IB Math AA HL · Unit E5: Differential Equations · Solutions

Syllabus AHL 5.17 (E5.1 – E5.6)考纲 AHL 5.17（E5.1 – E5.6）AA HL

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1AE5.1 Verification of Solution[4 marks]

Verify that $y = e^{2x}$ is a solution of $\dfrac{dy}{dx} = 2y$.验证 $y = e^{2x}$ 是 $\dfrac{dy}{dx} = 2y$ 的解。

Conclusion:结论： $\dfrac{dy}{dx} = 2 e^{2x} = 2y$ for all $x \in \mathbb{R}$. Hence $y = e^{2x}$ is a solution. $\checkmark$$\dfrac{dy}{dx} = 2 e^{2x} = 2y$ 对所有 $x \in \mathbb{R}$ 成立。故 $y = e^{2x}$ 为解。$\checkmark$

(a) Differentiate the candidate A1

$\dfrac{dy}{dx} = \dfrac{d}{dx}\bigl(e^{2x}\bigr) = 2 e^{2x}$ by the chain rule.

(b) Compute the RHS A1

$2y = 2 \cdot e^{2x} = 2 e^{2x}$.

(c) Match LHS to RHS M1·A1

LHS $= 2 e^{2x} = $ RHS for every $x \in \mathbb{R}$. The equality holds identically (not just at one point), so $y = e^{2x}$ satisfies the differential equation. Therefore $y = e^{2x}$ is a solution.

Verify $\ne$ derive. Verification questions hand you the candidate and ask only that you check it. Do not "solve" the DE from scratch — that wastes time and risks a different constant. The two-line ritual is: differentiate the candidate to get the LHS, plug the candidate into the RHS, and check identical for all $x$. The "for all $x$" phrase carries the final A1; without it you have only shown equality at unspecified points.

(a) 对候选解求导 A1

由链式法则 $\dfrac{dy}{dx} = \dfrac{d}{dx}\bigl(e^{2x}\bigr) = 2 e^{2x}$。

(b) 计算右侧 A1

$2y = 2 \cdot e^{2x} = 2 e^{2x}$。

(c) 比较两侧 M1·A1

左 $= 2 e^{2x} =$ 右，对每个 $x \in \mathbb{R}$ 成立。等式恒成立（非个别点），故 $y = e^{2x}$ 满足该微分方程，是其解。

验证 $\ne$ 推导。验证题已给出候选解，只需检验。不要重新"解"DE——浪费时间且常数可能不同。两行套路：对候选求导得左，候选代入右，确认对每个 $x$ 恒等。"对所有 $x$"这一表述决定最后那个 A1；不写就只是某些点等于，并未证明是解。

Q2MEDIUMPaper 1AE5.2 Separable + Initial Condition[5 marks]

Solve $\dfrac{dy}{dx} = xy$, $y(0) = 1$ explicitly.显式求解 $\dfrac{dy}{dx} = xy$、$y(0) = 1$。

Answer:答案： $y = e^{x^{2}/2}$

(a) Separate A1

Divide by $y$ (valid since $y(0) = 1 \ne 0$, and the solution will remain non-zero by uniqueness): $\dfrac{dy}{y} = x \, dx$.

(b) Integrate both sides M1·A1

$\displaystyle \int \frac{dy}{y} = \int x \, dx \;\Longrightarrow\; \ln|y| = \tfrac{x^{2}}{2} + C$.

(c) Solve explicitly, apply IC M1·A1

Exponentiate: $|y| = e^{C} \, e^{x^{2}/2}$, so $y = A \, e^{x^{2}/2}$ with $A = \pm e^{C} \in \mathbb{R} \setminus \{0\}$. Apply $y(0) = 1$: $1 = A \cdot e^{0} = A$, so $A = 1$. Hence $y = e^{x^{2}/2}$.

The constant $A = \pm e^{C}$ absorbs the sign. Step (b) keeps $\ln|y|$ — the absolute value is mandatory. At step (c) the standard move is to introduce $A := \pm e^{C}$ as a single non-zero real constant, which collapses two sign cases into one symbol. Forgetting the absolute value at (b) is harmless if you do introduce $A$ at (c), but losing both costs an A1. After applying the IC, $A$ is fixed, and the absolute value disappears for free.

(a) 分离变量 A1

两边除以 $y$（合法：$y(0) = 1 \ne 0$，由唯一性解恒不为零）：$\dfrac{dy}{y} = x \, dx$。

(b) 两边积分 M1·A1

$\displaystyle \int \frac{dy}{y} = \int x \, dx \;\Longrightarrow\; \ln|y| = \tfrac{x^{2}}{2} + C$。

(c) 显式解出并代初值 M1·A1

取指数：$|y| = e^{C} \, e^{x^{2}/2}$，故 $y = A \, e^{x^{2}/2}$（$A = \pm e^{C} \in \mathbb{R} \setminus \{0\}$）。代 $y(0) = 1$：$1 = A \cdot e^{0} = A$，得 $A = 1$。故 $y = e^{x^{2}/2}$。

$A = \pm e^{C}$ 把符号吸收。(b) 步保留 $\ln|y|$——绝对值必写。(c) 步标准操作是引入 $A := \pm e^{C}$ 作为单一非零实常数，两种符号情形合为一个记号。若 (b) 忘写绝对值但 (c) 引入了 $A$，无碍；二者皆漏丢 A1。代入初值后 $A$ 定，绝对值自然消失。

Q3MEDIUMPaper 1AE5.3 Integrating Factor (Linear)[6 marks]

General solution of $\dfrac{dy}{dx} + 2y = 4$ by integrating factor.用积分因子法求 $\dfrac{dy}{dx} + 2y = 4$ 的通解。

Answer:答案： $y = 2 + C e^{-2x}$

(a) Identify $P,Q$ and form $\mu$ M1·A1

Standard form: $P(x) = 2$, $Q(x) = 4$. Integrating factor: $$ \mu(x) \;=\; e^{\int P \, dx} \;=\; e^{\int 2 \, dx} \;=\; e^{2x}. $$ (We absorb the constant of integration into $\mu$ since any non-zero scalar multiple of $\mu$ works.)

(b) Multiply, recognise LHS, integrate M1·A1·A1

Multiply through: $$ e^{2x}\,\dfrac{dy}{dx} + 2 e^{2x}\, y \;=\; 4 e^{2x}. $$ The LHS is $\dfrac{d}{dx}\bigl(e^{2x}\, y\bigr)$ by the product rule (this is the whole point of $\mu$): $$ \dfrac{d}{dx}\bigl(e^{2x}\, y\bigr) \;=\; 4 e^{2x}. $$ Integrate both sides w.r.t. $x$: $$ e^{2x}\, y \;=\; \int 4 e^{2x} \, dx \;=\; 2 e^{2x} + C. $$

(c) Solve for $y$ A1

Divide by $e^{2x}$: $y = 2 + C e^{-2x}$.

The integrating factor turns the DE into a product-rule derivative. The whole reason for $\mu = e^{\int P \, dx}$ is that it forces $\mu' = P \mu$, which makes $(\mu y)' = \mu y' + \mu' y = \mu(y' + P y)$ — the LHS of the original DE times $\mu$. So Step (b) is mechanical: every linear first-order DE, once multiplied by $\mu$, becomes $(\mu y)' = \mu Q$, then a single integration finishes it. Common slip: forgetting to write the $+ C$ at the integration, which costs an A1 because the answer is no longer the general solution. Second slip: dropping $\mu$ on the RHS during the multiplication step — re-write the entire equation explicitly.

(a) 辨识 $P,Q$，求 $\mu$ M1·A1

标准形式：$P(x) = 2$、$Q(x) = 4$。积分因子： $$ \mu(x) \;=\; e^{\int P \, dx} \;=\; e^{\int 2 \, dx} \;=\; e^{2x}. $$ （积分常数并入 $\mu$，因为 $\mu$ 的任何非零标量倍都可用。）

(b) 两边乘 $\mu$、识别左侧、积分 M1·A1·A1

两边乘 $\mu$： $$ e^{2x}\,\dfrac{dy}{dx} + 2 e^{2x}\, y \;=\; 4 e^{2x}. $$ 由乘积法则左侧即 $\dfrac{d}{dx}\bigl(e^{2x}\, y\bigr)$（此为 $\mu$ 的核心目的）： $$ \dfrac{d}{dx}\bigl(e^{2x}\, y\bigr) \;=\; 4 e^{2x}. $$ 对 $x$ 积分两边： $$ e^{2x}\, y \;=\; \int 4 e^{2x} \, dx \;=\; 2 e^{2x} + C. $$

(c) 解出 $y$ A1

除以 $e^{2x}$：$y = 2 + C e^{-2x}$。

积分因子把 DE 变成乘积法则的导数。$\mu = e^{\int P \, dx}$ 的设计目的就是使 $\mu' = P \mu$，从而 $(\mu y)' = \mu y' + \mu' y = \mu(y' + P y)$——即原 DE 左侧乘以 $\mu$。因此 (b) 步机械：任何一阶线性 DE 乘 $\mu$ 后变为 $(\mu y)' = \mu Q$，一次积分搞定。常见失误一：积分时忘加 $+ C$——答案不再是通解，丢一个 A1。失误二：乘 $\mu$ 时漏掉右侧乘——把整方程显式写出来即可避免。

Q4HARDPaper 1AE5.2 IVP with $\cos x$[5 marks]

Solve $\dfrac{dy}{dx} = y \cos x$, $y(0) = 2$ explicitly.显式求解 $\dfrac{dy}{dx} = y \cos x$、$y(0) = 2$。

Answer:答案： $y = 2 e^{\sin x}$

(a) Separate & integrate M1·A1

$\dfrac{dy}{y} = \cos x \, dx$, so $\ln|y| = \sin x + C$.

(b) Exponentiate A1

$|y| = e^{\sin x + C} = e^{C} \, e^{\sin x}$, hence $y = A \, e^{\sin x}$ with $A = \pm e^{C}$.

(c) Apply $y(0) = 2$ M1·A1

$2 = A \cdot e^{\sin 0} = A \cdot e^{0} = A$. So $A = 2$ and $y = 2 e^{\sin x}$.

Sanity check: oscillation, not blow-up. Since $\sin x$ is bounded in $[-1, 1]$, the solution $y = 2 e^{\sin x}$ oscillates between $2 e^{-1} \approx 0.736$ and $2 e^{1} \approx 5.44$. This is the qualitative signature of a separable DE whose RHS coefficient $\cos x$ averages to zero over a period — no net growth or decay, just oscillation. Contrast with Q2's $y = e^{x^{2}/2}$, where the coefficient $x$ has unbounded primitive, so $y$ blows up. Reading the asymptotic behaviour from the integrand of $P(x)\,dx$ is a quick exam check before you commit your answer.

(a) 分离 & 积分 M1·A1

$\dfrac{dy}{y} = \cos x \, dx$，故 $\ln|y| = \sin x + C$。

(b) 取指数 A1

$|y| = e^{\sin x + C} = e^{C} \, e^{\sin x}$，故 $y = A \, e^{\sin x}$（$A = \pm e^{C}$）。

(c) 代入 $y(0) = 2$ M1·A1

$2 = A \cdot e^{\sin 0} = A$。故 $A = 2$，$y = 2 e^{\sin x}$。

合理性核查：振荡而非爆炸。因 $\sin x \in [-1, 1]$ 有界，解 $y = 2 e^{\sin x}$ 在 $2 e^{-1} \approx 0.736$ 与 $2 e^{1} \approx 5.44$ 之间振荡。这是可分离 DE 右侧系数 $\cos x$ 一个周期内均值为零的典型表现——无净增长或衰减，纯振荡。对比 Q2 中 $y = e^{x^{2}/2}$：系数 $x$ 的原函数无界，故 $y$ 爆炸。考前由 $P(x)\,dx$ 的被积函数读出渐进行为，是落笔前的快速核查。

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BE5.6 Newton's Law of Cooling[11 marks]

Coffee $T(t)$, ambient $22$, $T(0)=95$, $T(5)=70$. Build & solve, find $k$, predict $T(15)$.咖啡 $T(t)$，环境 $22$，$T(0)=95$、$T(5)=70$。建模并求解，求 $k$，预测 $T(15)$。

Answers:答案： (a) $\dfrac{dT}{dt} = -k(T - 22)$ · (b) $T = 22 + 73 e^{-kt}$ · (c) $k = \tfrac{1}{5}\ln(73/48) \approx 0.0838$ · (d) $T(15) \approx 43^{\circ}$

(a) Set up the DE M1·A1

Newton's law of cooling: the rate of cooling is proportional to the temperature gap. With $k > 0$ and ambient $T_{\text{env}} = 22$: $$ \dfrac{dT}{dt} \;=\; -k\,(T - 22). $$ The minus sign forces cooling (since $T > 22$ initially, $\frac{dT}{dt} < 0$, as required).

(b) Solve in terms of $k,t$ M1·A1·A1·A1

Substitute $u = T - 22$, so $\frac{du}{dt} = \frac{dT}{dt}$. The DE becomes $\frac{du}{dt} = -k u$, a pure exponential-decay equation. Separating: $$ \frac{du}{u} \;=\; -k \, dt \;\Longrightarrow\; \ln|u| \;=\; -k t + C \;\Longrightarrow\; u \;=\; A e^{-kt}. $$ Translate back: $T - 22 = A e^{-kt}$, i.e. $T(t) = 22 + A e^{-kt}$. Apply $T(0) = 95$: $95 = 22 + A \cdot 1$, so $A = 73$. Therefore $$ T(t) \;=\; 22 + 73 \, e^{-kt}. $$

(c) Find $k$ from $T(5) = 70$ M1·A1·A1

$70 = 22 + 73 \, e^{-5k} \;\Longrightarrow\; 73 \, e^{-5k} = 48 \;\Longrightarrow\; e^{-5k} = \dfrac{48}{73} \;\Longrightarrow\; -5k = \ln\!\left(\dfrac{48}{73}\right)$. Hence $$ k \;=\; -\tfrac{1}{5}\ln\!\left(\tfrac{48}{73}\right) \;=\; \tfrac{1}{5}\ln\!\left(\tfrac{73}{48}\right). $$ Numerically $k \approx \tfrac{1}{5}(0.4187) \approx 0.0838$ ($3$ sf).

(d) Predict $T(15)$ M1·A1

$T(15) = 22 + 73 \, e^{-15 k}$. Note $e^{-15k} = \bigl(e^{-5k}\bigr)^{3} = \bigl(48/73\bigr)^{3}$. Compute $(48/73)^{3} \approx (0.6575)^{3} \approx 0.2842$. So $T(15) \approx 22 + 73 \cdot 0.2842 \approx 22 + 20.7 \approx 42.7$, rounded to nearest degree: $T(15) \approx 43^{\circ}$.

Substitute first, then exponentiate. The cleanest cooling-law solution uses $u = T - T_{\text{env}}$ to reduce the inhomogeneous DE $T' = -k(T - T_{\text{env}})$ to the homogeneous $u' = -ku$ — and then the answer $T = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-kt}$ falls out. Memorise that template line: once you spot the words "Newton's law of cooling", write it down with the data substituted (ambient, $T_{0}$), and only then start solving for $k$. A bonus exam trick: $e^{-15k} = (e^{-5k})^{3}$ keeps the arithmetic exact through to the final substitution, avoiding compounding rounding error. Three-significant-figure answers in IB-style cooling problems must come from no fewer than $4$-sf intermediate values.

(a) 建立 DE M1·A1

牛顿冷却律：冷却速率正比于温差。$k > 0$、环境 $T_{\text{env}} = 22$： $$ \dfrac{dT}{dt} \;=\; -k\,(T - 22). $$ 负号确保冷却（初始 $T > 22$ 时 $\frac{dT}{dt} < 0$，符合实际）。

(b) 用 $k, t$ 解出 M1·A1·A1·A1

令 $u = T - 22$，则 $\frac{du}{dt} = \frac{dT}{dt}$，DE 化为 $\frac{du}{dt} = -k u$，纯指数衰减。分离： $$ \frac{du}{u} \;=\; -k \, dt \;\Longrightarrow\; \ln|u| \;=\; -k t + C \;\Longrightarrow\; u \;=\; A e^{-kt}. $$ 还原：$T - 22 = A e^{-kt}$，即 $T(t) = 22 + A e^{-kt}$。代 $T(0) = 95$：$95 = 22 + A$，故 $A = 73$。因此 $$ T(t) \;=\; 22 + 73 \, e^{-kt}. $$

(c) 由 $T(5) = 70$ 求 $k$ M1·A1·A1

$70 = 22 + 73 \, e^{-5k} \;\Longrightarrow\; 73 \, e^{-5k} = 48 \;\Longrightarrow\; e^{-5k} = \dfrac{48}{73} \;\Longrightarrow\; -5k = \ln\!\left(\dfrac{48}{73}\right)$。故 $$ k \;=\; -\tfrac{1}{5}\ln\!\left(\tfrac{48}{73}\right) \;=\; \tfrac{1}{5}\ln\!\left(\tfrac{73}{48}\right). $$ 数值 $k \approx \tfrac{1}{5}(0.4187) \approx 0.0838$（3 位有效数字）。

(d) 预测 $T(15)$ M1·A1

$T(15) = 22 + 73 \, e^{-15 k}$。$e^{-15k} = \bigl(e^{-5k}\bigr)^{3} = \bigl(48/73\bigr)^{3} \approx 0.2842$。故 $T(15) \approx 22 + 73 \cdot 0.2842 \approx 42.7$，取整 $T(15) \approx 43^{\circ}$。

先代换、再取指数。冷却律最干净的解法用 $u = T - T_{\text{env}}$ 把非齐次 $T' = -k(T - T_{\text{env}})$ 化为齐次 $u' = -ku$——答案 $T = T_{\text{env}} + (T_{0} - T_{\text{env}}) e^{-kt}$ 自然落出。背好这一模板：看到"牛顿冷却律"立刻把环境与 $T_{0}$ 代入写出，再解 $k$。考场小技巧：$e^{-15k} = (e^{-5k})^{3}$ 保留精确分数到最后代入，避免误差累积。IB 风格冷却题的 3 位有效数字答案需要中间值至少 4 位有效数字。

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2E5.4 Euler's Method (Two Steps)[7 marks]

Euler with $h = 0.1$ on $y' = x + y$, $y(0) = 1$. Approximate $y(0.2)$; compare to exact $y = 2 e^{x} - x - 1$.$y' = x + y$、$y(0) = 1$，步长 $h = 0.1$ 用欧拉法近似 $y(0.2)$；与精确解 $y = 2 e^{x} - x - 1$ 比较。

Answers:答案： (b) $y_{1} = 1.1$ · (c) $y(0.2) \approx y_{2} = 1.22$ · (d) exact $1.2428$, Euler under-estimates精确 $1.2428$，欧拉偏低

(a) Euler step A1

$f(x, y) = x + y$. Step formula: $y_{n+1} = y_{n} + 0.1\,(x_{n} + y_{n})$.

(b) Step 1 ($x_{0}=0, y_{0}=1$) M1·A1

$f(0, 1) = 0 + 1 = 1$. $y_{1} = 1 + 0.1 \cdot 1 = 1.1$. New point $(x_{1}, y_{1}) = (0.1, 1.1)$.

(c) Step 2 ($x_{1}=0.1, y_{1}=1.1$) M1·A1

$f(0.1, 1.1) = 0.1 + 1.1 = 1.2$. $y_{2} = 1.1 + 0.1 \cdot 1.2 = 1.1 + 0.12 = 1.22$. Approximation: $y(0.2) \approx 1.22$.

(d) Compare to exact M1·A1

Exact: $y(0.2) = 2 e^{0.2} - 0.2 - 1 = 2(1.22140\ldots) - 1.2 = 2.44281 - 1.2 = 1.24281$, i.e. $y(0.2) = 1.2428$ ($4$ dp). Euler gives $1.22 < 1.24$: Euler under-estimates by about $0.023$.

Euler's sign of error tracks concavity. Euler's tangent-line step approximates the curve by its tangent at the current point. If the true solution is concave up ($y'' > 0$), the tangent lies below the curve, so Euler under-estimates. Here $y' = x + y$, so $y'' = 1 + y' = 1 + x + y > 0$ for $x, y \ge 0$ — strictly concave up, so under-estimation is forced, matching the numbers. Conversely, on a concave-down arc Euler would over-estimate. This is one of the highest-scoring single sentences in a Paper 2 Euler question: "$y'' > 0$, hence Euler under-estimates" earns the R1 reasoning mark cleanly.

(a) 欧拉步 A1

$f(x, y) = x + y$。步进公式：$y_{n+1} = y_{n} + 0.1\,(x_{n} + y_{n})$。

(b) 第 1 步（$x_{0}=0, y_{0}=1$）M1·A1

$f(0, 1) = 1$。$y_{1} = 1 + 0.1 \cdot 1 = 1.1$。新点 $(0.1, 1.1)$。

(c) 第 2 步（$x_{1}=0.1, y_{1}=1.1$）M1·A1

$f(0.1, 1.1) = 1.2$。$y_{2} = 1.1 + 0.1 \cdot 1.2 = 1.22$。近似：$y(0.2) \approx 1.22$。

(d) 与精确比较 M1·A1

精确：$y(0.2) = 2 e^{0.2} - 0.2 - 1 = 2.44281 - 1.2 = 1.24281$，即 $y(0.2) = 1.2428$（4 位小数）。欧拉值 $1.22 < 1.24$：欧拉偏低，误差约 $0.023$。

欧拉误差符号由凹凸决定。欧拉切线步用当前点切线近似曲线。若真解上凹（$y'' > 0$），切线在曲线下方，故欧拉偏低。本题 $y' = x + y$，$y'' = 1 + y' = 1 + x + y > 0$（$x, y \ge 0$），严格上凹，必偏低，与数值吻合。反之下凹弧上欧拉偏高。Paper 2 欧拉题最高效的一句：写"$y'' > 0$，故欧拉偏低"即拿下 R1 推理分。

Q7HARDPaper 2E5.5 Slope Field Interpretation[10 marks]

Slope field $y' = g(x)$, slopes $\{+4, +1, 0, +1, +4\}$ at $x = -2, -1, 0, 1, 2$. Identify $g$; solve; particular through $(0, 1)$; describe curve; sketch curve through $(0, -1)$.斜率场 $y' = g(x)$，$x = -2, -1, 0, 1, 2$ 处斜率 $\{+4, +1, 0, +1, +4\}$。辨识 $g$、求解、过 $(0,1)$ 特解、描述形状、画过 $(0,-1)$ 的曲线。

Answers:答案： (a) $g(x) = x^{2}$ · (b) $y = \tfrac{x^{3}}{3} + C$ · (c) $y = \tfrac{x^{3}}{3} + 1$ · (e) vertical shift竖直平移

(a) Identify $g(x)$ M1·A1·A1

Slopes are $4, 1, 0, 1, 4$ at $x = -2, -1, 0, 1, 2$ respectively. These match $x^{2}$ exactly: $(-2)^{2} = 4$, $(-1)^{2} = 1$, $0^{2} = 0$, $1^{2} = 1$, $2^{2} = 4$. The slopes are even in $x$ (mirrored across $x = 0$) and quadratic in magnitude. Hence $g(x) = x^{2}$ and the DE is $\dfrac{dy}{dx} = x^{2}$.

(b) Integrate directly M1·A1

$\displaystyle y = \int x^{2} \, dx = \dfrac{x^{3}}{3} + C$, $C \in \mathbb{R}$.

(c) Particular solution through $(0, 1)$ M1·A1

$1 = \dfrac{0^{3}}{3} + C = C$. So $C = 1$ and $y = \dfrac{x^{3}}{3} + 1$.

(d) Shape on $[-2, 2]$ A1·A1

$y = \tfrac{x^{3}}{3} + 1$ has $y' = x^{2} \ge 0$ with $y'(0) = 0$, so the curve has a horizontal tangent at $(0, 1)$: a stationary point of inflection (neither local max nor min, because $y'$ does not change sign — it touches $0$ from above). $y'' = 2x$ changes sign at $x = 0$: concave down on $x < 0$, concave up on $x > 0$. The function $y - 1 = x^{3}/3$ is odd in $x$, so the graph is point-symmetric about $(0, 1)$: rotating the graph $180°$ around the centre $(0, 1)$ maps it to itself.

(e) Second curve through $(0, -1)$ A1

Set $-1 = \tfrac{0^{3}}{3} + C$, so $C = -1$ and the new curve is $y = \tfrac{x^{3}}{3} - 1$. Since the slope field depends only on $x$ (not $y$), every solution curve is a vertical translate of every other: $y_{1}(x) - y_{2}(x) = $ constant. The new curve is the curve from (c) shifted down by $2$.

$y' = g(x)$ slope fields are "all-vertical-translates". When the RHS depends only on $x$, the solutions form a one-parameter family $\{y = G(x) + C : C \in \mathbb{R}\}$ where $G' = g$. Visually, every grid column has identical slope marks (exactly the tabulated property here) and every solution curve is a vertical translate of every other. As soon as the RHS depends on $y$, this breaks: e.g. for $y' = y$ the solutions $y = A e^{x}$ are not vertical translates (different growth rates). When reading a slope field, the first test is: are columns identical? If yes, integrate directly; if no, you need separation, partial fractions, or another method. Part (d) tests vocabulary as much as calculus — "stationary inflection" (not "minimum"), "odd symmetry about the IC point" (not "even"), and "concavity changes sign at $x = 0$" (not "concave up everywhere"). Each phrase corresponds to a separate diagnostic of $y'$ and $y''$ and earns its own A1; the structure of $y'$ alone (a perfect square $x^{2}$) signals the inflection.

(a) 辨识 $g(x)$ M1·A1·A1

$x = -2, -1, 0, 1, 2$ 处斜率分别为 $4, 1, 0, 1, 4$，恰为 $x^{2}$ 的值：$(-2)^{2} = 4$、$(-1)^{2} = 1$、$0$、$1$、$4$。斜率关于 $x$ 偶（关于 $x = 0$ 镜像）、按平方增长。故 $g(x) = x^{2}$，DE 为 $\dfrac{dy}{dx} = x^{2}$。

(b) 直接积分 M1·A1

$\displaystyle y = \int x^{2} \, dx = \dfrac{x^{3}}{3} + C$，$C \in \mathbb{R}$。

(c) 过 $(0, 1)$ 的特解 M1·A1

$1 = \dfrac{0^{3}}{3} + C = C$。故 $C = 1$，$y = \dfrac{x^{3}}{3} + 1$。

(d) $[-2, 2]$ 上的形状 A1·A1

$y = \tfrac{x^{3}}{3} + 1$：$y' = x^{2} \ge 0$，$y'(0) = 0$，所以曲线在 $(0, 1)$ 处有水平切线，构成驻点拐点（既非极大也非极小，因 $y'$ 不变号——仅在 $x = 0$ 处触零）。$y'' = 2x$ 在 $x = 0$ 处变号：$x < 0$ 下凹，$x > 0$ 上凹，凹凸性在 $x = 0$ 处变号。$y - 1 = x^{3}/3$ 关于 $x$ 为奇函数，故图像关于 $(0, 1)$ 中心对称：以 $(0, 1)$ 为中心旋转 $180°$ 后与自身重合。

(e) 过 $(0, -1)$ 的另一条解曲线 A1

由 $-1 = 0 + C$ 得 $C = -1$，新曲线 $y = \tfrac{x^{3}}{3} - 1$。因斜率场仅依赖 $x$（不依赖 $y$），所有解曲线互为竖直平移：$y_{1}(x) - y_{2}(x) = $ 常数。新曲线为 (c) 中曲线向下平移 $2$ 单位。

$y' = g(x)$ 型斜率场 = "所有解互为竖直平移"。当 RHS 只依赖 $x$ 时，解构成一参数族 $\{y = G(x) + C : C \in \mathbb{R}\}$（$G' = g$）。视觉上：每列斜率短线完全相同（恰为本题表格性质），且任两解互为竖直平移。一旦 RHS 含 $y$ 即破坏此性质：如 $y' = y$ 的解 $y = A e^{x}$ 增长率不同，非平移。读斜率场首先核查：列是否相同？是则直接积分；否则需分离、部分分式或他法。(d) 同时考查词汇与微积分——"驻点拐点"对应 $y'$ 取零但不变号；"关于 $(0, 1)$ 的奇对称"对应 $y - 1 = x^{3}/3$ 为奇；"凹凸性在 $x = 0$ 处变号"对应 $y'' = 2x$ 变号。三短语各对应 $y'$ 或 $y''$ 的一个独立诊断，各得 A1。

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3E5.6 Logistic Growth via Partial Fractions[15 marks]

Logistic: $M = 1000$, $\frac{dP}{dt} = kP(M - P)$, $P(0) = 100$, $P(5) = 250$. Build, find $k$, limit.逻辑斯蒂：$M = 1000$、$\frac{dP}{dt} = kP(M - P)$、$P(0) = 100$、$P(5) = 250$。建模、求 $k$、求极限。

Answers:答案： (c) $P(t) = \dfrac{1000}{1 + 9 e^{-1000 k t}}$ · (d) $k = \tfrac{1}{5000}\ln 3 \approx 2.20 \times 10^{-4}$ · (e) $\lim P = 1000$

(a) Partial fraction identity M1·A1·A1

Combine the RHS over a common denominator: $$ \frac{1}{M}\!\left(\frac{1}{P} + \frac{1}{M - P}\right) \;=\; \frac{1}{M} \cdot \frac{(M - P) + P}{P(M - P)} \;=\; \frac{1}{M} \cdot \frac{M}{P(M - P)} \;=\; \frac{1}{P(M - P)}. \;\checkmark $$ Or, by undetermined coefficients: write $\dfrac{1}{P(M - P)} = \dfrac{A}{P} + \dfrac{B}{M - P}$, clear denominators to get $1 = A(M - P) + B P$. Set $P = 0$: $1 = AM$, $A = 1/M$. Set $P = M$: $1 = BM$, $B = 1/M$. Hence the identity.

(b) Separate & integrate M1·A1·A1·A1

From $\dfrac{dP}{dt} = k P(M - P)$, separate: $$ \frac{dP}{P(M - P)} \;=\; k \, dt. $$ Use (a) on the LHS: $$ \frac{1}{M}\!\left(\frac{1}{P} + \frac{1}{M - P}\right) dP \;=\; k \, dt. $$ Integrate both sides (noting $\int \frac{dP}{M - P} = -\ln|M - P|$): $$ \frac{1}{M}\bigl(\ln|P| - \ln|M - P|\bigr) \;=\; k t + C', $$ i.e. $\dfrac{1}{M}\ln\!\left|\dfrac{P}{M - P}\right| = k t + C'$, or $$ \ln\!\left|\frac{P}{M - P}\right| \;=\; M k t + C, $$ with $C = M C'$ a new constant.

(c) Solve explicitly, apply $P(0) = 100$ M1·A1·A1·A1

Exponentiate: $\dfrac{P}{M - P} = D \, e^{M k t}$ with $D = \pm e^{C}$. Solve for $P$: $$ P = (M - P) D e^{M k t} \;\Longrightarrow\; P\bigl(1 + D e^{M k t}\bigr) = M D e^{M k t} \;\Longrightarrow\; P = \frac{M D e^{M k t}}{1 + D e^{M k t}}. $$ Divide top and bottom by $D e^{M k t}$: $$ P(t) \;=\; \frac{M}{D^{-1} e^{-M k t} + 1} \;=\; \frac{M}{1 + B e^{-M k t}} \quad \text{with } B = D^{-1}. $$ Apply $P(0) = 100$ with $M = 1000$: $100 = \dfrac{1000}{1 + B}$, so $1 + B = 10$, $B = 9$. Hence $$ P(t) \;=\; \frac{1000}{1 + 9\, e^{-1000 k t}}. $$

(d) Find $k$ from $P(5) = 250$ M1·A1

$250 = \dfrac{1000}{1 + 9 e^{-5000 k}} \;\Longrightarrow\; 1 + 9 e^{-5000 k} = 4 \;\Longrightarrow\; e^{-5000 k} = \dfrac{1}{3}$. So $-5000 k = -\ln 3$, i.e. $$ k \;=\; \frac{\ln 3}{5000} \;\approx\; \frac{1.0986}{5000} \;\approx\; 2.20 \times 10^{-4} \quad \text{($3$ sf).} $$

(e) Long-run limit A1·R1

As $t \to \infty$, $1000 k t \to \infty$, so $e^{-1000 k t} \to 0$, hence $P(t) \to \dfrac{1000}{1 + 0} = 1000$. $$ \lim_{t \to \infty} P(t) \;=\; 1000 = M. $$ Biological meaning: the population approaches the carrying capacity $M = 1000$, the maximum number of deer the reserve can support, but never exceeds it. The growth slows as $P$ nears $M$ because the factor $(M - P)$ in the DE shrinks to $0$.

The logistic solution has a memorable normal form. The cleanest writeup of any logistic problem is to derive the standard form $P(t) = \dfrac{M}{1 + B e^{-M k t}}$ once, identifying $B$ from $P(0)$ via $B = \dfrac{M - P(0)}{P(0)}$ (here $B = \dfrac{1000 - 100}{100} = 9$, matching). Then every subsequent question — find $k$, find $t$ at half-capacity, find the inflection — is a one-line plug-in. The inflection of a logistic curve sits at $P = M/2$ (where $dP/dt$ is maximum, since $P(M - P)$ is maximised at $P = M/2$); the time of inflection is $t^{*} = \dfrac{\ln B}{M k}$. With our numbers: $P^{*} = 500$ at $t^{*} = \dfrac{\ln 9}{1000 \cdot \ln 3 / 5000} = \dfrac{5 \ln 9}{\ln 3} = \dfrac{5 \cdot 2 \ln 3}{\ln 3} = 10$ years. The carrying-capacity asymptote and the inflection at half-capacity are the two Paper 3 "extension" questions that almost always follow.

(a) 部分分式恒等式 M1·A1·A1

右侧通分： $$ \frac{1}{M}\!\left(\frac{1}{P} + \frac{1}{M - P}\right) \;=\; \frac{1}{M} \cdot \frac{(M - P) + P}{P(M - P)} \;=\; \frac{1}{M} \cdot \frac{M}{P(M - P)} \;=\; \frac{1}{P(M - P)}. \;\checkmark $$ 或用待定系数：$\dfrac{1}{P(M - P)} = \dfrac{A}{P} + \dfrac{B}{M - P}$，去分母得 $1 = A(M - P) + B P$。$P = 0$：$A = 1/M$；$P = M$：$B = 1/M$。

(b) 分离 & 积分 M1·A1·A1·A1

由 $\dfrac{dP}{dt} = k P(M - P)$ 分离： $$ \frac{dP}{P(M - P)} \;=\; k \, dt. $$ 左侧用 (a)： $$ \frac{1}{M}\!\left(\frac{1}{P} + \frac{1}{M - P}\right) dP \;=\; k \, dt. $$ 两边积分（$\int \frac{dP}{M - P} = -\ln|M - P|$）： $$ \frac{1}{M}\bigl(\ln|P| - \ln|M - P|\bigr) \;=\; k t + C', $$ 即 $\dfrac{1}{M}\ln\!\left|\dfrac{P}{M - P}\right| = k t + C'$，亦即 $$ \ln\!\left|\frac{P}{M - P}\right| \;=\; M k t + C, $$ 其中 $C = M C'$。

(c) 显式求解，代 $P(0) = 100$ M1·A1·A1·A1

取指数：$\dfrac{P}{M - P} = D \, e^{M k t}$（$D = \pm e^{C}$）。解 $P$： $$ P = (M - P) D e^{M k t} \;\Longrightarrow\; P\bigl(1 + D e^{M k t}\bigr) = M D e^{M k t} \;\Longrightarrow\; P = \frac{M D e^{M k t}}{1 + D e^{M k t}}. $$ 分子分母同除 $D e^{M k t}$： $$ P(t) \;=\; \frac{M}{1 + B e^{-M k t}} \quad (B = D^{-1}). $$ 代 $P(0) = 100$、$M = 1000$：$100 = \dfrac{1000}{1 + B}$，$B = 9$。故 $$ P(t) \;=\; \frac{1000}{1 + 9\, e^{-1000 k t}}. $$

(d) 由 $P(5) = 250$ 求 $k$ M1·A1

$250 = \dfrac{1000}{1 + 9 e^{-5000 k}} \;\Longrightarrow\; 1 + 9 e^{-5000 k} = 4 \;\Longrightarrow\; e^{-5000 k} = \dfrac{1}{3}$。故 $$ k \;=\; \frac{\ln 3}{5000} \;\approx\; 2.20 \times 10^{-4}. $$ （3 位有效数字。）

(e) 长期极限 A1·R1

$t \to \infty$ 时 $1000 k t \to \infty$，$e^{-1000 k t} \to 0$，故 $P(t) \to \dfrac{1000}{1 + 0} = 1000$。 $$ \lim_{t \to \infty} P(t) \;=\; 1000 = M. $$ 生物意义：种群趋向承载量 $M = 1000$，即保护区可支持的最大鹿群规模，但永不超过。当 $P$ 接近 $M$ 时因子 $(M - P) \to 0$，增长趋缓。

逻辑斯蒂解有一个值得背诵的标准形式。逻辑斯蒂题最干净的写法是一次性导出 $P(t) = \dfrac{M}{1 + B e^{-M k t}}$，由 $P(0)$ 用 $B = \dfrac{M - P(0)}{P(0)}$ 读出（本题 $B = \dfrac{900}{100} = 9$，吻合）。此后所有问——求 $k$、半承载时刻、拐点——皆为一行代入。逻辑斯蒂曲线的拐点在 $P = M/2$（$P(M - P)$ 在 $P = M/2$ 最大，故 $dP/dt$ 最大），拐点时刻 $t^{*} = \dfrac{\ln B}{M k}$。代入本题：$P^{*} = 500$ 时 $t^{*} = \dfrac{\ln 9}{1000 \cdot \ln 3 / 5000} = \dfrac{5 \cdot 2 \ln 3}{\ln 3} = 10$ 年。承载量渐近线与半承载拐点是 Paper 3 几乎必跟的两个延伸问。