Unit D1: First-Order ODEs: Separable and Linear单元 D1：一阶常微分方程：可分离与线性方程

Consider $y' = y(1-y)$. The slope depends only on $y$ (the equation is autonomous), so segments are constant along horizontal lines. Setting $f=0$ gives equilibria $y=0$ and $y=1$.

For $0 < y < 1$ we have $y' > 0$, so solutions increase toward $y=1$. For $y > 1$ we have $y' < 0$, so they decrease toward $y=1$. Thus $y=1$ is stable and $y=0$ is unstable. The field predicts the logistic S-curve before any integration.

考虑 $y' = y(1-y)$。斜率只依赖于 $y$（方程是自治的，autonomous），因此沿水平线段斜率不变。令 $f=0$ 得平衡解（equilibrium）$y=0$ 和 $y=1$。

在 $0 < y < 1$ 时 $y' > 0$，解向 $y=1$ 递增；在 $y > 1$ 时 $y' < 0$，解向 $y=1$ 递减。因此 $y=1$ 是稳定的，$y=0$ 是不稳定的。方向场在任何积分之前就预测了 logistic S 形曲线。

Classify the equilibria of $y' = (y-1)(y-3)^2$. The zeros of $f(y)=(y-1)(y-3)^2$ are $y=1$ and $y=3$. Test the sign of $f$ on each interval. For $y<1$, take $y=0$: $f=(-1)(9)=-9<0$. For $10$. For $y>3$, take $y=4$: $f=(3)(1)=3>0$.

At $y=1$ the sign goes from negative to positive, so arrows point away on both sides: $y=1$ is unstable. At $y=3$ the sign stays positive on both sides, so trajectories approach from below but leave from above: $y=3$ is semistable. The squared factor is exactly why $y=3$ is one-sided rather than fully stable, a subtlety the sign chart exposes that the linearization $f'(3)=0$ alone cannot.

对 $y' = (y-1)(y-3)^2$ 的平衡解分类。$f(y)=(y-1)(y-3)^2$ 的零点为 $y=1$ 和 $y=3$。逐区间检验 $f$ 的符号：$y<1$ 时取 $y=0$，得 $f=(-1)(9)=-9<0$；$1<y<3$ 时取 $y=2$，得 $f=(1)(1)=1>0$；$y>3$ 时取 $y=4$，得 $f=(3)(1)=3>0$。

在 $y=1$ 处符号由负变正，箭头向外：$y=1$ 不稳定。在 $y=3$ 处符号始终为正，轨迹从下方趋近但从上方离开：$y=3$ 是半稳定的（semistable）。二次因子正是 $y=3$ 只具有单侧稳定的原因——这是符号图所揭示的微妙之处，线性化 $f'(3)=0$ 单独无法给出此信息。

Verify that $y = \tan(x + C)$ solves $y' = 1 + y^2$, and find the interval of definition through $(0,0)$. Differentiate: $y' = \sec^2(x+C)$. The Pythagorean identity gives $\sec^2\theta = 1 + \tan^2\theta$, so $y' = 1 + \tan^2(x+C) = 1 + y^2$, which confirms the equation.

The data $y(0)=0$ forces $\tan(C)=0$, so $C=0$ and $y=\tan x$. Tangent is differentiable only between consecutive vertical asymptotes, so the solution through the origin lives on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. The same formula $\tan(x+C)$ describes a different solution on a different interval for each value of $C$, which is the point of the earlier remark that the interval of definition is part of the answer.

验证 $y = \tan(x + C)$ 满足 $y' = 1 + y^2$，并求过 $(0,0)$ 的解的定义区间。对其求导：$y' = \sec^2(x+C)$。利用勾股恒等式 $\sec^2\theta = 1 + \tan^2\theta$，得 $y' = 1 + \tan^2(x+C) = 1 + y^2$，确认方程成立。

初始条件 $y(0)=0$ 要求 $\tan(C)=0$，即 $C=0$，$y=\tan x$。正切函数仅在相邻垂直渐近线之间可微，因此过原点的解存在于 $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$。对每一个 $C$ 值，公式 $\tan(x+C)$ 在不同区间描述不同的解，这正说明了定义区间是答案的一部分。

Solve $y' = x\,y$ with $y(0)=3$. Separate and integrate:

Apply $y(0)=3$: $A = 3$. Hence $y = 3\,e^{x^2/2}$, valid for all $x$. The discarded case $y\equiv 0$ does not meet the data.

求解 $y' = x\,y$，$y(0)=3$。分离并积分：

代入 $y(0)=3$：$A=3$，故 $y = 3\,e^{x^2/2}$，对所有 $x$ 成立。被舍去的情况 $y\equiv 0$ 不满足初始条件。

Suppose $h(y)\neq 0$ on the range of interest and let $H$ be an antiderivative of $1/h$, so $H'(y)=1/h(y)$. Define $G$ with $G'(x)=g(x)$. If $y(x)$ solves $y'=g(x)h(y)$, then

Two functions with equal derivatives on an interval differ by a constant, so $H(y(x)) = G(x) + C$. This is precisely the implicit relation produced by separating and integrating, which proves the method rather than merely justifying the formal symbol pushing.

设 $h(y)\neq 0$ 且 $H$ 是 $1/h$ 的原函数，$G'(x)=g(x)$。若 $y(x)$ 满足 $y'=g(x)h(y)$，则

同一区间上导数相等的两个函数相差常数，因此 $H(y(x)) = G(x)+C$，这正是分离积分后得到的隐式关系，从而证明了该方法，而不仅仅是对形式操作的合理化。

Solve $y' = \dfrac{x(y^2-1)}{1}$, that is $y' = x(y^2-1)$, with $y(0)=2$. The right side factors, so separate for $y\neq\pm 1$:

Partial fractions give $\dfrac{1}{y^2-1} = \dfrac{1}{2}\left(\dfrac{1}{y-1} - \dfrac{1}{y+1}\right)$, so integrating both sides,

Apply $y(0)=2$: $\dfrac{1}{3}=A$, so $\dfrac{y-1}{y+1} = \dfrac{1}{3}e^{x^2}$. Solving for $y$ gives $y = \dfrac{3+e^{x^2}}{3-e^{x^2}}$. The division by $y^2-1$ discarded the constants $y\equiv 1$ and $y\equiv -1$; these are singular solutions of the ODE but neither meets the data, so they are recorded and set aside.

求解 $y' = x(y^2-1)$，$y(0)=2$。右端可分解，对 $y\neq\pm 1$ 分离变量：

部分分式得 $\dfrac{1}{y^2-1} = \dfrac{1}{2}\left(\dfrac{1}{y-1}-\dfrac{1}{y+1}\right)$，两端积分，

代入 $y(0)=2$：$\dfrac{1}{3}=A$，故 $\dfrac{y-1}{y+1}=\dfrac{1}{3}e^{x^2}$，解出 $y=\dfrac{3+e^{x^2}}{3-e^{x^2}}$。对 $y^2-1$ 的除法丢失了奇异解 $y\equiv 1$ 和 $y\equiv -1$；两者均不满足初始条件，记录后排除。

Solve $\dfrac{dy}{dx} = e^{x-y}$. The exponent splits as $e^{x-y}=e^x e^{-y}$, which is exactly the product form $g(x)h(y)$ with $g(x)=e^x$ and $h(y)=e^{-y}$. Separate:

Hence $y = \ln(e^{x} + C)$, valid wherever $e^x + C > 0$. If $C \ge 0$ the solution exists for all $x$; if $C < 0$ it exists only for $x > \ln(-C)$, where the logarithm's argument is positive. Recognizing the hidden product structure is the whole difficulty, since nothing about $e^{x-y}$ looks separable until the exponent is split.

求解 $\dfrac{dy}{dx} = e^{x-y}$。指数拆分为 $e^{x-y}=e^x e^{-y}$，恰好是乘积形式 $g(x)h(y)$，其中 $g(x)=e^x$，$h(y)=e^{-y}$。分离变量：

故 $y = \ln(e^{x}+C)$，在 $e^x+C>0$ 处成立。若 $C\ge 0$ 解对所有 $x$ 存在；若 $C<0$ 解仅在 $x>\ln(-C)$ 处存在。识别隐藏的乘积结构是全部难点，因为 $e^{x-y}$ 在指数未分拆时看起来并不可分离。

Solve $y' + \dfrac{1}{x}\,y = x$ for $x > 0$. Here $p = 1/x$, so $\mu = e^{\int dx/x} = e^{\ln x} = x$. Multiply through by $x$:

Therefore $y = \dfrac{x^2}{3} + \dfrac{C}{x}$. The term $C/x$ is the general solution of the associated homogeneous equation, and $x^2/3$ is one particular solution.

求解 $y' + \dfrac{1}{x}\,y = x$（$x>0$）。此处 $p=1/x$，故 $\mu = e^{\int dx/x} = e^{\ln x} = x$。全式乘以 $x$：

因此 $y = \dfrac{x^2}{3} + \dfrac{C}{x}$。其中 $C/x$ 是对应齐次方程的通解（general solution），$x^2/3$ 是一个特解（particular solution）。

We seek $\mu(x)$ so that $\mu(y'+py)$ is the derivative of $\mu y$. Expanding, $(\mu y)' = \mu y' + \mu' y$, which equals $\mu y' + \mu p y$ exactly when $\mu' = p\,\mu$. This is itself a separable equation:

Any nonzero constant multiple of this $\mu$ works, so we take the simplest representative. Multiplying the original equation by $\mu$ and integrating $(\mu y)' = \mu q$ gives the boxed formula, which shows the method is forced rather than guessed.

寻求 $\mu(x)$ 使 $\mu(y'+py)$ 成为 $\mu y$ 的导数。展开，$(\mu y)' = \mu y' + \mu' y$ 在 $\mu' = p\,\mu$ 时恰好等于 $\mu y' + \mu p y$。$\mu' = p\mu$ 本身就是可分离方程（separable）：

此 $\mu$ 的任意非零常数倍均可，取最简代表即可。将原方程乘以 $\mu$ 后积分 $(\mu y)'=\mu q$，得到框内公式，说明该方法是被迫的而非猜测的。

Solve $y' + (\tan x)\,y = \sec x$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Here $p = \tan x$, so

Multiply through by $\sec x$. The left side collapses to a derivative and the right side becomes $\sec^2 x$:

The integrating factor turns a messy coefficient into a clean single integration. Note that $\mu=\sec x$ is positive on the chosen interval, so no absolute value survives into the final answer.

在 $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ 上求解 $y' + (\tan x)\,y = \sec x$。这里 $p=\tan x$，

全式乘以 $\sec x$，左端化为导数，右端为 $\sec^2 x$：

积分因子将系数复杂的方程转化为干净的单次积分。注意 $\mu=\sec x$ 在所选区间上为正，因此绝对值不出现在最终答案中。

Solve $y' + y = q(x)$, $y(0)=0$, where $q(x)=1$ for $0\le x \le 1$ and $q(x)=0$ for $x>1$. The integrating factor is $\mu = e^{x}$ throughout, since $p=1$. On $[0,1]$, $(e^x y)' = e^x$, so $e^x y = e^x - 1$ using $y(0)=0$, giving $y = 1 - e^{-x}$.

At $x=1$ this reaches $y(1) = 1 - e^{-1}$. For $x>1$ the forcing is zero, so $(e^x y)' = 0$ and $e^x y = K$. Continuity at $x=1$ requires $e\cdot(1-e^{-1}) = K$, so $K = e - 1$ and $y = (e-1)e^{-x}$ for $x>1$.

The solution is continuous but its derivative jumps at $x=1$ because $q$ is discontinuous there. Solving on each piece and matching values at the breakpoint is the standard recipe for piecewise forcing, and it previews the step-function machinery of the Laplace transform.

求解 $y' + y = q(x)$，$y(0)=0$，其中 $q(x)=1$（$0\le x\le 1$），$q(x)=0$（$x>1$）。积分因子始终为 $\mu=e^x$。在 $[0,1]$ 上，$(e^x y)'=e^x$，利用 $y(0)=0$ 得 $e^x y=e^x-1$，即 $y=1-e^{-x}$。

在 $x=1$ 时值为 $y(1)=1-e^{-1}$。当 $x>1$ 时强迫项为零，$(e^x y)'=0$，$e^x y=K$。连续性要求 $e\cdot(1-e^{-1})=K$，所以 $K=e-1$，$y=(e-1)e^{-x}$（$x>1$）。

解是连续的，但导数在 $x=1$ 处有跳跃，因为 $q$ 在此不连续。逐段求解然后在断点匹配值是处理分段强迫项的标准方法，也预示了 Laplace 变换中阶跃函数的机理。

Solve $y' + 2y = 4x$, $y(0) = 1$. With $p=2$, $\mu = e^{2x}$, so $(e^{2x}y)' = 4x e^{2x}$. Integrating by parts,

Thus $y = 2x - 1 + Ce^{-2x}$. The data $y(0)=1$ gives $1 = -1 + C$, so $C = 2$ and $y = 2x - 1 + 2e^{-2x}$, valid for all real $x$.

求解 $y' + 2y = 4x$，$y(0)=1$。$p=2$，$\mu=e^{2x}$，故 $(e^{2x}y)'=4x e^{2x}$。分部积分，

故 $y = 2x-1+Ce^{-2x}$。由 $y(0)=1$：$1=-1+C$，即 $C=2$，$y=2x-1+2e^{-2x}$，对所有实数 $x$ 成立。

Solve $y' = \dfrac{x}{y}$, $y(1)=2$, and state the interval of definition. Separate and integrate:

Apply $y(1)=2$: $4 = 1 + K$, so $K=3$ and $y^2 = x^2 + 3$. The initial value is positive, so take the positive root: $y = \sqrt{x^2+3}$. The radicand $x^2+3$ is always positive, so this solution is differentiable for all real $x$. Choosing the correct branch of the square root is the step students miss; the sign of $y_0$ dictates it.

求解 $y'=\dfrac{x}{y}$，$y(1)=2$，并指明定义区间。分离并积分：

代入 $y(1)=2$：$4=1+K$，故 $K=3$，$y^2=x^2+3$。初始值为正，取正根：$y=\sqrt{x^2+3}$。被开方数 $x^2+3$ 恒正，解对所有实数 $x$ 可微。选取平方根的正确分支是学生常遗漏的步骤；$y_0$ 的符号决定了分支。

Solve $y' + 2xy = 1$, $y(0)=0$. The integrating factor is $\mu = e^{\int 2x\,dx} = e^{x^2}$, so $(e^{x^2}y)' = e^{x^2}$. The antiderivative of $e^{x^2}$ is not elementary, so write the answer with a definite integral anchored at the initial point:

This is a perfectly valid closed-form solution; the definite integral from $x_0=0$ automatically encodes the initial condition, since at $x=0$ the integral is zero and so is $y$. When the integrating-factor integral has no elementary antiderivative, the definite-integral form is the answer, not a failure to solve.

求解 $y' + 2xy = 1$，$y(0)=0$。积分因子为 $\mu=e^{x^2}$，故 $(e^{x^2}y)'=e^{x^2}$。$e^{x^2}$ 的原函数不是初等函数，因此以锚定在初始点的定积分写出答案：

这是完全合法的解析解；从 $x_0=0$ 开始的定积分自动编码了初始条件，因为 $x=0$ 时积分为零，$y$ 也为零。当积分因子积分没有初等原函数时，定积分形式就是答案，而非求解失败。

For the linear problem $y' + p(x)y = q(x)$, $y(x_0)=y_0$, suppose $p$ and $q$ are continuous on an open interval $I$ containing $x_0$. The integrating factor $\mu(x) = \exp\!\left(\int_{x_0}^{x} p(s)\,ds\right)$ is defined, positive, and differentiable on all of $I$. Multiplying and integrating from $x_0$ to $x$ gives

Existence is immediate: this formula defines a differentiable function on the entire interval $I$ and satisfies both the equation and the data. Uniqueness follows because any solution $z$ must satisfy $(\mu z)' = \mu q$, so $\mu z$ is determined up to its value at $x_0$, which the data fixes. Therefore $\mu z$ equals $\mu y$, and dividing by the nonzero $\mu$ gives $z = y$. Unlike the nonlinear case, where existence is only local, a linear IVP is solvable on the full interval where $p$ and $q$ are continuous; no finite-time blow-up can occur inside that interval.

对线性问题 $y'+p(x)y=q(x)$，$y(x_0)=y_0$，设 $p$ 和 $q$ 在含 $x_0$ 的开区间 $I$ 上连续。积分因子 $\mu(x)=\exp\!\left(\int_{x_0}^{x}p(s)\,ds\right)$ 在整个 $I$ 上有定义、为正且可微。从 $x_0$ 到 $x$ 乘以并积分得

存在性显然：此公式在整个区间 $I$ 上定义了一个可微函数，同时满足方程与初始条件。唯一性成立，因为任意解 $z$ 必须满足 $(\mu z)'=\mu q$，故 $\mu z$ 由其在 $x_0$ 处的值唯一确定，而数据固定了这个值。与非线性情形不同，线性 IVP 可在 $p$ 和 $q$ 连续的整个区间上求解；在该区间内不会发生有限时间爆破。

Consider $y' = y^{1/3}$ with $y(0) = 0$. Here $f = y^{1/3}$ is continuous, but $\partial f/\partial y = \tfrac13 y^{-2/3}$ blows up at $y=0$, so the Lipschitz condition fails there. Separating for $y>0$,

Both this curve and the constant $y \equiv 0$ satisfy the IVP, so uniqueness fails exactly because $\partial f/\partial y$ is not continuous at the initial point.

考虑 $y' = y^{1/3}$，$y(0)=0$。$f=y^{1/3}$ 连续，但 $\partial f/\partial y = \tfrac13 y^{-2/3}$ 在 $y=0$ 处爆破，Lipschitz 条件在此失效。对 $y>0$ 分离变量，

此曲线与常数解 $y\equiv 0$ 均满足初值问题，唯一性正是因为 $\partial f/\partial y$ 在初始点不连续而失效。

Take $y' = y^2$, $y(0)=1$. Both $f=y^2$ and $\partial f/\partial y = 2y$ are continuous everywhere, so a unique solution exists near $x=0$. Separating,

The solution exists only on $(-\infty, 1)$ and diverges as $x \to 1^-$. This shows the interval of existence is genuinely local: smoothness of $f$ does not extend the solution past the singularity, illustrating why the theorem only claims existence on some interval.

取 $y'=y^2$，$y(0)=1$。$f=y^2$ 和 $\partial f/\partial y=2y$ 处处连续，故 $x=0$ 附近存在唯一解。分离变量，

解仅在 $(-\infty,1)$ 上存在，当 $x\to 1^-$ 时发散。这表明存在区间确实是局部的：$f$ 的光滑性不能将解延伸过奇点，说明了为什么定理只断言在某个区间上存在解。

For $y' = 2\sqrt{|y|}$, count the solutions through $(0,0)$. Here $f = 2\sqrt{|y|}$ is continuous, but $\partial f/\partial y = 1/\sqrt{|y|}$ (for $y>0$) is unbounded as $y\to 0$, so Picard's hypothesis fails at the origin and uniqueness is in doubt. The constant $y\equiv 0$ is one solution. Separating for $y>0$ gives $\sqrt{y}=x+c$, so $y=(x+c)^2$ for $x\ge -c$.

For any $a\ge 0$ we can glue: $y=0$ on $(-\infty,a]$ and $y=(x-a)^2$ on $[a,\infty)$ is differentiable everywhere, including at $x=a$ where both pieces have value and slope zero, and it solves the IVP. Since $a$ is arbitrary, there are infinitely many solutions through the origin. This is the dramatic failure mode the Lipschitz condition rules out.

对 $y'=2\sqrt{|y|}$，计算过 $(0,0)$ 的解的个数。$f=2\sqrt{|y|}$ 连续，但 $\partial f/\partial y=1/\sqrt{|y|}$（$y>0$）在 $y\to 0$ 时无界，Picard 假设在原点失效，唯一性存疑。常数解 $y\equiv 0$ 是一个。对 $y>0$ 分离变量得 $\sqrt{y}=x+c$，即 $y=(x+c)^2$（$x\ge -c$）。

对任意 $a\ge 0$，可拼接：$y=0$（$x\le a$），$y=(x-a)^2$（$x\ge a$）处处可微，包括 $x=a$ 处两段值和斜率均为零，且满足初值问题。由于 $a$ 任意，过原点有无穷多个解。这是 Lipschitz 条件所排除的剧烈失效模式。

The existence half of the theorem is not magic; it is a fixed-point construction. Integrating $y'=f(x,y)$ from $x_0$ to $x$ turns the IVP into the equivalent integral equation

Define the Picard iterates by $y_0(x)=y_0$ and $y_{n+1}(x) = y_0 + \int_{x_0}^{x} f(t,y_n(t))\,dt$. When $f$ is continuous and Lipschitz in $y$ with constant $L$ on a rectangle, one shows by induction that the successive differences satisfy $|y_{n+1}(x)-y_n(x)| \le \dfrac{M L^{n}|x-x_0|^{n+1}}{(n+1)!}$, where $M$ bounds $|f|$. Because $\sum \dfrac{L^n h^{n+1}}{(n+1)!}$ converges (it is dominated by $\tfrac{1}{L}e^{Lh}$), the iterates form a uniformly Cauchy sequence and converge to a limit $y$ that satisfies the integral equation, hence the IVP.

Concretely, take $y'=y$, $y(0)=1$. Then $y_0=1$, $y_1 = 1+\int_0^x 1\,dt = 1+x$, $y_2 = 1+\int_0^x(1+t)\,dt = 1+x+\tfrac{x^2}{2}$, and in general $y_n = \sum_{k=0}^{n}\tfrac{x^k}{k!}$. The iterates are exactly the partial sums of $e^x$, so Picard iteration reproduces the known solution and shows where the exponential series comes from. The Lipschitz constant controls the rate, which is why continuity of $\partial f/\partial y$ (a Lipschitz supply) is the precise hypothesis that buys uniqueness.

存在性定理并非魔法，而是不动点构造。将 $y'=f(x,y)$ 从 $x_0$ 到 $x$ 积分，把初值问题化为等价的积分方程

定义 Picard 迭代：$y_0(x)=y_0$，$y_{n+1}(x)=y_0+\int_{x_0}^{x}f(t,y_n(t))\,dt$。当 $f$ 在矩形上连续且对 $y$ 满足 Lipschitz 常数为 $L$ 的条件时，可归纳证明相邻迭代的差满足 $|y_{n+1}(x)-y_n(x)|\le\dfrac{ML^n|x-x_0|^{n+1}}{(n+1)!}$，其中 $M$ 是 $|f|$ 的上界。因为 $\sum\dfrac{L^n h^{n+1}}{(n+1)!}$ 收敛（由 $\tfrac{1}{L}e^{Lh}$ 控制），迭代序列一致 Cauchy，收敛到满足积分方程的极限 $y$，从而满足初值问题。

具体地，取 $y'=y$，$y(0)=1$：$y_0=1$，$y_1=1+x$，$y_2=1+x+\tfrac{x^2}{2}$，$y_n=\sum_{k=0}^{n}\tfrac{x^k}{k!}$。迭代恰好是 $e^x$ 的部分和，Picard 迭代重现了已知解，并揭示了指数级数的来源。Lipschitz 常数控制收敛速率，这正是 $\partial f/\partial y$ 连续（提供 Lipschitz 条件）是保证唯一性的精确假设的原因。

A tank holds $100$ liters of water with $5$ kg of dissolved salt. Brine at $2$ kg per liter enters at $3$ liters per minute, and the well-mixed solution leaves at $3$ liters per minute. Let $A(t)$ be the kilograms of salt. The volume is constant at $100$, so

This is linear: $A' + 0.03A = 6$, with $\mu = e^{0.03t}$. The general solution is $A = 200 + Ce^{-0.03t}$. Using $A(0)=5$ gives $C = -195$, so $A(t) = 200 - 195e^{-0.03t}$, approaching the steady state $200$ kg.

一个容器装有 $100$ 升水，溶有 $5$ 千克盐。浓度为 $2$ 千克/升的盐水以 $3$ 升/分钟流入，充分混合后的溶液以 $3$ 升/分钟流出。设 $A(t)$ 为盐的千克数。体积恒为 $100$，故

这是线性方程（linear first-order）：$A' + 0.03A = 6$，$\mu=e^{0.03t}$。通解（general solution）为 $A=200+Ce^{-0.03t}$。由 $A(0)=5$ 得 $C=-195$，故 $A(t)=200-195e^{-0.03t}$，趋向稳态 $200$ 千克。

A cup of coffee at $90^\circ\mathrm{C}$ sits in a $20^\circ\mathrm{C}$ room. After $5$ minutes it has cooled to $60^\circ\mathrm{C}$. Find its temperature after $15$ minutes. With $T_{\text{env}}=20$, the law is $T' = -k(T-20)$. Let $u = T-20$, so $u' = -ku$ and $u = u_0 e^{-kt}$. Here $u_0 = 90-20 = 70$, giving $T = 20 + 70e^{-kt}$.

Use the $5$-minute data: $60 = 20 + 70e^{-5k}$, so $e^{-5k} = \tfrac{40}{70} = \tfrac47$ and $k = \tfrac15\ln\tfrac74$. Then

Substituting $u=T-T_{\text{env}}$ to reduce the cooling law to pure exponential decay is the move that makes the arithmetic clean. Notice we never needed the numerical value of $k$ on its own; the ratio $e^{-5k}=\tfrac47$ raised to the third power did all the work.

一杯 $90^\circ\mathrm{C}$ 的咖啡放在 $20^\circ\mathrm{C}$ 的房间里。$5$ 分钟后冷却至 $60^\circ\mathrm{C}$。求 $15$ 分钟后的温度。$T_\mathrm{env}=20$，方程为 $T'=-k(T-20)$。令 $u=T-20$，则 $u'=-ku$，$u=u_0 e^{-kt}$，$u_0=70$，故 $T=20+70e^{-kt}$。

利用 $5$ 分钟的数据：$60=20+70e^{-5k}$，故 $e^{-5k}=\tfrac{40}{70}=\tfrac47$，$k=\tfrac15\ln\tfrac74$。则

令 $u=T-T_\mathrm{env}$ 将冷却定律化为纯指数衰减，使计算简洁。注意我们无需单独求出 $k$ 的数值；比值 $e^{-5k}=\tfrac47$ 的三次方完成了所有工作。

A population grows by $P' = 0.5\,P\!\left(1 - \dfrac{P}{1000}\right)$ with $P(0)=100$. This is separable. Using partial fractions on $\dfrac{1}{P(1-P/1000)}$,

Exponentiating, $\dfrac{P}{1000-P} = A e^{0.5t}$. The data $P(0)=100$ gives $\dfrac{100}{900} = A$, so $A=\tfrac19$. Solving for $P$,

As $t\to\infty$, $P\to 1000$, the carrying capacity, matching the stable equilibrium found from the phase line in Section 1. The S-shaped curve rises fastest when $P=500$, exactly half the capacity, which is where $P'$ is maximal.

种群按 $P'=0.5\,P\!\left(1-\dfrac{P}{1000}\right)$，$P(0)=100$ 增长。这是可分离方程（separable）。对 $\dfrac{1}{P(1-P/1000)}$ 部分分式分解，

指数化，$\dfrac{P}{1000-P}=Ae^{0.5t}$。由 $P(0)=100$ 得 $\dfrac{100}{900}=A$，即 $A=\tfrac19$，解出 $P$：

当 $t\to\infty$ 时，$P\to 1000$，即环境容纳量，与第一节相线分析给出的稳定平衡一致。S 形曲线在 $P=500$ 时增长最快，恰好是容纳量的一半，即 $P'$ 最大的位置。

Solve $y' + y = y^2$. Here $n=2$, so let $v = y^{1-2} = y^{-1}$, giving $v' = -y^{-2}y'$. Dividing the equation by $y^2$,

This linear equation has $\mu = e^{-x}$, so $(e^{-x}v)' = -e^{-x}$, giving $e^{-x}v = e^{-x} + C$, hence $v = 1 + Ce^{x}$. Returning to $y = 1/v$ yields $y = \dfrac{1}{1 + Ce^{x}}$.

求解 $y'+y=y^2$。$n=2$，令 $v=y^{1-2}=y^{-1}$，则 $v'=-y^{-2}y'$。将方程除以 $y^2$，

此线性方程 $\mu=e^{-x}$，$(e^{-x}v)'=-e^{-x}$，得 $e^{-x}v=e^{-x}+C$，故 $v=1+Ce^x$。回代 $y=1/v$，得 $y=\dfrac{1}{1+Ce^x}$。

Suppose $y' = F(y/x)$. Let $v = y/x$, so $y = vx$ and $y' = v + x v'$. Substituting,

which is separable. For example $y' = (x+y)/x = 1 + y/x$ gives $F(v) = 1 + v$, so $F(v)-v = 1$ and $dv = dx/x$, hence $v = \ln|x| + C$ and $y = x\ln|x| + Cx$. The substitution converts the ratio structure into pure separation.

设 $y'=F(y/x)$。令 $v=y/x$，则 $y=vx$，$y'=v+xv'$。代入，

这是可分离方程（separable）。例如 $y'=(x+y)/x=1+y/x$ 给出 $F(v)=1+v$，$F(v)-v=1$，$dv=dx/x$，故 $v=\ln|x|+C$，$y=x\ln|x|+Cx$。换元将比值结构转化为纯分离变量。

Solve $y' + \dfrac{2}{x}\,y = \dfrac{y^3}{x^2}$ for $x>0$, with $y(1)=1$. This is Bernoulli with $n=3$, $p=2/x$, $q=1/x^2$. Let $v = y^{1-3} = y^{-2}$, so $v' = -2y^{-3}y'$. Dividing the equation by $y^3$ and multiplying by $-2$,

This is linear with $\mu = e^{\int -4\,dx/x} = e^{-4\ln x} = x^{-4}$. Then $(x^{-4}v)' = -2x^{-6}$, so $x^{-4}v = \tfrac{2}{5}x^{-5} + C$, giving $v = \tfrac{2}{5x} + Cx^4$. Since $v=y^{-2}$, the data $y(1)=1$ gives $v(1)=1 = \tfrac25 + C$, so $C=\tfrac35$ and

The branch is positive because $y(1)=1>0$. The Bernoulli substitution converts a nonlinear equation into the routine linear machinery of Section 3.

在 $x>0$ 上求解 $y'+\dfrac{2}{x}\,y=\dfrac{y^3}{x^2}$，$y(1)=1$。Bernoulli 方程，$n=3$，$p=2/x$，$q=1/x^2$。令 $v=y^{-2}$，$v'=-2y^{-3}y'$。将方程除以 $y^3$ 再乘以 $-2$，

线性方程，$\mu=e^{\int -4\,dx/x}=x^{-4}$。$(x^{-4}v)'=-2x^{-6}$，故 $x^{-4}v=\tfrac25 x^{-5}+C$，$v=\tfrac{2}{5x}+Cx^4$。由 $y(1)=1$ 得 $v(1)=1=\tfrac25+C$，$C=\tfrac35$，

分支取正是因为 $y(1)=1>0$。Bernoulli 换元将非线性方程转化为第三节的标准线性机制。

Solve $y' = \dfrac{x^2 + y^2}{xy}$ for $x>0$. The right side is constant along rays through the origin: dividing top and bottom by $x^2$ gives $y' = \dfrac{1 + (y/x)^2}{y/x} = F(v)$ with $v=y/x$. Using $y' = v + xv'$,

Integrating, $\tfrac12 v^2 = \ln|x| + C_1$, so $v^2 = 2\ln x + C$. Returning to $v = y/x$ gives $\dfrac{y^2}{x^2} = 2\ln x + C$, that is $y^2 = x^2(2\ln x + C)$. The cancellation $\tfrac{1+v^2}{v}-v = \tfrac1v$ is the small algebra step that makes the equation separable; without it the substitution would look stuck.

在 $x>0$ 上求解 $y'=\dfrac{x^2+y^2}{xy}$。右端沿过原点的射线为常数：分子分母均除以 $x^2$ 得 $y'=\dfrac{1+(y/x)^2}{y/x}=F(v)$，$v=y/x$。利用 $y'=v+xv'$，

积分得 $\tfrac12 v^2=\ln|x|+C_1$，$v^2=2\ln x+C$。回代 $v=y/x$ 得 $\dfrac{y^2}{x^2}=2\ln x+C$，即 $y^2=x^2(2\ln x+C)$。消去步骤 $\tfrac{1+v^2}{v}-v=\tfrac1v$ 是使方程可分离的小代数技巧；若未做此化简，换元后的方程看似无法继续。