Embed Size (px)
DESCRIPTION
math
Citation preview
SECTION 15.5 Series Solutions of Differential Equations 1125
15.5S E C T I O N Series Solutions of Differential Equations
Power Series Solution of a Differential Equation • Approximation by Taylor Series
Power Series Solution of a Differential EquationWe conclude this chapter by showing how power series can be used to solve certaintypes of differential equations. We begin with the general power series solutionmethod.
Recall from Chapter 8 that a power series represents a function f on an interval ofconvergence, and that you can successively differentiate the power series to obtain aseries for and so on. These properties are used in the power series solutionmethod demonstrated in the first two examples.
EXAMPLE 1 Power Series Solution
Use a power series to solve the differential equation
Solution Assume that is a solution. Then, Substitutingfor and you obtain the following series form of the differential equation.(Note that, from the third step to the fourth, the index of summation is changed toensure that occurs in both sums.)
Now, by equating coefficients of like terms, you obtain the recursion formulawhich implies that
This formula generates the following results.
. . .
. . .
Using these values as the coefficients for the solutionseries, you have
y 5 o`
n50 2na0
n! xn 5 a0 o
`
n50 s2xdn
n!5 a0e
2x.
25a0
5!24a0
4!23a0
3!22a0
22a0a0
a5a4a3a2a1a0
n ≥ 0.an11 52an
n 1 1,
sn 1 1dan11 5 2an,
o`
n50 sn 1 1dan11x
n 5 o`
n50 2anxn
o`
n51 nanxn21 5 o
`
n50 2anxn
o`
n51 nanxn21 2 2 o
`
n50 anxn 5 0
y9 2 2y 5 0
xn
22y,y9y9 5 onanxn21.y 5 oanxn
y9 2 2y 5 0.
f9, f 0,
1126 CHAPTER 15 Differential Equations
In Example 1, the differential equation could be solved easily without using aseries. The differential equation in Example 2 cannot be solved by any of the methodsdiscussed in previous sections.
EXAMPLE 2 Power Series Solution
Use a power series to solve the differential equation
Solution Assume that is a solution. Then you have
Substituting for and y in the given differential equation, you obtain the fol-lowing series.
To obtain equal powers of x, adjust the summation indices by replacing n by inthe left-hand sum, to obtain
By equating coefficients, you have from whichyou obtain the recursion formula
and the coefficients of the solution series are as follows.
Thus, you can represent the general solution as the sum of two series—one for theeven-powered terms with coefficients in terms of and one for the odd-poweredterms with coefficients in terms of
The solution has two arbitrary constants, and as you would expect in the general solution of a second-order differential equation.
a1,a0
5 a0 o`
k50 s21dkx2k
2ksk!d 1 a1 o`
k50
s21dkx2k11
3 ? 5 ? 7 . . . s2k 1 1d
y 5 a011 2x2
21
x4
2 ? 42 . . .2 1 a11x 2
x3
31
x5
3 ? 52 . . .2
a1.a0
a2k11 5s21dk a1
3 ? 5 ? 7 . . . s2k 1 1d a2k 5s21dk a0
2 ? 4 ? 6 . . . s2kd 5s21dka0
2ksk!d
::
a7 5 2a5
75 2
a1
3 ? 5 ? 7 a6 5 2
a4
65 2
a0
2 ? 4 ? 6
a5 5 2a3
55
a1
3 ? 5 a4 5 2
a2
45
a0
2 ? 4
a3 5 2a1
3 a2 5 2
a0
2
n ≥ 0,an12 5 2sn 1 1d
sn 1 2dsn 1 1d an 5 2an
n 1 2,
sn 1 2dsn 1 1dan12 5 2sn 1 1dan,
o`
n50 sn 1 2dsn 1 1dan12xn 5 2 o
`
n50 sn 1 1danxn.
n 1 2
o`
n52 nsn 2 1danxn22 5 2 o
`
n50 sn 1 1danxn
o`
n52nsn 2 1danxn22 1 o
`
n50 nanxn 1 o
`
n50 anxn 5 0
y0, xy9,
y0 5 o`
n52 nsn 2 1danxn22.xy9 5 o
`
n51 nanxn,y9 5 o
`
n51 nanxn21,
o`
n50anx
n
y0 1 xy9 1 y 5 0.
x y
0.0 1.0000
0.1 1.1057
0.2 1.2264
0.3 1.3691
0.4 1.5432
0.5 1.7620
0.6 2.0424
0.7 2.4062
0.8 2.8805
0.9 3.4985
1.0 4.3000
SECTION 15.5 Series Solutions of Differential Equations 1127
In Exercises 1–6, verify that the powerseries solution of the dif ferential equation is equivalent to the solution found usingthe techniques in Sections 5.7 and 15.1–15.4.
1. 2.
3. 4.
5. 6.
In Exercises 7–10, use powerseries to solve the differentialequation and find the interval of convergence of the series.
7. 8.
9. 10.
In Exercises 11 and 12, find the first three terms of each of thepower series representing independent solutions of the differen-tial equation.
11. 12.
In Exercises 13 and 14, use Taylor’s Theorem to find the seriessolution of the differential equation under the specified initialconditions. Use n terms of the series to approximate y for thegiven value of x and compare the result with the approximationgiven by Euler’s Method for
13.
14. y9 2 2xy 5 0, ys0d 5 1, n 5 4, x 5 1
y9 1 s2x 2 1dy 5 0, ys0d 5 2, n 5 5, x 512 ,
Dx 5 0.1.
y0 1 x2y 5 0sx2 1 4dy0 1 y 5 0
y0 2 xy9 2 y 5 0y0 2 xy9 5 0
y9 2 2xy 5 0y9 1 3xy 5 0
y0 1 k2y 5 0y0 1 4y 5 0
y0 2 k2y 5 0y0 2 9y 5 0
y9 2 ky 5 0y9 2 y 5 0
E X E R C I S E S F O R S E C T I O N 15 . 5
Approximation by Taylor SeriesA second type of series solution method involves a differential equation with initialconditionsand makes use of Taylor series, as given in Section 8.10.
EXAMPLE 3 Approximation by Taylor Series
Use a Taylor series to find the series solution of
given the initial condition when Then, use the first six terms of thisseries solution to approximate values of y for
Solution Recall from Section 8.10 that, for
Because and you obtain the following.
Therefore, you can approximate the values of the solution from the series
Using the first six terms of this series, you can compute values for y in the intervalas shown in the table at the left.0 ≤ x ≤ 1,
5 1 1 x 112
x2 143!
x3 1144!
x4 1665!
x5 1 . . . .
y 5 ys0d 1 y9s0dx 1y0 s0d
2! x2 1
y999s0d3!
x3 1ys4ds0d
4! x4 1
ys5ds0d5!
x5 1 . . .
ys5ds0d 5 28 1 32 1 6 5 66 ys5d 5 2yys4d 1 8y9y999 1 6sy0 d2
ys4ds0d 5 8 1 6 5 14 ys4d 5 2yy999 1 6y9y0
y999s0d 5 2 1 2 5 4 y999 5 2yy0 1 2sy9d2
y0 s0d 5 2 2 1 5 1 y9 5 2yy9 2 1
y9s0d 5 1 y9 5 y2 2 x
ys0d 5 1
y9 5 y2 2 x,ys0d 5 1
y 5 ys0d 1 y9s0dx 1y0 s0d
2! x2 1
y999s0d3!
x3 1 . . . .
c 5 0,
0 ≤ x ≤ 1.x 5 0.y 5 1
y9 5 y2 2 x
1128 CHAPTER 15 Differential Equations
15. Investigation Consider the differential equation with initial conditions and
(a) Find the solution of the differential equation using the techniques of Section 15.3.
(b) Find the series solution of the differential equation.
(c) The figure shows the graph of the solution of the differential equation and the third-degree and fifth-degreepolynomial approximations of the solution. Identify each.
16. Consider the differential equation with the initialconditions and
(See Exercise 9.)
(a) Find the series solution satisfying the initial conditions.
(b) Use a graphing utility to graph the third-degree and fifth-degree series approximations of the solution. Identify theapproximations.
(c) Identify the symmetry of the solution.
In Exercises 17 and 18, use Taylor’s Theorem to find the seriessolution of the differential equation under the specified initialconditions. Use n terms of the series to approximate y for thegiven value of x.
17.
18.
In Exercises 19–22, verify that the series converges to the givenfunction on the indicated interval. (Hint: Use the given differ-ential equation.)
19.
Differential equation:
20.
Differential equation:
21.
Differential equation:
22.
Differential equation:
23. Find the first six terms in the series solution of Airy’ s equationy0 2 xy 5 0.
s1 2 x2dy0 2 xy9 5 0
o`
n50
s2nd!x2n11
s2nn!d2s2n 1 1d 5 arcsin x, s21, 1d
sx2 1 1dy0 1 2xy9 5 0
o`
n50 s21dnx2n11
2n 1 15 arctan x, s21, 1d
y0 1 y 5 0
o`
n50 s21dnx2n
s2nd! 5 cos x, s2`, `d
y9 2 y 5 0
o`
n50 xn
n!5 ex, s2`, `d
y0 2 2xy9 1 y 5 0, ys0d 5 1, y9s0d 5 2, n 5 8, x 512
y0 2 2xy 5 0, ys0d 5 1, y9s0d 5 23, n 5 6, x 514
y9s0d 5 2.ys0d 5 0
y0 2 xy9 5 0
x
y
1 2
3
2
1
y9s0d 5 6.ys0d 5 29y 5 0y0 1
In Exercises 1–4, classify the differential equation according totype and order.
1. 2.
3. 4.
In Exercises 5 and 6, use the given differential equation and itsdir ection field.
5.
(a) Sketch several solution curves for the differential equationon the direction field.
(b) Find the general solution of the differential equation.Compare the result with the sketches from part (a).
6.
(a) Sketch several solution curves for the differential equationon the direction field.
(b) When is the rate of change of the solution greatest? When isit least?
(c) Find the general solution of the differential equation.Compare the result with the sketches from part (a).
x
y
2
2
−2
−2
dydx
5 !1 2 y 2
x
y
1
4
3
2
321 4−4 −3 −2
−4
−3
−2
−1
dydx
5yx
s y0 d2 1 4y9 5 0y0 1 3y9 2 10 5 0
yy0 5 x 1 12ut2
5 c2 2ux2
R E V I E W E X E R C I S E S F O R C H A P T E R 15
REVIEW EXERCISES 1129
In Exercises 7–10, match the differential equation with its solution.
7. (a)
8. (b)
9. (c)
10. (d)
In Exercises 11–32, find the general solution of the first-orderdif ferential equation.
11. 12.
13. 14.
15. 16.
17.
18.
19.
20.21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
In Exercises 33–40, find the particular solution of the differen-tial equation that satisfies the boundary condition.
33.
34.
35.
36.
37.
38.
39.
40.
In Exercises 41 and 42, find the orthogonal trajectories of thegiven family and sketch several members of each family.
41.
42.
43. Snow Removal Assume that the rate of change in the number of miles s of road cleared per hour by a snowplow isinversely proportional to the height h of snow.
(a) Write and solve the differential equation to find sas a func-tion of h.
(b) Find the particular solution if miles when inches and miles when inches
44. Growth Rate Let x and y be the sizes of two internal organsof a particular mammal at time t. Empirical data indicate thatthe relative growth rates of these two organs are equal, andhence we have
Solve this differential equation, writing y as a function of x.
45. Population Growth The rate of growth in the number N ofdeer in a state park varies jointly over time t as N and where is the estimated limiting size of the herd. WriteN as a function of t if when and when
46. Population Growth The rate of growth in the number N ofelk in a game preserve varies jointly over time t (in years) as Nand where 300 is the estimated limiting size of theherd.
(a) Write and solve the differential equation for the populationmodel if when and when
(b) Use a graphing utility to graph the direction field of the differential equation and the particular solution of part (a).
(c) At what time is the population increasing most rapidly?
(d) If 400 elk had been placed in the preserve initially, use thedirection field to describe the change in the population overtime.
t 5 1.N 5 75t 5 0N 5 50
300 2 N
t 5 4.N 5 200t 5 0N 5 100
L 5 500L 2 N,
1x dxdt
51y dydt
.
s2 ≤ h ≤ 15d.h 5 10s 5 12h 5 2
s 5 25
y 2 2x 5 C
sx 2 Cd2 1 y 2 5 C2
ys4d 5 2
2xy9 2 y 5 x3 2 x
ys0d 53s1 1 ed
1 2 e
y9 5 x2y 2 2 9x2
ys2d 5 0
s2x 1 y 2 3d dx 1 sx 2 3y 1 1d dy 5 0
ys0d 5 2
s1 1 yd ln s1 1 yd dx 1 dy 5 0
ys22d 5 25
yexydx 1 xexy dy 5 0
ys1d 5 10
xdy 5 sx 1 y 1 2d dx
ys1d 5 2
y9 12yx
5 2x9y5
ys0d 5 4
y9 2 2y 5 ex
y9 5 y 1 2xs y 2 ex dxy9 2 ay 5 bx4
x3yy9 5 x4 1 3x2y 2 1 y4
s1 1 x2d dy 5 s1 1 y 2d dx
2xdx 1 2ydy 5 sx2 1 y 2d dx
yy9 1 y 2 5 1 1 x2
xy9 1 y 5 sin x
x 1 yy9 5 !x2 1 y 2
y9 5 2x!1 2 y 2
sx 2 y 2 5d dx 2 sx 1 3y 2 2d dy 5 0
ydx 2 sx 1 !xy d dy 5 0
dy 5 sy tan x 1 2ex d dx
3x2y2 dx 1 s2x3y 1 x3y4d dy 5 0
s2x 2 2y3 1 yd dx 1 sx 2 6xy 2d dy 5 0
s y 1 x3 1 xy 2d dx 2 xdy 5 0
s10x 1 8y 1 2d dx 1 s8x 1 5y 1 2d dy 5 0
dydx
23yx2 5
1x2
dydx
2yx
5xy
dydx
2 3x2y 5 ex3y9 22yx
5y9
x
dydx
1 xy 5 2ydydx
2yx
5 2 1 !x
y 5 Ce4xy0 1 4y 5 0
y 5 C1 cos 2x 1 C2 sin 2xy0 2 4y 5 0
y 5 4x 1 Cy9 2 4y 5 0
y 5 C1e2x 1 C2e
22xy9 2 4 5 0
Solution Differential Equation
1130 CHAPTER 15 Differential Equations
47. Slope The slope of a graph is given by Find the equation of the graph if the graph passes through thepoint Use a graphing utility to graph the solution.
48. Investment Let be the amount in a fund earning interestat an annual rate r compounded continuously. If a continuouscash flow of P dollars per year is withdrawn from the fund, therate of change of A is given by the differential equation
where when Solve this differential equation forA as a function of t.
49. Investment A retired couple plans to withdraw P dollars peryear from a retirement account of $500,000 earning 10% compounded continuously. Use the result of Exercise 48 and agraphing utility to graph the function A for each of the follow-ing continuous annual cash flows. Use the graphs to describewhat happens to the balance in the fund for each of the cases.
(a)
(b)
(c)
50. Investment Use the result of Exercise 48 to find the timenecessary to deplete a fund earning 14% interest compoundedcontinuously if and
In Exercises 51–54, find the particular solution of the differen-tial equation that satisfies the initial conditions. Use a graphingutility to graph the solution.
51.
52.
53.
54.
In Exercises 55–60, find the general solution of the second-orderdif ferential equation.
55.
56.
57.
58.
59.
60.
In Exercises 61–64, find the particular solution of the differen-tial equation that satisfies the initial conditions.
61.
62.
63.
64.
Vibrating Spring In Exercises 65 and 66, describe the motionof a 64-pound weight suspended on a spring. Assume that theweight stretches the spring feet from its natural position.
65. The weight is pulled foot below the equilibrium position andreleased.
66. The weight is pulled foot below the equilibrium position andreleased. The motion takes place in a medium that furnishes adamping force of magnitude speed at all times.
67. Investigation The differential equation
models the motion of a weight suspended on a spring.
(a) Solve the differential equation and use a graphing utility tograph the solution for each of the assigned quantities for b,k, and
(i)
(ii)
(iii)
(iv)
(b) Describe the effect of increasing the resistance to motion b.
(c) Explain how the motion of the object would change if astiffer spring (increased k) were used.
(d) Matching the input and natural frequencies of a system isknown as resonance. In which case of part (a) does thisoccur, and what is the result?
68. Think About It Explain how you can find a particular solu-tion of the differential equation
by observation.
In Exercises 69 and 70, find the series solution of the differen-tial equation.
69.
70. y0 1 3xy9 2 3y 5 0
sx 2 4dy9 1 y 5 0
y0 1 4y9 1 6y 5 30
b 5 1, k 5 2, Fstd 5 0
b 5 0.1, k 5 2, Fstd 5 0
b 5 0, k 5 2, Fstd 5 24 sins2!2 tdb 5 0, k 5 1, Fstd 5 24 sin pt
Fstd.
y9s0d 5 0ys0d 512
,832
y0 1 by9 1 ky 58
32 Fstd,
18
12
12
43
ys0d 5 2, y9s0d 5103y0 1 3y9 5 6x
ys0d 5 6, y9s0d 5 26y0 1 4y 5 cos x
ys0d 5 0, y9s0d 5 0y0 1 25y 5 ex
ys0d 5 2, y9s0d 5 0y0 2 y9 2 6y 5 54
Initial Conditions Differential Equation
y0 1 2y9 1 y 51
x2ex
y0 2 2y9 1 y 5 2xex
y0 1 5y9 1 4y 5 x2 1 sin 2x
y0 1 y 5 2 cos x
y0 1 2y 5 e2x 1 x
y0 1 y 5 x3 1 x
ys1d 5 4, ys2d 5 0y0 1 2y9 1 5y 5 0
ys0d 5 2, y9s0d 5 0y0 1 2y9 2 3y 5 0
ys0d 5 2, y9s0d 5 27y0 1 4y9 1 5y 5 0
ys0d 5 0, y9s0d 5 3y0 2 y9 2 2y 5 0
Initial Conditions Differential Equation
P 5 $200,000.A0 5 $1,000,000
P 5 $60,000
P 5 $50,000
P 5 $40,000
t 5 0.A 5 A0
dAdt
5 rA 2 P
Astd
s0, 1d.
y9 5 sin x 2 0.5y.
