PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 20 marks不可使用计算器 · 简答题 · 20 分
Section A · Short ResponseA 节 · 简答题
Show every step of separation, integration, or substitution. Always carry the constant of integration $C$ through to the final line and apply any initial condition explicitly. Write $\ln|y|$ when integrating $\tfrac{1}{y}$, then drop the absolute value only when you absorb the sign into $A = \pm e^{C}$. No calculator permitted.分离、积分、代换的每一步都要写出。积分常数 $C$ 要带到最后一行，再显式代入初值条件。对 $\tfrac{1}{y}$ 积分时写 $\ln|y|$，仅当把符号并入 $A = \pm e^{C}$ 后才可去掉绝对值。不可使用计算器。
Q1EASYPaper 1AE5.1 Verification of Solution[4 marks]
Verify that $y = e^{2x}$ is a solution of the differential equation $\dfrac{dy}{dx} = 2y$.验证 $y = e^{2x}$ 是微分方程 $\dfrac{dy}{dx} = 2y$ 的解。
(a)Compute $\dfrac{dy}{dx}$ from the proposed solution.由所给候选解计算 $\dfrac{dy}{dx}$。[1]
(b)Compute $2y$ from the proposed solution.由所给候选解计算 $2y$。[1]
(c)Conclude that $\dfrac{dy}{dx} = 2y$ holds for all $x \in \mathbb{R}$, and state explicitly that $y = e^{2x}$ is therefore a solution.说明 $\dfrac{dy}{dx} = 2y$ 对所有 $x \in \mathbb{R}$ 成立，并显式写出 $y = e^{2x}$ 因此为该方程的解。[2]
Solve the initial-value problem $\dfrac{dy}{dx} = xy$, $y(0) = 1$, giving $y$ explicitly as a function of $x$.求解初值问题 $\dfrac{dy}{dx} = xy$、$y(0) = 1$，将 $y$ 显式表为 $x$ 的函数。
(a)Separate the variables to obtain $\dfrac{dy}{y} = x \, dx$.分离变量得 $\dfrac{dy}{y} = x \, dx$。[1]
(b)Integrate both sides and write the implicit general solution, keeping the constant $C$.两边积分，写出含常数 $C$ 的隐式通解。[2]
(c)Solve for $y$ explicitly, then apply $y(0) = 1$ to determine the constant.显式解出 $y$，再代入 $y(0) = 1$ 求常数。[2]
Find the general solution of $\dfrac{dy}{dx} + 2y = 4$ using the integrating-factor method.用积分因子法求 $\dfrac{dy}{dx} + 2y = 4$ 的通解。
(a)Identify $P(x)$ and $Q(x)$ in the standard form $\dfrac{dy}{dx} + P(x) y = Q(x)$, then compute the integrating factor $\mu(x) = e^{\int P \, dx}$.在标准形式 $\dfrac{dy}{dx} + P(x) y = Q(x)$ 中辨识 $P(x)$ 与 $Q(x)$，并求积分因子 $\mu(x) = e^{\int P \, dx}$。[2]
(b)Multiply through by $\mu$, recognise the LHS as $\dfrac{d}{dx}\bigl(\mu y\bigr)$, and integrate both sides.两边乘以 $\mu$，将左侧识别为 $\dfrac{d}{dx}\bigl(\mu y\bigr)$，并对两边积分。[3]
(c)Solve for $y$ to write the general solution, leaving the arbitrary constant as $C$.解出 $y$，写出通解，常数记作 $C$。[1]
Q4HARDPaper 1AE5.2 Initial-Value Problem with $\cos x$[5 marks]
Solve the initial-value problem $\dfrac{dy}{dx} = y \cos x$, $y(0) = 2$, giving $y$ explicitly.求解初值问题 $\dfrac{dy}{dx} = y \cos x$、$y(0) = 2$，将 $y$ 显式给出。
(a)Separate variables and integrate both sides, writing $\ln|y| = \sin x + C$.分离变量并对两边积分，写出 $\ln|y| = \sin x + C$。[2]
(b)Exponentiate, absorbing the sign into $A = \pm e^{C}$, to obtain $y = A e^{\sin x}$.取指数，把符号并入 $A = \pm e^{C}$，得 $y = A e^{\sin x}$。[1]
(c)Apply $y(0) = 2$ to determine $A$ and state the particular solution.代入 $y(0) = 2$ 求 $A$，写出特解。[2]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分
Section B · Extended ResponseB 节 · 长答题
Model the situation as a first-order DE before any algebra. Name every constant (the rate $k$, the ambient temperature $T_{\text{env}}$, the initial value $T_{0}$) at the moment you introduce it. When the model is Newton's law of cooling, exponentiate cleanly: the form $T(t) = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-kt}$ should appear before you plug in any data.先把情境建为一阶 DE，再做代数。每个常数（速率 $k$、环境温度 $T_{\text{env}}$、初值 $T_{0}$）引入时立即命名。模型为牛顿冷却律时，先把解写成 $T(t) = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-kt}$ 的标准形式，再代入任何数据。
Q5HARDPaper 1BE5.6 Newton's Law of Cooling[11 marks]
A cup of coffee at temperature $T(t)$ degrees (with $t$ in minutes) is placed in a room at ambient temperature $T_{\text{env}} = 22$. The initial temperature is $T(0) = 95$, and after $5$ minutes the temperature has fallen to $T(5) = 70$. Newton's law of cooling models the situation.咖啡温度 $T(t)$ 度（$t$ 以分钟计）置于环境温度 $T_{\text{env}} = 22$ 的房间。初始温度 $T(0) = 95$，$5$ 分钟后温度降至 $T(5) = 70$。用牛顿冷却律建模。
(a)Write down the differential equation governing $T(t)$, with $k > 0$ a constant to be found.写出 $T(t)$ 所满足的微分方程，其中 $k > 0$ 为待求常数。[2]
(b)By substituting $u = T - 22$ (or otherwise), solve the DE to express $T(t)$ in terms of $k$ and $t$, using the initial condition $T(0) = 95$.用代换 $u = T - 22$（或其他方法）求解 DE，将 $T(t)$ 表为 $k$ 与 $t$ 的函数，并用初值 $T(0) = 95$。[4]
(c)Use $T(5) = 70$ to show that $k = \dfrac{1}{5} \ln\!\left(\dfrac{73}{48}\right)$ exactly, and give a decimal approximation correct to $3$ significant figures.用 $T(5) = 70$ 证明 $k = \dfrac{1}{5} \ln\!\left(\dfrac{73}{48}\right)$（精确值），并给出保留 $3$ 位有效数字的近似值。[3]
(d)Hence approximate $T(15)$ to the nearest degree.由此把 $T(15)$ 近似到最近整数度。[2]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 17 marks可使用计算器 · 混合题型 · 17 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. For Euler's method, tabulate columns $n,\, x_{n},\, y_{n},\, f(x_{n}, y_{n}),\, h \cdot f,\, y_{n+1}$. Every cell scores a method mark, so do not collapse the table to a single line. For slope-field questions, read off the algebraic form of $f(x, y)$ before sketching any solution curve.需要图形计算器（GDC）。欧拉法须列表，列项为 $n,\, x_{n},\, y_{n},\, f(x_{n}, y_{n}),\, h \cdot f,\, y_{n+1}$。每个单元格都有方法分，故不要把表压缩为一行。斜率场题目先读出 $f(x, y)$ 的代数形式，再画解曲线。
Use Euler's method with step size $h = 0.1$ on the initial-value problem $\dfrac{dy}{dx} = x + y$, $y(0) = 1$, to approximate $y(0.2)$.对初值问题 $\dfrac{dy}{dx} = x + y$、$y(0) = 1$，用步长 $h = 0.1$ 的欧拉法近似 $y(0.2)$。
(a)Set up the Euler step formula $y_{n+1} = y_{n} + h \cdot f(x_{n}, y_{n})$ for this problem, identifying $f(x, y)$.写出本题的欧拉步进公式 $y_{n+1} = y_{n} + h \cdot f(x_{n}, y_{n})$，并明确 $f(x, y)$。[1]
(b)Carry out step $1$: compute $f(x_{0}, y_{0})$ and $y_{1}$.完成第 $1$ 步：算 $f(x_{0}, y_{0})$ 和 $y_{1}$。[2]
(c)Carry out step $2$: compute $f(x_{1}, y_{1})$ and $y_{2}$. State your approximation to $y(0.2)$.完成第 $2$ 步：算 $f(x_{1}, y_{1})$ 和 $y_{2}$。给出 $y(0.2)$ 的近似值。[2]
(d)The exact solution is $y(x) = 2 e^{x} - x - 1$. Compute $y(0.2)$ to $4$ decimal places, and state whether Euler over- or under-estimates here.精确解为 $y(x) = 2 e^{x} - x - 1$。算出 $y(0.2)$ 保留 $4$ 位小数，并说明欧拉值此处偏高还是偏低。[2]
Q7HARDPaper 2E5.5 Slope Field Interpretation[10 marks]
The slope field below corresponds to a first-order differential equation of the form $\dfrac{dy}{dx} = g(x)$ for some function $g$ depending on $x$ only (slopes do not vary in the $y$-direction). At each grid point $(x, y)$ a short segment of slope $g(x)$ is drawn. Selected slopes are tabulated to the right of the grid.下方斜率场对应一阶微分方程 $\dfrac{dy}{dx} = g(x)$，函数 $g$ 仅与 $x$ 有关（斜率沿 $y$ 方向不变）。每个网格点 $(x, y)$ 画一段斜率为 $g(x)$ 的短线段。部分斜率值列在表中。
x: -2 -1 0 +1 +2
slope: +4 +1 0 +1 +4
(at every y on each column)
Observation: slope is even in x, zero on the y-axis,
and grows quadratically away from x = 0.x： -2 -1 0 +1 +2
斜率：+4 +1 0 +1 +4
（每列对所有 y 都相同）
观察：斜率关于 x 偶函数，y 轴上为 0，
离 x = 0 越远按平方增长。
(a)Identify the differential equation, i.e. give an explicit formula for $g(x)$, and justify your choice from the tabulated slopes.辨识该微分方程，即给出 $g(x)$ 的显式表达式，并用表中斜率说明理由。[3]
(b)Solve the DE by direct integration to obtain the general solution $y(x)$, leaving an arbitrary constant.直接积分求通解 $y(x)$，保留任意常数。[2]
(c)Find the particular solution passing through the point $(0, 1)$, and write it down explicitly.求通过点 $(0, 1)$ 的特解，并显式写出。[2]
(d)Describe the shape of the solution curve through $(0, 1)$ for $-2 \le x \le 2$ in one or two sentences, using the phrases "stationary point of inflection", "odd symmetry about $(0, 1)$", and "concavity changes sign at $x = 0$".用一两句话描述通过 $(0, 1)$ 的解曲线在 $-2 \le x \le 2$ 上的形状，使用"驻点拐点"、"关于 $(0, 1)$ 的奇对称"、"凹凸性在 $x = 0$ 处变号"三个短语。[2]
(e)For the same slope field, sketch (or describe in words) a second solution curve that passes through $(0, -1)$. State how it relates to the curve in (c).在同一斜率场上，画出（或用文字描述）通过 $(0, -1)$ 的另一条解曲线。说明它与 (c) 的曲线的关系。[1]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分
Paper 3 · HL Extended Problem第三卷 · HL 长题探究
A graphing calculator is required. Method marks dominate. For partial-fraction decompositions inside a separable DE, show both the algebraic identity (cleared denominators) and the resulting two-term integrand before integrating. Apply the initial condition only after the most general closed form has been written down.需要图形计算器（GDC）。方法分占主导。在可分离 DE 中遇到部分分式分解，需先写出代数恒等式（去分母后），再写出两项被积函数，最后积分。初值条件应在最一般闭式写出后再代入。
Q8HARDPaper 3E5.6 Logistic Growth via Partial Fractions[15 marks]
A wildlife reserve has a carrying capacity of $M = 1000$ deer. The deer population $P(t)$ (with $t$ in years) is modelled by the logistic equation $\dfrac{dP}{dt} = k P (M - P)$ for a constant $k > 0$. At $t = 0$ the population is $P(0) = 100$.某野生动物保护区的鹿群承载量 $M = 1000$ 头。鹿群规模 $P(t)$（$t$ 以年计）满足逻辑斯蒂方程 $\dfrac{dP}{dt} = k P (M - P)$，常数 $k > 0$。$t = 0$ 时 $P(0) = 100$。
(a)Show that $\dfrac{1}{P(M - P)} = \dfrac{1}{M}\!\left(\dfrac{1}{P} + \dfrac{1}{M - P}\right)$ by clearing denominators and matching coefficients (or by direct algebra).通过去分母后比较系数（或直接代数运算）证明 $\dfrac{1}{P(M - P)} = \dfrac{1}{M}\!\left(\dfrac{1}{P} + \dfrac{1}{M - P}\right)$。[3]
(b)Separate variables in the logistic DE and use part (a) to integrate both sides. Write the implicit general solution, keeping a single combined constant $C$.在逻辑斯蒂 DE 中分离变量，用 (a) 对两边积分。写出隐式通解，合并为单个常数 $C$。[4]
(c)Solve the implicit equation for $P$, then apply $P(0) = 100$ to express $P(t)$ explicitly in the standard logistic form $P(t) = \dfrac{M}{1 + B e^{-M k t}}$ for a constant $B$ that you determine.由隐式方程解出 $P$，再用 $P(0) = 100$ 把 $P(t)$ 显式写为标准逻辑斯蒂形式 $P(t) = \dfrac{M}{1 + B e^{-M k t}}$，并确定常数 $B$。[4]
(d)Observation: $P(5) = 250$. Use this datum together with your formula from (c) to find $k$ exactly (in closed form using a logarithm), and then to $3$ significant figures.观测：$P(5) = 250$。结合 (c) 的公式求 $k$ 的精确闭式（含一个对数），并给出保留 $3$ 位有效数字的近似值。[2]
(e)State, with brief reasoning, $\displaystyle\lim_{t \to \infty} P(t)$ and explain its biological meaning.写出 $\displaystyle\lim_{t \to \infty} P(t)$ 并简要说明理由及其生物学意义。[2]