Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: exact equations, integrating factors, Bernoulli and homogeneous substitution, autonomous equations, logistic growth, mixing models, Newton cooling第 1 至 7 节：恰当方程、积分因子、Bernoulli 换元与齐次换元、自治方程、逻辑斯谛增长、混合模型、牛顿冷却定律CALC IV
Test exactness and find the general solution for: (a) $(y^2+2xy)\,dx+(2xy+x^2)\,dy=0$; (b) $(3x^2y)\,dx+(x^3+2y)\,dy=0$; (c) show $(2xy+y^2)\,dx+(x^2-y)\,dy=0$ is not exact.验证恰当性并求通解：(a) $(y^2+2xy)\,dx+(2xy+x^2)\,dy=0$；(b) $(3x^2y)\,dx+(x^3+2y)\,dy=0$；(c) 证明 $(2xy+y^2)\,dx+(x^2-y)\,dy=0$ 不是恰当方程。
Set $M=y^2+2xy$ and $N=2xy+x^2$. Compute the mixed partials: $M_y=2y+2x$ and $N_x=2y+2x$. (M1) Since $M_y=N_x$ on all of $\mathbb{R}^2$, the equation is exact.令 $M=y^2+2xy$，$N=2xy+x^2$。计算混合偏导数：$M_y=2y+2x$，$N_x=2y+2x$。(M1) 由于在整个 $\mathbb{R}^2$ 上 $M_y=N_x$，该方程是恰当方程。
Integrate $M$ in $x$ (holding $y$ fixed): $F=\displaystyle\int(y^2+2xy)\,dx=xy^2+x^2y+g(y)$. Differentiate in $y$ and match $F_y=N$: $2xy+x^2+g'(y)=2xy+x^2$, so $g'(y)=0$ and $g$ is a constant absorbed into $C$. (A1)对 $M$ 关于 $x$ 积分（固定 $y$）：$F=\displaystyle\int(y^2+2xy)\,dx=xy^2+x^2y+g(y)$。对 $y$ 求偏导并匹配 $F_y=N$：$2xy+x^2+g'(y)=2xy+x^2$，故 $g'(y)=0$，$g$ 为常数可并入 $C$。(A1)
$$ F(x,y)=xy^2+x^2y=C. $$Verification: $\nabla F=(y^2+2xy,\,2xy+x^2)=(M,N)$. (A1)验证：$\nabla F=(y^2+2xy,\,2xy+x^2)=(M,N)$。(A1)
$M=3x^2y$, $M_y=3x^2$; $N=x^3+2y$, $N_x=3x^2$. Equal, so exact. (M1)$M=3x^2y$，$M_y=3x^2$；$N=x^3+2y$，$N_x=3x^2$。二者相等，故为恰当方程。(M1)
$F=\displaystyle\int 3x^2y\,dx=x^3y+g(y)$. Then $F_y=x^3+g'(y)=x^3+2y$, giving $g'(y)=2y$ and $g(y)=y^2$. (A1)$F=\displaystyle\int 3x^2y\,dx=x^3y+g(y)$。则 $F_y=x^3+g'(y)=x^3+2y$，得 $g'(y)=2y$，$g(y)=y^2$。(A1)
$$ F(x,y)=x^3y+y^2=C. $$Verification: $F_x=3x^2y=M$ and $F_y=x^3+2y=N$. (A1)验证：$F_x=3x^2y=M$，$F_y=x^3+2y=N$。(A1)
$M=2xy+y^2$, $M_y=2x+2y$; $N=x^2-y$, $N_x=2x$. (M1) Then$M=2xy+y^2$，$M_y=2x+2y$；$N=x^2-y$，$N_x=2x$。(M1) 则
$$ M_y-N_x=(2x+2y)-2x=2y\ne 0 $$in general, so the equation is not exact. (A1) The non-zero obstruction $2y$ signals that no potential exists as written; an integrating factor would be needed.一般情况下不为零，故该方程不是恰当方程。(A1) 非零障碍项 $2y$ 说明如原式所写不存在势函数，需引入积分因子。
Solve $(3x^2y+2xy+y^3)\,dx+(x^3+x^2+3xy^2)\,dy=0$, $y(1)=2$. Integrate $N$ in $y$ first.求解 $(3x^2y+2xy+y^3)\,dx+(x^3+x^2+3xy^2)\,dy=0$，$y(1)=2$。先对 $N$ 关于 $y$ 积分。
Set $M=3x^2y+2xy+y^3$ and $N=x^3+x^2+3xy^2$. Differentiate: (M1)令 $M=3x^2y+2xy+y^3$，$N=x^3+x^2+3xy^2$。求偏导数：(M1)
$$ M_y=3x^2+2x+3y^2, \qquad N_x=3x^2+2x+3y^2. $$They agree everywhere, so the equation is exact. (A1)二者处处相等，故该方程是恰当方程。(A1)
Integrate $N=x^3+x^2+3xy^2$ in $y$ (holding $x$ fixed): (M1)对 $N=x^3+x^2+3xy^2$ 关于 $y$ 积分（固定 $x$）：(M1)
$$ F=\int N\,dy=x^3y+x^2y+xy^3+h(x). $$(M1 for carrying out the integration correctly) Differentiate $F$ in $x$ and match $F_x=M$: (M1)（M1 为正确完成积分）对 $F$ 关于 $x$ 求偏导并匹配 $F_x=M$：(M1)
$$ F_x=3x^2y+2xy+y^3+h'(x)=M=3x^2y+2xy+y^3. $$Therefore $h'(x)=0$, so $h$ is a constant absorbed into $C$. (A1) The potential is $F(x,y)=x^3y+x^2y+xy^3$. (A1)故 $h'(x)=0$，$h$ 为常数可并入 $C$。(A1) 势函数为 $F(x,y)=x^3y+x^2y+xy^3$。(A1)
Verification: $F_y=x^3+x^2+3xy^2=N$ and $F_x=3x^2y+2xy+y^3=M$. Both match.验证：$F_y=x^3+x^2+3xy^2=N$，$F_x=3x^2y+2xy+y^3=M$，两者均吻合。
The general solution is $x^3y+x^2y+xy^3=C$. Apply $y(1)=2$: (M1)通解为 $x^3y+x^2y+xy^3=C$。代入 $y(1)=2$：(M1)
$$ F(1,2)=1\cdot2+1\cdot2+1\cdot8=12. $$(A1) The particular solution is (A1)(A1) 特解为 (A1)
$$ x^3y+x^2y+xy^3=12. $$$(3xy+y^2)\,dx+(x^2+xy)\,dy=0$: (a) show not exact and compute $(M_y-N_x)/N$; (b) find $\mu(x)$ and verify new exactness; (c) solve.$(3xy+y^2)\,dx+(x^2+xy)\,dy=0$：(a) 证明非恰当并计算 $(M_y-N_x)/N$；(b) 求 $\mu(x)$ 并验证新方程的恰当性；(c) 求解。
$M=3xy+y^2$, $M_y=3x+2y$; $N=x^2+xy$, $N_x=2x+y$. (M1) The obstruction is$M=3xy+y^2$，$M_y=3x+2y$；$N=x^2+xy$，$N_x=2x+y$。(M1) 障碍项为
$$ M_y-N_x=(3x+2y)-(2x+y)=x+y. $$Since $M_y\ne N_x$, the equation is not exact. (A1) Divide by $N$:由于 $M_y\ne N_x$，该方程不是恰当方程。(A1) 除以 $N$：
$$ \frac{M_y-N_x}{N}=\frac{x+y}{x^2+xy}=\frac{x+y}{x(x+y)}=\frac{1}{x}, $$which depends on $x$ alone, so an integrating factor $\mu(x)$ exists. (A1)结果仅依赖于 $x$，故存在积分因子 $\mu(x)$。(A1)
$\mu=e^{\int(1/x)\,dx}=e^{\ln x}=x$. (M1·A1) Multiply through by $\mu=x$:$\mu=e^{\int(1/x)\,dx}=e^{\ln x}=x$。(M1·A1) 方程两边乘以 $\mu=x$：
$$ \tilde{M}=x(3xy+y^2)=3x^2y+xy^2, \qquad \tilde{N}=x(x^2+xy)=x^3+x^2y. $$Check: $\tilde{M}_y=3x^2+2xy$ and $\tilde{N}_x=3x^2+2xy$. (M1) Equal, so the multiplied equation is exact. (A1)验证：$\tilde{M}_y=3x^2+2xy$，$\tilde{N}_x=3x^2+2xy$。(M1) 二者相等，故乘以积分因子后的方程是恰当方程。(A1)
Integrate $\tilde{N}=x^3+x^2y$ in $y$: (M1)对 $\tilde{N}=x^3+x^2y$ 关于 $y$ 积分：(M1)
$$ F=\int(x^3+x^2y)\,dy=x^3y+\tfrac{1}{2}x^2y^2+h(x). $$Match $F_x=3x^2y+xy^2+h'(x)=\tilde{M}=3x^2y+xy^2$, so $h'(x)=0$ and $h$ is constant. (A1) The general solution is匹配 $F_x=3x^2y+xy^2+h'(x)=\tilde{M}=3x^2y+xy^2$，故 $h'(x)=0$，$h$ 为常数。(A1) 通解为
$$ x^3y+\tfrac{1}{2}x^2y^2=C, \quad\text{or equivalently}\quad 2x^3y+x^2y^2=C. $$Verification: $\nabla F=(3x^2y+xy^2,\,x^3+x^2y)=(\tilde{M},\tilde{N})$, confirming the potential. (A1)验证：$\nabla F=(3x^2y+xy^2,\,x^3+x^2y)=(\tilde{M},\tilde{N})$，势函数正确。(A1)
(a) Prove $M_y=N_x$ is necessary for exactness using $F_{xy}=F_{yx}$. (b) Show $g'(y)=N-\partial_y\int M\,dx$ is independent of $x$ when $M_y=N_x$, so $F$ can be constructed.(a) 利用 $F_{xy}=F_{yx}$ 证明 $M_y=N_x$ 是方程为恰当方程的必要条件。(b) 在 $M_y=N_x$ 时，证明 $g'(y)=N-\partial_y\int M\,dx$ 与 $x$ 无关，从而 $F$ 可以被构造。
Suppose $F$ exists with $F_x=M$ and $F_y=N$. Since $M$ and $N$ have continuous second partials, $F$ has continuous second partials, and Clairaut's theorem guarantees $F_{xy}=F_{yx}$. (M1) Therefore设 $F$ 存在，且 $F_x=M$，$F_y=N$。由于 $M$ 和 $N$ 具有连续的二阶偏导数，$F$ 也具有连续的二阶偏导数，Clairaut 定理保证 $F_{xy}=F_{yx}$。(M1) 因此
$$ M_y=\frac{\partial M}{\partial y}=\frac{\partial F_x}{\partial y}=F_{xy}=F_{yx}=\frac{\partial F_y}{\partial x}=\frac{\partial N}{\partial x}=N_x. $$(A1) Hence $M_y=N_x$ is a necessary condition: if a potential exists, the mixed partials must agree. (R1)(A1) 故 $M_y=N_x$ 是必要条件：若势函数存在，则混合偏导数必须相等。(R1)
Now assume $M_y=N_x$ on a simply connected region. Integrate $M$ in $x$ (holding $y$ fixed): (M1)现设 $M_y=N_x$ 在单连通区域上成立。对 $M$ 关于 $x$ 积分（固定 $y$）：(M1)
$$ F(x,y)=\int M(x,y)\,dx+g(y). $$The unknown $g(y)$ arises because $y$ was held fixed during integration. Differentiate $F$ in $y$ and require $F_y=N$: (M1)未知函数 $g(y)$ 的出现是因为积分时 $y$ 被固定。对 $F$ 关于 $y$ 求偏导并要求 $F_y=N$：(M1)
$$ F_y=\frac{\partial}{\partial y}\int M\,dx+g'(y)=N \implies g'(y)=N-\frac{\partial}{\partial y}\int M\,dx. $$To integrate $g'(y)$ in $y$ we must confirm the right side does not depend on $x$. Differentiate with respect to $x$: (A1)为了对 $g'(y)$ 关于 $y$ 积分，须确认右侧不依赖于 $x$。对 $x$ 求偏导：(A1)
$$ \frac{\partial}{\partial x}g'(y)=\frac{\partial N}{\partial x}-\frac{\partial}{\partial x}\frac{\partial}{\partial y}\int M\,dx=N_x-\frac{\partial^2}{\partial x\,\partial y}\int M\,dx=N_x-M_y=0, $$where the last step uses the hypothesis $M_y=N_x$ and the fact that differentiation and integration interchange (by continuity). (R1) Since $\partial_x(g'(y))=0$, the expression for $g'$ truly depends only on $y$, so $g(y)=\displaystyle\int g'(y)\,dy$ is well-defined by a single integration in $y$, and $F$ is constructed. (A1)最后一步利用了假设 $M_y=N_x$ 以及微分与积分可交换（由连续性保证）。(R1) 由于 $\partial_x(g'(y))=0$，$g'$ 的表达式确实仅依赖于 $y$，故 $g(y)=\displaystyle\int g'(y)\,dy$ 由单次关于 $y$ 的积分确定，$F$ 的构造完成。(A1)
$y'+\frac{1}{x}y=xy^3$, $y(1)=1$, $x>0$: (a) derive the linear $v$-equation from $v=y^{-2}$; (b) solve by integrating factor; (c) revert to $y$ and find the particular solution with its domain.$y'+\frac{1}{x}y=xy^3$，$y(1)=1$，$x>0$：(a) 由 $v=y^{-2}$ 推导关于 $v$ 的线性方程；(b) 用积分因子法求解；(c) 回代得 $y$ 并求特解及其定义域。
Here $n=3$. Divide the equation $y'+\tfrac{1}{x}y=xy^3$ through by $y^3$ (assuming $y\ne0$): (M1)此处 $n=3$。方程 $y'+\tfrac{1}{x}y=xy^3$ 两边除以 $y^3$（设 $y\ne0$）：(M1)
$$ y^{-3}y'+\frac{1}{x}y^{-2}=x. $$Set $v=y^{1-n}=y^{-2}$. By the chain rule, $v'=-2y^{-3}y'$, so $y^{-3}y'=v'/(-2)$. (M1) Substitute: $v'/(-2)+(1/x)v=x$, which gives (A1)令 $v=y^{1-n}=y^{-2}$。由链式法则，$v'=-2y^{-3}y'$，故 $y^{-3}y'=v'/(-2)$。(M1) 代入：$v'/(-2)+(1/x)v=x$，得 (A1)
$$ v'-\frac{2}{x}v=-2x. $$This is linear in $v$; the factor $(1-n)=1-3=-2$ appears both as the coefficient of the $p$-term and on the right side, as the Bernoulli formula predicts. (A1)该方程关于 $v$ 是线性的；因子 $(1-n)=1-3=-2$ 同时出现在 $p$ 项的系数和右侧，与 Bernoulli 公式的预测一致。(A1)
The equation $v'-(2/x)v=-2x$ has $p=-2/x$. The integrating factor is (M1)方程 $v'-(2/x)v=-2x$ 中 $p=-2/x$。积分因子为 (M1)
$$ \mu=e^{\int(-2/x)\,dx}=e^{-2\ln x}=x^{-2}. $$Multiply through by $x^{-2}$: $\displaystyle\frac{d}{dx}(x^{-2}v)=x^{-2}\cdot(-2x)=-\frac{2}{x}$. (A1) Integrate both sides: (M1)方程两边乘以 $x^{-2}$：$\displaystyle\frac{d}{dx}(x^{-2}v)=x^{-2}\cdot(-2x)=-\frac{2}{x}$。(A1) 对两边积分：(M1)
$$ x^{-2}v=\int-\frac{2}{x}\,dx=-2\ln x+C. $$Therefore $v=x^2(C-2\ln x)$. (A1)故 $v=x^2(C-2\ln x)$。(A1)
Since $v=y^{-2}$, we have $y^{-2}=x^2(C-2\ln x)$, so $y^2=\dfrac{1}{x^2(C-2\ln x)}$. (M1) Apply $y(1)=1$: $1=\dfrac{1}{1^2(C-0)}$ gives $C=1$. Therefore由 $v=y^{-2}$，得 $y^{-2}=x^2(C-2\ln x)$，故 $y^2=\dfrac{1}{x^2(C-2\ln x)}$。(M1) 代入 $y(1)=1$：$1=\dfrac{1}{1^2(C-0)}$，得 $C=1$。因此
$$ y(x)=\frac{1}{x\sqrt{1-2\ln x}}. $$The solution is valid where $1-2\ln x>0$, i.e. $\ln x<\tfrac12$, i.e. $x
$y'=y(y-2)(4-y)$: (a) find and classify equilibria via $f'(y^*)$; (b) draw a phase line; (c) give $\lim_{t\to\infty}y(t)$ for $y(0)=3$.$y'=y(y-2)(4-y)$：(a) 用 $f'(y^*)$ 求平衡点并分类；(b) 画相线；(c) 给出 $y(0)=3$ 时 $\lim_{t\to\infty}y(t)$ 的值。
Set $f(y)=y(y-2)(4-y)=0$. The equilibria are $y^*=0$, $y^*=2$, and $y^*=4$. (M1)令 $f(y)=y(y-2)(4-y)=0$。平衡点为 $y^*=0$、$y^*=2$ 和 $y^*=4$。(M1)
Expand to differentiate: $f(y)=y(y-2)(4-y)=y(4y-y^2-8+2y)=6y^2-y^3-8y$, so $f'(y)=12y-3y^2-8$. (M1) Evaluate at each equilibrium: (A1)展开后求导：$f(y)=y(y-2)(4-y)=y(4y-y^2-8+2y)=6y^2-y^3-8y$，故 $f'(y)=12y-3y^2-8$。(M1) 在各平衡点处求值：(A1)
$$ f'(0)=-8<0\;(\text{stable}), \quad f'(2)=24-12-8=4>0\;(\text{unstable}), \quad f'(4)=48-48-8=-8<0\;(\text{stable}). $$So $y^*=0$ and $y^*=4$ are sinks; $y^*=2$ is a source. (A1)故 $y^*=0$ 和 $y^*=4$ 是稳定平衡点（汇）；$y^*=2$ 是不稳定平衡点（源）。(A1)
Check the sign of $f$ on each interval using a test point: (M1)用测试点检验 $f$ 在各区间上的符号：(M1)
Phase line: arrows converge toward $0$ from $(-\infty,0)\cup(0,2)$ (note: from below $0$ arrows go up toward $0$, from above $0$ until $2$ arrows go down toward $0$), and arrows converge toward $4$ from $(2,4)\cup(4,\infty)$. The unstable equilibrium $y^*=2$ separates the two basins of attraction. (A1)相线：来自 $(-\infty,0)\cup(0,2)$ 的箭头收敛于 $0$（注意：从 $0$ 以下箭头向上指向 $0$，从 $0$ 到 $2$ 之间箭头向下指向 $0$），来自 $(2,4)\cup(4,\infty)$ 的箭头收敛于 $4$。不稳定平衡点 $y^*=2$ 将两个吸引盆分隔开。(A1)
Since $y(0)=3\in(2,4)$ and $f>0$ on this interval, $y$ is increasing and cannot cross the equilibrium $y^*=4$ (by uniqueness). (M1) All solutions starting in $(2,4)$ are attracted to the stable equilibrium $y^*=4$, so由于 $y(0)=3\in(2,4)$ 且 $f$ 在该区间上 $f>0$，$y$ 单调递增且不能越过平衡点 $y^*=4$（由唯一性保证）。(M1) 所有从 $(2,4)$ 出发的解均被稳定平衡点 $y^*=4$ 吸引，故
$$ \lim_{t\to\infty}y(t)=4. $$(A1)(A1)
$P'=0.4P(1-P/500)$, $P(0)=50$: (a) state $K$, $k$, equilibria and classify them; (b) separate variables and derive $P/(1-P/K)=Ae^{kt}$; (c) solve for $P(t)$; (d) find $\lim_{t\to\infty}P(t)$ and confirm it equals the stable equilibrium.$P'=0.4P(1-P/500)$，$P(0)=50$：(a) 写出 $K$、$k$、平衡点并分类；(b) 分离变量并推导 $P/(1-P/K)=Ae^{kt}$；(c) 求 $P(t)$；(d) 求 $\lim_{t\to\infty}P(t)$ 并确认其等于稳定平衡点。
Reading off the equation: carrying capacity $K=500$ and intrinsic growth rate $k=0.4$. Setting $f(P)=0.4P(1-P/500)=0$ gives equilibria $P=0$ (unstable, since $f'(0)=0.4>0$) and $P=500$ (stable, since $f'(500)=-0.4<0$). (A1)从方程中读出：环境容量 $K=500$，固有增长率 $k=0.4$。令 $f(P)=0.4P(1-P/500)=0$，得平衡点 $P=0$（不稳定，因为 $f'(0)=0.4>0$）和 $P=500$（稳定，因为 $f'(500)=-0.4<0$）。(A1)
Rewrite: $\dfrac{dP}{P(1-P/500)}=0.4\,dt$. (M1) Decompose the left integrand by partial fractions, writing $\dfrac{1}{P(1-P/500)}=\dfrac{1}{P}+\dfrac{1/500}{1-P/500}$ (verified by common denominator): (M1)改写为：$\dfrac{dP}{P(1-P/500)}=0.4\,dt$。(M1) 用部分分式分解左侧被积函数，写成 $\dfrac{1}{P(1-P/500)}=\dfrac{1}{P}+\dfrac{1/500}{1-P/500}$（通分可验证）：(M1)
$$ \int\left(\frac{1}{P}+\frac{1/500}{1-P/500}\right)dP=\int0.4\,dt. $$ $$ \ln|P|-\ln\!\left|1-\frac{P}{500}\right|=0.4t+C_1. $$(A1) Exponentiate: $\dfrac{P}{1-P/500}=e^{C_1}e^{0.4t}=A\,e^{0.4t}$. (A1)(A1) 取指数：$\dfrac{P}{1-P/500}=e^{C_1}e^{0.4t}=A\,e^{0.4t}$。(A1)
At $t=0$: $A=\dfrac{P_0}{1-P_0/K}=\dfrac{50}{1-50/500}=\dfrac{50}{0.9}=\dfrac{500}{9}$. (M1) Solve $\dfrac{P}{1-P/500}=\dfrac{500}{9}e^{0.4t}$ for $P$: cross-multiply and rearrange,当 $t=0$ 时：$A=\dfrac{P_0}{1-P_0/K}=\dfrac{50}{1-50/500}=\dfrac{50}{0.9}=\dfrac{500}{9}$。(M1) 从 $\dfrac{P}{1-P/500}=\dfrac{500}{9}e^{0.4t}$ 中解出 $P$：交叉相乘并整理，
$$ P=\frac{K P_0}{P_0+(K-P_0)e^{-kt}}=\frac{500\cdot50}{50+450\,e^{-0.4t}}=\frac{500}{1+9e^{-0.4t}}. $$(A1) Verification at $t=0$: $P=500/(1+9)=50$. Correct.(A1) 验证 $t=0$ 时：$P=500/(1+9)=50$，正确。
As $t\to\infty$, $e^{-0.4t}\to0$, so $P(t)\to500=K$. (A1) This confirms $P=500$ is the stable equilibrium: any positive initial population is attracted to the carrying capacity, consistent with the linearised stability finding $f'(500)=-0.4<0$.当 $t\to\infty$ 时，$e^{-0.4t}\to0$，故 $P(t)\to500=K$。(A1) 这证实 $P=500$ 是稳定平衡点：任何正的初始种群都被环境容量吸引，与线性化稳定性分析结果 $f'(500)=-0.4<0$ 一致。
Tank: 200 L pure water, inflow 0.05 kg/L at 4 L/min, outflow 4 L/min. $A(t)$ = salt mass. (a) set up $A'+pA=q$; (b) solve with $A(0)=0$; (c) find $A(30)$ and $\lim_{t\to\infty}A(t)$.水槽：200 L 纯水，0.05 kg/L 盐水以 4 L/min 流入，4 L/min 流出。$A(t)$ 为盐的质量。(a) 建立 $A'+pA=q$；(b) 以 $A(0)=0$ 求解；(c) 求 $A(30)$ 和 $\lim_{t\to\infty}A(t)$。
Apply the balance law: $A'=(\text{rate in})-(\text{rate out})$. Rate in: $0.05\times4=0.2\,\text{kg/min}$. Rate out: $(A/200)\times4=A/50\,\text{kg/min}$. (M1) Therefore应用守恒律：$A'=(\text{流入速率})-(\text{流出速率})$。流入速率：$0.05\times4=0.2\,\text{kg/min}$。流出速率：$(A/200)\times4=A/50\,\text{kg/min}$。(M1) 因此
$$ A'+\frac{1}{50}A=0.2, \quad p=\frac{1}{50},\quad q=0.2. \quad\text{(A1)}$$$\mu=e^{t/50}$. Multiply: $(e^{t/50}A)'=0.2e^{t/50}$. (M1) Integrate: $e^{t/50}A=10e^{t/50}+C$, so $A=10+Ce^{-t/50}$. Apply $A(0)=0$: $C=-10$. (A1)$\mu=e^{t/50}$。两边乘以 $\mu$：$(e^{t/50}A)'=0.2e^{t/50}$。(M1) 积分：$e^{t/50}A=10e^{t/50}+C$，故 $A=10+Ce^{-t/50}$。代入 $A(0)=0$：$C=-10$。(A1)
$$ A(t)=10\!\left(1-e^{-t/50}\right). $$Verification: $A'=0.2e^{-t/50}$ and $0.2-A/50=0.2e^{-t/50}$. Matches.验证：$A'=0.2e^{-t/50}$，$0.2-A/50=0.2e^{-t/50}$，两者吻合。
$A(30)=10(1-e^{-30/50})=10(1-e^{-0.6})\approx10(1-0.5488)=10\times0.4512\approx4.51\,\text{kg}$. (A1)$A(30)=10(1-e^{-30/50})=10(1-e^{-0.6})\approx10(1-0.5488)=10\times0.4512\approx4.51\,\text{kg}$。(A1)
$\displaystyle\lim_{t\to\infty}A(t)=10\,\text{kg}$. (A1) Physical interpretation: at steady state the salt concentration in the tank equals $10/200=0.05\,\text{kg/L}$, which is exactly the inflow concentration. The tank equilibrates to the incoming brine strength, as the balance law requires (rate in equals rate out at steady state).$\displaystyle\lim_{t\to\infty}A(t)=10\,\text{kg}$。(A1) 物理解释：稳态时槽内盐的浓度为 $10/200=0.05\,\text{kg/L}$，恰好等于流入浓度。槽内浓度趋近于流入盐水的浓度，这与守恒律的要求一致（稳态时流入速率等于流出速率）。
Coffee cools from 90°C, room at 20°C; after 10 min it is 70°C. (a) solve for $T(t)$; (b) find $k$; (c) find time to reach 50°C; (d) state $\lim_{t\to\infty}T(t)$ and link to autonomous stability.咖啡从 90°C 冷却，室温 20°C；10 分钟后降至 70°C。(a) 求 $T(t)$；(b) 求 $k$；(c) 求降至 50°C 所需时间；(d) 写出 $\lim_{t\to\infty}T(t)$ 并联系自治方程稳定性。
Let $u=T-20$; then $u'=-ku$, giving $u=70e^{-kt}$. (M1)令 $u=T-20$，则 $u'=-ku$，得 $u=70e^{-kt}$。(M1)
$$ T(t)=20+70\,e^{-kt}. $$Verification: $T(0)=20+70=90$. (A1)验证：$T(0)=20+70=90$。(A1)
$70=20+70e^{-10k}$, so $e^{-10k}=5/7$ and (M1)$70=20+70e^{-10k}$，故 $e^{-10k}=5/7$，(M1)
$$ k=\frac{1}{10}\ln\frac{7}{5}. $$Numerical value: $k\approx0.03365\,\text{min}^{-1}$. (A1)数值：$k\approx0.03365\,\text{min}^{-1}$。(A1)
Set $T(t^*)=50$: $50=20+70e^{-kt^*}$, so $30=70e^{-kt^*}$ and $e^{-kt^*}=\dfrac{3}{7}$. (M1) Taking the natural log:令 $T(t^*)=50$：$50=20+70e^{-kt^*}$，故 $30=70e^{-kt^*}$，$e^{-kt^*}=\dfrac{3}{7}$。(M1) 取自然对数：
$$ kt^*=\ln\frac{7}{3} \implies t^*=\frac{\ln(7/3)}{k}=\frac{10\ln(7/3)}{\ln(7/5)}. $$Numerically: $\ln(7/3)\approx0.8473$, $\ln(7/5)\approx0.3365$, so $t^*\approx8.473/0.3365\approx25.2\,\text{min}$. (A1)数值计算：$\ln(7/3)\approx0.8473$，$\ln(7/5)\approx0.3365$，故 $t^*\approx8.473/0.3365\approx25.2\,\text{min}$。(A1)
Since $k>0$, $e^{-kt}\to0$ as $t\to\infty$, so $\displaystyle\lim_{t\to\infty}T(t)=20=T_a$. (A1)由于 $k>0$，当 $t\to\infty$ 时 $e^{-kt}\to0$，故 $\displaystyle\lim_{t\to\infty}T(t)=20=T_a$。(A1)
This matches the autonomous stability analysis: rewrite $T'=-k(T-20)=f(T)$. The equilibrium is $T^*=20$, and $f'(T^*)=-k<0$, so $T^*$ is a stable equilibrium (a sink). Any initial temperature is attracted to room temperature, which is exactly what the formula $T=20+70e^{-kt}$ confirms. (A1)这与自治方程的稳定性分析一致：改写为 $T'=-k(T-20)=f(T)$。平衡点为 $T^*=20$，$f'(T^*)=-k<0$，故 $T^*$ 是稳定平衡点（汇）。任何初始温度都被室温吸引，这正是公式 $T=20+70e^{-kt}$ 所证实的。(A1)
$2xy\,dx+(y^2-x^2)\,dy=0$: (a) find $\mu(y)$ via $(N_x-M_y)/M$; (b) multiply, verify exactness, solve; (c) confirm the same family via the homogeneous substitution $w=x/y$.$2xy\,dx+(y^2-x^2)\,dy=0$：(a) 通过 $(N_x-M_y)/M$ 求 $\mu(y)$；(b) 乘以积分因子，验证恰当性，求解；(c) 通过齐次换元 $w=x/y$ 确认相同的曲线族。
$M=2xy$, $M_y=2x$; $N=y^2-x^2$, $N_x=-2x$. Since $M_y\ne N_x$, not exact. Test the $y$-only formula: (M1)$M=2xy$，$M_y=2x$；$N=y^2-x^2$，$N_x=-2x$。由于 $M_y\ne N_x$，非恰当方程。检验仅关于 $y$ 的公式：(M1)
$$ \frac{N_x-M_y}{M}=\frac{-4x}{2xy}=\frac{-2}{y}. $$Depends only on $y$, so $\mu=e^{\int(-2/y)\,dy}=y^{-2}$. (A1)仅依赖于 $y$，故 $\mu=e^{\int(-2/y)\,dy}=y^{-2}$。(A1)
Multiply by $y^{-2}$: $\tilde{M}=2x/y$, $\tilde{N}=1-x^2/y^2$. Check: $\tilde{M}_y=-2x/y^2=\tilde{N}_x$. Exact. (M1) Integrate $\tilde{M}$ in $x$: $F=x^2/y+g(y)$. Match $F_y=1-x^2/y^2$ gives $g'(y)=1$, $g=y$. (A1)两边乘以 $y^{-2}$：$\tilde{M}=2x/y$，$\tilde{N}=1-x^2/y^2$。验证：$\tilde{M}_y=-2x/y^2=\tilde{N}_x$，恰当方程。(M1) 对 $\tilde{M}$ 关于 $x$ 积分：$F=x^2/y+g(y)$。匹配 $F_y=1-x^2/y^2$ 得 $g'(y)=1$，$g=y$。(A1)
$$ x^2+y^2=Cy. $$Rewrite the equation treating $x$ as a function of $y$: $dx/dy=-(y^2-x^2)/(2xy)=(x^2-y^2)/(2xy)$. This is a homogeneous equation with the ratio $w=x/y$. Set $x=wy$, so $dx/dy=w+y\,dw/dy$: (M1)将方程改写为以 $x$ 作为 $y$ 的函数：$dx/dy=-(y^2-x^2)/(2xy)=(x^2-y^2)/(2xy)$。这是含比值 $w=x/y$ 的齐次方程。令 $x=wy$，则 $dx/dy=w+y\,dw/dy$：(M1)
$$ w+y\frac{dw}{dy}=\frac{w^2y^2-y^2}{2wy^2}=\frac{w^2-1}{2w} \implies y\frac{dw}{dy}=\frac{w^2-1}{2w}-w=\frac{-w^2-1}{2w}. $$Separate: $\dfrac{2w}{w^2+1}\,dw=-\dfrac{dy}{y}$. Integrate: $\ln(w^2+1)=-\ln y+C_1$. Substitute $w=x/y$:分离变量：$\dfrac{2w}{w^2+1}\,dw=-\dfrac{dy}{y}$。积分：$\ln(w^2+1)=-\ln y+C_1$。代入 $w=x/y$：
$$ \ln\!\left(\frac{x^2}{y^2}+1\right)+\ln y=C_1 \implies \ln\!\left(\frac{x^2+y^2}{y}\right)=C_1 \implies \frac{x^2+y^2}{y}=C. $$This is $x^2+y^2=Cy$, identical to the exact-method result. (A1)即 $x^2+y^2=Cy$，与恰当方程法的结果完全一致。(A1)