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CHAPTER 3
LINEAR EQUATIONS OF HIGHER ORDER SECTION 3.1 INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS In this section the central ideas of the theory of linear differential equations are introduced and illustrated concretely in the context of second-order equations. These key concepts include superposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2), linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). This discussion of second-order equations serves as preparation for the treatment of nth order linear equations in Section 3.2. Although the concepts in this section may seem somewhat abstract to students, the problems set is quite tangible and largely computational. In each of Problems 1–16 the verification that y1 and y2 satisfy the given differential equation is a routine matter. As in Example 2, we then impose the given initial conditions on the general solution y c1 y1 c2 y2 . This yields two linear equations that determine the values of the constants c1 and c2 . 1.
Imposition of the initial conditions y 0 0 , y 0 5 on the general solution y x c1e x c2e x yields the two equations c1 c2 0 , c1 c2 5 with solution c1
5 , 2
5 5 c2 . Hence the desired particular solution is y x e x e x . 2 2 2.
Imposition of the initial conditions y 0 1 , y 0 15 on the general solution
y x c1e3 x c2 e 3 x yields the two equations c1 c2 1 , 3c1 3c2 15 , with solution
c1 2 , c2 3 . Hence the desired particular solution is y x 2e3 x 3e 3 x . 3.
Imposition of the initial conditions y 0 3 , y 0 8 on the general solution
y x c1 cos 2 x c2 sin 2 x yields the two equations c1 3 , 2c2 8 with solution c1 3 ,
c2 4 . Hence the desired particular solution is y x 3cos 2 x 4sin 2 x . 4.
Imposition of the initial conditions y 0 10 , y 0 10 on the general solution
y x c1 cos5 x c2 sin 5 x yields the two equations c1 10 , 5c2 10 with solution
c1 3 , c2 4 . Hence the desired particular solution is y x 10cos5 x 2sin 5 x . 167 Copyright © 2015 Pearson Education, Inc.
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168
5.
INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS
Imposition of the initial conditions y 0 1 , y 0 0 on the general solution
y x c1e x c2 e 2 x yields the two equations c1 c2 1 , c1 2c2 0 with solution
c1 2 , c2 1 . Hence the desired particular solution is y x 2e x e 2 x . 6.
Imposition of the initial conditions y 0 7 , y 0 1 on the general solution
y x c1e 2 x c2 e 3 x yields the two equations c1 c2 7 , 2c1 3c2 1 with solution
c1 4 , c2 3 . Hence the desired particular solution is y x 4e 2 x 3e 3 x . 7.
Imposition of the initial conditions y 0 2 , y 0 8 on the general solution
y x c1 c2 e x yields the two equations c1 c2 2 , c2 8 with solution c1 6 ,
c2 8 . Hence the desired particular solution is y x 6 8e x . 8.
Imposition of the initial conditions y 0 4 , y 0 2 on the general solution
y ( x ) c1 c2e3 x yields the two equations c1 c2 4 , 3c2 2 with solution c1 c2 9.
14 , 3
2 1 . Hence the desired particular solution is y x 14 2e3 x . 3 3
Imposition of the initial conditions y 0 2 , y 0 1 on the general solution
y x c1e x c2 xe x yields the two equations c1 2 , c1 c2 1 with solution
c1 2 c2 1 . Hence the desired particular solution is y x 2e x xe x . 10.
Imposition of the initial conditions y 0 3 , y 0 13 on the general soution
y x c1e5 x c2 xe5 x yields the two equations c1 3 , 5c1 c2 13 with solution c1 3 ,
c2 2 . Hence the desired particular solution is y x 3e5 x 2 xe5 x . 11.
Imposition of the initial conditions y (0) 0 , y 0 5 on the general solution
y x c1e x cos x c2e x sin x yields the two equations c1 0 , c1 c2 5 with solution
c1 0 , c2 5 . Hence the desired particular solution is y x 5e x sin x . 12.
Imposition of the initial conditions y 0 2 , y 0 0 on the general solution
y x c1e 3 x cos 2 x c2 e 3 x sin 2 x yields the two equations c1 2 , 3c1 2c2 5 with
solution c1 2 , c2 3 . Hence the desired particular solution is y x e 3 x 2 cos 2 x 3sin 2 x .
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Section 3.1 169
13.
Imposition of the initial conditions y 1 3 , y 1 1 on the general solution
y x c1 x c2 x 2 yields the two equations c1 c2 3 , c1 2c2 1 with solution c1 5 ,
c2 2 . Hence the desired particular solution is y x 5 x 2 x 2 . 14.
Imposition of the initial conditions y 2 10 , y 2 15 on the general solution c2 3c 10 , 4c1 2 15 with solution 16 8 16 c1 3 , c2 16 . Hence the desired particular solution is y x 3x 2 3 . x y x c1 x 2 c2 x 3 yields the two equations 4c1
15.
Imposition of the initial conditions y 1 7 , y 1 2 on the general solution
y x c1 x c2 x ln x yields the two equations c1 7 , c1 c2 2 with solution c1 7 ,
c2 5 . Hence the desired particular solution is y x 7 x 5 x ln x . 16.
Imposition of the initial conditions y 1 2 , y 1 3 on the general solution
y x c1 cos ln x c2 sin ln x yields the two equations c1 2 , c2 3 . Hence the de-
sired particular solution is y x 2 cos ln x 3sin ln x .
c c 2 c c 1 c 0 unless either c 0 or c 1 . , then y y 2 2 2 x x x2 x
17.
If y
18.
If y cx 3 , then yy cx 3 6cx 6c 2 x 4 6 x 4 unless c 2 1 .
19.
x 3 2 x 1 2 x 3 2 If y 1 x , then yy y 1 x 0. 4 2 4
20.
Linearly dependent, because f x cos2 x sin 2 x g x .
21.
3 2 3 2 Linearly independent, because x x x if x 0 , whereas x x x if x 0 .
22.
Linearly independent, because 1 x c 1 x would require that c 1 with x 0 , but
2
2
c 0 with x 1 . Thus there is no such constant c.
23.
Linearly independent, because f x g x if x 0 , whereas f x g x if x 0 .
24.
Linearly dependent, because g x 2 f x .
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25.
INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS
f x e x sin x and g x e x cos x are linearly independent, because f x kg x
would imply that sin x k cos x , whereas sin x and cos x are linearly independent, as noted in Example 3. 26.
To see that f x and g x are linearly independent, assume that f x cg x , and then substitute both x 0 and x
27.
2
.
The operator notation used elsewhere in this chapter is convenient here. Let L y denote y py qy . Then L yc 0 and L y p f , so L yc y p 0 f f .
28.
If y x 1 c1 cos x c2 sin x , then y x c1 sin x c2 cos x , so the initial conditions y 0 y 0 1 yield c1 2 , c2 1 . Hence y x 1 2 cos x sin x .
29.
There is no contradiction, because if the given differential equation is divided by x 2 to 4 get the form in Equation (8) in the text, then the resulting functions p x and x 6 q x 2 are not continuous at x 0 . x
30.
(a) y1 x 3 and y2 x 3 are linearly independent because x 3 c x 3 would require that c 1 with x 1 , but c 1 with x 1 .
(b) For x 0 , W y1 , y2 0 because y2 y1 . For x 0 , W y1 , y2
x3
x3
3x 2
3 x 2
0
because y2 y1 . At x 0 , y2 has left- and right-hand derivatives both equal to zero, so that W y1 , y2
0 0 0 once again. Thus W y1 , y2 is identically zero. 0 0
The fact that W y1 , y2 0 everywhere does not contradict Theorem 3, because when the given equation is written in the required form, namely y cient functions p x
3 3 y 2 y 0 , the coeffix x
3 3 and q x 2 are not continuous at x 0 . x x
31.
W y1 , y2 2 x vanishes at x 0 , whereas if y1 and y2 were (linearly independent) solutions of an equation y py qy 0 with p and q both continuous on an open interval I containing x 0 , then Theorem 3 would imply that W 0 on I.
32.
(a) Because W y1 y2 y1 y2 , we have Copyright © 2015 Pearson Education, Inc.
Section 3.1 171
A
dW A y1 y2 y1 y2 y1y2 y1 y2 dx y1 Ay2 y2 Ay1
y1 By2 Cy2 y2 By1 Cy1 B y1 y2 y1 y2 BW y1 , y2 ,
and thus A x
dW B x W x . dx
(b) The differential equation for W x found in a can be rewritten as
dW B x W x 0 , since A x is never zero. We can solve this equation as a linear dx A x
B x first-order equation: Multiplying by the integrating factor x exp dx gives A x B x dW B x B x exp dx exp dx W x 0 , A x dx A x A x or B x d dx W x 0 , exp dx A x
or B x exp dx W x K , A x B x dx , as desired. where K is a constant. Finally we find W K exp A x (c) Because the exponential factor is never zero.
In Problems 33–42 we give the characteristic equation, its roots, and the corresponding general solution. 33.
r 2 3r 2 0 ; r 1, 2 ; y x c1e x c2 e 2 x
34.
r 2 2r 15 0 ; r 3, 5 ; y x c1e 5 x c2e3 x
35.
r 2 5r 0 ; r 0, 5 ; y x c1 c2 e 5 x
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INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS
3 ; y x c1 c2 e 3x 2 2
36.
2r 2 3r 0 ; r 0,
37.
1 2r 2 r 1 0 ; r 1, ; y x c1e x 2 c2e x 2
38.
1 3 4r 2 8r 3 0 ; r , ; y x c1e x 2 c2e 3x 2 2 2
39.
4r 2 4r 1 0 ; r
1 (repeated); y x c1 c2 x e x 2 2
40.
9r 2 12r 4 0 ; r
2 (repeated); y x c1 c2 x e 2 x 3 3
41.
4 5 6r 2 7r 20 0 ; r , ; y x c1e 4 x 3 c2e5x 2 3 2
42.
4 3 35r 2 r 12 0 ; r , ; y x c1e 4 x 7 c2 e3x 5 7 5
In Problems 43–48 we first write and simplify the equation with the indicated characteristic roots, and then write the corresponding differential equation. 43.
r 0 r 10 r 2 10r 0 ;
44.
r 10 r 10 r 2 100 0 ;
45.
r 10 r 10 r 2 20r 100 0 ;
46.
r 10 r 100 r 2 110r 1000 0 ;
47.
r 0 r 0 r 2 0 ;
48.
r 1 2 r 1 2 r 2 2 r 1 0 ; y 2 y y 0
49.
The solution curve with y 0 1 , y 0 6 is y x 8e x 7e 2 x . We find that
y 10 y 0
y 100 y 0 y 20 y 100 y 0 y 110 y 1000 y 0
y 0
7 4 16 7 16 . It follows that y ln , y x 0 when x ln , so that e x and e 2 x 4 7 49 4 7
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Section 3.1 173
7 16 so the high point on the curve is ln , (0.56, 2.29) , which looks consistent with 4 7 Fig. 3.1.6. 50.
The two solution curves satisfying y 0 a and y 0 b , as well as y 0 1 , are given by y 2a 1 e x a 1 e 2 x y 2b 1 e x b 1 e 2 x . Subtraction, followed by division by a b , gives 2e x e 2 x , so it follows that x ln 2 . Now substitution in either formula gives y 2 , so the common point of intersection is ln 2, 2 .
51.
(a) The substitution v ln x gives
dy dy dv 1 dy . dx dv dx x dv Then another differentiation using the chain and product rules gives y
d2y dx 2 d dy dx dx d 1 dy dx x dv
y
1 dy 1 d dy x 2 dv x dx dv
1 dy 1 d dy dv x 2 dv x dv dv dx
1 dy 1 d 2 y . x 2 dv x 2 dv 2 Substitution of these expressions for y and y into Eq. (21) in the text then yields immediately the desired Eq. (23):
d2y dy a 2 b a cy 0 . dv dv (b) If the roots r1 and r2 of the characteristic equation of Eq. (23) are real and distinct, then a general solution of the original Euler equation is y x c1e r1v c2 e r2v c1 ev c2 e v c1 x r1 c2 x r2 . r1
r2
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174
52.
53.
INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS
dy 2 y 0 , whose characteristic dv 2 equation r 2 1 0 has roots r1 1 and r2 1 . Because e v x , the corresponding genc eral solution is y c1e v c2e v c1 x 2 . x The substitution v ln x yields the converted equation
dy 2 dy 12 y 0 , whose chardv 2 dv acteristic equation r 2 r 12 0 has roots r1 4 and r2 3 . Because ev x , the cor-
The substitution v ln x yields the converted equation
responding general solution is y c1e 4 v c2e3v c1 x 4 c2 x 3 . dy 2 dy 4 3 y 0 , whose 2 dv dv 3 1 characteristic equation 4r 2 4r 3 0 has roots r1 and r2 . Because ev x , 2 2 3v /2 v /2 3/2 c2 e c1 x c2 x1/2 . the corresponding general solution is y c1e
54.
The substitution v ln x yields the converted equation 4
55.
The substitution v ln x yields the converted equation
dy 2 0 , whose characteristic dv 2 equation r 2 0 has repeated roots r1 , r2 0 . Because v ln x , the corresponding general solution is y c1 c2 v c1 c2 ln x .
56.
dy 2 dy 4 4 y 0 , whose char2 dv dv 2 acteristic equation r 4r 4 0 has roots r1 , r2 2 . Because ev x , the corresponding
The substitution v ln x yields the converted equation general solution is y c1e 2 v c2ve 2 v x 2 c1 c2 ln v .
SECTION 3.2 GENERAL SOLUTIONS OF LINEAR EQUATIONS Students should check each of Theorems 1 through 4 in this section to see that, in the case n 2 , it reduces to the corresponding theorem in Section 3.1. Similarly, the computational problems for this section largely parallel those for the previous section. By the end of Section 3.2 students should understand that, although we do not prove the existence-uniqueness theorem now, it provides the basis for everything we do with linear differential equations. The linear combinations listed in Problems 1–6 were discovered “by inspection”—that is, by trial and error.
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Section 3.2 175
1.
5 8 2 x 3x 2 1 5 x 8 x 2 0 for all x. 2 3
2.
4 5 5 2 3x 2 1 10 15 x 2 0 for all x.
3.
1 0 0 sin x 0 e x 0 for all x.
4.
17 17 1 17 2sin 2 x 3cos2 x 0 for all x, because sin 2 x cos2 x 1 . 2 3
5.
1 17 34 cos2 x 17 cos 2 x 0 for all x, because 2 cos2 x 1 cos 2 x .
6.
1 e x 1 cosh x 1 sinh x 0 , because cosh x sinh x
7.
8.
1 x e e x and 2
1 x e e x . 2
1 x x2 W 0 1 2 x 2 is nonzero everywhere. 0 0 2 ex
e2 x
e3 x
W ex ex
2e 2 x 4e 2 x
3e3 x 2e6 x is never zero. 9e 3 x
9.
W e x cos2 x sin 2 x e x is never zero.
10.
W x 7e x x 1 x 4 is nonzero for x 0 .
11.
W x 3e 2 x is nonzero if x 0 .
12.
W x 2 2 cos2 ln x 2sin 2 ln x 2 x 2 is nonzero for x 0 .
In each of Problems 13-20 we first form the general solution y x c1 y1 x c2 y2 x c3 y3 x , then calculate y x and y x , and finally impose the given initial conditions to determine the values of the coefficients c1 , c2 , c3 .
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13.
GENERAL SOLUTIONS OF LINEAR EQUATIONS
Imposition of the initial conditions y 0 1 , y 0 2 , y 0 0 on the general solu-
tion y x c1e x c2e x c3e 2 x yields the three equations
c1 c2 c3 1, c1 c2 2c3 2, c1 c2 4c3 0 , with solution c1 y x 14.
4 1 , c2 0 , c3 . Hence the desired particular solution is given by 3 3
1 4e x e 2 x . 3
Imposition of the initial conditions y 0 0 , y 0 0 , y 0 3 on the general solu-
tion y x c1e x c2e 2 x c3e3 x yields the three equations
c1 c2 c3 1, c1 2c2 3c3 2, c1 4c2 9c3 0 , 3 3 , c2 3 , c3 . Hence the desired particular solution is given by 2 2 3 3 y x e x 3e 2 x e3 x . 2 2
with solution c1
15.
Imposition of the initial conditions y 0 2 , y 0 0 , y 0 0 on the general solu-
tion y x c1e x c2 xe x c3 x 2 e x yields the three equations
c1 2, c1 c2 0, c1 2c2 2c3 0 , with solution c1 2 , c2 2 , c3 1 . Hence the desired particular solution is given by y x 2 2 x x2 ex .
16.
Imposition of the initial conditions y 0 1 , y 0 4 , y 0 0 on the general solu-
tion y x c1e x c2e 2 x c3 xe 2 x yields the three equations
c1 c2 1, c1 2c2 c3 4, c1 4c2 4c3 0 with solution c1 12 , c2 13 , c3 10 . Hence the desired particular solution is given by y x 12e x 13e 2 x 10 xe 2 x .
17.
Imposition of the initial conditions y 0 3 , y 0 1 , y 0 2 on the general solu-
tion y x c1 c2 cos 3x c3 sin 3x yields the three equations c1 c3 3, 3c3 1, 9c2 2
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Section 3.2 177
29 2 1 , c2 , c3 . Hence the desired particular solution is given 9 9 3 29 2 1 by y x cos 3x sin 3x . 9 9 3 with solution c1
18.
Imposition of the initial conditions y 0 1 , y 0 0 , y 0 0 on the general solu-
tion y x e x c1 c2 cos x c3 sin x yields the three equations c1 c2 1, c1 c2 c3 0, c1 2c3 0
with solution c1 2 , c2 1 , c3 1 . Hence the desired particular solution is given by y x e x 2 cos x sin x .
19.
Imposition of the initial conditions y 1 6 , y 1 14 , y 1 22 on the general solu-
tion y x c1 x c2 x 2 c3 x 3 yields the three equations
c1 c2 c3 6, c1 2c2 3c3 14, 2c2 6c3 22 , with solution c1 1 , c2 2 , c3 3 . Hence the desired particular solution is given by y x x 2 x 2 3x 3 .
20.
Imposition of the initial conditions y 1 1 , y 1 5 , y 1 11 on the general solu-
tion y x c1 x c2 x 2 c3 x 2 ln x yields the three equations
c1 c2 1, c1 2c2 c3 5, 6c2 5c3 11 , with solution c1 2 , c2 1 , c3 1 . Hence the desired particular solution is given by y x 2 x x 2 x 2 ln x .
In each of Problems 21-24 we first form the general solution y x yc x y p x c1 y1 x c2 y2 x y p x ,
then calculate y x , and finally impose the given initial conditions to determine the values of the coefficients c1 and c2 . 21.
Imposition of the initial conditions y 0 2 , y 0 2 on the general solution
y x c1 cos x c2 sin x 3 x yields the two equations c1 2 , c2 3 2 with solution
c1 2 , c2 5 . Hence the desired particular solution is given by y x 2 cos x 5sin x 3x .
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22.
GENERAL SOLUTIONS OF LINEAR EQUATIONS
Imposition of the initial conditions y 0 0 , y 0 10 on the general solution
y x c1e 2 x c2e 2 x 3 yields the two equations c1 c2 3 0 , 2c1 2c2 10 with so-
lution c1 4 , c2 1 . Hence the desired particular solution is given by y x 4 e 2 x e 2 x 3 .
23.
Imposition of the initial conditions y 0 3 , y 0 11 on the general solution
y x c1e x c2e3 x 2 yields the two equations c1 c2 2 3 , c1 3c2 11 with solu-
tion c1 1 , c2 4 . Hence the desired particular solution is given by y x e x 4e 3 x 2 .
24.
Imposition of the initial conditions y 0 4 , y 0 8 on the general solution
y x c1e x cos x c2e x cos x x 1 yields the two equations c1 1 4 , c1 c2 1 8
with solution c1 3 , c2 4 . Hence the desired particular solution is given by y x e x 3cos x 4 sin x x 1 .
25.
Ly L y1 y2 Ly1 Ly2 f g
26.
(a) y1 2 and y2 3x (b) y y1 y2 2 3x
27.
The equations c1 c2 x c3 x 2 10, c2 2c3 x 0, 2c3 0 (the latter two obtained by successive differentiation of the first one) evidently imply that c1 c2 c3 0 .
28.
If we differentiate the equation c0 c1 x c2 x 2 cn x n 0 repeatedly, n times in succession, the result is the system c0 c1 x c2 x 2 cn x n 0 c1 2c2 x ncn x n 1 0
(n 1)! cn 1 n ! cn x 0 n ! cn 0 of n 1 equations in the n 1 coefficients c0 , c1 , c2 , , cn . The last equation implies that cn 0 , whereupon the preceding equation gives cn 1 0 , and so forth. Thus it follows that all of the coefficients must vanish.
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Section 3.2 179
29.
If c0e rx c1 xe rx cn x n e rx 0 , then division by e rx yields c0 c1 x cn x n 0 , so the result of Problem 28 applies.
30.
When the equation x 2 y 2 xy 2 y 0 is rewritten in standard form
2 2 y y 2 y 0 , x x 2 2 we see that the coefficient functions p1 and p2 x 2 are not continuous at x x x 0 . Thus the hypotheses of Theorem 3 are not satisfied. 31.
(a) Substitution of x a in the differential equation gives y a py a q a . (b) If y 0 1 and y 0 0 , then the equation y 2 y 5 y 0 implies that y 0 2 y 0 5 y 0 5 .
32.
Let the functions y1 , y2 , , yn be chosen as indicated. Then evaluation at x a of the
k 1 st
derivative of the equation c1 y1 c2 y2 cn yn 0 yields ck 0 . Thus
c1 c2 cn 0 , so the functions are linearly independent. 33.
This follows from the fact that 1
1
1
a a2
b b2
c b a c b c a c2
when a, b, and c are distinct, which can be verified by expanding both sides of the equation.
r x , and neither V nor exp r x vanishes.
34.
W f1 , f 2 , , f n V exp
36.
If y vy1 , then substitution of the derivatives y vy1 vy1 , y vy1 2vy1 vy1 in the differential equation y py qy 0 gives
n
n
i 1 i
i 1 i
vy1 2vy1 vy1 p vy1 vy1 q vy1 0 , or v y1 py1 qy1 vy1 2vy1 pvy1 0 . But the terms within brackets vanish because y1 is a solution, and this leaves the equation y1v 2 y1 py1 v 0 .
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GENERAL SOLUTIONS OF LINEAR EQUATIONS
We can solve this by separating variables and integrating:
v y 2 1 p leads to v y1
ln v 2 ln y1 p x dx ln C , or v x
C p x dx e , y12
or
e p x dx v x C dx K . 2 y1 With C 1 and K 0 this gives the second solution e p x dx y2 x y1 x dx . 2 y 1 37.
When we substitute y vx 3 in the given differential equation and simplify, we get the v 1 separable equation xv v 0 , which we write as . Integrating gives v x A ln v ln x ln A , and then solving for v leads to v , or finally v x A ln x B . x With A 1 and B 0 we get v x ln x , and thus y2 x x 3 ln x .
38.
When we substitute y vx 3 in the given differential equation and simplify, we get the v 7 separable equation xv 7v 0 , which we write as . Integrating gives v x A ln v 7 ln x ln A , and then solving for v leads to v 7 , or finally x A 1 1 v x 6 B . With A 6 and B 0 we get v x 6 , and hence y2 x 3 . 6x x x
39.
When we substitute y ve x 2 in the given differential equation and simplify, we eventually get the simple equation v 0 , with general solution v x Ax B . With A 1 and B 0 we get v x x , and hence y2 x xe x 2 .
40.
When we substitute y vx in the given differential equation and simplify, we get the v separable equation v v 0 , which we write as 1 . Integrating gives v
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Section 3.2 181
ln v x ln A , and then solving for v leads to v Ae x , or finally v x Ae x B .
With A 1 and B 0 we get v x e x , and hence y2 x xe x . 41.
When we substitute y ve x in the given differential equation and simplify, we get the v x 1 . Inteseparable equation 1 x v xv 0 , which we write as 1 v 1 x 1 x grating gives ln v x ln 1 x ln A , and then solving for v leads to v A 1 x e x , or finally v x A 1 x e x dx A 2 x e x B . With A 1
and B 0 we get v x 2 x e x , and hence y2 x 2 x . 42.
When we substitute y vx in the given differential equation and simplify, we get the separable equation x x 2 1 v 2v , which we write as
v 2 2 1 1 . 2 v x x 1 x 1 x 1 x Integrating gives ln v 2 ln x ln 1 x ln 1 x ln A , A 1 x 2
1 A 2 1 , or finally x x 1 1 v x A x B . With A 1 and B 0 we get v x x , and hence x x 2 y2 x x 1 .
and then solving for v leads to v
43.
2
When we substitute y vx in the given differential equation and simplify, we get the
separable equation x x 2 1 v 2 4 x 2 v , which we write using the method of partial fractions as v 2 4 x2 2 1 1 . 2 v x x 1 x 1 x 1 x
Integrating gives ln v 2 ln x ln 1 x ln 1 x ln A ,
and then solving for v leads to v
1 A 1 1 A 2 , 2 x x x 2 1 2 1 x 1 x 2
or finally
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GENERAL SOLUTIONS OF LINEAR EQUATIONS
1 1 1 v x A ln 1 x ln 1 x B . 2 x 2 1 1 1 With A 1 and B 0 we get v x ln 1 x ln 1 x , and hence x 2 2 x 1 x y2 x 1 ln . 2 1 x 44.
When we substitute y vx 1/2 cos x in the given differential equation and simplify, we eventually get the separable equation cos x v 2 sin x v , which we write as v 2sin x . Integrating gives v cos x ln v 2 ln cos x ln A ln sec 2 x ln A ,
and then solving for v leads to v A sec2 x , or finally v x A tan x B . With A 1 and B 0 we get v x tan x , and hence y2 x tan x x 1/2 cos x x 1/2 sin x .
SECTION 3.3 HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS This is a purely computational section devoted to the single most widely applicable type of higher order differential equations—linear ones with constant coefficients. In Problems 1–20, we first write the characteristic equation and list its roots, then give the corresponding general solution of the given differential equation. Explanatory comments are included only when the solution of the characteristic equation is not routine. 1.
r 2 4 r 2 r 2 0 ; r 2, 2 ; y x c1e 2 x c2e 2 x
2.
3 2 r 2 3r r 2r 3 0 ; r 0, ; y x c1 c2e3 x 2 2
3.
r 2 3r 10 r 5 r 2 0 ; r 5, 2 ; y x c1e 2 x c2 e 5 x
4.
2 r 2 7 r 3 2 r 1 r 3 0 ; r
5.
r 2 6r 9 r 3 0 ; r 3 (repeated); y x c1e 3 x c2 xe 3 x
6.
r 2 5r 5 0 ; r
1 ,3 ; y x c1e x 2 c2e3 x 2
2
5 5 5 ; y x c1e 2 2
5
x
c2e
5 5 x 2
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Section 3.3 183
3 (repeated); y x c1e3 x 2 c2 xe3 x 2 2
7.
4r 2 12r 9 2r 3 0 ; r
8.
r 2 6r 13 0 ; r
6 16 3 2i ; y x e3 x c1 cos 2 x c2 sin 2 x 2
9.
r 2 8r 25 0 ; r
8 36 4 3i ; y x e 4 x c1 cos 3x c2 sin 3 x 2
10.
3 5r 4 3r 3 r 3 5r 3 0 ; r 0,0,0, ; y x c1 c2 x c3 x 2 c4 e 3 x 5 5
11.
r 4 8r 3 16r 2 r 2 r 4 0 ; r 0,0, 4, 4 ; y x c1 c2 x c3e 4 x c4 xe4 x
12.
r 4 3r 3 3r 2 r r r 1 0 ; r 0,1,1,1 ; y x c1 c2 e x c3 xe x c4 x 2 e x
13.
2 2 2 9r 3 12r 2 4r r 3r 2 0 ; r 0, , ; y x c1 c2e 2 x 3 c3 xe 2 x 3 3 3
14.
r 4 3r 2 4 r 2 1 r 2 4 0 ; r 1,1, 2i ; y x c1e x c2e x c3 cos 2 x c4 sin 2 x
15.
4r 4 8r 2 16 r 2 4 r 2 r 2 0 ; r 2, 2, 2, 2 ;
2
2
3
2
2
2
y x c1e 2 x c2 xe 2 x c3e 2 x c4 xe 2 x
16.
r 4 18r 2 81 r 2 9 0 ; r 3i, 3i ; y x c1 c2 x cos 3x c3 c4 x sin 3x
17.
6r 4 11r 2 4 2r 2 1 3r 2 4 0 ; r
2
y x c1 cos
i 2i ; , 2 3
x x 2x 2x c2 sin c3 cos c4 sin 2 2 3 3
18.
r 4 16 r 2 4 r 2 4 0 ; r 2, 2, 2i ; y x c1e 2 x c2 e 2 x c3 cos 2 x c4 sin 2 x
19.
Factoring by grouping gives r 3 r 2 r 1 r r 2 1 r 2 1 r 1 r 1 0 ; 2
r 1, 1, 1 ; y x c1e x c2 e x c3 xe x .
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20.
HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
r 4 2r 3 3r 2 2r 1 r 2 r 1 0 ; 2
1 3i 1 3i , ; 2 2
3 x 2 3 y x e x 2 c1 c2 x cos x e c3 c4 x sin x 2 2 21.
Imposition of the initial conditions y 0 7 , y 0 11 on the general solution
y x c1e x c2e3 x yields the two equations c1 c2 7 , c1 3c2 11 with solution
c1 5 , c2 2 . Hence the desired particular solution is y x 5e x 2e3 x . 22.
Imposition of the initial conditions y 0 3 , y 0 4 on the general solution x x c1 c2 y x e x 3 c1 cos c2 sin 4 with yields the two equations c1 3 , 3 3 3 3 solution c1 3 , c2 5 3 . Hence the desired particular solution is x x y x e x 3 3cos 5 3 sin . 3 3
23.
Imposition of the initial conditions y 0 3 , y 0 1 on the general solution
y x e3 x c1 cos 4 x c2 sin 4 x yields the two equations c1 3 , 3c1 4c2 1 with solu-
tion c1 3 , c2 2 . Hence the desired particular solution is y x e3 x 3cos 4 x 2 sin 4 x .
24.
Imposition of the initial conditions y 0 1 , y 0 1 , y 0 3 on the general solu-
tion y x c1 c2e 2 x c3e x 2 yields the three equations c1 c2 c3 1, 2c2
c3 c 1, 4c2 3 3 , 2 4
7 1 with solution c1 , c2 , c3 4 . Hence the desired particular solution is 2 2 7 1 2x y x e 4e x 2 . 2 2 25.
Imposition of the initial conditions y 0 1 , y 0 0 , y 0 1 on the general solu-
tion y x c1 c2 x c3e 2 x 3 yields the three equations c1 c3 1, c2
2c3 0, 3
4c3 1, 9
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Section 3.3 185
13 3 9 , c2 , c3 . Hence the desired particular solution is 4 2 4 13 3 9 2 x 3 . y x x e 4 2 4
with solution c1
26.
Imposition of the initial conditions y 0 1 , y 0 1 , y 0 3 on the general solu-
tion y x c1 c2e 5 x c3 xe 5 x yields the three equations
c1 c2 3, 5c2 c3 4, 25c2 10c3 5 , with solution c1 y x 27.
24 9 , c2 , c3 5 . Hence the desired particular solution is 5 5
24 9 5 x e 5 xe 5 x . 5 5
First we spot the root r 1 . Then long division of the polynomial r 3 3r 2 4 by r 1
yields the quadratic factor r 2 4r 4 r 2 , with roots r 2, 2 . Hence the gen2
eral solution is y x c1e x c2 e 2 x c3 xe 2 x .
28.
First we spot the root r 2 . Then long division of the polynomial 2r 3 r 2 5r 2 by the factor r 2 yields the quadratic factor 2 r 2 3r 1 2 r 1 r 1 , with roots 1 r 1, . Hence the general solution is y x c1e 2 x c2 e x c3e x 2 . 2
29.
First we spot the root r 3 . Then long division of the polynomial r 3 27 by r 3 3 3 3 . Hence the general soluyields the quadratic factor r 2 3r 9 , with roots r i 2 2 3 3 3 3 tion is y x c1e 3 x e3 x 2 c2 cos x c3 sin x . 2 2
30.
First we spot the root r 1 . Then long division of the polynomial r 4 r 3 r 2 3r 6 by r 1 yields the cubic factor r 3 2r 2 3r 6 . Next we spot the root r 2 , and another long division yields the quadratic factor r 2 3 , with roots r i 3 . Hence the general solution is y x c1e x c2 e2 x c3 cos 3x c4 sin 3x .
31.
The characteristic equation r 3 3r 2 4r 8 0 has the evident root r 1 , and long divi2 sion then yields the quadratic factor r 2 4r 8 r 2 4 , corresponding to the complex conjugate roots 2 2i . Hence the general solution is y x c1e x e 2 x c2 cos 2 x c3 sin 2 x .
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32.
HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
The characteristic equation r 4 r 3 3r 2 5r 2 0 has the root r 2 , as is readily found by trial and error, and long division then yields the factorization 3 r 2 r 1 0 . Thus we obtain the general solution y x c1e 2 x c2 c3 x c4 x 2 e x .
33.
Knowing that y e3 x is one solution, we divide the characteristic polynomial r 3 3r 2 54 by r 3 and get the quadratic factor r 2 6r 18 r 3 9 . Hence the 2
general solution is y x c1e3 x e 3 x c2 cos 3x c3 sin 3 x . 34.
Knowing that y e 2 x 3 is one solution, we divide the characteristic polynomial 3r 3 2r 2 12r 8 by 3r 2 and get the quadratic factor r 2 4 . Hence the general solution is y x c1e 2 x 3 c2 cos 2 x c3 sin 2 x .
35.
The fact that y cos 2 x is one solution tells us that r 2 4 is a factor of the characteristic polynomial 6r 4 5r 3 25r 2 20r 4 . Then long division yields the quadratic factor 1 1 6r 2 5r 1 3r 1 2 r 1 , with roots r , . Hence the general solution is 2 3 x 2 x 3 y x c1e c2e c3 cos 2 x c4 sin 2 x .
36.
The fact that y e x sin x is one solution tells us that r 1 1 r 2 2r 2 is a factor 2
of the characteristic polynomial 9r 3 11r 2 4r 14 . Then long division yields the linear factor 9 r 7 . Hence the general solution is y x c1e7 x 9 e x c2 cos x c3 sin x . 37.
The characteristic equation is r 4 r 3 r 3 r 1 0 , so the general solution is
y x A Bx Cx 2 De x . Imposition of the given initial conditions yields the equa-
tions A D 18, B D 12, 2C D 13, D 7
with solution A 11 , B 5 , C 3 , D 7 . Hence the desired particular solution is y x 11 5 x 3 x 2 7e x . 38.
Given that r 5 is one characteristic root, we divide r 5 into the characteristic polynomial r 3 5r 2 100r 500 and get the remaining factor r 2 100 . Thus the general solution is y x Ae5 x B cos10 x C sin10 x . Imposition of the given initial conditions yields the equations A B 0, 5 A 10C 10, 25 A 100 B 250 ,
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Section 3.3 187
with solution A 2 , B 2 , C 0 . Hence the desired particular solution is y x 2e5 x 2 cos10 x . 39.
40.
The characteristic polynomial is r 2 r 3 6r 2 12r 8 , so the differential equation is y 6 y 12 y 8 y 0 . 3
The characteristic polynomial is r 2 r 2 4 r 3 2r 2 4r 8 , so the differential
equation is y 2 y 4 y 8 y 0 . 41.
The characteristic polynomial is r 2 4 r 2 4 r 4 16 , so the differential equation is y (4) 16 y 0 .
42.
The characteristic polynomial is r 2 4 r 6 12 r 4 48r 2 64 , so the differential 3
equation is y (6) 12 y (4) 48 y 64 y 0 . 43.
(a) Given a complex number z x iy we define r to be x 2 y 2 and to bethe x y unique angle satisfying cos , sin , and . Then Euler’s formula r r gives
y x rei r cos i sin r i x iy z . r r (b) 4 4ei0 ; 2 2ei ; 3i 3ei 2 ; 1 i 2ei 4 ; 1 i 3 2ei2
3
(c) Because 2 2i 3 4e i 3 , the square roots of 2 2i 3 are 2e i 6 . Likewise, because 2 2i 3 4ei2 3 , the square roots of 2 2i 3 are 2ei 3 . 44.
(a) x (b) x
45.
i i 2 4 2 1 3 i i, 2i 2 2 2i
2i 2
2
43
24 i i,3i 2
The characteristic polynomial is the quadratic polynomial of Problem 44(b). Hence the general solution is y x c1e ix c2e3ix c1 cos x i sin x c2 cos 3x i sin 3x .
46.
The characteristic polynomial is r 2 ir 6 r 2i r 3i , so the general solution is
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HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
y x c1e3ix c2e 2ix c1 cos 3 x i sin 3x c2 cos 2 x i sin 2 x .
47.
y x c1e
48.
The characteristic roots are r 2 2i 3 1 i 3 , so the general solution is
1i 3 x
c2e
1i 3 x
The general solution is y x Ae x Be x Ce x , where
c1e x cos 3 x i sin 3 x c2 e x cos 3 x i sin 3 x .
1 i 3 and 2
1 i 3 . Imposition of the given initial conditions yields the equations 2 A B C 17 A B C 0 A 2 B 2C 0
1 that we solve for A B C . Thus the desired particular solution is given by 3 1 1i 3 x 2 1i 3 x 2 y x e x e e , which (using Euler’s relation) reduces to the given 3 real-valued solution. 49.
We adopt the same strategy as was used in Problem 48. The general solution is y x Ae 2 x Be x C cos x D sin x . Imposition of the given initial conditions yields the equations
A B C
0
2 A B D 0 4A B C 0 8 A B D 30 that we solve for A 2 , B 5 , C 3 , and D 9 . Thus y x 2e 2 x 5e x 3cos x 9sin x .
50.
If x 0 , then the differential equation is y y 0 , with general solution y A cos x B sin x . But if x 0 , then it is y y 0 , with general solution y C cosh x D sinh x . To satisfy the initial conditions y1 0 1 , y1 0 0 we choose A C 1 and B D 0 . But to satisfy the initial conditions y2 0 0 , y2 0 1 we
choose A C 0 and B D 1 . The corresponding solutions are defined by cos x, x 0 sin x, x 0 . y1 x ; and y2 x cosh x, x 0 sinh x, x 0
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Section 3.3 189
Examination of left- and right-hand derivatives at x 0 shows not only that y1 x and y2 x are differentiable at x 0 , but that y1 and y2 are in fact continuous there.
51.
In the solution of Problem 51 in Section 3.1 we showed that the substitution v ln x dy 1 dy d2y 1 dy 1 d 2 y and y 2 2 2 2 . A further differentiation usgives y dx x dv x dv dx x dv ing the chain rule gives d 3 y 2 dy 3 d 2 y 1 d 3 y . dx 3 x 3 dv x 3 dv 2 x 3 dv 3 Substitution of these expressions for y , y , and y into the third-order Euler equation ax 3 y bx 2 y cxy dy 0 , together with collection of coefficients, yields the desired constant-coefficient equation y
a
d3y d2y dy b 3 a c b 2a d y 0 . 3 2 dv dv dv
In Problems 52 through 58 we list first the transformed constant-coefficient equation, then its characteristic equation and roots, and finally the corresponding general solution with v ln x and e v x . 52.
d2y 9 y 0 ; r 2 9 0 ; r 3i ; 2 dv y x c1 cos 3v c2 sin 3v c1 cos 3ln x c2 sin 3ln x
53.
d2y dy 6 25 y 0 ; r 2 6r 25 0 ; r 3 4i ; 2 dv dv y x e 3v c1 cos 4v c2 sin 4v x 3 c1 cos 4 ln x c2 sin 4 ln x
54.
d3y d2y 3 0 ; r 3 3r 2 0 ; r 0,0, 3 ; dv 3 dv 2 y x c1 c2 v c3e 3v c1 c2 ln x c3 x 3
55.
d3y d2y dy 4 2 4 0 ; r 3 4r 2 4r 0 ; r 0, 2, 2 ; 3 dv dv dv y x c1 c2e 2 v c3ve 2 v c1 x 2 c2 c3 ln x
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56.
HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
d3y 0 ; r 3 0 ; r 0,0,0 ; 3 dv y x c1 c2v c3v 2 c1 c2 ln x c3 ln x
57.
d3y d2y dy 5 5 0 ; r 3 4r 2 4r 0 ; r 0,3 3 ; 3 2 dv dv dv y x c1 c2e
58.
2
3 3 v
c3ve
3 3 v
c1 x 3 c2 x
3
c3 x
3
d3y d2y dy 3 3 y 0 ; r 3 3r 2 3r 1 0 ; r 1, 1, 1 ; 3 2 dv dv dv 2 y x c1e v c2ve v c3v 2e v x 1 c1 c2 ln x c3 ln x
SECTION 3.4 MECHANICAL VIBRATIONS In this section we discuss four types of free motion of a mass on a spring—undamped, underdamped, critically damped, and overdamped. However, the undamped and underdamped cases—in which actual oscillations occur—are emphasized because they are both the most interesting and the most important cases for applications. 1.
Frequency: 0
2 2 k 16 1 sec 2 rad sec Hz ; period: P 0 2 m 4
2.
Frequency 0
2 2 48 4 k 8 rad sec Hz ; period: P sec 0 8 4 m 0.75
3.
The spring constant is k
15 N 75 N m . The solution of 3x 75 x 0 with 0.20 m
x 0 0 and x 0 10 is x t 2sin 5t . Thus the amplitude is 2 m, the frequency
is 0
4.
2 75 2.5 k 5rad sec Hz , and the period is sec . 5 m 3
1 kg and k 9 N 0.25m=36 N m , we find that 0 12 rad sec . The solu4 tion of x 144 x 0 with x 0 1 and x 0 5 is (a) With m
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Section 3.4 191
x t cos12
5 13 12 5 13 sin12t cos12t sin 2t cos 12t , 12 12 13 13 12
5 5.8884 rad . 12 13 2 (b) C 0.5236sec . 1.0833m and T 12 12 where 2 tan 1
GM . AcR2 cording to Equation (6) in the text, the (circular) frequency of a (linearized) pendulum g GM 2 L is given by 2 2 , so its period is p . 2 R L R L GM
5.
The gravitational acceleration at distance R from the center of the earth is g
6.
If the pendulum in the clock executes n cycles per day (86400 sec) at Paris, then its peri86400 od is p1 sec . At the equatorial location it takes 24 hr 2 min 40sec 86560sec n 86560 sec . Now let for the same number of cycles, so its period there is p1 n R1 3956 mi be the Earth’s “radius” at Paris, and R2 its “radius” at the equator. Then p R substitution in the equation 1 1 of Problem 5 (with L1 L2 ) yields R2 3963.33mi . p2 R2 Thus this (rather simplistic) calculation gives 7.33mi as the thickness of the Earth’s equatorial bulge.
7.
The period equation p 3960 100.10 3960 x 100 yields x 1.9795mi 10.450ft for the altitude of the mountain.
8.
Let n be the number of cycles required for a correct clock with unknown pendulum length L1 and period p1 to register 24 hrs 86400sec , so np1 86400 . The given clock with length L2 30in and period p2 loses 10 min 600sec per day, so np2 87000 . Then the formula of Problem 5 yields
L1 p np 86400 , so 1 1 L2 p2 np2 87000
2
86400 L1 30 29.59 in . 87000 9.
Designating x t as in the suggestion, we see that the mass is subject to a restorative force FS kx together with the force of gravity W mg . We also assume that the mass is subject to a damping force FR cx . Applying Newton’s law then gives
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MECHANICAL VIBRATIONS
mx kx mg cx , or mx cx kx mg . Finally, substituting y x s0 , so that x y s0 and thus x y and x y , yields my cy k y s0 mg , or
my cy ky mg ks0 , which is Equation (5) with F t assuming the constant value mg ks0 . 10.
The mass of the buoy is given by m r 2 h , and the net downward force on the buoy is F r 2hg r 2 g x . (Note that the depth x t is taken to be positive.) Therefore Newton’s second law ma F gives
r 2h x r 2hg r 2 g x , which simplifies to x
g xg. h
The complementary function for this equation is xc t c1 cos 0t c2 sin 0t , where g . A particular solution is given by x p t A , where A is a constant, and subh stituting into the differential equation shows that A h . Thus the general solution of the differential equation is
0
x t xc t x p t c1 cos 0t c2 sin 0t h .
Applying the initial conditions x 0 x 0 0 gives c1 h and c2 0 . All told, the motion of the buoy is given by x t h cos 0t h .
Thus the buoy undergoes simple harmonic motion about an equilibrium of xe h . Further, with the given numerical values of , h, and g, the amplitude of oscillation is
h 100cm and the period is p
11.
2
0
2 h 2 2.01sec . g g h
The differential equation from Problem 10 must be modified to reflect the fact that the weight density of water is 62.4 lb ft 3 (as opposed to 1g cm3 in the cgs system). Thus the weight of water displaced by the buoy is given by 62.4 r 2 x . Moreover, the mass and weight of the buoy are given to be 3.125 slugs and 100 lb, respectively. Applying 62.4 2 ma F then gives 3.125 x 100 62.4 r 2 x , or x r x 32 . The frequency 3.125 62.4 of the oscillations of the buoy is therefore 0 , where 0 r . Since the fre3.125 2
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Section 3.4 193
quency of the buoy’s motion is observed to be the two to conclude that r 0.8
1 2
4 cycles 0.4 cycles sec , we can equate 10sec
62.4 r 0.4 , which gives 3.125
3.125 0.3173ft 3.8in . 62.4 3
12.
GM r m GMm r yields Fr r. (a) Substitution of M r M in Fr 2 r R3 R GM g (b) Because 3 , the equation mr Fr yields the differential equation R R g r r 0 . R (c) The solution of this equation with r 0 R and r 0 0 is r t R cos 0t , where g . Hence, with g 32.2 ft sec 2 and R 3960 5280ft , we find that the period R of the particles simple harmonic motion is
0
p
2
0
2
R 5063.10sec 84.38 min . g
mv 2 R on the satellite just offsets the weight mg of the satellite at the surface of the earth. Thus
(d) The orbital velocity v of such a satellite must be such that the centrifugal force
mv 2 mg , which implies that R v gR 32.2 ft sec 2 3960 5280ft 2.5947 104 ft sec 1.7691 104 mi hr . Because the circumference of the earth is 2 R , the period of the satellite’s orbit is 2 R R , which is equal to the period of the particle found in part c. This is not a 2 g gR coincidence. Imagine that at time t 0 the satellite is directly over the hole in the earth at the top of Figure 3.4.13, and that its orbit proceeds in a clockwise direction. We found in part c that the distance r of the particle from the center of the earth is r t R cos 0t . The key observation is that 0t is the angle drawn clockwise from the vertical to the ra-
dius vector of the satellite at time t; thus, the distance r t is simply the vertical compo-
nent of the satellite’s position. It follows that r t completes one cycle through the earth
(and back) in the same length of time required for the satellite to complete one orbit around the earth.
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MECHANICAL VIBRATIONS
(e) The particle passes through the center of the earth when r t R cos 0t 0 , that is,
when 0t
2
, or t
. At this time the speed of the particle is 20
g gR 1.7691 104 mi hr . r t R0 sin 0t R0 sin 0 R 20 R (f) In part d we found the orbital velocity to be v gR , in agreement with part e. Again this is not a concidence. The vertical component of the satellite’s velocity vector v t at any given time t is equal to the speed r t of the particle at that time. At the
moment when the particle passes through the center of earth, the satellite is travelling straight downward, and hence v t is vertical. Therefore the orbital velocity v of the
satellite, which is the magnitude of v t , is equal to the speed of the particle at this moment.
13.
2 1 (a) The characteristic equation 10r 2 9 r 2 5r 2 2 r 1 0 has roots r , . 5 2 When we impose the initial conditions x 0 0 , x 0 5 on the general solution x t c1e 2 t 5 c2 e t 2 we get the particular solution x t 50 e 2 t 5 e t 2 .
(b) The derivative x t 25e t 2 20e 2 t 5 5e 2 t 5 5e t 10 4 0 when
5 2.23144 . Hence the mass’s farthest distance to the right is given by 4 5 512 4.096 . x 10ln 4 125
t 10ln
14.
(a) The characteristic equation 25r 2 10r 226 5r 1 152 0 has roots 2
1 15i 1 3i . When we impose the initial conditions x 0 20 , x 0 41 on 5 5 the general solution x t e t 5 A cos 3t B sin 3t we get A 20 , B 15 . The correr
sponding particular solution is given by x t e t 5 20cos 3t 15sin 3t 25e t 5 cos 3t , where tan 1
3 0.6435 . 4
(b) Thus the oscillations are “bounded” by the curves x 25e t 5 , and the pseudoperiod 2 (because 3 ). of oscillation is T 3
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Section 3.4 195
In Problems 15-21 the graph of the damped motion x t , that is, with the dashpot attached, is shown as a solid line; the graph of the corresponding undamped motion u t is dashed.
1 2 r 3r 4 0 has roots r 2, 4 . When 2 we impose the initial conditions x 0 2 , x 0 0 on the general solution With damping: The characteristic equation
15.
x t c1e 2 t c2e 4 t we get the particular solution x t 4e 2 t 2e 4 t that describes
overdamped motion. 1 2 r 4 0 has roots r 2i 2 . When 2 we impose the initial conditions x 0 2 , x 0 0 on the general solution
Without damping: The characteristic equation
u t A cos 2 2t B sin 2 2t we get the particular solution u t 2 cos 2 2t . Problem 15
Problem 16
2
x
2
x
0
−2
0
−2 0
1
2
3
0
1
t
16.
2
t
With damping: The characteristic equation 3r 2 30r 63 0 has roots r 3, 7 . When we impose the initial conditions x 0 2 , x 0 2 on the general solution
x t c1e 3t c2e 7 t we get the particular solution x t 4e 3t 2e 7 t that describes
overdamped motion. Without damping: The characteristic equation 3r 2 63 0 has roots r i 21 . When we impose the initial conditions x 0 2 , x 0 2 on the general solution
u t A cos
21t B sin
u t 2 cos
21t
21t we get the particular solution 2 sin 21
21t 2
22 cos 21
21t 0.2149 .
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MECHANICAL VIBRATIONS
With damping: The characteristic equation r 2 8r 16 0 has roots r 4, 4 . When we impose the initial conditions x 0 5 , x 0 10 on the general solution
17.
x t c1 c2t e 4 t we get the particular solution x t 5e 4 t 2t 1 that describes crit-
ically damped motion. Without damping: The characteristic equation r 2 16 0 has roots r 4i . When we impose the initial conditions x 0 5 , x 0 10 on the general solution u t A cos 4t B sin 4t we get the particular solution
5 5 u t 5cos 4t sin 4t 5 cos 4t 5.8195 . 2 2 Problem 17
Problem 18
5 1
x
x
0
−1 −5 0
1
2
0
1
t
18.
t
With damping: The characteristic equation 2r 2 12r 50 0 has roots r 3 4i . When we impose the initial conditions x 0 0 , x 0 8 on the general solution x t e 3t A cos 4t B sin 4t we get the particular solution
3 x t 2e 3t sin 4t 2e 3t cos 4t 2 that describes underdamped motion. Without damping: The characteristic equation 2r 2 50 0 has roots r 5i . When we impose the initial conditions x 0 0 , x 0 8 on the general solution u t A cos5t B sin 5t we get the particular solution
8 8 3 u t sin 5t cos 5t 5 5 2
.
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2
Section 3.4 197
5 The characteristic equation 4r 2 20r 169 0 has roots r 6i . When we impose 2 the initial conditions x 0 4 , x 0 16 on the general solution
19.
x t e 5t 2 A cos 6t B sin 6t we get the particular solution
13 1 313 e 5t 2 cos 6t 0.8254 x t e 5t 2 4 cos 6t sin 6t 3 3 that describes underdamped motion. Without damping: The characteristic equation 4r 2 169 0 has roots r
we impose the initial conditions x 0 4 , x 0 16 on the general solution
13 i . When 2
13 13 u t A cos t B sin t we get the particular solution 2 2 13t 32 13t 4 13 u t 4 cos 233 cos t 0.5517 . sin 2 13 2 13 2 Problem 19
Problem 20 5
4
x
x
−4
−5 0
1
2
0
t
20.
1
t
With damping: The characteristic equation 2r 2 16r 40 0 has roots r 4 2i . When we impose the initial conditions x 0 5 , x 0 4 on the general solution x t e 4 t A cos 2t B sin 2t we get the particular solution
x t e 4 t 5cos 2t 12sin 2t 13e 4 t cos 2t 1.1760
that describes underdamped motion.
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2
198
MECHANICAL VIBRATIONS
Without damping: The characteristic equation 2r 2 40 0 has roots r 2 5i . When we impose the initial conditions x 0 5 , x 0 4 on the general solution
u t A cos 2 5t B sin 2 5t we get the particular solution
u t 5cos 2 5t 21.
2 129 sin 2 5t cos 2 5t 0.1770 . 5 5
With damping: The characteristic equation r 2 10r 125 0 has roots r 5 10i . When we impose the initial conditions x 0 6 , x 0 50 on the general solution x t e 5t A cos10t B sin10t we get the particular solution
x t e 5t 6 cos10t 8sin10t 10e 5t cos 10t 0.9273
that describes underdamped motion. Without damping: The characteristic equation r 2 125 0 has roots r 5 5i . When we impose the initial conditions x 0 6 , x 0 50 on the general solution
u t 6cos 5 5 t 2 5 sin 5 5 t 2
u t A cos 5 5 t B sin 5 5 t we get the particular solution
14 cos 5 5 t 0.6405 .
Problem 21 6
x
−6 0
1
t
22.
12 0.375slug , c 3lb-sec ft , and k 24 lb ft , the differential equation 32 is equivalent to 3x 24 x 192 x 0 . The characteristic equation 3r 2 24r 192 0 has roots r 4 4 3i . When we impose the initial conditions x 0 1 , x 0 0 on (a) With m
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Section 3.4 199
the general solution x t e 4 t A cos 4 3 t B sin 4 3 t we get the particular solu tion
1 sin 4 3 t x t e 4 t cos 4 3 t 3 2 4 t 3 1 e cos 4 3 t sin 4 3 t 2 3 2 2 4 t e cos 4 3 t . 6 3
(b) The time-varying amplitude is
the phase angle is
23.
6
2 1.15ft , the frequency is 4 3 6.93rad sec , and 3
.
(a) With m 100slug we get
k . But we are given that 100
80cycles min 2 1min 60sec
8 , 3
and equating the two values yields k 7018lb ft . (b) With 1 2
Hence p 30.
78 cycles sec , Equation (21) in the text yields c 372.31lb ft sec . 60
c 1.8615 . Finally e pt 0.01 gives t 2.47sec . 2m
In the underdamped case we have x t e pt A cos 1t B sin 1t
and x t pe pt A cos 1t B sin 1t e pt A1 sin 1t B1 cos 1t .
The conditions x 0 x0 , x 0 v0 yield the equations A x0 and pA B1 v0 , whence B 31.
v0 px0
1
.
The binomial series
1 x
1x
1 2!
x2
1 2 3!
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x3
200
MECHANICAL VIBRATIONS
converges if x 1 . (See, for instance, Section 10.8 of Edwards and Penney, Calculus: c2 1 in Eq. and x 4mk 2 (22) of Section 3.4 in the differential equations text, the binomial series gives
Early Transcendentals, 7th edition, Pearson, 2008.) With
1 02 p 2 32.
k c2 k c2 k c2 c4 c2 1 1 1 0 . m 4m 2 m 4mk m 8mk 128m 2k 2 8 mk
If x t Ce pt cos 1t , then x t pCe pt cos 1t C1e pt sin 1t 0
yields tan 1t
p
1
. Because the tangent function is periodic with period and
the local maxima and minima of x t are interlaced, successive maxima are separated by a distance of 33.
2
1
.
If x1 x t1 and x2 x t2 are two successive local maxima, then 1t2 1t1 2 , and so x1 Ce pt1 cos 1t1 and x2 Ce pt2 cos 1t2 Ce pt2 cos 1t1 . Hence x x1 2 p e pt1 t2 and therefore ln 1 p t1 t2 . x2 1 x2
34.
With t1 0.34 and t2 1.17 we first use the equation 1t2 1t1 2 from Problem 33 2 to calculate 1 7.57 rad sec . Next, with x1 6.73 and x2 1.46 , the result of 0.83 1 6.73 Problem 33 yields p ln 1.84 . Then Equation (16) in this section gives 0.83 1.46 100 c 2mp 2 1.84 11.51lb-sec ft , and finally Equation (22) yields 32 4m 212 c 2 k 189.68lb ft . 4m
35.
The characteristic equation r 2 2r 1 0 has roots r 1, 1 . When we impose the initial conditions x 0 0 , x 1 0 on the general solution x t c1 c2t e t we get the particular solution x1 t te t .
36.
The characteristic equation r 2 2r 1 102 n 0 has roots r 1 10 n . When we impose the initial conditions x 0 0 , x 1 0 on the general solution Copyright © 2015 Pearson Education, Inc.
Section 3.4 201
x t c1 exp 1 10 n t c2 exp 1 10 n t we get the equations
1 10 c 1 10 c n
c1 c2 0,
n
1
2
1
with solution c1 2n 15n , c2 2n 15n . This gives the particular solution exp 10 n t exp 10 n t x2 t 10 e 10n e t sinh 10 n t . 2 n
37.
t
The characteristic equation r 2 2r 1 102 n 0 has roots r 1 10 n i . When we impose the initial conditions x 0 0 , x 1 0 on the general solution x t e t A cos 10 n t B sin 10 n t we get the equations c1 0 , c1 10 n c2 1 with solution c1 0 , c2 10n . This gives the particular solution x3 t 10n e t sin 10 n t .
38.
This follows from lim x2 t lim10 e sinh 10 t te lim t
n
n
n
t
n
n
sinh 10 n t 10 n t
te t
and lim x3 t lim10 e sin 10 t te lim n
n
t
n
n
using the fact that lim 0
sin
t
n
lim 0
sinh
sin 10 n t 10 n t
te t ,
0 (by L’Hôpital’s rule, for instance).
SECTION 3.5 NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS The method of undetermined coefficients is based on “educated guessing”. If we can guess correctly the form of a particular solution of a nonhomogeneous linear equation with constant coefficients, then we can determine the particular solution explicitly by substitution in the given differential equation. It is pointed out at the end of Section 3.5 that this simple approach is not always successful—in which case the method of variation of parameters is available if a complementary function is known. However, undetermined coefficients does turn out to work well with a surprisingly large number of the nonhomogeneous linear differential equations that arise in elementary scientific applications.
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NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
In each of Problems 1-20 we give first the form of the trial solution y trial , then the equations in the coefficients we get when we substitute y trial into the differential equation and collect like terms, and finally the resulting particular solution y p . 1 3x e . 25
1.
y trial Ae3 x ; 25 A 1 ; y p
2.
y trial A Bx ; 2 A B 4 , 2 B 3 ; y p
3.
y trial A cos 3x B sin 3x ; 15 A 3B 0 , 3 A 15B 2 ; y p
4.
4 1 y trial Ae x Bxe x ; 9 A 12 B 0 , 9 B 3 , y p e x xe x . 9 3
5.
First we substitute
1 5 6 x . 4 1 5 cos 3x sin 3x . 39 39
1 cos 2 x for sin 2 x on the right-hand side of the differential equa2 tion, leading to y trial A B cos 2 x C sin 2 x , and then 1 1 A , 3B 2C , 2 B 3C 0 ; 2 2 yp
6.
7.
8.
1 3 1 cos 2 x sin 2 x . 2 26 13
y trial A Bx Cx 2 ; 7 A 4 B 4C 0 , 7 B 8C 0 , 7C 1 ; y p
4 8 1 x x2 . 343 49 7
e x e x for sinh x on the right-hand side of the differential equatio, 2 1 1 1 1 1 leading to y trial Ae x Be x ; 3 A , 3B ; y p e x e x sinh x . (Note 2 2 6 6 3 that according to Rule 1 in the text, we could also have started with y trial A cosh x B sinh x .) First we substitute
e 2 x e 2 x is part of the complementary function 2 1 1 yc c1e 2 x c2e 2 x , leading to y trial x Ae2 x Be 2 x ; A , B ; 8 8
First we note that cosh 2 x
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Section 3.5 203
1 1 1 y p x e 2 x e 2 x x sinh 2 x . (As an extension of Rule 2 in the text, we could al8 8 4 so have started with y trial x A cosh 2 x B sinh 2 x .) 9.
First we note that e x is part of the complementary function yc c1e x c2e 3 x . Then y trial A x B Cx e x , and then
3 A 1 4 B 2C 0 8C 1 ;
1 1 1 y p x x ex . 3 16 8 10.
First we note the duplication with the complementary function yc c1 cos 3x c2 sin 3x . Then y trial x A cos 3 x B sin 3 x ; 6 B 2 ; 6 A 3 ;
1 1 1 y p x cos 3x sin 3 x 2 x sin 3x 3x cos 3 x . 2 3 6 11.
First we note the duplication with the complementary function yc c1 c2 cos 2 x c3 sin 2 x . Then y trial x A Bx ; 4 A 1 , 8B 3 ; 1 3 1 y p x x 3x 2 2 x . 4 8 8
12.
First we note the duplication with the complementary function yc c1 c2 cos x c3 sin x . 1 Then y trial Ax x B cos x C sin x ; A 2 , 2 B 0 , 2C 1 ; y p 2 x x sin x . 2
13.
y trial e x A cos x B sin x ; 7 A 4 B 0 , 4 A 7 B 1 ; y p
14.
First we note the duplication with the complementary function yc c1 c2 x e x c3 c4 x e x . Then y trial x 2 A Bx e x ; 8 A 24 B 0 , 24 B 1 ;
1 x e 7sin x 4 cos x . 65
1 1 x 1 y p x2 xe 3x 2e x x 3e x . 8 24 24 15.
This is something of a trick problem. We cannot solve the characteristic equation r 5 5r 4 1 0 to find the complementary function, but we can see that the complementary function contains no constant term (why?). Hence we can take y trial A , leading immediately to the particular solution y p 17 .
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204
16.
NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
y trial A B Cx Dx 2 e3 x ; 9 A 5, 18 B 6C 2 D 0, 18C 12 D 0, 18 D 2 ;
yp 17.
5 1 2 1 1 x x 2 e3 x 45 e3 x 6 xe3 x 9 x 2e3 x . 9 81 27 9 81
First we note the duplication with the complementary function yc c1 cos x c2 sin x . Then y trial x A Bx cos x C Dx sin x ; 2 B 2C 0 4 D 1 2 A 2 D 1 4 B 0 ;
1 1 1 y p x cos x x sin x x 2 sin x x cos x . 4 4 4 18.
First we note the duplication with the complementary function yc c1e x c2e x c3e 2 x c4e2 x . Then y trial x Ae x x B Cx e 2 x ; 6 A 1, 12 B 38C 0, 24C 1 ;
1 2x 1 1 19 y p x ex x xe 24 xe x 19 xe2 x 6 x 2e 2 x . 144 6 144 24 19.
First we note the duplication with the part c1 c2 x of the complementary function (which corresponds to the factor r 2 of the characteristic polynomial). Then y trial x 2 A Bx Cx 2 ; 4 A 12 B 1, 12 B 48C 0, 24C 3 ;
1 1 5 1 y p x 2 x x 2 10 x 2 4 x 3 x 4 . 8 8 4 2 20.
First we note that the characteristic polynomial r 3 r has the zero r 1 corresponding to the duplicating part e x of the complementary function. Then: y trial A x Be x ; 1 A 7 ; 3B 1 ; y p 7 xe x . 3
In Problems 21-30 we list first the complementary function yc , then the initially proposed trial function yi , and finally the actual trial function y p , in which duplication with the complementary function has been eliminated. 21.
yc e x c1 cos x c2 sin x ; yi e x A cos x B sin x ; y p x e x A cos x B sin x
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Section 3.5 205
22.
yc c1 c2 x c3 x 2 c4e x c5e x ; yi A Bx Cx 2 De x ; y p x 3 A Bx Cx 2 x De x
23.
yc c1 cos 2 x c2 sin 2 x ; yi A Bx cos 2 x C Dx sin 2 x ; y p x A Bx cos 2 x C Dx sin 2 x
24.
yc c1 c2e 3 x c3e4 x ; yi A Bx C Dx e 3 x ; y p x A Bx x C Dx e 3 x
25.
yc c1e x c2e 2 x ; yi A Bx e x C Dx e 2 x ; y p x A Bx e x x C Dx e 2 x
26.
yc e3 x c1 cos 2 x c2 sin 2 x ; yi A Bx e3 x cos 2 x C Dx e3 x sin 2 x ;
y p x A Bx e3 x cos 2 x C Dx e3 x sin 2 x 27.
yc c1 cos x c2 sin x c3 cos 2 x c4 sin 2 x ;
yi A cos x B sin x C cos 2 x D sin 2 x ;
y p x A cos x B sin x C cos 2 x D sin 2 x 28.
yc c1 c2 x c3 cos 3x c4 sin 3x ;
yi A Bx Cx 2 cos 3x D Ex Fx 2 sin 3x ;
y p x A Bx Cx 2 cos 3x D Ex Fx 2 sin 3 x 29.
yc c1 c2 x c3 x 2 e x c4e2 x c5e 2 x ; yi A Bx e x Ce 2 x De 2 x ; yp x 3 A Bx e x x Ce2 x x De 2 x
30.
yc c1 c2 x e x c3 c4 x e x ; yi y p A Bx Cx 2 cos x D Ex Fx 2 sin x
In Problems 31-40 we list first the complementary function yc , the trial solution y tr for the method of undetermined coefficients, and the corresponding general solution yg yc y p , where y p results from determining the coefficients in y tr so as to satisfy the given nonhomogeneous differential equation. Then we list the linear equations obtained by imposing the given initial conditions, and finally the resulting particular solution y x .
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31.
NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
yc c1 cos 2 x c2 sin 2 x ; y tr A Bx ; yg c1 cos 2 x c2 sin 2 x
y ( x ) cos 2 x (3/ 4)sin 2 x x / 2 32.
33.
34.
1 1 yc c1e x c2e 2 x ; y tr Ae x ; yg c1e x c2 e 2 x e x ; c1 c2 0 , 6 6 1 5 8 1 c1 2c2 3 ; y x e x e 2 x e x 6 2 3 6 1 yc c1 cos 3x c2 sin 3x ; y tr A cos 2 x B sin 2 x ; yg c1 cos 3x c2 sin 3x sin 2 x ; 5 2 2 1 c1 1 , 3c2 0 , y x cos 3x sin 3x sin 2 x 5 15 5 yc c1 cos x c2 sin x ; y tr x A cos x B sin x ; yg c1 cos x c2 sin x c1 1 , c2 1 , y x cos x sin x
35.
1 x sin x ; 2
1 x sin x 2
yc e x c1 cos x c2 sin x ; y tr A Bx ; yg e x c1 cos x c2 sin x 1
c1 c2 36.
x 1 ; c1 1 , 2c2 2 ; 2 2
x ; c1 1 3 , 2
1 5 x 0 ; y x e x 2 cos x sin x 1 2 2 2
yc c1 c2 x c3e 2 x c4 e2 x ; y tr x 2 A Bx Cx 2 x2 x4 16 48 1 c1 c3 c4 1, c2 2c3 2c4 1, 4c3 4c4 1, 8c3 8c4 1 8 yg c1 c2 x c3e 2 x c4 e2 x
y x 37.
39 5 3 11 1 1 4 x e 2 x e 2 x x 2 x 32 4 64 64 16 48
yc c1 c2e x c3 xe x ; y tr x A x 2 B Cx e x yg c1 c2 e x c3 xe x x
1 2 x 1 3 x xe xe 2 6
c1 c2 0, c2 c3 1 0, c2 2c3 1 1 1 1 y x 4 x e x 4 3x x 2 x 3 2 6
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Section 3.5 207
38.
yc e x c1 cos x c2 sin x ; y tr A cos 3x B sin 3x
yg e x c1 cos x c2 sin x c1
6 7 cos 3x sin 3x 85 85
6 21 2, c1 c2 0 185 85
197 7 176 6 cos x sin x cos 3x sin 3x y x e x 85 85 85 85 39.
yc c1 c2 x c3e x ; y tr x 2 A Bx x Ce x x2 x3 xe x 2 6
yg c1 c2 x c3e x
c1 c3 1, c2 c3 1 0, c3 3 1 y x 40.
1 18 18 x 3x 2 x 3 4 x e x 6
yc c1e x c2 e x c3 cos x c4 sin x;
ytr A
yg c1e x c2 e x c3 cos x c4 sin x 5
c1 c2 c3 5 0, c1 c2 c4 0, c1 c2 c3 0, c1 c2 c4 0 y x 41.
1 5e x 5e x 10cos x 20 4
The trial solution y tr A Bx Cx 2 Dx 3 Ex 4 Fx 5 leads to the equations 2A
B 2C 6 D 2 B 2C 6 D 2C 3D 2 D
24 E 24 E 12 E 4E 2 E
120 F 60 F 20 F 5F 2 F
0 0 0 0 0 8
that are readily solved by back-substitution. The resulting particular solution is y x 255 450 x 30 x 2 20 x 3 10 x 4 4 x 5 .
42.
The characteristic equation r 4 r 3 r 2 r 2 0 has roots r 1, 2 , and i , so the complementary function is yc c1e x c2 e2 x c3 cos x c4 sin x . We find that the coefficients satisfy the equations Copyright © 2015 Pearson Education, Inc.
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NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
c1 c1 c1 c1
255 c2 c3 c4 450 c2 60 4c2 c3 c4 120 8c2
0 0 . 0 0
Solution of this system gives finally the particular solution y yc yp , where yp is the particular solution of Problem 41 and yc 10e x 35e 2 x 210cos x 390sin x . 43.
(a) Applying Euler’s formula gives
cos 3x i sin 3x cos x i sin x cos3 x 3i cos2 x sin x 3cos x sin 2 x i sin 3 x . 3
When we equate real parts we get the equation cos3 x 3 cos x 1 cos2 x 4 cos3 x 3cos x and readily solve for cos3 x 43 cos x 14 cos 3x . The formula for sin 3 x is derived similarly by equating imaginary parts in the first equation above. (b) Upon substituting the trial solution y p A cos x B sin x C cos 3x D sin 3x in the differential equation y 4 y 43 cos x 14 cos 3x , we find that 1 1 A , B 0, C , D 0 . 4 20 The resulting general solution is 1 1 y x c1 cos 2 x c2 sin 2 x cos x cos 3x . 4 20 44.
We use the identity sin x sin 3x 12 cos 2 x 12 cos 4 x , and hence substitute the trial solution y p A cos 2 x B sin 2 x C cos 4 x D sin 4 x in the differential equation y y y 12 cos 2 x 12 cos 4 x . We find that 3 1 15 2 , B , C , D . 26 13 482 241 The resulting general solution is A
3 3 1 1 y x e x 2 c1 cos x c2 sin x 3cos 2 x 2sin 2 x 15cos 4 x 4sin 4 x . 2 2 26 482 45.
We substitute sin 4 x
1 1 1 2 1 cos 2 x 1 2 cos 2 x cos2 2 x 3 4 cos 2 x cos 4 x 4 4 8
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Section 3.5 209
on the right-hand side of the differential equation, and then substitute the trial solution y p A cos 2 x B sin 2 x C cos 4 x D sin 4 x E . We find that 1 1 1 . , B 0, C , D 0, E 10 56 24 The resulting general solution is A
y c1 cos 3x c2 sin 3x 46.
1 1 1 cos 2 x cos 4 x . 24 10 56
By the formula for cos3 x in Problem 43, the differential equation can be written as 3 1 x cos x x cos 3x . 4 4 The complementary solution is yc c1 cos x c2 sin x , so we substitute the trial solution y y
y p x A Bx cos x C Dx sin x E Fx cos 3x G Hx sin 3 x . We find that 3 3 1 3 , B C 0, D , E 0, F , G , H 0. 16 16 32 128 Hence the general solution is given by y yc y1 y2 , where A
y1
1 3x cos x 3x 2 sin x and 16
y2
1 3sin 3x 4 x cos 3x . 128
In Problems 47–49 we list the independent solutions y1 and y2 of the associated homogeneous
equation, their Wronskian W W y1 , y2 , the coefficient functions
y x f x y x f x u1 x 2 dx and u2 x 1 dx W x W x in the particular solution y p u1 y1 u2 y2 of Eq. (32) in the text, and finally y p itself. 47.
4 2 y1 e 2 x , y2 e x , W e 3 x , u1 e3 x , u2 2e 2 x , y p e x 3 3
48.
x 1 1 y1 e 2 x , y2 e 4 x , W 6e2 x , u1 , u2 e 6 x , y p 6 x 1 e 2 x 2 12 12
49.
y1 e 2 x , y2 xe 2 x , W e 4 x , u1 x 2 , u2 2 x , y p x 2 e 2 x .
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50.
NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
The complementary function is y1 c1 cosh 2 x c2 sinh 2 x , so the Wronskian is W 2 cosh 2 2 x 2sinh 2 2 x 2 , so when we solve Equations (31) simultaneously for u1 u2 , integrate each, and substitute in y p y1u1 y2u2 , the result is y p cosh 2 x
1 1 sinh 2 x sinh 2 x dx sinh 2 x cosh 2 x sinh 2 x dx . 2 2
Using the identities 2sinh 2 x cosh 2 x 1 and 2sinh x cosh x sinh 2 x , we evaluate the integrals and find that yp
1 4 x cosh 2 x sinh 4 x cosh 2 x cosh 4 x sinh 2 x . 16
Using the identity cosh 4 x sinh 2 x sinh 4 x cosh 2 x sinh 2 x 4 x sinh 2 x
we can reduce y p to simply yp 51.
1 4 x cosh 2 x sinh 2 x . 16
y1 cos 2 x , y2 sin 2 x , W 2 . Liberal use of trigonometric sum and product identities yields u1
1 1 cos5x 5cos x and u2 sin 5 x 5sin x . 20 20
Thus 1 1 cos5 x 5cos x cos 2 x sin 5 x 5sin x sin 2 x 20 20 1 cos5 x cos 2 x sin 5 x sin 2 x 5 cos x cos 2 x sin x sin 2 x 20 1 cos 3x 5cos 3 x 20 1 cos 3 x ! 5
yp
52.
y1 cos 3x , y2 sin 3x , W 3 ; u1
1 1 6 x sin 6 x , u2 1 cos 6 x ; 36 36
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Section 3.5 211
1 1 6 x sin 6 x cos 3x 1 cos 6 x sin 3x 36 36 1 6 x cos 3x sin 3x cos 6 x sin 3 x sin 6 x cos 3 x 36 1 6 x cos 3x sin 3 x sin 3x 36 1 x cos 3x. 6
yp
53.
2 2 2 y1 cos 3x , y2 sin 3x , W 3 ; u1 tan 3x , so u1 ln cos 3x ; u2 , so 3 9 3 2 u2 x . Thus 3 2 2 y p cos 3x ln cos 3x sin 3x x 9 3 2 3x sin 3x cos 3x ln cos 3x . 9
54.
y1 cos x , y2 sin x , W 1 ; u1 csc x , so u1 ln csc x cot x ; u2 cos x csc 2 x , so u2 csc x . Thus y p ln csc x cot x cos x csc x sin x 1 cos x ln csc x cot x .
55.
y1 cos 2 x , y2 sin 2 x , W 2 ; 1 1 1 cos 2 x 1 u1 sin 2 x sin 2 x sin 2 x sin 2 x cos 2 x sin 2 x , 2 2 2 4 1 so u1 2 cos 2 x cos2 2 x ; 16
1 1 1 cos 2 x 1 1 u2 sin 2 x cos 2 x cos 2 x cos 2 x cos2 2 x 2 cos 2 x 1 cos 4 x , 2 2 2 4 8 1 sin 4 x so u2 sin 2 x x . Thus 8 4
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212
NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
yp
1 1 sin 4 x 2 cos 2 x cos2 2 x cos 2 x sin 2 x x sin 2 x 16 8 4
1 1 2 3 2 2 cos 2 x cos 2 x 2 sin 2 x 2 x sin 2 x sin 4 x sin 2 x 16 2
1 1 3 2 2 x sin 2 x cos 2 x sin 4 x sin 2 x , 16 2
because the single-underlined terms sum to 2. The double-underlined terms reduce to 1 1 cos3 2 x sin 4 x sin 2 x cos3 2 x 2sin 2 x cos 2 x sin 2 x 2 2 3 cos 2 x sin 2 2 x cos 2 x cos3 2 x 1 cos2 2 x cos 2 x cos 2 x. 1 2 2 x sin 2 x cos 2 x . However, because cos 2 x is a 16 solution of the associated homogeneous equation y y 0 —that is, because cos 2x is part of the complimentary function yc —we can in fact omit the cos 2 x term from y p , Therefore we can take y p
leading to the simpler version y p
56.
y1 e 2 x , y2 e 2 x , W 4 ; u1 yp
57.
1 1 x sin 2 x . 8
1 1 3x 1 e3 x ; u2 x 1 e x ; 36 4
1 3x 2 e x . 9
With y1 x , y2 x 1 , and f x 72 x 3 , Equations (31) in the text take the form xu1 x 1u2 0, u1 x 2u2 72 x 3.
Upon multiplying the second equation by x and then adding, we readily solve first for u1 36 x 3 , so u1 9 x 4 ; then u2 x 2u1 , so u2 6 x 6 . It follows that y p y1u1 y2u2 x 9 x 4 x 1 6 x 6 3 x 5 . 58.
Here it is important to remember that in order to apply the method of variation of parameters, the differential equation must be written in standard form with leading coefficient 1. We therefore rewrite the given equation with complementary function yc c1 x 2 c2 x 3 as y
4 6 y 2 y x . x x
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Section 3.5 213
Thus f x x and W x 4 , so simultaneous solution of Equations (31) as in Problem 50 (followed by integration of u1 and u2 ) yields 1 y p x 2 x 3 x x 4 dx x 3 x 2 x x 4dx x 2 dx x 3 dx x 3 ln x 1 . x 59.
y1 x 2 , y2 x 2 ln x , W x 3 , f x x 2 ; u1 x ln x , u2 x ; y p
60.
y1 x1 2 , y2 x 3 2 , f x 2 x 2 3 , W x ; u1
61.
y1 cos ln x , y2 sin ln x , W
u2
ln x cos ln x x u1
1 4 x . 4
12 5 6 72 x , u2 12 x 1 6 ; y p x 4 3 . 5 5
ln x sin ln x and 1 ln x , f x 2 ; from u1 x x x
integration by parts yields
ln x sin ln x dx x
cos ln x ln x
sin ln x ln x dx x
cos ln x dx cos ln x ln x sin ln x x
and cos ln x ln x dx x x sin ln x sin ln x ln x dx sin ln x ln x cos ln x . x
u2
ln x cos ln x dx
Thus y p u1 y1 u2 y2 cos ln x ln x sin ln x cos ln x sin ln x ln x cos ln x sin ln x cos2 ln x sin 2 ln x ln x sin ln x cos ln x cos ln x sin ln x
ln x !
62.
y1 x , y2 1 x 2 , W x 2 1 , f x 1 . From u1 method of partial fractions yield u1
1 x2 , long division and the 1 x2
1 x2 1 1 1 x dx 1 dx x ln ; 2 1 x 1 x 1 x 1 x
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NONHOMOGENEOUS EQUATIONS AND UNDETERMINED COEFFICIENTS
likewise u2
x 1 gives u2 ln x 2 1 . Altogether 2 x 1 2
y p u1 y1 u2 y2 x 2 x ln 63.
1 x 1 1 x 2 ln x 2 1 . 1 x 2
This is simply a matter of solving the equations in (31) for the derivatives u1
y2 x f x y x f x and u2 1 , W x W x
integrating each, and then substituting the results in (32). 64.
Here we have y1 x cos x , y2 x sin x , W x 1 , and f x 2sin x , so (33) gives y p x cos x sin x 2sin x dx sin x cos x 2sin x dx cos x 1 cos 2 x dx sin x 2 sin x cos x dx cos x x sin x cos x sin x sin 2 x x cos x sin x cos2 x sin 2 x
x cos x sin x. We can drop the term sin x because it satisfies the associated homogeneous equation y y 0 , leaving simply y p x x cos x , as desired.
SECTION 3.6 FORCED OSCILLATIONS AND RESONANCE 1.
Trial of x A cos 2t yields the particular solution x p 2 cos 2t . (Can you see that because the differential equation contains no first-derivative term, there is no need to include a sin 2t term in the trial solution?) Hence the general solution is x t c1 cos 3t c2 sin 3t 2 cos 2t .
The initial conditions imply that c1 2 and c2 0 , so x t 2 cos 2t 2 cos 3t . The figure shows the graph of x t .
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Section 3.6 215
Problem 1
Problem 2
x
x
t
2.
t
Trial of x A sin 3t yields the particular solution x p sin 3t . Then we impose the initial conditions x 0 x 0 0 on the general solution x t c1 cos 2t c2 sin 2t sin 3t
3 and find that x t sin 2t sin 3t . The figure shows the graph of x t . 2 3.
First we apply the method of undetermined coefficients with trial solution x A cos5t B sin 5t to find the particular solution 4 3 x p 3cos5t 4sin 5t 5 cos5t sin 5t 5cos 5t , 5 5 4 where tan 1 0.9273 . Hence the general solution is 3 x t c1 cos10t c2 sin10t 3cos5t 4sin 5t .
The initial conditions x 0 375 , x 0 0 now yield c1 372 and c2 2 , so the part of the solution with frequency 10 is xc 372 cos10t 2sin10t 2 372 cos10t sin10t 138388 138388 138388 138388 cos 10t , where 2 tan 1 graph of x t .
1 6.2778 is a fourth-quadrant angle. The figure shows the 186
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216
FORCED OSCILLATIONS AND RESONANCE
Problem 4
Problem 3 2π 5
2π
x
x
4.
t
t
Noting that there is no first-derivative term, we try x A cos 4t and find the particular solution x p 10cos 4t . Then imposition of the initial conditions on the general solution x t c1 cos5t c2 sin 5t 10cos 4t yields x t 10cos5t 18sin 5t 10 cos 4t 2 5cos5t 9sin 5t 10cos 4t
5 9 cos5t sin 5t 10cos 4t 2 106 106 106 2 106 cos 5t 10 cos 4t ,
where tan 1 of x t . 5.
9 2.0779 is a second-quadrant angle. The figure shows the graph 5
F0 . Then imposition of the k m 2 initial conditions x 0 x0 , x 0 0 on the general solution Substitution of the trial solution x C cos t gives C
x t c1 cos 0t c2 sin 0t C cos t
(where 0 k m ) gives the particular solution x t x0 C cos 0t C cos t . 6.
F0 cos 0t , which is the m F0 same as Eq. (13) in the text, and therefore has the particular solution x p t sin 0t 2m0 First, let's write the differential equation in the form x 02 x
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Section 3.6 217
given in Eq. (14). When we impose the initial conditions x 0 0 , x 0 v0 on the general solution x t c1 cos 0t c2 sin 0t we find that c1 0 , c2 lem is x t
v0
0
F0 t sin 0t , 2m0
. The resulting resonance solution of our initial value prob-
2mv0 F0t sin 0t . 2m0
In Problems 7–10 we give first the trial solution x p involving undetermined coefficients A and B, then the equations that determine these coefficients, and finally the resulting steady periodic solution xsp . In each case the figure shows the graphs of xsp t and the adjusted forcing function F1 (t ) F t m .
x p A cos 3t B sin 3t ; 5 A 12 B 10 , 12 A 5B 0 .
7.
xsp t
50 120 10 5 12 10 cos 3t sin 3t cos 3t sin 3t cos 3t , 169 169 13 13 13 13
where tan 1
12 1.9656 , a 2nd-quadrant angle. 5
Problem 7
Problem 8
x sp xsp
x
x
F
F 1
t
8.
1
t
x p A cos5t B sin 5t ; 20 A 15B 4 , 15 A 20 B 0 . xsp t
16 12 4 4 3 4 cos5t sin 5t cos5t sin 5t cos 5t , 125 125 25 5 5 25
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FORCED OSCILLATIONS AND RESONANCE
where 2 tan 1
3 5.6397 , a 4th-quadrant angle. 4
x p A cos10t 10sin10t ; 199 A 20 B 0 , 20 A 199 B 3 .
9.
60 597 cos10t sin10t 40001 40001 3 20 199 cos10t sin10t 40001 40001 40001 3 cos 10t , 40001
xsp t
where tan 1
199 4.6122 , a 3rd-quadrant angle. 20
Problem 9
Problem 10
x sp
x sp
x
x
F
1
F
t
10.
1
t
xp A cos10t 10sin 5t ; 97 A 30 B 8 , 30 A 97 B 6 . 956 342 cos10t sin10t 10309 10309 2 257725 478 171 10 cos10t sin10t 61 cos 10t , 10309 257725 257725 793
xsp t
where tan 1
171 3.4851 , a 3rd-quadrant angle. 478
Each solution in Problems 11–14 has two parts. For the first part, we give first the trial solution x p involving undetermined coefficients A and B, then the equations that determine these coefficients, and finally the resulting steady periodic solution xsp . For the second part, we give first
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Section 3.6 219
the general solution x t involving the coefficients c1 and c2 in the transient solution, then the equations that determine these coefficients, and finally the resulting transient solution xtr , so that x t xtr t xsp t satisfies the given initial conditions. For each problem, the graph shows
the graphs of both x t and xsp t .
xp A cos 3t B sin 3t ; 4 A 12 B 10 , 12 A 4 B 0 .
11.
1 3 10 1 3 10 xsp t cos 3t sin 3t cos 3t sin 3t cos 3t , 4 4 4 10 4 10 where tan 1 3 1.8925 , a 2nd-quadrant angle. 1 9 x t e 2 t c1 cos t c2 sin t xsp t ; c1 0 , 2c1 c2 0 . 4 4 7 50 2 t 1 7 5 1 cos t sin t 2e 2 t cos t , xtr t e 2 t cos t sin t e 4 4 50 4 50 4 where 2 tan 1 7 4.8543 , a 4th-quadrant angle. Problem 11
Problem 12
x s p( t )
x (t)
x (t) x
x
x s p( t )
t
12.
t
x p A cos5t B sin 5t ; 12 A 30 B 0 , 30 A 12 B 10 . xsp t
25 10 5 29 5 2 5 cos5t sin 5t cos5t sin 5t cos 5t , 87 87 87 29 29 3 29
2 3.5221 , a 3rd-quadrant angle. 5 25 50 0 , 3c1 2c2 0. x t e 3t c1 cos 2t c2 sin 2t xsp t ; c1 87 87
where tan 1
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FORCED OSCILLATIONS AND RESONANCE
125 50 cos 2t sin 2t xtr t e 3t 174 174 25 29 3t 2 5 cos 2t sin 2t , e 174 29 29 25 3t e cos 2t , 6 29
where tan 1 13.
5 1.1903 , a 1st-quadrant angle. 2
x p A cos10t B sin10t ; 74 A 20 B 600 , 20 A 74 B 0 . 11100 3000 cos10t sin10t 1469 1469 300 37 10 cos10t sin10t 1469 1469 1469 300 cos 10t , 1469
xsp t
10 2.8776 , a 2nd-quadrant angle. 37 11100 30000 . x t e t c1 cos5t c2 sin 5t xsp t ; c1 10 , c1 5c2 1469 1469
where tan 1
et xtr t 25790cos5t 842sin 5t 1469 2 166458266 t 12895 421 e cos5t sin 5t 1469 166458266 166458266 113314 t e cos 5t , 1469 421 where 2 tan 1 6.2505 , a 4th-quadrant angle. 12895 2
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Section 3.6 221
Problem 13
Problem 14
x (t)
x s p( t )
x
x
x s p( t )
x (t)
t
14.
t
x p A cos t B sin t ; 24 A 8 B 200 , 8 A 24 B 520 . 22 1 xsp t cos t 22sin t 485 cos t sin t 485 cos t , 485 485 where tan 1 22 1.5254 , a 1st-quadrant angle. x t e 4 t c1 cos 3t c2 sin 3t xsp t ; c1 1 30 , 4c1 3c2 22 10 .
xtr t e 4 t 31cos 3t 52sin 3t 31 52 3665e 4 t cos 3t sin 3t 3665 3665 3665e 4 t cos 3t , where tan 1
52 4.1748 , a 3rd-quadrant angle. 31
In Problems 15-18 we substitute x t A cos t B sin t into the differential equation mx cx kx F0 cos t with the given numerical values of m, c, k, and F0 . We give first the equations in A and B that result upon collection of coefficients of cos t and sin t , followed by the values of A and B that we get by solving these equations. Finally, C
A2 B 2
gives the amplitude of the resulting forced oscillations as a function of the forcing frequency , and we show the graph of the function C .
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FORCED OSCILLATIONS AND RESONANCE
2 2 2
2 A 2 B 2 , 2 A 2 B 0 ; A 2
15.
2
4
2
C
4
, B
4 ; 4 4
begins with C 0 1 and steadily decreases as increases. Hence 4 4 there is no practical resonance frequency. Problem 15
Problem 16
1
2
C
C 1
0
16.
0
5
0
10
5 A 4 B 10 , 4 A 5 B 0 ; 2
2
C
A
0
5
10
10 5 2
25 6 2
4
, B
40 ; 25 6 2 4
10
begins with C 0 2 and steadily decreases as increases. 25 6 2 4 Hence there is no practical resonance frequency. 17.
45 A 6 B 50 , 6 A 45 B 0 ; 2
2
A
50 45 2
2025 54 2 4
,
300 50 ; C . So, to find the maximum value of 2 4 2025 54 2025 54 2 4 C , we calculate its derivative B
C
100 27 2
2025 54 2 4
3/2
.
Hence the practical resonance frequency (where the derivative vanishes) is 27 3 3 .
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Section 3.6 223
Problem 17
Problem 18
0.4 1
C
C
0
18.
0
10
0
20
0
25
650 A 10 B 100 , 10 A 650 B 0 ; 2
2
A
100 650 2
50
422500 1200 2 4
,
1000 100 ; C . So, to find the maximum 2 4 422500 1200 422500 1200 2 4 value of C , we calculate its derivative B
C
200 600 2
422500 1200
2
4 3/2
.
Hence the practical resonance frequency (where the derivative vanishes) is 600 10 6 . 19.
m 100 32 slugs and k 1200 lb/ft, so the critical frequency is
0 20.
k 384 384 rad sec Hz 3.12 Hz . m 2
Let the machine have mass m. Then the force F mg 9.8 m (the machine’s weight) 1 causes a displacement of x 0.5 cm m , so Hooke's law F kx gives 200 1 mg k ; that is, the spring constant is k 200mg N/m. Hence the resonance fre200 quency is
k 200 g 200 9.8 44.27 rad sec 7.05 Hz , m
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FORCED OSCILLATIONS AND RESONANCE
which is about 423 rpm (revolutions per minute). 21.
If is the angular displacement from the vertical, then the (essentially horizontal) displacement of the mass is x L , so twice its total energy (KE + PE) is m x kx 2 2mgh mL2 kL2 2 2mgL 1 cos C . 2
2
k g Differentiation, substitution of sin , and simplification yield 0 , so m L k g 0 . m L 22.
Let x denote the displacement of the mass from its equilibrium position, v x its velocity, and v a the angular velocity of the pulley. Then conservation of energy yields 1 2 1 2 1 2 mv I kx mgx C . 2 2 2 When we differentiate both sides with respect to t and simplify the result, we get the differential equation I m 2 x kx mg . a Hence
23.
k I m 2 a
.
(a) In units of ft-lb-sec we have m 1000 and k 10000 , so 0 10 rad sec 0.50 Hz . (b) We are given that 2 2.25 2.79 rad sec , and the equation mx kx F t
1 simplifies to x 10 x 2 sin t . When we substitute x t A sin t we find that the 4 amplitude is A 24.
2
4 10 2
0.8854 ft 10.63 in .
By the identity of Problem 43 in Section 3.5, the differential equation is 1 3 mx kx F0 cos t cos 3t . 4 4 Hence resonance occurs when either or 3 equals 0 k m , that is, when either 0 or 0 3 .
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Section 3.6 225
25.
Substitution of the trial solution x A cos t B sin t in the differential equation followed by collection of coefficients as usual yields the equations
k m A c B 0, c A k m B F with coefficient determinant k m c and solution 2
2
0
2 2
A
2
1 1 c F0 , B k m 2 F0 .
Hence x t
where C
26.
F0 k m 2 c sin t cos t C sin t ,
c 1 1 , and cos F0 , sin k m 2 . 1
Let G0 E02 F02 and
k m c 2
2
. Then
xsp t E0 cos(t ) F0 sin(t ) E F G0 0 cos t 0 sin t G0 G0 G0 cos cos t sin sin t , or xsp t G0 cos t , where tan F0 E0 . The desired formula now results when we substitute the value of defined above. 27.
The derivative of C
F0
k m c 2 2
C
(a) Therefore, if c 2 ccr
2
F0
2
2
2
is given by
c 2km 2 m k m c 2
2 2
2
2 3/2
.
2km , it is clear from the numerator that C 0 for
all , so that C steadily decreases as increases. (b) If instead c 2 2km , however, then the numerator (and hence C ) vanishes when
m
k c2 k 0 . 2 m 2m m
Calculation then shows that Copyright © 2015 Pearson Education, Inc.
226
FORCED OSCILLATIONS AND RESONANCE
C m
16 F0m 3 c 2 2km
c 4km c 3
2 3/2
0,
so it follows from the second-derivative test that C m is a local maximum value. 28.
(a) The given differential equation corresponds to Equation (17) with F0 mA 2 . It therefore follows from Equation (21) that the amplitude of the steady periodic vibrations at frequency is
C
F0
k m 2 c 2
2
mA 2
k m 2 c 2
2
.
(b) Now we calculate C
mA 2k 2 2mk c 2 2
k m 2 2 c 2 and we see that the numerator vanishes when
3/2
,
2k 2 k 2mk k 0 . 2 2 2mk c m 2mk c m
29.
We need only substitute E0 ac and F0 ak in the result of Problem 26.
30.
When we substitute the values 2 v L , m 800 , k 7 104 , c 3000 , L 10 , and a 0.05 in the formula of Problem 29, simplify, and square, we get the function Csq v
25 9 2v 2 122500
16 16 4v 4 64375 2v 2 76562500
2
giving the square of the amplitude C (in meters) as a function of the velocity v (in meters per second). Differentiation gives Csq v
50 2v 9 4v 4 245000 2v 2 535937500
16 4v 4 64375 2v 2 76562500
2
.
Because the principal factor in the numerator is quadratic in v 2 it is easy to solve the equation Csq v 0 to find where the maximum amplitude occurs; we find that the only positive solution is v 14.36 m sec 32.12 mi hr . The corresponding amplitude of the car's vibrations is
Csq 14.36 0.1364 m 13.64 cm .
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Section 3.7 227
SECTION 3.7 ELECTRICAL CIRCUITS 1.
With E t 0 we have the simple exponential equation 5 I 25I 0 , whose solution with I 0 4 is I t 4e 5t .
2.
With E t 100 we have the simple linear equation 5 I 25 I 100 , whose solution with I 0 0 is I t 4 1 e 5t .
3.
Now the differential equation is 5 I 25 I 100 cos 60t . Substitution of the trial solution I p A cos 60t B sin 60t yields Ip
4 cos 60t 12sin 60t . 145
The complementary function is I c Ce 5t ; the solution with I 0 0 is I t 4.
4 cos 60t 12sin 60t e 5t . 145
The solution of the initial value problem 2 I 40 I 100e 10t , I 0 0 is
I t 5 e 10 t e 20 t . To find the maximum current we solve the equation I t 50e 10t 100e 20t 50e 20t e 20t 2 0
for t
ln 2 ln 2 5 . Then I max I . 10 10 4
5.
The linear equation I 10 I 50e 10t cos 60t has integrating factor e10t . The result5 ing general solution is I t e 10t sin 60t C . To satisfy the initial condition 6 5 I 0 0 , we take C 0 and get I t e 10t sin 60t . 6
6.
Substitution of the trial solution I A cos 60t B sin 60t in the differential equation I 10 I 30cos 60t 40sin 60t gives the equations 10 A 60 B 30 , 60 A 10 B 40 21 22 with solution A , B . The resulting steady periodic solution is 37 37
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ELECTRICAL CIRCUITS
I sp
1 5 cos 60t , 21cos 60t 22sin 60t 37 37
where tan 1
7.
22 2.3329 , a 2nd-quadrant angle. 21
t 1 Q E0 has integrating factor e RC . The C t E0 RCt RC resulting solution with Q 0 0 is Q t E0C 1 e . Then I t Q t e . R
(a) The linear differential equation RQ
(b) These solutions make it obvious that lim Q t E0C and lim I t 0 . t
8.
(a) The linear equation Q 5Q 10e 5t has integrating factor e5t . The resulting solution with Q 0 0 is Q t 10te 5t , so I t Q t 10 1 5t e 5t . (b) I t 0 when t
9.
t
1 1 , so Qmax Q 2e 1 . 5 5
Substitution of the trial solution Q A cos120t B sin120t into the differential equation 200Q 4000Q 100cos120t yields the equations 4000 A 24000 B 100, 24000 A 4000 B 0
1 3 , B . The complementary function is Qc ce 20t , and im1480 740 position of the initial condition Q 0 0 yields the solution with solution A
Q t
1 cos120t 6sin120t e20t . 1480
The current function is then 1 36cos120t 6sin120t e 20 t . 74 Thus the steady-periodic current is I t Q t
I sp
6 6 37 6 1 3 cos120t sin120t cos 120t 6cos120t sin120t 74 74 37 37 37
1 (with 2 tan 1 ), so the steady-state amplitude is 6 10.
3 . 37
Substitution of the trial solution Q A cos t B sin t into the differential equation 1 RQ Q E0 cos t yields the equations C Copyright © 2015 Pearson Education, Inc.
Section 3.7 229
1 1 A rB E0 , r A B 0 , C C with solution A
E0C E0 RC 2 , B , 1 2 R 2C 2 1 2 R 2C 2
so
E0C cos t RC sin t 1 2 R 2C 2 E0C 1 RC = cos t sin t 1 2 R 2C 2 1 2 R 2 C 2 1 2 R 2C 2 E0C cos t , 1 2 R 2C 2
Qsp t
where tan 1 RC , a 1st-quadrant angle. In Problems 11-16, we give first the trial solution I p A cos t B sin t , then the equations in A and B that we get upon substituting this trial solution into the RLC equation 1 LI RI I E t , and finally the resulting steady periodic solution. C 11.
I p A cos 2t B sin 2t ; A 6 B 10 , 6 A B 0 ; I sp t
10 60 10 1 6 10 cos 2t sin 2t cos 2t sin 2t sin 2t , 37 37 37 37 37 37
where 2 tan 1 12.
I p A cos10t B sin10t ; A 4 B 2 , 4 A B 0 ; I sp t
2 8 2 1 4 2 cos10t sin10t cos10t sin10t sin 10t , 17 17 17 17 17 17
where 2 tan 1 13.
1 6.1180 , a 4th-quadrant angle. 6
1 6.0382 , a 4th-quadrant angle. 4
I p A cos5t B sin 5t ; 3 A 2 B 0 , 2 A 3B 20 ; I sp t
40 60 20 2 3 20 cos5t sin 5t cos5t sin 5t sin 5t , 13 13 13 13 13 13
where 2 tan 1
2 5.6952 , a 4th-quadrant angle. 3 Copyright © 2015 Pearson Education, Inc.
230
14.
ELECTRICAL CIRCUITS
I p A cos100t B sin100t ; 249 A 25B 200 , 25 A 249 B 150 ; I sp t
25 921cos100t 847sin100t 31313 25 1565650 921 847 cos100t sin100t 31313 1565650 1565650
0.9990sin 100t ,
where tan 1 15.
921 0.8272 , a 1st-quadrant angle. 847
I p A cos 60 t B sin 60 t ;
1000 36 A 30 B 33 , 15 A 500 18 B 0 ; 33 250 9 495 A , B ; 250000 17775 324 2 250000 17775 324 2
2
2
2
2
4
I sp t I 0 sin 60 t , where I 0
2
33 2 250000 17775 2 324 4
4
0.1591 and
500 18 2 2 tan 4.8576 . 15 1
16.
I p A cos 377t B sin 377t ; 132129 A 47125B 226200, 47125 A 132129 B 0 ;
A
14943789900 5329837500 ; , B 9839419133 9839419133
I sp t I 0 sin 377t , where I 0
tan 1
25583220000 1.6125 and 9839419133
132129 1.2282 . 47215
In each of Problems 17–22, the first step is to substitute the given RLC parameters, the initial values I 0 and Q 0 , and the voltage E t into Eq. (16) and solve for the remaining initial value I 0
1 1 E (0) RI (0) Q (0) . L C
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(*)
Section 3.7 231
17.
With I 0 0 and Q 0 5 , Equation (*) gives I 0 75 . The solution of the RLC
equation 2 I 16 I 50 I 0 with these initial conditions is I t 25e 4 t sin 3t . 18.
Our differential equation to solve is
2 I 60 I 400 I 100e t . 50 t e by substituting the trial solution Ae t ; the We find the particular solution I p 171 general solution is I t c1e 10t c2e 20t
50 t e . 171
The initial conditions are I 0 0 and I 0 50 , the latter found by substituting 1 400 , I 0 Q 0 0 , and E 0 100 into Equation (*). ImposiC tion of these initial values on the general solution above yields the equations L 2 , R 60, ,
50 50 0, 10c1 20c2 50 , 171 171 50 100 with solution c1 , c2 . This gives the solution 9 19 c1 c2
I t 19.
50 19e 10 t 18e 20t e t . 171
Now our differential equation to solve is
2 I 60 I 400 I 1000e 10t . We find the particular solution I p 50te 10t by substituting the trial solution Ate 10t ; the general solution is I t c1e 10 t c2 e 20t 50te 10 t .
The initial conditions are I 0 0 and I 0 150 , the latter found by substituting 1 400 , I 0 0 , Q 0 1 , and E 0 100 into Equation (*). ImC position of these initial values on the general solution above yields the equations
L 2 , R 60 ,
c1 c2 0, 10c1 20c2 50 150 , with solution c1 10 , c2 10 . Thus we get the solution I t 10e 20 t 10e 10 t 50te 10 t .
20.
The differential equation 10 I 30 I 50 I 100 cos 2t has transient solution
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ELECTRICAL CIRCUITS
11 11 I tr t e 3t /2 c1 cos t c2 sin t , 2 2 and in Problem 11 we found the steady periodic solution I sp t
10 cos 2t 6sin 2t . 37
When we impose the initial conditions I 0 I 0 0 on the general solution I t I tr t I sp t , we get the equations
10 3 11 120 0, c1 c2 0, 37 2 2 37 10 270 . The figure shows the graphs of I t and I sp t . with solution c1 , c2 37 37 11 c1
Problem 20
Problem 21
Isp
Isp
I
I
I = I s p+ I t r
I = I s p+ I t r
t
21.
t
The differential equation 10 I 20 I 100 I 1000sin 5t has transient solution I tr t e t c1 cos 3t c2 sin 3t , and in Problem 13 we found the steady periodic solution 20 2 cos5t 3sin 5t . When we impose the initial conditions I 0 0 , 13 I 0 10 on the general solution I t I tr t I sp t , we get the equations
I sp t
40 300 0, c1 3c2 10 , 13 13 40 470 with solution c1 , c2 . The figure shows the graphs of I t and I sp t . 13 39 c1
22.
The differential equation 2 I 100 I 200000 I 6600 cos 60 t has transient solution Copyright © 2015 Pearson Education, Inc.
Section 3.7 233
I tr t e 25t c1 cos 25 159t c2 sin 25 159t , and in Problem 15 we found the steady periodic solution I sp t A cos 60 t B sin 60 t 0.157444 cos 60 t 0.023017 sin 60 t
(with the exact values of A and B given there). When we impose the initial conditions I 0 I 0 0 on the general solution I t I tr t I sp t , we find (with the aid of a computer algebra system) that c1 c2
33 250 9 2
250000 17775 2 324 4 11 159 250 9 2
0.157444,
53 250000 17775 2 324 4
0.026249.
The figure shows the graphs of I t and I sp t . Problem 22
I = I s p+ I t r
0.2
I
0
Isp
−0.2
0
0.1
t
23.
The LC equation LI ical frequency 0
24.
25.
1 I 0 has general solution I t c1 cos 0t c2 sin 0t with critC
1 . LC
R R2 4 L C We need only observe that the roots r both necessarily have nega2L tive real parts. According to Eq. (8) in the text, the amplitude of the steady periodic current is
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ELECTRICAL CIRCUITS
E0 1 R2 L C
2
.
Because the radicand in the denominator is a sum of squares, it is obvious that the de1 1 nominator is least when L 0 , that is, when . C LC
SECTION 3.8 ENDPOINT PROBLEMS AND EIGENVALUES The material on eigenvalues and endpoint problems in Section 3.8 can be considered optional at this point in a first course. It will not be needed until we discuss boundary value problems in the last three sections of Chapter 9 and in Chapter 10. However, after the concentration thus far on initial value problems, the inclusion of this section can give students a view of a new class of problems that have diverse and important applications (as illustrated by the subsection on the whirling string). If Section 3.8 is not covered at this point in the course, then it can be inserted just prior to Section 9.5. 1.
If 0 , then y 0 implies that y x A Bx . The endpoint conditions y 0 0
and y 1 0 yield B 0 and A 0 , respectively. Hence 0 is not an eigenvalue. If 2 0 , then the general solution of y 2 y 0 is y x A cos x B sin x ,
so y x A sin x B cos x .
Then y 0 0 yields B 0 , so y x A cos x . Next y 1 0 implies that cos 0 , so is an odd multiple of
2
. Hence the positive eigenvalues are
2 2n 1 2 , with associated eigenfunctions cos 2n 1 x , for
2
4
2.
n 1, 2,3, .
If 0 , then y 0 implies that y x A Bx . The endpoint conditions
y 0 y 0 imply only that B 0 , so 0 0 is an eigenvalue with associated ei-
genfunction y0 x 1 .
If 2 0 , then the general solution of y 2 y 0 is y x A cos x B sin x .
Then Copyright © 2015 Pearson Education, Inc.
Section 3.8 235
y x A sin x B cos x ,
so y 0 0 implies that B 0 . Next, y 0 implies that is an integral multiple of . Hence the positive eigenvalues are n 2 , with associated eigenfunctions cos nx , for n 1, 2,3, . 3.
Much as in Problem 1 we see that 0 is not an eigenvalue. Suppose that 2 0 , so y x A cos x B sin x .
Then the conditions y y 0 yield A cos B sin 0,
A cos B sin 0 .
It follows that A cos 0 B sin . Hence either A 0 and B 0 with an even
, or A 0 and B 0 with an odd multiple of
. Thus the eigenval2 nx n2 ues are if n is even, and , n 1, 2,3, , and the nth eigenfunction is yn x cos 2 4 nx if n is odd. yn x sin 2
multiple of
4.
2
Just as in Problem 2, 0 0 is an eigenvalue with associated eigenfunction y0 x 1 . If
2 0 and y x A cos x B sin x ,
then the equations y A sin B cos 0,
y A sin B cos 0
yield A sin B cos 0 . If A 0 and B 0 , then cos 0 , so must be an odd multiple of
2
. If A 0 and B 0 , then sin 0 , so must be an even multi-
n2 ple of . Therefore the positive eigenvalues are , n 1, 2,3, , with associated ei4 2 nx nx genfunctions yn x cos if n is even and yn x sin if n is odd. 2 2 5.
If 2 0 and y x A cos x B sin x ,
so that y x A sin x B cos x ,
Copyright © 2015 Pearson Education, Inc.
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ENDPOINT PROBLEMS AND EIGENVALUES
then the conditions y 2 y 2 0 yield A cos 2 B sin 2 0,
A sin 2 B cos 2 0 .
It follows either that A B and cos 2 sin 2 , or that A B and cos 2 sin 2 . The former occurs if 2
5 9
, , , 4 4 4
and the latter if 2
3 7 11 , , , 4 4 4
Hence the nth eigenvalue is 2 2n 1 2 n n , n 1, 2,3, , 2
64
and the associated eigenfunction is cos n x sin n x, n odd . yn x cos n x sin n x, n even
6.
(a) If 0 and y x A Bx , then y 0 B 0 , so y x A . But then y 1 y 1 A 0 also, so 0 is not an eigenvalue.
(b) If 2 0 and y x A cos x B sin x ,
then y x A sin x B cos x ,
so that y 0 B 0 . Hence B 0 , so y x A cos x . Then y 1 y 1 A cos sin 0 ,
so must be a positive root of the equation tan 7.
1
.
(a) If 0 and y x A Bx , then y 0 A 0 , so y x Bx . But then y 1 y 1 2 B 0 , so A B 0 and 0 is not an eigenvalue.
(b) If 2 0 and y x A cos x B sin x ,
then y 0 A 0 , so y x B sin x . Hence
Copyright © 2015 Pearson Education, Inc.
Section 3.8 237
y 1 y 1 B sin cos 0 ,
so must be a positive root of the equation tan , and hence the abscissa of a point of intersection of the lines y tan z and y z . We see from the figure below 2n 1 and lies closer and closer to that n lies just to the right of the vertical line z 2 this line as n gets larger and larger. Problem 7
β
β β
y
3
y= z
2
1
α
1
α
2
y = −z α3 α4
z
8.
(a) If 0 and y x A Bx , then y 0 A 0 , so y x Bx . But then
y 1 y 1 says only that B B . Hence 0 0 is an eigenvalue with associated eigen-
function y0 x x .
(b) If 2 0 and y x A cos x B sin x , then y 0 A 0 , so
y x B sin x . Then y 1 y 1 says that B sin B cos , so must be a posi-
tive root of the equation tan , and hence the abscissa of a point of intersection of the lines y tan z and y z . We see from the figure above that n lies just to the left of the vertical line z
2n 1 2
and lies closer and closer to this line as n gets larger
and larger. 9.
If y y 0 and 2 0 , then y x Ae x Be x . Then y 0 A B 0 , so
B A and therefore y x A e x e x . Hence y L A e L e L 0 . But
0 and e L e L 0 , so A 0 . Thus 2 is not an eigenvalue.
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238
10.
ENDPOINT PROBLEMS AND EIGENVALUES
If 2 0 , then the general solution of y y 0 is y x A cosh x B sinh x . Then y 0 0 implies that A 0 , so y x sinh x (or a nonzero multiple thereof).
Next, y 1 y 1 sinh cosh 0
implies that tanh . But the graph of y tanh lies in the first and third quadrants, while the graph of y lies in the second and fourth quadrants. It follows that the only solution of tanh is 0 , and hence that our eigenvalue problem has no negative eigenvalues. 11.
If 2 0 , then the general solution of y y 0 is y x A cosh x B sinh x . Then y 0 0 implies that B 0 , y x cosh x so (or a nonzero multiple thereof).
Next, y 1 y 1 cosh sinh 0
implies that tanh
1
. But the graph of y tanh lies in the first and third quad-
rants, while the graph of y the only solution of tanh negative eigenvalues. 12.
1
1
lies in the second and fourth quadrants. It follows that is 0 , and hence that our eigenvalue problem has no
(a) If 0 and y x A Bx , then y y means that A B A B , so
B 0 and y x A . But then y y implies nothing about A. Hence 0 0
is an eigenvalue with y0 x 1 .
(b) If 2 0 and y x Ae x Be x , then the conditions y y and y y yield the equations
Ae Be Ae Be , Ae Be Ae Be . Addition of these equations yields 2 Ae 2 Ae . Since e e because 0 , it follows that A 0 . Similarly B 0 . Thus there are no negative eigenvalues. (c) If 2 0 and y x A cos x B sin x ,
then the endpoint conditions yield the equations A cos B sin A cos B sin , A sin B cos A sin B cos . Copyright © 2015 Pearson Education, Inc.
Section 3.8 239
The first equation implies that B sin 0 , the second that A sin 0 . If A and B are not both zero, then it follows that sin 0 , so n , an integer. In this case A and B are both arbitrary. Thus cos nx and sin nx are two different eigenfunctions associated with the single eigenvalue n 2 . 13.
(a) With 1 , the general solution of y 2 y y 0 is y x Ae x Bxe x . But then y 0 A 0 and y 1 e 1 A B 0 . Hence 1 is not an eigenvalue.
(b) If 1 , then the equation y 2 y y 0 has characteristic equation
r 2 2r 0 . This equation has the two distinct real roots
2 4 4 ; call them r 2
and s. Then the general solution is y x Ae rx Be sx ,
and the conditions y 0 y 1 0 yield the equations A B 0,
Ae r Be s 0 .
If A, B 0 , then it follows that e r e s . But r s , so there is no eigenvalue 1 . (c) If 1 , then let 1 2 , so 1 2 . Then the characteristic equation
r 2 2r r 1 2 0 2
has roots 1 i , so y x e x A cos x B sin x .
Now y 0 A 0 , so y x Ae x sin x . Next, y 1 Ae 1 sin 0 , so n with n an integer. Thus the nth positive eigenvalue is n n 2 2 1 . Because 1 2 , the eigenfunction associated with n is yn x e x sin n x .
14.
If 1 2 , then we first impose the condition y 0 0 on the solution y x e x A cos x B sin x
found in Problem 13, and find that A 0 . Hence y x Be x sin x , so that y x B e x sin x e x cos x , so the condition y 1 0 yields sin cos 0 , that is, tan . 15.
(a) The endpoint conditions are
y 0 y 0 y L y 3 L 0 .
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Solutions Manual for Differential Equations Computing and Modeling and Differential Equations and Boundary Value Pro Full Download: https://downloadlink.org/p/solutions-manual-for-differential-equations-computing-and-modeling-and-differential-e 240 ENDPOINT PROBLEMS AND EIGENVALUES
With these conditions, four successive integrations as in Example 5 yield the indicated shape function y x . (b) The maximum value ymax of y x on the closed interval 0.L must occur either at
an interior point where y x 0 or at one of the endpoints x 0 and x L . Now y x k 4 x 3 12 Lx 2 12 L2 x 4kx x 2 3Lx 3L2 ,
w and the quadratic factor has no real zero. Hence x 0 is the only zero 24 EI of y x . But y 0 0 , so it follows that ymax y L .
where k
16.
(a) The endpoint conditions are y 0 y 0 0 and y L y L 0 . (b) The derivative
y x k 4 x 3 6 Lx 2 2 L2 x 2kx 2 x L x L L vanishes at x 0, , L . Because y 0 y L 0 , the argument of Problem 15(b) im2 L plies that ymax y . 2 17.
If y x k x 4 2 Lx 3 L3 x with k
w , then 24 EI
y x k 4 x 3 6 Lx 2 L3 0 L that we can verify by inspection. Now long division of the cubic 2 4 x 3 6 Lx 2 L3 by 2 x L yields the quadratic factor 2 x 2 2 Lx L2 whose zeros
has the solution x
1 3 L L 2 L 12 L2 both lie outside the interval 0, L . Thus x is, indeed, the 2 4 2 only zero of y x 0 in this interval.
18.
(a) The endpoint conditions are y 0 y 0 0 and y L y L 0 . (b) The only zero of the derivative
y x 2kx 8 x 2 15Lx 6 L2 interior to the interval 0, L is xm
15
the argument of Problem 15(b) that ymax
33 L
, and y 0 y L 0 , so it follows by 16 y xm .
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