Differential Equations · ⁨미분방정식⁩

A differential equation 微分方程 relates a function to its derivatives. Many real situations are described by a rate: "the population grows at a rate proportional to its size" becomes $\dfrac{dP}{dt}=kP$. Setting up the equation from a verbal description – identifying what changes and what it is proportional to – is the first skill.

A solution is a function that satisfies the equation. To verify a proposed solution, differentiate it and substitute into the equation, checking that both sides agree. A general solution contains a constant $C$; a particular solution fixes $C$ from a condition.

A slope field 斜率场 draws a short line segment at many points, each with the slope $\dfrac{dy}{dx}$ the equation gives there. It pictures the family of solution curves without solving. To sketch one, evaluate the right-hand side at each grid point and draw a segment of that slope.

A solution curve follows the segments like a boat following a current. From a slope field you can sketch the particular solution through a given point, describe long-run behavior, and locate where solutions level off (slopes near zero) – reasoning about solutions purely from the picture.

Euler's method 欧拉方法 approximates a solution numerically by stepping along the slope field. Starting from a known point, take a small step $\Delta x$ and update:

Repeat for each step. Smaller steps give a better approximation. (This is a BC-only technique.)

Exam skill: be able to carry out two or three Euler steps by hand from a table, and know that Euler's method under- or over-estimates depending on the solution's concavity.

Worked example. Approximate $y(1)$ for $\dfrac{dy}{dx}=x+y$, $y(0)=1$, with step $\Delta x=0.5$. Step 1: slope at $(0,1)$ is $0+1=1$, so $y(0.5)\approx 1+1(0.5)=1.5$. Step 2: slope at $(0.5,1.5)$ is $0.5+1.5=2$, so $y(1)\approx 1.5+2(0.5)=2.5$.

A separable 可分离 differential equation can be written with all the $y$'s on one side and all the $x$'s on the other, then integrated:

This produces the general solution (with a $+C$), the main analytic method for solving differential equations in this course.

Worked example. Solve $\dfrac{dy}{dx}=xy$ with $y(0)=2$. Separating, $\int\frac{dy}{y}=\int x\,dx$ gives $\ln|y|=\frac{x^2}{2}+C$, so $y=Ae^{x^2/2}$. The condition $y(0)=2$ gives $A=2$, so $y=2e^{x^2/2}$.

An initial condition 初始条件 (a known point, e.g. $y(0)=5$) pins down the constant $C$. Solve for the general solution, substitute the condition to find $C$, then write the particular solution. Watch the domain – a particular solution is valid only on the interval containing the initial point.

The equation $\dfrac{dy}{dt}=ky$ says the rate of change is proportional to the amount – giving exponential growth or decay 指数增长. Separating variables yields

with $k>0$ for growth and $k<0$ for decay. This models unrestricted population growth, radioactive decay, and continuously compounded interest.

Real growth is limited by resources, so the logistic model 逻辑斯蒂模型 adds a carrying capacity 环境容纳量 $L$:

Growth is nearly exponential when $P$ is small, slows as $P$ approaches $L$, and stops at $P=L$. Key BC facts: the population levels off at $L$ ($\lim_{t\to\infty}P=L$), and it grows fastest when $P=\tfrac{L}{2}$ (the inflection point of the S-shaped curve). You are expected to read $L$ and the fastest-growth value directly from the equation.

Worked example. For $\dfrac{dP}{dt}=0.05\,P\!\left(1-\dfrac{P}{2000}\right)$, the carrying capacity is $L=2000$ (the population levels off there), and growth is fastest when $P=\dfrac{L}{2}=1000$ – both read straight off the equation, no solving needed.

• 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.