Companion to the University-Style Practice Set大学风格练习题配套解答
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
For each equation, state (i) the order and (ii) linear or nonlinear, with standard form or the nonlinear term: (a) $xy' + 2y = \sin x$; (b) $y'' - 3y' + 2y = e^{x}$; (c) $dy/dx = y^{2} - x$.对于每个方程，说明 (i) 阶数，(ii) 线性或非线性，给出标准形式或非线性项：(a) $xy' + 2y = \sin x$；(b) $y'' - 3y' + 2y = e^{x}$；(c) $dy/dx = y^{2} - x$。
The equation involves $y$ and $y'$ to the first power only, with no products of $y$ and $y'$, so it is linear. (M1) Dividing by $x$ (for $x \ne 0$) puts it in standard form:方程中 $y$ 和 $y'$ 均以一次幂出现，且无 $y$ 与 $y'$ 的乘积，故为线性方程。(M1) 两边除以 $x$（$x \ne 0$）得标准形式：
$$ y' + \frac{2}{x}\,y = \frac{\sin x}{x}. $$Order: 1. Linear. Standard form: $P(x) = \tfrac{2}{x}$, $Q(x) = \tfrac{\sin x}{x}$. (A1)阶数：1。线性。标准形式：$P(x) = \tfrac{2}{x}$，$Q(x) = \tfrac{\sin x}{x}$。(A1)
The equation contains $y''$, making it order 2. (M1) It is linear because $y$, $y'$, $y''$ all appear to the first power with no products among them. The equation is already in standard form for a second-order linear ODE. (A1)方程含 $y''$，故为 2 阶。(M1) 由于 $y$、$y'$、$y''$ 均以一次幂出现且无相互乘积，故为线性方程。方程已为二阶线性常微分方程的标准形式。(A1)
The highest derivative is $dy/dx$, so the order is 1. (M1) The term $y^{2}$ is a nonlinear function of the dependent variable $y$; a linear ODE would require all such terms to be linear (first power, no products). Therefore the equation is nonlinear. (A1)最高阶导数为 $dy/dx$，故阶数为 1。(M1) 项 $y^{2}$ 是因变量 $y$ 的非线性函数；线性常微分方程要求所有此类项均为一次幂且无乘积。因此该方程为非线性方程。(A1)
Solve by separating variables: (a) $dy/dx = 2xy$, $y(0) = 3$; (b) $dy/dx = x/y$, $y(0) = 4$.用分离变量法求解：(a) $dy/dx = 2xy$，$y(0) = 3$；(b) $dy/dx = x/y$，$y(0) = 4$。
Separate: $\dfrac{dy}{y} = 2x\,dx$ (valid for $y \ne 0$). (M1) Integrate both sides:分离变量：$\dfrac{dy}{y} = 2x\,dx$（$y \ne 0$ 时有效）。(M1) 两边积分：
$$ \ln|y| = x^{2} + C_{1} \implies y = Ce^{x^{2}}, \quad C \ne 0. $$(M1·A1) Apply $y(0) = 3$: $3 = Ce^{0} = C$. (A1) So $y = 3e^{x^{2}}$.(M1·A1) 代入 $y(0) = 3$：$3 = Ce^{0} = C$。(A1) 故 $y = 3e^{x^{2}}$。
Verification:验证： $y' = 6xe^{x^{2}} = 2x \cdot 3e^{x^{2}} = 2xy$. Initial condition: $y(0) = 3e^{0} = 3$. Both satisfied.$y' = 6xe^{x^{2}} = 2x \cdot 3e^{x^{2}} = 2xy$。初始条件：$y(0) = 3e^{0} = 3$。两者均满足。
Separate: $y\,dy = x\,dx$. (M1) Integrate both sides:分离变量：$y\,dy = x\,dx$。(M1) 两边积分：
$$ \frac{y^{2}}{2} = \frac{x^{2}}{2} + C \implies y^{2} = x^{2} + K. $$(M1·A1) Apply $y(0) = 4$: $16 = 0 + K$, so $K = 16$. (A1) Since $y(0) = 4 > 0$, take the positive root: $y = \sqrt{x^{2}+16}$.(M1·A1) 代入 $y(0) = 4$：$16 = 0 + K$，故 $K = 16$。(A1) 因 $y(0) = 4 > 0$，取正根：$y = \sqrt{x^{2}+16}$。
Verification:验证： $y' = \tfrac{x}{\sqrt{x^{2}+16}} = \tfrac{x}{y}$. Initial condition: $y(0) = \sqrt{16} = 4$. Both satisfied.$y' = \tfrac{x}{\sqrt{x^{2}+16}} = \tfrac{x}{y}$。初始条件：$y(0) = \sqrt{16} = 4$。两者均满足。
Solve using the integrating-factor method: (a) $y' - \tfrac{2}{x}y = x^{2}$, $x > 0$; (b) $y' + y\tan x = \sec x$, $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$, $y(0) = 0$.用积分因子法求解：(a) $y' - \tfrac{2}{x}y = x^{2}$，$x > 0$；(b) $y' + y\tan x = \sec x$，$-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$，$y(0) = 0$。
Already in standard form with $P(x) = -2/x$. (M1) Compute the integrating factor:方程已为标准形式，$P(x) = -2/x$。(M1) 计算积分因子：
$$ \mu = e^{\int -2/x\,dx} = e^{-2\ln x} = x^{-2} \quad (x > 0). $$(M1) Multiply through by $\mu = x^{-2}$:(M1) 两边乘以 $\mu = x^{-2}$：
$$ \left(x^{-2}y\right)' = x^{-2} \cdot x^{2} = 1. $$(A1) Integrate: $x^{-2}y = x + C$. (A1) Hence $y = x^{3} + Cx^{2}$.(A1) 积分得：$x^{-2}y = x + C$。(A1) 故 $y = x^{3} + Cx^{2}$。
Verification:验证： $y' = 3x^{2} + 2Cx$. Then $y' - \tfrac{2}{x}y = 3x^{2} + 2Cx - \tfrac{2}{x}(x^{3} + Cx^{2}) = 3x^{2} + 2Cx - 2x^{2} - 2Cx = x^{2}$. Correct.$y' = 3x^{2} + 2Cx$。则 $y' - \tfrac{2}{x}y = 3x^{2} + 2Cx - \tfrac{2}{x}(x^{3} + Cx^{2}) = 3x^{2} + 2Cx - 2x^{2} - 2Cx = x^{2}$。正确。
Standard form: $P(x) = \tan x$, $Q(x) = \sec x$. (M1) Integrating factor:标准形式：$P(x) = \tan x$，$Q(x) = \sec x$。(M1) 积分因子：
$$ \mu = e^{\int \tan x\,dx} = e^{\ln|\sec x|} = \sec x \quad \text{on } \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right). $$(M1) Multiply through by $\sec x$:(M1) 两边乘以 $\sec x$：
$$ (\sec x\cdot y)' = \sec x \cdot \sec x = \sec^{2}x. $$(A1) Integrate: $y\sec x = \tan x + C$, so $y\sec x = \tan x + C$. Apply $y(0) = 0$: $0 \cdot \sec 0 = 0 + C$, so $C = 0$. (A1) Hence $y\sec x = \tan x$, giving $y = \tan x\cos x = \sin x$.(A1) 积分得：$y\sec x = \tan x + C$。代入 $y(0) = 0$：$0 \cdot \sec 0 = 0 + C$，故 $C = 0$。(A1) 从而 $y\sec x = \tan x$，得 $y = \tan x\cos x = \sin x$。
Verification:验证： $y' = \cos x$. Then $y' + y\tan x = \cos x + \sin x \cdot \tfrac{\sin x}{\cos x} = \tfrac{\cos^{2}x + \sin^{2}x}{\cos x} = \sec x$. Correct. The closed form is $y = \sin x$.$y' = \cos x$。则 $y' + y\tan x = \cos x + \sin x \cdot \tfrac{\sin x}{\cos x} = \tfrac{\cos^{2}x + \sin^{2}x}{\cos x} = \sec x$。正确。闭合形式为 $y = \sin x$。
For $dy/dt = y(2-y)$: (a) find all equilibria; (b) determine stability via $\partial f/\partial y$ at each equilibrium; (c) long-run behaviour of solutions starting at $y(0) = 1$ and $y(0) = 3$.对于 $dy/dt = y(2-y)$：(a) 求所有平衡点；(b) 通过在各平衡点处求 $\partial f/\partial y$ 判断稳定性；(c) 从 $y(0) = 1$ 和 $y(0) = 3$ 出发的解的长期行为。
Equilibria occur where $f(y) = y(2-y) = 0$. (M1) This gives $y = 0$ or $y = 2$. (A1)平衡点满足 $f(y) = y(2-y) = 0$。(M1) 得 $y = 0$ 或 $y = 2$。(A1)
Compute $f'(y) = \dfrac{\partial}{\partial y}[y(2-y)] = 2 - 2y$. (M1)计算 $f'(y) = \dfrac{\partial}{\partial y}[y(2-y)] = 2 - 2y$。(M1)
At $y = 0$: $f'(0) = 2 > 0$, so $y = 0$ is an unstable equilibrium (small perturbations grow). At $y = 2$: $f'(2) = 2 - 4 = -2 < 0$, so $y = 2$ is a stable equilibrium (small perturbations decay). (A1)在 $y = 0$ 处：$f'(0) = 2 > 0$，故 $y = 0$ 为不稳定平衡点（小扰动增大）。在 $y = 2$ 处：$f'(2) = 2 - 4 = -2 < 0$，故 $y = 2$ 为稳定平衡点（小扰动衰减）。(A1)
For $y(0) = 1$: since $0 < 1 < 2$ we have $f(1) = 1 > 0$, so $y$ is increasing and, since $y = 2$ is stable and acts as an upper bound, $y(t) \to 2$ as $t \to \infty$. (M1)对于 $y(0) = 1$：因 $0 < 1 < 2$，有 $f(1) = 1 > 0$，故 $y$ 递增，又因 $y = 2$ 为稳定平衡点且为上界，故 $y(t) \to 2$（$t \to \infty$）。(M1)
For $y(0) = 3$: since $3 > 2$ we have $f(3) = 3(-1) = -3 < 0$, so $y$ is decreasing and again $y(t) \to 2$ as $t \to \infty$. (A1) Both solutions are attracted to the stable equilibrium $y = 2$.对于 $y(0) = 3$：因 $3 > 2$，有 $f(3) = 3(-1) = -3 < 0$，故 $y$ 递减，同样 $y(t) \to 2$（$t \to \infty$）。(A1) 两个解均被稳定平衡点 $y = 2$ 吸引。
For $y' + P(x)y = Q(x)$: (a) show $\mu$ must satisfy $\mu' = P\mu$ and hence $\mu = e^{\int P\,dx}$; (b) integrate $(\mu y)' = \mu Q$ from $x_0$ to $x$ to derive the explicit solution formula and state why it is unique.对于 $y' + P(x)y = Q(x)$：(a) 证明 $\mu$ 必须满足 $\mu' = P\mu$，从而 $\mu = e^{\int P\,dx}$；(b) 将 $(\mu y)' = \mu Q$ 从 $x_0$ 积分到 $x$，推导出显式解公式，并说明解唯一的原因。
Multiply both sides of $y' + P(x)y = Q(x)$ by an unknown positive function $\mu(x)$:将 $y' + P(x)y = Q(x)$ 两边乘以未知正函数 $\mu(x)$：
$$ \mu y' + \mu P y = \mu Q. $$(M1) We want the left side to be $(\mu y)' = \mu' y + \mu y'$. Matching $\mu y' + \mu P y$ with $\mu y' + \mu' y$ requires:(M1) 我们希望左边变为 $(\mu y)' = \mu' y + \mu y'$。将 $\mu y' + \mu P y$ 与 $\mu y' + \mu' y$ 比较，需要：
$$ \mu' y = \mu P y \implies \mu' = P(x)\,\mu. $$(M1) This is a separable equation for $\mu$:(M1) 这是关于 $\mu$ 的可分离方程：
$$ \frac{d\mu}{\mu} = P(x)\,dx \implies \ln|\mu| = \int P(x)\,dx \implies \mu = e^{\int P(x)\,dx}. $$(A1) We choose the positive solution (no absolute value, no integration constant needed because any nonzero constant would cancel). With this $\mu$, the equation becomes $(\mu y)' = \mu Q$. (A1)(A1) 取正解（不含绝对值，积分常数可省略，因任何非零常数均可约去）。有了此 $\mu$，方程化为 $(\mu y)' = \mu Q$。(A1)
Integrate both sides of $(\mu y)' = \mu(t) Q(t)$ from $x_0$ to $x$: (M1)将 $(\mu y)' = \mu(t) Q(t)$ 两边从 $x_0$ 积分到 $x$：(M1)
$$ \mu(x)\,y(x) - \mu(x_0)\,y_0 = \int_{x_0}^{x} \mu(t)\,Q(t)\,dt. $$(M1) Solve for $y(x)$: (A1)(M1) 解出 $y(x)$：(A1)
$$ y(x) = \frac{1}{\mu(x)}\!\left(y_0\,\mu(x_0) + \int_{x_0}^{x}\mu(t)\,Q(t)\,dt\right). $$(A1) Uniqueness: every step is reversible. Any solution of the IVP satisfies $(\mu y)' = \mu Q$ (since $\mu \ne 0$), and integrating a continuous function from $x_0$ to $x$ gives a unique antiderivative (by the Fundamental Theorem of Calculus). Hence the formula gives the only solution on $I$.(A1) 唯一性：每步均可逆。初值问题的任意解均满足 $(\mu y)' = \mu Q$（因 $\mu \ne 0$），而对连续函数从 $x_0$ 到 $x$ 积分，由微积分基本定理给出唯一的原函数。故该公式给出 $I$ 上的唯一解。
Picard theorem: if $f$ and $\partial f/\partial y$ are continuous near $(x_0,y_0)$, the IVP $y'=f(x,y)$, $y(x_0)=y_0$ has a unique local solution. (a) $y'=y^{1/3}$, $y(0)=0$: find two solutions and identify which hypothesis fails; (b) $y'=y^2$, $y(0)=1$: solve and state the maximal interval; (c) $y'=\sqrt{|y|}$, $y(0)=0$: show the hypothesis fails and verify the one-parameter family of solutions.皮卡定理：若 $f$ 和 $\partial f/\partial y$ 在 $(x_0,y_0)$ 附近连续，则初值问题 $y'=f(x,y)$，$y(x_0)=y_0$ 在 $x_0$ 某邻域内有唯一局部解。(a) $y'=y^{1/3}$，$y(0)=0$：求两个解并指出哪个假设失效；(b) $y'=y^2$，$y(0)=1$：求解并给出最大存在区间；(c) $y'=\sqrt{|y|}$，$y(0)=0$：证明假设失效并验证单参数族解。
Let $f(x,y) = y^{1/3}$. First, $y \equiv 0$ satisfies $y' = 0 = 0^{1/3}$, so it is a solution. (M1) For a second solution, try $y = \bigl(\tfrac{2}{3}x\bigr)^{3/2}$ for $x \ge 0$. Then设 $f(x,y) = y^{1/3}$。首先，$y \equiv 0$ 满足 $y' = 0 = 0^{1/3}$，故为一个解。(M1) 对于第二个解，尝试 $y = \bigl(\tfrac{2}{3}x\bigr)^{3/2}$（$x \ge 0$）。则
$$ y' = \frac{3}{2}\cdot\left(\frac{2}{3}x\right)^{1/2}\cdot\frac{2}{3} = \left(\frac{2}{3}x\right)^{1/2}, $$and $y^{1/3} = \bigl[(\tfrac{2}{3}x)^{3/2}\bigr]^{1/3} = (\tfrac{2}{3}x)^{1/2}$. (A1) Both sides agree, confirming the second solution. The Picard hypothesis that fails is the continuity of $\partial f/\partial y = \tfrac{1}{3}y^{-2/3}$, which blows up as $y \to 0$, precisely at the initial point $(0,0)$. (R1)且 $y^{1/3} = \bigl[(\tfrac{2}{3}x)^{3/2}\bigr]^{1/3} = (\tfrac{2}{3}x)^{1/2}$。(A1) 两边相等，确认了第二个解。失效的皮卡假设是 $\partial f/\partial y = \tfrac{1}{3}y^{-2/3}$ 的连续性，当 $y \to 0$ 时其爆破，恰在初始点 $(0,0)$ 处。(R1)
Separate: $dy/y^{2} = dx$, so $-1/y = x + C$. (M1) With $y(0) = 1$: $-1 = 0 + C$, giving $C = -1$. Hence $y = \tfrac{1}{1-x}$. (A1)分离变量：$dy/y^{2} = dx$，得 $-1/y = x + C$。(M1) 代入 $y(0) = 1$：$-1 = 0 + C$，故 $C = -1$。从而 $y = \tfrac{1}{1-x}$。(A1)
As $x \to 1^{-}$, $y \to +\infty$: the solution blows up in finite time. The Picard theorem only guarantees existence on some interval around $x_0$; it says nothing about how large that interval is. Even though $f = y^{2}$ and $\partial f/\partial y = 2y$ are smooth for all $(x,y)$, the solution grows without bound before reaching $x = 1$. The maximal interval of existence is $(-\infty, 1)$. (R1)当 $x \to 1^{-}$ 时，$y \to +\infty$：解在有限时间内爆破。皮卡定理仅保证在 $x_0$ 的某个邻域上存在解；对区间大小无任何保证。即使 $f = y^{2}$ 和 $\partial f/\partial y = 2y$ 对所有 $(x,y)$ 均光滑，解在到达 $x = 1$ 之前仍会无界增长。最大存在区间为 $(-\infty, 1)$。(R1)
Here $f(x,y) = \sqrt{|y|}$. Compute $\partial f/\partial y = \tfrac{1}{2\sqrt{|y|}}$ for $y \ne 0$; this is undefined at $y = 0$, so $\partial f/\partial y$ is not continuous at $(0,0)$. (M1) The Picard uniqueness hypothesis fails.此处 $f(x,y) = \sqrt{|y|}$。计算 $\partial f/\partial y = \tfrac{1}{2\sqrt{|y|}}$（$y \ne 0$）；该量在 $y = 0$ 处无定义，故 $\partial f/\partial y$ 在 $(0,0)$ 处不连续。(M1) 皮卡唯一性假设失效。
Consider the family: for any $a \ge 0$, define $y_a(x) = 0$ for $x \le a$ and $y_a(x) = \tfrac{1}{4}(x-a)^{2}$ for $x > a$. (M1)考虑如下一族解：对任意 $a \ge 0$，定义 $y_a(x) = 0$（$x \le a$），$y_a(x) = \tfrac{1}{4}(x-a)^{2}$（$x > a$）。(M1)
Verification: for $x \le a$, $y_a = 0$ so $y_a' = 0 = \sqrt{|0|}$. For $x > a$, $y_a' = \tfrac{1}{2}(x-a)$ and $\sqrt{|y_a|} = \sqrt{\tfrac{1}{4}(x-a)^{2}} = \tfrac{1}{2}(x-a)$ (since $x > a$). (A1) Both match. At $x = a$: $y_a(a) = 0$ and $y_a'(a^{+}) = 0$, so the function is differentiable and solves the ODE everywhere. Each $y_a$ satisfies $y(0) = 0$ because $0 < a$ or $a = 0$ both give $y_a(0) = 0$. (A1)验证：当 $x \le a$ 时，$y_a = 0$，故 $y_a' = 0 = \sqrt{|0|}$。当 $x > a$ 时，$y_a' = \tfrac{1}{2}(x-a)$，$\sqrt{|y_a|} = \sqrt{\tfrac{1}{4}(x-a)^{2}} = \tfrac{1}{2}(x-a)$（因 $x > a$）。(A1) 两边吻合。在 $x = a$ 处：$y_a(a) = 0$，$y_a'(a^{+}) = 0$，故函数在处处可微且满足常微分方程。每个 $y_a$ 均满足 $y(0) = 0$，因为 $0 < a$ 或 $a = 0$ 均给出 $y_a(0) = 0$。(A1)
For $y' + 2xy = e^{-x^{2}}\cos x$, $y(0)=1$: (a) show $\mu = e^{x^2}$ and derive $y = e^{-x^2}(1+\int_0^x \cos t\,dt)$; (b) state the maximal interval, simplify to $y = e^{-x^2}(1+\sin x)$, and verify by differentiating.对于 $y' + 2xy = e^{-x^{2}}\cos x$，$y(0)=1$：(a) 证明 $\mu = e^{x^2}$，推导 $y = e^{-x^2}(1+\int_0^x \cos t\,dt)$；(b) 给出最大存在区间，化简为 $y = e^{-x^2}(1+\sin x)$，并通过求导验证。
Standard form: $P(x) = 2x$, $Q(x) = e^{-x^{2}}\cos x$. Integrating factor: (M1)标准形式：$P(x) = 2x$，$Q(x) = e^{-x^{2}}\cos x$。积分因子：(M1)
$$ \mu = e^{\int 2x\,dx} = e^{x^{2}}. $$Multiply through by $e^{x^{2}}$:两边乘以 $e^{x^{2}}$：
$$ e^{x^{2}}y' + 2xe^{x^{2}}y = e^{x^{2}}\cdot e^{-x^{2}}\cos x = \cos x. $$(M1) The left side is $(e^{x^{2}}y)'$. Integrate from $0$ to $x$:(M1) 左边为 $(e^{x^{2}}y)'$。从 $0$ 到 $x$ 积分：
$$ e^{x^{2}}y(x) - e^{0}y(0) = \int_{0}^{x}\cos t\,dt. $$(A1) Since $y(0) = 1$:(A1) 因 $y(0) = 1$：
$$ y(x) = e^{-x^{2}}\!\left(1 + \int_{0}^{x}\cos t\,dt\right). $$(A1) Check: $y(0) = e^{0}(1 + 0) = 1$. Correct.(A1) 验证：$y(0) = e^{0}(1 + 0) = 1$。正确。
Since $P(x) = 2x$ and $Q(x) = e^{-x^{2}}\cos x$ are both continuous on all of $\mathbb{R}$, the Picard theorem (or linear ODE theory) guarantees the solution exists on $(-\infty, \infty)$. (M1) The maximal interval of existence is $(-\infty, \infty)$. (A1)由于 $P(x) = 2x$ 和 $Q(x) = e^{-x^{2}}\cos x$ 在整个 $\mathbb{R}$ 上均连续，皮卡定理（或线性常微分方程理论）保证解在 $(-\infty, \infty)$ 上存在。(M1) 最大存在区间为 $(-\infty, \infty)$。(A1)
Evaluate the integral: $\int_{0}^{x}\cos t\,dt = \sin x - \sin 0 = \sin x$. So the closed form is: (M1)计算积分：$\int_{0}^{x}\cos t\,dt = \sin x - \sin 0 = \sin x$。故闭合形式为：(M1)
$$ y = e^{-x^{2}}(1 + \sin x). $$Verification: differentiate using the product rule:验证：用乘积法则求导：
$$ y' = -2x e^{-x^{2}}(1+\sin x) + e^{-x^{2}}\cos x. $$Then $y' + 2xy = -2xe^{-x^{2}}(1+\sin x) + e^{-x^{2}}\cos x + 2xe^{-x^{2}}(1+\sin x) = e^{-x^{2}}\cos x$. (A1) The ODE is satisfied exactly.则 $y' + 2xy = -2xe^{-x^{2}}(1+\sin x) + e^{-x^{2}}\cos x + 2xe^{-x^{2}}(1+\sin x) = e^{-x^{2}}\cos x$。(A1) 常微分方程恰好满足。
Tea at $95^{\circ}\text{C}$ in a $20^{\circ}\text{C}$ room cools to $65^{\circ}\text{C}$ after 10 min. (a) Derive $T(t) = 20 + 75e^{-kt}$; (b) find $k$ exactly and as a decimal; (c) find $t^*$ when $T = 40^{\circ}\text{C}$.$95^{\circ}\text{C}$ 的茶在 $20^{\circ}\text{C}$ 的房间中，10 分钟后冷却至 $65^{\circ}\text{C}$。(a) 推导 $T(t) = 20 + 75e^{-kt}$；(b) 精确求 $k$ 并给出近似值；(c) 求 $T = 40^{\circ}\text{C}$ 时的 $t^*$。
Let $u = T - 20$. Then $u' = T'$ and Newton's law becomes $u' = -ku$. (M1) This is separable with solution $u = Ae^{-kt}$. The initial condition $T(0) = 95$ gives $u(0) = 75$, so $A = 75$. (M1) Therefore:令 $u = T - 20$。则 $u' = T'$，牛顿冷却定律变为 $u' = -ku$。(M1) 此方程可分离，解为 $u = Ae^{-kt}$。初始条件 $T(0) = 95$ 给出 $u(0) = 75$，故 $A = 75$。(M1) 因此：
$$ T(t) = 20 + 75e^{-kt}. $$(A1)(A1)
Apply $T(10) = 65$: (M1)代入 $T(10) = 65$：(M1)
$$ 65 = 20 + 75e^{-10k} \implies 45 = 75e^{-10k} \implies e^{-10k} = \frac{45}{75} = \frac{3}{5}. $$(M1) Taking logarithms: $-10k = \ln(3/5)$, so:(M1) 取对数：$-10k = \ln(3/5)$，故：
$$ k = -\frac{1}{10}\ln\!\frac{3}{5} = \frac{1}{10}\ln\!\frac{5}{3} = \frac{\ln 5 - \ln 3}{10}. $$(A1) Decimal: $k \approx \tfrac{0.5108}{10} \approx 0.0511$ min$^{-1}$. (A1)(A1) 近似值：$k \approx \tfrac{0.5108}{10} \approx 0.0511$ min$^{-1}$。(A1)
Set $T(t^{*}) = 40$: (M1)令 $T(t^{*}) = 40$：(M1)
$$ 40 = 20 + 75e^{-kt^{*}} \implies e^{-kt^{*}} = \frac{20}{75} = \frac{4}{15}. $$(A1) Take logarithms: $-kt^{*} = \ln(4/15)$, so $t^{*} = \tfrac{\ln(15/4)}{k} = \tfrac{10\ln(15/4)}{\ln(5/3)}$. (A1) Numerically: $t^{*} \approx \tfrac{10 \times 1.3218}{0.5108} \approx 25.9$ min, so approximately $26$ minutes.(A1) 取对数：$-kt^{*} = \ln(4/15)$，故 $t^{*} = \tfrac{\ln(15/4)}{k} = \tfrac{10\ln(15/4)}{\ln(5/3)}$。(A1) 数值上：$t^{*} \approx \tfrac{10 \times 1.3218}{0.5108} \approx 25.9$ 分钟，约为 $26$ 分钟。
200 L tank, pure water initially, 3 g/L brine in at 4 L/min, well-mixed out at 4 L/min. (a) Write the ODE in standard form and state $A(0)$; (b) solve to get $A(t) = 600(1-e^{-t/50})$; (c) steady state, physical explanation, and time when concentration reaches 2 g/L.200 L 槽，初始为纯水，3 g/L 盐水以 4 L/min 流入，充分混合后以 4 L/min 流出。(a) 写出标准形式的常微分方程并给出 $A(0)$；(b) 求解得 $A(t) = 600(1-e^{-t/50})$；(c) 稳态，物理解释，以及浓度达到 2 g/L 的时刻。
Volume is constant at 200 L since inflow rate equals outflow rate. (M1) Rate in: $3\,\text{g/L} \times 4\,\text{L/min} = 12\,\text{g/min}$. Rate out: $\tfrac{A(t)}{200}\,\text{g/L} \times 4\,\text{L/min} = \tfrac{A}{50}\,\text{g/min}$. (M1) Therefore:由于流入速率等于流出速率，体积恒为 200 L。(M1) 流入速率：$3\,\text{g/L} \times 4\,\text{L/min} = 12\,\text{g/min}$。流出速率：$\tfrac{A(t)}{200}\,\text{g/L} \times 4\,\text{L/min} = \tfrac{A}{50}\,\text{g/min}$。(M1) 因此：
$$ A' = 12 - \frac{A}{50} \implies A' + \frac{1}{50}A = 12, \quad A(0) = 0. $$(A1)(A1)
$P(t) = 1/50$, so $\mu = e^{t/50}$. (M1) Multiply: $(e^{t/50}A)' = 12e^{t/50}$. Integrate:$P(t) = 1/50$，故 $\mu = e^{t/50}$。(M1) 乘以积分因子：$(e^{t/50}A)' = 12e^{t/50}$。积分：
$$ e^{t/50}A = 600e^{t/50} + C. $$(M1) Hence $A = 600 + Ce^{-t/50}$. Apply $A(0) = 0$: $0 = 600 + C$, so $C = -600$. (A1)(M1) 故 $A = 600 + Ce^{-t/50}$。代入 $A(0) = 0$：$0 = 600 + C$，故 $C = -600$。(A1)
$$ A(t) = 600\!\left(1 - e^{-t/50}\right). $$As $t \to \infty$, $e^{-t/50} \to 0$ and $A \to 600$ g. (M1) Physical explanation: at steady state, outflow concentration equals inflow concentration (3 g/L), so the tank holds $3\,\text{g/L} \times 200\,\text{L} = 600\,\text{g}$.当 $t \to \infty$ 时，$e^{-t/50} \to 0$，$A \to 600$ 克。(M1) 物理解释：稳态时，流出浓度等于流入浓度（3 g/L），故槽中含盐 $3\,\text{g/L} \times 200\,\text{L} = 600\,\text{g}$。
Concentration 2 g/L corresponds to $A = 2 \times 200 = 400$ g. Solve $600(1 - e^{-t/50}) = 400$: $e^{-t/50} = 1/3$, so $t = 50\ln 3 \approx 54.9 \approx 55$ min. (A1)浓度 2 g/L 对应 $A = 2 \times 200 = 400$ 克。解 $600(1 - e^{-t/50}) = 400$：$e^{-t/50} = 1/3$，故 $t = 50\ln 3 \approx 54.9 \approx 55$ 分钟。(A1)
For $dy/dx = x/(1-y^2)$, $y(0) = 0$: (a) separate and integrate to $y - y^3/3 = x^2/2$, noting excluded values; (b) explain why the solution cannot extend to all of $\mathbb{R}$ and find the maximal interval; (c) explain geometrically why the solution terminates before $y = 1$.对于 $dy/dx = x/(1-y^2)$，$y(0) = 0$：(a) 分离变量并积分得 $y - y^3/3 = x^2/2$，注明排除值；(b) 解释为何解不能延拓到整个 $\mathbb{R}$，并求最大存在区间；(c) 从几何角度解释为何解在 $y = 1$ 之前终止。
Separate variables (assuming $y \ne \pm 1$): $(1 - y^{2})\,dy = x\,dx$. (M1) Integrate both sides:分离变量（假设 $y \ne \pm 1$）：$(1 - y^{2})\,dy = x\,dx$。(M1) 两边积分：
$$ \int(1 - y^{2})\,dy = \int x\,dx \implies y - \frac{y^{3}}{3} = \frac{x^{2}}{2} + C. $$(M1) Apply $y(0) = 0$: $0 - 0 = 0 + C$, so $C = 0$. (A1) The implicit solution is:(M1) 代入 $y(0) = 0$：$0 - 0 = 0 + C$，故 $C = 0$。(A1) 隐式解为：
$$ y - \frac{y^{3}}{3} = \frac{x^{2}}{2}. $$(A1) The values $y = \pm 1$ were excluded when we divided by $1 - y^{2}$; they must be checked separately. Substituting $y = 1$: $1 - 1/3 = 2/3 \ne x^{2}/2$ for $x = 0$, so $y \equiv 1$ does not satisfy the initial condition.(A1) 我们除以 $1 - y^{2}$ 时排除了 $y = \pm 1$；须单独检验。代入 $y = 1$：$1 - 1/3 = 2/3 \ne x^{2}/2$（$x = 0$ 时），故 $y \equiv 1$ 不满足初始条件。
By the Picard theorem, since $f = x/(1-y^{2})$ and $\partial f/\partial y = 2xy/(1-y^{2})^{2}$ are continuous near $(0,0)$ (where $y = 0 \ne \pm 1$), a unique local solution exists. (M1)由皮卡定理，由于 $f = x/(1-y^{2})$ 和 $\partial f/\partial y = 2xy/(1-y^{2})^{2}$ 在 $(0,0)$ 附近连续（此处 $y = 0 \ne \pm 1$），唯一局部解存在。(M1)
The implicit function theorem requires $\partial/\partial y(y - y^3/3) = 1 - y^2 \ne 0$ to solve for $y$ as a function of $x$; this fails when $y = \pm 1$. As $y$ increases from $0$ (since $y' = x/(1-y^2) > 0$ for $x > 0$ near $0$), the solution reaches $y = 1$ when $1 - 1/3 = x^{2}/2$, i.e. $x^{2} = 4/3$, so $x = 2/\sqrt{3}$. (M1·A1) The solution is valid on the interval $(-2/\sqrt{3},\,2/\sqrt{3})$; beyond this $x$-range the solution cannot be continued as a single-valued function.隐函数定理要求 $\partial/\partial y(y - y^3/3) = 1 - y^2 \ne 0$ 以将 $y$ 表示为 $x$ 的函数；当 $y = \pm 1$ 时此条件失效。由于 $y$ 从 $0$ 开始递增（$x > 0$ 且 $x$ 接近 $0$ 时 $y' = x/(1-y^2) > 0$），解在 $1 - 1/3 = x^{2}/2$ 即 $x^{2} = 4/3$ 时达到 $y = 1$，故 $x = 2/\sqrt{3}$。(M1·A1) 解在区间 $(-2/\sqrt{3},\,2/\sqrt{3})$ 上有效；超出此 $x$ 范围，解无法作为单值函数延续。
The right-hand side $f(x,y) = x/(1-y^{2})$ is undefined on the horizontal lines $y = 1$ and $y = -1$. (M1) On these lines, the slope field has no defined slope; approaching these lines from below, the slope $y' \to +\infty$ (for $x > 0$), so the solution curve becomes arbitrarily steep. (A1) A smooth solution cannot cross a line of vertical tangents: the direction field is singular there, and the curve must terminate at or before $y = 1$. (R1)右端函数 $f(x,y) = x/(1-y^{2})$ 在水平线 $y = 1$ 和 $y = -1$ 上无定义。(M1) 在这些直线上，斜率场无定义斜率；从下方趋近这些直线时，斜率 $y' \to +\infty$（$x > 0$ 时），故解曲线变得任意陡峭。(A1) 光滑解不能穿过竖直切线所在的直线：方向场在那里奇异，曲线必须在到达 $y = 1$ 之前或恰好在该处终止。(R1)