📚 IB Mathematics: Differential Equations – Key Points Review | IB数学：微分方程考点精讲
Differential equations are a cornerstone of the IB higher-level calculus syllabus, connecting the rate of change of a quantity to the quantity itself and thereby modeling a vast range of real-world phenomena. This article distills the essential techniques you need to master: from classifying equations and solving first-order ODEs by separation, integrating factor, and homogeneous substitution, to numerical Euler’s method, second-order linear equations with constant coefficients, the method of undetermined coefficients, and key applications.
微分方程是IB高阶微积分的核心部分，它将量的变化率与量自身联系起来，从而能够模拟各种各样的现实世界现象。本文凝练了您必须掌握的核心技巧：从方程分类、用分离变量法、积分因子法和齐次代换求解一阶常微分方程，到欧拉数值方法、二阶常系数线性微分方程、待定系数法以及关键应用。
1. What is a Differential Equation? | 什么是微分方程？
A differential equation (DE) is an equation that contains an unknown function and one or more of its derivatives. For example, the equation dy/dx = 2x states that the derivative of y with respect to x is 2x, and its solution is the family of functions y = x² + C. Differential equations allow us to describe processes involving change: population growth, radioactive decay, Newton’s law of cooling, and motion under resistance.
微分方程是包含未知函数及其一个或多个导数的方程。例如，方程 dy/dx = 2x 表明 y 关于 x 的导数为 2x，其解为函数族 y = x² + C。微分方程让我们能够描述涉及变化的过程：人口增长、放射性衰变、牛顿冷却定律以及阻力下的运动。
- An ordinary differential equation (ODE) involves functions of a single independent variable.
- 常微分方程涉及单个自变量的函数。
- A partial differential equation (PDE) involves multiple independent variables, but IB only covers ODEs.
- 偏微分方程涉及多个自变量，但IB仅考查常微分方程。
2. Order and General vs. Particular Solutions | 阶数与通解、特解
The order of a differential equation is the highest derivative that appears. A first-order ODE contains only the first derivative, e.g. dy/dx = x + y. A second-order ODE contains the second derivative, e.g. d²y/dx² + 3 dy/dx + 2y = 0. The general solution includes arbitrary constants whose number equals the order of the equation. A particular solution is obtained by applying initial conditions or boundary conditions to determine these constants.
微分方程的阶数是其中出现的最高阶导数。一阶常微分方程只包含一阶导数，例如 dy/dx = x + y。二阶常微分方程包含二阶导数，例如 d²y/dx² + 3 dy/dx + 2y = 0。通解包含任意常数，常数的个数等于方程的阶数。特解是通过施加初始条件或边界条件确定这些常数后得到的。
- For y’ = 2x, general solution: y = x² + C.
- 对 y’ = 2x，通解为：y = x² + C。
- If y(0) = 3, then C = 3 and the particular solution is y = x² + 3.
- 若 y(0) = 3，则 C = 3，特解为 y = x² + 3。
3. Separable First-Order ODEs | 可分离变量的一阶微分方程
A first-order ODE is separable if it can be written as g(y) dy = f(x) dx. The solution is obtained by integrating both sides: ∫ g(y) dy = ∫ f(x) dx. Always remember to include the constant of integration on one side. After integrating, attempt to write y explicitly as a function of x if possible.
一阶常微分方程若可写为 g(y) dy = f(x) dx，则称为可分离的。通过对两边积分求得解：∫ g(y) dy = ∫ f(x) dx。请务必记得在一边加上积分常数。积分后，如果可能，尝试把 y 显式写成 x 的函数。
Example: dy/dx = xy → (1/y) dy = x dx → ln|y| = x²/2 + C → y = A e^(x²/2)
范例：dy/dx = xy → (1/y) dy = x dx → ln|y| = x²/2 + C → y = A e^(x²/2)。
- Remember to check for constant solutions (e.g. y = 0) that might be lost when dividing by g(y).
- 记得检查那些在除以 g(y) 时可能丢失的常数解（例如 y = 0）。
4. Integrating Factor Method | 积分因子法
A first-order linear ODE has the form dy/dx + P(x)y = Q(x). To solve it, compute the integrating factor μ(x) = e^(∫P(x) dx). Multiply the entire equation by μ(x), which transforms the left side into the derivative of μ(x)y. Then integrate both sides with respect to x to obtain the general solution.
一阶线性常微分方程具有形式 dy/dx + P(x)y = Q(x)。求解时，先计算积分因子 μ(x) = e^(∫P(x) dx)。将整个方程乘以 μ(x)，左边将变为 μ(x)y 的导数。然后两边对 x 积分即可得到通解。
μ(x) = e^(∫P(x) dx) ⇒ d/dx [μ(x) y] = μ(x) Q(x)
- Example: y’ + 2xy = x. Here P(x)=2x, μ = e^(x²). Then d/dx (e^(x²) y) = x e^(x²). Integrate: e^(x²) y = (1/2) e^(x²) + C ⇒ y = 1/2 + C e^(−x²).
- 范例：y’ + 2xy = x。此处 P(x)=2x，μ = e^(x²)。则 d/dx (e^(x²) y) = x e^(x²)。积分得：e^(x²) y = (1/2) e^(x²) + C ⇒ y = 1/2 + C e^(−x²)。
5. Homogeneous First-Order ODEs | 齐次一阶微分方程
A first-order ODE is called homogeneous if it can be written as dy/dx = F(y/x). The substitution v = y/x (so that y = v x and dy/dx = v + x dv/dx) transforms the equation into a separable ODE in v and x. After solving for v, substitute back v = y/x to obtain the solution in terms of y and x.
若一阶常微分方程可以写成 dy/dx = F(y/x) 的形式，则称其为齐次方程。通过代换 v = y/x（即 y = v x 且 dy/dx = v + x dv/dx）可将方程化为关于 v 和 x 的可分离方程。解出 v 后，再代回 v = y/x 即得 y 关于 x 的解。
Example: dy/dx = (x + y)/x = 1 + y/x ⇒ v + x dv/dx = 1 + v ⇒ x dv/dx = 1 ⇒ dv = dx/x
- After integration: v = ln|x| + C ⇒ y/x = ln|x| + C ⇒ y = x ln|x| + Cx.
- 积分后：v = ln|x| + C ⇒ y/x = ln|x| + C ⇒ y = x ln|x| + Cx。
6. Substitutions Reducible to Separable Form | 可化为可分离形式的代换
Equations of the form dy/dx = f(ax + by + c) can be tackled by letting u = ax + by + c. Differentiating with respect to x gives du/dx = a + b dy/dx, which allows us to replace dy/dx. The resulting ODE in u and x becomes separable: du/dx = a + b f(u). Integrate to find u, then back-substitute to obtain y.
形如 dy/dx = f(ax + by + c) 的方程可通过令 u = ax + by + c 来处理。对 x 求导得 du/dx = a + b dy/dx，从而可以替换掉 dy/dx。得到的关于 u 和 x 的常微分方程变为可分离的：du/dx = a + b f(u)。积分求出 u，再回代即得 y。
- Example: dy/dx = (2x + y + 1)². Let u = 2x + y + 1 ⇒ du/dx = 2 + dy/dx ⇒ dy/dx = du/dx − 2 = u². Thus du/dx = u² + 2.
- 范例：dy/dx = (2x + y + 1)²。令 u = 2x + y + 1 ⇒ du/dx = 2 + dy/dx ⇒ dy/dx = du/dx − 2 = u²。于是 du/dx = u² + 2。
7. Euler’s Method for Numerical Solutions | 欧拉方法求数值解
Euler’s method provides a numerical approximation to the solution of a first-order ODE y’ = f(x,y) with initial condition y(x₀) = y₀, using a step size h. The iterative formula is: xₙ₊₁ = xₙ + h, yₙ₊₁ = yₙ + h·f(xₙ, yₙ). Starting from the initial point, the method constructs a polygonal path that approximates the true solution curve.
欧拉方法为求解一阶常微分方程 y’ = f(x,y) 在初始条件 y(x₀) = y₀ 下的解提供了一种数值近似，步长为 h。迭代公式为：xₙ₊₁ = xₙ + h，yₙ₊₁ = yₙ + h·f(xₙ, yₙ)。从初始点出发，该方法构建一条折线来逼近真实的解曲线。
- The smaller the step size h, the more accurate the approximation, but more steps are needed.
- 步长 h 越小，近似越精确，但需要更多的步数。
- IB questions often ask you to carry out a few steps, fill a table, and discuss the error.
- IB 考题常要求进行若干步迭代、填表和讨论误差。
8. Second-Order Linear Homogeneous ODEs with Constant Coefficients | 常系数二阶齐次线性微分方程
These equations take the form a y” + b y’ + c y = 0, where a, b, c are constants. To solve, form the auxiliary equation a r² + b r + c = 0 and find its roots r₁ and r₂. The form of the general solution depends on the discriminant Δ = b² − 4ac.
这类方程具有形式 a y” + b y’ + c y = 0，其中 a, b, c 为常数。求解时，写出辅助方程 a r² + b r + c = 0 并求出其根 r₁ 和 r₂。通解的形式取决于判别式 Δ = b² − 4ac。
| Case | Roots | General Solution |
|---|---|---|
| Δ > 0 | Two distinct real r₁, r₂ | y = C₁ e^(r₁x) + C₂ e^(r₂x) |
| Δ = 0 | Repeated real root r | y = (C₁ + C₂ x) e^(rx) |
| Δ < 0 | Complex α ± iβ | y = e^(αx) (C₁ cos βx + C₂ sin βx) |
中文表格：情况、根、通解。Δ>0：两个不等实根 r₁, r₂ → y = C₁ e^(r₁x) + C₂ e^(r₂x)；Δ=0：重实根 r → y = (C₁ + C₂ x) e^(rx)；Δ<0：共轭复根 α ± iβ → y = e^(αx) (C₁ cos βx + C₂ sin βx)。
- Initial conditions are applied after writing the general solution to determine the constants C₁, C₂.
- 在写出通解后应用初始条件来确定常数 C₁ 和 C₂。
9. Non-Homogeneous Second-Order ODEs and Undetermined Coefficients | 非齐次二阶常微分方程与待定系数法
A non-homogeneous linear ODE has the form a y” + b y’ + c y = f(x). Its general solution is the sum of the complementary function y_c (solution of the homogeneous equation) and a particular integral y_p. For certain f(x) – polynomials, exponentials, sines/cosines – we can guess a suitable form for y_p with undetermined coefficients and substitute into the DE to find them.
非齐次线性常微分方程具有形式 a y” + b y’ + c y = f(x)。其通解为余函数 y_c（齐次方程的通解）与一个特解 y_p 之和。对于某些特定的 f(x)——多项式、指数函数、正弦/余弦函数——我们可以猜测一个含有待定系数的 y_p 的适当形式，并将其代入方程来确定这些系数。
If f(x) = P(x), try y_p = polynomial of same degree.
If f(x) = k e^(mx), try y_p = A e^(mx) (or A x e^(mx) if e^(mx) already solves the homogeneous equation).
If f(x) = k sin(ωx) or k cos(ωx), try y_p = A sin(ωx) + B cos(ωx).
- When the standard trial form already appears in the complementary function, multiply the trial y_p by x (or x²) until it differs from y_c.
- 当标准试探形式已经在余函数中出现时，将试探 y_p 乘以 x（或 x²），直至它与 y_c 不再重复。
10. Applications: Growth, Decay, and Newton’s Law of Cooling | 应用：增长、衰变和牛顿冷却定律
Exponential growth and decay are modeled by dP/dt = k P, with solution P(t) = P₀ e^(kt). If k > 0, growth; if k < 0, decay. For Newton’s law of cooling, the rate of change of temperature T is proportional to the difference from the ambient temperature Tₐ: dT/dt = −k (T − Tₐ). The solution is T(t) = Tₐ + (T₀ − Tₐ) e^(−kt), where T₀ is the initial temperature. These models are frequently tested, requiring you to set up the DE, solve it, and interpret constants or half-life.
指数增长和衰减模型为 dP/dt = k P，其解为 P(t) = P₀ e^(kt)。若 k > 0，则为增长；若 k < 0，则为衰减。对于牛顿冷却定律，温度 T 的变化率与跟环境温度 Tₐ 的差成正比：dT/dt = −k (T − Tₐ)。解为 T(t) = Tₐ + (T₀ − Tₐ) e^(−kt)，其中 T₀ 为初始温度。这些模型经常被考查，要求您建立微分方程、求解并解释常数或半衰期。
11. Applications: Motion with Resistance | 应用：有阻力的运动
For a falling object experiencing air resistance proportional to velocity, the net force is mg − kv, leading to m dv/dt = mg − kv. This is a first-order linear ODE. Solving gives v(t) = (mg/k) (1 − e^(−kt/m)) + v₀ e^(−kt/m), approaching terminal velocity mg/k as t → ∞. Similar setups appear for objects propelled through resistive media, and IB problems may ask for velocity as a function of time or distance.
对于受到与速度成正比的空气阻力的下落物体，合外力为 mg − kv，从而得到 m dv/dt = mg − kv。这是一个一阶线性常微分方程。求解得 v(t) = (mg/k)(1 − e^(−kt/m)) + v₀ e^(−kt/m)，当 t → ∞ 时趋近于终极速度 mg/k。类似的设定也可用于通过阻力介质推进的物体，IB 试题可能会要求求出速度关于时间或距离的函数。
12. Slope Fields and Qualitative Analysis | 斜率场与定性分析
A slope field (or direction field) is a graphical representation of the first-order ODE y’ = f(x,y). At a grid of points (x,y), short line segments are drawn with slope f(x,y). This visual tool helps predict the behavior of solutions without solving the DE analytically. You can sketch solution curves that follow the field, identify equilibrium solutions, and analyze stability. Euler’s method essentially follows the local slope from step to step.
斜率场（方向场）是一阶常微分方程 y’ = f(x,y) 的图形表示。在网格点 (x,y) 处，画出斜率为 f(x,y) 的短线段。这一可视化工具帮助在不解析求解的情况下预测解的行为。您可以画出跟随场的解曲线、识别平衡解并分析稳定性。欧拉方法本质上就是一步一步地沿着局部斜率前进。
- Equilibrium solutions occur where f(x,y) = 0, giving horizontal slopes.
- 平衡解出现在 f(x,y) = 0 处，此处斜率为水平。
- IB may provide a slope field and ask you to draw a solution passing through a given point.
- IB 可能会提供一个斜率场，并要求绘制经过给定点的解曲线。
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