Sections 1 to 7: classifying ODEs, separable equations, first-order linear equations and integrating factors, IVPs, direction fields and equilibria, existence-uniqueness, and first-order modelsCALC IV第 1 至 7 节：常微分方程的分类、可分离方程、一阶线性方程与积分因子、初值问题、方向场与平衡点、存在唯一性定理、一阶模型CALC IV
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PART I · CORE TECHNIQUESComputational fluency · 28 marks计算能力 · 28 分
Classifying and Solving First-Order ODEs一阶常微分方程的分类与求解
Show all working. Put linear equations in standard form $y' + P(x)y = Q(x)$ before computing the integrating factor. State the method (separation, integrating factor) at the start of each solution.写出完整解题过程。在计算积分因子之前，先将线性方程化为标准形式 $y' + P(x)y = Q(x)$。在每题解答开始处注明所用方法（分离变量法、积分因子法）。
Q1MEDIUMCOREclassifying ODEs: order, linearity, and normal form常微分方程的分类：阶数、线性性与标准形式[6 marks]
For each equation, state (i) the order and (ii) whether it is linear or nonlinear. If it is linear, write it in standard form $y' + P(x)y = Q(x)$; if it is nonlinear, identify the term that breaks linearity.对于每个方程，说明 (i) 阶数，以及 (ii) 是否为线性方程。若为线性方程，将其化为标准形式 $y' + P(x)y = Q(x)$；若为非线性方程，指出破坏线性性的项。
(a) $x\,y' + 2y = \sin x$ [2]
(b) $y'' - 3y' + 2y = e^{x}$ [2]
(c) $\dfrac{dy}{dx} = y^{2} - x$ [2]
Q2MEDIUMCOREseparable equations with initial value problems带初值条件的可分离变量方程[8 marks]
Solve each initial value problem by separating variables. Express $y$ explicitly as a function of $x$ and verify your solution satisfies both the ODE and the initial condition.用分离变量法求解各初值问题。将 $y$ 显式表示为 $x$ 的函数，并验证解满足常微分方程和初始条件。
Q3HARDCOREfirst-order linear equations via the integrating factor用积分因子法求解一阶线性方程[8 marks]
Solve each equation using the integrating-factor method. Put the equation in standard form first, compute $\mu = e^{\int P(x)\,dx}$, and write the general solution.用积分因子法求解各方程。先将方程化为标准形式，计算 $\mu = e^{\int P(x)\,dx}$，再写出通解。
(a) $y' - \dfrac{2}{x}\,y = x^{2},\quad x > 0$ [4]
(b) $y' + y\tan x = \sec x,\quad -\dfrac{\pi}{2} < x < \dfrac{\pi}{2},\quad y(0) = 0$ [4]
Q4MEDIUMCOREdirection fields, equilibria, and qualitative behaviour方向场、平衡点与定性行为[6 marks]
(a)Find all equilibrium (constant) solutions.求所有平衡（常数）解。[2]
(b)Determine the stability of each equilibrium. Justify your answer by evaluating $\dfrac{\partial f}{\partial y}$ at each equilibrium, where $f(y) = y(2-y)$.判断每个平衡点的稳定性。通过在各平衡点处计算 $\dfrac{\partial f}{\partial y}$（其中 $f(y) = y(2-y)$）来论证你的答案。[2]
(c)Without solving the ODE, describe the long-run behaviour of solutions starting at (i) $y(0) = 1$ and (ii) $y(0) = 3$.不求解常微分方程，描述初始条件为 (i) $y(0) = 1$ 和 (ii) $y(0) = 3$ 的解的长期行为。[2]
PART II · DEFINITIONS AND PROOFRigorous arguments · 26 marks严密论证 · 26 分
Deriving the Integrating Factor and the Existence-Uniqueness Theorem推导积分因子与存在唯一性定理
These items are graded on the logic of the argument, not just the final answer. In proofs, state each hypothesis before invoking the conclusion. In the existence-uniqueness questions, check the Picard hypotheses explicitly.这些题目按论证逻辑评分，不仅看最终答案。在证明中，先陈述每个假设再得出结论。在存在唯一性问题中，明确验证皮卡定理的各个假设条件。
Q5HARDPROOFderiving the integrating-factor formula推导积分因子公式[8 marks]
Consider the first-order linear equation $y' + P(x)\,y = Q(x)$, where $P$ and $Q$ are continuous on an open interval $I$.考虑一阶线性方程 $y' + P(x)\,y = Q(x)$，其中 $P$ 和 $Q$ 在开区间 $I$ 上连续。
(a)We seek a function $\mu(x) > 0$ such that multiplying both sides by $\mu$ converts the left side into an exact derivative $\bigl(\mu\,y\bigr)'$. By expanding $\bigl(\mu\,y\bigr)'$ and matching terms, show that $\mu$ must satisfy the separable equation $\mu' = P(x)\,\mu$, and hence that $\mu(x) = e^{\int P(x)\,dx}$ works.我们寻找函数 $\mu(x) > 0$，使得方程两边乘以 $\mu$ 后，左边变为恰当导数 $\bigl(\mu\,y\bigr)'$。通过展开 $\bigl(\mu\,y\bigr)'$ 并比较各项，证明 $\mu$ 必须满足可分离方程 $\mu' = P(x)\,\mu$，从而 $\mu(x) = e^{\int P(x)\,dx}$ 满足条件。[4]
(b)Using the integrating factor found in (a), integrate both sides of $\bigl(\mu\,y\bigr)' = \mu\,Q$ from $x_{0}$ to $x$ (with initial condition $y(x_{0}) = y_{0}$) to derive the explicit solution formula
$$ y(x) = \frac{1}{\mu(x)}\left(y_{0}\,\mu(x_{0}) + \int_{x_{0}}^{x}\mu(t)\,Q(t)\,dt\right). $$
State why this solution is unique on $I$.利用 (a) 中求得的积分因子，将 $\bigl(\mu\,y\bigr)' = \mu\,Q$ 两边从 $x_{0}$ 积分到 $x$（初始条件 $y(x_{0}) = y_{0}$），推导出显式解公式
$$ y(x) = \frac{1}{\mu(x)}\left(y_{0}\,\mu(x_{0}) + \int_{x_{0}}^{x}\mu(t)\,Q(t)\,dt\right). $$
说明此解在 $I$ 上唯一的原因。[4]
Q6HARDPROOFPicard existence-uniqueness theorem: hypotheses and failures皮卡存在唯一性定理：假设条件与失效情形[10 marks]
The Picard theorem states: if $f(x,y)$ and $\partial f/\partial y$ are both continuous in a rectangle containing $(x_{0},y_{0})$, then the IVP $y' = f(x,y)$, $y(x_{0}) = y_{0}$ has a unique solution on some interval around $x_{0}$.皮卡定理：若 $f(x,y)$ 和 $\partial f/\partial y$ 在包含 $(x_{0},y_{0})$ 的矩形区域内均连续，则初值问题 $y' = f(x,y)$，$y(x_{0}) = y_{0}$ 在 $x_{0}$ 某邻域内有唯一解。
(a)For the IVP $y' = \sqrt[3]{y}$, $y(0) = 0$: verify that $y \equiv 0$ is one solution. Find a second solution of the form $y = \bigl(\tfrac{2}{3}x\bigr)^{3/2}$ (for $x \ge 0$) by substituting into the ODE. Which Picard hypothesis fails, and why?对于初值问题 $y' = \sqrt[3]{y}$，$y(0) = 0$：验证 $y \equiv 0$ 是一个解。通过代入常微分方程，验证 $y = \bigl(\tfrac{2}{3}x\bigr)^{3/2}$（$x \ge 0$）是第二个解。皮卡定理的哪个假设失效，为什么？[3]
(b)Solve the IVP $y' = y^{2}$, $y(0) = 1$ by separating variables. State the maximal interval of existence. Explain why the solution does not exist for all $x > 0$, even though $f = y^{2}$ and $\partial f/\partial y = 2y$ are smooth everywhere.用分离变量法求解初值问题 $y' = y^{2}$，$y(0) = 1$。给出解的最大存在区间。解释为何即使 $f = y^{2}$ 和 $\partial f/\partial y = 2y$ 在处处光滑，解仍不能对所有 $x > 0$ 存在。[3]
(c)Verify that $f(x,y) = \sqrt{|y|}$ fails the Picard uniqueness hypothesis at $(0,0)$, and exhibit two distinct solutions to $y' = \sqrt{|y|}$, $y(0) = 0$: the trivial solution $y \equiv 0$ and a family of solutions of the form
$$ y(x) = \begin{cases} 0, & x \le a, \\ \tfrac{1}{4}(x-a)^{2}, & x > a, \end{cases} $$
for any $a \ge 0$. Confirm that each member of this family satisfies the IVP.验证 $f(x,y) = \sqrt{|y|}$ 在 $(0,0)$ 处不满足皮卡唯一性假设，并给出初值问题 $y' = \sqrt{|y|}$，$y(0) = 0$ 的两个不同解：零解 $y \equiv 0$ 以及形如
$$ y(x) = \begin{cases} 0, & x \le a, \\ \tfrac{1}{4}(x-a)^{2}, & x > a, \end{cases} $$
的一族解（$a \ge 0$ 任意）。验证该族中每个成员均满足初值问题。[4]
Q7HARDPROOFIVP with definite-integral solution and interval of validity含定积分解的初值问题及其有效区间[8 marks]
(a)Show that the integrating factor is $\mu(x) = e^{x^{2}}$. Multiply through by $\mu$ and integrate both sides from $0$ to $x$ to express the solution as
$$ y(x) = e^{-x^{2}}\!\left(1 + \int_{0}^{x}\cos t\,dt\right). $$
Verify this satisfies $y(0) = 1$.证明积分因子为 $\mu(x) = e^{x^{2}}$。两边乘以 $\mu$ 后从 $0$ 到 $x$ 积分，将解表示为
$$ y(x) = e^{-x^{2}}\!\left(1 + \int_{0}^{x}\cos t\,dt\right). $$
验证此解满足 $y(0) = 1$。[4]
(b)State, with justification, the maximal interval of existence of this solution. Then simplify the integral to obtain the closed form $y(x) = e^{-x^{2}}(1 + \sin x)$, and verify it satisfies the original ODE by differentiating.给出并论证此解的最大存在区间。然后化简积分，得到闭合形式 $y(x) = e^{-x^{2}}(1 + \sin x)$，并通过求导验证其满足原常微分方程。[4]
PART III · APPLICATIONS AND SYNTHESISExtended problems · 28 marks综合应用题 · 28 分
First-Order Models and Synthesis一阶模型与综合应用
Set up the governing ODE from the physical description before solving. Carry exact values through intermediate steps. Each model question expects you to state the ODE, solve it with the given initial condition, and answer the specific physical question asked.在求解前，先根据物理描述建立方程。在中间步骤中保持精确值。每道建模题要求写出常微分方程，用给定初始条件求解，并回答具体的物理问题。
Q8HARDAPPLIEDNewton's law of cooling: setup, IVP, and prediction牛顿冷却定律：建模、初值问题与预测[10 marks]
Newton's law of cooling states that the rate of change of temperature $T(t)$ of an object is proportional to the difference between $T$ and the ambient temperature $T_{\text{env}}$:
$$ \frac{dT}{dt} = -k\,(T - T_{\text{env}}), \quad k > 0. $$
A cup of tea is made at $95\,^{\circ}\text{C}$ and left in a room at $20\,^{\circ}\text{C}$. After $10$ minutes the tea has cooled to $65\,^{\circ}\text{C}$.牛顿冷却定律：物体温度 $T(t)$ 的变化率与 $T$ 和环境温度 $T_{\text{env}}$ 之差成正比：
$$ \frac{dT}{dt} = -k\,(T - T_{\text{env}}), \quad k > 0. $$
一杯茶在 $95\,^{\circ}\text{C}$ 时沏好，置于 $20\,^{\circ}\text{C}$ 的房间中。$10$ 分钟后，茶冷却至 $65\,^{\circ}\text{C}$。
(a)Let $u(t) = T(t) - 20$. Show that $u$ satisfies $u' = -ku$ and hence that $T(t) = 20 + 75\,e^{-kt}$. State the initial condition used.令 $u(t) = T(t) - 20$。证明 $u$ 满足 $u' = -ku$，从而 $T(t) = 20 + 75\,e^{-kt}$。写出所用的初始条件。[3]
(b)Use the condition $T(10) = 65$ to find $k$ exactly. Leave your answer in logarithmic form, and give a decimal approximation to three significant figures.利用条件 $T(10) = 65$ 精确求 $k$。答案保留对数形式，并给出三位有效数字的近似值。[4]
(c)Find the time $t^{*}$ at which the tea reaches $40\,^{\circ}\text{C}$. Express your answer exactly and as a decimal to the nearest minute.求茶降温至 $40\,^{\circ}\text{C}$ 的时刻 $t^{*}$。给出精确值，并精确到最近的整分钟的近似值。[3]
Q9HARDAPPLIEDmixing tank model: linear ODE and steady-state behaviour混合槽模型：线性常微分方程与稳态行为[8 marks]
A tank initially contains $200\,\text{L}$ of pure water. Brine containing $3\,\text{g/L}$ of salt flows in at $4\,\text{L/min}$, and the well-mixed solution flows out at $4\,\text{L/min}$. Let $A(t)$ be the amount of salt (in grams) at time $t$ (in minutes).一个槽最初装有 $200\,\text{L}$ 纯水。浓度为 $3\,\text{g/L}$ 的盐水以 $4\,\text{L/min}$ 的流量流入，充分混合后的溶液以 $4\,\text{L/min}$ 的流量流出。设 $A(t)$ 为 $t$ 时刻（分钟）槽中盐的质量（克）。
(a)Write down the rate-in minus rate-out equation for $A'(t)$ and put it in the standard form $A' + P(t)\,A = Q(t)$. State the initial condition.写出 $A'(t)$ 的流入速率减去流出速率的方程，并化为标准形式 $A' + P(t)\,A = Q(t)$。写出初始条件。[3]
(b)Solve the IVP using the integrating factor and show that $A(t) = 600\left(1 - e^{-t/50}\right)$.用积分因子法求解初值问题，证明 $A(t) = 600\left(1 - e^{-t/50}\right)$。[3]
(c)State the steady-state (equilibrium) amount of salt and explain briefly why it equals what it does from a physical perspective. Find the time at which the salt concentration in the tank first reaches $2\,\text{g/L}$.给出盐的稳态（平衡）质量，并从物理角度简要解释其值。求槽中盐浓度首次达到 $2\,\text{g/L}$ 的时刻。[2]
Q10HARDAPPLIEDseparable IVP: explicit solution and interval of validity可分离变量初值问题：显式解与有效区间[10 marks]
(a)Separate variables and integrate both sides. Show that the implicit solution satisfies $y - \dfrac{y^{3}}{3} = \dfrac{x^{2}}{2}$, and identify any values of $y$ that were excluded when separating.分离变量并两边积分。证明隐式解满足 $y - \dfrac{y^{3}}{3} = \dfrac{x^{2}}{2}$，并指出分离变量时排除的 $y$ 值。[4]
(b)The Picard theorem guarantees a unique local solution. Explain why the explicit function $y(x)$ solving this IVP cannot be extended to all of $\mathbb{R}$: find the value $x_{1} > 0$ at which the solution must blow up or cease to be defined, and state the maximal interval of existence.皮卡定理保证存在唯一局部解。解释为何求解此初值问题的显式函数 $y(x)$ 无法延拓到整个 $\mathbb{R}$：求解爆破或无定义的 $x_{1} > 0$，并给出最大存在区间。[3]
(c)The right-hand side $f(x,y) = x/(1-y^{2})$ is undefined on the lines $y = \pm 1$. Use this to explain geometrically (in terms of the direction field) why the solution curve starting at $(0,0)$ must terminate before reaching $y = 1$.右端函数 $f(x,y) = x/(1-y^{2})$ 在直线 $y = \pm 1$ 上无定义。由此从几何角度（利用方向场）解释，为何从 $(0,0)$ 出发的解曲线必须在到达 $y = 1$ 之前终止。[3]