Chapter 09 Differential Equations
微分方程
【Vocabulary · 词汇梳理】
paration of variable [ˌpəˈreɪʃn əv ˈveriəbl] 变量分离
[引]exponential growth rate [ˌekspəˈnenʃl ɡroʊθ reɪt]指数增长
be proportional to [bi prəˈpɔːrʃənl tu] 成比例
colony [ˈkɑːləni]菌落
radioactive [ˌreɪdioʊˈæktɪv] 放射性的
【导图】
A.BASIC DEFINITIONS 基本定义
A differential equation (d.e.) is any equation involving a derivative. In §E of Chapter 5, we solved some
simple differential equation. For example, we were given the velocity at time t of a particle moving along the x-axis (Example 54, Chapter 5):
v(t)=dx
dt
=4t3−3t2 (1)
From this we found the antiderivative:
x=∫(4t3−3t2)dt=t4−t3+C (2)
If the initial position (at time t=0) of the particle is x=3, then
x(0)=0−0+C=3
And c=3, so the solution to the initial-value problem is
x(t)=t4−t3+3 (3)
A solution of is any function that satisfies it. We e from (2) above that (1) has an infinite number of solutions, one for each real value of C. We call the family of functions (2) the general solution of (1). With the given initial condition x(0)=3, we determined C, thus finding the unique solution (3). This is called the particular solution.
Note that the particular solution must not only satisfy the differential equation and the initial condition, but the function must also be differentiable on an interval that contains the initial point. Features such as vertical tangents or asymptotes restrict the domain of the solution. Therefore, even
when they are defined by the same algebraic reprentation, particular solutions with different initial points may have different domains. Determining the proper domain is an important part of finding the particular solution.
A1. Rate of Change 变化率
A differential equation contains derivatives. A derivative gives information about the rate of change of a function. For example:
(1) If P is the size of a population at time t, then we can interpret
dP
dt
=0.0325P
As saying that at any time t the rate at which the population is growing is proportional (3.25%)to its size at that time.
(2) dQ
dt
=−(0.000275)Q tell us that at time t the rate at which the quantity Q is decreasing is proportional (0.0275%)to the quantity existing at that time.
(3) In psychology, one typical stimulus-respon situation, known as logarithmic respon, is that in which the respon y changes at a rate inverly proportional to the strength of the stimulus x. This is expresd neatly by the differential equation
零零落落
dy dx =
跳音记号
k
x
(k a constant)
神的九十亿个名字
If we suppo, further, that there is no respon when x=x0, then we have the condition y= 0when x=x0.
(4) We are familiar with
a=d2s
低碳环保作文dt2
=−32 ft/c2
for the acceleration due to gravity acting on an object at a height s above ground level at time t. The acceleration is the rate of change of the object’s velocity.
B.SLOPE FIELDS 斜率场
The slope field of is bad on the fact that can be interpreted as a statement about the slopes of its solution curves.
Example 1
dy
dx
=y tells us that at any point (x,y)on a solution curve the slope of the curve is equal to its y-coordinate. Since says that y is a function who derivative is also y, we know that
y=e x
is a solution. In fact, y=Ce x is a solution of for every constant C,since y′= Ce x=y. y′=y says that at any point where y=1, say (0,1)or (1,1)or (5,1), the slope of the solution curve is 1; at any point where y=3, say (0,3), (ln3,3), or (π,3), the slope equals 3; and so on.
(a)
(b)
(c)
and some gments of slope 3 at veral points where y =3. In Figure 09-1b we e the curve of x Figure 09-1c is the slope field for dy
dx =y . Slopes at many points are reprented by small gments of the tangents at tho points. The small gments approximate the solution curves. If we start at any point in the slope field and move so that the slope gments are always tangent to our motion, we will trace a solution curve. The slope field, as mentioned above, cloly approximates the family of solutions.
Example 2
The slope field for dy
dx =1
x is shown in Figure 09-2(a)
(a) Carefully draw the solution curve that pass through the point (1,0.5). (b) Find the general solution for the equation. Solution: (a)
In Figure 09-2b we started at the point (1,0.5), then moved from gment to gment drawing the curve to which the gments were tangent. The particular solution curve shown is the member of the family of solution curves
y =ln x +C
That goes through the point (1,0.5).
小油菜的功效与作用
(b)
Since we already know that, if dy
dx =1
x
, then y=∫1
x
dx=ln x+C, we are assured of having
found the correct general solution in (b)
In Figure 09-2c we have drawn veral particular solution curves of the Note that the vertical distance between any pair of curves is constant.
(a)(b)(c)
Example 3
黄金投资分析师Match each slope field in Figure 09-3 with the from the following t. Find the general solution for The particular solution that goes through (0,0)has been sketched in
(A) y′=cos x(B) dy
dx =2x(C) dy
dx
=3x2−3(D) y′=−2π
(a)(b)
(c)(d)
(A) goes with Figure 09-3c. The solution curves in the family y=sin x+C are quite obvious.
Figure 09- 2
(B) goes with Figure 09-3a. The general solution is the family of parabolas y=x2+C.
(C) goes with Figure 09-3b. The general solution is the family of cubics y=x3−3x+C.
逝青春(D) goes with Figure 09-3d; The general solution is the family of lines y=−π
2
x+C. Example 4
(a) Verify that relations of the form x2+y2=r2are solutions of dy
initialization
dx =−x
y
.
(b) Using the slope field in Figure 09-4 and your answer to (a), find the particular solution to given in (a) that contains point (4,−3).
Solution:
(a)
By differentiation equation x2+y2=r2implicitly, we get 2x+2y dy
dx
=0,from which
dy dx =−x
y
, which is the
(b)
x2+y2=r2describes circles centered at the origin. For initial point (4,−3) , 42+
(−3)2=25. So x2+y2=25. However, this is not the particular solution.
A particular solution must be differentiable on an interval containing the initial point. This circle is not differentiable at (−5,0)and (5,0). ( shows dy
dx
undefined when y=0, and the slope field shows vertical tangents along the x-axis.) Hence, the particular solution includes only the micircle in quadrants Ⅲand Ⅳ.
Solving x2+y2=25for y yields y=±√25−x2. The particular solution through point (4,−3)is y=−√25−x2with domain −5<x<5.
Figure 09- 4
