📚 Applications of Differential Equations | 微分方程的应用
Differential equations form a vital bridge between calculus and real-world modelling in IB Mathematics. They allow us to describe how quantities change – from population growth and temperature decay to mechanical vibrations and mixing solutions. In the IB syllabus, you are expected to construct differential equations from descriptions, solve them analytically using separation of variables or integrating factors, apply numerical methods, and interpret the meaning of solutions in context. This article explores the classic applications that appear most often in IB examinations, with paired explanations to strengthen your biliterate understanding.
微分方程是 IB 数学中连接微积分与现实世界建模的重要桥梁。它们帮助我们描述数量如何变化——从人口增长、温度衰减到机械振动和混合溶液。IB 大纲要求你能够根据文字描述建立微分方程，用分离变量法或积分因子法求解，应用数值方法，并在具体情境中解释解的意义。本文逐一探讨 IB 考试中最常见的经典应用，并配以中英双语解说，帮助你深化理解。
1. Modelling with Differential Equations | 用微分方程建模
hange. Write dy/dx = f(x,y) where f captures the physical law or assumption. For instance, 'the rate of population growth is proportional to the current population' ntly ask you to formulate the differential equation from a word problem before solving it.
要为一个实际问题建模，首先要确定变化量 y 以及影响其变化率的因素，写出 dy/dx = f(x,y)，其中 f 体现了物理规律或假设。例如，“人口增长速度与当前人口成正比” 转化为 dP/dt = kP。初始条件 y(x₀) = y₀ 则确定了唯一的特解。IB 试题经常要求你先根据文字题写出微分方程，再求解。
, integrate both sides and add the constant of integration. Many applications lead to exponential or logarithmic solutions. When an explicit analytic solution is impossible, IB may ask you to sketch the slope field or apply Euler's method to approximate values.
方程建立后，IB 中最常用的解析方法是分离变量法。若方程能整理成 g(y) dy = h(x) dx，则两边积分并加上积分常数。许多应用会导出指数或对数形式的解。当无法得到解析解时，IB 可能要求你绘制斜率场，或使用欧拉法进行数值近似。
2. Exponential Growth and Decay | 指数增长与衰减
uantity grows without bound; if k < 0, it decays to zero. The general solution is y = y₀ ekt 0. This model underpins population biology, radioactive decay, continuously compounded interest, and drug elimination from the bloodstream.
最简单的一阶微分方程是 dy/dt = k y，k 为常数。若 k > 0，数量无限增长；若 k < 0，则衰减至零。通解为 y = y₀ ekt，y₀ 为 t = 0 时的初始值。该模型是种群生物学、放射性衰变、连续复利以及药物从血液中消除等问题的基础。
y = y₀ ekt
decay, the half-life t₁/₂ is the time taken for half the substance to decay. From y = y₀ ekt with k < 0, setting y = y₀/2 gives t₁/₂ = ln 2 / |k| / k. Typical IB problems provide data points and ask you to find k, then make predictions.
在放射性衰变中，半衰期 t₁/₂ 是物质衰变一半所需的时间。由 y = y₀ ekt 且 k < 0，令 y = y₀/2 可得 t₁/₂ = ln 2 / |k|。类似地，指数增长中的倍增时间为 ln 2 / k。典型的 IB 题目会给出数据点，要求你求出 k，然后进行预测。
3. Logistic Growth Model | 逻辑斯蒂增长模型
hes K, the rate slows to zero. The solution is an S-shaped curve given by P = K / (1 + A e-r t), where A = (K - P₀) / P₀. This model is widely used in ecology and epidemiology.
当资源有限时，指数增长并不现实。逻辑斯蒂模型引入了环境容纳量 K：dP/dt = r P (1 - P/K)。当 P 较小时，增长近乎指数型；随着 P 趋近 K，增长率趋于零。其解为 S 形曲线，表达式为 P = K / (1 + A e-r t)，其中 A = (K - P₀) / P₀。该模型广泛应用于生态学和流行病学。
t reaches its peak. You might also be asked to sketch the solution family and analyse long-term behaviour.
IB 考题可能要求你验证给定函数满足微分方程，计算某一时刻的人口，或确定增长率最大的时刻。拐点出现在 P = K/2 处，此时增长率 dP/dt 达到最大。考题还可能要求你绘制解曲线族，并分析长期行为。
4. Newton's Law of Cooling | 牛顿冷却定律
Newton's Law of Cooling states that the rate of change of an object' k > 0. Solving by separation of variables gives the explicit solution T = Tₐ + (T₀ - Tₐ) e-k t, with T₀ being the initial temperature.
牛顿冷却定律指出，物体温度 T 的变化率与 T 和环境温度 Tₐ 之差成正比。微分方程为 dT/dt = -k (T - Tₐ)，k > 0。通过分离变量法解得 T = Tₐ + (T₀ - Tₐ) e-k t，其中 T₀ 为初始温度。
This application often appears in IB examinations together with data – e.g., a hot liquid cooling in a room. You may need to find the constant k using two temperature readings at known times, then predict the temperature after another interval or determine the time when a certain temperature is reached. Remember to check that temperatures are in consistent units and that Tₐ is constant.
该应用经常与数据一起出现在 IB 试题中，比如一杯热液体在室内冷却。你可能需要利用两个已知时刻的温度读数求出常数 k，然后预测另一时间后的温度，或确定达到某温度的时刻。要注意保持温度单位一致，并确认环境温度 Tₐ 恒定。
5. Simple Harmonic Motion | 简谐运动
or small angles. The general solution is x = A cos(ω t + φ), with amplitude A and phase φ.
简谐运动 (SHM) 由二阶微分方程 d²x/dt² = -ω² x 控制，其中 x 为位移，ω 为角频率。当回复力与负位移成正比时，如弹簧振子或小角度单摆，就会得到该方程。其通解为 x = A cos(ω t + φ)，振幅为 A，相位为 φ。
x = A cos(ω t + φ)
In IB problems, you may be given initial conditions x(0) and v(0) = dx/dt| ting between expressions. Graphical interpretation of SHM solutions is also tested, especially linking the differential equation to features of the trig graph.
在 IB 问题中，可能给出初始条件 x(0) 和 v(0) = dx/dt|₀ 以确定 A 和 φ。周期为 T = 2π/ω，最大速率为 ω A。你需要能熟练地将解写成正弦形式，并进行表达式的转换。考题还会测试对 SHM 解的图形解释，特别是将微分方程与三角函数的图像特征联系起来。
6. Damped Harmonic Motion | 阻尼谐运动
e get three cases: overdamped (Δ > 0), critically damped (Δ = 0), and underdamped (Δ < 0).
当加入与速度成正比的阻尼力时，方程变为 d²x/dt² + 2β dx/dt + ω₀² x = 0，其中 β 为阻尼系数，ω₀ 为自然频率。辅助方程 r² + 2β r + ω₀² = 0 决定了运动形态。根据判别式 Δ = 4(β² - ω₀²)，有三种情况：过阻尼 (Δ > 0)、临界阻尼 (Δ = 0) 和欠阻尼 (Δ < 0)。
In the underdamped case, the solution is x = e-β t (A cos(ω₁ t) + B sin(ω₁ t)), where ω₁ = √ω₀² - β². This represents oscillation with an exponentially decaying
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