普通微分方程入门：科学与工程的解析方法

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"这是一本面向数学、物理科学和工程专业本科生的《微分方程》教材，旨在介绍如何分析、图形化和近似求解微分方程。书中涵盖了常微分方程的理论、求解方法及其应用，包括使用拉普拉斯变换和幂级数方法等常见技巧。作者是Russell Herman，并且该书基于Tufte-book LaTeX文档类进行了格式调整。此外，自2014年夏天以来，这本书的版本就已经在网上发布。" 在微分方程的学习中，我们首先会接触到一阶微分方程。这部分内容包括自由落体问题，它常用来引出微分方程的概念。第一部分详细讲解了以下主题： 1.1 自由落体：这是物理学中的一个经典例子，用来演示如何用微分方程描述物体在重力作用下的运动。 1.2 第一阶微分方程： - 分离变量法：这种方法适用于形如dy/dx = f(x)g(y)的方程，通过分离变量将方程转化为两个积分，从而找到解。 - 线性一阶方程：这类方程通常形式为dy/dx + p(x)y = q(x)，可以通过乘以适当的积分因子来解。 - 准确微分方程：如果微分方程的左端可以表示为某个函数关于x和y的全微分，则该方程被称为准确微分方程，可以通过找到一个潜在的势函数来求解。 1.3 应用： - 增长与衰减：如细菌增长模型或放射性物质衰变，微分方程可以描述这些过程的变化。 - 牛顿冷却定律：用于描述物体温度随时间的变化，常用于热力学问题。 - 终端速度：在空气阻力或其他阻力影响下，物体下落或上升达到稳定速度的状态。 - 混合问题：例如化学反应中的溶质混合，可以通过微分方程来预测浓度随时间的变化。 - 正交轨迹：对于一组曲线，其正交轨迹是使得每条曲线上任一点与原曲线族正交的曲线。 - 追踪曲线*：涉及到物体追踪另一物体路径的问题，例如导弹追踪目标的运动。 1.4 其他一阶方程： - 贝努利方程*：这类方程具有非线性项，可以通过变换变量将其转化为线性方程。这本书深入浅出地介绍了微分方程的基础知识，不仅包含了理论，还提供了多种实际问题的应用示例，适合初学者和专业人士学习和参考。书中还可能涉及更高级的主题，如高阶微分方程、数值方法和偏微分方程，这些都是理解和解决科学与工程问题的关键工具。

2 differential equations

fact that some letters are used to represent constants. We will come back to

the more general form after we see how to solve the differential equation.

Consider

y(t) = 5. (1.2)

Recalling that the second derivative is just the derivative of a derivative, we

can rewrite this equation as





= 5. (1.3)

This tells us that the derivative of dy/dt is 5. Can you think of a function

whose derivative is 5? (Do not forget that the independent variable is t.) Yes,

the derivative of 5t with respect to t is 5. Is this the only function whose

derivative is 5? No! You can also differentiate 5t + 1, 5t + π, 5t − 6, etc.

In general, the derivative of 5t + C is 5, where C is an arbitrary integration

constant.

So, Equation (1.2) can be reduced to

= 5t + C. (1.4)

Now we ask if you know a function whose derivative is 5t + C. Well, you

might be able to do this one in your head, but we just need to recall the

Fundamental Theorem of Calculus, which relates integrals and derivatives.

Thus, we have

y(t) =

+ Ct + D, (1.5)

where D is a second integration constant.

Equation (1.5) gives the solution to the original differential equation. That

means that when the solution is placed into the differential equation, both

sides of the differential equation give the same expression. You can always

check your answer to a differential equation by showing that your solution

satisﬁes the equation. In this case we have

y(t) =



+ Ct + D



(5t + C) = 5.

Therefore, Equation (1.5) gives the general solution of the differential equa-

tion.

We also see that there are two arbitrary constants, C and D. Picking

any values for these gives a whole family of solutions. As we will see, the

equation

y(t) = 5 is a linear second order ordinary differential equation.

The general solution of such an equation always has two arbitrary constants.

Let’s return to the free fall problem. We solve it the same way. The only

difference is that we can replace the constant 5 with the constant −g. So, we

ﬁnd that

= −gt + C, (1.6)

and

y(t) = −

+ Ct + D. (1.7)

first order differential equations 3

Once you get down the process, it only takes a line or two to solve.

There seems to be a problem. Imagine dropping a ball that then un-

dergoes free fall. We just determined that there are an inﬁnite number of

solutions for the position of the ball at any time! Well, that is not possi-

ble. Experience tells us that if you drop a ball you expect it to behave the

same way every time. Or does it? Actually, you could drop the ball from

anywhere. You could also toss it up or throw it down. So, there are many

ways you can release the ball before it is in free fall producing many dif-

ferent paths, y(t). That is where the constants come in. They have physical

meanings.

If you set t = 0 in the equation, then you have that y(0) = D. Thus, D

gives the initial position of the ball. Typically, we denote initial values with

a subscript. So, we will write y(0) = y

. Thus, D = y

That leaves us to determine C. It appears at ﬁrst in Equation (1.6). Recall

that

, the derivative of the position, is the vertical velocity, v(t). It is

positive when the ball moves upward. We will denote the initial velocity

v(0) = v

. Inserting t = 0 in Equation (1.6), we ﬁnd that

y(0) = C. This

implies that C = v(0) = v

Putting this all together, we have the physical form of the solution for

free fall as

y(t) = −

+ v

t + y

. (1.8)

Doesn’t this equation look familiar? Now we see that the inﬁnite family of

solutions consists of free fall resulting from initially dropping a ball at po-

sition y

with initial velocity v

. The conditions y(0) = y

and

y(0) = v

are

called the initial conditions. A solution of a differential equation satisfying

a set of initial conditions is often called a particular solution. Specifying the

initial conditions results in a unique solution.

So, we have solved the free fall equation. Along the way we have be-

gun to see some of the features that will appear in the solutions of other

problems that are modeled with differential equation. Throughout the book

we will see several applications of differential equations. We will extend

our analysis to higher dimensions, in which we case will be faced with so-

called partial differential equations, which involve the partial derivatives of

functions of more that one variable.

But are we done with free fall? Not at all! We can relax some of the

conditions that we have imposed. We can add air resistance. We will visit

this problem later in this chapter after introducing some more techniques.

We can also provide a horizontal component of motion, leading to projectile

motion.

h(t)

Figure 1.1: Free fall far from the Earth

from a height h(t) from the surface.

Finally, we should also note that free fall at constant g only takes place

near the surface of the Earth. What if a tile falls off the shuttle far from the

surface of the Earth? It will also fall towards the Earth. Actually, the tile also

has a velocity component in the direction of the motion of the shuttle. So,

it would not necessarily take radial path downwards. For now, let’s ignore

that component.

4 differential equations

To look at this problem in more detail, we need to go to the origins of the

acceleration due to gravity. This comes out of Newton’s Law of Gravitation.

Consider a mass m at some distance h( t) from the surface of the (spheri-

cal) Earth. Letting M and R be the Earth’s mass and radius, respectively,

Newton’s Law of Gravitation states that

ma = F

h(t)

= −G

(R + h(t))

. (1.9)

Thus, we arrive at a differential equationHere G = 6.6730 × 10

−11

−1

−2

is the Universal Gravitational Constant,

M = 5.9736 ×10

kg and R = 6371 km

are the Earth’s mass and mean radius,

respectively. For h << R, GM/R

≈ g.

h(t)

= −

(R + h(t))

. (1.10)

This equation is not as easy to solve. We will leave it as a homework exercise

for the reader.

1.2 First Order Differential Equations

Before moving on, we first define an n-th order ordinary differential

equation. It is an equation for an unknown function y(x) that expresses an-th order ordinary differential equation

relationship between the unknown function and its ﬁrst n derivatives. One

could write this generally as

F(y

(n)

(x), y

(n−1)

(x), . . . , y

(x), y(x), x) = 0. (1.11)

Here y

(n)

(x) represents the nth derivative of y(x).

An initial value problem consists of the differential equation plus theInitial value problem.

values of the ﬁrst n −1 derivatives at a particular value of the independent

variable, say x

(n−1)

) = y

n−1

, y

(n−2)

) = y

n−2

, . . . , y(x

) = y

. (1.12)

A linear nth order differential equation takes the formLinear nth order differential equation

(x)y

(n)

(x) + a

n−1

(x)y

(n−1)

(x) + . . . + a

(x)y

(x) + a

(x)y(x)) = f (x).

(1.13)

If f (x) ≡ 0, then the equation is said to be homogeneous, otherwise it is

called nonhomogeneous.Homogeneous and nonhomogeneous

equations.

Typically, the ﬁrst differential equations encountered are ﬁrst order equa-

tions. A ﬁrst order differential equation takes the formFirst order differential equation

F(y

, y, x) = 0. (1.14)

There are two common ﬁrst order differential equations for which one can

formally obtain a solution. The ﬁrst is the separable case and the second is

a ﬁrst order equation. We indicate that we can formally obtain solutions, as

one can display the needed integration that leads to a solution. However,

the resulting integrals are not always reducible to elementary functions nor

does one obtain explicit solutions when the integrals are doable.

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