Differential equations
Introduced| English | Meaning |
|---|---|
| decay/dɪˈkeɪ/ | A decrease in a quantity over time, often modelled by an exponential relationship. |
| differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ | An equation involving a function and one or more of its derivatives. |
| rate of change/reɪt ɒv tʃeɪndʒ/ | The change in one quantity divided by the corresponding change in another. |
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | A specified starting value used to determine a particular solution of an equation or model. |
| separable/ˈsepərəbl/ | Describing a differential equation that can be rearranged with each variable and its differential on a separate side. |
| constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ | An arbitrary constant included in an indefinite integral to represent the family of antiderivatives. |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | A specific solution of an equation, often one selected by initial or boundary conditions. |
| general solution/ˈdʒenərəl səˈluːʃn/ | A family of solutions containing arbitrary constants or parameters that represents all solutions of the specified type. |
The equation that predicts the future
- How fast does a population grow? How quickly does a radioactive substance decay 衰减? How does a cup of coffee cool?
- All these questions lead to differential equations 微分方程 — equations that link a quantity to its rate of change 变化率. Solving them predicts the future.
What is a differential equation?
- A differential equation links a quantity to its rate of change.
- Example: $\dfrac{dy}{dx} = ky$ says the rate of growth is proportional to the current value — exponential growth.
Real-world example. Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - T_{\text{room}})$. The rate of cooling is proportional to how much hotter the object is than its surroundings.
The constant gives a family of curves; an initial condition 初始条件 picks out one
The slope field
dy/dx = a y
A differential equation gives the slope everywhere — the solution is the curve that follows those slopes from y₀.
A differential equation links a quantity to its rate of change.
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That is the defining feature: the equation involves both y and dy/dx (or higher derivatives).
Separable 可分离变量 equations
- For a separable first-order equation: put all $y$ terms on one side, all $x$ terms on the other, then integrate both sides.
- Example: $\dfrac{dy}{dx} = xy \Rightarrow \dfrac{1}{y}\,dy = x\,dx \Rightarrow \ln|y| = \dfrac{x^2}{2} + C$.
Don't forget the constant of integration 积分常数. When you integrate both sides, add $+C$ to one side. This constant determines which particular solution 特解 you get.
To solve a separable first-order differential equation, you:
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Separate the variables, then integrate each side.
General vs particular solutions
- The general solution 通解 contains a constant (a whole family of curves).
- An initial condition (a known value) fixes the constant → the particular solution.
An initial condition is used to:
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A known value fixes the arbitrary constant, turning the general solution into the particular one.
The general solution of a differential equation contains an arbitrary constant.
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Integrating introduces a constant, so the general solution is a whole family of curves.
Worked example
- Solve $\dfrac{dy}{dx} = xy$ with $y = 1$ at $x = 0$.
- Separate: $\dfrac{1}{y}\,dy = x\,dx$. Integrate: $\ln|y| = \dfrac{x^2}{2} + C$.
- Apply $y(0) = 1$: $\ln 1 = 0 + C \Rightarrow C = 0$.
- So $\ln y = \dfrac{x^2}{2} \Rightarrow y = e^{x^2/2}$.
allow_no_figure: Differential equations are solved algebraically — the solution process (separation of variables, integration) is the key content, not a diagram.
- Solve a first order differential equation to model real-life rates of change.
Solve dy/dx = xy with y(0) = 1. At x = 2, y = e^(x²/2). Find y (2 dp).
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y = e^(x²/2). At x = 2: y = e² ≈ 7.39.
If ln y = x²/2 + C and y(0) = 1, what is C?
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ln 1 = 0 + C, so C = 0.
You've got it
- solve a separable equation by splitting variables and integrating both sides
- the general solution has a constant; an initial condition gives the particular solution
- a differential equation links a quantity to its rate of change
Write from the particular solution y equals 3 e to the 2x
- dy/dx = 2y and y = 3 when x = 0.
Solve dy/dx = 2y with y = 3 when x = 0.
At x = 0 the exponential is 1, so A equals the given value 3.
What is the particular solution?
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The initial value fixes A at 3.