Unit D2: First-Order Models and Exact Equations第D2单元：一阶模型与恰当方程

A body at temperature $T(t)$ sits in surroundings held at $T_a$. Newton's law states that the rate of cooling is proportional to the temperature difference. Model and solve.

温度为 $T(t)$ 的物体置于温度恒为 $T_a$ 的环境中。牛顿冷却定律（Newton's law of cooling）指出：冷却速率与温差成正比。建立模型并求解。

The verbal rule becomes $T' = -k(T - T_a)$ with $k > 0$. This is linear and separable. Let $u = T - T_a$, so $u' = T'$ and $u' = -ku$, giving $u = u_0 e^{-kt}$.

文字规律化为 $T' = -k(T - T_a)$，$k > 0$。此方程为一阶线性且可分离变量。令 $u = T - T_a$，则 $u' = T'$，$u' = -ku$，得 $u = u_0 e^{-kt}$。

$$ T(t) = T_a + (T_0 - T_a)\,e^{-kt}. $$

As $t \to \infty$ the exponential decays and $T \to T_a$: the body approaches ambient temperature, exactly as physical intuition demands.

当 $t \to \infty$ 时指数项衰减，$T \to T_a$：物体趋近于环境温度，完全符合物理直觉。

A mass $m$ falls under gravity $g$ with air resistance proportional to velocity, $F_{\text{drag}} = -bv$ with $b > 0$. Model the velocity $v(t)$, solve, and interpret the long-run behavior.

质量为 $m$ 的物体在重力 $g$ 下下落，空气阻力与速度成正比：$F_{\text{drag}} = -bv$，$b > 0$。建立速度 $v(t)$ 的模型，求解并解释长期行为。

Newton's second law $m v' = mg - bv$ gives the first-order linear equation

牛顿第二定律 $m v' = mg - bv$ 给出一阶线性方程

$$ v' + \frac{b}{m}\,v = g, \qquad v(0) = v_0. $$

Separate variables, or notice the constant-coefficient form. With $k = b/m$, the integrating factor is $e^{kt}$ and $(e^{kt}v)' = g\,e^{kt}$, so $e^{kt}v = \tfrac{g}{k}e^{kt} + C$. Solving for $v$ and applying $v(0) = v_0$,

分离变量，或注意到常系数形式。令 $k = b/m$，积分因子为 $e^{kt}$，$(e^{kt}v)' = g\,e^{kt}$，故 $e^{kt}v = \tfrac{g}{k}e^{kt} + C$。解出 $v$ 并代入 $v(0) = v_0$，

$$ v(t) = \frac{mg}{b} + \left(v_0 - \frac{mg}{b}\right)e^{-bt/m}. $$

As $t \to \infty$ the exponential vanishes and $v \to mg/b$, the terminal velocity. This is exactly the equilibrium where drag balances gravity, $bv = mg$, so the body stops accelerating.

当 $t \to \infty$ 时指数项消失，$v \to mg/b$，即终端速度。这正是阻力与重力平衡 $bv = mg$ 的平衡解，物体停止加速。

Without solving, describe the solutions of $y' = y - t$ that pass through the line $y = t$.

不求解，描述 $y' = y - t$ 中穿过直线 $y = t$ 的解的行为。

The slope at any point is $f(t,y) = y - t$. On the line $y = t$ the slope is $0$, so every solution crossing that line does so with a horizontal tangent. Above the line, $y > t$ gives $y' > 0$ (solutions rise); below it, $y < t$ gives $y' < 0$ (solutions fall).

任意点处的斜率为 $f(t,y) = y - t$。在直线 $y = t$ 上斜率为 $0$，故每条穿过该线的解曲线在此处有水平切线。线上方 $y > t$ 时 $y' > 0$（解上升）；线下方 $y < t$ 时 $y' < 0$（解下降）。

The line $y = t + 1$ is itself a solution, since substituting $y = t+1$ gives $y' = 1$ and $y - t = 1$, an identity. Solutions above this line curve upward and away from it, while those below are funneled toward it for a while before bending away. Reading slopes alone reveals the linear solution $y = t+1$ acts as a separatrix, all visible before any integration.

直线 $y = t + 1$ 本身是一个解：代入 $y = t+1$ 得 $y' = 1$，$y - t = 1$，构成恒等式。此线上方的解曲线向上弯离，下方的解曲线先被吸引再弯离。仅读斜率便可发现线性解 $y = t+1$ 充当分界线（separatrix），无需积分即可看出。

A tank holds $100$ liters of pure water. Brine with $2$ grams of salt per liter flows in at $5$ liters per minute; the stirred mixture flows out at $5$ liters per minute. Find $A(t)$, the grams of salt at time $t$.

水箱盛有 $100$ 升纯水。浓度为每升 $2$ 克的盐水以每分钟 $5$ 升的速率流入；搅拌后的混合液以每分钟 $5$ 升流出。求时刻 $t$ 时的盐量 $A(t)$（克）。

Volume stays at $100$ liters. Rate in is $(2)(5) = 10$ grams per minute. Rate out is $(A/100)(5) = A/20$ grams per minute.

体积保持 $100$ 升不变。流入速率为 $(2)(5) = 10$ 克/分钟；流出速率为 $(A/100)(5) = A/20$ 克/分钟。

$$ A' = 10 - \frac{A}{20}, \qquad A(0) = 0. $$

Rewrite as $A' + \tfrac{1}{20}A = 10$. The integrating factor is $\mu = e^{t/20}$, so $(e^{t/20}A)' = 10\,e^{t/20}$. Integrate:

改写为 $A' + \tfrac{1}{20}A = 10$。积分因子为 $\mu = e^{t/20}$，故 $(e^{t/20}A)' = 10\,e^{t/20}$。积分：

$$ e^{t/20}A = 200\,e^{t/20} + C \implies A(t) = 200 + C e^{-t/20}. $$

Apply $A(0) = 0$: $0 = 200 + C$, so $C = -200$.

代入 $A(0) = 0$：$0 = 200 + C$，故 $C = -200$。

$$ A(t) = 200\left(1 - e^{-t/20}\right). $$

As $t \to \infty$, $A \to 200$ grams, which is the steady state where inflow concentration $2$ g/L fills the entire $100$ L tank.

当 $t \to \infty$ 时，$A \to 200$ 克，即稳态：流入浓度 $2$ 克/升充满整个 $100$ 升水箱。

A tank starts with $200$ L of brine containing $10$ kg of salt. Pure water flows in at $4$ L/min and the well-stirred mixture flows out at $6$ L/min. Find the salt $A(t)$ while the tank still holds liquid.

水箱初始盛有含 $10$ 千克盐的 $200$ 升盐水。纯水以 $4$ 升/分钟流入，混合液以 $6$ 升/分钟流出。在水箱未排空期间求盐量 $A(t)$。

Now inflow and outflow differ, so the volume drains: $V(t) = 200 + (4 - 6)t = 200 - 2t$, valid for $0 \le t < 100$. The inflow is pure water, so the rate in is $0$. The rate out is the concentration $A/V$ times the outflow rate $6$:

流入与流出不等，体积减少：$V(t) = 200 + (4 - 6)t = 200 - 2t$，对 $0 \le t < 100$ 有效。流入为纯水，故流入速率为 $0$。流出速率为浓度 $A/V$ 乘以体积速率 $6$：

$$ A' = 0 - \frac{A}{200 - 2t}\cdot 6 = -\frac{6A}{200 - 2t} = -\frac{3A}{100 - t}. $$

This is linear and separable. Separate: $\dfrac{dA}{A} = -\dfrac{3}{100 - t}\,dt$. Integrate, using $\int \frac{dt}{100 - t} = -\ln|100 - t|$:

方程为一阶线性且可分离。分离变量：$\dfrac{dA}{A} = -\dfrac{3}{100 - t}\,dt$。积分，用 $\int \frac{dt}{100 - t} = -\ln|100 - t|$：

$$ \ln|A| = 3\ln|100 - t| + C_1 \implies A = C(100 - t)^3. $$

Apply $A(0) = 10$: $10 = C(100)^3$, so $C = 10/10^6 = 10^{-5}$. Hence

代入 $A(0) = 10$：$10 = C(100)^3$，故 $C = 10^{-5}$。从而

$$ A(t) = 10^{-5}(100 - t)^3, \qquad 0 \le t < 100. $$

The salt falls to $0$ as the tank empties at $t = 100$, which is correct: with pure water in and brine out, all the salt is eventually flushed.

当 $t = 100$ 时水箱排空，盐量降至 $0$，这是正确的：纯水流入、盐水流出，最终所有盐都被冲走。

A $500$ L tank of pure water receives brine at $0.4$ kg/L flowing in at $10$ L/min, with equal outflow. Without solving the full IVP, find the eventual salt content, then confirm it is the equilibrium.

$500$ 升纯水水箱以每分钟 $10$ 升的速率接收浓度为 $0.4$ 千克/升的盐水，流出速率相等。不求解完整初值问题，直接求最终盐量，然后验证它是平衡解。

The model is $A' + \tfrac{10}{500}A = (0.4)(10) = 4$, that is $A' + 0.02A = 4$. The steady state is the constant solution where $A' = 0$:

模型为 $A' + \tfrac{10}{500}A = (0.4)(10) = 4$，即 $A' + 0.02A = 4$。稳态是 $A' = 0$ 时的常数解：

$$ 0 = 4 - 0.02A^{*} \implies A^{*} = 200 \text{ kg}. $$

The full solution is $A(t) = 200 + Ce^{-0.02t}$; whatever the start, the transient $Ce^{-0.02t}$ decays and $A \to 200$. The steady concentration is $200/500 = 0.4$ kg/L, equal to the inflow concentration, exactly as conservation predicts when the tank is fully replaced by inflowing brine.

完整解为 $A(t) = 200 + Ce^{-0.02t}$；无论初始值如何，暂态项 $Ce^{-0.02t}$ 衰减，$A \to 200$。稳态浓度为 $200/500 = 0.4$ 千克/升，等于流入浓度，与守恒定律在盐水完全替换后的预测完全一致。

Suppose $M_y = N_x$ on a simply connected region. We build $F$ explicitly. Integrate $M$ in $x$, holding $y$ fixed:

设 $M_y = N_x$ 在单连通区域上成立。我们显式构造 $F$。固定 $y$，对 $M$ 关于 $x$ 积分：

$$ F(x,y) = \int M(x,y)\,dx + g(y), $$

where the constant of integration is an unknown function $g(y)$ because $y$ was held fixed. Differentiating in $y$ and matching $F_y = N$:

积分常数是未知函数 $g(y)$，因为 $y$ 被固定。对 $y$ 求导并令 $F_y = N$：

$$ F_y = \frac{\partial}{\partial y}\int M\,dx + g'(y) = N \implies g'(y) = N - \frac{\partial}{\partial y}\int M\,dx. $$

The right side is a function of $y$ alone: its $x$-derivative is $N_x - \partial_x \partial_y \int M\,dx = N_x - M_y = 0$ by hypothesis. So $g$ can be recovered by a single integration in $y$, and $F$ exists. This is the same logic that recovers a scalar potential for a conservative field.

右端仅是 $y$ 的函数：其对 $x$ 的偏导为 $N_x - \partial_x \partial_y \int M\,dx = N_x - M_y = 0$（由假设）。故 $g$ 可通过对 $y$ 的一次积分恢复，$F$ 存在。这与从保守场中恢复标量势的逻辑完全相同。

Solve $(2xy + 3)\,dx + (x^2 - 1)\,dy = 0$.

求解 $(2xy + 3)\,dx + (x^2 - 1)\,dy = 0$。

Here $M = 2xy + 3$ and $N = x^2 - 1$. Test: $M_y = 2x$ and $N_x = 2x$, equal, so the equation is exact.

此处 $M = 2xy + 3$，$N = x^2 - 1$。检验：$M_y = 2x$，$N_x = 2x$，相等，故方程为恰当方程。

Integrate $M$ in $x$: $F = x^2 y + 3x + g(y)$. Then $F_y = x^2 + g'(y)$ must equal $N = x^2 - 1$, so $g'(y) = -1$ and $g(y) = -y$.

对 $M$ 关于 $x$ 积分：$F = x^2 y + 3x + g(y)$。则 $F_y = x^2 + g'(y)$ 须等于 $N = x^2 - 1$，故 $g'(y) = -1$，$g(y) = -y$。

$$ F(x,y) = x^2 y + 3x - y = C. $$

This implicit relation is the general solution; an initial condition fixes $C$.

此隐式关系为通解（general solution）；初始条件确定 $C$ 的值。

Solve $(3x^2 y + 2xy + y^3)\,dx + (x^3 + x^2 + 3xy^2)\,dy = 0$ with $y(0) = 1$.

求解 $(3x^2 y + 2xy + y^3)\,dx + (x^3 + x^2 + 3xy^2)\,dy = 0$，初始条件 $y(0) = 1$。

Set $M = 3x^2 y + 2xy + y^3$ and $N = x^3 + x^2 + 3xy^2$. Test exactness:

设 $M = 3x^2 y + 2xy + y^3$，$N = x^3 + x^2 + 3xy^2$。检验恰当性：

$$ M_y = 3x^2 + 2x + 3y^2, \qquad N_x = 3x^2 + 2x + 3y^2. $$

They agree, so the equation is exact. This time integrate $N$ in $y$ (holding $x$ fixed), which is the cleaner route here:

两者相等，方程为恰当方程。本次先对 $N$ 关于 $y$ 积分（固定 $x$），这里是更简洁的路径：

$$ F = \int N\,dy = x^3 y + x^2 y + xy^3 + h(x). $$

Now match $F_x = M$: differentiating gives $F_x = 3x^2 y + 2xy + y^3 + h'(x)$, which must equal $M = 3x^2 y + 2xy + y^3$. Hence $h'(x) = 0$ and $h$ is a constant absorbed into $C$:

令 $F_x = M$：求导得 $F_x = 3x^2 y + 2xy + y^3 + h'(x)$，须等于 $M = 3x^2 y + 2xy + y^3$。故 $h'(x) = 0$，$h$ 为常数并入 $C$：

$$ F(x,y) = x^3 y + x^2 y + xy^3 = C. $$

Apply $y(0) = 1$: every term has a factor of $x$, so $F(0,1) = 0 = C$. The particular solution is the implicit relation $x^3 y + x^2 y + xy^3 = 0$, equivalently $x\big(x^2 y + xy + y^3\big) = 0$.

代入 $y(0) = 1$：每项都含因子 $x$，故 $F(0,1) = 0 = C$。特解（particular solution）为隐式关系 $x^3 y + x^2 y + xy^3 = 0$，即 $x\big(x^2 y + xy + y^3\big) = 0$。

Decide whether $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$ is exact.

判断 $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$ 是否为恰当方程。

With $M = 3xy + y^2$, $N = x^2 + xy$: $M_y = 3x + 2y$ while $N_x = 2x + y$. Since $3x + 2y \ne 2x + y$ in general, the equation is not exact, and the level-curve method does not apply as written.

设 $M = 3xy + y^2$，$N = x^2 + xy$：$M_y = 3x + 2y$，$N_x = 2x + y$。一般而言 $3x + 2y \ne 2x + y$，故方程不是恰当方程，等值线方法不可直接应用。

The remedy belongs to Section 4: test $(M_y - N_x)/N = (x + y)/(x^2 + xy) = (x+y)/\big(x(x+y)\big) = 1/x$, a function of $x$ alone, so the integrating factor $\mu = e^{\int (1/x)\,dx} = x$ restores exactness. The lesson is that failing the test is not a dead end, it is the signal to look for an integrating factor.

补救方法见第4节：检验 $(M_y - N_x)/N = (x + y)/(x^2 + xy) = 1/x$，仅为 $x$ 的函数，故积分因子 $\mu = e^{\int (1/x)\,dx} = x$ 可恢复恰当性。教训是：检验失败并非死路，而是寻找积分因子的信号。

We seek $\mu(x)$ so that multiplying $y' + p y = q$ by $\mu$ makes the left side a single derivative $(\mu y)'$. Expand the target:

我们寻求 $\mu(x)$，使得用 $\mu$ 乘以 $y' + p y = q$ 后，左端变为单一导数 $(\mu y)'$。展开目标：

$$ (\mu y)' = \mu y' + \mu' y. $$

Multiplying the equation by $\mu$ gives $\mu y' + \mu p y = \mu q$. For the left side to equal $(\mu y)'$ we need the $y$ coefficients to match: $\mu' = \mu p$. This is itself a separable equation:

方程两端乘以 $\mu$ 得 $\mu y' + \mu p y = \mu q$。为使左端等于 $(\mu y)'$，需 $y$ 的系数匹配：$\mu' = \mu p$。这本身是一个可分离方程：

$$ \frac{d\mu}{\mu} = p\,dx \implies \ln|\mu| = \int p\,dx \implies \mu = e^{\int p\,dx}. $$

With this $\mu$, the equation collapses to $(\mu y)' = \mu q$, and integrating once recovers the solution formula above.

选定此 $\mu$ 后，方程化为 $(\mu y)' = \mu q$，积分一次即恢复上述求解公式。

Solve $y' + \dfrac{2}{x}\,y = x^3$ for $x > 0$.

求解 $x > 0$ 时的 $y' + \dfrac{2}{x}\,y = x^3$。

Here $p = 2/x$, so $\int p\,dx = 2\ln x$ and $\mu = e^{2\ln x} = x^2$. Multiply through:

此处 $p = 2/x$，故 $\int p\,dx = 2\ln x$，$\mu = e^{2\ln x} = x^2$。两端乘以积分因子：

$$ (x^2 y)' = x^2 \cdot x^3 = x^5. $$

Integrate: $x^2 y = \tfrac{1}{6}x^6 + C$, so

积分：$x^2 y = \tfrac{1}{6}x^6 + C$，故

$$ y(x) = \frac{x^4}{6} + \frac{C}{x^2}. $$

Solve $y' + y = \cos x$ with $y(0) = 0$.

求解 $y' + y = \cos x$，初始条件 $y(0) = 0$。

Here $p = 1$, so $\mu = e^{\int 1\,dx} = e^{x}$. Multiply through: $(e^{x}y)' = e^{x}\cos x$. Integrate the right side by parts twice (a standard cycle):

此处 $p = 1$，故 $\mu = e^x$。两端乘以 $e^x$：$(e^x y)' = e^x \cos x$。对右端进行两次分部积分（标准循环）：

$$ \int e^{x}\cos x\,dx = \frac{e^{x}(\cos x + \sin x)}{2} + C. $$

Therefore $e^{x}y = \tfrac{1}{2}e^{x}(\cos x + \sin x) + C$, and dividing by $e^{x}$,

故 $e^x y = \tfrac12 e^x(\cos x + \sin x) + C$，除以 $e^x$，

$$ y(x) = \frac{\cos x + \sin x}{2} + C e^{-x}. $$

Apply $y(0) = 0$: $0 = \tfrac{1}{2}(1 + 0) + C$, so $C = -\tfrac12$. The solution is $y = \tfrac12(\cos x + \sin x) - \tfrac12 e^{-x}$. The transient $-\tfrac12 e^{-x}$ decays, leaving the steady oscillation $\tfrac12(\cos x + \sin x)$, the periodic response to periodic forcing.

代入 $y(0) = 0$：$0 = \tfrac12(1+0) + C$，故 $C = -\tfrac12$。解为 $y = \tfrac12(\cos x + \sin x) - \tfrac12 e^{-x}$。暂态项 $-\tfrac12 e^{-x}$ 衰减，留下稳态振荡 $\tfrac12(\cos x + \sin x)$，即周期驱动的周期响应。

Solve $(x^2 + y^2 + x)\,dx + xy\,dy = 0$.

求解 $(x^2 + y^2 + x)\,dx + xy\,dy = 0$。

With $M = x^2 + y^2 + x$, $N = xy$: $M_y = 2y$, $N_x = y$, so the equation is not exact. Test for an $x$-only factor:

设 $M = x^2 + y^2 + x$，$N = xy$：$M_y = 2y$，$N_x = y$，故方程非恰当。检验仅关于 $x$ 的因子：

$$ \frac{M_y - N_x}{N} = \frac{2y - y}{xy} = \frac{1}{x}, $$

a function of $x$ alone, so $\mu = e^{\int (1/x)\,dx} = x$. Multiply through by $x$:

仅为 $x$ 的函数，故 $\mu = x$。两端乘以 $x$：

$$ (x^3 + xy^2 + x^2)\,dx + x^2 y\,dy = 0. $$

Now $\tilde M_y = 2xy = \tilde N_x$, exact. Integrate $\tilde N = x^2 y$ in $y$: $F = \tfrac12 x^2 y^2 + h(x)$. Match $F_x = xy^2 + h'(x) = x^3 + xy^2 + x^2$, so $h'(x) = x^3 + x^2$ and $h = \tfrac14 x^4 + \tfrac13 x^3$. The solution is

现在 $\tilde M_y = 2xy = \tilde N_x$，为恰当方程。对 $\tilde N = x^2 y$ 关于 $y$ 积分：$F = \tfrac12 x^2 y^2 + h(x)$。令 $F_x = xy^2 + h'(x) = x^3 + xy^2 + x^2$，故 $h'(x) = x^3 + x^2$，$h = \tfrac14 x^4 + \tfrac13 x^3$。解为

$$ F(x,y) = \tfrac12 x^2 y^2 + \tfrac14 x^4 + \tfrac13 x^3 = C. $$

Start from $y' + p y = q y^n$. Divide by $y^n$ (assuming $y \ne 0$):

从 $y' + p y = q y^n$ 出发，两端除以 $y^n$（假设 $y \ne 0$）：

$$ y^{-n} y' + p\,y^{1-n} = q. $$

Let $v = y^{1-n}$. By the chain rule $v' = (1-n)y^{-n}y'$, so $y^{-n}y' = \dfrac{v'}{1-n}$. Substitute:

令 $v = y^{1-n}$。由链式法则 $v' = (1-n)y^{-n}y'$，故 $y^{-n}y' = \dfrac{v'}{1-n}$。代入：

$$ \frac{v'}{1-n} + p\,v = q \implies v' + (1-n)p\,v = (1-n)q. $$

This is linear in $v$, solvable by the integrating factor $\mu = e^{\int (1-n)p\,dx}$. After finding $v$, revert with $y = v^{1/(1-n)}$.

这是关于 $v$ 的线性方程，用积分因子 $\mu = e^{\int (1-n)p\,dx}$ 求解。求出 $v$ 后，用 $y = v^{1/(1-n)}$ 还原。

Solve $y' + y = y^2$ (here $p = 1$, $q = 1$, $n = 2$).

求解 $y' + y = y^2$（此处 $p = 1$，$q = 1$，$n = 2$）。

Set $v = y^{1-2} = y^{-1}$. The reduced equation is $v' + (1-2)(1)v = (1-2)(1)$, that is $v' - v = -1$.

令 $v = y^{1-2} = y^{-1}$。化简后方程为 $v' + (1-2)(1)v = (1-2)(1)$，即 $v' - v = -1$。

Integrating factor $\mu = e^{-x}$: $(e^{-x}v)' = -e^{-x}$, so $e^{-x}v = e^{-x} + C$ and $v = 1 + C e^{x}$.

积分因子 $\mu = e^{-x}$：$(e^{-x}v)' = -e^{-x}$，故 $e^{-x}v = e^{-x} + C$，$v = 1 + Ce^x$。

$$ y = \frac{1}{v} = \frac{1}{1 + C e^{x}}. $$

Solve $y' + \dfrac{1}{x}y = x\,y^3$ for $x > 0$, with $y(1) = 1$.

求解 $x > 0$ 时 $y' + \dfrac{1}{x}y = x\,y^3$，初始条件 $y(1) = 1$。

Here $p = 1/x$, $q = x$, $n = 3$. Set $v = y^{1-3} = y^{-2}$. The reduced linear equation is $v' + (1-n)p\,v = (1-n)q$, that is

此处 $p = 1/x$，$q = x$，$n = 3$。令 $v = y^{1-3} = y^{-2}$。化简线性方程为 $v' + (1-n)p\,v = (1-n)q$，即

$$ v' + (1 - 3)\frac{1}{x}v = (1 - 3)x \implies v' - \frac{2}{x}v = -2x. $$

The integrating factor is $\mu = e^{\int (-2/x)\,dx} = e^{-2\ln x} = x^{-2}$. Multiply through: $(x^{-2}v)' = -2x \cdot x^{-2} = -2/x$. Integrate:

积分因子为 $\mu = x^{-2}$。两端乘以 $x^{-2}$：$(x^{-2}v)' = -2/x$。积分：

$$ x^{-2}v = -2\ln x + C \implies v = x^2(C - 2\ln x). $$

Revert with $v = y^{-2}$, so $y^2 = 1/v$ and

用 $v = y^{-2}$ 还原，故 $y^2 = 1/v$，

$$ y(x) = \frac{1}{x\sqrt{C - 2\ln x}}. $$

Apply $y(1) = 1$: $1 = 1/\sqrt{C - 0}$, so $C = 1$, giving $y = 1/\big(x\sqrt{1 - 2\ln x}\,\big)$, valid while $1 - 2\ln x > 0$.

代入 $y(1) = 1$：$1 = 1/\sqrt{C}$，故 $C = 1$，得 $y = 1/\big(x\sqrt{1 - 2\ln x}\,\big)$，在 $1 - 2\ln x > 0$ 时有效。

Solve $y' = \dfrac{x^2 + y^2}{xy}$ for $x > 0$.

求解 $x > 0$ 时的 $y' = \dfrac{x^2 + y^2}{xy}$。

Divide numerator and denominator by $x^2$ to expose the ratio $v = y/x$: $y' = \dfrac{1 + (y/x)^2}{y/x} = \dfrac{1 + v^2}{v}$. With $y = vx$ and $y' = v + xv'$,

分子分母同除以 $x^2$，令 $v = y/x$：$y' = \dfrac{1 + v^2}{v}$。令 $y = vx$，$y' = v + xv'$，

$$ v + x\frac{dv}{dx} = \frac{1 + v^2}{v} \implies x\frac{dv}{dx} = \frac{1 + v^2}{v} - v = \frac{1}{v}. $$

This is separable: $v\,dv = \dfrac{dx}{x}$. Integrate: $\tfrac12 v^2 = \ln x + C_1$, so $v^2 = 2\ln x + C$. Revert $v = y/x$:

方程为可分离变量：$v\,dv = \dfrac{dx}{x}$。积分：$\tfrac12 v^2 = \ln x + C_1$，故 $v^2 = 2\ln x + C$。用 $v = y/x$ 还原：

$$ \frac{y^2}{x^2} = 2\ln x + C \implies y^2 = x^2(2\ln x + C). $$

Classify the equilibria of the logistic model $y' = ky(1 - y/K)$ with $k, K > 0$.

对逻辑斯谛模型 $y' = ky(1 - y/K)$（$k, K > 0$）的平衡解进行分类。

Set $f(y) = ky(1 - y/K) = 0$. The equilibria are $y^{*} = 0$ and $y^{*} = K$. Differentiate $f$:

令 $f(y) = ky(1 - y/K) = 0$，平衡解为 $y^{*} = 0$ 和 $y^{*} = K$。对 $f$ 求导：

$$ f'(y) = k\left(1 - \frac{2y}{K}\right). $$

At $y^{*} = 0$: $f'(0) = k > 0$, so the extinction state is unstable. At $y^{*} = K$: $f'(K) = k(1 - 2) = -k < 0$, so the carrying capacity is stable.

在 $y^{*} = 0$ 处：$f'(0) = k > 0$，故灭绝状态不稳定。在 $y^{*} = K$ 处：$f'(K) = -k < 0$，故环境容纳量 $K$ 稳定。

Hence any positive initial population is drawn toward $K$, the conclusion every ecologist expects from the logistic model.

故任何正初始种群都会被吸引趋向 $K$，这是每位生态学家对逻辑斯谛模型所期望的结论。

The phase line gave the qualitative answer; here is the exact solution of $y' = ky(1 - y/K)$, $y(0) = y_0$, which confirms it. Separate variables:

相线给出了定性结论；这里求 $y' = ky(1 - y/K)$，$y(0) = y_0$ 的精确解来验证。分离变量：

$$ \frac{dy}{y(1 - y/K)} = k\,dt. $$

Decompose by partial fractions. Writing $\dfrac{1}{y(1 - y/K)} = \dfrac{1}{y} + \dfrac{1/K}{1 - y/K}$, integration gives

用部分分数分解：$\dfrac{1}{y(1 - y/K)} = \dfrac{1}{y} + \dfrac{1/K}{1 - y/K}$，积分得

$$ \ln|y| - \ln\left|1 - \frac{y}{K}\right| = kt + C_1 \implies \frac{y}{1 - y/K} = A e^{kt}. $$

Solve for $y$ and fix $A$ from $y(0) = y_0$, namely $A = \dfrac{y_0}{1 - y_0/K}$. After simplifying,

解出 $y$，由 $y(0) = y_0$ 确定 $A = \dfrac{y_0}{1 - y_0/K}$。化简后，

$$ y(t) = \frac{K\,y_0}{y_0 + (K - y_0)e^{-kt}}. $$

As $t \to \infty$ the exponential vanishes and $y \to K$, confirming the stable equilibrium found from $f'(K) = -k < 0$. The qualitative phase-line prediction and the exact formula agree.

当 $t \to \infty$ 时指数项消失，$y \to K$，验证了由 $f'(K) = -k < 0$ 得出的稳定平衡解。相线定性预测与精确公式完全吻合。

Find and classify the equilibria of $y' = f(y) = y(1 - y)(y - 3)$.

求并分类 $y' = f(y) = y(1 - y)(y - 3)$ 的平衡解。

Set $f(y) = 0$: the equilibria are $y^{*} = 0,\ 1,\ 3$. Read the sign of $f$ on each interval, or differentiate. Expanding, $f(y) = -y^3 + 4y^2 - 3y$, so $f'(y) = -3y^2 + 8y - 3$.

令 $f(y) = 0$，平衡解为 $y^{*} = 0, 1, 3$。在各区间读取 $f$ 的符号，或求导。展开 $f(y) = -y^3 + 4y^2 - 3y$，故 $f'(y) = -3y^2 + 8y - 3$。

$$ f'(0) = -3 < 0\ (\text{stable}), \quad f'(1) = -3 + 8 - 3 = 2 > 0\ (\text{unstable}), \quad f'(3) = -27 + 24 - 3 = -6 < 0\ (\text{stable}). $$

So $0$ and $3$ are sinks and $1$ is a source. A solution starting in $(1,3)$ rises to $3$; one starting in $(0,1)$ falls to $0$. The unstable equilibrium $y = 1$ is the threshold (a separatrix) dividing the two basins of attraction.

故 $0$ 和 $3$ 为汇，$1$ 为源。从 $(1,3)$ 出发的解趋向 $3$；从 $(0,1)$ 出发的解趋向 $0$。不稳定平衡 $y = 1$ 是阈值（分界线），将两个吸引域分开。

Let $y^{*}$ be an equilibrium of $y' = f(y)$, so $f(y^{*}) = 0$. Study a small perturbation $u(t) = y(t) - y^{*}$. Differentiating, $u' = y' = f(y^{*} + u)$. Taylor-expand $f$ about $y^{*}$, using $f(y^{*}) = 0$:

设 $y^{*}$ 是 $y' = f(y)$ 的平衡解，$f(y^{*}) = 0$。研究小扰动 $u(t) = y(t) - y^{*}$。求导得 $u' = y' = f(y^{*} + u)$。在 $y^{*}$ 处对 $f$ 做Taylor展开（利用 $f(y^{*}) = 0$）：

$$ f(y^{*} + u) = f(y^{*}) + f'(y^{*})\,u + O(u^2) = f'(y^{*})\,u + O(u^2). $$

For small $u$ the higher-order terms are negligible and the perturbation obeys the linear equation $u' \approx f'(y^{*})\,u$, whose solution is $u(t) \approx u(0)\,e^{f'(y^{*})\,t}$. Therefore:

对于小 $u$，高阶项可忽略，扰动满足线性方程 $u' \approx f'(y^{*})\,u$，解为 $u(t) \approx u(0)\,e^{f'(y^{*})\,t}$。因此：

$$ f'(y^{*}) < 0 \implies u \to 0 \ (\text{stable}), \qquad f'(y^{*}) > 0 \implies |u| \to \infty \ (\text{unstable}). $$

The borderline case $f'(y^{*}) = 0$ is inconclusive at this order: the $O(u^2)$ term decides, and the equilibrium may be semi-stable (attracting from one side, repelling from the other). This is exactly the eigenvalue criterion of linear stability theory in one dimension.

临界情况 $f'(y^{*}) = 0$ 在此阶无法判断：由 $O(u^2)$ 项决定，平衡解可能是半稳定的（一侧吸引，另一侧排斥）。这正是一维线性稳定性理论中的特征值判据。

Consider $y' = \sqrt{|y|}$ with $y(0) = 0$. The right side $f(y) = \sqrt{|y|}$ is continuous, but $\partial f/\partial y = \tfrac{1}{2\sqrt{y}}$ blows up at $y = 0$, so the Lipschitz hypothesis fails there.

考虑 $y' = \sqrt{|y|}$，$y(0) = 0$。右端 $f(y) = \sqrt{|y|}$ 连续，但 $\partial f/\partial y = \tfrac{1}{2\sqrt{y}}$ 在 $y = 0$ 处趋于无穷，Lipschitz条件在此失败。

One obvious solution is the constant $y \equiv 0$. But separating variables for $y > 0$ gives $\int y^{-1/2}\,dy = \int dt$, so $2\sqrt{y} = t$, that is $y = t^2/4$ for $t \ge 0$. Both satisfy the IVP:

一个显然的解是常数 $y \equiv 0$。但对 $y > 0$ 分离变量得 $\int y^{-1/2}\,dy = \int dt$，即 $y = t^2/4$（$t \ge 0$）。两者均满足初值问题：

$$ y(t) = 0 \quad\text{and}\quad y(t) = \frac{t^2}{4} \quad (t \ge 0). $$

In fact infinitely many solutions exist, each staying at $0$ until some time $a$ and then leaving along $(t-a)^2/4$. Uniqueness genuinely fails because the theorem's hypothesis is violated. This is the standard cautionary example for why the smoothness condition matters.

事实上存在无穷多个解，每个解在某时刻 $a$ 之前停留在 $0$，之后沿 $(t-a)^2/4$ 离开。唯一性真正失败，因为定理假设被违反。这是说明光滑性条件重要性的标准警示例题。

Solve $y\,dx + (2x - y e^{y})\,dy = 0$, which is not exact.

求解 $y\,dx + (2x - y e^{y})\,dy = 0$（非恰当方程）。

With $M = y$, $N = 2x - ye^{y}$: $M_y = 1$, $N_x = 2$, so $M_y \ne N_x$. Try a factor in $y$: $(N_x - M_y)/M = (2 - 1)/y = 1/y$, a function of $y$ alone. So $\mu = e^{\int (1/y)\,dy} = y$.

设 $M = y$，$N = 2x - ye^y$：$M_y = 1$，$N_x = 2$，故 $M_y \ne N_x$。尝试仅关于 $y$ 的因子：$(N_x - M_y)/M = 1/y$，故 $\mu = y$。

Multiply through by $y$: $y^2\,dx + (2xy - y^2 e^{y})\,dy = 0$. Now $\tilde M = y^2$, $\tilde N = 2xy - y^2 e^{y}$, and $\tilde M_y = 2y = \tilde N_x$: exact.

两端乘以 $y$：$y^2\,dx + (2xy - y^2 e^y)\,dy = 0$。此时 $\tilde M_y = 2y = \tilde N_x$：为恰当方程。

Integrate $\tilde M$ in $x$: $F = xy^2 + g(y)$. Match $F_y = 2xy + g'(y) = 2xy - y^2 e^{y}$, so $g'(y) = -y^2 e^{y}$. Integrate by parts: $g(y) = -(y^2 - 2y + 2)e^{y}$.

对 $\tilde M$ 关于 $x$ 积分：$F = xy^2 + g(y)$。令 $F_y = 2xy + g'(y) = 2xy - y^2 e^y$，故 $g'(y) = -y^2 e^y$。分部积分：$g(y) = -(y^2 - 2y + 2)e^y$。

$$ F(x,y) = xy^2 - (y^2 - 2y + 2)e^{y} = C. $$

The existence half of the theorem is not magic; it is a constructive fixed-point argument. Integrate the IVP $y' = f(t,y)$, $y(t_0) = y_0$ from $t_0$ to $t$ to convert it into the equivalent integral equation

定理中的存在性部分并非魔法，而是一个构造性的不动点论证。将初值问题 $y' = f(t,y)$，$y(t_0) = y_0$ 从 $t_0$ 积到 $t$，转化为等价的积分方程

$$ y(t) = y_0 + \int_{t_0}^{t} f\big(s, y(s)\big)\,ds. $$

Define the Picard iterates $y_0(t) = y_0$ and $y_{n+1}(t) = y_0 + \int_{t_0}^{t} f(s, y_n(s))\,ds$. The Lipschitz condition on $f$ in $y$ makes this map a contraction on a small time interval, so the iterates converge to a unique fixed point, which is the solution.

定义皮卡迭代：$y_0(t) = y_0$，$y_{n+1}(t) = y_0 + \int_{t_0}^{t} f(s, y_n(s))\,ds$。$f$ 关于 $y$ 的 Lipschitz 条件使该映射在小时间区间上为压缩映射，故迭代收敛到唯一不动点，即方程的解。

Concretely, for $y' = y$, $y(0) = 1$: the iterates are $y_1 = 1 + \int_0^t 1\,ds = 1 + t$, then $y_2 = 1 + \int_0^t (1 + s)\,ds = 1 + t + \tfrac{t^2}{2}$, then $y_3 = 1 + t + \tfrac{t^2}{2} + \tfrac{t^3}{6}$. The pattern is the partial sums of $e^{t}$, and $y_n \to e^{t}$, the known solution. The contraction guarantees both existence and uniqueness in one stroke.

以 $y' = y$，$y(0) = 1$ 为例：$y_1 = 1 + t$，$y_2 = 1 + t + \tfrac{t^2}{2}$，$y_3 = 1 + t + \tfrac{t^2}{2} + \tfrac{t^3}{6}$。规律正是 $e^t$ 的部分和，$y_n \to e^t$，即已知解。压缩映射一举保证存在性与唯一性。

Solve $2xy\,dx + (y^2 - x^2)\,dy = 0$, illustrating that one equation can be attacked by several methods.

求解 $2xy\,dx + (y^2 - x^2)\,dy = 0$，说明同一方程可用多种方法求解。

Test exactness: $M = 2xy$, $N = y^2 - x^2$, $M_y = 2x$, $N_x = -2x$, not exact. Seek a $y$-only factor: $(N_x - M_y)/M = (-2x - 2x)/(2xy) = -2/y$, a function of $y$ alone, so $\mu = e^{\int (-2/y)\,dy} = y^{-2}$. Multiply through:

检验恰当性：$M = 2xy$，$N = y^2 - x^2$，$M_y = 2x$，$N_x = -2x$，非恰当。寻找仅关于 $y$ 的积分因子：$(N_x - M_y)/M = -2/y$，故 $\mu = y^{-2}$。两端乘以 $\mu$：

$$ \frac{2x}{y}\,dx + \frac{y^2 - x^2}{y^2}\,dy = 0. $$

Now $\tilde M = 2x/y$, $\tilde N = 1 - x^2/y^2$, and $\tilde M_y = -2x/y^2 = \tilde N_x$, exact. Integrate $\tilde M$ in $x$: $F = x^2/y + h(y)$. Match $F_y = -x^2/y^2 + h'(y) = 1 - x^2/y^2$, so $h'(y) = 1$ and $h = y$. The solution is

此时 $\tilde M_y = -2x/y^2 = \tilde N_x$：为恰当方程。对 $\tilde M$ 关于 $x$ 积分：$F = x^2/y + h(y)$。令 $F_y = -x^2/y^2 + h'(y) = 1 - x^2/y^2$，故 $h'(y) = 1$，$h = y$。解为

$$ F(x,y) = \frac{x^2}{y} + y = C \implies x^2 + y^2 = Cy, $$

a family of circles, recognizable as the homogeneous-equation answer too. The same curve emerges no matter which method you choose, a reassuring consistency check.

这是一族圆，与齐次方程的结果一致。无论选用哪种方法，得到的曲线相同，这是令人放心的自洽性验证。