📚 Solving Question 9: Differential Equations in Real-World Contexts | 解答第9题：微分方程在实际问题中的应用
In AQA A-Level Mathematics, Question 9 in the pure mathematics paper frequently focuses on differential equations, especially in the context of real-world modelling. This type of question tests your ability to translate a written description into a differential equation, solve it using separation of variables, apply initial conditions, and interpret the result. A typical mark scheme rewards clear working, correct integration, and careful substitution. In this article, we will work through a representative example step by step and explore the essential techniques you need to succeed.
在AQA A-Level数学中，纯数学试卷的第9题常常聚焦于微分方程，尤其是在实际建模的背景下。这类题目考查你将文字描述转化为微分方程、用分离变量法求解、应用初始条件并解释结果的能力。典型的评分标准会奖励清晰的步骤、正确的积分以及仔细的代入。本文将逐步讲解一个代表性例题，并探讨你需要掌握的关键技巧。
1. Understanding the Question | 理解题意
Consider this typical AQA-style Question 9: “The number of bacteria in a culture grows at a rate proportional to the number present. Initially there are 1000 bacteria, and after 2 hours there are 3000. Find the time taken for the population to double.” Before writing any equations, we must identify the key elements of the problem. The independent variable is time t, measured in hours. The dependent variable is the number of bacteria, which we can denote by N(t). The phrase “grows at a rate proportional to the number present” tells us that dN/dt is directly proportional to N, not to t or any other quantity.
考虑以下典型的AQA风格第9题：“培养皿中细菌数量以与当前数量成正比的速率增长。最初有1000个细菌，2小时后有3000个。求细菌数量翻倍所需时间。”在写任何方程之前，我们必须识别题目中的关键要素。自变量是时间t，以小时为单位。因变量是细菌数量，我们可以用N(t)表示。短语“以与当前数量成正比的速率增长”告诉我们dN/dt与N成正比，而不是与t或其他量成正比。
It is also important to note the initial condition N(0) = 1000 and the additional condition N(2) = 3000. These will allow us to determine the arbitrary constants in the general solution. We are asked to find the doubling time, meaning the time at which N = 2000. By clearly listing the given information, we set a solid foundation for the mathematical work ahead.
同时要注意初始条件N(0) = 1000和额外条件N(2) = 3000。这些条件将帮助我们确定通解中的任意常数。题目要求我们求倍增时间，即N = 2000的时刻。通过清晰列出已知信息，我们为后续的数学推导打下了坚实基础。
2. Setting Up the Differential Equation | 建立微分方程
The statement “rate proportional to the number present” can be written directly as a mathematical relationship. Let k be the positive constant of proportionality. Since the population is growing, dN/dt is positive, and we have the differential equation:
“与当前数量成正比”这句话可以直接写成数学关系。设k为正的比例常数。由于种群在增长，dN/dt为正，因此我们有微分方程：
dN/dt = k N
This is a first-order ordinary differential equation. It is also a separable differential equation, meaning that we can rearrange it so that all terms involving N are on one side and all terms involving t are on the other. Notice that the equation is linear and does not involve any extra terms. Many real-world problems, such as radioactive decay, population growth, and Newton’s law of cooling, lead to similar relationships. In each case, the rate of change of a quantity is proportional to the quantity itself.
这是一个一阶常微分方程。同时它也是一个可分离变量的微分方程，也就是说，我们能够重新排列它，使所有含N的项在一边，所有含t的项在另一边。注意这个方程是线性的，并且不包含任何额外项。许多实际问题，例如放射性衰变、种群增长和牛顿冷却定律，都会导出类似的关系。每种情况下，一个量的变化率都与这个量本身成正比。
When setting up the equation, be careful to define the variable N clearly and state the units. In an exam, writing “Let N(t) be the number of bacteria at time t hours” earns you a mark and removes ambiguity. It also helps you keep track of the meaning of each term in your later working.
建立方程时，务必明确定义变量N并说明单位。在考试中，写出“设N(t)为t小时后的细菌数量”能够得分并消除歧义。它还能帮助你在后续步骤中保持各项含义清晰。
3. Solving the Differential Equation | 解微分方程
To solve dN/dt = k N, we separate the variables. Treat dN/dt as a ratio of differentials, and rearrange to get:
为了解dN/dt = k N，我们分离变量。将dN/dt视为微分的商，重排得到：
(1/N) dN = k dt
Now we integrate both sides. The left-hand side is the natural logarithm of N, because the derivative of ln N with respect to N is 1/N. Since N represents a positive quantity, we do not need the absolute value sign. The right-hand side is simply kt plus an arbitrary constant of integration. Thus:
现在我们对两边积分。左边是N的自然对数，因为ln N对N的导数是1/N。由于N表示正数，我们不需要绝对值符号。右边是kt加上一个任意积分常数。因此：
ln N = k t + C
We can make N the subject of the equation by exponentiating both sides. Let A = eᶜ (using the base e), so the general solution becomes:
我们可以通过两边取指数来将N表示为其余量的函数。令A = eᶜ，所以通解变为：
N = A e^(kt)
Here A is a positive constant, often called the initial value because when t = 0, N = A. This exponential form is characteristic of quantities that grow or decay at a rate proportional to their current value. In the next section, we will use the given conditions to find the precise values of A and k.
这里A是一个正常数，通常称为初值，因为当t = 0时，N = A。这种指数形式是那些以与其当前值成正比的速率增长或衰减的量的典型特征。在下一节中，我们将利用给定条件求出A和k的准确值。
4. Applying Initial Conditions | 应用初始条件
The initial condition N(0) = 1000 is the first piece of information we substitute into the general solution. Setting t = 0 gives N = A e^(0) = A. Therefore A = 1000. If we had forgotten the constant of integration, this substitution would have been impossible, which is why marking schemes frequently award a method mark for writing the general solution with a constant. Now we use the second condition N(2) = 3000 to find k. Substitute t = 2 and N = 3000 into N = 1000 e^(kt):
初始条件N(0) = 1000是我们代入通解的第一条信息。令t = 0得到N = A e^(0) = A。因此A = 1000。如果我们忘记了积分常数，这次代入就不可能完成，这也是评分标准通常会给写出含常数的通解一个方法分的原因。现在我们用第二个条件N(2) = 3000求k。将t = 2和N = 3000代入N = 1000 e^(kt)：
3000 = 1000 e^(2k)
Divide both sides by 1000 to obtain 3 = e^(2k). Taking the natural logarithm of both sides gives ln 3 = 2k, so k = (½) ln 3. We can simplify the particular solution by using the laws of exponents: e^((½ ln 3)t) = (e^(ln 3))^(t/2) = 3^(t/2). Hence the number of bacteria at time t is:
两边除以1000得3 = e^(2k)。两边取自然对数得ln 3 = 2k，所以k = (½) ln 3。我们可以利用指数法则简化特解：e^((½ ln 3)t) = (e^(ln 3))^(t/2) = 3^(t/2)。因此t时刻的细菌数量为：
N = 1000 · 3^(t/2)
This form is easier to interpret and avoids writing a complicated exponent. It shows that the population triples every 2 hours, which is consistent with the given information: after 2 hours, the population goes from 1000 to 3000. Keeping your solution in this simple form also makes subsequent calculations, such as finding the doubling time, much more manageable.
这种形式更容易解释，也避免了书写复杂的指数。它表明种群每2小时增长为原来的3倍，这与给定信息一致：2小时后，种群从1000增长到3000。将解保持在这种简单形式，也会使后续计算（例如求倍增时间）更加容易。
5. Finding the Doubling Time | 求倍增时间
The doubling time is the value of t for which N = 2000. Substitute N = 2000 into the particular solution:
倍增时间是使N = 2000的t值。将N = 2000代入特解：
2000 = 1000 · 3^(t/2)
Divide both sides by 1000 to get 2 = 3^(t/2). To bring the variable t down from the exponent, take the natural logarithm of both sides:
两边除以1000得2 = 3^(t/2)。为了将变量t从指数上移下来，两边取自然对数：
ln 2 = (t/2) ln 3
Now solve for t by multiplying both sides by 2 and then dividing by ln 3:
现在两边乘以2再除以ln 3来解t：
t = (2 ln 2) / (ln 3)
Using a calculator, ln 2 ≈ 0.6931 and ln 3 ≈ 1.0986, so t ≈ (2 × 0.6931) / 1.0986 ≈ 1.26 hours. This is approximately 1 hour and 16 minutes. Always check that your answer is positive and sensible. Since the population is increasing, the doubling time must be positive, and it should be less than the 2 hours needed to triple, which it is. We could express the answer in hours or minutes depending on the question; here hours is natural, but you may convert to minutes if asked.
使用计算器，ln 2 ≈ 0.6931，ln 3 ≈ 1.0986，所以t ≈ (2 × 0.6931) / 1.0986 ≈ 1.26小时。这大约是1小时16分钟。始终检查你的答案是否为正且合理。由于种群在增长，倍增时间必须为正，而且应该小于翻三倍所需的2小时，确实如此。我们可以根据题目要求以小时或分钟表示答案；这里使用小时是自然的，但如果题目要求，你也可以转换为分钟。
6. General Strategy for Separable Differential Equations | 可分离变量微分方程的一般策略
The method used in this example can be generalised to any first-order separable differential equation of the form dy/dx = f(x)g(y). The key steps are the same, and a disciplined approach will help you avoid careless errors in an exam. Below is a step-by-step checklist that you can apply to Question 9 and beyond.
本例题中使用的方法可以推广到任何形如dy/dx = f(x)g(y)的一阶可分离微分方程。关键步骤是相同的，有条理的方法能帮助你在考试中避免粗心错误。下面是一个可应用于第9题及后续题目的逐步检查清单。
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Write down the differential equation clearly, defining all variables.
清晰写出微分方程，定义所有变量。
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Separate the variables so that all terms involving y are on one side and all terms involving x are on the other.
分离变量，使所有含y的项在一边，所有含x的项在另一边。
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Integrate both sides with respect to the relevant variable. Do not forget an arbitrary constant C.
对两边关于相应变量积分。不要忘记任意常数C。
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If possible, rearrange the solution to make y the subject.
如有可能，重排解的形式，使y成为主题。
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Use any initial or boundary conditions to determine the constants.
利用初始条件或边界条件确定常数。
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Interpret your final expression in the context of the problem and state appropriate units.
在问题情境中解释最终表达式，并注明合适的单位。
To see how these steps apply more generally, consider a standard table of operations:
为了更一般地理解这些步骤，请看下面一个标准操作表：
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