课程： 5.2 抗变性和异异等

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选择节古代物和异异等

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古代物和异异等
Differential equations are important parts of the study of economics, biology, engineering and physics. Because differential equations can describe and decay, they are very useful in modeling the decay of radioactive isotopes, the population growth of species or the change in investment return over time. Did you know that many of the equations that describe major concepts in physics and engineering are differential equations? How is Newton's Second Law (“force equals mass times acceleration”) a differential equation ?
::差异方程式是经济学、生物学、工程学和物理学研究的重要部分。由于差异方程式可以描述和衰变,因此在模拟放射性同位素衰减、物种人口增长或投资回报随时间变化方面非常有用。你知道描述物理学和工程学主要概念的许多方程式是差异方程式吗?牛顿的第二法(“力等于质量时间加速”)如何是差异方程式?

Differential Equations
::差异等量

Recall from the previous concept that the determination of an was introduced as an integral equation
::回顾以前的概念,即确定某一项是作为一个整体等式提出的

$\begin{aligned} \int f (x) d x = F (x) + C . \end{aligned}$

::f(x)dx=F(x)+C。

Here we have a function $\begin{aligned} f (x) \end{aligned}$ that we integrate to find the antiderivative function $\begin{aligned} F (x) \end{aligned}$ such that $\begin{aligned} F^{'} (x) = f (x) \end{aligned}$ . The equation $\begin{aligned} F^{'} (x) = f (x) \end{aligned}$ is called a differential equation.
::这里我们整合了一个 f( x) 函数, 以找到 F( x) 的反降解函数 F( x) , 即 F_( x) = f( x) 。 F_( x) = f( x) 等式被称为差异方程。

Problems in calculus are often presented as differential equations.
::微积分问题往往被作为差别方程提出。

Solve the general differential equation $\begin{aligned} f^{'} (x) = x^{\frac{2}{3}} + \sqrt{x} \end{aligned}$ .
::解决一般差分方程式 f(x)=x23+x。

We solve the equation by integrating the right side of the equation and have
::我们通过将等式的右侧结合,解决等式问题,

$\begin{aligned} f (x) = \int f^{'} (x) d x = \int x^{\frac{2}{3}} d x + \int \sqrt{x} d x . \end{aligned}$

:xxxxxxx23dxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

We can integrate both terms using the power rule, first noting that $\begin{aligned} \sqrt{x} = x^{\frac{1}{2}} \end{aligned}$ , and have
::我们可以使用权力规则, 将这两个术语合并, 首先指出 x=x12, 并且

$\begin{aligned} f (x) = \int x^{\frac{2}{3}} d x + \int x^{\frac{1}{2}} d x = \frac{3}{5} x^{\frac{5}{3}} + \frac{2}{3} x^{\frac{3}{2}} + C . \end{aligned}$

::f(xx)x23dxx12dx=35x53+23x32+C。

Notice the constant of integration $\begin{aligned} C \end{aligned}$ is included in the above general solution because without further information there are an infinite number of solutions. More problem information is needed in order to pick a specific value for $\begin{aligned} C \end{aligned}$ .
::请注意,整合C的常态包含在上述一般解决办法中,因为如果没有进一步的信息,就会有无限数量的解决方案。需要更多的问题信息,才能为C确定具体价值。

Suppose we wish to solve the following equation:
::假设我们希望解决以下等式:

$\begin{aligned} f^{'} (x) = e^{3 x} - 6 \sqrt{x} . \end{aligned}$
::f_(x)=e3x-6x。

We can solve the equation by integration and we have
::我们可以通过一体化解决等式,我们有

$\begin{aligned} f (x) = \frac{1}{3} e^{3 x} - 4 x^{\frac{3}{2}} + C . \end{aligned}$
:x)=13e3x-4x32+C。

We note that there are an infinite number of solutions. In some applications, we would like to designate exactly one solution. In order to do so, we need to impose a condition on the function $\begin{aligned} f \end{aligned}$ . We can do this by specifying the value of $\begin{aligned} f \end{aligned}$ for a particular value of $\begin{aligned} x \end{aligned}$ .
::我们注意到,有无限数量的解决办法,在某些应用中,我们希望指定一个确切的解决方法,为了这样做,我们需要对功能f规定一个条件。我们可以为此规定f对x特定值的价值。

In this problem, suppose that we add the condition that $\begin{aligned} f (0) = 1 \end{aligned}$ . This will specify exactly one value of $\begin{aligned} C \end{aligned}$ and thus one particular solution of the original equation:
::在此问题上,假设我们增加一个条件,即f(0)=1。这将确切地说明C的一个值,从而说明原始方程的一个特定解决办法:

Substituting $\begin{aligned} f (0) = 1 \end{aligned}$ into our general solution $\begin{aligned} f (x) = \frac{1}{3} e^{3 x} - 4 x^{\frac{3}{2}} + C \end{aligned}$ gives
::将 f( 0) =1 替换为我们的一般溶解 f( x) = 13e3x- 4x32+C 给予

$\begin{aligned} f (0) & = 1 = \frac{1}{3} e^{3 (0)} - 4 (0)^{\frac{3}{2}} + C o r \\ C & = 1 - \frac{1}{3} = \frac{2}{3} . \end{aligned}$

::f(0)=1=13e3(0)-4(0)32+C或C=1-13=23。

Hence the solution $\begin{aligned} f (x) = \frac{1}{3} e^{3 x} - 4 x^{\frac{3}{2}} + \frac{2}{3} \end{aligned}$ is the particular solution of the original equation $\begin{aligned} f^{'} (x) = e^{3 x} - 6 \sqrt{x} \end{aligned}$ satisfying the initial condition $\begin{aligned} f (0) = 1 \end{aligned}$ .
::因此,f(x)=13e3x-4x32+23的溶液是满足初始条件f(0)=1的原始方程式f*(x)=e3x-6x的特殊溶液。

We now can think of other problems that can be stated as differential equations with initial conditions. Consider the following problem:
::我们现在可以考虑其它问题,这些问题可以称为有初步条件的差别方程式。

Suppose the graph of $\begin{aligned} f (x) \end{aligned}$ includes the point (2, 6) and that the slope of the tangent line to $\begin{aligned} f (x) \end{aligned}$ at any point $\begin{aligned} x \end{aligned}$ is given by the expression $\begin{aligned} 3 x + 4 \end{aligned}$ . Find $\begin{aligned} f (- 2) \end{aligned}$ .
::假设f(x) 的图形包含点( 2, 6) , 3x+4 的表达式给出正切线的斜度到任何点 x 的 f(x) 的 f(x) 。 find f(-2) 。

We can re-state the problem in terms of a differential equation that satisfies an initial condition.
::我们可以用满足最初条件的差别方程式来重述问题。

$\begin{aligned} f^{'} (x) = 3 x + 4 \end{aligned}$ with $\begin{aligned} f (2) = 6 \end{aligned}$ .
::f_(x)=3x+4,f(2)=6。

By integrating the right side of the differential equation we have $\begin{aligned} f (x) = \frac{3}{2} x^{2} + 4 x + C \end{aligned}$ as the general solution.
::通过整合差异方程的右侧,我们有f(x)=32x2+4x+C作为一般解决办法。

Substituting the condition that $\begin{aligned} f (2) = 6 \end{aligned}$ gives
::替换 f(2)=6 给出的条件

$\begin{aligned} f (x) & = \frac{3}{2} x^{2} + 4 x + C \\ 6 & = \frac{3}{2} (2)^{2} + 4 (2) + C \\ = 6 + 8 + C \\ C & = - 8 \\ f (x) & = \frac{3}{2} x^{2} + 4 x + C \end{aligned}$

::f(x)=32x2+4x+C6=32(2)2+4(2)+C=6+8+CC=8f(x)=32x2+4x+C

Hence $\begin{aligned} f (x) = \frac{3}{2} x^{2} + 4 x - 8 \end{aligned}$ is the particular solution of the original equation $\begin{aligned} f^{'} (x) = 3 x + 4 \end{aligned}$ satisfying the initial condition $\begin{aligned} f (2) = 6 \end{aligned}$ .
::因此 f(x) = 32x2+4x-8 是满足初始条件f(2)=6的原始方程式f*(x)=3x+4的特殊解决方案。

Finally, since we are interested in the value $\begin{aligned} f (- 2) \end{aligned}$ , we obtain:
::最后,由于我们对f(-2)价值感兴趣,我们获得了:

$\begin{aligned} f (- 2) = - 10. \end{aligned}$
:至2)______________________________________________________________________________________

Examples
::实例

Example 1
::例1

Earlier, you were asked to explain how Newton's Second Law (“force equals mass times acceleration”) is a differential equation.
::早些时候,有人要求你解释牛顿的第二法(“力等于质量时间加速度”)如何是一个差别方程式。

Newton’s Second Law is simply: $\begin{aligned} F = m a = m \frac{d v}{d t} = m \frac{d^{2} x}{d t^{2}} \end{aligned}$ where the acceleration $\begin{aligned} a \end{aligned}$ can be expressed as the first derivative of velocity, or the second derivative of position.
::牛顿的第二法则很简单: F=ma=mdvdt=md2xdt2,其中加速度可以表示为速度的第一个衍生物,或位置的第二个衍生物。

Example 2
::例2

Solve the general differential equation $\begin{aligned} f^{'} (x) = x^{2} - 4 \end{aligned}$ with the condition that $\begin{aligned} f (1) = - 3 \end{aligned}$ .
::解决一般差分方程 f(x)=x2-4, 条件是 f(1)3 。

We solve the equation by integrating the right side of the equation and have
::我们通过将等式的右侧结合,解决等式问题,

$\begin{aligned} f (x) = \int f^{'} (x) d x = \int x^{2} d x - \int 4 d x . \end{aligned}$

::f(x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\dx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\D\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\d\\\\\\\\\\\\\\\\\dxdxdxdxdxdxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\4dx\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

We can integrate both terms and have
::我们可以将两个术语结合起来,并且

$\begin{aligned} f (x) = \int x^{2} d x - \int 4 d x = \frac{x^{3}}{3} - 4 x + C \end{aligned}$

:xx) x2dx4dx=x33-4x+C

Notice the constant of integration $\begin{aligned} C \end{aligned}$ is included in the above general solution, but there is more information that we have not used: $\begin{aligned} f (1) = - 3 \end{aligned}$ . The solution is now shaped by the constraint or initial condition to be $\begin{aligned} f (1) = \frac{(1)^{3}}{3} - 4 (1) + C = - 3 \end{aligned}$ which is satisfied if $\begin{aligned} C = \frac{2}{3} \end{aligned}$ .
::通知中,整合的常数C包含在上述一般解决办法中,但有更多信息我们没有使用:f(1)3。解决办法现在受以下制约或初始条件的制约或初始条件影响,即(1)=(1)33-4(1)+C3,如果C=23,则满足这一条件。

The particular solution to the differential equation is $\begin{aligned} f (x) = \frac{x^{3}}{3} - 4 x + \frac{2}{3} \end{aligned}$ .
::差别方程的具体解决办法是f(x)=x33-4x+23。

Review
::回顾

For #1-4, solve the differential equation for $\begin{aligned} f (x) \end{aligned}$ .
::对于 # 1-4, 解析 f( x) 的差别方程。

$\begin{aligned} f^{'} (x) = 4 x^{3} - 3 x^{2} + x - 3 \end{aligned}$ .
::f*(x)=4x3-3x2+x-3。

$\begin{aligned} f^{'} (x) = 2 e^{2 x} - 2 \sqrt{x} \end{aligned}$ .
::f_(x)=2e2x-2x。

$\begin{aligned} f^{'} (x) = \sin x - \frac{1}{e^{x}} \end{aligned}$ .
::f_(xx)=sin_x_1ex。

$\begin{aligned} f^{''} (x) = (2 + x) \sqrt{x} \end{aligned}$ .
::f*(x)=(2+x)x。

For #5-9, solve the differential equation for $\begin{aligned} f (x) \end{aligned}$ given the initial condition.
::对于# 5- 9, 根据初始条件, 解析 f( x) 的差别方程。

$\begin{aligned} f^{'} (x) = 6 x^{5} - 4 x^{2} + \frac{7}{3} \end{aligned}$ and $\begin{aligned} f (1) = 4 \end{aligned}$ .
::f_(x)=6x5-4x2+73和f(1)=4。

$\begin{aligned} f^{'} (x) = 3 x^{2} + e^{2 x} \end{aligned}$ and $\begin{aligned} f (0) = 3 \end{aligned}$ .
::f_(x)=3x2+e2x和f(0)=3。

$\begin{aligned} f^{'} (x) = \sqrt[3]{x^{2}} - \frac{1}{x^{2}} \end{aligned}$ and $\begin{aligned} f (1) = 3 \end{aligned}$ .
::f_(x)=x23-1x2和f(1)=3。

$\begin{aligned} f^{'} (x) = - 4 x^{3} + 12 x^{2} - 8 x \end{aligned}$ and $\begin{aligned} f (2) = 7 \end{aligned}$ .
::f*(x) 4x3+12x2-8x 和 f(2)= 7。

$\begin{aligned} f^{'} (x) = (2 \cos x - \sin x), - \frac{π}{2} \leq x \leq \frac{π}{2} \end{aligned}$ , and $\begin{aligned} f (\frac{π}{3}) = \sqrt{3} + \frac{1}{2} \end{aligned}$ .
::f*(x) = (2cosx-sinx) , 2x2, 和 f(%3) = 3+12 。

Suppose the graph of $\begin{aligned} f \end{aligned}$ includes the point (-2, 4) and that the slope of the tangent line to $\begin{aligned} f \end{aligned}$ at $\begin{aligned} x \end{aligned}$ is $\begin{aligned} - 2 x + 4 \end{aligned}$ . Find $\begin{aligned} f (5) \end{aligned}$ .
::假设 f 的图形包含点 (-2, 4) , 正切线的斜度为 x x x 的 f , 则为-2x+ 4。查找 f(5) 。

For #11-15, find the function $\begin{aligned} f \end{aligned}$ that satisfies the given conditions.
::对于#11-15, 找到符合特定条件的函数 f 。

$\begin{aligned} f^{''} (x) = \sin x - e^{- 2 x} \end{aligned}$ with $\begin{aligned} f^{'} (0) = \frac{5}{2} \end{aligned}$ and $\begin{aligned} f (0) = 0 \end{aligned}$ .
::f*(x) =sinx-e-2x 和 f*(0) = 52 和 f(0) = 0。

$\begin{aligned} f^{''} (x) = \frac{1}{\sqrt{x}} \end{aligned}$ with $\begin{aligned} f^{'} (4) = 7 \end{aligned}$ and $\begin{aligned} f (4) = 25 \end{aligned}$ .
::f*(x) = 1x, 加上 f`(4)=7 和 f(4)=25。

$\begin{aligned} f^{''} (x) = e^{3 x} + 7 \end{aligned}$ with $\begin{aligned} f^{'} (0) = 2 \end{aligned}$ and $\begin{aligned} f (0) = 4 \end{aligned}$ .
::f*(x) =e3x+7, 加上 f*(0) =2 和 f(0) = 4。

$\begin{aligned} f^{''} (x) = 2 \sec^{2} x \tan x \end{aligned}$ with $\begin{aligned} f^{'} (0) = 5 \end{aligned}$ and $\begin{aligned} f (0) = 7 \end{aligned}$ .
::f(x) = 2sec2 xtanx 和 f(0) = 7

$\begin{aligned} f^{''} (x) = 3 x^{2} + 2 x + 7 \end{aligned}$ with $\begin{aligned} f^{'} (2) = 0 \end{aligned}$ and $\begin{aligned} f (0) = 1 \end{aligned}$ .
::f(x) = 3x2+2x+7, 加上 f- (2)= 0 和 f( 0) = 1 。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

古代物和异异等

Differential Equations ::差异等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Differential Equations
::差异等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)