Differential Equations

A differential equation 微分方程 is an equation that relates a function to its own derivatives. It describes a situation by its rate of change. For example, "the rate of change of a quantity is proportional 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 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 – differing by a 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 without solving.

Each short segment has the slope $\dfrac{dy}{dx}$ at that point; a solution curve threads through, staying tangent to the field.

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 to them.

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; horizontal rows of equal slope suggest the rate depends only on $y$ (or 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:

Add the constant of integration once (on the $x$-side). This gives the general solution 通解.

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 $y=2e^{x^2/2}$.

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. There is exactly one solution through a given point. 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).

The constant gives a family of curves; an initial condition picks out one

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 also models motion along a line. 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 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.

• 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.