微分方程

A differential equation 微分方程 is an equation that relates a function to its own derivatives 导数. It describes a situation by its rate of change 变化率. For example, "the rate of change of a quantity is proportional 成正比 to its size" becomes

Learning to translate a sentence about a rate into a differential equation is the first skill of this unit.

A solution 解 of a differential equation is a function that makes it true. You can verify 验证 a proposed solution by differentiating it and substituting into the equation: if both sides match, it is a solution. Note a differential equation usually has infinitely many solutions – a whole family of solutions 解族 – indexed by an arbitrary constant 任意常数, which need not be an additive constant.

A slope field 斜率场 draws the differential equation as a grid of short segments: at each point $(x,y)$ the segment has slope $\dfrac{dy}{dx}$ evaluated there. To sketch one, plug several points into the right-hand side and draw a small segment with that slope at each. The picture shows the shape of the solution curves 解曲线 without solving.

Exam skill. A common part gives a portion of a slope field and asks you to sketch the particular solution through a given point – start at the point and follow the segments, staying tangent 相切 to them.

Worked example. Sketch the slope field of $\dfrac{dy}{dx}=x-y$ at a few points. Substitute each point into $x-y$:

point $(x,y)$	$(0,0)$	$(1,0)$	$(0,1)$	$(1,1)$	$(2,1)$	$(-1,1)$
slope $x-y$	$0$	$1$	$-1$	$0$	$1$	$-2$

Draw a flat segment at $(0,0)$ and $(1,1)$, a segment rising at $45^{\circ}$ at $(1,0)$ and $(2,1)$, and so on. Every point on the line $y=x$ has slope $0$, and every point on $y=x-1$ has slope $1$.

Solutions to a differential equation are functions or families of functions. Read a slope field to reason about behavior: where the segments are flat ($\tfrac{dy}{dx}=0$) the solution is momentarily level; where they steepen the solution rises or falls faster; equal slopes across every horizontal row suggest dependence only on $y$; equal slopes down every vertical column suggest dependence only on $x$.

Many exam differential equations are solved by separation of variables 分离变量法. If $\dfrac{dy}{dx}$ factors into a function of $x$ times a function of $y$, move all $y$'s to one side and all $x$'s to the other, then integrate both sides where the divided factor is nonzero:

Add the constant of integration 积分常数 once (on the $x$-side). This gives the general solution 通解. Also check constant solutions 常数解 with $h(y)=0$ separately; division by $h(y)$ can lose them.

Worked example. Solve $\dfrac{dy}{dx}=xy$ with the initial condition $y(0)=2$. Separate and integrate:

Applying $y(0)=2$ gives $A=2$, so the particular solution is $y=2e^{x^2/2}$.

Common mistakes. Adding the constant at the end: $y=e^{x^2/2}+C$ is wrong, because the constant enters when you integrate and becomes the multiplier $A=e^{C}$. Trying to separate an equation such as $\dfrac{dy}{dx}=x+y$: a sum cannot be written as $g(x)\,h(y)$.

An initial condition 初始条件 – a known point $(x_0,y_0)$ – pins down one curve from the family. Substitute it to solve for the constant $C$; the result is the particular solution 特解. Uniqueness 唯一性 requires conditions: for $y'=F(x,y)$, continuity 连续性 of $F$ and its partial derivative 偏导数 with respect to $y$ near the initial point is a sufficient local condition. Watch for domain restrictions 定义域限制: keep the branch that contains the initial point (for example, the correct sign of a square root).

Worked example. Solve $\dfrac{dy}{dx}=\dfrac{x}{y}$ with $y(0)=-3$.

Substitute $x=0$, $y=-3$ now, before solving for $y$: $\tfrac92=0+C$, so $C=\tfrac92$ and $y^2=x^2+9$. Taking the square root gives $y=\pm\sqrt{x^2+9}$; the initial value $y(0)=-3$ is negative, so the particular solution is

Writing $y=+\sqrt{x^2+9}$ is the classic error: it fails the initial condition.

The most important model is exponential growth and decay 指数增长与衰减. The equation $\dfrac{dy}{dt}=ky$ (rate proportional to size) has the solution:

where $y_0$ is the value at $t=0$. Here $k>0$ gives growth and $k<0$ gives decay. You can derive this by separation of variables ($\int \tfrac{dy}{y} = \int k\,dt$), and it models processes like population growth and radioactive decay 放射性衰变. In Newton's law of cooling 牛顿冷却定律 with fixed ambient temperature, apply this exponential model to the temperature difference from the surroundings. A follow-up part may ask for $\dfrac{d^2y}{dt^2}$: differentiate $\dfrac{dy}{dt}=ky$ again (using $\dfrac{dy}{dt}=ky$) to express it in terms of $y$.

Worked example. A sample decays by $\dfrac{dy}{dt}=-0.10\,y$ (in years) from $y_0=50\ \text{g}$. The solution is $y=50e^{-0.10t}$, so after $10$ years $y=50e^{-1}=18.4\ \text{g}$. Its half-life 半衰期 solves $25=50e^{-0.10t}$, i.e. $e^{-0.10t}=\tfrac12$, giving $t=\dfrac{\ln 2}{0.10}=6.9\ \text{years}$ – independent of the starting amount.

Worked example: Newton's law of cooling. Coffee at $90^{\circ}$C is left in a room at $20^{\circ}$C. Its temperature $T$ obeys $\dfrac{dT}{dt}=-k\,(T-20)$, with $t$ in minutes. After $5$ minutes it is $70^{\circ}$C. When does it reach $40^{\circ}$C?

The temperature difference $D=T-20$ decays exponentially: $D=70e^{-kt}$ (the start value is $90-20=70$), so

Use $T(5)=70$: $50=70e^{-5k}$, so $k=\dfrac{\ln(7/5)}{5}\approx 0.0673$ per minute. For $T=40$:

Common error: writing $T=90e^{-kt}$, which would cool the coffee towards $0^{\circ}$C instead of towards room temperature.

• Solve a separable equation by getting all $y$ on one side and all $x$ on the other, then integrating both sides (add $+C$ once).
• Use the initial condition to find $C$ (a particular solution).
• Sketch or read a slope field: the little segments show $\tfrac{dy}{dx}$ at each point, and a solution curve follows them.
• Recognise exponential models $\tfrac{dy}{dt}=ky\Rightarrow y=Ce^{kt}$ (growth/decay).
• A differential equation gives the slope — you must integrate to recover the function.