Unformatted text preview: 5E18(pp 11761185) 1/19/06 3:42 PM Page 1176 CHAPTER 18
The charge in an electric
circuit is governed by the
differential equations that
we solve in Section 18.3. SecondOrder Differential Equations 5E18(pp 11761185) 1/19/06 3:42 PM Page 1177 The basic ideas of differential equations were explained in
Chapter 10; there we concentrated on ﬁrstorder equations. In this
chapter we study secondorder linear differential equations and
learn how they can be applied to solve problems concerning the
vibrations of springs and electric circuits. We will also see how
inﬁnite series can be used to solve differential equations.  18.1 SecondOrder Linear Equations
A secondorder linear differential equation has the form 1 P͑x͒ d 2y
dy
ϩ R͑x͒y G͑x͒
ϩ Q͑x͒
dx 2
dx where P, Q, R, and G are continuous functions. We saw in Section 10.1 that equations of
this type arise in the study of the motion of a spring. In Section 18.3 we will further pursue this application as well as the application to electric circuits.
In this section we study the case where G͑x͒ 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a secondorder linear
homogeneous differential equation is 2 P͑x͒ d 2y
dy
ϩ Q͑x͒
ϩ R͑x͒y 0
2
dx
dx If G͑x͒ 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 18.2.
Two basic facts enable us to solve homogeneous linear equations. The ﬁrst of these says
that if we know two solutions y1 and y2 of such an equation, then the linear combination
y c1 y1 ϩ c2 y2 is also a solution.
3 Theorem If y1͑x͒ and y2͑x͒ are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function y͑x͒ c1 y1͑x͒ ϩ c2 y2͑x͒
is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have P͑x͒y1Љ ϩ Q͑x͒y1 ϩ R͑x͒y1 0
Ј
and P͑x͒y2Љ ϩ Q͑x͒yЈ ϩ R͑x͒y2 0
2 1177 5E18(pp 11761185) 1178 ❙❙❙❙ 1/19/06 3:42 PM Page 1178 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS Therefore, using the basic rules for differentiation, we have
P͑x͒yЉ ϩ Q͑x͒yЈ ϩ R͑x͒y
P͑x͒͑c1 y1 ϩ c2 y2͒Љ ϩ Q͑x͒͑c1 y1 ϩ c2 y2͒Ј ϩ R͑x͒͑c1 y1 ϩ c2 y2͒
P͑x͒͑c1 y1Љ ϩ c2 y2Љ͒ ϩ Q͑x͒͑c1 yЈ ϩ c2 yЈ͒ ϩ R͑x͒͑c1 y1 ϩ c2 y2͒
1
2
c1͓P͑x͒y1Љ ϩ Q͑x͒y1 ϩ R͑x͒y1͔ ϩ c2 ͓P͑x͒y2Љ ϩ Q͑x͒y2 ϩ R͑x͒y2͔
Ј
Ј
c1͑0͒ ϩ c2͑0͒ 0
Thus, y c1 y1 ϩ c2 y2 is a solution of Equation 2.
The other fact we need is given by the following theorem, which is proved in more
advanced courses. It says that the general solution is a linear combination of two linearly
independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple
of the other. For instance, the functions f ͑x͒ x 2 and t͑x͒ 5x 2 are linearly dependent,
but f ͑x͒ e x and t͑x͒ xe x are linearly independent.
4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and P͑x͒
is never 0, then the general solution is given by y͑x͒ c1 y1͑x͒ ϩ c2 y2͑x͒
where c1 and c2 are arbitrary constants.
Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution.
In general, it is not easy to discover particular solutions to a secondorder linear equation. But it is always possible to do so if the coefﬁcient functions P, Q, and R are constant
functions, that is, if the differential equation has the form
5 ayЉ ϩ byЈ ϩ cy 0 where a, b, and c are constants and a 0.
It’s not hard to think of some likely candidates for particular solutions of Equation 5 if
we state the equation verbally. We are looking for a function y such that a constant times
its second derivative yЉ plus another constant times yЈ plus a third constant times y is equal
to 0. We know that the exponential function y e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: yЈ re rx. Furthermore, yЉ r 2e rx.
If we substitute these expressions into Equation 5, we see that y e rx is a solution if
ar 2e rx ϩ bre rx ϩ ce rx 0
͑ar 2 ϩ br ϩ c͒e rx 0 or But e rx is never 0. Thus, y e rx is a solution of Equation 5 if r is a root of the equation
6 ar 2 ϩ br ϩ c 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ayЉ ϩ byЈ ϩ cy 0. Notice that it is an algebraic equation that is obtained
from the differential equation by replacing yЉ by r 2, yЈ by r, and y by 1. 5E18(pp 11761185) 1/19/06 3:42 PM Page 1179 SECTION 18.1 SECONDORDER LINEAR EQUATIONS ❙❙❙❙ 1179 Sometimes the roots r1 and r 2 of the auxiliary equation can be found by factoring. In
other cases they are found by using the quadratic formula:
r1 7 Ϫb ϩ sb 2 Ϫ 4ac
2a r2 Ϫb Ϫ sb 2 Ϫ 4ac
2a We distinguish three cases according to the sign of the discriminant b 2 Ϫ 4ac.
b2 Ϫ 4ac Ͼ 0
In this case the roots r1 and r 2 of the auxiliary equation are real and distinct, so y1 e r 1 x
and y2 e r 2 x are two linearly independent solutions of Equation 5. (Note that e r 2 x is not a
constant multiple of e r 1 x.) Therefore, by Theorem 4, we have the following fact.
CASE I ■ If the roots r1 and r 2 of the auxiliary equation ar 2 ϩ br ϩ c 0 are real and
unequal, then the general solution of ayЉ ϩ byЈ ϩ cy 0 is
8 y c1 e r 1 x ϩ c2 e r 2 x  In Figure 1 the graphs of the basic solutions
f ͑x͒ e 2 x and t͑x͒ eϪ3 x of the differential
equation in Example 1 are shown in black and
red, respectively. Some of the other solutions,
linear combinations of f and t , are shown
in blue.
8 EXAMPLE 1 Solve the equation yЉ ϩ yЈ Ϫ 6y 0.
SOLUTION The auxiliary equation is r 2 ϩ r Ϫ 6 ͑r Ϫ 2͒͑r ϩ 3͒ 0
whose roots are r 2, Ϫ3. Therefore, by (8) the general solution of the given differential equation is 5f+g y c1 e 2x ϩ c2 eϪ3x f+5g
f+g
f g _1 gf fg
_5 1 We could verify that this is indeed a solution by differentiating and substituting into the
differential equation.
EXAMPLE 2 Solve 3 FIGURE 1 d 2y
dy
ϩ
Ϫ y 0.
dx 2
dx SOLUTION To solve the auxiliary equation 3r 2 ϩ r Ϫ 1 0 we use the quadratic formula:
r Ϫ1 Ϯ s13
6 Since the roots are real and distinct, the general solution is
y c1 e (Ϫ1ϩs13 ) x͞6 ϩ c2 e (Ϫ1Ϫs13 ) x͞6
b 2 Ϫ 4 ac 0
In this case r1 r2 ; that is, the roots of the auxiliary equation are real and equal. Let’s
denote by r the common value of r1 and r 2. Then, from Equations 7, we have
CASE II 9 ■ rϪ b
2a so 2ar ϩ b 0 5E18(pp 11761185) 1180 ❙❙❙❙ 1/19/06 3:42 PM Page 1180 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS We know that y1 e rx is one solution of Equation 5. We now verify that y2 xe rx is also
a solution:
ay2Љ ϩ byЈ ϩ cy2 a͑2re rx ϩ r 2xe rx ͒ ϩ b͑e rx ϩ rxe rx ͒ ϩ cxe rx
2
͑2ar ϩ b͒e rx ϩ ͑ar 2 ϩ br ϩ c͒xe rx
0͑e rx ͒ ϩ 0͑xe rx ͒ 0
The ﬁrst term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary
equation. Since y1 e rx and y2 xe rx are linearly independent solutions, Theorem 4 provides us with the general solution.
If the auxiliary equation ar 2 ϩ br ϩ c 0 has only one real root r, then the
general solution of ayЉ ϩ byЈ ϩ cy 0 is
10 y c1 e rx ϩ c2 xe rx  Figure 2 shows the basic solutions
f ͑x͒ eϪ3x͞2 and t͑x͒ xeϪ3x͞2 in
Example 3 and some other members of the
family of solutions. Notice that all of them
approach 0 as x l ϱ.
fg 8
5f+g f+g gf g
_5 FIGURE 2 SOLUTION The auxiliary equation 4r 2 ϩ 12r ϩ 9 0 can be factored as ͑2r ϩ 3͒2 0
so the only root is r Ϫ 3 . By (10) the general solution is
2 f _2 EXAMPLE 3 Solve the equation 4yЉ ϩ 12yЈ ϩ 9y 0. y c1 eϪ3x͞2 ϩ c2 xeϪ3x͞2 f+5g
2 b 2 Ϫ 4ac Ͻ 0
In this case the roots r1 and r 2 of the auxiliary equation are complex numbers. (See Appendix G for information about complex numbers.) We can write
CASE III ■ r1 ␣ ϩ i r 2 ␣ Ϫ i where ␣ and  are real numbers. [In fact, ␣ Ϫb͑͞2a͒,  s4ac Ϫ b 2͑͞2a͒.] Then,
using Euler’s equation
e i cos ϩ i sin
from Appendix G, we write the solution of the differential equation as
y C1 e r 1 x ϩ C2 e r 2 x C1 e ͑␣ϩi͒x ϩ C2 e ͑␣Ϫi͒x
C1 e ␣ x͑cos  x ϩ i sin  x͒ ϩ C2 e ␣ x͑cos  x Ϫ i sin  x͒
e ␣ x ͓͑C1 ϩ C2 ͒ cos  x ϩ i͑C1 Ϫ C2 ͒ sin  x͔
e ␣ x͑c1 cos  x ϩ c2 sin  x͒
where c1 C1 ϩ C2 , c2 i͑C1 Ϫ C2͒. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We summarize the discussion as follows. 5E18(pp 11761185) 1/19/06 3:42 PM Page 1181 SECTION 18.1 SECONDORDER LINEAR EQUATIONS ❙❙❙❙ 1181 If the roots of the auxiliary equation ar 2 ϩ br ϩ c 0 are the complex numbers r1 ␣ ϩ i, r 2 ␣ Ϫ i, then the general solution of ayЉ ϩ byЈ ϩ cy 0
is
11 y e ␣ x͑c1 cos  x ϩ c2 sin  x͒  Figure 3 shows the graphs of the solutions in Example 4, f ͑x͒ e 3 x cos 2x and
t͑x͒ e 3 x sin 2x, together with some linear
combinations. All solutions approach 0
as x l Ϫϱ. EXAMPLE 4 Solve the equation yЉ Ϫ 6yЈ ϩ 13y 0.
SOLUTION The auxiliary equation is r 2 Ϫ 6r ϩ 13 0. By the quadratic formula, the roots are
r 3
f+g g fg 6 Ϯ s36 Ϫ 52
6 Ϯ sϪ16
3 Ϯ 2i
2
2 By (11) the general solution of the differential equation is f _3 y e 3x͑c1 cos 2x ϩ c2 sin 2x͒ 2 InitialValue and BoundaryValue Problems
_3 An initialvalue problem for the secondorder Equation 1 or 2 consists of ﬁnding a solution y of the differential equation that also satisﬁes initial conditions of the form FIGURE 3 y͑x 0 ͒ y0 yЈ͑x 0 ͒ y1 where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and
P͑x͒ 0 there, then a theorem found in more advanced books guarantees the existence
and uniqueness of a solution to this initialvalue problem. Examples 5 and 6 illustrate the
technique for solving such a problem.
EXAMPLE 5 Solve the initialvalue problem yЉ ϩ yЈ Ϫ 6y 0
 Figure 4 shows the graph of the solution of
the initialvalue problem in Example 5. Compare
with Figure 1. y͑0͒ 1 yЈ͑0͒ 0 SOLUTION From Example 1 we know that the general solution of the differential equa tion is
y͑x͒ c1 e 2x ϩ c2 eϪ3x 20 Differentiating this solution, we get
yЈ͑x͒ 2c1 e 2x Ϫ 3c2 eϪ3x
To satisfy the initial conditions we require that
_2 FIGURE 4 0 2 12 y͑0͒ c1 ϩ c2 1 13 yЈ͑0͒ 2c1 Ϫ 3c2 0 From (13) we have c2 2 c1 and so (12) gives
3
c1 ϩ 2 c1 1
3 c1 3
5 Thus, the required solution of the initialvalue problem is
y 3 e 2x ϩ 2 eϪ3x
5
5 c2 2
5 5E18(pp 11761185) 1182 ❙❙❙❙ 1/19/06 3:43 PM Page 1182 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS ■ The solution to Example 6 is graphed in
Figure 5. It appears to be a shifted sine curve
and, indeed, you can verify that another way of
writing the solution is
y s13 sin͑x ϩ ͒ where tan 2
3 EXAMPLE 6 Solve the initialvalue problem yЉ ϩ y 0 y͑0͒ 2 yЈ͑0͒ 3 SOLUTION The auxiliary equation is r 2 ϩ 1 0, or r 2 Ϫ1, whose roots are Ϯi. Thus ␣ 0,  1, and since e 0x 1, the general solution is 5 y͑x͒ c1 cos x ϩ c2 sin x
_2π 2π yЈ͑x͒ Ϫc1 sin x ϩ c2 cos x Since
the initial conditions become y͑0͒ c1 2 _5 FIGURE 5 yЈ͑0͒ c2 3 Therefore, the solution of the initialvalue problem is
y͑x͒ 2 cos x ϩ 3 sin x
A boundaryvalue problem for Equation 1 consists of ﬁnding a solution y of the differential equation that also satisﬁes boundary conditions of the form
y͑x 0 ͒ y0 y͑x 1 ͒ y1 In contrast with the situation for initialvalue problems, a boundaryvalue problem does
not always have a solution.
EXAMPLE 7 Solve the boundaryvalue problem yЉ ϩ 2yЈ ϩ y 0 y͑0͒ 1 y͑1͒ 3 SOLUTION The auxiliary equation is r 2 ϩ 2r ϩ 1 0 ͑r ϩ 1͒2 0 or whose only root is r Ϫ1. Therefore, the general solution is
y͑x͒ c1 eϪx ϩ c2 xeϪx
The boundary conditions are satisﬁed if
y͑0͒ c1 1
 Figure 6 shows the graph of the solution of
the boundaryvalue problem in Example 7. y͑1͒ c1 eϪ1 ϩ c2 eϪ1 3
The ﬁrst condition gives c1 1, so the second condition becomes 5 eϪ1 ϩ c2 eϪ1 3
_1 5 Solving this equation for c2 by ﬁrst multiplying through by e, we get
1 ϩ c2 3e _5 FIGURE 6 so c2 3e Ϫ 1 Thus, the solution of the boundaryvalue problem is
y eϪx ϩ ͑3e Ϫ 1͒xeϪx 5E18(pp 11761185) 1/19/06 3:43 PM Page 1183 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1183 Ј
Ј
Summary: Solutions of ayЈЈ ϩ byЈ ϩ c ϭ 0 Roots of ar 2 ϩ br ϩ c 0 General solution
y c1 e r 1 x ϩ c2 e r 2 x
y c1 e rx ϩ c2 xe rx
y e ␣ x͑c1 cos  x ϩ c2 sin  x͒ r1, r2 real and distinct
r1 r2 r
r1, r2 complex: ␣ Ϯ i  18.1
1–13 Exercises
20. 2yЉ ϩ 5yЈ Ϫ 3y 0, Solve the differential equation.  y͑0͒ 1, y͑͞4͒ Ϫ3, yЈ͑0͒ 4
yЈ͑͞4͒ 4 1. yЉ Ϫ 6yЈ ϩ 8y 0 2. yЉ Ϫ 4yЈ ϩ 8y 0 21. yЉ ϩ 16y 0, 3. yЉ ϩ 8yЈ ϩ 41y 0 4. 2yЉ Ϫ yЈ Ϫ y 0 22. yЉ Ϫ 2yЈ ϩ 5y 0, y͑͒ 0, yЈ͑͒ 2 5. yЉ Ϫ 2yЈ ϩ y 0 6. 3yЉ 5yЈ 23. yЉ ϩ 2yЈ ϩ 2y 0, y͑0͒ 2, yЈ͑0͒ 1 7. 4yЉ ϩ y 0 8. 16yЉ ϩ 24yЈ ϩ 9y 0 24. yЉ ϩ 12yЈ ϩ 36y 0,
■ 9. 4yЉ ϩ yЈ 0 10. 9yЉ ϩ 4y 0 11. d 2y
dy
Ϫ2
Ϫy0
dt 2
dt 13. 25–32 d 2y
dy
ϩ
ϩy0
dt 2
dt ■ ■ 14–16 ■ ■ 12. d 2y
dy
Ϫ6
ϩ 4y 0
dt 2
dt ■ ■ ■ ■ ■ ■ ■  17–24  ■ ■ ■ ■ ■ ■ ■ ■ Solve the initialvalue problem. 17. 2yЉ ϩ 5yЈ ϩ 3y 0,
18. yЉ ϩ 3y 0, 19. 4 yЉ Ϫ 4 yЈ ϩ y 0,  18.2 y͑0͒ 3, y͑0͒ 1, ■ ■ ■ ■ ■ ■ y͑͒ Ϫ4 y͑0͒ 1, 30. yЉ Ϫ 6yЈ ϩ 9y 0, ■ ■ y͑1͒ 2 y͑0͒ 1, y͑0͒ 2, 29. yЉ Ϫ 6yЈ ϩ 25y 0, ■ ■ ■ y͑3͒ 0 y͑͒ 5 y͑0͒ 1,
y͑0͒ 1,
y͑0͒ 2, y͑͒ 2
y͑1͒ 0
y͑ ͞2͒ 1
y͑͒ 1 y͑0͒ 0,
■ ■ ■ ■ ■ 33. Let L be a nonzero real number. d y
dy
Ϫ2
ϩ 5y 0
dx 2
dx
■ 28. yЉ ϩ 100 y 0, ■ yЈ͑1͒ 1 ■ y͑0͒ 3, 32. 9yЉ Ϫ 18yЈ ϩ 10 y 0, 2 ■ ■ Solve the boundaryvalue problem, if possible. 31. yЉ ϩ 4yЈ ϩ 13y 0, d y
dy
Ϫ
Ϫ 2y 0
dx 2
dx y͑1͒ 0, ■ 27. yЉ Ϫ 3yЈ ϩ 2y 0, ■ d 2y
dy
Ϫ8
ϩ 16y 0
15.
dx 2
dx ■  ■ 26. yЉ ϩ 2yЈ 0, 2 16. ■ 25. 4 yЉ ϩ y 0, Graph the two basic solutions of the differential equation
;
and several other solutions. What features do the solutions have in
common?
14. 6 ■ yЈ͑0͒ Ϫ4 34. If a, b, and c are all positive constants and y͑x͒ is a solution yЈ͑0͒ 3 y͑0͒ 1, ■ (a) Show that the boundaryvalue problem yЉ ϩ y 0,
y͑0͒ 0, y͑L͒ 0 has only the trivial solution y 0 for
the cases 0 and Ͻ 0.
(b) For the case Ͼ 0, ﬁnd the values of for which this problem has a nontrivial solution and give the corresponding
solution. yЈ͑0͒ Ϫ1.5 of the differential equation ayЉ ϩ byЈ ϩ cy 0, show that
lim x l ϱ y͑x͒ 0. Nonhomogeneous Linear Equations
In this section we learn how to solve secondorder nonhomogeneous linear differential equations with constant coefﬁcients, that is, equations of the form
1 ayЉ ϩ byЈ ϩ cy G͑x͒ 5E18(pp 11761185) 1184 ❙❙❙❙ 1/19/06 3:43 PM Page 1184 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS where a, b, and c are constants and G is a continuous function. The related homogeneous
equation
2 ayЉ ϩ byЈ ϩ cy 0 is called the complementary equation and plays an important role in the solution of the
original nonhomogeneous equation (1).
3 Theorem The general solution of the nonhomogeneous differential equation (1)
can be written as y͑x͒ yp͑x͒ ϩ yc͑x͒
where yp is a particular solution of Equation 1 and yc is the general solution of the
complementary Equation 2.
Proof All we have to do is verify that if y is any solution of Equation 1, then y Ϫ yp is a solution of the complementary Equation 2. Indeed
a͑y Ϫ yp ͒Љ ϩ b͑y Ϫ yp ͒Ј ϩ c͑y Ϫ yp ͒ ayЉ Ϫ aypЉ ϩ byЈ Ϫ byЈ ϩ cy Ϫ cyp
p
͑ayЉ ϩ byЈ ϩ cy͒ Ϫ ͑aypЉ ϩ byp ϩ cyp ͒
Ј
t͑x͒ Ϫ t͑x͒ 0
We know from Section 18.1 how to solve the complementary equation. (Recall that the
solution is yc c1 y1 ϩ c2 y2 , where y1 and y2 are linearly independent solutions of Equation 2.) Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp . There are two methods for
ﬁnding a particular solution: The method of undetermined coefﬁcients is straightforward
but works only for a restricted class of functions G. The method of variation of parameters
works for every function G but is usually more difﬁcult to apply in practice. The Method of Undetermined Coefficients
We ﬁrst illustrate the method of undetermined coefﬁcients for the equation
ayЉ ϩ byЈ ϩ cy G͑x͒
where G͑x) is a polynomial. It is reasonable to guess that there is a particular solution
yp that is a polynomial of the same degree as G because if y is a polynomial, then
ayЉ ϩ byЈ ϩ cy is also a polynomial. We therefore substitute yp͑x͒ a polynomial (of the
same degree as G ) into the differential equation and determine the coefﬁcients.
EXAMPLE 1 Solve the equation yЉ ϩ yЈ Ϫ 2y x 2.
SOLUTION The auxiliary equation of yЉ ϩ yЈ Ϫ 2y 0 is r 2 ϩ r Ϫ 2 ͑r Ϫ 1͒͑r ϩ 2͒ 0
with roots r 1, Ϫ2. So the solution of the complementary equation is
yc c1 e x ϩ c2 eϪ2x 5E18(pp 11761185) 1/19/06 3:43 PM Page 1185 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1185 Since G͑x͒ x 2 is a polynomial of degree 2, we seek a particular solution of the form
yp͑x͒ Ax 2 ϩ Bx ϩ C
Then yp 2Ax ϩ B and ypЉ 2A so, substituting into the given differential equation, we
Ј
have
͑2A͒ ϩ ͑2Ax ϩ B͒ Ϫ 2͑Ax 2 ϩ Bx ϩ C͒ x 2
or Ϫ2Ax 2 ϩ ͑2A Ϫ 2B͒x ϩ ͑2A ϩ B Ϫ 2C͒ x 2 Polynomials are equal when their coefﬁcients are equal. Thus
 Figure 1 shows four solutions of the differential equation in Example 1 in terms of the particular solution yp and the functions f ͑x͒ e x
and t͑x͒ eϪ2 x.
8 2A Ϫ 2B 0 2A ϩ B Ϫ 2C 0 The solution of this system of equations is
A Ϫ1
2 yp+2f+3g
yp+3g Ϫ2A 1 B Ϫ1
2 C Ϫ3
4 A particular solution is therefore yp+2f _3 3 yp͑x͒ Ϫ 1 x 2 Ϫ 1 x Ϫ 3
2
2
4 yp and, by Theorem 3, the general solution is _5 y yc ϩ yp c1 e x ϩ c2 eϪ2x Ϫ 1 x 2 Ϫ 1 x Ϫ 3
2
2
4 FIGURE 1 If G͑x͒ (the right side of Equation 1) is of the form Ce k x, where C and k are constants,
then we take as a trial solution a function of the same form, yp͑x͒ Ae k x, because the
derivatives of e k x are constant multiples of e k x.
EXAMPLE 2 Solve yЉ ϩ 4y e 3x.
 Figure 2 shows solutions of the differential
equation in Example 2 in terms of yp and the
functions f ͑x͒ cos 2x and t͑x͒ sin 2x.
Notice that all solutions approach ϱ as x l ϱ
and all solutions resemble sine functions when x
is negative.
4 SOLUTION The auxiliary equation is r 2 ϩ 4 0 with roots Ϯ2i, so the solution of the complementary equation is
yc͑x͒ c1 cos 2x ϩ c2 sin 2x
For a particular solution we try yp͑x͒ Ae 3x. Then yp 3Ae 3x and ypЉ 9Ae 3x. SubstiЈ
tuting into the differential equation, we have
9Ae 3x ϩ 4͑Ae 3x ͒ e 3x yp+f+g yp+g
yp _4
yp+f
_2 FIGURE 2 1
so 13Ae 3x e 3x and A 13 . Thus, a particular solution is 2 1
yp͑x͒ 13 e 3x and the general solution is
1
y͑x͒ c1 cos 2x ϩ c2 sin 2x ϩ 13 e 3x If G͑x͒ is either C cos kx or C sin kx, then, because of the rules for differentiating the
sine and cosine functions, we take as a trial particular solution a function of the form
yp͑x͒ A cos kx ϩ B sin kx 5E18(pp 11861195) 1186 ❙❙❙❙ 1/19/06 3:44 PM Page 1186 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Solve yЉ ϩ yЈ Ϫ 2y sin x.
SOLUTION We try a particular solution yp͑x͒ A cos x ϩ B sin x
yp ϪA sin x ϩ B cos x
Ј Then ypЉ ϪA cos x Ϫ B sin x so substitution in the differential equation gives
͑ϪA cos x Ϫ B sin x͒ ϩ ͑ϪA sin x ϩ B cos x͒ Ϫ 2͑A cos x ϩ B sin x͒ sin x
͑Ϫ3A ϩ B͒ cos x ϩ ͑ϪA Ϫ 3B͒ sin x sin x or
This is true if
Ϫ3A ϩ B 0 and ϪA Ϫ 3B 1 The solution of this system is
1
A Ϫ 10 3
B Ϫ 10 so a particular solution is
1
3
yp͑x͒ Ϫ 10 cos x Ϫ 10 sin x In Example 1 we determined that the solution of the complementary equation is
yc c1 e x ϩ c2 eϪ2x. Thus, the general solution of the given equation is
1
y͑x͒ c1 e x ϩ c2 eϪ2x Ϫ 10 ͑cos x ϩ 3 sin x͒ If G͑x͒ is a product of functions of the preceding types, then we take the trial solution to be a product of functions of the same type. For instance, in solving the differential
equation
yЉ ϩ 2yЈ ϩ 4y x cos 3x
we would try
yp͑x͒ ͑Ax ϩ B͒ cos 3x ϩ ͑Cx ϩ D͒ sin 3x
If G͑x͒ is a sum of functions of these types, we use the easily veriﬁed principle of superposition, which says that if yp1 and yp2 are solutions of
ayЉ ϩ byЈ ϩ cy G1͑x͒ ayЉ ϩ byЈ ϩ cy G2͑x͒ respectively, then yp1 ϩ yp2 is a solution of
ayЉ ϩ byЈ ϩ cy G1͑x͒ ϩ G2͑x͒
EXAMPLE 4 Solve yЉ Ϫ 4y xe x ϩ cos 2x.
SOLUTION The auxiliary equation is r 2 Ϫ 4 0 with roots Ϯ2, so the solution of the com plementary equation is yc͑x͒ c1 e 2x ϩ c2 eϪ2x. For the equation yЉ Ϫ 4y xe x we try
yp1͑x͒ ͑Ax ϩ B͒e x 5E18(pp 11861195) 1/19/06 3:44 PM Page 1187 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1187 Ј
Љ
Then yp1 ͑Ax ϩ A ϩ B͒e x, yp1 ͑Ax ϩ 2A ϩ B͒e x, so substitution in the equation
gives
͑Ax ϩ 2A ϩ B͒e x Ϫ 4͑Ax ϩ B͒e x xe x
͑Ϫ3Ax ϩ 2A Ϫ 3B͒e x xe x or Thus, Ϫ3A 1 and 2A Ϫ 3B 0, so A Ϫ 1 , B Ϫ 2 , and
3
9
yp1͑x͒ (Ϫ 1 x Ϫ 2 )e x
3
9
For the equation yЉ Ϫ 4y cos 2x, we try
yp2͑x͒ C cos 2x ϩ D sin 2x
 In Figure 3 we show the particular solution
yp yp1 ϩ yp 2 of the differential equation in
Example 4. The other solutions are given in terms
of f ͑x͒ e 2 x and t͑x͒ eϪ2 x.
5 Substitution gives
Ϫ4C cos 2x Ϫ 4D sin 2x Ϫ 4͑C cos 2x ϩ D sin 2x͒ cos 2x
Ϫ8C cos 2x Ϫ 8D sin 2x cos 2x or yp+2f+g Therefore, Ϫ8C 1, Ϫ8D 0, and yp+g
yp+f
_4 yp2͑x͒ Ϫ 1 cos 2x
8 1 yp By the superposition principle, the general solution is
_2 y yc ϩ yp1 ϩ yp2 c1 e 2x ϩ c2 eϪ2x Ϫ ( 1 x ϩ 2 )e x Ϫ 1 cos 2x
3
9
8 FIGURE 3 Finally we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation and therefore can’t be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or by x 2
if necessary) so that no term in yp͑x͒ is a solution of the complementary equation.
EXAMPLE 5 Solve yЉ ϩ y sin x.
SOLUTION The auxiliary equation is r 2 ϩ 1 0 with roots Ϯi, so the solution of the com plementary equation is
yc͑x͒ c1 cos x ϩ c2 sin x
Ordinarily, we would use the trial solution
yp͑x͒ A cos x ϩ B sin x
but we observe that it is a solution of the complementary equation, so instead we try
yp͑x͒ Ax cos x ϩ Bx sin x
Then yp͑x͒ A cos x Ϫ Ax sin x ϩ B sin x ϩ Bx cos x
Ј
ypЉ͑x͒ Ϫ2A sin x Ϫ Ax cos x ϩ 2B cos x Ϫ Bx sin x 5E18(pp 11861195) 1188 ❙❙❙❙ 1/19/06 3:44 PM Page 1188 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS  The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4. Substitution in the differential equation gives
ypЉ ϩ yp Ϫ2A sin x ϩ 2B cos x sin x 4 so A Ϫ 1 , B 0, and
2
_2π 2π yp͑x͒ Ϫ 1 x cos x
2 yp The general solution is
_4 y͑x͒ c1 cos x ϩ c2 sin x Ϫ 1 x cos x
2 FIGURE 4 We summarize the method of undetermined coefﬁcients as follows:
1. If G͑x͒ e kxP͑x͒, where P is a polynomial of degree n, then try yp͑x͒ e kxQ͑x͒, where Q͑x͒ is an nthdegree polynomial (whose coefﬁcients are determined by
substituting in the differential equation.)
2. If G͑x͒ e kxP͑x͒ cos mx or G͑x͒ e kxP͑x͒ sin mx, where P is an nthdegree polynomial, then try
yp͑x͒ e kxQ͑x͒ cos mx ϩ e kxR͑x͒ sin mx
where Q and R are nthdegree polynomials.
Modiﬁcation: If any term of yp is a solution of the complementary equation, multiply yp
by x (or by x 2 if necessary).
EXAMPLE 6 Determine the form of the trial solution for the differential equation yЉ Ϫ 4yЈ ϩ 13y e 2x cos 3x.
SOLUTION Here G͑x͒ has the form of part 2 of the summary, where k 2, m 3, and
P͑x͒ 1. So, at ﬁrst glance, the form of the trial solution would be yp͑x͒ e 2x͑A cos 3x ϩ B sin 3x͒
But the auxiliary equation is r 2 Ϫ 4r ϩ 13 0, with roots r 2 Ϯ 3i, so the solution
of the complementary equation is
yc͑x͒ e 2x͑c1 cos 3x ϩ c2 sin 3x͒
This means that we have to multiply the suggested trial solution by x. So, instead, we
use
yp͑x͒ xe 2x͑A cos 3x ϩ B sin 3x͒ The Method of Variation of Parameters
Suppose we have already solved the homogeneous equation ayЉ ϩ byЈ ϩ cy 0 and written the solution as
4 y͑x͒ c1 y1͑x͒ ϩ c2 y2͑x͒ where y1 and y2 are linearly independent solutions. Let’s replace the constants (or parame 5E18(pp 11861195) 1/19/06 3:44 PM Page 1189 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1189 ters) c1 and c2 in Equation 4 by arbitrary functions u1͑x͒ and u2͑x͒. We look for a particular solution of the nonhomogeneous equation ayЉ ϩ byЈ ϩ cy G͑x͒ of the form
yp͑x͒ u1͑x͒y1͑x͒ ϩ u2͑x͒y2͑x͒ 5 (This method is called variation of parameters because we have varied the parameters c1
and c2 to make them functions.) Differentiating Equation 5, we get
6 yp ͑u1 y1 ϩ u2 y2 ͒ ϩ ͑u1 y1 ϩ u2 y2 ͒
Ј
Ј
Ј
Ј
Ј Since u1 and u2 are arbitrary functions, we can impose two conditions on them. One condition is that yp is a solution of the differential equation; we can choose the other condition
so as to simplify our calculations. In view of the expression in Equation 6, let’s impose the
condition that
u1 y1 ϩ u2 y2 0
Ј
Ј 7 Then ypЉ u1 y1 ϩ u2 y2 ϩ u1 y1Љ ϩ u2 y2Љ
Ј Ј
Ј Ј Substituting in the differential equation, we get
a͑u1 y1 ϩ u2 y2 ϩ u1 y1Љ ϩ u2 y2Љ͒ ϩ b͑u1 y1 ϩ u2 y2 ͒ ϩ c͑u1 y1 ϩ u2 y2 ͒ G
Ј Ј
Ј Ј
Ј
Ј
or
8 u1͑ay1Љ ϩ by1 ϩ cy1 ͒ ϩ u2͑ay2Љ ϩ by2 ϩ cy2 ͒ ϩ a͑u1 y1 ϩ u2 y2 ͒ G
Ј
Ј
Ј Ј
Ј Ј But y1 and y2 are solutions of the complementary equation, so
ay1Љ ϩ by1 ϩ cy1 0
Ј and ay2Љ ϩ by2 ϩ cy2 0
Ј and Equation 8 simpliﬁes to
9 a͑u1 y1 ϩ u2 y2 ͒ G
Ј Ј
Ј Ј Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2 .
Ј
Ј
After solving this system we may be able to integrate to ﬁnd u1 and u2 and then the particular solution is given by Equation 5.
EXAMPLE 7 Solve the equation yЉ ϩ y tan x, 0 Ͻ x Ͻ ͞2.
SOLUTION The auxiliary equation is r 2 ϩ 1 0 with roots Ϯi, so the solution of yЉ ϩ y 0 is c1 sin x ϩ c2 cos x. Using variation of parameters, we seek a solution
of the form
yp͑x͒ u1͑x͒ sin x ϩ u2͑x͒ cos x
Then yp ͑u1 sin x ϩ u2 cos x͒ ϩ ͑u1 cos x Ϫ u2 sin x͒
Ј
Ј
Ј Set
10 uЈ sin x ϩ uЈ cos x 0
1
2 5E18(pp 11861195) ❙❙❙❙ 1190 1/19/06 3:45 PM Page 1190 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS ypЉ u1 cos x Ϫ u2 sin x Ϫ u1 sin x Ϫ u2 cos x
Ј
Ј Then For yp to be a solution we must have
ypЉ ϩ yp uЈ cos x Ϫ uЈ sin x tan x
1
2 11 Solving Equations 10 and 11, we get
u1͑sin 2x ϩ cos 2x͒ cos x tan x
Ј
u1 sin x
Ј u1͑x͒ Ϫcos x (We seek a particular solution, so we don’t need a constant of integration here.) Then,
from Equation 10, we obtain
u2 Ϫ
Ј sin x
sin 2x
cos 2x Ϫ 1
u1 Ϫ
Ј
cos x Ϫ sec x
cos x
cos x
cos x
u2͑x͒ sin x Ϫ ln͑sec x ϩ tan x͒  Figure 5 shows four solutions of the
differential equation in Example 7. So 2.5 (Note that sec x ϩ tan x Ͼ 0 for 0 Ͻ x Ͻ ͞2.) Therefore
yp͑x͒ Ϫcos x sin x ϩ ͓sin x Ϫ ln͑sec x ϩ tan x͔͒ cos x
Ϫcos x ln͑sec x ϩ tan x͒ π
2 0
yp and the general solution is _1 y͑x͒ c1 sin x ϩ c2 cos x Ϫ cos x ln͑sec x ϩ tan x͒ FIGURE 5  18.2 Exercises
12. 2yЉ ϩ 3yЈ ϩ y 1 ϩ cos 2x 1–10  Solve the differential equation or initialvalue problem
using the method of undetermined coefﬁcients. 1. yЉ ϩ 3yЈ ϩ 2y x 2 2. yЉ ϩ 9y e 3x 3. yЉ Ϫ 2yЈ sin 4x ■ 4. yЉ ϩ 6yЈ ϩ 9y 1 ϩ x
Ϫx 7. yЉ ϩ y e x ϩ x 3, y͑0͒ 2, 8. yЉ Ϫ 4y e x cos x,
9. yЉ Ϫ yЈ xe x, y͑0͒ 2, 10. yЉ ϩ yЈ Ϫ 2y x ϩ sin 2x,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 15. yЉ ϩ 9 yЈ 1 ϩ xe 9x
16. yЉ ϩ 3yЈ Ϫ 4 y ͑x 3 ϩ x͒e x
17. yЉ ϩ 2 yЈ ϩ 10 y x 2eϪx cos 3x yЈ͑0͒ 0
■ ■ 18. yЉ ϩ 4y e 3x ϩ x sin 2x
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■  Graph the particular solution and several other solutions.
What characteristics do these solutions have in common? 19–22  Solve the differential equation using (a) undetermined
coefﬁcients and (b) variation of parameters. 11. 4yЉ ϩ 5yЈ ϩ y e x 19. yЉ ϩ 4y x ; 11–12 ■ 14. yЉ ϩ 9 yЈ xeϪx cos x yЈ͑0͒ 1 ■ ■ 13. yЉ ϩ 9 y e 2x ϩ x 2 sin x yЈ͑0͒ 2 y͑0͒ 1, ■  Write a trial solution for the method of undetermined
coefﬁcients. Do not determine the coefﬁcients. yЈ͑0͒ 0 y͑0͒ 1, ■ 13–18 6. yЉ ϩ 2yЈ ϩ y xeϪx 5. yЉ Ϫ 4yЈ ϩ 5y e ■ 20. yЉ Ϫ 3yЈ ϩ 2y sin x ■ 5E18(pp 11861195) 1/19/06 3:45 PM Page 1191 SECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS 21. yЉ Ϫ 2yЈ ϩ y e 2x
22. yЉ Ϫ yЈ e
■ ■ ■ 25. yЉ Ϫ 3yЈ ϩ 2y x
■ ■ ■ ■ ■ ■ ■ ■ 1191 1
1 ϩ eϪx 26. yЉ ϩ 3yЈ ϩ 2y sin͑e x ͒ ■ 1
x 23–28  Solve the differential equation using the method of variation of parameters. 27. yЉ Ϫ y 23. yЉ ϩ y sec x, 0 Ͻ x Ͻ ͞2 28. yЉ ϩ 4yЈ ϩ 4y 24. yЉ ϩ y cot x, 0 Ͻ x Ͻ ͞2  18.3 ❙❙❙❙ ■ ■ ■ ■ eϪ2x
x3
■ ■ ■ ■ ■ ■ ■ ■ Applications of SecondOrder Differential Equations
Secondorder linear differential equations have a variety of applications in science and
engineering. In this section we explore two of them: the vibration of springs and electric
circuits. Vibrating Springs
We consider the motion of an object with mass m at the end of a spring that is either vertical (as in Figure 1) or horizontal on a level surface (as in Figure 2).
In Section 6.4 we discussed Hooke’s Law, which says that if the spring is stretched (or
compressed) x units from its natural length, then it exerts a force that is proportional to x :
0 equilibrium
position restoring force Ϫkx x m where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals
mass times acceleration), we have m x FIGURE 1 1
equilibrium position
m
0 FIGURE 2 x m d 2x
Ϫkx
dt 2 or m d 2x
ϩ kx 0
dt 2 This is a secondorder linear differential equation. Its auxiliary equation is mr 2 ϩ k 0
with roots r Ϯ i, where sk͞m. Thus, the general solution is
x͑t͒ c1 cos t ϩ c2 sin t x which can also be written as
x͑t͒ A cos͑ t ϩ ␦͒
where sk͞m (frequency) A sc 2 ϩ c 2
1
2
cos ␦ c1
A (amplitude) sin ␦ Ϫ c2
A ͑␦ is the phase angle͒ (See Exercise 17.) This type of motion is called simple harmonic motion. 5E18(pp 11861195) 1192 ❙❙❙❙ 1/19/06 3:45 PM Page 1192 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS EXAMPLE 1 A spring with a mass of 2 kg has natural length 0.5 m. A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length
of 0.7 m and then released with initial velocity 0, ﬁnd the position of the mass at any
time t .
SOLUTION From Hooke’s Law, the force required to stretch the spring is k͑0.2͒ 25.6
so k 25.6͞0.2 128. Using this value of the spring constant k, together with m 2
in Equation 1, we have
2 d 2x
ϩ 128x 0
dt 2 As in the earlier general discussion, the solution of this equation is
x͑t͒ c1 cos 8t ϩ c2 sin 8t 2 We are given the initial condition that x͑0͒ 0.2. But, from Equation 2, x͑0͒ c1.
Therefore, c1 0.2. Differentiating Equation 2, we get
xЈ͑t͒ Ϫ8c1 sin 8t ϩ 8c2 cos 8t
Since the initial velocity is given as xЈ͑0͒ 0, we have c2 0 and so the solution is
x͑t͒ 1 cos 8t
5 Damped Vibrations m We next consider the motion of a spring that is subject to a frictional force (in the case of
the horizontal spring of Figure 2) or a damping force (in the case where a vertical spring
moves through a ﬂuid as in Figure 3). An example is the damping force supplied by a
shock absorber in a car or a bicycle.
We assume that the damping force is proportional to the velocity of the mass and acts
in the direction opposite to the motion. (This has been conﬁrmed, at least approximately,
by some physical experiments.) Thus FIGURE 3 damping force Ϫc dx
dt where c is a positive constant, called the damping constant. Thus, in this case, Newton’s
Second Law gives
m d 2x
dx
restoring force ϩ damping force Ϫkx Ϫ c
dt 2
dt or
3 m d 2x
dx
ϩc
ϩ kx 0
dt 2
dt 5E18(pp 11861195) 1/19/06 3:45 PM Page 1193 SECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1193 Equation 3 is a secondorder linear differential equation and its auxiliary equation is
mr 2 ϩ cr ϩ k 0. The roots are
r1 4 Ϫc ϩ sc 2 Ϫ 4mk
2m r2 Ϫc Ϫ sc 2 Ϫ 4mk
2m According to Section 18.1 we need to discuss three cases.
x CASE I c 2 Ϫ 4 mk Ͼ 0 (overdamping) ■ In this case r1 and r 2 are distinct real roots and
0 x c1 e r1 t ϩ c2 e r2 t t x 0 t Since c, m, and k are all positive, we have sc 2 Ϫ 4mk Ͻ c, so the roots r1 and r 2 given by
Equations 4 must both be negative. This shows that x l 0 as t l ϱ. Typical graphs of
x as a function of t are shown in Figure 4. Notice that oscillations do not occur. (It’s possible for the mass to pass through the equilibrium position once, but only once.) This is
because c 2 Ͼ 4mk means that there is a strong damping force (highviscosity oil or grease)
compared with a weak spring or small mass.
c 2 Ϫ 4mk 0 (critical damping)
This case corresponds to equal roots
CASE II FIGURE 4 Overdamping ■ r1 r 2 Ϫ c
2m and the solution is given by
x ͑c1 ϩ c2 t͒eϪ͑c͞2m͒t
It is similar to Case I, and typical graphs resemble those in Figure 4 (see Exercise 12), but
the damping is just sufﬁcient to suppress vibrations. Any decrease in the viscosity of the
ﬂuid leads to the vibrations of the following case.
c 2 Ϫ 4mk Ͻ 0 (underdamping)
Here the roots are complex: CASE III ■ ͮ r1
c
Ϫ
Ϯ i
r2
2m x x=Ae– (c/ 2m)t where
0 t x=_Ae– (c/ 2m)t
FIGURE 5 Underdamping s4mk Ϫ c 2
2m The solution is given by
x eϪ͑c͞2m͒t͑c1 cos t ϩ c2 sin t͒
We see that there are oscillations that are damped by the factor eϪ͑c͞2m͒t. Since c Ͼ 0 and
m Ͼ 0, we have Ϫ͑c͞2m͒ Ͻ 0 so eϪ͑c͞2m͒t l 0 as t l ϱ. This implies that x l 0 as t l ϱ;
that is, the motion decays to 0 as time increases. A typical graph is shown in Figure 5. 5E18(pp 11861195) 1194 ❙❙❙❙ 1/19/06 3:45 PM Page 1194 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS EXAMPLE 2 Suppose that the spring of Example 1 is immersed in a ﬂuid with damping
constant c 40. Find the position of the mass at any time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m͞s.
SOLUTION From Example 1 the mass is m 2 and the spring constant is k 128, so the
differential equation (3) becomes 2 d 2x
dx
ϩ 40
ϩ 128x 0
dt 2
dt
d 2x
dx
ϩ 64x 0
2 ϩ 20
dt
dt or The auxiliary equation is r 2 ϩ 20r ϩ 64 ͑r ϩ 4͒͑r ϩ 16͒ 0 with roots Ϫ4
and Ϫ16, so the motion is overdamped and the solution is
x͑t͒ c1 eϪ4t ϩ c2 eϪ16t  Figure 6 shows the graph of the position
function for the overdamped motion in Example 2. We are given that x͑0͒ 0, so c1 ϩ c2 0. Differentiating, we get 0.03 xЈ͑t͒ Ϫ4c1 eϪ4t Ϫ 16c2 eϪ16t
xЈ͑0͒ Ϫ4c1 Ϫ 16c2 0.6 so 0 1.5 Since c2 Ϫc1 , this gives 12c1 0.6 or c1 0.05. Therefore
x 0.05͑eϪ4t Ϫ eϪ16t ͒ FIGURE 6 Forced Vibrations
Suppose that, in addition to the restoring force and the damping force, the motion of the
spring is affected by an external force F͑t͒. Then Newton’s Second Law gives
m d 2x
restoring force ϩ damping force ϩ external force
dt 2
Ϫkx Ϫ c dx
ϩ F͑t͒
dt Thus, instead of the homogeneous equation (3), the motion of the spring is now governed
by the following nonhomogeneous differential equation: 5 m d 2x
dx
ϩc
ϩ kx F͑t͒
dt 2
dt The motion of the spring can be determined by the methods of Section 18.2. 5E18(pp 11861195) 1/19/06 3:45 PM Page 1195 SECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1195 A commonly occurring type of external force is a periodic force function
F͑t͒ F0 cos 0 t where 0 sk͞m In this case, and in the absence of a damping force (c 0), you are asked in Exercise 9 to
use the method of undetermined coefﬁcients to show that 6 x͑t͒ c1 cos t ϩ c2 sin t ϩ F0
2 cos 0 t
m͑ Ϫ 0 ͒
2 If 0 , then the applied frequency reinforces the natural frequency and the result is
vibrations of large amplitude. This is the phenomenon of resonance (see Exercise 10). Electric Circuits
R switch
L
E
C In Sections 10.3 and 10.6 we were able to use ﬁrstorder separable and linear equations to
analyze electric circuits that contain a resistor and inductor (see Figure 5 on page 639 or
Figure 4 on page 671) or a resistor and capacitor (see Exercise 29 on page 673). Now that
we know how to solve secondorder linear equations, we are in a position to analyze the
circuit shown in Figure 7. It contains an electromotive force E (supplied by a battery or
generator), a resistor R, an inductor L, and a capacitor C, in series. If the charge on the
capacitor at time t is Q Q͑t͒, then the current is the rate of change of Q with respect
to t : I dQ͞dt. As in Section 10.6, it is known from physics that the voltage drops across
the resistor, inductor, and capacitor are FIGURE 7 RI L dI
dt Q
C respectively. Kirchhoff’s voltage law says that the sum of these voltage drops is equal to
the supplied voltage:
L dI
Q
ϩ RI ϩ
E͑t͒
dt
C Since I dQ͞dt, this equation becomes 7 L d 2Q
dQ
1
ϩR
ϩ
Q E͑t͒
dt 2
dt
C which is a secondorder linear differential equation with constant coefﬁcients. If the charge
Q0 and the current I 0 are known at time 0, then we have the initial conditions
Q͑0͒ Q0 QЈ͑0͒ I͑0͒ I 0 and the initialvalue problem can be solved by the methods of Section 18.2. 5E18(pp 11961204) 1196 ❙❙❙❙ 1/19/06 3:27 PM Page 1196 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS A differential equation for the current can be obtained by differentiating Equation 7
with respect to t and remembering that I dQ͞dt :
L d 2I
dI
1
ϩ
I EЈ͑t͒
2 ϩ R
dt
dt
C EXAMPLE 3 Find the charge and current at time t in the circuit of Figure 7 if R 40 ⍀, L 1 H, C 16 ϫ 10Ϫ4 F, E͑t͒ 100 cos 10t, and the initial charge and current are
both 0. SOLUTION With the given values of L, R, C, and E͑t͒, Equation 7 becomes 8 d 2Q
dQ
ϩ 40
ϩ 625Q 100 cos 10t
dt 2
dt The auxiliary equation is r 2 ϩ 40r ϩ 625 0 with roots
r Ϫ40 Ϯ sϪ900
Ϫ20 Ϯ 15i
2 so the solution of the complementary equation is
Qc͑t͒ eϪ20t ͑c1 cos 15t ϩ c2 sin 15t͒
For the method of undetermined coefﬁcients we try the particular solution
Qp͑t͒ A cos 10t ϩ B sin 10t
Then Qp͑t͒ Ϫ10A sin 10t ϩ 10B cos 10t
Ј
QpЉ͑t͒ Ϫ100A cos 10t Ϫ 100B sin 10t Substituting into Equation 8, we have
͑Ϫ100A cos 10t Ϫ 100B sin 10t͒ ϩ 40͑Ϫ10A sin 10t ϩ 10B cos 10t͒
ϩ 625͑A cos 10t ϩ B sin 10t͒ 100 cos 10t
or ͑525A ϩ 400B͒ cos 10t ϩ ͑Ϫ400A ϩ 525B͒ sin 10t 100 cos 10t Equating coefﬁcients, we have
525A ϩ 400B 100
Ϫ400A ϩ 525B 0 21A ϩ 16B 4
or
or Ϫ16A ϩ 21B 0 84
64
The solution of this system is A 697 and B 697 , so a particular solution is
1
Qp͑t͒ 697 ͑84 cos 10t ϩ 64 sin 10t͒ and the general solution is
4
Q͑t͒ Qc͑t͒ ϩ Qp͑t͒ eϪ20t ͑c1 cos 15t ϩ c2 sin 15t͒ ϩ 697 ͑21 cos 10t ϩ 16 sin 10t͒ 5E18(pp 11961204) 1/19/06 3:28 PM Page 1197 SECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1197 Imposing the initial condition Q͑0͒ 0, we get
Q͑0͒ c1 ϩ 697 0
84 84
c1 Ϫ 697 To impose the other initial condition we ﬁrst differentiate to ﬁnd the current:
I dQ
eϪ20t ͓͑Ϫ20c1 ϩ 15c2 ͒ cos 15t ϩ ͑Ϫ15c1 Ϫ 20c2 ͒ sin 15t͔
dt
40
ϩ 697 ͑Ϫ21 sin 10t ϩ 16 cos 10t͒ I͑0͒ Ϫ20c1 ϩ 15c2 ϩ 640 0
697 464
c2 Ϫ 2091 Thus, the formula for the charge is
4
697 Q͑t͒ ͫ ͬ eϪ20t
͑Ϫ63 cos 15t Ϫ 116 sin 15t͒ ϩ ͑21 cos 10t ϩ 16 sin 10t͒
3 and the expression for the current is
1
I͑t͒ 2091 ͓eϪ20t͑Ϫ1920 cos 15t ϩ 13,060 sin 15t͒ ϩ 120͑Ϫ21 sin 10t ϩ 16 cos 10t͔͒ In Example 3 the solution for Q͑t͒ consists of two parts. Since eϪ20t l 0 as
t l ϱ and both cos 15t and sin 15t are bounded functions,
NOTE 1 ■ 0.2
4
Qc͑t͒ 2091 eϪ20t͑Ϫ63 cos 15t Ϫ 116 sin 15t͒ l 0 Qp 0 Q 1.2 as t l ϱ So, for large values of t ,
4
Q͑t͒ Ϸ Qp͑t͒ 697 ͑21 cos 10t ϩ 16 sin 10t͒ _0.2 FIGURE 8 and, for this reason, Qp͑t͒ is called the steady state solution. Figure 8 shows how the graph
of the steady state solution compares with the graph of Q in this case.
NOTE 2
Comparing Equations 5 and 7, we see that mathematically they are identical.
This suggests the analogies given in the following chart between physical situations that,
at ﬁrst glance, are very different.
■ 5
7 d 2x
dx
ϩ c
ϩ kx F͑t͒
dt 2
dt
d 2Q
dQ
1
L
ϩR
ϩ
Q E͑t͒
dt 2
dt
C
m Spring system
x
dx͞dt
m
c
k
F͑t͒ displacement
velocity
mass
damping constant
spring constant
external force Electric circuit
Q
I dQ͞dt
L
R
1͞C
E͑t͒ charge
current
inductance
resistance
elastance
electromotive force We can also transfer other ideas from one situation to the other. For instance, the steady
state solution discussed in Note 1 makes sense in the spring system. And the phenomenon
of resonance in the spring system can be usefully carried over to electric circuits as electrical resonance. 5E18(pp 11961204) 1198 ❙❙❙❙ 1/19/06 3:28 PM Page 1198 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS  18.3 Exercises
12. Consider a spring subject to a frictional or damping force. 1. A spring with a 3kg mass is held stretched 0.6 m beyond its (a) In the critically damped case, the motion is given by
x c1 ert ϩ c2 tert. Show that the graph of x crosses the
taxis whenever c1 and c2 have opposite signs.
(b) In the overdamped case, the motion is given by
x c1e r t ϩ c2 e r t, where r1 Ͼ r2. Determine a condition on
the relative magnitudes of c1 and c2 under which the graph
of x crosses the taxis at a positive value of t. natural length by a force of 20 N. If the spring begins at its
equilibrium position but a push gives it an initial velocity of
1.2 m͞s, ﬁnd the position of the mass after t seconds.
2. A spring with a 4kg mass has natural length 1 m and is main 1 tained stretched to a length of 1.3 m by a force of 24.3 N. If the
spring is compressed to a length of 0.8 m and then released
with zero velocity, ﬁnd the position of the mass at any time t. 13. A series circuit consists of a resistor with R 20 ⍀, an induc tor with L 1 H, a capacitor with C 0.002 F, and a 12V
battery. If the initial charge and current are both 0, ﬁnd the
charge and current at time t. 3. A spring with a mass of 2 kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m
beyond its natural length. The spring is stretched 1 m beyond
its natural length and then released with zero velocity. Find the
position of the mass at any time t. 14. A series circuit contains a resistor with R 24 ⍀, an inductor 4. A spring with a mass of 3 kg has damping constant 30 and ; spring constant 123.
(a) Find the position of the mass at time t if it starts at the
equilibrium position with a velocity of 2 m͞s.
(b) Graph the position function of the mass. ; a voltage of E͑t͒ 12 sin 10t. Find the charge at time t.
16. The battery in Exercise 14 is replaced by a generator producing critical damping.
6. For the spring in Exercise 4, ﬁnd the damping constant that ; 7. A spring has a mass of 1 kg and its spring constant is k 100.
The spring is released at a point 0.1 m above its equilibrium
position. Graph the position function for the following values
of the damping constant c: 10, 15, 20, 25, 30. What type of
damping occurs in each case? ; 8. A spring has a mass of 1 kg and its damping constant is
c 10. The spring starts from its equilibrium position with a
velocity of 1 m͞s. Graph the position function for the following
values of the spring constant k: 10, 20, 25, 30, 40. What type
of damping occurs in each case?
9. Suppose a spring has mass m and spring constant k and let sk͞m. Suppose that the damping constant is so small
that the damping force is negligible. If an external force
F͑t͒ F0 cos 0 t is applied, where 0 , use the method
of undetermined coefﬁcients to show that the motion of the
mass is described by Equation 6. 10. As in Exercise 9, consider a spring with mass m, spring con stant k, and damping constant c 0, and let sk͞m.
If an external force F͑t͒ F0 cos t is applied (the applied
frequency equals the natural frequency), use the method of
undetermined coefﬁcients to show that the motion of the mass
is given by x͑t͒ c1 cos t ϩ c2 sin t ϩ ͑F0 ͑͞2m͒͒t sin t.
11. Show that if 0 , but ͞ 0 is a rational number, then the
motion described by Equation 6 is periodic. with L 2 H, a capacitor with C 0.005 F, and a 12V battery. The initial charge is Q 0.001 C and the initial current
is 0.
(a) Find the charge and current at time t.
(b) Graph the charge and current functions. 15. The battery in Exercise 13 is replaced by a generator producing 5. For the spring in Exercise 3, ﬁnd the mass that would produce would produce critical damping. 2 ; a voltage of E͑t͒ 12 sin 10t.
(a) Find the charge at time t.
(b) Graph the charge function.
17. Verify that the solution to Equation 1 can be written in the form x͑t͒ A cos͑ t ϩ ␦͒. 18. The ﬁgure shows a pendulum with length L and the angle from the vertical to the pendulum. It can be shown that , as a
function of time, satisﬁes the nonlinear differential equation
t
d 2
ϩ sin 0
dt 2
L
where t is the acceleration due to gravity. For small values of
we can use the linear approximation sin Ϸ and then the
differential equation becomes linear.
(a) Find the equation of motion of a pendulum with length 1 m
if is initially 0.2 rad and the initial angular velocity is
d͞dt 1 rad͞s.
(b) What is the maximum angle from the vertical?
(c) What is the period of the pendulum (that is, the time to
complete one backandforth swing)?
(d) When will the pendulum ﬁrst be vertical?
(e) What is the angular velocity when the pendulum is vertical? ¨ L 5E18(pp 11961204) 1/19/06 3:29 PM Page 1199 SECTION 18.4 SERIES SOLUTIONS  18.4 ❙❙❙❙ 1199 Series Solutions
Many differential equations can’t be solved explicitly in terms of ﬁnite combinations of
simple familiar functions. This is true even for a simplelooking equation like
yЉ Ϫ 2xyЈ ϩ y 0 1 But it is important to be able to solve equations such as Equation 1 because they arise from
physical problems and, in particular, in connection with the Schrödinger equation in quantum mechanics. In such a case we use the method of power series; that is, we look for a
solution of the form
y f ͑x͒ ϱ ͚cx
n n c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и n0 The method is to substitute this expression into the differential equation and determine the
values of the coefﬁcients c0 , c1, c2 , . . . . This technique resembles the method of undetermined coefﬁcients discussed in Section 18.2.
Before using power series to solve Equation 1, we illustrate the method on the simpler
equation yЉ ϩ y 0 in Example 1. It’s true that we already know how to solve this equation by the techniques of Section 18.1, but it’s easier to understand the power series
method when it is applied to this simpler equation.
EXAMPLE 1 Use power series to solve the equation yЉ ϩ y 0.
SOLUTION We assume there is a solution of the form
2 y c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и ϱ ͚cx
n n n0 We can differentiate power series term by term, so
ϱ ͚ nc x yЈ c1 ϩ 2c2 x ϩ 3c3 x 2 ϩ и и и n nϪ1 n1 3 yЉ 2c2 ϩ 2 и 3c3 x ϩ и и и ϱ ͚ n͑n Ϫ 1͒c x
n nϪ2 n2 In order to compare the expressions for y and yЉ more easily, we rewrite yЉ as follows:
 By writing out the ﬁrst few terms of (4), you
can see that it is the same as (3). To obtain (4)
we replaced n by n ϩ 2 and began the summation at 0 instead of 2. yЉ 4 ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c nϩ2 xn n0 Substituting the expressions in Equations 2 and 4 into the differential equation, we
obtain
ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c xn ϩ nϩ2 n0 ϱ ͚cx
n n 0 n0 or
ϱ 5 ͚ ͓͑n ϩ 2͒͑n ϩ 1͒c nϩ2 n0 ϩ cn ͔x n 0 5E18(pp 11961204) 1200 ❙❙❙❙ 1/19/06 3:30 PM Page 1200 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS If two power series are equal, then the corresponding coefﬁcients must be equal. Therefore, the coefﬁcients of x n in Equation 5 must be 0:
͑n ϩ 2͒͑n ϩ 1͒cnϩ2 ϩ cn 0
cnϩ2 Ϫ 6 cn
͑n ϩ 1͒͑n ϩ 2͒ n 0, 1, 2, 3, . . . Equation 6 is called a recursion relation. If c0 and c1 are known, this equation allows
us to determine the remaining coefﬁcients recursively by putting n 0, 1, 2, 3, . . . in
succession.
Put n 0: c2 Ϫ c0
1ؒ2 Put n 1: c3 Ϫ c1
2ؒ3 Put n 2: c4 Ϫ c2
c0
c0
3ؒ4
1ؒ2ؒ3ؒ4
4! Put n 3: c5 Ϫ c3
c1
c1
4ؒ5
2ؒ3ؒ4ؒ5
5! Put n 4: c6 Ϫ c4
c0
c0
Ϫ
Ϫ
5ؒ6
4! 5 ؒ 6
6! Put n 5: c7 Ϫ c5
c1
c1
Ϫ
Ϫ
6ؒ7
5! 6 ؒ 7
7! By now we see the pattern:
For the even coefficients, c2n ͑Ϫ1͒n c0
͑2n͒! For the odd coefficients, c2nϩ1 ͑Ϫ1͒n c1
͑2n ϩ 1͒! Putting these values back into Equation 2, we write the solution as
y c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ c5 x 5 ϩ и и и ͩ c0 1 Ϫ
c0 ͩ ϩ c1 x Ϫ
ϱ ͚ ͑Ϫ1͒ n n0 ͪ x2
x4
x6
x 2n
ϩ
Ϫ
ϩ и и и ϩ ͑Ϫ1͒n
ϩ иии
2!
4!
6!
͑2n͒! ͪ x3
x5
x7
x 2nϩ1
ϩ
Ϫ
ϩ и и и ϩ ͑Ϫ1͒n
ϩ иии
3!
5!
7!
͑2n ϩ 1͒! ϱ
x 2n
x 2nϩ1
ϩ c1 ͚ ͑Ϫ1͒n
͑2n͒!
͑2n ϩ 1͒!
n0 Notice that there are two arbitrary constants, c0 and c1. 5E18(pp 11961204) 1/19/06 3:31 PM Page 1201 SECTION 18.4 SERIES SOLUTIONS ❙❙❙❙ 1201 NOTE 1
We recognize the series obtained in Example 1 as being the Maclaurin series
for cos x and sin x. (See Equations 12.10.16 and 12.10.15.) Therefore, we could write the
solution as
■ y͑x͒ c0 cos x ϩ c1 sin x
But we are not usually able to express power series solutions of differential equations in
terms of known functions.
EXAMPLE 2 Solve yЉ Ϫ 2xyЈ ϩ y 0.
SOLUTION We assume there is a solution of the form y ϱ ͚cx
n n n0 yЈ Then ϱ ͚ nc x
n nϪ1 n1 yЉ and ϱ ͚ n͑n Ϫ 1͒c x
n nϪ2 n2 ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c nϩ2 xn n0 as in Example 1. Substituting in the differential equation, we get
ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c x n Ϫ 2x nϩ2 n0 ϱ nϩ2 xn Ϫ n0 ͚ n1 2ncn x n n n1 nϩ2 n0 ϩ ϱ ͚cx
n n 0 n 0 n0 ϱ ͚ ͓͑n ϩ 2͒͑n ϩ 1͒c 2ncn x n nϪ1 ͚ 2nc x ϱ ϱ ͚ n n1 ͚ ͑n ϩ 2͒͑n ϩ 1͒c ϱ ϱ ͚ nc x n ϩ ϱ ͚cx
n n0 Ϫ ͑2n Ϫ 1͒cn ͔x n 0 n0 This equation is true if the coefﬁcient of x n is 0:
͑n ϩ 2͒͑n ϩ 1͒cnϩ2 Ϫ ͑2n Ϫ 1͒cn 0
7 cnϩ2 2n Ϫ 1
cn
͑n ϩ 1͒͑n ϩ 2͒ n 0, 1, 2, 3, . . . We solve this recursion relation by putting n 0, 1, 2, 3, . . . successively in Equation 7:
Put n 0: c2 Ϫ1
c0
1ؒ2 Put n 1: c3 1
c1
2ؒ3 Put n 2: c4 3
3
3
c2 Ϫ
c0 Ϫ c0
3ؒ4
1ؒ2ؒ3ؒ4
4! Put n 3: c5 5
1ؒ5
1ؒ5
c3
c1
c1
4ؒ5
2ؒ3ؒ4ؒ5
5! 5E18(pp 11961204) 1202 ❙❙❙❙ 1/19/06 3:32 PM Page 1202 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS Put n 4: c6 7
3ؒ7
3ؒ7
c4 Ϫ
c0 Ϫ
c0
5ؒ6
4! 5 ؒ 6
6! Put n 5: c7 9
1ؒ5ؒ9
1ؒ5ؒ9
c5
c1
c1
6ؒ7
5! 6 ؒ 7
7! Put n 6: c8 11
3 ؒ 7 ؒ 11
c6 Ϫ
c0
7ؒ8
8! Put n 7: c9 13
1 ؒ 5 ؒ 9 ؒ 13
c7
c1
8ؒ9
9! In general, the even coefﬁcients are given by
c2n Ϫ 3 ؒ 7 ؒ 11 ؒ и и и ؒ ͑4n Ϫ 5͒
c0
͑2n͒! and the odd coefﬁcients are given by
c2nϩ1 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒
c1
͑2n ϩ 1͒! The solution is
y c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ и и и ͩ c0 1 Ϫ
ͪ 1 2
3 4
3ؒ7 6
3 ؒ 7 ؒ 11 8
x Ϫ
x Ϫ
x Ϫ
x Ϫ иии
2!
4!
6!
8! ͩ ϩ c1 x ϩ ͪ 1 3
1ؒ5 5
1ؒ5ؒ9 7
1 ؒ 5 ؒ 9 ؒ 13 9
x ϩ
x ϩ
x ϩ
x ϩ иии
3!
5!
7!
9! or ͩ y c0 1 Ϫ 8 ϱ
1 2
3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n
x Ϫ ͚
x
2!
͑2n͒!
n2 ͩ ϩ c1 x ϩ ϱ ͚ n1 ͪ 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1
x
͑2n ϩ 1͒! ͪ NOTE 2
In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a
solution.
NOTE 3
Unlike the situation of Example 1, the power series that arise in the solution of
Example 2 do not deﬁne elementary functions. The functions
■ ■ y1͑x͒ 1 Ϫ
and ϱ
1 2
3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n
x Ϫ ͚
x
2!
͑2n͒!
n2 y2͑x͒ x ϩ ͚ ϱ n1 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1
x
͑2n ϩ 1͒! 5E18(pp 11961204) 1/19/06 3:32 PM Page 1203 CHAPTER 18 REVIEW 2 1203 are perfectly good functions but they can’t be expressed in terms of familiar functions. We
can use these power series expressions for y1 and y2 to compute approximate values of the
functions and even to graph them. Figure 1 shows the ﬁrst few partial sums T0, T2, T4, . . .
(Taylor polynomials) for y1͑x͒, and we see how they converge to y1 . In this way we can
graph both y1 and y2 in Figure 2. T¸
2 _2 ❙❙❙ T¡¸ NOTE 4 ■ If we were asked to solve the initialvalue problem _8 yЉ Ϫ 2xyЈ ϩ y 0 FIGURE 1 c0 y͑0͒ 0 ﬁ
_2.5 c1 yЈ͑0͒ 1 2.5 This would simplify the calculations in Example 2, since all of the even coefﬁcients would
be 0. The solution to the initialvalue problem is › _15  18.4
 ϱ ͚ y͑x͒ x ϩ FIGURE 2 n1 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1
x
͑2n ϩ 1͒! Exercises
11. yЉ ϩ x 2 yЈ ϩ x y 0, Use power series to solve the differential equation. 1. yЈ Ϫ y 0 2. yЈ x y 3. yЈ x 2 y 4. ͑x Ϫ 3͒yЈ ϩ 2y 0 5. yЉ ϩ x yЈ ϩ y 0 6. yЉ y ■ ■ 10. yЉ ϩ x 2 y 0, y͑0͒ 1, y͑0͒ 1, 18 Review yЈ͑0͒ 0 ■ CONCEPT CHECK 1. (a) Write the general form of a secondorder homogeneous linear differential equation with constant coefﬁcients.
(b) Write the auxiliary equation.
(c) How do you use the roots of the auxiliary equation to solve
the differential equation? Write the form of the solution for
each of the three cases that can occur.
2. (a) What is an initialvalue problem for a secondorder differ ential equation?
(b) What is a boundaryvalue problem for such an equation?
3. (a) Write the general form of a secondorder nonhomogeneous linear differential equation with constant coefﬁcients. ■ y͑0͒ 0,
■ ■ yЈ͑0͒ 1
■ ■ ■ ■ y͑0͒ 1 yЈ͑0͒ 0 is called a Bessel function of order 0.
(a) Solve the initialvalue problem to ﬁnd a power series
expansion for the Bessel function.
(b) Graph several Taylor polynomials until you reach one that
looks like a good approximation to the Bessel function on
the interval ͓Ϫ5, 5͔. ; yЈ͑0͒ 0 ■ x 2 yЉ ϩ x yЈ ϩ x 2 y 0 8. yЉ x y
9. yЉ Ϫ x yЈ Ϫ y 0, ■ 12. The solution of the initialvalue problem 7. ͑x 2 ϩ 1͒yЉ ϩ x yЈ Ϫ y 0  yЈ͑0͒ 1 we would observe from Theorem 11.10.5 that 15 1–11 y͑0͒ 0 ■ (b) What is the complementary equation? How does it help
solve the original differential equation?
(c) Explain how the method of undetermined coefﬁcients
works.
(d) Explain how the method of variation of parameters works.
4. Discuss two applications of secondorder linear differential equations.
5. How do you use power series to solve a differential equation? ■ 5E18(pp 11961204) 1204 ❙❙❙❙ 1/19/06 3:33 PM Page 1204 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS ■ TRUEFALSE QUIZ 3. The general solution of yЉ Ϫ y 0 can be written as Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. y c1 cosh x ϩ c2 sinh x 1. If y1 and y2 are solutions of yЉ ϩ y 0, then y1 ϩ y2 is also a solution of the equation. 4. The equation yЉ Ϫ y e x has a particular solution of the form 2. If y1 and y2 are solutions of yЉ ϩ 6yЈ ϩ 5y x, then yp Ae x c1 y1 ϩ c2 y2 is also a solution of the equation. ■ 1–10  EXERCISES yЉ Ϫ x yЈ Ϫ 2y 0 2. yЉ ϩ 4yЈ ϩ 13y 0 17. A series circuit contains a resistor with R 40 ⍀, an inductor with L 2 H, a capacitor with C 0.0025 F, and a 12V battery. The initial charge is Q 0.01 C and the initial current
is 0. Find the charge at time t. 3. yЉ ϩ 3y 0
4. 4yЉ ϩ 4yЈ ϩ y 0 d 2y
dy
Ϫ4
ϩ 5y e 2x
dx 2
dx 18. A spring with a mass of 2 kg has damping constant 16, and a force of 12.8 N keeps the spring stretched 0.2 m beyond its
natural length. Find the position of the mass at time t if it
starts at the equilibrium position with a velocity of 2.4 m͞s. d 2y
dy
Ϫ 2y x 2
6.
2 ϩ
dx
dx
7.
8. 19. Assume that the earth is a solid sphere of uniform density with d 2y
dy
Ϫ2
ϩ y x cos x
dx 2
dx
d 2y
ϩ 4 y sin 2 x
dx 2 mass M and radius R 3960 mi. For a particle of mass m
within the earth at a distance r from the earth’s center, the
gravitational force attracting the particle to the center is
Fr d 2y
dy
Ϫ
Ϫ 6y 1 ϩ eϪ2x
9.
dx 2
dx
d 2y
ϩ y csc x,
10.
dx 2
■ ■ 11–14 ■  ■ 0 Ͻ x Ͻ ͞2
■ y͑1͒ 3, 12. yЉ Ϫ 6yЈ ϩ 25y 0,
13. yЉ Ϫ 5yЈ ϩ 4y 0,
14. 9yЉ ϩ y 3x ϩ e Ϫx,
■ ■ ■ ■ ■ ■ ■ ■ ■ Solve the initialvalue problem. 11. yЉ ϩ 6yЈ 0, ■ ■ ■ 16. Use power series to solve the equation Solve the differential equation. 1. yЉ Ϫ 2yЈ Ϫ 15y 0 5. ■ yЈ͑1͒ 12 y͑0͒ 2,
y͑0͒ 0,
y͑0͒ 1, ■ ■ ■ yЈ͑0͒ 1
yЈ͑0͒ 2
■ ■ 15. Use power series to solve the initialvalue problem yЉ ϩ x yЈ ϩ y 0 y͑0͒ 0 where G is the gravitational constant and Mr is the mass of the
earth within the sphere of radius r .
ϪGMm
(a) Show that Fr
r.
R3
(b) Suppose a hole is drilled through the earth along a diameter. Show that if a particle of mass m is dropped from rest,
at the surface, into the hole, then the distance y y͑t͒ of
the particle from the center of the earth at time t is given by
yЉ͑t͒ Ϫk 2 y͑t͒ yЈ͑0͒ 1
■ ϪGMr m
r2 yЈ͑0͒ 1 ■ ■ where k 2 GM͞R 3 t͞R.
(c) Conclude from part (b) that the particle undergoes simple
harmonic motion. Find the period T.
(d) With what speed does the particle pass through the center
of the earth? ...
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This note was uploaded on 02/08/2010 for the course M 340L taught by Professor Lay during the Spring '10 term at École Normale Supérieure.
 Spring '10
 Lay
 Linear Algebra, Algebra

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