Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 7.1 - 7.8主题 7.1 - 7.8AB
+ Extensions: 7.5 Euler's Method · 7.9 Logistic Models+ 扩展：7.5 欧拉法 · 7.9 逻辑斯谛模型BC
A population grows at a rate proportional to the current population. Which differential equation models this situation?某种群的增长速率与当前种群数量成正比。哪个微分方程描述了这一情形？
"Rate proportional to the current population" means $\dfrac{dP}{dt}$ equals a constant $k$ times $P$ itself, not a constant, not $\tfrac{1}{P}$, and not the independent variable $t$. (M1)"速率与当前种群数量成正比"意味着 $\dfrac{dP}{dt}$ 等于常数 $k$ 乘以 $P$ 本身，而非常数、非 $\tfrac{1}{P}$、也非自变量 $t$。(M1)
This is exactly $\dfrac{dP}{dt}=kP$. (A1)这正是 $\dfrac{dP}{dt}=kP$。(A1)
Which of the following functions is a solution to $\dfrac{dy}{dx}=2y$?下列哪个函数是 $\dfrac{dy}{dx}=2y$ 的解？
For $y=e^{2x}$: $\dfrac{dy}{dx}=2e^{2x}=2y$, so the equation holds identically. (M1)对 $y=e^{2x}$：$\dfrac{dy}{dx}=2e^{2x}=2y$，方程恒成立。(M1)
Checking the rest: $y=2x$ gives $y'=2\ne 2y=4x$; $y=x^{2}$ gives $y'=2x\ne 2x^{2}$; $y=\sin(2x)$ gives $y'=2\cos(2x)\ne 2\sin(2x)$. Only (C) survives. (A1)检验其余选项：$y=2x$ 给出 $y'=2\ne 2y=4x$；$y=x^{2}$ 给出 $y'=2x\ne 2x^{2}$；$y=\sin(2x)$ 给出 $y'=2\cos(2x)\ne 2\sin(2x)$。只有 (C) 成立。(A1)
A cup of coffee cools at a rate proportional to the difference between its temperature $T$ and a room temperature of $70$°F. Which equation models this?一杯咖啡的冷却速率与其温度 $T$ 和室温 $70$°F 之差成正比。哪个方程描述了这一情形？
The described difference is $T-70$ (temperature minus room temperature); "proportional to" that difference means $\dfrac{dT}{dt}=k(T-70)$ for some constant $k$. (M1)所描述的差为 $T-70$（温度减室温）；"与该差成正比"意味着 $\dfrac{dT}{dt}=k(T-70)$，其中 $k$ 为某常数。(M1)
This matches (B); note that $k<0$ is what makes the coffee actually cool (since $T>70$ initially), but the sign of $k$ is not fixed by the wording of the model itself. (A1)这与 (B) 相符；注意 $k<0$ 才能使咖啡真正降温（因初始 $T>70$），但模型本身的措辞并未固定 $k$ 的符号。(A1)
The slope field shown could represent which differential equation?图示斜率场可能代表哪个微分方程？
Within each horizontal row (fixed $y$), every marked segment has the same slope, and the slope grows steeper moving up the page (to greater $y$); across a single row the segments do not change as $x$ varies. (M1)在每个水平行（$y$ 固定）内，所有标出的线段斜率相同，且斜率随图像上移（$y$ 增大）而变陡；在同一行内，线段不随 $x$ 变化。(M1)
A slope that depends only on $y$, and increases as $y$ increases, matches $\dfrac{dy}{dx}=y$. Option (A) would vary within a row (it depends on $x$); options (C) and (D) both mix $x$ and $y$, so they would also vary across a fixed-$y$ row. Only (B) is constant along each row. (A1)斜率仅依赖 $y$ 且随 $y$ 增大而增大，与 $\dfrac{dy}{dx}=y$ 相符。选项 (A) 会在同一行内变化（依赖 $x$）；选项 (C)、(D) 均混合了 $x$ 与 $y$，同样会在固定 $y$ 的一行内变化。只有 (B) 在每行内保持不变。(A1)
For which differential equation are all line segments in the same horizontal row of a slope field identical?对于哪个微分方程，斜率场中同一水平行的所有线段完全相同？
A horizontal row fixes $y$ and lets $x$ vary. The segments in that row are identical exactly when the slope formula does not change as $x$ changes, i.e. when $\dfrac{dy}{dx}$ is a function of $y$ alone. (M1)水平行固定 $y$，$x$ 变化。当斜率公式随 $x$ 变化不变时，即当 $\dfrac{dy}{dx}$ 仅是 $y$ 的函数时，该行内线段完全相同。(M1)
Of the options, only $\dfrac{dy}{dx}=y$ has no $x$ on the right side; $x$, $x+y$, and $\sin x$ all change value as $x$ varies along a fixed row. (A1)选项中只有 $\dfrac{dy}{dx}=y$ 右端不含 $x$；$x$、$x+y$、$\sin x$ 均会随固定行内 $x$ 的变化而改变。(A1)
If $\dfrac{dy}{dx}=\dfrac{x}{y}$, which is a general solution?若 $\dfrac{dy}{dx}=\dfrac{x}{y}$，下列哪个是通解？
Cross-multiply: $y\,dy=x\,dx$, then integrate: $\dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+C'$. (M1)交叉相乘：$y\,dy=x\,dx$，再积分：$\dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+C'$。(M1)
Multiplying by $2$ and renaming the constant gives $y^{2}=x^{2}+C$, matching (A). (A1)两边乘以 $2$ 并重新命名常数，得 $y^{2}=x^{2}+C$，与 (A) 相符。(A1)
Let $y$ be the solution to $\dfrac{dy}{dx}=2xy$ with $y(0)=3$. Then $y(1)=$设 $y$ 是满足 $\dfrac{dy}{dx}=2xy$，且 $y(0)=3$ 的解，则 $y(1)=$
Separate: $\dfrac{dy}{y}=2x\,dx$, integrate: $\ln|y|=x^{2}+C$, so $y=Ae^{x^{2}}$ for $A=e^{C}$. Using $y(0)=3$ gives $A=3$, so $y=3e^{x^{2}}$. (M1)分离变量：$\dfrac{dy}{y}=2x\,dx$，积分：$\ln|y|=x^{2}+C$，故 $y=Ae^{x^{2}}$（$A=e^{C}$）。由 $y(0)=3$ 得 $A=3$，即 $y=3e^{x^{2}}$。(M1)
Then $y(1)=3e^{1}=3e$. (A1)故 $y(1)=3e^{1}=3e$。(A1)
Let $y$ be the solution to $\dfrac{dy}{dx}=y^{2}$ with $y(0)=1$. The largest open interval containing $0$ on which the solution is defined is设 $y$ 是满足 $\dfrac{dy}{dx}=y^{2}$，且 $y(0)=1$ 的解，该解定义在包含 $0$ 的最大开区间是
Separate: $\dfrac{dy}{y^{2}}=dx$, integrate: $-\dfrac{1}{y}=x+C$. Using $y(0)=1$: $-1=C$, so $-\dfrac{1}{y}=x-1$, i.e. $y=\dfrac{1}{1-x}$. (M1)分离变量：$\dfrac{dy}{y^{2}}=dx$，积分：$-\dfrac{1}{y}=x+C$。由 $y(0)=1$：$-1=C$，故 $-\dfrac{1}{y}=x-1$，即 $y=\dfrac{1}{1-x}$。(M1)
This function is undefined at $x=1$ (denominator vanishes) and finite everywhere else; the largest open interval containing $x=0$ that avoids $x=1$ is $(-\infty,1)$. (A1)该函数在 $x=1$ 处无定义（分母为零），其余各处均有限；包含 $x=0$ 且避开 $x=1$ 的最大开区间为 $(-\infty,1)$。(A1)
A quantity $Q$ satisfies $\dfrac{dQ}{dt}=0.05\,Q$ and $Q(0)=200$. Then $Q(t)=$某量 $Q$ 满足 $\dfrac{dQ}{dt}=0.05\,Q$，且 $Q(0)=200$，则 $Q(t)=$
Any equation of the form $\dfrac{dQ}{dt}=kQ$ has general solution $Q(t)=Q_{0}e^{kt}$, obtained by separating and integrating exactly as in Q1/Q2. Here $k=0.05$. (M1)任何形如 $\dfrac{dQ}{dt}=kQ$ 的方程通解为 $Q(t)=Q_{0}e^{kt}$，可如 Q1/Q2 一样通过分离变量并积分得到。此处 $k=0.05$。(M1)
Substituting $Q_{0}=200$: $Q(t)=200\,e^{0.05t}$. (A1)代入 $Q_{0}=200$：$Q(t)=200\,e^{0.05t}$。(A1)
A radioactive substance decays so that $\dfrac{dN}{dt}=-0.04\,N$, where $t$ is in years. To the nearest year, the half-life is某放射性物质的衰变满足 $\dfrac{dN}{dt}=-0.04\,N$，其中 $t$ 以年为单位。精确到最近的整年，其半衰期为
The solution is $N(t)=N_{0}e^{-0.04t}$. Half-life is the time $t$ at which $N(t)=\tfrac12 N_{0}$: $\tfrac12=e^{-0.04t}\Rightarrow -0.04t=\ln\!\left(\tfrac12\right)=-\ln 2$. (M1)解为 $N(t)=N_{0}e^{-0.04t}$。半衰期是使 $N(t)=\tfrac12 N_{0}$ 的时刻 $t$：$\tfrac12=e^{-0.04t}\Rightarrow -0.04t=\ln\!\left(\tfrac12\right)=-\ln 2$。(M1)
So $t=\dfrac{\ln 2}{0.04}\approx\dfrac{0.6931}{0.04}\approx 17.33$ years, which rounds to $17$ years. (A1)故 $t=\dfrac{\ln 2}{0.04}\approx\dfrac{0.6931}{0.04}\approx 17.33$ 年，四舍五入为 $17$ 年。(A1)
Which of the following is NOT a solution to $\dfrac{dy}{dx}=y$?下列哪个函数不是 $\dfrac{dy}{dx}=y$ 的解？
$y=e^{x}\Rightarrow y'=e^{x}=y$ (works); $y=2e^{x}\Rightarrow y'=2e^{x}=y$ (works); $y=-e^{x}\Rightarrow y'=-e^{x}=y$ (works). (M1)$y=e^{x}\Rightarrow y'=e^{x}=y$（成立）；$y=2e^{x}\Rightarrow y'=2e^{x}=y$（成立）；$y=-e^{x}\Rightarrow y'=-e^{x}=y$（成立）。(M1)
But $y=e^{2x}\Rightarrow y'=2e^{2x}=2y\ne y$ (fails, since it actually solves a different equation, $y'=2y$). (A1)但 $y=e^{2x}\Rightarrow y'=2e^{2x}=2y\ne y$（不成立，因它实际上是另一方程 $y'=2y$ 的解）。(A1)
Let $y=f(x)$ be the solution to $\dfrac{dy}{dx}=x+y$ with $f(1)=2$. The tangent line to $y=f(x)$ at $x=1$ is used to approximate $f(1.2)$. The approximation is设 $y=f(x)$ 是满足 $\dfrac{dy}{dx}=x+y$，且 $f(1)=2$ 的解，利用 $y=f(x)$ 在 $x=1$ 处的切线近似 $f(1.2)$，近似值为
The DE gives the slope of the tangent line directly at $x=1$: $f'(1)=1+f(1)=1+2=3$. (M1)该微分方程直接给出 $x=1$ 处切线的斜率：$f'(1)=1+f(1)=1+2=3$。(M1)
The tangent-line (local linearization) approximation is $f(1.2)\approx f(1)+f'(1)(1.2-1)=2+3(0.2)=2.6$. (A1)切线（局部线性化）近似为 $f(1.2)\approx f(1)+f'(1)(1.2-1)=2+3(0.2)=2.6$。(A1)
Which of the following is a general solution to $\dfrac{dy}{dx}=\dfrac{y}{x}$?下列哪个是 $\dfrac{dy}{dx}=\dfrac{y}{x}$ 的通解？
Separate: $\dfrac{dy}{y}=\dfrac{dx}{x}\Rightarrow\ln|y|=\ln|x|+C$, which is exactly option (B), the implicit general solution straight from integrating both sides. (M1)分离变量：$\dfrac{dy}{y}=\dfrac{dx}{x}\Rightarrow\ln|y|=\ln|x|+C$，这正是选项 (B)，是两边积分后直接得到的隐式通解。(M1)
Exponentiating (B): $|y|=e^{C}|x|$, i.e. $y=Ax$ for a constant $A=\pm e^{C}$, which is exactly option (A) with the constant renamed. So (A) and (B) describe the same family of solutions, just written implicitly versus explicitly, and (C) is correct. (A1)对 (B) 取指数：$|y|=e^{C}|x|$，即 $y=Ax$（$A=\pm e^{C}$ 为常数），这正是重新命名常数后的选项 (A)。故 (A)、(B) 描述的是同一解族，只是分别以隐式和显式书写，(C) 正确。(A1)
A bacterial culture has $1000$ bacteria. After $4$ hours, the culture has $4000$ bacteria. If the growth rate is proportional to the current population, how many bacteria are present after $6$ hours?某细菌培养液初始含有 $1000$ 个细菌，$4$ 小时后增至 $4000$ 个。若增长速率与当前数量成正比，$6$ 小时后共有多少个细菌？
$P(t)=1000\,e^{kt}$. Using $P(4)=4000$: $e^{4k}=4$. Rather than solving for $k$ itself, note $P(6)=1000\,e^{6k}=1000\left(e^{4k}\right)^{3/2}$. (M1)$P(t)=1000\,e^{kt}$。由 $P(4)=4000$：$e^{4k}=4$。无需单独求出 $k$，注意到 $P(6)=1000\,e^{6k}=1000\left(e^{4k}\right)^{3/2}$。(M1)
So $P(6)=1000\cdot 4^{3/2}=1000\cdot 8=8000$. (A1)故 $P(6)=1000\cdot 4^{3/2}=1000\cdot 8=8000$。(A1)
Consider $\dfrac{dy}{dx}=x-y$. The slope of the line segment in the slope field at the point $(2,1)$ is考虑 $\dfrac{dy}{dx}=x-y$，斜率场在点 $(2,1)$ 处线段的斜率为
The slope at any point $(x,y)$ is just the value of the right-hand side there: at $(2,1)$, $x=2$ and $y=1$. (M1)任意点 $(x,y)$ 处的斜率即为该点处方程右端的值：在 $(2,1)$ 处，$x=2$，$y=1$。(M1)
$\dfrac{dy}{dx}=x-y=2-1=1$. (A1)$\dfrac{dy}{dx}=x-y=2-1=1$。(A1)
If $\dfrac{dy}{dx}=\dfrac{x}{y^{2}}$ with $y(0)=1$, then $y$ when $x=2$ equals若 $\dfrac{dy}{dx}=\dfrac{x}{y^{2}}$，且 $y(0)=1$，则 $x=2$ 时 $y$ 的值为
Separate: $y^{2}\,dy=x\,dx$, integrate: $\dfrac{y^{3}}{3}=\dfrac{x^{2}}{2}+C$. Using $y(0)=1$: $\dfrac{1}{3}=C$. (M1)分离变量：$y^{2}\,dy=x\,dx$，积分：$\dfrac{y^{3}}{3}=\dfrac{x^{2}}{2}+C$。由 $y(0)=1$：$\dfrac{1}{3}=C$。(M1)
At $x=2$: $\dfrac{y^{3}}{3}=\dfrac{4}{2}+\dfrac{1}{3}=2+\dfrac{1}{3}=\dfrac{7}{3}$, so $y^{3}=7$, giving $y=\sqrt[3]{7}$. (A1)当 $x=2$ 时：$\dfrac{y^{3}}{3}=\dfrac{4}{2}+\dfrac{1}{3}=2+\dfrac{1}{3}=\dfrac{7}{3}$，故 $y^{3}=7$，$y=\sqrt[3]{7}$。(A1)
If $y$ satisfies $\dfrac{dy}{dx}=ye^{x}$ with $y(0)=1$, then $y(\ln 2)=$若 $y$ 满足 $\dfrac{dy}{dx}=ye^{x}$，且 $y(0)=1$，则 $y(\ln 2)=$
Separate: $\dfrac{dy}{y}=e^{x}\,dx$, integrate: $\ln|y|=e^{x}+C$. Using $y(0)=1$: $\ln 1=e^{0}+C\Rightarrow 0=1+C\Rightarrow C=-1$, so $\ln y=e^{x}-1$. (M1)分离变量：$\dfrac{dy}{y}=e^{x}\,dx$，积分：$\ln|y|=e^{x}+C$。由 $y(0)=1$：$\ln 1=e^{0}+C\Rightarrow 0=1+C\Rightarrow C=-1$，故 $\ln y=e^{x}-1$。(M1)
At $x=\ln 2$: $\ln y=e^{\ln 2}-1=2-1=1$, so $y=e^{1}=e$. (A1)当 $x=\ln 2$ 时：$\ln y=e^{\ln 2}-1=2-1=1$，故 $y=e^{1}=e$。(A1)
Consider $\dfrac{dy}{dt}=y(2-y)$. The equilibrium solutions are考虑 $\dfrac{dy}{dt}=y(2-y)$，其平衡解为
Equilibrium (constant) solutions occur exactly where $\dfrac{dy}{dt}=0$ for all $t$: $y(2-y)=0$. (M1)平衡（常数）解恰好出现在使 $\dfrac{dy}{dt}=0$ 对所有 $t$ 成立之处：$y(2-y)=0$。(M1)
This factors to $y=0$ or $2-y=0$, giving the two equilibria $y=0$ and $y=2$. (A1)因式分解得 $y=0$ 或 $2-y=0$，即两个平衡解 $y=0$ 和 $y=2$。(A1)
Full credit requires complete algebraic work: separation of variables, antiderivatives, $+C$, use of initial conditions, and correct solving for $y$.满分需展示完整代数步骤：分离变量、求不定积分、写 $+C$、代入初始条件，以及正确解出 $y$。
Consider the differential equation $\dfrac{dy}{dx}=\dfrac{x}{y}$, where $y>0$.考虑微分方程 $\dfrac{dy}{dx}=\dfrac{x}{y}$，其中 $y>0$。
Cross-multiply: $y\,dy=x\,dx$. (M1)交叉相乘：$y\,dy=x\,dx$。(M1)
$$ \int y\,dy=\int x\,dx \;\Longrightarrow\; \frac{y^{2}}{2}=\frac{x^{2}}{2}+C' \;\Longrightarrow\; y^{2}=x^{2}+C. $$(A1)
Substitute $x=0,\ y=2$ into the general solution: $2^{2}=0^{2}+C\Rightarrow C=4$, so $y^{2}=x^{2}+4$. (M1)代入 $x=0,\ y=2$：$2^{2}=0^{2}+C\Rightarrow C=4$，即 $y^{2}=x^{2}+4$。(M1)
Since $y>0$ is given, take the positive square root: $y=\sqrt{x^{2}+4}$. (A1)因题设 $y>0$，取正平方根：$y=\sqrt{x^{2}+4}$。(A1)
The expression under the radical, $x^{2}+4$, is at least $4$ for every real $x$: it is never zero or negative, so the square root is defined for every $x$. (A1)根号内的表达式 $x^{2}+4$ 对任意实数 $x$ 均不小于 $4$：从不为零或负，故平方根对所有 $x$ 均有定义。(A1)
Hence the domain of the particular solution is all real numbers, $(-\infty,\infty)$, with no restriction. (R1)故该特解的定义域为全体实数 $(-\infty,\infty)$，不受任何限制。(R1)
Consider the differential equation $\dfrac{dy}{dx}=2xy$.考虑微分方程 $\dfrac{dy}{dx}=2xy$。
Substitute each $(x,y)$ pair directly into $\dfrac{dy}{dx}=2xy$: (M1)将每个 $(x,y)$ 直接代入 $\dfrac{dy}{dx}=2xy$：(M1)
| $(x,y)$ | $(-1,-1)$ | $(0,-1)$ | $(1,-1)$ | $(-1,0)$ | $(0,0)$ | $(1,0)$ | $(-1,1)$ | $(0,1)$ | $(1,1)$ |
|---|---|---|---|---|---|---|---|---|---|
| slope | $2$ | $0$ | $-2$ | $0$ | $0$ | $0$ | $-2$ | $0$ | $2$ |
All three points along $y=0$ have slope $0$ (horizontal segments); segments in the top and bottom rows tilt oppositely because $2xy$ flips sign with $y$. (A1)所有 $y=0$ 上的三点斜率均为 $0$（水平线段）；顶行与底行的线段方向相反，因为 $2xy$ 随 $y$ 变号。(A1)
Separate: $\dfrac{dy}{y}=2x\,dx$. (M1)分离变量：$\dfrac{dy}{y}=2x\,dx$。(M1)
$$ \int\frac{dy}{y}=\int 2x\,dx \;\Longrightarrow\; \ln|y|=x^{2}+C \;\Longrightarrow\; y=Ae^{x^{2}}. $$Using $f(0)=1$: $1=Ae^{0}=A$, so $A=1$. (A1)由 $f(0)=1$：$1=Ae^{0}=A$，故 $A=1$。(A1)
The particular solution is $y=e^{x^{2}}$. (A1)故特解为 $y=e^{x^{2}}$。(A1)
$f(1)=e^{1^{2}}=e$. (A1)$f(1)=e^{1^{2}}=e$。(A1)
A tank initially contains $500$ gallons of water. Water leaks out at a rate proportional to the amount remaining: $\dfrac{dW}{dt}=k\,W$, where $t$ is in minutes. After $20$ minutes, the tank contains $400$ gallons.水箱初始含有 $500$ 加仑水，水以与剩余水量成正比的速率漏出，满足 $\dfrac{dW}{dt}=k\,W$，$t$ 以分钟为单位。$20$ 分钟后，水箱中剩余 $400$ 加仑。
$\dfrac{dW}{dt}=kW$ separates to $\dfrac{dW}{W}=k\,dt$, integrating to $W(t)=W_{0}e^{kt}$. (M1)$\dfrac{dW}{dt}=kW$ 分离变量为 $\dfrac{dW}{W}=k\,dt$，积分得 $W(t)=W_{0}e^{kt}$。(M1)
With $W_{0}=500$: $W(t)=500e^{kt}$. (A1)代入 $W_{0}=500$：$W(t)=500e^{kt}$。(A1)
$500e^{20k}=400\Rightarrow e^{20k}=0.8\Rightarrow 20k=\ln(0.8)$. (M1)$500e^{20k}=400\Rightarrow e^{20k}=0.8\Rightarrow 20k=\ln(0.8)$。(M1)
$k=\dfrac{\ln(0.8)}{20}\approx\dfrac{-0.22314}{20}\approx -0.0112$ (four decimal places). (A1)$k=\dfrac{\ln(0.8)}{20}\approx\dfrac{-0.22314}{20}\approx -0.0112$（精确到小数点后四位）。(A1)
Since $60=3\times 20$, write $W(60)=500\,e^{60k}=500\left(e^{20k}\right)^{3}=500(0.8)^{3}$ using the exact ratio from part (b) rather than the rounded $k$. (M1)因 $60=3\times 20$，可写 $W(60)=500\,e^{60k}=500\left(e^{20k}\right)^{3}=500(0.8)^{3}$，使用 (b) 中的精确比值而非四舍五入后的 $k$。(M1)
$W(60)=500(0.512)=256$ gallons (nearest gallon). (A1)$W(60)=500(0.512)=256$ 加仑（精确到整加仑）。(A1)
The rate is given directly by the DE itself: $\dfrac{dW}{dt}=k\,W$. At $t=60$: $\dfrac{dW}{dt}=k\cdot W(60)\approx(-0.011157)(256)$. (M1)该速率由微分方程本身直接给出：$\dfrac{dW}{dt}=k\,W$。当 $t=60$ 时：$\dfrac{dW}{dt}=k\cdot W(60)\approx(-0.011157)(256)$。(M1)
$\dfrac{dW}{dt}\approx -2.856$ gallons per minute: water is leaking out at about $2.86$ gal/min at $t=60$. (A1)$\dfrac{dW}{dt}\approx -2.856$ 加仑/分钟：$t=60$ 时水以约 $2.86$ 加仑/分钟的速率漏出。(A1)
Consider the differential equation $\dfrac{dy}{dx}=\dfrac{2y}{x+1}$.考虑微分方程 $\dfrac{dy}{dx}=\dfrac{2y}{x+1}$。
For $y=C(x+1)^{2}$: $y'=2C(x+1)$. (M1)对 $y=C(x+1)^{2}$：$y'=2C(x+1)$。(M1)
The right side of the DE is $\dfrac{2y}{x+1}=\dfrac{2C(x+1)^{2}}{x+1}=2C(x+1)=y'$: the two sides match identically for any $C$, so $y=C(x+1)^{2}$ is verified as a solution (for $x\ne -1$). (A1)方程右端为 $\dfrac{2y}{x+1}=\dfrac{2C(x+1)^{2}}{x+1}=2C(x+1)=y'$：对任意 $C$ 两边恒等，故 $y=C(x+1)^{2}$（$x\ne -1$）被验证为解。(A1)
Substitute $x=0,\ y=3$ into $y=C(x+1)^{2}$: $3=C(1)^{2}=C$. (M1)将 $x=0,\ y=3$ 代入 $y=C(x+1)^{2}$：$3=C(1)^{2}=C$。(M1)
So $f(x)=3(x+1)^{2}$. (A1)故 $f(x)=3(x+1)^{2}$。(A1)
The differential equation's right side, $\dfrac{2y}{x+1}$, is undefined at $x=-1$ (division by zero); a solution to the DE (as opposed to the formula $3(x+1)^2$ considered on its own) can therefore only be claimed on an interval that avoids $x=-1$. (M1)该微分方程右端 $\dfrac{2y}{x+1}$ 在 $x=-1$ 处无定义（除以零）；因此该微分方程的解（区别于单独考虑的公式 $3(x+1)^2$）只能在避开 $x=-1$ 的区间上成立。(M1)
Since $x=0$ lies to the right of $x=-1$, the largest open interval containing $x=0$ on which the differential equation (and hence the solution to it) is defined is $(-1,\infty)$. (R1)因 $x=0$ 位于 $x=-1$ 右侧，包含 $x=0$ 且使该微分方程（及其解）有定义的最大开区间是 $(-1,\infty)$。(R1)
Differentiate $f(x)=3(x+1)^{2}$ directly: $f'(x)=6(x+1)$, so $f''(x)=6$ for all $x$. (M1)直接对 $f(x)=3(x+1)^{2}$ 求导：$f'(x)=6(x+1)$，故对所有 $x$ 有 $f''(x)=6$。(M1)
So $f''(0)=6$. (A1)故 $f''(0)=6$。(A1)
A biologist studies a fish population $P(t)$ in a lake, where $t$ is in years. The population is modeled by the differential equation $\dfrac{dP}{dt}=0.1\,P\left(1-\dfrac{P}{1000}\right)$. (You are not required to solve this differential equation.)某生物学家研究湖中鱼类种群 $P(t)$，$t$ 以年为单位，种群由微分方程 $\dfrac{dP}{dt}=0.1\,P\left(1-\dfrac{P}{1000}\right)$ 建模。（不要求求解该微分方程。）
$\dfrac{dP}{dt}=0.1(400)\left(1-\dfrac{400}{1000}\right)=40(0.6)=24$. (M1)$\dfrac{dP}{dt}=0.1(400)\left(1-\dfrac{400}{1000}\right)=40(0.6)=24$。(M1)
In context: when the population is $400$ fish, it is increasing at a rate of $24$ fish per year. (A1)在语境中：当种群为 $400$ 条鱼时，其增长速率为每年 $24$ 条。(A1)
$\dfrac{dP}{dt}=0$ requires $0.1P\left(1-\dfrac{P}{1000}\right)=0$, so $P=0$ or $P=1000$. (M1)$\dfrac{dP}{dt}=0$ 要求 $0.1P\left(1-\dfrac{P}{1000}\right)=0$，故 $P=0$ 或 $P=1000$。(M1)
$P=0$ represents extinction (no fish, population stays at zero); $P=1000$ is the carrying capacity, the population level the lake can sustain indefinitely. (A1)$P=0$ 代表灭绝（无鱼，种群保持为零）；$P=1000$ 是承载容量，即湖泊可持续维持的种群水平。(A1)
Using $P(0)=400$ and $P'(0)=24$ from part (a): $P(2)\approx P(0)+P'(0)\cdot 2=400+24(2)$. (M1)利用 (a) 中的 $P(0)=400$ 和 $P'(0)=24$：$P(2)\approx P(0)+P'(0)\cdot 2=400+24(2)$。(M1)
$P(2)\approx 400+48=448$. (A1)$P(2)\approx 400+48=448$。(A1)
Expand: $\dfrac{dP}{dt}=0.1P-0.0001P^{2}$. Differentiate both sides with respect to $t$, treating $P$ as a function of $t$: $\dfrac{d^{2}P}{dt^{2}}=\left(0.1-0.0002P\right)\dfrac{dP}{dt}$. (M1)展开：$\dfrac{dP}{dt}=0.1P-0.0001P^{2}$。对两边关于 $t$ 求导（将 $P$ 视为 $t$ 的函数）：$\dfrac{d^{2}P}{dt^{2}}=\left(0.1-0.0002P\right)\dfrac{dP}{dt}$。(M1)
At $P=400$: $\dfrac{dP}{dt}=24$ (part a), so $\dfrac{d^{2}P}{dt^{2}}=(0.1-0.0002(400))(24)=(0.1-0.08)(24)=(0.02)(24)=0.48>0$. (A1)当 $P=400$ 时：$\dfrac{dP}{dt}=24$（(a) 中已求得），故 $\dfrac{d^{2}P}{dt^{2}}=(0.1-0.0002(400))(24)=(0.1-0.08)(24)=(0.02)(24)=0.48>0$。(A1)
Since $\dfrac{d^{2}P}{dt^{2}}>0$ at $t=0$, the graph of $P$ is concave up there, so the tangent line lies below the true curve: the approximation in part (c) is an underestimate of $P(2)$. (R1)因 $t=0$ 处 $\dfrac{d^{2}P}{dt^{2}}>0$，$P$ 的图像在该处凹向上，故切线位于真实曲线下方：(c) 中的近似是 $P(2)$ 的低估。(R1)
Let $y=f(x)$ be the solution to $\dfrac{dy}{dx}=x+y$ with $f(0)=1$. Use Euler's method with two steps of equal size $h=0.5$ starting at $x=0$ to approximate $f(1)$.设 $y=f(x)$ 是满足 $\dfrac{dy}{dx}=x+y$，且 $f(0)=1$ 的解，从 $x=0$ 出发，用步长 $h=0.5$ 的两步欧拉法近似 $f(1)$。
Step 1, from $(x_{0},y_{0})=(0,1)$: slope $=x_{0}+y_{0}=0+1=1$, so $y_{1}=y_{0}+h\cdot\text{slope}=1+0.5(1)=1.5$ at $x_{1}=0.5$. (M1)第一步，从 $(x_{0},y_{0})=(0,1)$ 出发：斜率 $=x_{0}+y_{0}=0+1=1$，故 $y_{1}=y_{0}+h\cdot\text{斜率}=1+0.5(1)=1.5$，此时 $x_{1}=0.5$。(M1)
Step 2, from $(0.5,1.5)$: slope $=0.5+1.5=2$, so $y_{2}=1.5+0.5(2)=2.5$ at $x_{2}=1$. Hence $f(1)\approx 2.500$. (A1)第二步，从 $(0.5,1.5)$ 出发：斜率 $=0.5+1.5=2$，故 $y_{2}=1.5+0.5(2)=2.5$，此时 $x_{2}=1$。故 $f(1)\approx 2.500$。(A1)
Suppose the solution $y=f(x)$ to a differential equation is concave up on the interval $[a,b]$. An Euler's-method approximation of $f(b)$ starting from $f(a)$ with positive step size $h$ will be设某微分方程的解 $y=f(x)$ 在区间 $[a,b]$ 上是上凸的（即凹形）。从 $f(a)$ 出发用正步长 $h$ 的欧拉法近似 $f(b)$，结果将
Each Euler step advances along the tangent line at the current point. On a concave-up curve ($f''>0$), every tangent line lies entirely below the curve except at the point of tangency itself. (M1)每一步欧拉迭代都沿当前点处的切线前进。在上凸曲线（$f''>0$）上，除切点本身外，每条切线均完全位于曲线下方。(M1)
Since every step lands below the true curve, and the next step's slope is then computed from that low point (compounding the underestimate), the Euler approximation is always less than $f(b)$. (A1)因每一步都落在真实曲线下方，且下一步的斜率又是从这一偏低点计算得出（使低估累积），故欧拉近似始终小于 $f(b)$。(A1)
A population $P(t)$ satisfies $\dfrac{dP}{dt}=0.04\,P\!\left(1-\dfrac{P}{500}\right)$. The carrying capacity and the population value at which $P$ is increasing fastest are, respectively,种群 $P(t)$ 满足 $\dfrac{dP}{dt}=0.04\,P\!\left(1-\dfrac{P}{500}\right)$，承载容量以及 $P$ 增长最快时的种群值分别为
Comparing to $\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{K}\right)$, here $k=0.04$ and $K=500$: the carrying capacity is $K=500$. (M1)与 $\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{K}\right)$ 比较，此处 $k=0.04$，$K=500$：承载容量为 $K=500$。(M1)
The logistic model grows fastest at half the carrying capacity, $P=\dfrac{K}{2}=\dfrac{500}{2}=250$. (A1)逻辑斯谛模型在承载容量的一半处增长最快，即 $P=\dfrac{K}{2}=\dfrac{500}{2}=250$。(A1)
Consider the differential equation $\dfrac{dy}{dx}=x-y$ with initial condition $y(0)=2$.考虑微分方程 $\dfrac{dy}{dx}=x-y$，初始条件为 $y(0)=2$。
| $n$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $x_n$ | $0$ | $0.5$ | $1$ |
| $y_n$ | $2$ | $1$ | $0.75$ |
| slope $x_n-y_n$ | $-2$ | $-0.5$ | $\cdot$ |
Step 1: slope at $(0,2)$ is $0-2=-2$, so $y_{1}=2+0.5(-2)=1$ at $x_{1}=0.5$. (M1)第一步：$(0,2)$ 处斜率为 $0-2=-2$，故 $y_{1}=2+0.5(-2)=1$，$x_{1}=0.5$。(M1)
Step 2: slope at $(0.5,1)$ is $0.5-1=-0.5$, so $y_{2}=1+0.5(-0.5)=0.75$ at $x_{2}=1$. (A1)第二步：$(0.5,1)$ 处斜率为 $0.5-1=-0.5$，故 $y_{2}=1+0.5(-0.5)=0.75$，$x_{2}=1$。(A1)
So $y(1)\approx 0.75$. (A1)故 $y(1)\approx 0.75$。(A1)
Differentiate $\dfrac{dy}{dx}=x-y$ implicitly with respect to $x$: $\dfrac{d^{2}y}{dx^{2}}=1-\dfrac{dy}{dx}=1-(x-y)=1-x+y$. (M1)对 $\dfrac{dy}{dx}=x-y$ 关于 $x$ 隐式求导：$\dfrac{d^{2}y}{dx^{2}}=1-\dfrac{dy}{dx}=1-(x-y)=1-x+y$。(M1)
Evaluating along the Euler path: at $(0,2)$, $1-0+2=3>0$; at $(0.5,1)$, $1-0.5+1=1.5>0$. Both sampled points give $\dfrac{d^{2}y}{dx^{2}}>0$, indicating the true solution curve is concave up on this interval. (A1)沿欧拉路径求值：在 $(0,2)$ 处，$1-0+2=3>0$；在 $(0.5,1)$ 处，$1-0.5+1=1.5>0$。两个采样点均给出 $\dfrac{d^{2}y}{dx^{2}}>0$，表明真实解曲线在此区间上凹向上。(A1)
Since the curve is concave up, each tangent-line (Euler) step lies below the true curve, so the approximation $y(1)\approx 0.75$ is an underestimate of the true value of $y(1)$. (R1)因曲线凹向上，每一步切线（欧拉）近似都落在真实曲线下方，故近似值 $y(1)\approx 0.75$ 是 $y(1)$ 真实值的低估。(R1)
Euler's method error grows with the size of each step, since a larger $h$ lets the tangent line drift farther from the true curve before the slope is recomputed; using four steps of $h=0.25$ instead of two steps of $h=0.5$ re-corrects the direction twice as often. (R1)欧拉法的误差随每步步长增大而增大，因为较大的 $h$ 会使切线在下一次重新计算斜率之前偏离真实曲线更远；用步长 $h=0.25$ 的四步而非 $h=0.5$ 的两步，重新校正方向的次数增加了一倍。(R1)
So the second student's approximation, with the smaller step size, will be closer to the true value of $y(1)$ (though, since the curve is still concave up throughout, it will still be an underestimate). (R1)故第二位同学用更小步长得到的近似值将更接近 $y(1)$ 的真实值（不过因曲线始终凹向上，该近似仍将是低估）。(R1)
A wildlife biologist models a deer population $P(t)$, where $t$ is in years, by $\dfrac{dP}{dt}=0.2\,P\!\left(1-\dfrac{P}{800}\right)$, with $P(0)=100$.某野生生物学家以 $\dfrac{dP}{dt}=0.2\,P\!\left(1-\dfrac{P}{800}\right)$ 建模鹿群数量 $P(t)$，$t$ 以年为单位，初始条件 $P(0)=100$。
Matching to $kP(1-P/K)$ gives carrying capacity $K=800$. (A1)与 $kP(1-P/K)$ 匹配，得承载容量 $K=800$。(A1)
Since $\dfrac{dP}{dt}=0$ at $P=0$ and $P=800$, and $\dfrac{dP}{dt}>0$ for $0
Growth is fastest at half the carrying capacity: $P=\dfrac{800}{2}=400$. (M1)增长在承载容量一半处最快：$P=\dfrac{800}{2}=400$。(M1)
$\dfrac{dP}{dt}\bigg|_{P=400}=0.2(400)\left(1-\dfrac{400}{800}\right)=80(0.5)=40$ deer per year. (A1)$\dfrac{dP}{dt}\bigg|_{P=400}=0.2(400)\left(1-\dfrac{400}{800}\right)=80(0.5)=40$ 只/年。(A1)
Expand: $\dfrac{dP}{dt}=0.2P-0.00025P^{2}$. Differentiate with respect to $t$: $\dfrac{d^{2}P}{dt^{2}}=(0.2-0.0005P)\dfrac{dP}{dt}$. (M1)展开：$\dfrac{dP}{dt}=0.2P-0.00025P^{2}$。关于 $t$ 求导：$\dfrac{d^{2}P}{dt^{2}}=(0.2-0.0005P)\dfrac{dP}{dt}$。(M1)
Substituting $\dfrac{dP}{dt}=0.2P(1-P/800)$ back in gives $\dfrac{d^{2}P}{dt^{2}}$ purely in terms of $P$: (A1)代回 $\dfrac{dP}{dt}=0.2P(1-P/800)$，得到仅以 $P$ 表示的 $\dfrac{d^{2}P}{dt^{2}}$：(A1)
$$ \frac{d^{2}P}{dt^{2}}=0.04\,P\left(1-\frac{P}{800}\right)\left(1-\frac{P}{400}\right). $$The factor $0.04P\left(1-\tfrac{P}{800}\right)>0$ throughout $0
Since $\dfrac{d^{2}P}{dt^{2}}$ changes sign at $P=400$, the graph of $P(t)$ has an inflection point there, confirming the fastest-growth value found in part (b). (R1)因 $\dfrac{d^{2}P}{dt^{2}}$ 在 $P=400$ 处变号，$P(t)$ 的图像在该处有拐点，印证了 (b) 中求得的最快增长值。(R1)
Starting from $P(0)=100$, the curve rises, concave up while $P<400$, passes through the inflection point at $P=400$, then continues rising but concave down while $400
Label the inflection at height $P=400$ and the horizontal asymptote at $P=800$ on the sketch. (A1)在图中标出高度 $P=400$ 处的拐点和 $P=800$ 处的水平渐近线。(A1)