Unit E5: Differential
Equations单元 E5：微分方程
A differential equation is an equation whose unknown is a function and which involves that function together with one or more of its derivatives. E5 covers the AHL-only toolkit for first-order DEs: verify a candidate solution, separate variables, multiply by an integrating factor, march numerically with Euler's method, read a slope field, and model the standard applications (growth, decay, Newton's law of cooling).微分方程是以函数为未知量、并把该函数与其若干阶导数联系起来的方程。E5 涵盖 AHL 专属的一阶微分方程工具集：验证候选解、分离变量、乘以积分因子、用欧拉法数值推进、读懂斜率场，以及标准建模应用（增长、衰减、牛顿冷却律）。
How to use this guide本指南使用说明
Differential equations are AHL-only territory. The IB Paper 3 exploration regularly opens with a DE that has to be classified (separable or linear), solved by hand for a particular initial condition, and then either compared against an Euler approximation or interpreted as a real-world model. Each section of E5 isolates one of those steps.微分方程是 AHL 专属内容。IB Paper 3 的探究题常以一个微分方程开篇：先分类（可分离或线性），用初始条件手解出特定解，再与欧拉近似比较或解读为实际模型。E5 各节分别聚焦其中一步。
Memorise two recipes. Separable: if $\tfrac{dy}{dx} = f(x) g(y)$, rearrange to $\tfrac{dy}{g(y)} = f(x) \, dx$ and integrate both sides. Linear: if $\tfrac{dy}{dx} + P(x) y = Q(x)$, the integrating factor is $\mu = e^{\int P \, dx}$ and $\tfrac{d}{dx}(\mu y) = \mu Q$. Do one worked problem of each end to end.
背两个套路。可分离：若 $\tfrac{dy}{dx} = f(x) g(y)$，整理为 $\tfrac{dy}{g(y)} = f(x) \, dx$ 两边积分。线性：若 $\tfrac{dy}{dx} + P(x) y = Q(x)$，积分因子 $\mu = e^{\int P \, dx}$ 且 $\tfrac{d}{dx}(\mu y) = \mu Q$。每类完整做一道例题。
Cleanly separate general solution (one arbitrary constant) from particular solution (constant fixed by an initial condition). Run Euler's method by hand for at least two steps and recognise that smaller step size $h$ improves accuracy but multiplies arithmetic. For applications, derive the cooling and logistic models from first principles instead of citing them.
把通解（含一个任意常数）与特解（由初始条件确定常数）严格区分。手算欧拉法至少两步，并理解步长 $h$ 越小精度越高、但运算量增加。应用题中由第一性原理推导冷却与逻辑斯蒂模型，不要直接背公式。
What is a Differential Equation何谓微分方程 HL AHL 5.17
- A differential equation (DE) is an equation involving an unknown function $y(x)$ together with one or more of its derivatives.
- The order of a DE is the order of the highest derivative that appears. First-order means only $\tfrac{dy}{dx}$ shows up.
- To solve a DE is to find an explicit formula $y = $ (expression in $x$), or an implicit relation $F(x, y) = 0$, that satisfies the equation.
- The general solution contains one arbitrary constant (for a first-order DE). A particular solution is the general solution with that constant fixed by an initial condition, such as $y(0) = 1$.
- To verify a candidate $y(x)$, substitute it into both sides of the DE and check that the two sides agree as functions of $x$.
- 微分方程（DE）是把未知函数 $y(x)$ 与其若干阶导数联系起来的方程。
- 阶数是方程中所出现的最高阶导数的阶数。一阶表示只出现 $\tfrac{dy}{dx}$。
- 解微分方程即求得显式 $y = $（$x$ 的表达式）或隐式关系 $F(x, y) = 0$ 满足该方程。
- 通解含一个任意常数（一阶 DE）。特解是用初始条件（如 $y(0) = 1$）把通解中常数确定后的解。
- 验证候选解 $y(x)$：代入方程两边，检查作为 $x$ 的函数是否相等。
Verify that $y = e^{x}$ is a solution of the differential equation $\tfrac{dy}{dx} = y$.验证 $y = e^{x}$ 是微分方程 $\tfrac{dy}{dx} = y$ 的解。
Differentiate the candidate. $\tfrac{dy}{dx} = \tfrac{d}{dx}(e^{x}) = e^{x}$.
对候选解求导。$\tfrac{dy}{dx} = \tfrac{d}{dx}(e^{x}) = e^{x}$。
Compare with the right-hand side. The DE says $\tfrac{dy}{dx} = y$. Substituting the candidate, $y = e^{x}$, so the right-hand side is $e^{x}$.
与右边比较。方程为 $\tfrac{dy}{dx} = y$。把候选解 $y = e^{x}$ 代入，右边为 $e^{x}$。
Conclusion. Left-hand side $e^{x}$ equals right-hand side $e^{x}$ for every $x$. Therefore $y = e^{x}$ satisfies the DE.
结论。左右两边对所有 $x$ 均为 $e^{x}$。故 $y = e^{x}$ 满足该方程。
Remark. $y = e^{x}$ is one particular solution. The general solution is $y = A e^{x}$ for any constant $A$, because differentiating $A e^{x}$ still gives $A e^{x}$. An initial condition such as $y(0) = 1$ would fix $A = 1$ and recover the candidate above.
注记。$y = e^{x}$ 只是一个特解。通解为 $y = A e^{x}$（$A$ 为任意常数），因为 $A e^{x}$ 的导数仍为 $A e^{x}$。初始条件 $y(0) = 1$ 可确定 $A = 1$，恰好得到上述候选解。
Going deeper: why first-order DEs need exactly one constant深入：一阶 DE 为何只含一个常数
Integrating a first-order DE involves exactly one antiderivative step, and every antiderivative carries an additive constant of integration. Higher-order DEs require more antiderivative steps and pick up more constants: a second-order DE has a general solution with two arbitrary constants, and so on. In E5 we work with first-order equations only, so one initial condition is always enough to pin down a unique particular solution.
解一阶微分方程恰好做一次反求导（积分），每次反求导都带出一个积分常数。更高阶 DE 需要更多次反求导，从而引入更多任意常数：二阶 DE 的通解含两个任意常数，依此类推。E5 只处理一阶方程，所以一个初始条件总足以唯一确定特解。
Separable Variables分离变量法 HL AHL 5.17
Three traps.
- If $g(y) = 0$ at some value of $y$, that constant function is also a solution. Check whether dividing by $g(y)$ lost a solution.
- $\int \tfrac{dy}{y} = \ln |y| + C$, not $\ln y + C$. The absolute value disappears only after exponentiating into $y = A e^{(\dots)}$ with $A$ a real constant that may be positive or negative.
- Constants combine: $\ln |y| = h(x) + C$ becomes $|y| = e^{C} e^{h(x)}$, then $y = A e^{h(x)}$ with $A = \pm e^{C} \neq 0$ (the case $A = 0$ comes from the lost constant solution $y = 0$).
三个陷阱。
- 若 $g(y) = 0$ 在某 $y$ 值成立，则该常值函数也是解。检查"除以 $g(y)$"是否丢解。
- $\int \tfrac{dy}{y} = \ln |y| + C$，不是 $\ln y + C$。绝对值在指数化为 $y = A e^{(\dots)}$（$A$ 可正可负）后才消失。
- 常数合并：$\ln |y| = h(x) + C$ 变为 $|y| = e^{C} e^{h(x)}$，再写作 $y = A e^{h(x)}$，其中 $A = \pm e^{C} \ne 0$（$A = 0$ 对应被丢掉的常值解 $y = 0$）。
Solve $\dfrac{dy}{dx} = x y$ subject to $y(0) = 1$.求解 $\dfrac{dy}{dx} = x y$，初始条件 $y(0) = 1$。
Identify the form. Here $f(x) = x$ and $g(y) = y$, so the equation is separable.
识别形式。$f(x) = x$，$g(y) = y$，方程可分离。
Separate.
分离变量。
$$ \frac{dy}{y} \;=\; x \, dx. $$Integrate both sides.
两边积分。
$$ \int \frac{dy}{y} \;=\; \int x \, dx \quad \Longrightarrow \quad \ln |y| \;=\; \frac{x^{2}}{2} + C. $$Solve for $y$. Exponentiate both sides:
解出 $y$。两边取指数：
$$ |y| \;=\; e^{C} e^{x^{2}/2} \quad \Longrightarrow \quad y \;=\; A \, e^{x^{2}/2}, \qquad A = \pm e^{C}. $$Apply the initial condition. $y(0) = 1$ gives $1 = A \cdot e^{0} = A$, so $A = 1$.
代入初始条件。$y(0) = 1$ 得 $1 = A \cdot e^{0} = A$，故 $A = 1$。
Particular solution.
特解。
$$ y \;=\; e^{x^{2}/2}. $$Check. $\tfrac{dy}{dx} = e^{x^{2}/2} \cdot x = x y$, and $y(0) = e^{0} = 1$. Both sides match the original DE.
检验。$\tfrac{dy}{dx} = e^{x^{2}/2} \cdot x = x y$，且 $y(0) = e^{0} = 1$。与原方程一致。
Integrating Factor for Linear First-Order Equations线性一阶方程的积分因子 HL AHL 5.17
The integrating factor. $$ \mu(x) \;=\; \exp\Bigl( \int P(x) \, dx \Bigr). $$ Multiplying the standard form by $\mu(x)$ collapses the left-hand side into a product-rule derivative: $$ \frac{d}{dx} \bigl( \mu(x) \, y \bigr) \;=\; \mu(x) \, Q(x). $$ Integrate both sides with respect to $x$ and divide by $\mu(x)$ to recover $y$.
Three-line workflow. Identify $P, Q$. Compute $\mu = e^{\int P \, dx}$ (no $+ C$ needed inside the exponent, since any constant just rescales $\mu$). Write $\mu y = \int \mu Q \, dx + C$ and solve for $y$.
积分因子。 $$ \mu(x) \;=\; \exp\Bigl( \int P(x) \, dx \Bigr). $$ 标准形式两边乘 $\mu(x)$ 后，左边化为乘积法则的导数： $$ \frac{d}{dx} \bigl( \mu(x) \, y \bigr) \;=\; \mu(x) \, Q(x). $$ 对 $x$ 积分两边、再除以 $\mu(x)$ 即得 $y$。
三行流程。识别 $P, Q$；算 $\mu = e^{\int P \, dx}$（指数内不必加 $+ C$，常数只缩放 $\mu$）；写 $\mu y = \int \mu Q \, dx + C$ 并解出 $y$。
Solve $\dfrac{dy}{dx} + 2 x y = 4 x$.求解 $\dfrac{dy}{dx} + 2 x y = 4 x$。
Identify $P$ and $Q$. $P(x) = 2 x$ and $Q(x) = 4 x$.
识别 $P$ 与 $Q$。$P(x) = 2 x$，$Q(x) = 4 x$。
Compute the integrating factor.
算积分因子。
$$ \mu(x) \;=\; \exp\Bigl( \int 2 x \, dx \Bigr) \;=\; e^{x^{2}}. $$Multiply through and recognise the product-rule derivative.
两边乘 $\mu$，识别乘积法则。
$$ e^{x^{2}} \frac{dy}{dx} + 2 x \, e^{x^{2}} y \;=\; 4 x \, e^{x^{2}} \quad \Longleftrightarrow \quad \frac{d}{dx}\bigl( e^{x^{2}} y \bigr) \;=\; 4 x \, e^{x^{2}}. $$Integrate both sides. Substitute $u = x^{2}$, so $du = 2 x \, dx$ and $4 x \, dx = 2 \, du$:
两边积分。取 $u = x^{2}$，则 $du = 2 x \, dx$，$4 x \, dx = 2 \, du$：
$$ \int 4 x \, e^{x^{2}} \, dx \;=\; \int 2 \, e^{u} \, du \;=\; 2 \, e^{u} + C \;=\; 2 \, e^{x^{2}} + C. $$Therefore
故
$$ e^{x^{2}} y \;=\; 2 \, e^{x^{2}} + C. $$Solve for $y$. Divide by $e^{x^{2}}$:
解出 $y$。除以 $e^{x^{2}}$：
$$ y \;=\; 2 + C \, e^{-x^{2}}. $$Check. $\tfrac{dy}{dx} = -2 C x \, e^{-x^{2}}$, and $2 x y = 4 x + 2 C x \, e^{-x^{2}}$. Adding gives $\tfrac{dy}{dx} + 2 x y = 4 x$, which is the original equation.
检验。$\tfrac{dy}{dx} = -2 C x \, e^{-x^{2}}$，$2 x y = 4 x + 2 C x \, e^{-x^{2}}$。相加得 $\tfrac{dy}{dx} + 2 x y = 4 x$，与原方程一致。
Going deeper: why the integrating factor works深入：积分因子为何奏效
We want a function $\mu(x)$ such that $\mu y' + \mu P y$ is the derivative of some product. The product rule gives $\tfrac{d}{dx}(\mu y) = \mu' y + \mu y'$. Matching coefficients on $y$ requires $\mu' = \mu P$, which is itself a separable DE for $\mu$. Solving $\tfrac{d \mu}{\mu} = P \, dx$ gives $\ln |\mu| = \int P \, dx$, hence $\mu = e^{\int P \, dx}$ (the absolute value and overall sign are absorbed because any non-zero scalar multiple of $\mu$ still works). By construction, multiplying the standard form by this $\mu$ makes the left-hand side an exact derivative.
我们希望找到 $\mu(x)$ 使 $\mu y' + \mu P y$ 能写成某乘积的导数。由乘积法则 $\tfrac{d}{dx}(\mu y) = \mu' y + \mu y'$，要让 $y$ 的系数对得上需 $\mu' = \mu P$，这本身是一个关于 $\mu$ 的可分离方程。解 $\tfrac{d \mu}{\mu} = P \, dx$ 得 $\ln |\mu| = \int P \, dx$，故 $\mu = e^{\int P \, dx}$（绝对值与正负号可吸收，因为 $\mu$ 的任一非零数乘仍可作积分因子）。如此构造下，标准形式乘 $\mu$ 后左边恰为某函数的导数。
Euler's Method (Numerical)欧拉法（数值方法） HL AHL 5.17
Error behaviour. Halving $h$ roughly halves the total error (Euler is first-order accurate). The trade-off is twice the arithmetic. For exam problems with two or three steps, hand computation is expected; show every $f(x_{n}, y_{n})$ value.
误差行为。$h$ 减半，总误差大致减半（欧拉法为一阶精度），但运算量翻倍。考题通常只要走两三步，要求手算；每个 $f(x_{n}, y_{n})$ 必须写出。
Use Euler's method with step size $h = 0.1$ to approximate $y(0.2)$ for the initial-value problem $\dfrac{dy}{dx} = x + y$, $y(0) = 1$.用步长 $h = 0.1$ 的欧拉法近似初值问题 $\dfrac{dy}{dx} = x + y$、$y(0) = 1$ 的 $y(0.2)$。
Setup. $f(x, y) = x + y$. Start at $(x_{0}, y_{0}) = (0, 1)$.
设置。$f(x, y) = x + y$。起点 $(x_{0}, y_{0}) = (0, 1)$。
Step 1: advance to $x_{1} = 0.1$.
第 1 步：推进到 $x_{1} = 0.1$。
$$ f(x_{0}, y_{0}) \;=\; 0 + 1 \;=\; 1, \qquad y_{1} \;=\; y_{0} + h \cdot f(x_{0}, y_{0}) \;=\; 1 + 0.1 \cdot 1 \;=\; 1.1. $$Step 2: advance to $x_{2} = 0.2$.
第 2 步：推进到 $x_{2} = 0.2$。
$$ f(x_{1}, y_{1}) \;=\; 0.1 + 1.1 \;=\; 1.2, \qquad y_{2} \;=\; y_{1} + h \cdot f(x_{1}, y_{1}) \;=\; 1.1 + 0.1 \cdot 1.2 \;=\; 1.22. $$Approximation. $y(0.2) \approx 1.22$.
近似值。$y(0.2) \approx 1.22$。
Remark. The exact solution to this initial-value problem is $y(x) = 2 e^{x} - x - 1$, giving $y(0.2) = 2 e^{0.2} - 1.2 \approx 1.2428$. The Euler estimate $1.22$ is low by about $0.023$. Halving $h$ to $0.05$ (four steps instead of two) would roughly halve that error.
注记。本题的精确解为 $y(x) = 2 e^{x} - x - 1$，故 $y(0.2) = 2 e^{0.2} - 1.2 \approx 1.2428$。欧拉估计 $1.22$ 偏小约 $0.023$。$h$ 减半到 $0.05$（改为四步）可大致把误差减半。
Slope (Direction) Fields斜率场（方向场） HL AHL 5.17
Reading a slope field.
- Horizontal segments ($f = 0$) mark equilibrium or turning behaviour. Constant solutions $y = c$ live where $f(x, c) = 0$ for all $x$.
- Steep upward segments ($f$ large positive) indicate fast growth; steep downward ($f$ large negative) indicate fast decay.
- If $f(x, y)$ depends only on $y$ (the DE is autonomous), all segments at the same $y$-level have the same slope: the field is invariant under horizontal shifts.
读图要点。
- 水平线段（$f = 0$）标记平衡或极值。常值解 $y = c$ 出现在 $f(x, c) = 0$ 对所有 $x$ 成立处。
- 陡峭向上线段（$f$ 大正）表示快速增长；陡峭向下（$f$ 大负）表示快速衰减。
- 若 $f(x, y)$ 只依赖 $y$（自治方程），同一 $y$ 高度上所有线段斜率相同：斜率场在水平方向平移不变。
For the DE $\dfrac{dy}{dx} = y - x$, compute the slope at the grid points $(0, 0)$, $(0, 1)$, $(1, 0)$, $(1, 1)$, $(2, 1)$, and describe how a solution curve passing through $(0, 1)$ behaves locally.对方程 $\dfrac{dy}{dx} = y - x$，在格点 $(0, 0)$、$(0, 1)$、$(1, 0)$、$(1, 1)$、$(2, 1)$ 求斜率，并描述过 $(0, 1)$ 的解曲线在局部的行为。
Tabulate slopes. Slope at $(x, y)$ equals $y - x$.
列表算斜率。$(x, y)$ 处的斜率为 $y - x$。
| $(x, y)$ | $(0, 0)$ | $(0, 1)$ | $(1, 0)$ | $(1, 1)$ | $(2, 1)$ |
|---|---|---|---|---|---|
| Slope $f(x, y)$ | $0$ | $1$ | $-1$ | $0$ | $-1$ |
Equilibrium line. Slope is zero whenever $y = x$. The line $y = x$ is a locus of horizontal segments. (It is not itself a solution: if $y = x$, then $\tfrac{dy}{dx} = 1$, not $0$. The line is the boundary between regions of increasing and decreasing slope.)
零斜率线。$y = x$ 时斜率为 $0$。直线 $y = x$ 是水平线段的轨迹。（它本身不是解：$y = x$ 给 $\tfrac{dy}{dx} = 1 \ne 0$。该直线只是增减分界。）
Behaviour through $(0, 1)$. At $(0, 1)$ the slope is $+1$, so the solution starts heading up-right at $45^{\circ}$. Just past $x = 1$ the curve approaches the line $y = x$ from above; once it dips below $y = x$, the slope turns negative and the curve begins to fall.
过 $(0, 1)$ 的局部行为。$(0, 1)$ 处斜率 $+1$，解曲线以 $45^{\circ}$ 向右上方出发。$x$ 越过 $1$ 后，曲线从上方接近直线 $y = x$；一旦跌落至 $y = x$ 之下，斜率变负，曲线开始下降。
Going deeper: matching a solution to a slope field深入：将解与斜率场匹配
A common exam task shows three or four candidate slope-field diagrams and asks which one corresponds to a given DE. The drill is: pick two or three test points where the DE is easy to evaluate (often along the axes), compute the slope, and check which diagram has matching segments. Look first at where the slope is zero (horizontal segments): that is the fastest discriminator between candidate fields.
常见题型给出三四张候选斜率场图，问哪一张对应给定方程。做法：选两三个易代入的测试点（通常在坐标轴上）计算斜率，再核对哪张图的线段方向相符。最先看的应是零斜率位置（水平线段），它是最快区分候选图的特征。
Applications: Growth, Decay, Cooling应用：增长、衰减、冷却 HL AHL 5.17
- Exponential growth or decay. $\tfrac{dy}{dt} = k y$ with $k$ constant. Separable. Solution $y(t) = y_{0} e^{k t}$. Positive $k$ gives growth; negative $k$ gives decay.
- Newton's law of cooling. $\tfrac{dT}{dt} = -k (T - T_{\text{env}})$ with $k > 0$ and $T_{\text{env}}$ the ambient temperature. Separable after the substitution $u = T - T_{\text{env}}$. Solution $T(t) = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-k t}$.
- Logistic growth. $\tfrac{dP}{dt} = k P (M - P)$ with $M > 0$ the carrying capacity. Separable; the $P$-side needs partial fractions. Solution $P(t) = \dfrac{M}{1 + B e^{-k M t}}$ for some constant $B$ fixed by the initial condition.
- 指数增长或衰减。$\tfrac{dy}{dt} = k y$（$k$ 为常数）。可分离。解 $y(t) = y_{0} e^{k t}$。$k > 0$ 增长，$k < 0$ 衰减。
- 牛顿冷却律。$\tfrac{dT}{dt} = -k (T - T_{\text{env}})$（$k > 0$，$T_{\text{env}}$ 为环境温度）。换元 $u = T - T_{\text{env}}$ 后可分离。解 $T(t) = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-k t}$。
- 逻辑斯蒂增长。$\tfrac{dP}{dt} = k P (M - P)$（$M > 0$ 为环境容量）。可分离；$P$ 一侧需部分分式。解 $P(t) = \dfrac{M}{1 + B e^{-k M t}}$，$B$ 由初始条件确定。
A cup of coffee at $T(0) = 95 \, ^{\circ}\text{C}$ sits in a room at constant temperature $T_{\text{env}} = 22 \, ^{\circ}\text{C}$. It cools according to Newton's law $\tfrac{dT}{dt} = -k (T - T_{\text{env}})$. Find an explicit formula for $T(t)$ in terms of $k$ and $t$.一杯咖啡 $T(0) = 95 \, ^{\circ}\text{C}$ 放在恒温 $T_{\text{env}} = 22 \, ^{\circ}\text{C}$ 的房间中，按牛顿冷却律 $\tfrac{dT}{dt} = -k (T - T_{\text{env}})$ 散热。求 $T(t)$ 关于 $k$ 与 $t$ 的显式表达。
Substitute $u = T - 22$. Then $\tfrac{du}{dt} = \tfrac{dT}{dt}$, and the DE becomes
换元 $u = T - 22$。则 $\tfrac{du}{dt} = \tfrac{dT}{dt}$，方程化为
$$ \frac{du}{dt} \;=\; -k \, u. $$Solve the reduced DE. Separate: $\tfrac{du}{u} = -k \, dt$. Integrate: $\ln |u| = -k t + C_{1}$, so $u = A e^{-k t}$ with $A$ a real constant.
解化简后的方程。分离：$\tfrac{du}{u} = -k \, dt$。积分：$\ln |u| = -k t + C_{1}$，故 $u = A e^{-k t}$，$A$ 为实常数。
Translate back. $T = u + 22 = 22 + A e^{-k t}$.
换回原变量。$T = u + 22 = 22 + A e^{-k t}$。
Apply the initial condition. $T(0) = 95$ gives $95 = 22 + A$, so $A = 73$.
代入初始条件。$T(0) = 95$ 得 $95 = 22 + A$，故 $A = 73$。
Final formula.
最终公式。
$$ T(t) \;=\; 22 + 73 \, e^{-k t}. $$Sanity checks. At $t = 0$, $T(0) = 22 + 73 = 95$. As $t \to \infty$, $T(t) \to 22$, the ambient temperature. Both consistent with the physical setup.
合理性检查。$t = 0$ 时 $T(0) = 22 + 73 = 95$。$t \to \infty$ 时 $T(t) \to 22$，即环境温度。物理意义一致。
Going deeper: deriving the logistic solution深入：推导逻辑斯蒂方程的解
Starting from $\tfrac{dP}{dt} = k P (M - P)$, separate: $$ \frac{dP}{P (M - P)} \;=\; k \, dt. $$ Decompose by partial fractions: $\tfrac{1}{P (M - P)} = \tfrac{1}{M} \bigl( \tfrac{1}{P} + \tfrac{1}{M - P} \bigr)$. Integrate both sides: $$ \frac{1}{M} \bigl( \ln |P| - \ln |M - P| \bigr) \;=\; k t + C_{1} \quad \Longrightarrow \quad \ln \Bigl| \frac{P}{M - P} \Bigr| \;=\; k M t + C_{2}. $$ Exponentiate: $\tfrac{P}{M - P} = D e^{k M t}$. Solve for $P$: $$ P \;=\; (M - P) D e^{k M t} \quad \Longrightarrow \quad P \bigl( 1 + D e^{k M t} \bigr) \;=\; M D e^{k M t} \quad \Longrightarrow \quad P \;=\; \frac{M}{1 + B e^{-k M t}}, $$ where $B = 1 / D$. As $t \to \infty$, $P \to M$, recovering the carrying-capacity behaviour.
由 $\tfrac{dP}{dt} = k P (M - P)$ 出发，分离： $$ \frac{dP}{P (M - P)} \;=\; k \, dt. $$ 部分分式：$\tfrac{1}{P (M - P)} = \tfrac{1}{M} \bigl( \tfrac{1}{P} + \tfrac{1}{M - P} \bigr)$。两边积分： $$ \frac{1}{M} \bigl( \ln |P| - \ln |M - P| \bigr) \;=\; k t + C_{1} \quad \Longrightarrow \quad \ln \Bigl| \frac{P}{M - P} \Bigr| \;=\; k M t + C_{2}. $$ 取指数：$\tfrac{P}{M - P} = D e^{k M t}$。解出 $P$： $$ P \;=\; (M - P) D e^{k M t} \quad \Longrightarrow \quad P \bigl( 1 + D e^{k M t} \bigr) \;=\; M D e^{k M t} \quad \Longrightarrow \quad P \;=\; \frac{M}{1 + B e^{-k M t}}, $$ 其中 $B = 1 / D$。$t \to \infty$ 时 $P \to M$，复现环境容量行为。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Test for separability first. If $\tfrac{dy}{dx}$ factors as $f(x) g(y)$, separation is the cleanest route. Worth checking even when the DE is also linear, because separation usually beats integrating factor for marks per minute.
- 先测可分离性。若 $\tfrac{dy}{dx}$ 能分解为 $f(x) g(y)$，分离法最干净。即便方程也是线性的，仍值得先试分离，因为通常分离法单位时间得分更高。
- If not separable, try the linear form. Reshape into $\tfrac{dy}{dx} + P(x) y = Q(x)$ and read off $P, Q$. The integrating factor $\mu = e^{\int P \, dx}$ then unlocks the equation.
- 若不可分离，尝试线性形式。整理为 $\tfrac{dy}{dx} + P(x) y = Q(x)$ 后读出 $P, Q$。积分因子 $\mu = e^{\int P \, dx}$ 即可破题。
- Write $+ C$ explicitly. Omitting the constant is a guaranteed mark loss. Show it appearing on the first integration step and then solve for it after the initial condition.
- $+ C$ 必须显式写出。漏写常数必扣分。在第一步积分时就写出，再由初始条件解出。
- Combine constants by renaming. $e^{C_{1}}$ becomes a new constant $A$; do not carry chains of $C_{1}, C_{2}, C_{3}$ through the algebra. One symbol per final constant is cleaner.
- 合并常数时重命名。$e^{C_{1}}$ 重命名为新常数 $A$；不要把 $C_{1}, C_{2}, C_{3}$ 链条带到最后。一个最终常数对应一个符号更清晰。
- For separable DEs, $\int \tfrac{dy}{y}$ is $\ln |y|$. Drop the absolute value only after exponentiating into $y = A e^{(\dots)}$ where $A$ ranges over all real numbers.
- 分离法中 $\int \tfrac{dy}{y}$ 写作 $\ln |y|$。取指数化为 $y = A e^{(\dots)}$（$A$ 可正可负可零）后才能去掉绝对值。
- For Euler, tabulate. Columns $n, x_{n}, y_{n}, f(x_{n}, y_{n}), h \cdot f, y_{n + 1}$. Every cell scores. Pure numerical answers lose method marks.
- 欧拉法务必列表。列 $n, x_{n}, y_{n}, f(x_{n}, y_{n}), h \cdot f, y_{n + 1}$。每格都计分；只写数字会丢方法分。
- Compare Euler to the exact solution when possible. Showing that $y_{\text{Euler}}$ undershoots or overshoots, and explaining the link to the concavity of the exact curve, earns commentary marks.
- 有可能时把欧拉值与精确解比较。说明 $y_{\text{Euler}}$ 偏低或偏高，并用精确解的凹凸性解释，可拿评注分。
- For slope-field questions, start at the zeros. Where $f(x, y) = 0$ marks horizontal segments, which is the fastest way to identify or rule out a candidate diagram.
- 斜率场题先看零点。$f(x, y) = 0$ 处是水平线段，是辨识或排除候选图最快的特征。
Flashcards闪卡
Unit E5 Practice Quiz单元 E5 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- HL State the order of a DE and identify whether a first-order DE is separable, linear, or neither说出 DE 的阶数，并判断一阶 DE 是可分离、线性还是都不是
- HL Verify a candidate function satisfies a given DE by substitution通过代入验证候选函数是否满足给定 DE
- HL Distinguish general solution (with arbitrary constant) from particular solution (initial condition applied)区分通解（含任意常数）与特解（已用初始条件）
- HL Solve a separable first-order DE end to end, including the initial-condition substitution完整求解可分离一阶 DE，包括代入初始条件
- HL Recognise that $\int \tfrac{dy}{y} = \ln |y| + C$ and convert into $y = A e^{(\dots)}$ form correctly用 $\int \tfrac{dy}{y} = \ln |y| + C$，并正确化为 $y = A e^{(\dots)}$ 形式
- HL Reshape a first-order DE into the linear standard form $y' + P(x) y = Q(x)$把一阶 DE 整理为线性标准形式 $y' + P(x) y = Q(x)$
- HL Compute the integrating factor $\mu = e^{\int P \, dx}$ and use it to solve a linear DE算积分因子 $\mu = e^{\int P \, dx}$，用以解线性 DE
- HL Apply Euler's method by hand for two or three steps, showing every $f(x_{n}, y_{n})$手算欧拉法两三步，每个 $f(x_{n}, y_{n})$ 必须写出
- HL Explain how step size $h$ affects accuracy and total error for Euler's method解释步长 $h$ 如何影响欧拉法的精度与总误差
- HL Read a slope field, identify horizontal segments, and sketch a solution curve through a given point读懂斜率场、识别水平线段、过给定点画出解曲线
- HL Model exponential growth or decay and solve the resulting DE for a given initial value建模指数增长或衰减，并对给定初值解出 DE
- HL Apply Newton's law of cooling, using the substitution $u = T - T_{\text{env}}$ to separate variables应用牛顿冷却律，用换元 $u = T - T_{\text{env}}$ 分离变量
- HL Recognise the logistic equation $\tfrac{dP}{dt} = k P (M - P)$ and solve it using partial fractions识别逻辑斯蒂方程 $\tfrac{dP}{dt} = k P (M - P)$，并用部分分式求解
IB Paper-Style PracticeIB 试卷风格练习
E5 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_E5_*.html with the bilingual built-in pattern.
E5 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_E5_*.html，采用双语内嵌格式。