📚 IB Mathematics: First-Order Linear Differential Equations with Variable Coefficients | IB数学：变系数一阶线性微分方程
In IB Mathematics Analysis and Approaches HL, first-order linear differential equations are a core topic in the calculus option. When the coefficients are not constant but functions of the independent variable, the equation takes the form dy/dx + P(x)y = Q(x). Solving these equations requires the method of integrating factors, a technique that transforms a non-exact equation into an exact one and enables direct integration.
在IB数学分析与方法（AA）HL课程中，一阶线性微分方程是微积分选修模块的核心内容。当系数不再是常数而是自变量的函数时，方程的一般形式为 dy/dx + P(x)y = Q(x)。求解此类方程需要用到积分因子法，这一技巧能将非恰当方程转化为恰当方程，从而通过直接积分求解。
1. Standard Form and General Strategy | 标准形式与总体策略
Every first-order linear differential equation with variable coefficients must first be rearranged into the standard form dy/dx + P(x)y = Q(x). The coefficient of dy/dx must be 1, and both P(x) and Q(x) are functions of x only. This normalization is essential before applying the integrating factor method.
任何变系数一阶线性微分方程都必须先整理成标准形式 dy/dx + P(x)y = Q(x)。dy/dx 的系数必须化为1，P(x) 和 Q(x) 都只是 x 的函数。这种标准化是在应用积分因子法之前必不可少的步骤。
Once in standard form, the integrating factor is defined as I(x) = e^(∫P(x)dx). Multiplying both sides of the equation by I(x) turns the left-hand side into the derivative of I(x)y, allowing the equation to be solved by a single integration.
一旦化为标准形式，积分因子定义为 I(x) = e^(∫P(x)dx)。将方程两边同时乘以 I(x)，左边就变成了 I(x)y 的导数形式，从而只需一次积分即可求解。
I(x) = e^(∫P(x)dx), then d/dx[I(x)·y] = I(x)·Q(x)
The general solution is obtained by integrating both sides, yielding y = (∫I(x)Q(x)dx + C) / I(x). In IB exams, problems typically require finding the particular solution given an initial condition such as y(x₀) = y₀.
通解通过对两边积分得到：y = (∫I(x)Q(x)dx + C) / I(x)。在IB考试中，题目通常要求利用初始条件 y(x₀) = y₀ 求出特解。
2. Deriving the Integrating Factor | 积分因子的推导
Why does multiplying by e^(∫P(x)dx) work? Consider the product rule: d/dx[u(x)y] = u'(x)y + u(x)y’. We want this to match u(x)[y’ + P(x)y]. This requires u'(x) = u(x)P(x), which is a separable differential equation.
为什么乘以 e^(∫P(x)dx) 有效？考虑乘积法则：d/dx[u(x)y] = u'(x)y + u(x)y’。我们希望这个结果等于 u(x)[y’ + P(x)y]。这要求 u'(x) = u(x)P(x)，这是一个可分离变量的微分方程。
Solving du/dx = P(x)u gives du/u = P(x)dx, integrating both sides yields ln|u| = ∫P(x)dx, and hence u = e^(∫P(x)dx). The constant of integration is customarily omitted because it cancels out in the final solution.
求解 du/dx = P(x)u，得到 du/u = P(x)dx，两边积分得 ln|u| = ∫P(x)dx，因此 u = e^(∫P(x)dx)。通常省略积分常数，因为它最终会约去。
In IB marking schemes, the first step is always to identify P(x) correctly. A common error is reading P(x) from an equation that is not in standard form, leading to a wrong integrating factor and an incorrect solution.
在IB评分标准中，第一步始终是正确识别 P(x)。一个常见错误是从未化为标准形式的方程中直接读取 P(x)，这会导致积分因子错误，进而得到错误的解。
3. Worked Example: Constant Coefficient Quick Check | 例题：常数系数快速检验
Although the focus is variable coefficients, consider dy/dx + 3y = 6 as a warm-up. Here P(x) = 3 and Q(x) = 6. The integrating factor is I(x) = e^(∫3dx) = e^(3x).
虽然本讲重点是变系数，但先考虑 dy/dx + 3y = 6 作为热身。这里 P(x) = 3，Q(x) = 6。积分因子为 I(x) = e^(∫3dx) = e^(3x)。
Multiplying through gives e^(3x)dy/dx + 3e^(3x)y = 6e^(3x). The left side is d/dx(e^(3x)y). Integrating both sides yields e^(3x)y = 2e^(3x) + C, so y = 2 + Ce^(−3x).
两边乘以 e^(3x) 得到 e^(3x)dy/dx + 3e^(3x)y = 6e^(3x)。左边正是 d/dx(e^(3x)y)。两边积分得 e^(3x)y = 2e^(3x) + C，因此 y = 2 + Ce^(−3x)。
This simple case verifies the method: the general solution combines a particular solution (y = 2) with the complementary function Ce^(−3x). The same structure appears for variable coefficients, though the integrals become more challenging.
这个简单例子验证了方法：通解是特解（y = 2）与余函数 Ce^(−3x) 的组合。对于变系数情况，结构完全相同，只是积分运算更具挑战性。
4. Variable Coefficient Example: P(x) = 2/x | 变系数示例：P(x) = 2/x
Solve dy/dx + (2/x)y = x² with x > 0. This equation already appears in standard form, with P(x) = 2/x and Q(x) = x².
求解 dy/dx + (2/x)y = x²，其中 x > 0。该方程已经是标准形式，其中 P(x) = 2/x，Q(x) = x²。
Compute the integrating factor: I(x) = e^(∫(2/x)dx) = e^(2ln|x|) = x². Notice that e^(2lnx) simplifies directly to x², using the identity a·ln|b| = ln|b^a|.
计算积分因子：I(x) = e^(∫(2/x)dx) = e^(2ln|x|) = x²。注意 e^(2lnx) 直接化简为 x²，这里用到了恒等式 a·ln|b| = ln|b^a|。
d/dx(x²y) = x⁴ → x²y = x⁵/5 + C → y = x³/5 + C/x²
If an initial condition y(1) = 1 is given, then 1 = 1/5 + C, so C = 4/5. The particular solution is y = x³/5 + 4/(5x²). This demonstrates how variable coefficients naturally produce polynomial and reciprocal terms in the solution.
若给定初值条件 y(1) = 1，则 1 = 1/5 + C，故 C = 4/5。特解为 y = x³/5 + 4/(5x²)。这一例子展示了变系数如何自然地产生多项式和倒数项。
5. The Role of P(x) in the Solution’s Domain | P(x) 与解的定义域
When integrating P(x), absolute values arise inside the logarithm, and this affects the domain. For example, if P(x) = 1/x, then ∫(1/x)dx = ln|x|, and the integrating factor is |x|. In practice, IB problems specify intervals, such as x > 0, to avoid sign ambiguities.
对 P(x) 积分时，对数内部会出现绝对值，这会影响定义域。例如，若 P(x) = 1/x，则 ∫(1/x)dx = ln|x|，积分因子为 |x|。实践中，IB题目会明确指定区间，如 x > 0，以避免符号歧义。
If no interval is given, the solution is generally valid on any interval not containing x = 0 where P(x) is discontinuous. Students should state the domain explicitly when the question asks for it.
若未给出区间，解通常在任何不包含 x = 0 的区间上有效，因为 P(x) 在该处不连续。当题目要求时，学生应明确写出解的定义域。
A careful approach: write the integrating factor as e^(ln|x|·k) = x^k for x > 0, or use |x|^k for full generality. Most IB questions restrict the domain so this subtlety is not examined excessively, but understanding it prevents surprises.
严谨的做法是：将积分因子写为 e^(ln|x|·k) = x^k（当 x > 0），或使用 |x|^k 保持一般性。大多数IB题目会限定定义域，因此这一细微之处不会被过度考察，但理解它能避免意外。
6. Recognising Exact Differentials | 识别恰当微分
After multiplying by the integrating factor, the left-hand side becomes d/dx[I(x)y]. To check correctness, differentiate I(x)y and verify it matches I(x)y’ + I(x)P(x)y. This is a powerful self-check during examinations.
乘以积分因子后，左边成为 d/dx[I(x)y]。为检验正确性，对 I(x)y 求导，验证其结果是否等于 I(x)y’ + I(x)P(x)y。这是考试中一个强有力的自查技巧。
For instance, with I(x) = x², d/dx(x²y) = 2xy + x²y’. The original equation multiplied by x² is x²y’ + 2xy = x⁴, which confirms the exactness.
例如，当 I(x) = x² 时，d/dx(x²y) = 2xy + x²y’。原方程乘以 x² 后为 x²y’ + 2xy = x⁴，这正好验证了恰当性。
This recognition helps students avoid the common mistake of integrating an expression that is not a total derivative, which would lead to incorrect results.
这种识别能帮助学生避免一个常见错误：对并非全导数的表达式直接积分，从而得出错误结果。
7. Variable Coefficient with Trigonometric P(x) | 含三角函数系数的变系数方程
Consider dy/dx + (tan x)y = cos x. Here P(x) = tan x and Q(x) = cos x. The integrating factor is I(x) = e^(∫tan x dx) = e^(−ln|cos x|) = sec x, assuming cos x > 0 on the interval.
考虑 dy/dx + (tan x)y = cos x。这里 P(x) = tan x，Q(x) = cos x。积分因子为 I(x) = e^(∫tan x dx) = e^(−ln|cos x|) = sec x，假设在区间上 cos x > 0。
Multiplying through, we obtain sec x·dy/dx + sec x·tan x·y = 1. The left side is d/dx(sec x · y) because the derivative of sec x is sec x·tan x. Then integrating gives sec x·y = x + C.
两边乘以 sec x，得到 sec x·dy/dx + sec x·tan x·y = 1。左边正是 d/dx(sec x·y)，因为 sec x 的导数为 sec x·tan x。积分后得到 sec x·y = x + C。
y = x·cos x + C·cos x
Letting C be an arbitrary constant, the solution family consists of curves that oscillate with the cosine envelope. If y(0) = 2, then 2 = 0 + C, so y = cos x(x + 2).
令 C 为任意常数，解族由随余弦包络振荡的曲线组成。若 y(0) = 2，则 2 = 0 + C，故 y = (x + 2)cos x。
8. Polynomial P(x): Linear Coefficient Case | 多项式系数：线性函数情形
Solve dy/dx + 2x·y = x. This appears in standard form with P(x) = 2x and Q(x) = x. The integrating factor is I(x) = e^(∫2x dx) = e^(x²).
求解 dy/dx + 2x·y = x。这是标准形式，其中 P(x) = 2x，Q(x) = x。积分因子为 I(x) = e^(∫2x dx) = e^(x²)。
Multiplying gives e^(x²)dy/dx + 2xe^(x²)y = xe^(x²). The left side is d/dx(e^(x²)y). Integrating: e^(x²)y = ∫xe^(x²)dx = ½ e^(x²) + C.
两边乘以 e^(x²) 得 e^(x²)dy/dx + 2xe^(x²)y = xe^(x²)。左边是 d/dx(e^(x²)y)。积分：e^(x²)y = ∫xe^(x²)dx = ½ e^(x²) + C。
Therefore the general solution is y = ½ + Ce^(−x²). The substitution u = x² simplifies the integral on the right; this technique frequently appears in IB when Q(x) contains related derivatives.
因此通解为 y = ½ + Ce^(−x²)。换元 u = x² 简化了右边的积分；这种技巧在IB中经常出现，尤其是当 Q(x) 包含相关导数因子时。
Notably, as x → ±∞, the exponential term decays to zero, and all solution curves approach the constant y = ½. This asymptotic behaviour is a useful check on the correctness of the solution.
值得注意的是，当 x → ±∞ 时，指数项衰减至零，所有解曲线趋近于常数 y = ½。这种渐近行为是检验解正确性的有效手段。
9. Solving for a Particular Solution Step by Step | 逐步求解特解
IB exam questions often provide an initial condition and require the full working. Suppose dy/dx + y/x = x, with y(1) = 0, for x > 0.
IB考试题通常给出初始条件并要求完整求解过程。假设 dy/dx + y/x = x，且 y(1) = 0，其中 x > 0。
Here P(x) = 1/x, so I(x) = e^(ln x) = x. Multiplying: x·dy/dx + y = x². The left side is d/dx(xy). Integrating: xy = ∫x²dx = x³/3 + C.
这里 P(x) = 1/x，故 I(x) = e^(ln x) = x。两边乘以 x：x·dy/dx + y = x²。左边是 d/dx(xy)。积分：xy = ∫x²dx = x³/3 + C。
Thus y = x²/3 + C/x. Applying y(1) = 0: 0 = 1/3 + C, so C = −1/3. The particular solution is y = (x² − 1)/(3x), or y = x/3 − 1/(3x).
因此 y = x²/3 + C/x。代入 y(1) = 0：0 = 1/3 + C，故 C = −1/3。特解为 y = (x² − 1)/(3x)，即 y = x/3 − 1/(3x)。
| Step | Action | Result |
| 1 | Write in standard form | dy/dx + (1/x)y = x |
| 2 | Find I(x) | I = x |
| 3 | Multiply through | d/dx(xy) = x² |
| 4 | Integrate | xy = x³/3 + C |
| 5 | Apply initial condition | C = −1/3 |
10. Common Pitfalls in IB Exams | IB考试中的常见陷阱
The most frequent error is omitting the constant of integration when computing the integrating factor. For example, writing e^(∫P dx) without considering the absolute value may be acceptable on a restricted domain, but skipping the integration of Q(x) after multiplying by I(x) invalidates the solution.
最常见的错误是计算积分因子时省略积分常数。例如，在限定定义域内写 e^(∫P dx) 而忽略绝对值或许可以接受，但乘以 I(x) 后遗漏对 Q(x) 的积分则会使解失效。
Another pitfall is confusing the roles of P(x) and Q(x). If the original equation is dy/dx = x − (2/x)y, it must be rewritten as dy/dx + (2/x)y = x. Failure to move all y-terms to the left leads to an incorrect P(x).
另一个陷阱是混淆 P(x) 和 Q(x) 的角色。如果原方程为 dy/dx = x − (2/x)y，必须改写为 dy/dx + (2/x)y = x。若未能将所有含 y 的项移到左侧，P(x) 就会出错。
Additionally, when the integrating factor simplifies (e.g., e^(ln x) = x), students sometimes forget that multiplication applies to Q(x) as well. Always multiply both sides, not just the left side.
此外，当积分因子化简（如 e^(ln x) = x）时，学生有时会忘记乘法同样作用于 Q(x)。务必两边同乘，而非仅乘左边。
11. Connection to Homogeneous Equations | 与齐次方程的联系
When Q(x) = 0, the differential equation reduces to dy/dx + P(x)y = 0. This is both separable and linear. The solution is y = Ce^(−∫P(x)dx), which is exactly the complementary function used in the general solution of the non-homogeneous equation.
当 Q(x) = 0 时，微分方程退化为 dy/dx + P(x)y = 0。该方程既可分离变量又属于线性方程。其解为 y = Ce^(−∫P(x)dx)，这正是非齐次方程通解中使用的余函数。
This relationship provides an alternative derivation. By setting y = C(x)e^(−∫P(x)dx), substituting into the non-homogeneous equation yields a differential equation for C'(x), which after integration gives the same result as the integrating factor method.
这种关系提供了一种替代推导。令 y = C(x)e^(−∫P(x)dx)，代入非齐次方程后得到关于 C'(x) 的微分方程，积分后得到与积分因子法完全相同的结果。
For IB students, understanding this connection strengthens conceptual mastery and offers a cross-check between two methods. However, the integrating factor method remains the standard expected in written examinations.
对于IB学生而言，理解这种联系能增强概念掌握，并提供两种方法之间的互验。不过，积分因子法仍是笔试中期望的标准方法。
12. Summary and Final Advice | 总结与最终建议
To solve a first-order linear differential equation with variable coefficients, always follow the same algorithm: write in standard form, compute the integrating factor, multiply through, recognise the exact derivative, integrate, and apply any initial condition.
求解变系数一阶线性微分方程，始终遵循同一算法：化为标准形式，计算积分因子，两边相乘，识别全导数，积分，并应用初始条件。
Practice with a variety of P(x) functions: constants, reciprocals, polynomials, and trigonometric functions. Each exposes a different integration skill, and IB exam questions often combine techniques such as substitution or partial fractions within the final integration step.
使用不同类型的 P(x) 进行练习：常数、倒数、多项式和三角函数。每一种都涉及不同的积分技巧，IB考试题常在最后一步积分中结合换元法或部分分式等技巧。
Finally, always check the domain and mention any restrictions when presenting your solution. A complete answer includes not just the algebraic formula but also a statement of where it is valid.
最后，始终检查定义域，并在呈现解时注明任何限制条件。一个完整的答案不仅包括代数公式，还应说明该解在何处有效。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply