3) = 2A 定出 k，最后才回答题目真正问的那个时刻。半衰期题同理，只是把 2A 换成 A/2。衰减题给出的 k 是负数，e^(kt) 实际写成 e^(-kt)，负号别漏在指数外面。作答时把 k 写成精确的 ln2 除以 3，比写 0.231 更稳妥，除非题目明确要求保留小数。

The form dy/dt = ky is the most common version, with positive k giving growth and negative k giving decay, and the solution is always y equals A times e to the kt. Here A is the value at t equals zero, and the sign of k sets the direction of the curve, both read straight off the structure. Questions often hide k behind a statement such as a quantity doubling every three hours, so write the general form, impose the doubling to find k, and only then answer the time being asked for. Leaving k as ln2 over 3 is safer than a rounded decimal unless a decimal is requested.

趋向极限值的模型 k(M - y)

dy/dt = k(M - y) 描述的是趋向一个上限的过程，M 是那个上限，y 越接近 M 变化越慢。分离变量后左边积出 ln|M - y|，这里会带一个负号，整理成 y = M - A e^(-kt) 之后曲线从初值单调爬向 M，永远到不了。很多学生把这类解写成 y = A e^(-kt) 就收手，等于丢掉了上限 M，整道题的结论全错。题目若问达到某个数值需要多久，把 y 代进解里，两边取对数就能算出时间。AQA 的 H8 要求解释解在实际情境中的含义并指出模型的局限，长期行为正是最常被问到的一句。写答案时用极限的说法，例如 t 增大时 y 趋于 M，比只给公式更贴题。

An equation of the form dy/dt = k times (M minus y) models a quantity approaching a ceiling, where M is that ceiling and the rate slows as y gets closer. Separating and integrating produces a logarithm of M minus y, with a minus sign that has to be tracked carefully, and the tidy result is y equals M minus A times e to the minus kt. Dropping the M and answering with a plain decaying exponential throws away the whole point of the question. The specification asks for interpretation in context, so a sentence about y tending to M as t grows is part of the answer rather than decoration.

从文字到方程 比例关系的翻译

建模题的第一句通常是文字的，例如变化率与现存量成正比，或者与现存量同环境温度之差成正比。前者直接给出 dy/dt = ky，后者给出 dy/dt = -k(y - T)，负号来自温度高于环境时在下降这个事实。写方程之前先给变量下定义，例如 y 表示现存量、t 表示小时，后面的每个数字都要放得进这个定义。很多学生以为负号可以随便放，结果解出来的曲线方向反了，后面的数值全部对不上。翻译这一步最好写出中间文字，先写清楚比例的对象是谁，再落成方程，卷面上能拿到建模分。单位也要跟着写，k 的单位由两边的量纲反推，写错单位往往说明方程本身写错了。

Modelling questions usually open with a sentence rather than a formula, for example that the rate of change is proportional to the amount present, or proportional to the difference from the surrounding temperature. The first translates to dy/dt = ky and the second to dy/dt = minus k times (y minus T), where the minus sign records that a body warmer than its surroundings cools down. Guessing the sign is a reliable way to end up with a curve bending the wrong way and every later number wrong. Writing the translation in words before the algebra earns method marks, and checking the units of k is a quick way to catch a reversed equation.

运动学里 v 与 a 的两个导数

运动学里 x、v、a 之间本来就是求导关系，v = dx/dt，a = dv/dt = d²x/dt²，所以给出加速度的表达式就是在给微分方程。若 a 是常数，两次积分会带出两个常数，正好由初始位移和初始速度定出。遇到变加速的题，先看加速度依赖的是时间、速度还是位移，那决定方程能不能分离。若阻力与速度成正比，方程是 dv/dt = -kv，解出速度随时间指数衰减，再积一次才得到位移，这一步很容易忘记回到题目真正要的量。AQA 的 H8 明确提到与运动学的联系，题目出现速度减半所需时间这类问法时，答案通常是对速度解取对数。位移的极限值直接用积分求，不必等 t 趋于无穷再讨论。

In kinematics the three quantities are linked by differentiation: velocity is the derivative of displacement and acceleration is the derivative of velocity, so any statement about acceleration is a differential equation in disguise. Constant acceleration needs two integrations and therefore two constants, fixed by the initial displacement and the initial velocity. If resistance is proportional to speed, the equation becomes dv/dt = minus kv and the velocity decays exponentially, but the displacement requires one more integration that students often forget. The specification links this content to kinematics, and a question asking for the time to halve a speed is answered by taking logarithms of the velocity expression.

解的长期行为与模型的局限

解出一个函数不等于回答完问题，题目最后常在问这个模型还有没有意义。指数增长的解在长时间里趋向无穷，现实中的种群、债务、传播人数都不可能无限增长，所以模型只在有限范围内有效。冷却模型的解趋向环境温度，这是合理的那一面，局限则在于把环境温度当成恒定。题目若要求给出更合理的模型，答一句让增长率随数量下降即可，不必真的解出新方程。写局限时给具体条件比写一句笼统的批评有用，例如指出超过某个时间之后预测值超出了物理上限。分数往往就落在这两三行文字上，公式算得再漂亮也不会替代它。

Producing a function is not the end of a question, since the last part usually asks whether the model still makes sense. An exponential solution grows without bound, while real populations, debts and case counts cannot, so the model only holds over a limited range. A cooling solution tends to the ambient temperature, which is the sensible part, and its weakness is treating that ambient temperature as fixed. Naming a concrete failure point, such as a predicted value exceeding a physical maximum after some time, scores better than a vague remark about limitations.

三个最常见的失分动作

翻遍真题，丢分的地方集中在三处，每一处都不是难点。一处是忘了写常数 C，积分一完成就直接解 y，答案天然少一个自由度。另一处是化简时把 e^c 写成 c，或者把负号丢在绝对值里。第三种是题目问 y 的表达式，卷面只写到 ln y = ... 就停笔。这三处错误在阅卷时都算表达不完整，方法分能拿到，答案分拿不到。写完最后一行，回头对照题目问的是 y、t 还是某个特定时刻的数值，能挡掉大部分无谓损失。

Across past papers, lost marks cluster in three places, and none of them is difficult. One is a missing constant of integration, which quietly removes one degree of freedom from every later line. Another is rewriting e to the c as plain c, or losing a minus sign on the way out of a modulus. The third is stopping at a logarithmic equation when the question asked for y. All three read as incomplete expression to a marker: method marks survive, the answer mark does not.

Edexcel AQA OCR 三份考纲的措辞差别

三个考试局对同一块内容的措辞不一样，读一遍原文比刷十道同类题更有用。Pearson Edexcel 9MA0 把微分方程放在纯数学第二年，考在 Paper 2，specification 的写法是可分离变量加上求特解。AQA 7357 把它并进积分那一节，编号 H7 要求解可分离的一阶方程并求出特解，H8 要求解释解的含义与局限。OCR H240 的 1.08k 表述几乎一样，也点名了先因式分解再分离的可能。三家都是三份两小时的卷子、每份 100 分，区别在内容编排的位置，复习时按各自考纲的措辞对答案更省时间。把三家的考纲条目各打印一页贴在错题本上，比记二手总结可靠。

The three boards phrase the same content differently, and one careful read of the wording beats ten similar practice questions. Pearson Edexcel 9MA0 places the topic in the second year of pure mathematics, examined in Paper 2, described as differential equations with separable variables including particular solutions. AQA 7357 folds it into the integration section, where H7 asks for analytical solutions of simple first order separable equations and H8 asks for interpretation and limitations. OCR H240 uses almost identical wording at 1.08k, including the note about factorising first, and all three boards set three two-hour papers of 100 marks each.

按方程结构分类刷题与自测

刷题时按方程的结构分堆，比按年份顺着做有效得多。一堆是右边本来就能写成 f(x) 乘 g(y) 的直接型，一堆是需要提取公因式才能分离的类型，一堆是趋向极限值的 k(M - y) 型，一堆是运动学里由加速度反推位移的题。每一堆做三到五道，写完自己做一次检查，把 y 代回原方程看两边是否恒等，看常数是否只剩一个，看答案是否是题目问的那个量。自测时把解在长时间下的行为讲出来，讲得出来才算真的懂了这一块。错题再按错误类型贴标签，符号错、常数错、漏写最后一行，三种标签分开统计。

Practice is more efficient when questions are grouped by the structure of the equation rather than worked through by year. One group has a right hand side that already factors into a function of x times a function of y, another needs a common factor extracted before separation, a third approaches a limiting value, and a fourth runs from acceleration back to displacement. Three to five questions per group is enough to fix the pattern. After each solution, substitute y back into the original equation, count the constants, and check that the final line answers the quantity actually asked for.

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