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. S econdOrder 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 d 2y
dx 2 Px Qx dy
dx Rxy Gx 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.
0, for all x, in Equation 1. Such equaIn this section we study the case where G x
tions are called homogeneous linear equations. Thus, the form of a secondorder linear
homogeneous differential equation is 2 Px d 2y
dx 2 Qx dy
dx Rxy 0 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 18.2.
If G x
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 c2 y2 is also a solution. 3 Theorem If y1 x and y2 x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function yx c1 y1 x c2 y2 x is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have P x y1
and Q x y1 R x y1 0 P x y2 Q x y2 R x y2 0 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
Pxy Qxy Rxy P x c1 y1 c2 y2 Q x c1 y1 c2 y2 R x c1 y1 c2 y2 P x c1 y1 c2 y2 Q x c1 y1 c2 y2 R x c1 y1 c2 y2 c1 P x y1 Q x y1 c1 0
Thus, y c1 y1 c2 0 R x y1 c2 P x y2 Q x y2 R x y2 0 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 x
x 2 and t x
5x 2 are linearly dependent,
x
x
but f x
e and t x
xe are linearly independent.
4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and P x
is never 0, then the general solution is given by yx c1 y1 x c2 y2 x 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 ay by 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 y plus another constant times y 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: y
re rx. Furthermore, y
r 2e rx.
rx
If we substitute these expressions into Equation 5, we see that y e is a solution if
ar 2e rx
ar 2 or
But e rx is never 0. Thus, y
6 bre rx ce rx 0 br c e rx 0 e rx is a solution of Equation 5 if r is a root of the equation
ar 2 br c 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation a y
by
c y 0. Notice that it is an algebraic equation that is obtained
from the differential equation by replacing y by r 2, y 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:
sb 2
2a b r1 7 4 ac b r2 sb 2
2a 4 ac We distinguish three cases according to the sign of the discriminant b 2 4 ac. b2 4 ac 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
unequal, then the general solution of ay
by
cy
8 c1 e r 1 x y  In Figure 1 the graphs of the basic solutions
fx
e 2 x and t x
e 3 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. EXAMPLE 1 Solve the equation y c 0 are real and c2 e r 2 x 6y 0. SOLUTION The auxiliary equation is r2
whose roots are r
tial equation is 8 y br
0 is 2, r 6 r 2r 3 0 3. Therefore, by (8) the general solution of the given differen 5f+g
f+5g
f+g
f g _1 gf fg
_5 c1 e 2 x y
1 3x c2 e 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
dx 2 dy
dx y 0. SOLUTION To solve the auxiliary equation 3r 2 r 1 0 we use the quadratic formula:
1 r s13
6 Since the roots are real and distinct, the general solution is
y c1 e ( 1 s13 ) x 6 c2 e ( 1 s13 ) x 6 b 2 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 ■ r b
2a so 2 ar b 0 5E18(pp 11761185) 1180 ❙❙❙❙ 1/19/06 3:42 PM Page 1180 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS e rx is one solution of Equation 5. We now verify that y2 We know that y1
a solution:
a y2 b y2 a 2re rx c y2 r 2xe rx
b e rx 2 ar
0 e rx b e rx
ar 2 0 xe rx r xe rx xe rx is also c xe rx c xe rx br 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 br c
general solution of ay
by
c y 0 is 0 has only one real root r, then the 10 c1 e rx y  Figure 2 shows the basic solutions
fx
e 3x 2 and t x
x e 3x 2 in
Example 3 and some other members of the
family of solutions. Notice that all of them
approach 0 as x l . EXAMPLE 3 Solve the equation 4 y 12 y SOLUTION The auxiliary equation 4r 2 c2 xe rx 9y 0. 9 0 can be factored as 2 0 12r
2r 3 fg 8
5f+g
_2 f+g 3
2 so the only root is r f gf g
_5 . By (10) the general solution is f+5g y c1 e 3x 2 3x 2 c2 xe 2 b 2 4 ac 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 ■ FIGURE 2 r1 i r2 where and are real numbers. [In fact,
using Euler’s equation
ei i
b 2a , cos s4ac b 2 2a .] Then, i sin from Appendix G, we write the solution of the differential equation as
y C1 e r 1 x
C1 e
e x e x x C2 e r 2 x
cos x C1 i sin x C2 cos x c1 cos x i C1 e x C2 e
C2 e i C1 x i x cos x i sin x C2 sin x c2 sin x where c1 C1 C2 , c2 i C1 C2 . 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 br c 0 are the complex numbers r1
i , r2
i , then the general solution of a y
by
cy 0
is
11 y  Figure 3 shows the graphs of the solue 3 x cos 2 x and
tions in Example 4, f x
3x
tx
e sin 2 x, together with some linear
combinations. All solutions approach 0
as x l
. 6y SOLUTION The auxiliary equation is r 2 fg c2 sin x 13 y
6r 0. 13 0. By the quadratic formula, the roots are
6 r
g c1 cos x EXAMPLE 4 Solve the equation y 3
f+g x e s36
2 52 6 s 16
2 3 2i By (11) the general solution of the differential equation is f _3 2 e 3x c1 cos 2 x y c2 sin 2 x 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 y x0 y0 y x0 y1 where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and
Px
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 y
 Figure 4 shows the graph of the solution of
the initialvalue problem in Example 5. Compare
with Figure 1. y 6y 0 y0 1 y0 SOLUTION From Example 1 we know that the general solution of the differential equa tion is
yx 20 c1 e 2 x c2 e 3x Differentiating this solution, we get
2c1 e 2 x yx 3c2 e 3x To satisfy the initial conditions we require that
_2 FIGURE 4 0 2 0 12 y0 c1 13 y0 2c1 From (13) we have c2 c2 1 3c2 0 2
31 c and so (12) gives c1 2
31 c 1 3
5 c1 c2 Thus, the required solution of the initialvalue problem is
y 3
5 e 2x 2
5 e 3x 2
5 5E18(pp 11761185) 1182 ❙❙❙❙ 1/19/06 3:43 PM 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 Page 1182 s13 sin x 2
3 where tan EXAMPLE 6 Solve the initialvalue problem y 0 y0 2π y0 3 1, and since e 1 0, or r 2
1, whose roots are
1, the general solution is 0x yx
_2π 2 SOLUTION The auxiliary equation is r 2 0, 5 y Since c1 cos x yx i. Thus c2 sin x c1 sin x c2 cos x the initial conditions become
y0 _5 FIGURE 5 c1 2 y0 c2 3 Therefore, the solution of the initialvalue problem is
yx 2 cos x 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
y x0 y0 y x1 y1 In contrast with the situation for initialvalue problems, a boundaryvalue problem does
not always have a solution.
EXAMPLE 7 Solve the boundaryvalue problem y 2y y 0 y0 1 y1 SOLUTION The auxiliary equation is r2
whose only root is r 2r 1 0 or r 1 2 0 1. Therefore, the general solution is
yx c1 e x c2 xe x The boundary conditions are satisﬁed if
y0
y1  Figure 6 shows the graph of the solution of
the boundaryvalue problem in Example 7. The ﬁrst condition gives c1 5 c1
c1 e 5 FIGURE 6 1 c2 e 3 1 c2 e 1 3 Solving this equation for c2 by ﬁrst multiplying through by e, we get
1 _5 1 1, so the second condition becomes
e _1 1 c2 3e so c2 3e Thus, the solution of the boundaryvalue problem is
y e x 3e 1 xe x 1 3 5E18(pp 11761185) 1/19/06 3:43 PM Page 1183 S ECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Summary: Solutions of ay Roots of ar 2 by c br c 1–13 General solution 0 1. y 6y 8y 3. y 8y 41y 5. y 2y y 7. 4 y y 20. 2 y 2. y 0 21. y 8. 16 y 0 y 0
0 24 y 9y 0, 22. y 5y 8y 16 y
2y 5y 0, y 0, y 23. y y 6. 3 y 0 4y 4. 2 y 0 5y 2y 2y 0, y0 2, y0 24. y 0 ■ 9. 4 y y 10. 9 y 0 11. d 2y
dt 2 2 13. d 2y
dt 2 dy
dt dy
dt y 12. 0 d 2y
dt 2 4y
6 0
dy
dt 25–32 4y 0 25. 4 y 0 16. ■ ■ ■ ■ ■ ■ 2 17. 2 y
18. y
19. 4 y  5y
■ 0 ■ ■ ■ ■ ■ ■ Solve the initialvalue problem.
5y
3y
4y  18.2 3y
0, 0, y0
y 0, ■ ■ 2y 0, y0 1, y3 0, y0 6y 25 y 0, 30. y 6y 9y 4y 13 y ■ 18 y
■ 0,
0, y0
y0 ■ y ■ ■ 5 1,
1, y0
0, 10 y
■ 2, 0 2, 2 y1 0 y 2 0, y0
■ y ■ 1 y 1 ■ ■ ■ 33. Let L be a nonzero real number. 0
■ ■ 2 32. 9 y 16 y dy
dx ■ 0 ■ 4 y1 ■ dy
8
dx ■ 17–24 2y 1 ■ y 3, y0 31. y dy
dx dy
dx 2 ■ ■ 1 y1 ■ 1, 29. y 2 ■ y0 100 y d 2y
15.
dx 2 0, y1 4
2 0, 28. y ■  dy
dx 2 0, 4 4 2y Graph the two basic solutions of the differential equation
;
and several other solutions. What features do the solutions have in
common?
14. 6 y 3, ■ y0 26. y y ■ 2 4 ■ 0, y 3y 14–16 1, y0 Solve the boundaryvalue problem, if possible. 27. y ■ y 36 y ■  0, 3y 12 y
■ ■ ■ c1 e r 1 x c2 e r 2 x
c1 e rx c2 xe rx
e x c1 cos x c2 sin x y
y
y Exercises Solve the differential equation.  1183 0 r1, r2 real and distinct
r1 r2 r
r1, r2 complex:
i  18.1 ❙❙❙❙ y0
1,
y0 3,
y0
1, y0 4 ■ (a) Show that the boundaryvalue problem y
y 0,
y0
0, y L
0 has only the trivial solution y 0 for
the cases
0 and
0.
(b) For the case
0, ﬁnd the values of for which this problem has a nontrivial solution and give the corresponding
solution.
34. If a, b, and c are all positive constants and y x is a solution 3
y0 1.5 of the differential equation ay
lim x l y x
0. by cy 0, show that 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 ay by cy Gx 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
ay 2 by 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 yx yp x yc x 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 yp is a solution of the complementary Equation 2. Indeed
ay yp by yp cy yp ay a yp by ay by cy tx tx b yp cy 0 a yp b yp c yp
c yp We know from Section 18.1 how to solve the complementary equation. (Recall that the
solution is yc c1 y1 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
ay by cy Gx where G x) 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
ay
by
c y is also a polynomial. We therefore substitute yp x
a polynomial (of the
same degree as G ) into the differential equation and determine the coefﬁcients.
EXAMPLE 1 Solve the equation y y 2y SOLUTION The auxiliary equation of y r2
with roots r 1, r y
2 x 2.
2y r 0 is 1r 2 0 2. So the solution of the complementary equation is
yc c1 e x c2 e 2x 5E18(pp 11761185) 1/19/06 3:43 PM Page 1185 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Since G x 1185 x 2 is a polynomial of degree 2, we seek a particular solution of the form
Ax 2 yp x
Then yp
have ❙❙❙❙ B and yp 2 Ax C 2 A so, substituting into the given differential equation, we
2 Ax 2 2A 2 Ax B 2 Ax 2 or Bx 2A 2B x Bx x2 B 2A C
2C x2 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 x
ex
and t x
e 2 x. 2A 1 2B 0 2A B 2C 0 The solution of this system of equations is 8 1
2 A yp+2f+3g
yp+3g 2A 1
2 B 3
4 C A particular solution is therefore yp+2f _3 3 1
2 yp x yp 1
2 x2 3
4 x and, by Theorem 3, the general solution is _5 y FIGURE 1 yc c1 e x yp 1
2 2x c2 e 1
2 x2 x 3
4 If G x (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, yp x
Ae k x, because the
kx
kx
derivatives of e are constant multiples of e .
EXAMPLE 2 Solve y e 3 x. 4y SOLUTION The auxiliary equation is r 2
 Figure 2 shows solutions of the differential
equation in Example 2 in terms of yp and the
functions f x
cos 2 x and t x
sin 2 x.
Notice that all solutions approach as x l
and all solutions resemble sine functions when x
is negative.
4 yc x c1 cos 2 x yp+g
yp _4 _2 FIGURE 2 so 13Ae 3x e 3x and A 1
13 4 Ae 3x 3Ae 3x and yp 9Ae 3x. Substi e 3x . Thus, a particular solution is 2 yp x yp+f 2i, so the solution of the c2 sin 2 x For a particular solution we try yp x
Ae 3x. Then yp
tuting into the differential equation, we have
9Ae 3x yp+f+g 0 with roots 4 complementary equation is 1
13 e 3x and the general solution is
yx c1 cos 2 x c2 sin 2 x 1
13 e 3x If G x is either C cos k x or C sin k x, then, because of the rules for differentiating the
sine and cosine functions, we take as a trial particular solution a function of the form
yp x A cos k x B sin k x 5E18(pp 11861195) 1186 ❙❙❙❙ 1/19/06 3:44 PM Page 1186 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Solve y y 2y sin x. SOLUTION We try a particular solution yp x
yp Then A sin x A cos x B sin x B cos x yp A cos x B sin x so substitution in the differential equation gives
A cos x B sin x A sin x B cos x
3A or 2 A cos x B cos x B sin x
3B sin x A sin x
sin x This is true if
3A B 0 and A 3B 1 The solution of this system is
1
10 A 3
10 B so a particular solution is
1
10 yp x 3
10 cos x sin x In Example 1 we determined that the solution of the complementary equation is
yc c1 e x c2 e 2 x. Thus, the general solution of the given equation is
c1 e x yx 1
10 2x c2 e cos x 3 sin x If G x 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
y 2y 4y x cos 3x we would try
yp x Ax B cos 3x Cx D sin 3x If G x 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
ay
respectively, then yp1 by cy ay by cy G2 x yp2 is a solution of
ay EXAMPLE 4 Solve y G1 x 4y xe x by cy G1 x G2 x cos 2 x.
2 SOLUTION The auxiliary equation is r
plementary equation is yc x
c1 e 2 x yp1 x 4 0 with roots 2, so the solution of the comc2 e 2 x. For the equation y
4 y xe x we try
Ax B ex 5E18(pp 11861195) 1/19/06 3:44 PM Page 1187 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Then yp1
gives Ax B e x, yp1 A Ax Ax or
Thus, 4 Ax 3A 1 and 2 A 3B 4y 5 ( 4 C cos 2 x 4 D sin 2 x 8C 1, 8D )e x D sin 2 x 4 C cos 2 x D sin 2 x cos 2 x 8 D sin 2 x cos 2 x 0, and
1
8 yp2 x 1 yp 2
9 x , and 8C cos 2 x yp+f
_4 1
3 2
9 ,B Substitution gives Therefore, yp+g xe x C cos 2 x or yp+2f+g 3B e x cos 2 x, we try
yp2 x  In Figure 3 we show the particular solution
yp yp1 yp 2 of the differential equation in
Example 4. The other solutions are given in terms
e 2 x and t x
of f x
e 2 x. xe x 1
3 0, so A
yp1 x For the equation y B ex 2A 3Ax 1187 B e x, so substitution in the equation 2A B ex 2A ❙❙❙❙ cos 2 x By the superposition principle, the general solution is
_2 y FIGURE 3 yc yp1 c1 e 2 x yp2 c2 e 2x (1x
3 2
9 )e x 1
8 cos 2 x 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 yp x is a solution of the complementary equation.
EXAMPLE 5 Solve y y sin x. SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of the com plementary equation is
yc x c1 cos x c2 sin x Ordinarily, we would use the trial solution
yp x A cos x B sin x but we observe that it is a solution of the complementary equation, so instead we try
yp x
Then yp x
yp x A cos x
2 A sin x A x cos x
A x sin x
A x cos x B x sin x
B sin x B x cos x 2 B cos x B x 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
yp 4
1
2 so A
_2π ,B yp 2 A sin x 2B cos x sin x 0, and 2π 1
2 yp x x cos x yp The general solution is
_4 yx FIGURE 4 c1 cos x c2 sin x 1
2 x cos x We summarize the method of undetermined coefﬁcients as follows:
e kxP x , where P is a polynomial of degree n, then try yp x
e kxQ x ,
where Q x is an nthdegree polynomial (whose coefﬁcients are determined by
substituting in the differential equation.) 1. If G x e kxP x cos m x or G x
polynomial, then try e kxP x sin m x, where P is an nthdegree 2. If G x e kxQ x cos m x yp x e kxR x sin m x 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 y 4y 13y e 2 x cos 3x. SOLUTION Here G x has the form of part 2 of the summary, where k Px 2, m 3, and 1. So, at ﬁrst glance, the form of the trial solution would be
e 2 x A cos 3x yp x But the auxiliary equation is r 2 4r
of the complementary equation is 13 0, with roots r e 2 x c1 cos 3x yc x B sin 3x
2 3i, so the solution c2 sin 3x This means that we have to multiply the suggested trial solution by x. So, instead, we
use
xe 2 x A cos 3x yp x B sin 3x The Method of Variation of Parameters
Suppose we have already solved the homogeneous equation ay
ten the solution as
4 yx c1 y1 x by cy 0 and writ c2 y2 x 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 u1 x and u2 x . We look for a particular solution of the nonhomogeneous equation ay
by
c y G x of the form
yp x 5 u1 x y1 x u2 x y2 x (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
yp 6 u1 y1 u2 y2 u1 y1 u2 y2 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 7 Then yp u1 y1 u2 y2
u2 y2 0
u1 y1 u2 y2 Substituting in the differential equation, we get
a u1 y1 u2 y2 u1 y1 u2 y2 b u1 y1 u2 y2 c u1 y1 u2 y2 G or
8 u1 a y1 b y1 c y1 u2 a y2 b y2 c y2 a u1 y1 u2 y2 G But y1 and y2 are solutions of the complementary equation, so
ay1 b y1 c y1 0 and ay2 b y2 c y2 0 and Equation 8 simpliﬁes to
a u1 y1 9 u2 y2 G Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2 .
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 y y SOLUTION The auxiliary equation is r 2 y
y 0 is c1 sin x
of the form tan x, 0 2. x 1 0 with roots i, so the solution of
c2 cos x. Using variation of parameters, we seek a solution
yp x Then yp u1 x sin x u2 x cos x u1 sin x u2 cos x u1 cos x Set
10 u1 sin x u2 cos x 0 u2 sin x 5E18(pp 11861195) ❙❙❙❙ 1190 1/19/06 3:45 PM Page 1190 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS yp Then u1 cos x u2 sin x u1 sin x u2 cos x For yp to be a solution we must have
yp 11 yp u1 cos x u2 sin x tan x Solving Equations 10 and 11, we get
u1 sin 2x
u1 cos 2x sin x cos x tan x
u1 x cos x (We seek a particular solution, so we don’t need a constant of integration here.) Then,
from Equation 10, we obtain  Figure 5 shows four solutions of the
differential equation in Example 7. So 2.5 (Note that sec x u2 x
tan x ln sec x x cos x sec x tan x 2.) Therefore cos x sin x sin x cos x ln sec x π
2 yp cos 2x 1
cos x sin x 0 for 0 yp x
0 sin 2x
cos x sin x
u1
cos x u2 ln sec x tan x cos x tan x and the general solution is _1 yx FIGURE 5  18.2 x2 1. y 3y 2y 3. y 2y 2. y x 5. y 4y 7. y y 8. y 4y e x cos x, 9. y y x e x, y0 10. y y 2y x 5y
ex ■ e
x 3, ■ 12. 2 y 2, 9y 2,
sin 2 x,
■ 2y y0 1, y0 ; 11–12 6y y0
y0 y 1 5y tan x x y ex 1 ■ 13. y cos 2 x
■ 9y e 2x 9y xe x cos 0
■ 9y 1 3y 4y ■ ■ ■ ■ ■ 2y 10 y x 2e 4y 3x 18. y
■ ■ x x3 17. y
■ ■ 9x 16. y y0 ■ x 2 sin x 15. y 1 ■ y ■ 14. y xe 2 1, 3y ■  Write a trial solution for the method of undetermined
coefﬁcients. Do not determine the coefﬁcients. x  Graph the particular solution and several other solutions.
What characteristics do these solutions have in common? 11. 4 y cos x ln sec x 13–18 0 y0 ■ ■ e 3x 6. y y0 ■ 9y 4. y sin 4 x ■ c2 cos x Exercises 1–10  Solve the differential equation or initialvalue problem
using the method of undetermined coefﬁcients. ■ c1 sin x ■ ■ e
■ xe ■ x ex
x cos 3 x x sin 2 x
■ ■ ■ ■ ■ ■ ■ 19–22  Solve the differential equation using (a) undetermined
coefﬁcients and (b) variation of parameters. 19. y 4y x 20. y 3y 2y sin x ■ 5E18(pp 11861195) 1/19/06 3:45 PM Page 1191 S ECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS 21. y 2y
y e2x y
x ❙❙❙❙ 1191 1 25. y 3y 2y 26. y 3y 2y sin e x 23–28  Solve the differential equation using the method of variation of parameters. 27. y y 23. y y sec x, 0 x 2 28. y 4y 4y e 2x
x3 24. y y cot x, 0 x 2 22. y
■ ■ e
■ ■  18.3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 e x 1
x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 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 :
m equilibrium
position 0 restoring force m x kx 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 x FIGURE 1
equilibrium position
m
0 FIGURE 2 m 1 x d 2x
dt 2 kx or m d 2x
dt 2 kx 0 This is a secondorder linear differential equation. Its auxiliary equation is mr 2
with roots r
i, where
sk m. Thus, the general solution is x c1 cos t xt c2 sin t which can also be written as
xt
where A cos sk m
A
cos (frequency) sc 2
1 c2
2 t c1
A (amplitude) sin c2
A is the phase angle (See Exercise 17.) This type of motion is called simple harmonic motion. k 0 5E18(pp 11861195) 1192 ❙❙❙❙ 1/19/06 3:45 PM Page 1192 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS E XAMPLE 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 k 0.2 25.6 so k 25.6 0.2 128. Using this value of the spring constant k, together with m
in Equation 1, we have
2 d 2x
dt 2 128 x 2 0 As in the earlier general discussion, the solution of this equation is
xt 2 c1 cos 8t c2 sin 8t We are given the initial condition that x 0
0.2. But, from Equation 2, x 0
Therefore, c1 0.2. Differentiating Equation 2, we get
xt 8c1 sin 8t Since the initial velocity is given as x 0 8c2 cos 8t 0, we have c2
1
5 xt c1. 0 and so the solution is cos 8t 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 c dx
dt where c is a positive constant, called the damping constant. Thus, in this case, Newton’s
Second Law gives
m d 2x
dt 2 restoring force damping force or
3 m d 2x
dt 2 c dx
dt kx 0 kx c dx
dt 5E18(pp 11861195) 1/19/06 3:45 PM Page 1193 S ECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1193 Equation 3 is a secondorder linear differential equation and its auxiliary equation is
mr 2 cr k 0. The roots are
r1 4 c sc 2
2m 4 mk c r2 sc 2
2m 4 mk According to Section 18.1 we need to discuss three cases.
x CASE I c2 4 mk 0 (overdamping)
In this case r1 and r 2 are distinct real roots and 0 ■ x t x 0 t c1 e r1 t c2 e r2 t Since c, m, and k are all positive, we have sc 2 4 mk 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 . 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 4 mk means that there is a strong damping force (highviscosity oil or grease)
compared with a weak spring or small mass.
c 2 4 mk 0 (critical damping)
This case corresponds to equal roots
CASE II FIGURE 4 Overdamping ■ r1 c
2m r2 and the solution is given by
x c1 c2 t e c 2m t 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 4 mk 0 (underdamping)
Here the roots are complex: CASE III ■ r1
r2 x x=Ae– (c/ 2m)t t x=_Ae– (c/ 2m)t FIGURE 5 Underdamping i s4 mk c 2
2m where
0 c
2m The solution is given by
x e c 2m t c1 cos t c2 sin t We see that there are oscillations that are damped by the factor e c 2 m t. Since c 0 and
m 0, we have c 2 m
0 so e c 2 m t l 0 as t l . This implies that x l 0 as t l ;
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 E XAMPLE 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 m s.
SOLUTION From Example 1 the mass is m
differential equation (3) becomes 2 d 2x
dt 2 40 d 2x
dt 2 or 2 and the spring constant is k dx
dt 20 128 x dx
dt 0 64 x 0 The auxiliary equation is r 2 20 r 64
r 4 r 16
and 16, so the motion is overdamped and the solution is
xt  Figure 6 shows the graph of the position
function for the overdamped motion in Example 2. We are given that x 0 0.03 0, so c1 c1 e
c2 0 1.5 4 c1 e x0 Since c2 4 c1 c1 , this gives 12c1 FIGURE 6 0.6 or c1
x c2 e 0 with roots 4 16 t 0. Differentiating, we get xt
so 4t 128, so the 0.05 e 4t 16 c2 e 16 c2 16 t 0.6 0.05. Therefore
4t e 16 t 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 F t . Then Newton’s Second Law gives
m d 2x
dt 2 restoring force
kx c dx
dt damping force external force Ft 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
dt 2 c dx
dt kx Ft 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
Ft F0 cos 0 t where sk m 0 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
xt 6 c1 cos t F0 c2 sin t 2 m 2
0 cos 0 t 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 Q t , then the current is the rate of change of Q with respect
to t : I dQ dt. 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
Since I 7 dI
dt RI Q
C Et dQ dt, this equation becomes L d 2Q
dt 2 R dQ
dt 1
Q
C Et 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
Q0 Q0 Q0 I0 I0 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 dQ dt :
L d 2I
dt 2 R dI
dt 1
I
C Et EXAMPLE 3 Find the charge and current at time t in the circuit of Figure 7 if R L 1 H, C
both 0. 16 10 4 40 ,
100 cos 10 t, and the initial charge and current are F, E t SOLUTION With the given values of L, R, C, and E t , Equation 7 becomes d 2Q
dt 2 8 40 The auxiliary equation is r 2 dQ
dt 40r 625 40 r 625Q 100 cos 10 t 0 with roots s 900
2 20 15i so the solution of the complementary equation is
Qc t e 20 t c1 cos 15t c2 sin 15t For the method of undetermined coefﬁcients we try the particular solution
Qp t
Then A cos 10 t B sin 10 t Qp t 10 A sin 10 t 10 B cos 10 t Qp t 100 A cos 10 t 100 B sin 10 t Substituting into Equation 8, we have
100 A cos 10 t 100 B sin 10 t 40 10 A sin 10 t 10 B cos 10 t 625 A cos 10 t
or 525 A 400 B cos 10 t 400 A B sin 10 t 525B sin 10 t 100 cos 10 t 100 cos 10 t Equating coefﬁcients, we have
525 A
400 A 400 B
525B The solution of this system is A
Qp t 100 21 A 0
84
697 and B 64
697 1
697 84 cos 10 t 16B 4 16 A or
or 21B 0 , so a particular solution is
64 sin 10 t and the general solution is
Qt Qc t Qp t e 20t c1 cos 15t c2 sin 15t 4
697 21 cos 10 t 16 sin 10 t 5E18(pp 11961204) 1/19/06 3:28 PM Page 1197 S ECTION 18.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS Imposing the initial condition Q 0
Q0 ❙❙❙❙ 1197 0, we get
84
697 c1 84
697 c1 0 To impose the other initial condition we ﬁrst differentiate to ﬁnd the current:
dQ
dt I e 20t 20c1
40
697 I0 20c1 15c2 cos 15t 21 sin 10 t
640
697 15c2 0 15c1 20c2 sin 15t 16 cos 10 t
c2 464
2091 Thus, the formula for the charge is
4
697 Qt e 20 t 63 cos 15t 3 116 sin 15t 21 cos 10 t 16 sin 10 t and the expression for the current is
1
2091 It NOTE 1 tl e 20 t 1920 cos 15t 13,060 sin 15t 120 21 sin 10 t 16 cos 10 t In Example 3 the solution for Q t consists of two parts. Since e
and both cos 15t and sin 15t are bounded functions,
■ 20 t l 0 as 0.2
Qp 0 Qc t Q 1.2 4
2091 e 63 cos 15t 116 sin 15t l 0 as t l So, for large values of t ,
Qt _0.2 20 t Qp t 4
697 21 cos 10 t 16 sin 10 t and, for this reason, Qp t 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. FIGURE 8 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
dt 2
d 2Q
L
dt 2
m dx
dt
dQ
R
dt
c kx Ft 1
Q
C Et Spring system
x
dx dt
m
c
k
Ft displacement
velocity
mass
damping constant
spring constant
external force Electric circuit
Q
I dQ dt
L
R
1C
Et 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 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 c2 e r t, where r1 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 m s, 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 inductor 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 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 m s.
(b) Graph the position function of the mass. ; a voltage of E t 6. For the spring in Exercise 4, ﬁnd the damping constant that 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 m s. 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 sk m. Suppose that the damping constant is so small
that the damping force is negligible. If an external force
, use the method
Ft
F0 cos 0 t is applied, where 0
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
sk m.
F0 cos t is applied (the applied
If an external force F t
frequency equals the natural frequency), use the method of
undetermined coefﬁcients to show that the motion of the mass
is given by x t
c1 cos t c2 sin t
F0 2m t sin t.
11. Show that if , but
0
0 is a rational number, then the
motion described by Equation 6 is periodic. 12 sin 10 t. Find the charge at time t. 16. The battery in Exercise 14 is replaced by a generator producing critical damping. ; 7. A spring has a mass of 1 kg and its spring constant is k 24 , an inductor
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 E t
12 sin 10 t.
(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 x t A cos t . 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
d2
dt 2 t
sin
L 0 where t is the acceleration due to gravity. For small values of
we can use the linear approximation sin
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
d dt 1 rad s.
(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
y 1 2 xy y 0 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 cn x n fx c0 c2 x 2 c1 x c3 x 3 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 y
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 y 0. y SOLUTION We assume there is a solution of the form y 2 c0 c2 x 2 c1 x c3 x 3 cn x n
n0 We can differentiate power series term by term, so
y c1 3c3 x 2 2 c2 x n cn x n 1 n1 3 y 2 c2 2 3c3 x nn 1 cn x n 2 n2 In order to compare the expressions for y and y more easily, we rewrite y 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 2 and began the summation at 0 instead of 2. y 4 n 1 cn 2 x n 2n n0 Substituting the expressions in Equations 2 and 4 into the differential equation, we
obtain
n 2n 1 cn 2 x n n0 cn x n 0 n0 or
n 5
n0 2n 1 cn 2 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
cn 6 1 cn n 2 2n
cn
1n 2 cn 2 n 0
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 c0
12 Put n 1: c3 c1
23 Put n 2: c4 Put n 3: c5 Put n 4: Put n 5: c2 c0 c0
4! 34 1234 c3 c1 45 2345 c1
5! c6 c4
56 c0
4! 5 6 c0
6! c7 c5
67 c1
5! 6 7 c1
7! By now we see the pattern:
For the even coefficients, c2 n For the odd coefficients, c2 n 1 1 1 c0
2n ! n c1 n 2n 1! Putting these values back into Equation 2, we write the solution as
y c0 c2 x 2 c1 x c0 1 x2
2!
c1 x c0 1
n0 n x4
4!
x3
3!
x 2n
2n ! c3 x 3 c4 x 4 c5 x 5 x6
6! 1 x5
5!
c1 x 2n
2n ! x7
7!
1 n0 n 1
n x 2n 1
2n 1 ! Notice that there are two arbitrary constants, c0 and c1. n x 2n 1
2n 1 ! 5E18(pp 11961204) 1/19/06 3:31 PM Page 1201 SECTION 18.4 SERIES SOLUTIONS ❙❙❙❙ 1201 N OTE 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
■ yx c0 cos x 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 y 2 xy y 0. SOLUTION We assume there is a solution of the form cn x n y
n0 n cn x n 1 nn y Then 1 cn x n n1 y and 2 n n2 1 cn 2 x n 2n n0 as in Example 1. Substituting in the differential equation, we get
n 1 cn 2 x n 2n 1 cn 2 x n 2n n1 n 2 ncn x n
n0 cn x n 0 cn x n 0 1 cn x n 0 n0 2 ncn x n n0 2 ncn x n 1 n1 n n1 n cn x n 2x n0 2n 1 cn n0 2n 2 n0 This equation is true if the coefﬁcient of x n is 0:
n
cn 7 2 2n n 1 cn 2 2n 1
cn
1n 2 We solve this recursion relation by putting n
Put n 0: c2 1: c3 2: c4 3
c2
34 Put n 3: c5 0 0, 1, 2, 3, . . . 0, 1, 2, 3, . . . successively in Equation 7: 1
c1
23 Put n n 1 cn 1
c0
12 Put n 2n 5
45 c3 3
c0
1234
15
c1
2345 3
c0
4!
15
c1
5! 5E18(pp 11961204) 1202 ❙❙❙❙ 1/19/06 3:32 PM Page 1202 CHAPTER 18 SECONDORDER DIFFERENTIAL EQUATIONS Put n 4: c6 7
c4
56 37
c0
4! 5 6 37
c0
6! Put n 5: c7 9
c5
67 Put n 6: c8 11
c6
78 3 7 11
c0
8! Put n 7: c9 13
c7
89 1 5 9 13
c1
9! 159
c1
5! 6 7 159
c1
7! In general, the even coefﬁcients are given by
3 7 11 c2 n 4n 5 c0 2n ! and the odd coefﬁcients are given by
159 c2 n 1 4n
1! 2n 3 c1 The solution is
y c0 c1 x c2 x 2
12
x
2! c0 1 c1 x c3 x 3 c4 x 4 34
x
4! 376
x
6! 13
x
3! 155
x
5! 3 7 11 8
x
8!
1597
x
7! 1 5 9 13 9
x
9! or
y 8 12
x
2! c0 1 37 4n n2 159 c1 x 5 2n !
2n n1 4n
1! x 2n
3 x 2n 1 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
■ ■ y1 x and y2 x 1 12
x
2! 37 n1 5 x 2n 2n ! n2 159 x 4n 2n 4n
1! 3 x 2n 1 5E18(pp 11961204) 1/19/06 3:32 PM Page 1203 ❙❙❙❙ C HAPTER 18 REVIEW 2 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 y1 x , and we see how they converge to y1 . In this way we can
graph both y1 and y2 in Figure 2. T¸
2 _2 1203 T¡¸ NOTE 4 ■ If we were asked to solve the initialvalue problem _8 y FIGURE 1 2 xy y 0 ﬁ c0 _2.5 y0 y0 1 y0 0 c1 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 yx FIGURE 2  y 3. y x2y 5. y xy 4n
1! 2n 3 y0 0, x 2n 1 Exercises
2. y y 6. y 1y 0
xy y 8. y xy y 0, 10. y x2y 0, y0 ■ y 2y x2y 18 Review 1, 1,
y0 y0 ; 0 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. xy
■ 0,
■ ■ ■ y0 1 ■ ■ ■ ■ xy x2y 0 y0 1 y0 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 5, 5 . 0 y0 ■ 12. The solution of the initialvalue problem 0 xy 9. y 3y ■ xy 4. x 0 x2y 11. y Use power series to solve the differential equation. 1. y 7. x 2 159 x
n1  18.4  0 we would observe from Theorem 11.10.5 that 15 1–11 y0 ■ (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 y 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.
1. If y1 and y2 are solutions of y y 0, then y1 ■ y y2 is also a solution of the equation. 4. The equation y y c1 cosh x
y 0 can be written as
c2 sinh x e x has a particular solution of the form 2. If y1 and y2 are solutions of y c1 y1 6y
5 y x, then
c2 y2 is also a solution of the equation. yp ■ 1–10  2y 15 y 0 2. y 4y 13 y 0 3. y 3y 4. 4 y
5. 4y dy
dx 2 4 d 2y
6.
dx 2 y 7. d 2y
dx 2 2 8. d 2y
dx 2 4y d 2y
9.
dx 2 ■ ■ 11–14 dy
dx  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 m s. x2 y x cos x 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 6y 1 csc x, e 0 ■ 2x x ■ ■ 2
■ ■ ■ ■ ■ ■ Solve the initialvalue problem. 11. y 6y 0, 12. y 6y 25 y 13. y 5y 4y 14. 9 y 18. A spring with a mass of 2 kg has damping constant 16, and a e 2x sin 2 x ■ y ■ y1 3, 0,
0,
e x, 3x
■ ■ y0
y0 2,
0, y0 ■ y1 ■ 1,
■ 12
y0
y0
y0
■ 1 2
■ ■ ■ xy y 0 y0 0 y0 1 ■ G Mr m
r2 where G is the gravitational constant and Mr is the mass of the
earth within the sphere of radius r .
G Mm
(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 y t of
the particle from the center of the earth at time t is given by
yt 1 15. Use power series to solve the initialvalue problem y 0 19. Assume that the earth is a solid sphere of uniform density with dy
dx
y 2y 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. 5y
2y xy 17. A series circuit contains a resistor with R 0 dy
dx dy
dx d 2y
10.
dx 2 y 0 2 ■ 16. Use power series to solve the equation Solve the differential equation. 1. y ■ EXERCISES Ae x k2y t where k 2 G M R 3 t R.
(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 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
 Fall '09
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