Differential Equations: Core Solution Methods & Past Paper Analysis | 微分方程考点：常见解法与历年真题

📚 Differential Equations: Core Solution Methods & Past Paper Analysis | 微分方程考点：常见解法与历年真题

Differential equations are a cornerstone of A-Level and IB mathematics, frequently appearing in both pure mathematics and applications papers. Mastering the standard solution techniques not only secures marks in exams but also builds intuition for modeling real-world phenomena.

微分方程是 A-Level 和 IB 数学的核心内容，在纯数学与应用题中频繁出现。掌握标准解法不仅能帮助你在考试中稳拿分数，更能培养对现实世界建模的直觉。

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

A differential equation is an equation that involves an unknown function and one or more of its derivatives. The order of a differential equation is the highest derivative present.

微分方程是包含未知函数及其一个或多个导数的方程。阶数是方程中出现的最高阶导数的阶数。

dy/dx = 2x is a first-order equation, while d²y/dx² − 3 dy/dx + 2y = 0 is second-order. A general solution contains arbitrary constants; a particular solution is obtained when initial or boundary conditions are applied.

例如，dy/dx = 2x 是一阶方程，而 d²y/dx² − 3 dy/dx + 2y = 0 是二阶方程。通解包含任意常数；当给定初始条件或边界条件后得到的解称为特解。

cients. Occasionally, homogeneous first-order equations appear in the form dy/dx = f(y/x).

考试题目通常分三类：可分离变量方程的求解、使用积分因子法、以及求解常系数二阶线性方程。偶尔也会出现形如 dy/dx = f(y/x) 的一阶齐次方程。

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2. Separable Differential Equations | 可分离变量方程

teps are to separate the variables (collect all y terms on one side and all x terms on the other) and then integrate both sides.

可分离变量方程可以写成 dy/dx = g(x)·h(y) 的形式。关键步骤是将变量分离（把所有含 y 的项放在一边，含 x 的项放在另一边），然后对两边分别积分。

General procedure:

一般步骤：

• Rewrite the equation as (1/h(y)) dy = g(x) dx.
• Integrate both sides: ∫(1/h(y)) dy = ∫g(x) dx.
• Add the constant of integration C and solve for y where possible.

• 将方程改写为 (1/h(y)) dy = g(x) dx。
• 两边积分：∫(1/h(y)) dy = ∫g(x) dx。
• 加上积分常数 C，并在可能的情况下解出 y。

Worked Example: Solve dy/dx = (x + 1)/y² given y(0) = 2.

例题：求解 dy/dx = (x + 1)/y²，已知 y(0) = 2。

Separate the variables:

分离变量：

∫y² dy = ∫(x + 1) dx

Integrating gives:

积分得到：

y³/3 = (x + 1)²/2 + C

Using y(0) = 2, we have 8/3 = 1/2 + C, so C = 13/6. Hence:

代入 y(0) = 2，得 8/3 = 1/2 + C，所以 C = 13/6。因此：

y = [ (3(x + 1)² + 13) / 2 ]^1/3

Always check whether the solution is valid for all x or only on a restricted interval—this is a common mark target in exams.

务必检查解是否对所有 x 有效，还是仅在一个受限区间内有效——这是考试中常见的得分点。

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3. First-Order Linear Equations: The Integrating Factor | 一阶线性方程：积分因子法

e solution method uses an integrating factor that turns the LHS into a perfect derivative.

一阶线性微分方程的标准形式为 dy/dx + P(x)y = Q(x)。解法通过积分因子将等式左边转化为一个完全的导数。

The integrating factor is:

积分因子为：

IF = e^(∫P(x) dx)

Multiplying through by the IF gives:

两边乘以积分因子后得到：

d/dx [ y · e^{∫P dx} ] = Q(x) · e^{∫P dx}

Then integrate both sides and solve for y. Do not forget the constant of integration!

然后两边积分并解出 y。千万不要忘记积分常数！

Worked Example: Solve dy/dx + 2y = e^3x.

例题：求解 dy/dx + 2y = e^3x。

Here P(x) = 2, so:

这里 P(x) = 2，因此：

IF = e^{∫2 dx} = e^2x

Multiplying through:

两边同乘：

d/dx ( y e^2x ) = e^5x

Integrate:

积分：

y e^2x = e^5x/5 + C

Finally:

最终：

y = e^3x/5 + C e^−2x

A common pitfall is misidentifying P(x) or forgetting to divide by the coefficient of dy/dx if it is not 1. Always rearrange into standard form first.

常见错误是识别错了 P(x)，或者当 dy/dx 的系数不为 1 时忘记先除掉。务必先将方程整理成标准形式。

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4. Homogeneous First-Order Equations | 一阶齐次方程

on y = vx (where v is a function of x) transforms it into a separable equation in v and x.

一阶齐次微分方程可以表示为 dy/dx = F(y/x)。令 y = vx（其中 v 是 x 的函数），可以将原方程转化为关于 v 和 x 的可分离变量方程。

Follow these steps:

步骤如下：

• Substitute y = vx, so dy/dx = v + x·dv/dx (via the product rule).
• Rewrite the equation in terms of v and x; it should now be separable.
• Integrate, then substitute v = y/x back to obtain the final solution.

• 代入 y = vx，由乘积法则得 dy/dx = v + x·dv/dx。
• 将方程改写为关于 v 和 x 的形式，此时应当可以分离变量。
• 积分后，代回 v = y/x 得到最终解。

Worked Example: Solve dy/dx = (x² + y²)/(xy).

例题：求解 dy/dx = (x² + y²)/(xy)。

Rewrite:

改写：

dy/dx = x/y + y/x = v + 1/v

Since y = vx, dy/dx = v + x·dv/dx, so:

因为 y = vx，dy/dx = v + x·dv/dx，所以：

x·dv/dx = 1/v

Separating and integrating:

分离并积分：

∫v dv = ∫(1/x) dx → v²/2 = ln|x| + C

Substituting back v = y/x:

代回 v = y/x：

y² = 2x²(ln|x| + C)

Homogeneous equations are less common in standard A-Level papers but appear frequently in Further Mathematics or IB HL exams.

齐次方程在标准 A-Level 试卷中出现较少，但在进阶数学（Further Mathematics）或 IB HL 考试中则较为常见。

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5. Second-Order Linear Equations with Constant Coefficients | 常系数二阶线性方程

he general form is a·d²y/dx² + b·dy/dx + c·y = 0. The solution depends on the roots of the characteristic equation:

一般形式为 a·d²y/dx² + b·dy/dx + c·y = 0。解的形式取决于特征方程的根：

ar² + br + c = 0

There are three cases:

有三种情况：

Roots of characteristic equation	General solution
Distinct real roots: r₁ ≠ r₂	y = A er₁x + B er₂x
Repeated real root: r	y = (A + Bx) erx
Complex roots: r = a ± bi	y = eax[ A cos(bx) + B sin(bx) ]

特征方程的根

通解

两个不同实根：r₁ ≠ r₂

y = A e^r₁x + B e^r₂x

重实根：r

y = (A + Bx) e^rx

复数根：r = a ± bi

y = e^ax[ A cos(bx) + B sin(bx) ]

Worked Example: Solve d²y/dx² − 5 dy/dx + 6y = 0.

例题：求解 d²y/dx² − 5 dy/dx + 6y = 0。

The characteristic equation is:

特征方程为：

r² − 5r + 6 = 0 → (r − 2)(r − 3) = 0

So r₁ = 2, r₂ = 3, giving the general solution:

因此 r₁ = 2，r₂ = 3，通解为：

y = A e^2x + B e^3x

When dealing with complex roots, do not forget that i² = −1, and the Euler relation e^iθ = cos θ + i sin θ underpins the trigonometric form of the solution.

处理复数根时，不要忘记 i² = −1，并且欧拉公式 e^iθ = cos θ + i sin θ 是三角函数形式解的基础。

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6. Initial Value Problems and Particular Solutions | 初值问题与特解

₁. These conditions allow us to determine the arbitrary constants in the general solution.

初值问题（IVP）给出诸如 y(x₀) = y₀ 以及可能的 y′(x₀) = y₁ 等条件。这些条件用于确定通解中的任意常数。

Worked Example: Solve d²y/dx² − 5 dy/dx + 6y = 0 with y(0) = 1 and y′(0) = 0.

例题：求解 d²y/dx² − 5 dy/dx + 6y = 0，已知 y(0) = 1，y′(0) = 0。

From the previous example, y = A e^2x + B e^3x. Then:

由上面例题，y = A e^2x + B e^3x。则：

y′ = 2A e^2x + 3B e^3x

Applying the conditions:

代入条件：

A + B = 1, 2A + 3B = 0

Solving gives A = 3 and B = −2. Hence the particular solution is:

解得 A = 3，B = −2。因此特解为：

y = 3e^2x − 2e^3x

For first-order equations, only one condition is needed; for second-order equations, two independent conditions are required. Missing a condition or confusing y′(0) with y(0) are classic errors.

一阶方程只需要一个条件；二阶方程则需要两个独立条件。漏掉一个条件或混淆 y′(0) 与 y(0) 是典型的失误。

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7. Real-World Applications and Modeling | 实际应用与建模

Differential equations are powerful tools for modeling exponential growth and decay, Newton's law of cooling, and simple harmonic motion. Exams often frame these as word problems requiring you to set up and solve the equation.

微分方程是建模指数增长与衰减、牛顿冷却定律以及简谐运动的强大工具。考试通常将这些内容设置为应用题，要求你建立并求解方程。

Exponential model: dP/dt = kP has the solution P = P₀ e^kt. If k > 0, it models population growth; if k < 0, it models radioactive decay.

指数模型：dP/dt = kP 的解为 P = P₀ e^kt。若 k > 0，表示种群增长；若 k < 0，表示放射性衰变。

Newton's law of cooling: /dt = −k(T − Tₐ), where Tₐ is the ambient temperature. The solution is T = Tₐ + (T₀ − Tₐ)e^−kt.

牛顿冷却定律：dT/dt = −k(T − Tₐ)，其中 Tₐ 为环境温度。解为 T = Tₐ + (T₀ − Tₐ)e^−kt。

Examiners expect you to identify the appropriate model from the wording, write the differential equation correctly, then solve it. Always state the units and the meaning of each constant in your final answer.

阅卷者期望你能从题干中识别出合适的模型、正确写出微分方程，然后求解。在最终答案中务必注明单位以及每个常数的实际意义。

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8. Past Exam Question Walkthrough | 历年真题精讲

Let us work through a typical past-paper question that combines several key skills.

下面我们来看一道结合多个关键技能的典型真题。

Question (adapted from A-Level Paper): tial equation is dS/dt = 3 − 0.05S, where S is the amount of salt in kg at time t minutes.

题目（改编自 A-Level 试卷）：一个容器装有盐溶液。在 t = 0 时，盐的量为 50 kg。盐以每分钟 3 kg 的恒定速率加入，同时混合物以与当前盐量成正比的速率被移除。微分方程为 dS/dt = 3 − 0.05S，其中 S 为 t 分钟时盐的量（单位 kg）。

Part (a): Solve this differential equation to find S in terms of t.

第（a）问：求解该微分方程，用 t 表示 S。

This is separable:

该方程可分离变量：

∫ dS/(3 − 0.05S) = ∫ dt

Using the substitution u = 3 − 0.05S (or direct integration):

令 u = 3 − 0.05S（或直接积分）：

−20 ln|3 − 0.05S| = t + C

Since S(0) = 50: −20 ln(3 − 2.5) = 0 + C → C = −20 ln(0.5). Therefore:

因为 S(0) = 50：−20 ln(3 − 2.5) = 0 + C → C = −20 ln(0.5)。因此：

S = 60 − 10 e^−t/20

Part (b): What is the limiting amount of salt as t → ∞?

第（b）问：当 t → ∞ 时盐量的极限是多少？

As t → ∞, e^−t/20 → 0, so S → 60 kg. This is the steady-state solution, obtained by setting dS/dt = 0.

当 t → ∞ 时，e^−t/20 → 0，所以 S → 60 kg。这就是稳态解，可通过令 dS/dt = 0 得到。

Notice how the examiner structures questions to test separation of variables, handling of natural logarithms, and the physical interpretation of limits—all in one problem.

请注意，出题者通过一道题同时考查了分离变量法、自然对数的处理以及对极限的物理解释。

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9. Common Mistakes and Exam Tips | 常见错误与应试技巧

Across hundreds of student exam scripts, a few mistakes appear again and again. Avoiding these can immediately boost your grade.

在大量学生答卷中，有几种错误反复出现。避免这些错误可以立刻提高你的分数。

• Forgetting the "+ C": Losing the constant of integration makes a general solution incomplete. If initial conditions are given, you must use them to find C.
• Ignoring absolute values: When integrating 1/y or 1/x, always write ln|y| and ln|x|. This matters for the domain of the solution.

• 忘记 "+ C"：遗漏积分常数会导致通解不完整。若给定了初始条件，必须用它们求出 C。
• 忽略绝对值：对 1/y 或 1/x 积分时，务必写成 ln|y| 和 ln|x|。这关系到解的定义域。

• Misapplying the integrating factor: Ensure the coefficient of dy/dx is 1 first. If it is not, divide the entire equation by that coefficient before computing the IF.
• Incorrect characteristic roots: When the quadratic has negative discriminant, do not try to force real roots—write the complex solution and use Euler's identity.

• 错误使用积分因子：确保 dy/dx 的系数为 1。若不是，先对整个方程除以该系数，再计算积分因子。
• 特征根求错：当判别式为负时，不要强行求实根——应写出复数形式的解并使用欧拉恒等式。

Finally, always check your solution by substituting it back into the original differential equation. Even a quick mental check can catch sign errors and algebraic slips.

最后，务必通过代回原微分方程来检验你的解。即使只是快速心算，也能发现符号错误和代数疏漏。

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