Solving Differential Equations | 求解微分方程

📚 Solving Differential Equations | 求解微分方程

A differential equation is an equation that involves an unknown function and one or more of its derivatives. In A-level Mathematics, the focus is on first-order ordinary differential equations, where the highest derivative appearing is dy/dx. Solving a differential equation means finding all functions y(x) that satisfy the given relationship, often expressed as a family of curves containing an arbitrary constant.

微分方程是包含未知函数及其一个或多个导数的方程。在 A-level 数学中，重点是一阶常微分方程，即出现的最高阶导数是 dy/dx。求解微分方程意味着找出所有满足给定关系的函数 y(x)，通常表示为一族含有任意常数的曲线。

1. What is a Differential Equation? | 什么是微分方程？

A first-order differential equation can usually be written in the form dy/dx = f(x, y). The order of a differential equation is the order of the highest derivative present. For example, dy/dx = 3x² is a first-order equation, while d²y/dx² + 5dy/dx + 6y = 0 is second-order. At A-level, we solve first-order equations using direct integration or separation of variables.

一阶微分方程通常可以写成 dy/dx = f(x, y) 的形式。微分方程的阶数是其中最高阶导数的阶数。例如，dy/dx = 3x² 是一阶方程，而 d²y/dx² + 5dy/dx + 6y = 0 是二阶方程。在 A-level 阶段，我们使用直接积分法或分离变量法来求解一阶方程。

Some equations are linear, such as dy/dx + P(x)y = Q(x), but at this level we mainly meet separable equations. You should be able to classify an equation and decide which method applies.

有些方程是线性的，例如 dy/dx + P(x)y = Q(x)，但在本阶段我们主要遇到的是可分离变量的方程。你应该能够判断方程的类型并选择合适的方法。

2. Verifying a Solution | 验证一个解

Before solving a differential equation, it is useful to know how to check whether a given function is a solution. To verify, differentiate the proposed function y(x) and substitute both y and dy/dx into the equation. If the equation is satisfied for all x in the domain, then the function is indeed a solution.

在求解微分方程之前，了解如何检验一个给定函数是否为解是很有用的。验证时，对提出的函数 y(x) 求导，并将 y 和 dy/dx 代入方程。如果方程在定义域内对所有 x 都成立，那么该函数确实是一个解。

Example: Show that y = Ae^{2x} satisfies dy/dx = 2y. Differentiating gives dy/dx = 2Ae^{2x}. Substitution into the right-hand side gives 2y = 2Ae^{2x}, so the equation holds. The function is a solution for any constant A.

示例：证明 y = Ae^{2x} 满足 dy/dx = 2y。求导得 dy/dx = 2Ae^{2x}。代入右边得 2y = 2Ae^{2x}，因此方程成立。对于任意常数 A，该函数都是解。

3. Direct Integration: dy/dx = f(x) | 直接积分法：dy/dx = f(x)

The simplest type of differential equation has the form dy/dx = f(x), where the derivative depends only on x. The solution is obtained by integrating both sides with respect to x: y = ∫ f(x) dx + C. The constant of integration C is essential because differentiation removes constants, and the general solution must represent all possible antiderivatives.

最简单的微分方程形式为 dy/dx = f(x)，其中导数只依赖于 x。通过对两边关于 x 积分得到解：y = ∫ f(x) dx + C。积分常数 C 是必不可少的，因为求导会消去常数，而通解必须表示所有可能的原函数。

Example: Solve dy/dx = 6x² – 4x + 3. Integrating gives y = ∫(6x² – 4x + 3) dx = 2x³ – 2x² + 3x + C. This is the general solution. If an additional condition is given, C can be determined.

示例：求解 dy/dx = 6x² – 4x + 3。积分得 y = ∫(6x² – 4x + 3) dx = 2x³ – 2x² + 3x + C。这是通解。如果给出附加条件，就可以确定 C。

4. Separation of Variables: dy/dx = g(y) | 分离变量法：dy/dx = g(y)

When the derivative depends only on y, the equation can be rearranged so that all y terms are on one side and all x terms on the other. Starting from dy/dx = g(y), we write dy/g(y) = dx, provided g(y) ≠ 0. Then integrate both sides: ∫ 1/g(y) dy = ∫ dx + C. After integration, solve for y if possible.

当导数只依赖于 y 时，可以将方程整理为所有 y 项在一侧、所有 x 项在另一侧。从 dy/dx = g(y) 出发，如果 g(y) ≠ 0，则写成 dy/g(y) = dx。然后两边积分：∫ 1/g(y) dy = ∫ dx + C。积分后，如果可能，解出 y。

Example: Solve dy/dx = 3y. Rearrange to dy/y = 3 dx. Integrating gives ln|y| = 3x + C. Exponentiating yields |y| = e^{3x+C} = e^C e^{3x}. Writing A = ±e^C gives y = Ae^{3x}, where A is an arbitrary constant.

示例：求解 dy/dx = 3y。移项得 dy/y = 3 dx。积分得 ln|y| = 3x + C。取指数得 |y| = e^{3x+C} = e^C e^{3x}。令 A = ±e^C，得到 y = Ae^{3x}，其中 A 是任意常数。

5. Separation of Variables: dy/dx = f(x)g(y) | 分离变量法：dy/dx = f(x)g(y)

The most general separable form at A-level is dy/dx = f(x)g(y). To solve, first check whether g(y) = 0 gives any constant solutions. Then, for g(y) ≠ 0, divide both sides by g(y) and multiply by dx to obtain (1/g(y)) dy = f(x) dx. Integrate both sides and solve for y if possible. If solving explicitly is difficult, an implicit solution is acceptable.

A-level 中最一般的可分离形式是 dy/dx = f(x)g(y)。求解时，首先检查 g(y) = 0 是否给出常数解。然后，对于 g(y) ≠ 0，两边除以 g(y) 并乘以 dx，得到 (1/g(y)) dy = f(x) dx。两边积分，如果可能则解出 y。如果难以显式求解，隐式解也是可以接受的。

Example: Solve dy/dx = x/y. Rewrite as y dy = x dx. Integrating gives y²/2 = x²/2 + C. Multiplying by 2 yields y² = x² + 2C, or y² = x² + K, where K = 2C is an arbitrary constant. This is the general solution in implicit form.

示例：求解 dy/dx = x/y。改写为 y dy = x dx。积分得 y²/2 = x²/2 + C。两边乘以 2 得 y² = x² + 2C，即 y² = x² + K，其中 K = 2C 是任意常数。这是隐式形式的通解。

6. Finding Particular Solutions Using Initial Conditions | 利用初始条件求特解

A general solution contains an arbitrary constant. To find a particular solution, use an initial condition, usually written as y(x₀) = y₀ or as a boundary condition such as y = 2 when x = 0. Substitute the given values of x and y into the general solution, then solve for the constant. This gives the unique solution that passes through the specified point.

通解包含一个任意常数。要求特解，需要使用初始条件，通常写作 y(x₀) = y₀ 或边界条件，例如当 x = 0 时 y = 2。将给定的 x 和 y 值代入通解，然后解出常数。这样就得到经过指定点的唯一解。

Example: Solve dy/dx = 2xy with the initial condition y(0) = 3. Separating variables gives dy/y = 2x dx. Integrating yields ln|y| = x² + C, so y = Ae^{x²}. Using y(0) = 3 gives 3 = Ae⁰ = A, hence the particular solution is y = 3e^{x²}.

示例：求解 dy/dx = 2xy，初始条件为 y(0) = 3。分离变量得 dy/y = 2x dx。积分得 ln|y| = x² + C，因此 y = Ae^{x²}。利用 y(0) = 3 得 3 = Ae⁰ = A，所以特解为 y = 3e^{x²}。

Always check that the initial condition is used in the correct place. Sometimes the equation may have a singular solution that is not captured by the general solution, but for A-level this is not usually required.

始终注意初始条件是否用在了正确的位置。有时方程可能存在通解未包含的奇异解，但在 A-level 中通常不要求考虑这一点。

7. Modelling with Differential Equations: Growth and Decay | 微分方程建模：增长与衰减

Many real-world situations are modelled by the exponential growth or decay equation dP/dt = kP, where P is a quantity such as population, amount of substance, or investment value, and k is a constant. The general solution is P = P₀e^{kt}, where P₀ is the initial value at t = 0. If k > 0, the quantity grows; if k < 0, it decays.

许多现实情境可以用指数增长或衰减方程 dP/dt = kP 来建模，其中 P 是数量，如人口、物质总量或投资价值，k 是常数。通解为 P = P₀e^{kt}，其中 P₀ 是 t = 0 时的初始值。如果 k > 0，数量增长；如果 k < 0，数量衰减。

Example: A population initially has 2000 individuals and doubles every 5 years. Let P = 2000e^{kt}. Since P(5) = 4000, we have 4000 = 2000e^{5k}, so 2 = e^{5k} and k = (ln 2)/5. The model is P = 2000e^{(ln 2 / 5)t}.

示例：某人口最初有 2000 人，每 5 年翻一番。设 P = 2000e^{kt}。由于 P(5) = 4000，有 4000 = 2000e^{5k}，因此 2 = e^{5k}，k = (ln 2)/5。模型为 P = 2000e^{(ln 2 / 5)t}。

In modelling questions, you may need to interpret k in context, find half-life, doubling time, or predict future values. Always state units and refer back to the physical situation.

在建模问题中，你可能需要解释 k 在情境中的含义，求半衰期、倍增时间或预测未来值。始终注明单位并联系物理情境。

8. Newton’s Law of Cooling and Other Contexts | 牛顿冷却定律及其他情境

Newton’s law of cooling states that the rate of change of temperature is proportional to the difference between the object’s temperature T and the surrounding temperature T_env. This gives dT/dt = -k(T – T_env), where k > 0. Letting θ = T – T_env transforms the equation into dθ/dt = -kθ, which is a standard exponential decay problem.

牛顿冷却定律指出，温度的变化率与物体温度 T 和环境温度 T_env 之间的差成正比。由此得到 dT/dt = -k(T – T_env)，其中 k > 0。令 θ = T – T_env，方程转化为 dθ/dt = -kθ，这是一个标准的指数衰减问题。

Example: A cup of coffee at 100°C is placed in a room at 20°C. After 10 minutes, its temperature is 60°C. Let θ = T – 20. Then θ₀ = 100 – 20 = 80, and θ(10) = 60 – 20 = 40. From θ = 80e^{-kt}, we have 40 = 80e^{-10k}, so e^{-10k} = 1/2, giving k = (ln 2)/10.

示例：一杯 100°C 的咖啡放在 20°C 的房间中。10 分钟后温度降为 60°C。令 θ = T – 20。则 θ₀ = 100 – 20 = 80，θ(10) = 60 – 20 = 40。由 θ = 80e^{-kt} 得 40 = 80e^{-10k}，因此 e^{-10k} = 1/2，解得 k = (ln 2)/10。

Other contexts include radioactive decay, chemical reaction rates, and water leaking from a tank. All these follow the same separation of variables approach after setting up the equation from the problem statement.

其他情境包括放射性衰变、化学反应速率以及水箱漏水等。在根据问题描述建立方程后，这些都可以用相同的分离变量法求解。

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One frequent mistake is forgetting the constant of integration. Always add C immediately after integrating. Another is incorrectly separating variables: ensure that every y term, including dy, is on one side and every x term, including dx, is on the other.

一个常见错误是忘记积分常数。积分后要立即加上 C。另一个错误是分离变量时出错：确保每个 y 项（包括 dy）在一侧，每个 x 项（包括 dx）在另一侧。

Dividing by g(y) can remove constant solutions. Before dividing, check whether g(y) = 0 gives any valid solutions, and include them if necessary. Be careful with logarithmic integration: use absolute value signs, so ∫ 1/y dy = ln|y| + C, not ln y + C.

除以 g(y) 可能会丢失常数解。在除法之前，检查 g(y) = 0 是否给出有效解，如有必要应将其包含在内。对数积分时要小心：使用绝对值符号，即 ∫ 1/y dy = ln|y| + C，而不是 ln y + C。

When applying an initial condition, substitute into the original integrated expression, not into the differentiated form. Finally, always check that your final answer actually satisfies the differential equation, especially in modelling questions where units and context matter.

应用初始条件时，应代入积分后的表达式，而不是代入求导后的形式。最后，始终检查最终答案是否确实满足微分方程，尤其是在建模问题中，单位和情境非常重要。

10. Worked Exam-Style Question | 考试风格例题精讲

Question: Solve the differential equation dy/dx = (y² + 1)/x for x > 0, given that y = 1 when x = 1. Express y in terms of x.

题目：求解微分方程 dy/dx = (y² + 1)/x，其中 x > 0，且当 x = 1 时 y = 1。将 y 用 x 表示。

Solution: Separate variables to get dy/(y² + 1) = dx/x. Integrating both sides gives arctan y = ln x + C, where C is a constant. Using the condition y(1) = 1: arctan 1 = ln 1 + C, so π/4 = 0 + C, hence C = π/4. Therefore arctan y = ln x + π/4. Taking the tangent of both sides gives the particular solution y = tan(ln x + π/4).

解答：分离变量得 dy/(y² + 1) = dx/x。两边积分得 arctan y = ln x + C，其中 C 是常数。利用条件 y(1) = 1：arctan 1 = ln 1 + C，即 π/4 = 0 + C，因此 C = π/4。所以 arctan y = ln x

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Differential Equations | 求解微分方程