Differential Equations
AP Calculus AB Topic 7 5:24 English narration · English + 中文 subtitles burned in
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Transcript
Many real situations are described not by a formula, but by a rate.
许多真实情境不是用公式描述的，而是用变化率来描述。
A population grows faster when it is larger.
人口越多，增长得越快。
A hot drink cools faster when it is hotter.
热饮越烫，冷却得越快。
An equation that links a quantity to its own rate of change is called a differential equation.
把一个量和它自身的变化率联系起来的方程，叫做微分方程。
The most famous one says: the rate of change is proportional to the size.
最著名的一个说：变化率与这个量的大小成正比。
Solving it tells us how the quantity behaves over time.
解出它，就能知道这个量随时间如何变化。
Today we model with differential equations — read them from slope fields, and solve them by separating the variables.
今天我们用微分方程来建模——从斜率场中读出它们，并通过分离变量来求解。
Let's begin.
让我们开始吧。
First, translating words into an equation.
首先，把文字翻译成方程。
"The rate of change of y is proportional to y" becomes d y d t equals k y.
"y 的变化率与 y 成正比"就变成 d y d t 等于 k y。
A solution is a function that makes the equation true — and you can verify one by differentiating it and substituting back.
解是一个使方程成立的函数——你可以通过求导并代回去来验证它。
Notice something important: a differential equation usually has infinitely many solutions — a whole family of curves, differing by a constant.
注意一件重要的事：微分方程通常有无穷多个解——一整族曲线，彼此相差一个常数。
A slope field turns the equation into a picture.
斜率场把方程变成一幅图。
At each point on the grid, we draw a tiny segment whose slope is the derivative, evaluated right there.
在网格上的每一点，我们画一小段线段， 它的斜率就是导数在那一点的取值。
We never solve anything — yet the shape of every solution appears.
我们什么都没有解——但每一个解的形状都显现出来了。
A solution curve simply threads through the field, staying tangent to the segments as it goes.
一条解曲线只是穿行在这个场中，一路保持与那些线段相切。
So how do you use one in the exam?
那么考试里怎么用它呢？
You are given a slope field and a starting point.
题目会给你一个斜率场和一个起点。
Put your pencil at that point, and follow the segments — always staying tangent to them.
把笔尖放在那一点， 沿着线段走——始终与它们保持相切。
That traces the particular solution through that point.
这就画出了经过该点的特解。
You can also read behaviour straight off the picture: where the segments are flat, the solution is momentarily level; where they steepen, it climbs or falls faster.
你也可以直接从图上读出行为：线段平坦的地方，解暂时是水平的； 线段变陡的地方，它上升或下降得更快。
Now let's actually solve one.
现在我们真正来解一个。
The main method is separation of variables.
主要方法是分离变量法。
It works when the derivative factors into a function of x, times a function of y.
当导数能分解成 x 的函数乘以 y 的函数时，它就适用。
Move every y to one side, and every x to the other.
把每一个 y 移到一边，每一个 x 移到另一边。
Then integrate both sides.
然后对两边积分。
Add the constant of integration just once.
积分常数只加一次。
That gives the general solution — the whole family at once.
这就给出了通解——一次得到整族解。
Let's do one.
我们来做一个。
Solve: the derivative of y with respect to x equals x times y, with y equal to two when x is zero.
解：y 对 x 的导数等于 x 乘以 y，且当 x 为零时 y 等于二。
Separate: d y over y, equals x d x.
分离：y 分之 d y，等于 x d x。
Integrate both sides: the natural log of y equals x squared over two, plus C.
对两边积分：y 的自然对数等于 二分之 x 平方，加 C。
Take exponentials, and y equals A times e to the x squared over two.
取指数，得到 y 等于 A 乘以 e 的 二分之 x 平方 次方。
Now use the initial condition — at x equals zero, y is two, so A is two.
现在用初始条件—— 当 x 等于零时 y 是二，所以 A 是二。
The particular solution is two e to the x squared over two.
特解就是 二 e 的 二分之 x 平方 次方。
Watch for domain restrictions: keep the branch that contains the initial point.
还要注意定义域限制：保留包含初始点的那一支。
One model matters more than any other.
有一个模型比其他任何模型都重要。
When the rate of change is proportional to the amount — the derivative equals k times y — the solution is exponential: y equals y-nought, times e to the k t.
当变化率与这个量的大小成正比时—— 导数等于 k 乘以 y——解就是指数形式：y 等于 y 零，乘以 e 的 k t 次方。
Here y-nought is the starting amount.
这里 y 零是初始量。
If k is positive you get growth; if k is negative, you get decay.
如果 k 为正，就得到增长；如果 k 为负，就得到衰减。
This single equation models populations, radioactive decay, and cooling drinks.
这一个方程就能刻画人口、放射性衰变，以及饮料的冷却。
Here is decay in action.
来看衰减的实际例子。
A sample starts at fifty grams, and decays at a rate of minus zero point one times its amount, per year.
一个样品从五十克开始，以每年负零点一乘以其自身量的速率衰减。
So y equals fifty, e to the minus zero point one t.
于是 y 等于五十，乘以 e 的 负零点一 t 次方。
After ten years, that is fifty over e — about eighteen point four grams.
十年后，那就是 e 分之五十—— 大约十八点四克。
And the half-life?
那半衰期呢？
Set the amount to twenty-five and solve: the half-life is the natural log of two, divided by zero point one — about six point nine years.
把量设为二十五并求解：半衰期等于二的自然对数， 除以零点一——大约六点九年。
Notice it does not depend on the starting amount at all.
注意它完全不依赖于初始量。
A follow-up part often asks for the second derivative of an exponential model, and it looks harder than it is.
后续小问经常会问指数模型的二阶导数，它看起来比实际难。
Start from d y d t equals k y, and differentiate the whole equation again: d two y d t squared equals k times d y d t.
从 d y d t 等于 k y 出发， 把整个方程再求一次导：d 二 y d t 平方 等于 k 乘以 d y d t。
Now substitute the original equation back in, because d y d t is k y.
现在把原方程代回去， 因为 d y d t 就是 k y。
So the second derivative is k squared y.
所以二阶导数是 k 平方 乘以 y。
Notice what that tells you: k squared is positive, so for a positive quantity the curve is always concave up, whether k means growth or decay.
注意它告诉你什么：k 平方为正， 所以对于一个正的量，曲线总是上凹的，无论 k 表示增长还是衰减。
The rate of growth is itself growing.
增长的速率本身也在增长。
Before you go, three marks to keep.
结束之前，三个要守住的分。
First, for a separable equation, get all the y's on one side and all the x's on the other, then integrate both sides — and add the constant just once.
第一，对可分离的方程，把所有 y 移到一边，所有 x 移到另一边， 再对两边积分——常数只加一次。
Second, use the initial condition to pin down that constant, which gives the particular solution.
第二，用初始条件定出那个常数，就得到特解。
Third, in a slope field, a solution curve follows the segments, staying tangent.
第三，在斜率场里，解曲线沿着线段走，始终相切。
Remember: a differential equation gives you the slope — you must integrate to recover the function.
记住：微分方程给你的是斜率—— 你必须积分才能还原出函数。